*1,926*
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*Year 2011*

*Table of contents : Cover......Page 1Title Page......Page 2Copyright Page......Page 3Acknowledgments......Page 14Contents......Page 6Preface......Page 121.2 Examples of Di.erential Equations......Page 181.3 Direction Fields......Page 252.1 Introduction......Page 322.2 First Order Linear Di.erential Equations......Page 362.3 Introduction to Mathematical Models......Page 482.4 Population Dynamics and Radioactive Decay......Page 582.5 First Order Nonlinear Di.erential Equations......Page 652.6 Separable First Order Equations......Page 712.7 Exact Differential Equations......Page 802.8 The Logistic Population Model......Page 872.9 Applications to Mechanics......Page 942.10 Euler’s Method......Page 106Review Exercises......Page 117Projects......Page 118Chapter 3 Second and Higher Order Linear Differential Equations......Page 1243.1 Introduction......Page 1253.2 The General Solution of Homogeneous Equations......Page 1323.3 Constant Coe.cient Homogeneous Equations......Page 1383.4 Real Repeated Roots; Reduction of Order......Page 1443.5 Complex Roots......Page 1493.6 Unforced Mechanical Vibrations......Page 1593.7 The General Solution of a Linear Nonhomogeneous Equation......Page 1713.8 The Method of Undetermined Coe.cients......Page 1753.9 The Method of Variation of Parameters......Page 1853.10 Forced Mechanical Vibrations, Electrical Networks, and Resonance......Page 1913.11 Higher Order Linear Homogeneous Di.erential Equations......Page 2053.12 Higher Order Homogeneous Constant Coe.cient Di.erential Equations......Page 2123.13 Higher Order Linear Nonhomogeneous Di.erential Equations......Page 218Projects......Page 2234.1 Introduction......Page 2304.2 Existence and Uniqueness......Page 2404.3 Homogeneous Linear Systems......Page 2454.4 Constant Coe.cient Homogeneous Systems; the Eigenvalue Problem......Page 2554.5 Real Eigenvalues and the Phase Plane......Page 2644.6 Complex Eigenvalues......Page 2734.7 Repeated Eigenvalues......Page 2834.8 Nonhomogeneous Linear Systems......Page 2944.9 Numerical Methods for Systems of Linear Di.erential Equations......Page 3054.10 The Exponential Matrix and Diagonalization......Page 317Review Exercises......Page 327Projects......Page 3285.1 Introduction......Page 3345.2 Laplace Transform Pairs......Page 3465.3 The Method of Partial Fractions......Page 3615.4 Laplace Transforms of Periodic Functions and System Transfer Functions......Page 3675.5 Solving Systems of Di.erential Equations......Page 3765.6 Convolution......Page 3855.7 The Delta Function and Impulse Response......Page 394Projects......Page 4026.1 Introduction......Page 4086.2 Equilibrium Solutions and Direction Fields......Page 4176.3 Conservative Systems......Page 4306.4 Stability......Page 4416.5 Linearization and the Local Picture......Page 4506.6 Two-Dimensional Linear Systems......Page 4656.7 Predator-Prey Population Models......Page 475Projects......Page 4837.1 Introduction......Page 4887.2 Euler’s Method, Heun’s Method, and the Modi.ed Euler’s Method......Page 4907.3 Taylor Series Methods......Page 4967.4 Runge-Kutta Methods......Page 510Appendix 1: Convergence of One-Step Methods......Page 523Appendix 2: Stability of One-Step Methods......Page 524Projects......Page 5278.1 Introduction......Page 5328.2 Series Solutions Near an Ordinary Point......Page 5448.3 The Euler Equation......Page 5538.4 Solutions Near a Regular Singular Point and the Method of Frobenius......Page 5598.5 The Method of Frobenius Continued: Special Cases and a Summary......Page 567Projects......Page 5789.1 Introduction......Page 5829.2 Heat Flow in a Thin Bar; Separation of Variables......Page 5879.3 Series Solutions......Page 5979.4 Calculating the Solution......Page 6069.5 Fourier Series......Page 6179.6 The Wave Equation......Page 6339.7 Laplace’s Equation......Page 6459.8 Higher-Dimensional Problems; Nonhomogeneous Equations......Page 658Project......Page 67210.1 Introduction......Page 67610.2 The Cauchy Problem......Page 67910.3 Existence and Uniqueness......Page 68510.4 The Method of Characteristics......Page 688Projects......Page 69611.1 Introduction......Page 69811.2 Existence and Uniqueness......Page 69911.3 Two-Point Boundary Value Problems for Linear Systems......Page 71011.4 Sturm-Liouville Boundary Value Problems......Page 722Project......Page 732Answers......Page 735C......Page 788E......Page 789F......Page 790I......Page 791M......Page 792P......Page 793R......Page 794T......Page 795Z......Page 796*

ELEMENTARY DIFFERENTIAL EQUATIONS with Boundary Value Problems Second Edition

WERNER

KOHLER

Virginia Tech

& LEE

JOHNSON

Virginia Tech

Boston

San Francisco

New York

London Toronto Sydney Tokyo Singapore Madrid Mexico City Munich Paris Cape Town Hong Kong Montreal

Publisher: Greg Tobin Senior Acquisitions Editor: William Hoffman Executive Project Manager: Christine O’Brien Editorial Assistant: Emily Portwood Managing Editor: Karen Wernholm Senior Production Supervisor: Peggy McMahon Senior Designer: Barbara T. Atkinson Digital Assets Manager: Jason Miranda Media Producer: Michelle Murray Marketing Manager: Phyllis Hubbard Marketing Coordinator: Celena Carr Senior Author Support/Technology Specialist: Joe Vetere Senior Prepress Supervisor: Caroline Fell Senior Manufacturing Buyer: Evelyn Beaton Cover and Text Design: Barbara T. Atkinson Production Coordination: Liﬂand et al., Bookmakers Composition and Illustration: Techsetters, Inc. Cover Image: Christo and Jeanne-Claude: The Gates, Central Park, New York City, 1979–2005. Photograph taken by Mark Harmel. Copyright 2005 Christo and Jeanne-Claude. About the Cover: The cover image captures the interplay of the imagination and mathematics. To the eyes of the artists, Christo and Jeanne-Claude, The Gates were fantasy to be experienced for a few weeks in the dead of winter in New York’s Central Park, and then dismantled. But with the content of a differential equations course in mind, one can imagine differential equations in the wind’s velocity that lift the orange fabric panels, or more simply as the area of ice melting on the pond’s surface in the background. But for the central ﬁgure in the image whose jacket takes on the hue of the orange panels, understanding comes in that subtle sort of change within change that is at once both artistic and at the essence of differential equations. Both the art and the mathematics require the participation of the curious. Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and Addison-Wesley was aware of a trademark claim, the designations have been printed in initial caps or all caps. Library of Congress Cataloging-in-Publication Data Kohler, W. E. (Werner E.), 1939– Elementary differential equations with boundary value problems / Werner Kohler and Lee Johnson.—2nd ed. p. cm. Includes index. ISBN 0-321-28835-1 (alk. paper) 1. Differential equations—Textbooks. 2. Boundary value problems—Textbooks. I. Johnson, Lee W. II. Title. QA371.K855 2003 515'.35—dc22 2005045849 Copyright © 2006 Pearson Education, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. For information on obtaining permission for use of material in this work, please submit a written request to Pearson Education, Inc., Rights and Contracts Department, 75 Arlington Street, Suite 300, Boston, MA 02116, fax your request to 617-848-7047, or e-mail at http://www.pearsoned.com/legal/permissions.htm. 1 2 3 4 5 6 7 8 9 10—VHP—09 08 07 06 05 Book ISBN 0-321-28835-1 Book/CD package ISBN 0-321-39850-5

We especially want to acknowledge our families for making it fun. Thanks to Abbie, Larry, Tom, Liz, Paul, Maggie, Cathy, Connie, Luke, Cano, Kiera, Mike, and baby Maya. Thanks to Rochelle, Eric, Mark, Ali, Hannah, Quinn, and Casey.

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Contents xi

Preface

Chapter 1

Introduction to Differential Equations

1.1 Introduction 1.2 Examples of Diﬀerential Equations 1.3 Direction Fields

Chapter 2

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

First Order Differential Equations

1

1 1 8

15

Introduction

15

First Order Linear Diﬀerential Equations

19

Introduction to Mathematical Models

31

Population Dynamics and Radioactive Decay

41

First Order Nonlinear Diﬀerential Equations

48

Separable First Order Equations

54

Exact Diﬀerential Equations

63

The Logistic Population Model

70

Applications to Mechanics

77

Euler’s Method

89 100

Review Exercises

101

Projects

Chapter 3

Second and Higher Order Linear Differential Equations

3.1 Introduction 3.2 The General Solution of Homogeneous Equations v

107

108 115

vi

Contents

3.3 3.4 3.5 3.6 3.7

Constant Coeﬃcient Homogeneous Equations

121

Real Repeated Roots; Reduction of Order

127

Complex Roots

132

Unforced Mechanical Vibrations

142

The General Solution of a Linear Nonhomogeneous Equation

154

3.8 The Method of Undetermined Coeﬃcients 3.9 The Method of Variation of Parameters 3.10 Forced Mechanical Vibrations, Electrical Networks, and Resonance

158 168 174

3.11 Higher Order Linear Homogeneous Diﬀerential Equations

188

3.12 Higher Order Homogeneous Constant Coeﬃcient Diﬀerential Equations

195

3.13 Higher Order Linear Nonhomogeneous Diﬀerential Equations Review Exercises

206

Projects

206

Chapter 4

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

201

First Order Linear Systems

213

Introduction

213

Existence and Uniqueness

223

Homogeneous Linear Systems

228

Constant Coeﬃcient Homogeneous Systems; the Eigenvalue Problem

238

Real Eigenvalues and the Phase Plane

247

Complex Eigenvalues

256

Repeated Eigenvalues

266

Nonhomogeneous Linear Systems

277

Contents

4.9 Numerical Methods for Systems of Linear Diﬀerential Equations

4.10 The Exponential Matrix and Diagonalization

311

Projects

5.1 5.2 5.3 5.4

Laplace Transforms

317

Introduction

317

Laplace Transform Pairs

329

The Method of Partial Fractions

344

Laplace Transforms of Periodic Functions and System Transfer Functions

350

5.5 Solving Systems of Diﬀerential Equations 5.6 Convolution 5.7 The Delta Function and Impulse Response

Chapter 6

359 368 377 385

Projects

6.1 6.2 6.3 6.4 6.5 6.6 6.7

300 310

Review Exercises

Chapter 5

288

Nonlinear Systems

Introduction Equilibrium Solutions and Direction Fields

391

391 400

Conservative Systems

413

Stability

424

Linearization and the Local Picture

433

Two-Dimensional Linear Systems

448

Predator-Prey Population Models

458

Projects

466

vii

viii

Contents

Chapter 7

Numerical Methods

7.1 Introduction 7.2 Euler’s Method, Heun’s Method, and the Modiﬁed Euler’s Method

7.3 Taylor Series Methods 7.4 Runge-Kutta Methods

471 473 479 493

Appendix 1: Convergence of One-Step Methods

506

Appendix 2: Stability of One-Step Methods

507

Projects

510

Chapter 8

8.1 8.2 8.3 8.4

471

Series Solutions of Linear Differential Equations

515

Introduction

515

Series Solutions Near an Ordinary Point

527

The Euler Equation

536

Solutions Near a Regular Singular Point and the Method of Frobenius

542

8.5 The Method of Frobenius Continued: Special Cases and a Summary

550

Projects

561

Chapter 9

9.1 9.2 9.3 9.4 9.5 9.6 9.7

Second Order Partial Differential Equations and Fourier Series

565

Introduction

565

Heat Flow in a Thin Bar; Separation of Variables

570

Series Solutions

580

Calculating the Solution

589

Fourier Series

600

The Wave Equation

616

Laplace’s Equation

628

Contents

9.8 Higher-Dimensional Problems; Nonhomogeneous Equations

641

Project

655

Chapter 10

10.1 10.2 10.3 10.4

First Order Partial Differential Equations and the Method of Characteristics

Introduction

659

The Cauchy Problem

662

Existence and Uniqueness

668

The Method of Characteristics

Chapter 11

671 679

Projects

11.1 11.2 11.3 11.4

659

Linear Two-Point Boundary Value Problems

681

Introduction

681

Existence and Uniqueness

682

Two-Point Boundary Value Problems for Linear Systems

693

Sturm-Liouville Boundary Value Problems

705

Project

715

Answers

A-1

Index

I-1

ix

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Preface This book is designed for the undergraduate differential equations course taken by students majoring in science and engineering. A year of calculus is the prerequisite. The main goal of the text is to help students integrate the underlying theory, solution procedures, and computational aspects of differential equations as seamlessly as possible. Since we want the text to be easy to read and understand, we discuss the theory as simply as possible and emphasize how to use it. When developing models, we try to guide the reader carefully through the physical principles underlying the mathematical model. We also emphasize the importance of common sense, intuition, and “back of the envelope’’ checks. When solving problems, we remind the student to ask “Does my answer make sense?’’ Where appropriate, examples and exercises ask the student to anticipate and subsequently interpret the physical content of their solution. (For example, “Should an equilibrium solution exist for this application? If so, why? What should its value be?’’) We believe that developing this mind-set is particularly important in resisting the temptation to accept almost any computer-generated output as correct. Chapters 9, 10, and 11, dealing with partial differential equations and boundary value problems, are self-contained; they can be covered in any order.

New Features As in the ﬁrst edition, we have made a determined effort to write a text that is easy to understand. In response to the suggestions of ﬁrst edition users and reviewers, this second edition offers even more support for both students and instructors.

• We have followed the advice of our reviewers to provide a more concise presentation. First order differential equations (linear and nonlinear) are now discussed in a single chapter, Chapter 2. Similarly, second order and higher order linear equations are discussed in one chapter, Chapter 3. The bulk of Chapter 3 develops the theory for the second order equations; the last three sections extend these ideas to higher order linear equations. • The introductory discussion for linear systems (Chapter 4) has been streamlined to reach the computational aspects of the theory as quickly as possible. • Chapter 11 has been shortened to focus solely on linear two-point boundary value problems (for second order scalar equations and systems). • We have included a review of core material in the form of a set of review exercises at the end of Chapters 2, 3, and 4. These exercises, consisting of initial value problems for ﬁrst order equations, higher order linear equations, and ﬁrst order linear systems, respectively, require the student to select as well as apply the appropriate solution technique developed in the chapter.

xi

xii

Preface

• We have added a number of new exercises, ranging from routine drill exercises to those with applications to a variety of different disciplines. Answers to the odd-numbered exercises are again given at the back of the text. • A brief look at boundary value problems appears as a project at the end of Chapter 3. This brief introductory overview of linear two-point boundary value problems highlights how these problems differ from their initial-value counterparts. • We have added projects. There are now short projects at the end of each chapter. Some of these are challenging applications. Others are intended to expand the student’s mathematical horizons, showing how the material in the chapter can be generalized. In certain applications, such as food processing, the project exposes the student to the mathematics aspects of current research.

A Multilevel Development of Certain Topics Numerical Methods. The basic ideas underlying numerical methods and their use in applications are presented early for both scalar problems and systems. In Chapters 2 and 4, after Euler’s method is developed, the route to more accurate algorithms is brieﬂy outlined. The Runge-Kutta algorithm is then offered as an example of such an improved algorithm; accompanying exercises allow the student to apply the algorithm and experience its increased accuracy at an introductory level. Chapter 7 subsequently builds upon this introduction, developing a comprehensive treatment of one-step methods. Phase Plane. An introduction to the phase plane is provided in Chapter 4 as the different solutions of the homogeneous constant coefﬁcient linear system are developed. These ideas are then revisited and extended in the Chapter 6 discussion of autonomous nonlinear systems. Boundary Value Problems. As previously mentioned, the brief introductory exposure to linear two-point boundary value problems is presented at the end of Chapter 3. The purpose here is to make the student aware of such problems and to point out how they differ from initial value problems. Chapter 11 provides a more thorough study of these problems.

Supplements The Student’s Solutions Manual (0-321-28837-8) contains detailed solutions to the odd-numbered problems. The Instructor’s Solutions Manual (0-321-28838-6) contains detailed solutions to most problems. The Online Technology Resource Manual includes suggestions for how to use a computer algebra system with the text. Speciﬁc instructions are given for MATLAB and Mathematica. It is available at http://www.aw-bc.com/kohler/.

Preface

xiii

Acknowledgments Many people helped and encouraged us in this effort. Besides the support provided by our families, we are especially thankful for the editorial and developmental assistance of William Hoffman, our editor at Pearson Addison-Wesley. We are very grateful to our reviewers, who made many insightful suggestions that improved the text: John Baxley William Beckner Jerry L. Bona Thomas Branson Dennis Brewer Peter Brinkmann Almut Burchard Donatella Danielli Charles Friedman Moses Glasner Weimin Han Harumi Hattori Jason Huffman Gregor Kovacic Yang Kuang Zhiqin Lu Peter Mucha John Neuberger Lei Ni Joan Remski Sinai Robins Robert Snider A. Shadi Tahvildar-Zadeh Jason Whitt Tamas Wiandt Jennifer Zhao Fangyang Zheng Kehe Zhu

Wake Forest University University of Texas at Austin University of Illinois, Chicago University of Iowa University of Arkansas University of Illinois, Urbana-Champaign University of Virginia Purdue University University of Texas at Austin Pennsylvania State University—University Park University of Iowa West Virginia University Georgia College & State University Rensselaer Polytechnic Institute Arizona State University University of California—Irvine Georgia Institute of Technology Northern Arizona University University of California—San Diego University of Michigan—Dearborn Temple University Virginia Polytechnic Institute & State University Rutgers University Rhodes College Rochester Institute of Technology University of Michigan—Dearborn Ohio State University State University of New York at Albany

Special thanks are due to Peter Mucha. His ongoing interest and constructive feedback helped us greatly during the revision process. Our good friend and colleague George Flick continued to provide us with examples of applications from the life sciences; we greatly appreciate his encouragement and assistance. Tiri Chinyoka and Ermira Cami helped us at Virginia Tech with proofreading and problem-checking. We also thank Jeremy Bourdon and our Virginia Tech colleague Terri Bourdon for revising the solutions manuals. We are very grateful for the professional expertise provided by the personnel at Addison-Wesley. Christine O’Brien was our project manager, and she made certain that we maintained our schedule. We are grateful to Peggy McMahon, our Production Supervisor, to Barbara Atkinson for the design of the text, to Jeanne Yost for her careful copyediting, to Rena Lam at Techsetters, Inc. for superb typesetting, and to Sally Liﬂand at Liﬂand et al., Bookmakers for her careful oversight and coordination of the revision process.

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ELEMENTARY DIFFERENTIAL EQUATIONS with Boundary Value Problems

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C H A P T E R

Introduction to Diﬀerential Equations

1

CHAPTER OVERVIEW 1.1

Introduction

1.2

Examples of Differential Equations

1.3

Direction Fields

1.1

Introduction Scientists and engineers develop mathematical models for physical processes as an aid to understanding and predicting the behavior of the processes. In this book we discuss mathematical models that help us understand, among other things, decay of radioactive substances, electrical networks, population dynamics, dispersion of pollutants, and trajectories of moving objects. Modeling a physical process often leads to equations that involve not only the physical quantity of interest but also some of its derivatives. Such equations are referred to as differential equations. In Section 1.2, we give some simple examples that show how mathematical models are derived. We also begin our study of differential equations by introducing the corresponding terminology and by presenting some concrete examples of differential equations. Section 1.3 introduces the idea of a direction ﬁeld for a differential equation. The concept of direction ﬁelds allows us to visualize, in geometric terms, the graphs of solutions of differential equations.

1.2

Examples of Diﬀerential Equations When we apply Newton’s second law of motion, ma = f , to an object moving in a straight line, we obtain a differential equation of the form my (t) = f (t, y(t), y (t)). 1

(1)

2

CHAPTER 1

Introduction to Differential Equations

In equation (1), y(t) represents the position, at time t, of the object. As expressed in equation (1), the product of mass m and acceleration y (t) is equal to the sum of the applied forces. The applied forces [the right-hand side of equation (1)] often depend on time t, position y(t), and velocity y (t). E X A M P L E

1

One of the simplest examples of linear motion is an object falling under the inﬂuence of gravity. Let y(t) represent the height of the object above the surface of the earth, and let g denote the constant acceleration due to gravity (32 ft/sec2 or 9.8 m/s2 ). See Figure 1.1. – mg

y(t)

FIGURE 1.1

The only force acting on the falling body is its weight. The body’s position, y(t), is governed by the differential equation y = −g.

Since the only force acting on the body is assumed to be its weight, W = mg, equation (1) reduces to my (t) = −mg, or y (t) = −g.

(2)

The negative sign appears on the right-hand side of the equation because the acceleration due to gravity is positive downward, while we assumed y to be positive in the upward direction. (Again, see Figure 1.1.) Equation (2) is solved easily by taking successive antiderivatives. The ﬁrst antiderivative gives the object’s velocity, y (t) = −gt + C1 . Another antidifferentiation gives the object’s position, y(t) = − 12 gt2 + C1 t + C2 . Here, C1 and C2 represent arbitrary constants of integration. ❖

Notice in Example 1 that the solution involves two undetermined constants. This means that, by itself, differential equation (2) does not completely specify the solution y(t). This makes sense physically since, to completely determine the motion, we also need some information about the initial state of the object. The arbitrary constants of integration that arise are often speciﬁed by prescribing velocity and position at some initial time, say t = 0. For example, if the object’s

1.2

Examples of Differential Equations

3

initial velocity is y (0) = v 0 and its initial position is y(0) = y0 , then we obtain a complete description of velocity and position: y (t) = −gt + v 0 ,

y(t) = − 12 gt2 + v 0 t + y0 .

Unless an application suggests otherwise, we normally use t to represent the independent variable and y to represent the dependent variable. Thus, in a typical differential equation, we are searching for a solution y(t). As is common in a mathematics text, we use a variety of notations to denote derivatives. For instance, we may use d2 y/dt2 instead of y (t) or d4 y/dt4 instead of y(4) (t). In addition, we often suppress the independent variable t and simply write y and y instead of y(t) and y (t). An example using this notation is the differential equation y +

1 y + t3 y = 5. t

E X A M P L E

2

Scientists have observed that radioactive materials have an instantaneous rate of decay (that is, a rate of decrease) that is proportional to the amount of material present. If Q(t) represents the amount of material present at time t, then dQ/dt is proportional to Q(t); that is, dQ = −kQ, k > 0. (3) dt The negative sign in equation (3) arises because Q is both positive and decreasing; that is, Q(t) > 0 and Q (t) < 0. Unlike equation (2), differential equation (3) cannot be solved by integrating the right-hand side, −kQ(t), because Q(t) is not known. Instead, equation (3) requires that we somehow ﬁnd a function Q(t) whose derivative, Q (t), is a constant multiple of Q(t). Recall that the exponential function has a derivative that is a constant multiple of itself. For example, if y = Ce−kt , then y = −kCe−kt = −ky. Therefore, we see that a solution of equation (3) is Q(t) = Ce−kt ,

(4)

where C can be any constant. ❖

As in Example 1, the differential equation by itself does not completely specify the solution. But setting t = 0 in (4) leads to Q(0) = C. Therefore, the quantity Q(t) given in equation (4) is completely determined once the amount of material initially present is speciﬁed.

The Form of an nth Order Differential Equation We now state the formal deﬁnition of a differential equation and point to some issues that need to be addressed. An equation of the form y(n) = f (t, y, y , . . . , y(n−1) ) is called an nth order ordinary differential equation.

(5)

4

CHAPTER 1

Introduction to Differential Equations

In equation (5), t is the independent variable, while y is the dependent variable. A solution of the differential equation (5) is any function y(t) that satisﬁes the equation on our t-interval of interest. For instance, Example 2 showed that Q(t) = Ce−kt is a solution of Q = −kQ for any value of the constant C; the t-interval of interest for Example 2 is typically the interval 0 ≤ t < ∞. The order of a differential equation is the order of the highest derivative that appears in the equation. For example, y = −g is a second order differential equation. Similarly, Q = −kQ is a ﬁrst order differential equation. The form of the nth order ordinary differential equation (5) is not the most general one. In particular, an nth order ordinary differential equation is any equation of the form G(t, y, y , y , . . . , y(n) ) = 0. For example, the following equation is classiﬁed as a second order ordinary differential equation: t2 sin y + y ln y = 1. Notice that it is not possible to rewrite this equation in the explicit form y = f (t, y, y ). In our study, however, we usually consider only equations of the form y(n) = f (t, y, y , y , . . . , y(n−1) ), where the nth derivative is given explicitly in terms of t, y, and lower order derivatives of y. Differential equation (5) is called ordinary because the equation involves just a single independent variable, t. This is in contrast to other equations called partial differential equations, which involve two or more independent variables. An example of a partial differential equation is the one-dimensional wave equation ∂ 2 u(x, t) ∂x2

−

∂ 2 u(x, t) ∂t2

= 0.

Here, the dependent variable u is a function of two independent variables, the spatial coordinate x and time t.

Initial Value Problems What we have seen about differential equations thus far raises some important questions that we will address throughout this book. One such question is “What constitutes a properly formulated problem?” Examples 1 and 2 illustrate that auxiliary initial conditions are required if the differential equation is to have a unique solution. The differential equation, together with the proper number of initial conditions, constitutes what is known as an initial value problem. For instance, an initial value problem associated with the falling object in Example 1 consists of the differential equation together with initial conditions specifying the object’s initial position and velocity: d2 y dt2

= −g,

y(0) = y0 ,

y (0) = v 0 .

1.2

Examples of Differential Equations

5

Similarly, an initial value problem associated with the radioactive decay process in Example 2 consists of the differential equation together with a speciﬁcation of the initial amount of the substance: dQ = −kQ, Q(0) = Q0 . dt These examples suggest that the number of initial conditions we need to specify must be equal to the order of the differential equation. When we address the question of properly formulating problems, it will be apparent that this is the case. Once we understand how to properly formulate the problem to be solved, the obvious next question is “How do we go about solving this problem?” Answering the two questions 1. How do we properly formulate the problem? 2. How do we solve the problem? is central to the study of differential equations.

Solving Initial Value Problems As Chapters 2, 3, and 4 show, certain special types of differential equations have formulas for the general solution. The general solution is an expression containing arbitrary constants (or parameters) that can be adjusted so as to give every solution of the equation. Finding the general solution is often the ﬁrst step in solving an initial value problem. The next example illustrates this idea. E X A M P L E

3

Consider the initial value problem y + 3y = 6t + 5,

y(0) = 3.

(6)

(a) Show, for any constant C, that y = Ce−3t + 2t + 1

(7)

is a solution of the differential equation y + 3y = 6t + 5. (b) Use expression (7) to solve the initial value problem (6). Solution: (a) Inserting expression (7) into the differential equation y + 3y = 6t + 5, we ﬁnd y + 3y = (Ce−3t + 2t + 1) + 3(Ce−3t + 2t + 1) = (−3Ce−3t + 2) + (3Ce−3t + 6t + 3) = 6t + 5. Therefore, for any value C, y = Ce−3t + 2t + 1 is a solution of y + 3y = 6t + 5. (b) Imposing the constraint y(0) = 3 on y(t) = Ce−3t + 2t + 1 leads to y(0) = C + 1 = 3. Therefore, C = 2, and a solution of the initial value problem is y = 2e−3t + 2t + 1. ❖

6

CHAPTER 1

Introduction to Differential Equations

We will show later that y = Ce−3t + 2t + 1 is the general solution of the differential equation in Example 3. A geometric interpretation is given in Figure 1.2, which shows graphs of the general solution for representative values of C. The solution whose graph passes through the point (t, y) = (0, 3) is the one that solves the initial value problem posed in Example 3. y C=3 6 C=2 4 (0, 3) 2 C=0 0.5

1

1.5

2

t

–2 C = –1 –4 C = –3 FIGURE 1.2 −3t

For any constant C, y = Ce + 2t + 1 is a solution of y + 3y = 6t + 5. Solution curves are displayed for several values of C. For C = 2, the curve passes through the point (t, y) = (0, 3); this is the solution of the initial value problem posed in Example 3.

EXERCISES Exercises 1–4: What is the order of the differential equation? 1. y + 3ty3 = 1

2. t4 y + y sin t = 6

3. ( y )3 + t5 sin y = y4

4. ( y )4 −

t2 =0 (y ) + 4 4

Exercises 5–8: For what value(s) of the constant k, if any, is y(t) a solution of the given differential equation? 5. y + 2y = 0,

y(t) = ekt

7. y + (sin 2t)y = 0,

y(t) = ek cos 2t

6. y − y = 0,

y(t) = ekt

8. y + y = 0,

y(t) = ke−t

2

9. (a) Show that y(t) = Cet is a solution of y − 2ty = 0 for any value of the constant C. (b) Determine the value of C needed for this solution to satisfy the initial condition y(1) = 2. 10. Solve the differential equation y = 2 by computing successive antiderivatives. What is the order of this differential equation? How many arbitrary constants arise in the antidifferentiation solution process? 11. (a) Show that y(t) = C1 sin 2t + C2 cos 2t is a solution of the differential equation y + 4y = 0, where C1 and C2 are arbitrary constants. (b) Find values of the constants C1 and C2 so that the solution satisﬁes the initial conditions y(π/4) = 3, y (π/4) = −2.

1.2

Examples of Differential Equations

7

12. Suppose y(t) = 2e−4t is the solution of the initial value problem y + ky = 0, y(0) = y0 . What are the constants k and y0 ? 13. Consider t > 0. For what value(s) of the constant c, if any, is y(t) = c/t a solution of the differential equation y + y2 = 0? 14. Let y(t) = −e−t + sin t be a solution of the initial value problem y + y = g(t), y(0) = y0 . What must the function g(t) and the constant y0 be? 15. Consider t > 0. For what value(s) of the constant r, if any, is y(t) = tr a solution of the differential equation t2 y − 2ty + 2y = 0? 16. Show that y(t) = C1 e2t + C2 e−2t is a solution of the differential equation y − 4y = 0, where C1 and C2 are arbitrary constants.

Exercises 17–18: Use the result of Exercise 16 to solve the initial value problem. 17. y − 4y = 0,

y(0) = 2, y (0) = 0

18. y − 4y = 0,

y(0) = 1, y (0) = 2

Exercises 19–20: Use the result of Exercise 16 to ﬁnd a function y(t) that satisﬁes the given conditions. 19. y − 4y = 0,

y(0) = 3, lim y(t) = 0 t→∞

20. y − 4y = 0,

y(0) = 10,

lim y(t) = 0

t→−∞

Exercises 21–22: The graph shows the solution of the given initial value problem. In each case, m is an integer. In Exercise 21, determine m, y0 , and y(t). In Exercise 22, determine m, t0 , and y(t). 21. y (t) = m + 1,

22. y (t) = mt,

y(1) = y0

y

y

2.5

2.5

2

2

1.5

1.5

1

1

0.5

0.5

–0.5

y(t0 ) = −1

0.5

1

1.5

2

–0.5

2.5

t

–0.5

0.5

1

1.5

2

2.5

t

–0.5

Figure for Exercise 21

Figure for Exercise 22

23. At time t = 0, an object having mass m is released from rest at a height y0 above the ground. Let g represent the (constant) gravitational acceleration. Derive an expression for the impact time (the time at which the object strikes the ground). What is the velocity with which the object strikes the ground? (Express your answers in terms of the initial height y0 and the gravitational acceleration g.) 24. A car, initially at rest, begins moving at time t = 0 with a constant acceleration down a straight track. If the car achieves a speed of 60 mph (88 ft/sec) at time t = 8 sec, what is the car’s acceleration? How far down the track will the car have traveled when its speed reaches 60 mph?

8

CHAPTER 1

1.3

Introduction to Differential Equations

Direction Fields Before beginning a systematic study of differential equations, we consider a geometric entity called a direction ﬁeld, which will assist in understanding the ﬁrst order differential equation y = f (t, y). A direction ﬁeld is a way of predicting the qualitative behavior of solutions of a differential equation. A good way to understand the idea of a direction ﬁeld is to recall the “iron ﬁlings” experiment that is often done in science classes to illustrate magnetism. In this experiment, iron ﬁlings (minute ﬁlaments of iron) are sprinkled on a sheet of cardboard, beneath which two magnets of opposite polarity are positioned. When the cardboard sheet is gently tapped, the iron ﬁlings align themselves so that their axes are tangent to the magnetic ﬁeld lines. At a given point on the sheet, the orientation of an iron ﬁling indicates the direction of the magnetic ﬁeld line. The totality of oriented iron ﬁlings gives a good picture of the ﬂow of magnetic ﬁeld lines connecting the two magnetic poles. Figure 1.3 illustrates this experiment. Iron filings

Cardboard sheet

Bar magnet

FIGURE 1.3

The orientation of iron ﬁlings gives a good picture of the ﬂow of magnetic ﬁeld lines connecting two magnetic poles.

The Direction Field for a Differential Equation What is the connection between the iron ﬁlings experiment illustrated in Figure 1.3 and a qualitative understanding of differential equations? From calculus we know that if we graph a differentiable function y(t), the slope of the curve at the point (t, y(t)) is y (t). If y(t) is a solution of a differential equation y = f (t, y), then we can calculate this slope by simply evaluating the right-hand side f (t, y) at the point (t, y(t)). For example, suppose y(t) is a solution of the equation y = 1 + 2ty

(1)

and suppose the graph of y(t) passes through the point (t, y) = (2, y(2)) = (2, −1). For differential equation (1), the right-hand side is f (t, y) = 1 + 2ty. Thus, we ﬁnd y (2) = f (2, y(2)) = f (2, −1) = 1 + 2(2)(−1) = −3. Even though we have not solved y = 1 + 2ty, the preceding calculation has taught us something about the speciﬁc solution y(t) passing through (t, y) = (2, −1): it is decreasing (with slope equal to −3) when it passes through the point (t, y) = (2, −1).

1.3

Direction Fields

9

To exploit this idea, suppose we systematically evaluate the right-hand side f (t, y) at a large number of points (t, y) throughout a region of interest in the ty-plane. At each point, we evaluate the function f (t, y) to determine the slope of the solution curve passing through that point. We then sketch a tiny line segment at that point, oriented with the given slope f (t, y). The resulting picture, called a direction ﬁeld, is similar to that illustrated in Figure 1.3. Using such a direction ﬁeld, we can create a good qualitative picture of the ﬂow of solution curves throughout the region of interest. E X A M P L E

1

(a) Sketch a direction ﬁeld for y = 1 + 2ty in the square −2 ≤ t ≤ 2, −2 ≤ y ≤ 2. (b) Using the direction ﬁeld, sketch your guess for the solution curve passing through the point P = (−2, 2). Also, using the direction ﬁeld, sketch your guess for the solution curve passing through the point Q = (0, −1). Solution: The direction ﬁeld for y = 1 + 2ty shown in Figure 1.4(a) was computer generated. There are a number of computer programs available for drawing direction ﬁelds. Figure 1.4(b) shows our guesses for the solutions of the initial value problems in part (b). y

y

2

2

1.5

1.5

1

1

0.5

0.5

t

t –2 –1.5 –1 –0.5 –0.5

0.5

1

1.5

2

–2

–1.5 –1

–0.5 –0.5

–1

–1

–1.5

–1.5

–2

–2

(a)

0.5

1

1.5

2

(b) FIGURE 1.4

(a) The direction ﬁeld for y = 1 + 2ty. (b) Using the direction ﬁeld, we have drawn our guess for the solution of y = 1 + 2ty, y(−2) = 2 and for the solution of y = 1 + 2ty, y(0) = −1.

❖

Isoclines The “method of isoclines” is helpful when you need to draw a direction ﬁeld by hand. An isocline of the differential equation y = f (t, y) is a curve of the form f (t, y) = c, where c is a constant. For example, consider the differential equation y = y − t 2 .

10

CHAPTER 1

Introduction to Differential Equations

In this case, curves of the form y − t2 = c are isoclines of the differential equation. (These curves, y = t2 + c, are parabolas opening upward. Each has its vertex on the y-axis.) Isoclines are useful because, at every point on an isocline, the associated direction ﬁeld ﬁlaments have the same slope, namely f (t, y) = c. (In fact, the word “isocline” means “equal inclination” or “equal slope.”) To carry out the method of isoclines, we ﬁrst sketch, for various values of c, the corresponding curves f (t, y) = c. Then, at representative points on these curves, we sketch direction ﬁeld ﬁlaments having slope f (t, y) = c. E X A M P L E

2

(a) Use the method of isoclines to sketch the direction ﬁeld for y = y − t. Restrict your direction ﬁeld to the square deﬁned by −2 ≤ t ≤ 2, −2 ≤ y ≤ 2. (b) Using the direction ﬁeld, sketch your guess for the solution curve passing through the point (−1, 12 ). Also, sketch your guess for the solution curve passing through the point (−1, − 12 ). Solution: For the equation y = y − t, lines of the form y = t + c are isoclines. In Figure 1.5(a) we have drawn the isoclines y = t + 3, y = t + 2, . . . , y = t − 2. At selected points along an isocline of the form y = t + c, we have drawn direction ﬁeld ﬁlaments, each having slope c. Figure 1.5(b) shows our guesses for the solutions of the initial value problems in part (b). In addition, note that the line y = t + 1 appears to be a solution curve. y c=3

y c=1

2

c=3

c=0

c=2

c=1

2

c=0

c=2 1.5

1.5 c = –1

1 0.5 –2 –1.5 –1 –0.5 –0.5

c = –1

1 0.5

0.5

1

1.5

2

c = –2

t

–2 –1.5 –1 –0.5 –0.5

–1

–1

–1.5

–1.5

–2

–2

(a)

0.5

1

1.5

2

t

c = –2

(b) FIGURE 1.5

(a) The method of isoclines was used to sketch the direction ﬁeld for y = y − t. (b) Using the direction ﬁeld, we have sketched our guesses for the solutions of the initial value problems in part (b) of Example 2.

❖

Direction Fields for Autonomous Equations The method of isoclines is particularly well suited for differential equations that have the special form y = f ( y).

(2)

1.3

Direction Fields

11

For equations of this form, the isoclines are horizontal lines. That is, if b is any number in the domain of f ( y), then the horizontal line y = b is an isocline of the equation y = f ( y). In particular, y has the same value f (b) all along the horizontal line y = b. Differential equations of the form (2), where the right-hand side does not depend explicitly on t, are called autonomous differential equations. An example of an autonomous differential equation is y = y2 − 3y. By contrast, the differential equation y = y + 2t is not autonomous. Autonomous differential equations are quite important in applications, and we study them in Chapter 2. As noted with respect to the autonomous equation y = f ( y), all the slopes of direction ﬁeld ﬁlaments along the horizontal line y = b are equal. This fact is illustrated in Figure 1.6, which shows the direction ﬁeld for the differential equation y = y(2 − y). For instance, the ﬁlaments along the line y = 1 all have slope equal to 1. Similarly, the ﬁlaments along the line y = 2 all have slope equal to 0. In fact, looking at Figure 1.6, the horizontal lines y = 0 and y = 2 appear to be solution curves for the differential equation y = y(2 − y). This is indeed the case, as we show in the next subsection. y 3 2.5 2 1.5 1 0.5

t –0.5

1

2

3

4

5

6

–1

FIGURE 1.6

The direction ﬁeld for the autonomous equation y = y(2 − y), together with portions of the graphs of some typical solutions. For an autonomous equation, the slopes are constant along horizontal lines.

Equilibrium Solutions Consider the autonomous differential equation y = y(2 − y) whose direction ﬁeld is shown in Figure 1.6. The horizontal lines y = 0 and y = 2 appear to be solution curves for this differential equation. In fact, by substituting either of the constant functions y(t) = 0 or y(t) = 2 into the differential equation, we see that it is a solution of y = y(2 − y). In general, consider the autonomous differential equation y = f ( y).

12

CHAPTER 1

Introduction to Differential Equations

If the real number β is a root of the equation f ( y) = 0, then the constant function y(t) = β is a solution of y = f ( y). Conversely, if the constant function y(t) = β is a solution of y = f (y), then β must be a root of f ( y) = 0. Constant solutions of an autonomous differential equation are known as equilibrium solutions.

REMARK: It is possible for differential equations that are not autonomous to have constant solutions. For example, y(t) = 0 is a solution of y = ty + sin y and y(t) = 1 is a solution of y = ( y − 1)t2 . We will refer to any constant solution of a differential equation (autonomous or not) as an equilibrium solution. E X A M P L E

3

Find the equilibrium solutions (if any) of y = y2 − 4y + 3. Solution:

The right-hand side of the differential equation is f ( y) = y2 − 4y + 3 = ( y − 1)( y − 3).

Therefore, the equilibrium solutions are the constant functions y(t) = 1 and y(t) = 3. ❖

EXERCISES Exercises 1–6: (a) State whether or not the equation is autonomous. (b) Identify all equilibrium solutions (if any). (c) Sketch the direction ﬁeld for the differential equation in the rectangular portion of the ty-plane deﬁned by −2 ≤ t ≤ 2, −2 ≤ y ≤ 2. 1. y = −y + 1

2. y = t − 1

3. y = sin y

4. y = y2 − y

5. y = −1

6. y = −ty

Exercises 7–9: (a) Determine and sketch the isoclines f ( t, y ) = c for c = −1, 0, and 1. (b) On each of the isoclines drawn in part (a), add representative direction ﬁeld ﬁlaments. 7. y = −y + 1

8. y = −y + t

9. y = y2 − t2

Exercises 10–13: Find an autonomous differential equation that possesses the speciﬁed properties. [Note: There are many possible solutions for each exercise.] 10. Equilibrium solutions at y = 0 and y = 2; y > 0 for 0 < y < 2; y < 0 for −∞ < y < 0 and 2 < y < ∞. 11. An equilibrium solution at y = 1; y < 0 for −∞ < y < 1 and 1 < y < ∞. 12. A differential equation with no equilibrium solutions and y > 0 for all y. 13. Equilibrium solutions at y = n/2, n = 0, ± 1, ± 2, . . . .

1.3

Direction Fields

13

Exercises 14–19: Consider the six direction ﬁeld plots shown. Associate a direction ﬁeld with each of the following differential equations. y

y 2

2

1.5

1.5

1

1

0.5

0.5

t

t –2

–1.5 –1 –0.5 –0.5

0.5

1

1.5

–2 –1.5 –1 –0.5 –0.5

2

–1

–1

–1.5

–1.5

–2

–2

0.5

1

Direction Field A

Direction Field B

y

y

2

2

1.5

1.5

1

1

0.5

0.5

1.5

2

1.5

2

1.5

2

t –2 –1.5 –1 –0.5 –0.5

0.5

1

1.5

2

t –2 –1.5 –1 –0.5 –0.5

–1

–1

–1.5

–1.5

–2

–2

0.5

1

Direction Field C

Direction Field D

y

y

2

2

1.5

1.5

1

1

0.5

0.5

t

t –2

–1.5 –1 –0.5 –0.5

0.5

1

1.5

2

–2 –1.5 –1 –0.5 –0.5

–1

–1

–1.5

–1.5

–2

–2

Direction Field E

0.5

1

Direction Field F Figure for Exercises 14 –21

14

CHAPTER 1

Introduction to Differential Equations 14. y = −y 1 17. y = − 2

15. y = −t + 1

16. y = y2 − 1 1 18. y = y + t 19. y = 1 + y2 20. For each of the six direction ﬁelds shown, assume we are interested in the solution that satisﬁes the initial condition y(0) = 0. Use the graphical information contained in the plots to roughly estimate y(1). 21. Repeat Exercise 20 with y(0) = 0 as before, but this time estimate y(−1).

C H A P T E R

2

First Order Diﬀerential Equations CHAPTER OVERVIEW 2.1

Introduction

2.2

First Order Linear Differential Equations

2.3

Introduction to Mathematical Models

2.4

Population Dynamics and Radioactive Decay

2.5

First Order Nonlinear Differential Equations

2.6

Separable First Order Equations

2.7

Exact Differential Equations

2.8

The Logistic Population Model

2.9

Applications to Mechanics

2.10

Euler’s Method

2.1

Introduction First order differential equations arise in modeling a wide variety of physical phenomena. In this chapter we study the differential equations that model applications such as population dynamics, radioactive decay, belt friction, and mixing and cooling. Chapter 2 has two main parts. The ﬁrst part, consisting of Sections 2.1–2.4, focuses on ﬁrst order linear differential equations and their applications. The second part, consisting of Sections 2.5–2.9, treats ﬁrst order nonlinear equations. The ﬁnal section, Section 2.10, introduces numerical techniques, such as Euler’s method and Runge-Kutta methods, that can be used to approximate the solution of a ﬁrst order differential equation. 15

16

CHAPTER 2

First Order Differential Equations

First Order Linear Differential Equations A differential equation of the form y + p(t)y = g(t)

(1)

is called a ﬁrst order linear differential equation. In equation (1), p(t) and g(t) are functions deﬁned on some t-interval of interest, a < t < b. If the function g(t) on the right-hand side of (1) is the zero function, then equation (1) is called homogeneous. If g(t) is not the zero function, then equation (1) is nonhomogeneous. A ﬁrst order equation that can be put into the form of equation (1) by algebraic manipulations is also called a ﬁrst order linear differential equation. For example, the following are ﬁrst order linear differential equations: (a) e−t y + 3ty = sin t

(b)

1 ln t y + t2 = . y y

A ﬁrst order differential equation that cannot be put into the form of equation (1) is called nonlinear. As we will see, it is possible to ﬁnd an explicit representation for the solution of a ﬁrst order linear equation. By contrast, most ﬁrst order nonlinear equations cannot be solved explicitly. In Sections 2.6 and 2.7, we will discuss solution techniques for certain special types of ﬁrst order nonlinear differential equations. E X A M P L E

1

Is the differential equation linear or nonlinear? If the equation is linear, decide whether it is homogeneous or nonhomogeneous. y + t3 = sin t (b) y = t2 y (c) (cos t)y + et y = sin t (d) (a) y = ty2 y Solution: (a) This equation is nonlinear because of the presence of the y2 term. (b) This equation is linear and homogeneous; it can be put in the form y − t2 y = 0. (c) This equation can be put into the form of equation (1), et y = tan t. cos t Therefore, the equation is linear and nonhomogeneous. (d) This equation can be rewritten as y +

y + (t3 − sin t) y = 0. Therefore, the equation is linear and homogeneous. ❖

Existence and Uniqueness for First Order Linear Initial Value Problems Before looking at how to solve a ﬁrst order linear equation, we want to address the following question: “What constitutes a properly formulated problem?” This question is answered by the following theorem, which we state now and prove in the Exercises at the end of Section 2.2.

2.1

Theorem 2.1

Introduction

17

Let p(t) and g(t) be continuous functions on the interval (a, b), and let t0 be in (a, b). Then the initial value problem y + p(t)y = g(t),

y(t0 ) = y0

has a unique solution on the entire interval (a, b).

Notice that the theorem states three conclusions. A solution exists, it is unique, and this unique solution exists on the entire interval (a, b). We will see that determining intervals of existence is considerably more complicated for nonlinear differential equations. The importance of Theorem 2.1 lies in the fact that it deﬁnes the framework within which we can construct solutions. In particular, suppose we are given a linear differential equation y + p(t)y = g(t) with coefﬁcient functions p(t) and g(t) that are continuous on (a, b). If we impose an initial condition of the form y(t0 ) = y0 , where a < t0 < b, the theorem tells us there is one and only one solution. Therefore, if we are able to construct a solution by using some technique we have discovered, the theorem guarantees that it is the only solution—there is no other solution we might have overlooked, one obtainable perhaps by a technique other than the one we are using. E X A M P L E

2

Consider the initial value problem y +

1 1 y= , t(t + 2) t−5

y(3) = 1.

What is the largest interval (a, b) on which Theorem 2.1 guarantees the existence of a unique solution? Solution: The coefﬁcient function p(t) = t−1 (t + 2)−1 has discontinuities at t = 0 and t = −2 but is continuous everywhere else. Similarly, q(t) = (t − 5)−1 has a discontinuity at t = 5 but is continuous for all other values t. Therefore, Theorem 2.1 guarantees that a unique solution exists on each of the following t-intervals: (−∞, −2),

(−2, 0),

(0, 5),

(5, ∞).

Since the initial condition is imposed at t = 3, we are guaranteed that a unique solution exists on the interval 0 < t < 5. (The solution might actually exist over a larger interval, but we cannot ascertain this without actually solving the initial value problem.) ❖

EXERCISES Exercises 1–10: Classify each of the following ﬁrst order differential equations as linear or nonlinear. If the equation is linear, decide whether it is homogeneous or nonhomogeneous. y − y cos t = t 2. y − sin t = ty2 3. 1. y − sin t = t2 y y

18

CHAPTER 2

First Order Differential Equations t2 + 1 y

4. y sin y = (t2 + 1)y

5. y sin t =

7. yy = t3 + y sin 3t

8. 2ty + ey y =

10.

6. 2ty + et y =

y t2 + 4

9.

y t +4 2

ty e3t = cos t + y (t4 + 2)y

y = cos t (t + 1)y 2

Exercises 11–14: Consider the following ﬁrst order linear differential equations. For each of the initial conditions, determine the largest interval a < t < b on which Theorem 2.1 guarantees the existence of a unique solution. 11. y +

t y = sin t t +1 2

(a) y(−2) = 1 12. y +

(b) y(0) = π

t y=0 t −4 2

(a) y(6) = 2 13. y +

(c) y(π) = 0

(b) y(1) = −1

(c) y(0) = 1

(d) y(−6) = 2

t

e t y= t−3 t2 − 4

(a) y(5) = 2 (d) y(−5) = 4 14. y + (t − 1)y = (a) y(3) = 0

(b) y(− 32 ) = 1 ln t + t−1 t−2 (b) y

(e)

1 2

y( 32 )

(c) y(0) = 0

=3

= −1

(c) y(− 12 ) = 1

(d) y(−3) = 2

2

15. If y(t) = 3et is known to be the solution of the initial value problem y + p(t)y = 0,

y(0) = y0 ,

what must the function p(t) and the constant y0 be? 16. (a) For what value of the constant C and exponent r is y = Ctr the solution of the initial value problem 2ty − 6y = 0,

y(−2) = 8?

(b) Determine the largest interval of the form (a, b) on which Theorem 2.1 guarantees the existence of a unique solution. (c) What is the actual interval of existence for the solution found in part (a)? 17. If p(t) is any function continuous on an interval of the form a < t < b and if t0 is any point lying within this interval, what is the unique solution of the initial value problem y + p(t)y = 0,

y(t0 ) = 0

on this interval? [Hint: If, by inspection, you can identify one solution of the given initial value problem, then Theorem 2.1 tells you that it must be the only solution.]

2.2

2.2

First Order Linear Differential Equations

19

First Order Linear Diﬀerential Equations In this section, we solve the ﬁrst order linear homogeneous differential equation y + p(t)y = 0,

(1)

and then we build on this result to solve the nonhomogeneous equation y + p(t)y = g(t).

Solving the Linear Homogeneous Equation Consider the homogeneous ﬁrst order linear equation y + p(t)y = 0, which we rewrite as y = −p(t)y.

(2)

We assume that p(t) is continuous on the t-interval of interest. To solve equation (2), we need to ﬁnd a function y(t) whose derivative is equal to −p(t) times y(t). Recall from calculus that d −P(t) = −P (t)e−P(t) . e dt The function y = e−P(t) has the property that y = −P (t)y. Therefore, if we choose a function P(t) such that P (t) = p(t), then y = e−P(t)

(3)

is a solution of y = −p(t)y. If P (t) = p(t), then P(t) is an antiderivative of p(t) and is usually denoted by the integral notation, P(t) = p(t) dt. So a solution of y = −p(t)y can be expressed as y = e−

p(t) dt

.

E X A M P L E

1

Find a solution of the differential equation y + 2ty = 0. Solution: For this linear equation, p(t) = 2t. For P(t) we can choose any convenient antiderivative of p(t). If we select P(t) = t2 , then, using (3), we obtain the solution y = e−t . 2

As a check, let y = e−t . Then y = −2te−t = −2ty. Thus, we have veriﬁed that 2 y = e−t is a solution of y + 2ty = 0. Figure 2.1 shows the direction ﬁeld for this differential equation, as well as a graph of the solution. 2

2

(continued)

20

CHAPTER 2

First Order Differential Equations (continued)

y 2 1.5 1 0.5

t –1

–0.5

0.5

1

–0.5 –1

FIGURE 2.1

The direction ﬁeld for the differential equation in Example 1 2 and the graph of a solution, y = e−t .

❖

The General Solution Equation (3) represents one solution of y + p(t)y = 0. But, in order to solve initial value problems, we need to develop a method for ﬁnding all the solutions. Observe that if we multiply solution (3) by any constant C, then the resulting function, y = Ce−P(t) ,

(4)

is also a solution. In fact (see Exercises 47–48), Theorem 2.1 can be used to show that every solution of y + p(t)y = 0 has the form (4) for some constant C. We call (4) the general solution of y + p(t)y = 0. E X A M P L E

2

Find the general solution of y + (cos t)y = 0. Solution: A convenient antiderivative for p(t) = cos t is P(t) = sin t. Thus, the general solution is y = Ce− sin t . ❖

REMARK: Let P(t) be an antiderivative of p(t). From calculus we know that any other antiderivative of p(t) has the form P(t) + K, where K is some constant. For instance, in Example 2, we chose P(t) = sin t as an antiderivative of p(t) = cos t. We could just as well have chosen P(t) = 2 + sin t as the antiderivative. In that case, the general solution would have had the form y = Ce−(2+sin t) = Ce−2 e− sin t = C1 e− sin t .

2.2

First Order Linear Differential Equations

21

In this expression, C is an arbitrary constant. Since C1 = Ce−2 , we can regard C1 as an arbitrary constant as well. Thus, no matter which antiderivative we choose, the general solution is still the product of an arbitrary constant and the function e− sin t .

Using the General Solution to Solve Initial Value Problems An initial value problem for a homogeneous ﬁrst order linear equation can be solved by ﬁrst forming the general solution y = Ce−P(t) and then choosing the constant C so as to satisfy the initial condition. E X A M P L E

3

Solve the initial value problem ty + 2y = 0,

y(1) = 5.

Solution: Notice that the differential equation is not in the standard form for a ﬁrst order linear equation. In order to use equation (4) to represent the general solution, we need to rewrite the differential equation as 2 y = 0, y(1) = 5. t As rewritten, p(t) = 2/t. A convenient antiderivative is 2 dt = 2 ln | t | = ln | t |2 = ln t2 . P(t) = t y +

Having an antiderivative P(t), we obtain the general solution 2

y = Ce−P(t) = Ce−ln t = Ct−2 . The initial condition y(1) = 5 requires that C = 5. Therefore, the unique solution of the initial value problem is y=

5 t2

. ❖

Example 3 illustrates a point about Theorem 2.1. The differential equation has a coefﬁcient function, p(t) = 2/t, that is not deﬁned and certainly not continuous at t = 0. Therefore, Theorem 2.1 cannot be used to guarantee that solutions exist across any interval containing t = 0. In fact, for this initial value problem, the solution, y(t) = 5/t2 , is not deﬁned at t = 0. However, if we change the initial condition in Example 3 to y(1) = 0, we ﬁnd that the solution is the zero function, y(t) = 0 (see Figure 2.2). Thus, even though this particular initial value problem does not satisfy the conditions of Theorem 2.1 on (−∞, ∞), it does in fact have a solution that is deﬁned for all t. It is important to realize that the failure of Theorem 2.1 to apply to an initial value problem does not imply that the solution must necessarily “behave badly.” The logical distinction is important. Theorem 2.1 asserts that if the hypotheses are satisﬁed, “good things will happen.” It does not assert that when the hypotheses are not satisﬁed, “bad things must happen.”

22

CHAPTER 2

First Order Differential Equations y (a) y(t) = 52 t

6

(b) y(t) = 12 t

4 2 (c) y(t) = 0 –0.5

0.5

1

1.5

2

2.5

3

3.5

t

–2 (d) y(t) = –2 t2 FIGURE 2.2

Some solutions of the problem ty + 2y = 0, y(1) = y0 , posed in Example 3. When y0 is nonzero, the solution is deﬁned only for t > 0. But if y0 = 0, the solution is the zero function and is deﬁned for all t.

First Order Linear Nonhomogeneous Equations We now solve the nonhomogeneous linear equation y + p(t)y = g(t). We assume that p(t) and g(t) are continuous on the t-interval of interest.

Integrating Factors As preparation for solving the nonhomogeneous equation, we reconsider the homogeneous equation from a slightly different point of view. In particular, the homogeneous equation has the form y + p(t)y = 0.

(5)

Let P(t) be some antiderivative of p(t), and deﬁne a new function μ(t) by μ(t) = eP(t) .

(6)

P(t)

The function μ(t) = e is called an integrating factor. We will shortly see the reason for this name. Note from equation (6) that μ (t) = P (t)eP(t) = p(t)μ(t).

(7a)

We multiply equation (5) by the integrating factor μ(t) to obtain a new equation, μ(t)y + μ(t)p(t)y = 0. From (7a), μ (t) = p(t)μ(t), and therefore μ(t)y + μ (t)y = 0.

(7b)

The left-hand side of equation (7b) is the derivative of a product and can be rewritten as d (μ(t)y(t)) = 0. (8) dt

2.2

First Order Linear Differential Equations

23

If the derivative of a function is identically zero, then the function must be constant. Therefore, equation (8) implies μ(t)y(t) = C, where C is a constant. Since μ(t)y(t) = C and since μ(t) = eP(t) is nonzero, we can solve for y(t): 1 C μ(t)

y(t) =

= Ce−P(t) . The derivation of equation (8) explains why the function μ(t) = eP(t) is called an “integrating factor.” That is, we multiply equation (5) by μ(t) to obtain the new equation (8), which can be integrated. Also note that this derivation leads to the same general solution for equation (5) that we found in equation (4), y = Ce−P(t) . The purpose of the derivation is to introduce the concept of an integrating factor.

Using an Integrating Factor to Solve the Nonhomogeneous Equation Now consider the nonhomogeneous equation y + p(t)y = g(t). If we multiply equation (9) by the integrating factor μ(t) = e μ(t)y + μ(t)p(t)y = μ(t)g(t). Since μ (t) = μ(t)p(t), we have

(9) P(t)

, we obtain

μ(t)y + μ (t)y = μ(t)g(t), or d (μ(t)y(t)) = μ(t)g(t). dt Integrating both sides gives μ(t)y(t) =

μ(t)g(t) dt + C,

where C is a constant and where μ(t)g(t) dt represents some particular antiderivative of μ(t)g(t). Solving for y(t), we are led to the general solution of the nonhomogeneous equation (9): −P(t) y=e (10) eP(t) g(t) dt + Ce−P(t) . REMARKS: 1. Don’t be confused by the notation. In particular, the terms e−P(t) and eP(t) in (10) do not cancel; eP(t) is part of the function eP(t) g(t) whose antiderivative must be determined. Once this antiderivative has been calculated, it is multiplied by the term e−P(t) . 2. Notice that the general solution given by (10) is the sum of two terms, e−P(t) eP(t) g(t) dt and Ce−P(t) . The ﬁrst term is some particular solution of the nonhomogeneous equation, while the second term represents the

24

CHAPTER 2

First Order Differential Equations

general solution of the homogeneous equation. We’ll see this same solution structure again when we study higher order linear equations and systems of linear equations. 3. Observe that the general solution contains only one arbitrary constant, C. This constant is determined by imposing an initial condition. 4. Although expression (10) is the general solution of the nonhomogeneous equation, you should not try to memorize it. Instead, remember the steps leading to (10). E X A M P L E

4

Find the general solution and then solve the initial value problem y + 2ty = 4t,

y(0) = 5.

Solution: For this differential equation, p(t) = 2t. An antiderivative is P(t) = t2 , and so an integrating factor is 2

μ(t) = et . Multiplying the differential equation by μ(t), we obtain et y + 2tet y = 4tet 2

2

2

(et y) = 4tet . 2

or

2

Therefore, 2

2

et y = 2et + C. Solving for y, we obtain the general solution y = 2 + Ce−t . 2

Imposing the initial condition y(0) = 5, we ﬁnd y = 2 + 3e−t . 2

The solution is graphed in Figure 2.3. y 6 5 4

y(t) = 2 + 3e–t

2

3 y=2

2 1 –3

–2

–1

1

2

3

t

–1

FIGURE 2.3

The solution of the problem posed in Example 4 is y = 2 + 3e−t . 2

❖

Example 4 illustrates the second remark following equation (10). The gen2 eral solution we found (namely y = 2 + Ce−t ) is the sum of some particular so-

2.2

First Order Linear Differential Equations

25

lution of the nonhomogeneous equation (namely the constant function y = 2) 2 and the general solution of the homogeneous equation (namely y = Ce−t ). Note that the initial condition was imposed on the general solution as the last step. This will always be the case.

Discontinuous Coefﬁcient Functions In some applications, physical conditions undergo abrupt changes. For example, a hot metal object might be plunged suddenly into a cooling bath, or we might throw a switch and abruptly change the source voltage in an electrical network. Such applications are often modeled by an initial value problem y + p(t)y = g(t),

y(a) = y0 ,

a ≤ t ≤ b,

where one or both of the functions p(t) and g(t) have a jump discontinuity at some point, say t = c, where a < c < b. In such cases, even though y (t) is not continuous at t = c, we expect on physical grounds that the solution y(t) is continuous at t = c. For these problems we ﬁrst solve the initial value problem on the interval a ≤ t < c; the solution y(t) will have a one-sided limit, lim y(t) = y(c− ).

t→c−

To complete the solution, we use the limiting value y(c− ) as the initial condition on the subinterval [c, b] and then solve a second initial value problem on [c, b]. E X A M P L E

5

Solve the following initial value problem on the interval 0 ≤ t ≤ 2: 1, 0 ≤ t < 1 y − y = g(t), y(0) = 0, where g(t) = −2, 1 ≤ t ≤ 2. Solution: The graph of g(t) is shown in Figure 2.4(a); it has a jump discontinuity at t = 1. On the interval [0, 1), the differential equation reduces to y − y = 1. The general solution is y(t) = Cet − 1. Imposing the initial condition, we obtain y(t) = et − 1, 0 ≤ t < 1. As t approaches 1 from the left, y(t) approaches the value e − 1. Therefore, to complete the solution process, we solve a second initial value problem, y − y = −2,

y(1) = e − 1,

The solution of this initial value problem is 3 t y(t) = 1 − e + 2, e

1 ≤ t ≤ 2.

1 ≤ t ≤ 2.

Combining the individual solutions of these two initial value problems, we obtain the solution for the entire interval 0 ≤ t ≤ 2: ⎧ t 0≤t0

22. (t2 + 4)y + 2ty = t2 (t2 + 4)

2.2 23. y + y = t

First Order Linear Differential Equations

27

24. y + 2y = cos 3t

25. Consider the three direction ﬁelds shown. Match each of the direction ﬁeld plots with one of the following differential equations: (a) y + y = 0

(b) y + t2 y = 0

(c) y − y = 0 y

y 2

2 1.5

1.5

1

1

0.5

0.5

t

t –1

–0.5

0.5

–1

1

–0.5

0.5

–0.5

–0.5

–1

–1

Direction Field 1

1

Direction Field 2 y 2 1.5 1 0.5

t –1

–0.5

0.5

1

–0.5 –1

Direction Field 3 Figure for Exercise 25

Exercises 26–27: The graph of the solution of the given initial value problem is known to pass through the (t, y) points listed. Determine the constants α and y0 . 26. y + α y = 0,

y(0) = y0 .

Solution graph passes through the points (1, 4) and (3, 1).

27. ty − α y = 0,

y(1) = y0 .

Solution graph passes through the points (2, 1) and (4, 4).

28. Following are four graphs of y(t) versus t, 0 ≤ t ≤ 10, corresponding to solutions of the four differential equations (a)–(d). Match the graphs to the differential equations. For each match, identify the initial condition, y(0). (a) 2y + y = 0

(c) 10y − (1 − cos 2t)y = 0

(b) y + (cos 2t)y = 0 (d) 10y − y = 0

28

CHAPTER 2

First Order Differential Equations y

y 2

3

1.8 2.5

1.6 1.4

2

1.2

1.5

0.8

1 0.6 0.4

1

0.2 1

2

3

4

5

6

7

8

9

10

t

1

2

3

Graph 1

4

5

6

7

8

9

10

7

8

9

10

t

Graph 2 y

y 5.5

5

5

4.5

4.5

4

4

3.5

3.5 3

3

2.5

2.5

2

2 1.5

1

2

3

4

5

6

7

8

9

10

t

1.5

1

Graph 3

2

3

4

5

6

t

Graph 4 Figure for Exercise 28

29. Antioxidants Active oxygen and free radicals are believed to be exacerbating factors in causing cell injury and aging in living tissue.1 These molecules also accelerate the deterioration of foods. Researchers are therefore interested in understanding the protective role of natural antioxidants. In the study of one such antioxidant (Hsian-tsao leaf gum), the antioxidation activity of the substance has been found to depend on concentration in the following way: dA(c) = k[A∗ − A(c)], dc

A(0) = 0.

In this equation, the dependent variable A is a quantitative measure of antioxidant activity at concentration c. The constant A∗ represents a limiting or equilibrium value of this activity, and k is a positive rate constant. (a) Let B(c) = A(c) − A∗ and reformulate the given initial value problem in terms of this new dependent variable, B. (b) Solve the new initial value problem for B(c) and then determine the quantity of interest, A(c). Does the activity A(c) ever exceed the value A∗ ? (c) Determine the concentration at which 95% of the limiting antioxidation activity is achieved. (Your answer is a function of the rate constant k.) 30. The solution of the initial value problem ty + 4y = αt2 , y(1) = − 13 is known to exist on −∞ < t < ∞. What is the constant α? 1

Lih-Shiuh Lai, Su-Tze Chou, and Wen-Wan Chao, “Studies on the Antioxidative Activities of Hsian-tsao (Mesona procumbens Hemsl) Leaf Gum,” J. Agric. Food Chem., Vol. 49, 2001, pp. 963–968.

2.2

First Order Linear Differential Equations

29

Exercises 31–33: In each exercise, the general solution of the differential equation y + p(t)y = g(t) is given, where C is an arbitrary constant. Determine the functions p(t) and g(t). 31. y(t) = Ce−2t + t + 1

2

32. y(t) = Cet + 2

33. y(t) = Ct−1 + 1,

t>0

Exercises 34–35: In each exercise, the unique solution of the initial value problem y + y = g(t), y(0) = y0 is given. Determine the constant y0 and the function g(t). 34. y(t) = e−t + t − 1

35. y(t) = −2e−t + et + sin t

Exercises 36–37: In each exercise, discuss the behavior of the solution y(t) as t becomes large. Does lim t→∞ y(t) exist? If so, what is the limit? 36. y + y + y cos t = 1 + cos t, y(0) = 3 37.

y − e−t + 2 = −2, y

y(0) = −2

38. The solution of the initial value problem y + y = e−t , y(0) = y0 has a maximum value of e−1 = 0.367 . . . , attained at t = 1. What is the initial condition y0 ? 39. Let y(t) be a nonconstant solution of the differential equation y + λy = 1, where λ is a real number. For what values of λ is lim t→∞ y(t) ﬁnite? What is the limit in this case?

Exercises 40–43: As in Example 5, ﬁnd a solution to the initial value problem that given interval [a, b]. 3t, 1 ≤ t ≤ 2 1 40. y + y = g(t), y(1) = 1; g(t) = t 0, 2 < t ≤ 3; sin t, 0≤t≤π 41. y + (sin t)y = g(t), y(0) = 3; g(t) = − sin t, π < t ≤ 2π ; ⎧ ⎨0, 0 ≤ t ≤ 1 p(t) = 1 42. y + p(t)y = 2, y(0) = 1; ⎩ , 1 < t ≤ 2; t ⎧ 2t − 1, 0 ≤ t ≤ 1 ⎪ ⎪ ⎪ ⎨ 0, 1 0. (5) dt We can obtain differential equation (5) by invoking the same basic conservation law, equation (1), that was used to derive the population model P (t) = (rb − rd )P(t) + M(t). Here, the birth rate rb is zero since no radioactive material is being created. The death rate constant rd has been replaced by k. Likewise, we are tacitly assuming that no material is being added or taken away, and therefore the migration rate M(t) is also zero. E X A M P L E

3

Initially, 50 mg of a radioactive substance is present. Five days later, the quantity has decreased to 43 mg. How much will remain after 30 days?

2.4

Solution:

Population Dynamics and Radioactive Decay

45

The general solution of equation (5) is Q(t) = Ce−kt ,

where t is measured in days. Imposing the initial condition, we obtain Q(t) = 50e−kt . As in Example 1, we use the fact that Q(5) = 43 mg to determine the decay rate k: k = − 15 ln

43 50

= 0.03016 . . . days−1 .

After 30 days, therefore, we expect to have Q(30) = 50e−k30 = 20.228 . . . mg. ❖

The half-life of a radioactive substance is the length of time it takes a given amount of the substance to be reduced to one half of its original amount. Thus, the half-life τ is deﬁned by the equation Q(t + τ ) = 12 Q(t). Since Q(t) = Ce−kt , this equation reduces to e−kτ = 0.5 and hence τ=

ln 2 . k

For example, the substance in Example 3 has a half-life of about ln 2 = 22.95 days. 0.0302 If we had 300 mg of the substance at some given time, we would have about 150 mg of the substance 22.95 days later and 75 mg of the substance after 45.9 days.

EXERCISES Assume the populations in Exercises 1–4 evolve according to the differential equation P = kP. 1. A colony of bacteria initially has 10,000,000 members. After 5 days, the population increases to 11,000,000. Estimate the population after 30 days. 2. How many days will it take the colony in Exercise 1 to double in size? 3. A colony of bacteria is observed to increase in size by 30% over a 2-week period. How long will the colony take to triple its initial size? 4. A colony of bacteria initially has 100,000 members. After 6 days, the population has decreased to 80,000. At that time, 50,000 new organisms are added to replenish its size. How many bacteria will be in the colony after an additional 6 days? 5. Initially, 100 g of a radioactive material is present. After 3 days, only 75 g remains. How much additional time will it take for radioactive decay to reduce the amount present to 30 g?

46

CHAPTER 2

First Order Differential Equations

6. Radioactive decay reduces an initial amount of material by 20% over a period of 90 days. What is the half-life of this material? 7. A radioactive material has a half-life of 2 weeks. After 5 weeks, 20 g of the material is seen to remain. How much material was initially present? 8. After 30 days of radioactive decay, 100 mg of a radioactive substance was observed to remain. After 120 days, only 30 mg of this substance was left. (a) How much of the substance was initially present? (b) What is the half-life of this radioactive substance? (c) How long will it take before only 1% of the original amount remains? 9. Initially, 100 g of material A and 50 g of material B were present. Material A is known to have a half-life of 30 days, while material B has a half-life of 90 days. At some later time it was observed that equal amounts of the two radioactive materials were present. When was this observation made? 10. The evolution of a population with constant migration rate M is described by the initial value problem dP = kP + M, dt

P(0) = P0 .

(a) Solve this initial value problem; assume k is constant. (b) Examine the solution P(t) and determine the relation between the constants k and M that will result in P(t) remaining constant in time and equal to P0 . Explain, on physical grounds, why the two constants k and M must have opposite signs to achieve this constant equilibrium solution for P(t). 11. Assume that the population of ﬁsh in an aquaculture farm can be modeled by the differential equation dP/dt = kP + M(t), where k is a positive constant. The manager wants to operate the farm in such a way that the ﬁsh population remains constant from year to year. The following two harvesting strategies are under consideration. Strategy I: Harvest the ﬁsh at a constant and continuous rate so that the population itself remains constant in time. Therefore, P(t) would be a constant and M(t) would be a negative constant; call it −M. (Refer to Exercise 10.) Strategy II: Let the ﬁsh population evolve without harvesting throughout the year, and then harvest the excess population at year’s end to return the population to its value at the year’s beginning. (a) Determine the number of ﬁsh harvested annually with each of the two strategies. Express your answer in terms of the population at year’s beginning; call it P0 . (Assume that the units of k are year−1 .) (b) Suppose, as in Example 2, that P0 = 500,000 ﬁsh and k = 0.0061 × 52 = 0.3172 year−1 . Assume further that Strategy I, with its steady harvesting and return, provides the farm with a net proﬁt of $0.75/ﬁsh while Strategy II provides a proﬁt of only $0.60/ﬁsh. Which harvesting strategy will ultimately prove more proﬁtable to the farm? 12. Assume that two colonies each have P0 members at time t = 0 and that each evolves with a constant relative birth rate k = rb − rd . For colony 1, assume that individuals migrate into the colony at a rate of M individuals per unit time. Assume that this immigration occurs for 0 ≤ t ≤ 1 and ceases thereafter. For colony 2, assume that a similar migration pattern occurs but is delayed by one unit of time; that is, individuals migrate at a rate of M individuals per unit time, 1 ≤ t ≤ 2. Suppose we are interested in comparing the evolution of these two populations over the time

2.4

Population Dynamics and Radioactive Decay

47

interval 0 ≤ t ≤ 2. The initial value problems governing the two populations are dP1 = kP1 + M1 (t), dt

P1 (0) = P0 ,

dP2 = kP2 + M2 (t), dt

P2 (0) = P0 ,

1, M1 (t) = 0, 0, M2 (t) = 1,

0≤t≤1 1 < t ≤ 2; 0≤t 0, which population is larger at time t = 2? What happens if k < 0? (c) Suppose that there is a ﬁxed number of individuals that can be introduced into a population at any time through migration and that the objective is to maximize the population at some ﬁxed future time. Do the calculations performed in this problem suggest a strategy (based on the relative birth rate) for accomplishing this? 13. Radiocarbon Dating Carbon-14 is a radioactive isotope of carbon produced in the upper atmosphere by radiation from the sun. Plants absorb carbon dioxide from the air, and living organisms, in turn, eat the plants. The ratio of normal carbon (carbon-12) to carbon-14 in the air and in living things at any given time is nearly constant. When a living creature dies, however, the carbon-14 begins to decrease as a result of radioactive decay. By comparing the amounts of carbon-14 and carbon12 present, the amount of carbon-14 that has decayed can therefore be ascertained. Let Q(t) denote the amount of carbon-14 present at time t after death. If we assume its behavior is modeled by the differential equation Q (t) = −kQ(t), then Q(t) = Q(0)e−kt . Knowing the half-life of carbon-14, we can determine the constant k. Given a specimen to be dated, we can measure its radioactive content and deduce Q(t). Knowing the amount of carbon-12 present enables us to determine Q(0). Therefore, we can use the solution of the differential equation Q(t) = Q(0)e−kt to deduce the age, t, of the radioactive sample. (a) The half-life of carbon-14 is nominally 5730 years. Suppose remains have been found in which it is estimated that 30% of the original amount of carbon-14 is present. Estimate the age of the remains. (b) The half-life of carbon-14 is not known precisely. Let us assume that its half-life is 5730 ± 30 years. Determine how this half-life uncertainty affects the age estimate you computed in (a); that is, what is the corresponding uncertainty in the age of the remains? (c) It is claimed that radiocarbon dating cannot be used to date objects older than about 60,000 years. To appreciate this practical limitation, compute the ratio Q(60,000)/Q(0), assuming a half-life of 5730 years. 14. Suppose that 50 mg of a radioactive substance, having a half-life of 3 years, is initially present. More of this material is to be added at a constant rate so that 100 mg of the substance is present at the end of 2 years. At what constant rate must this radioactive material be added? 15. Iodine-131, a ﬁssion product created in nuclear reactors and nuclear weapons explosions, has a half-life of 8 days. If 30 micrograms of iodine-131 is detected in a tissue site 3 days after ingestion of the radioactive substance, how much was originally present? 16. U-238, the dominant isotope of natural uranium, has a half-life of roughly 4 billion years. Determine how long it takes for a sample to be reduced in amount by 1% through radioactive decay.

48

CHAPTER 2

2.5

First Order Differential Equations

First Order Nonlinear Diﬀerential Equations Thus far we have studied ﬁrst order linear differential equations, equations of the form y + p(t)y = g(t). We now consider ﬁrst order nonlinear differential equations. The term “nonlinear differential equation” encompasses all differential equations that are not linear. In particular, a ﬁrst order nonlinear differential equation has the form y = f (t, y), where f (t, y) = −p(t)y + g(t). Three examples of ﬁrst order nonlinear differential equations are (a) y = t2 + y2

(b) y = t + cos y

(c) y =

t . y

Nonlinear differential equations arise in many models of physical phenomena, such as population dynamics inﬂuenced by environmental constraints and one-dimensional motion in the presence of air resistance. We’ll consider such applications in Sections 2.8 and 2.9. Because the set of nonlinear differential equations is so diverse, the type of theoretical statement that can be made about the behavior of their solutions is less comprehensive than that made in Theorem 2.1 for linear equations. In addition, unlike the situation for linear equations, we cannot derive a general solution procedure that applies to the entire class of nonlinear equations. We therefore concentrate on certain subclasses of nonlinear differential equations for which solution procedures do exist.

Existence and Uniqueness We begin our study of nonlinear equations by considering questions of existence and uniqueness for initial value problems. In particular, given the initial value problem y = f (t, y),

y(t0 ) = y0 ,

(1)

what conditions on the function f (t, y) guarantee that problem (1) has a unique solution? On what t-interval does this unique solution exist? The answers to these questions provide a framework within which we can work. For example, if we use some special technique to ﬁnd a solution of problem (1), then it is essential to know whether the solution we found is the only solution. In fact, if the initial value problem does not have a unique solution, then it probably is not a good mathematical model for the physical phenomenon under consideration. Existence and uniqueness are also important considerations if we need to use numerical methods to approximate a solution. For example, if a numerical solution “blows up,” we want to know whether this behavior arises from inaccuracies in the numerical method or correctly depicts the behavior of the solution.

2.5

First Order Nonlinear Differential Equations

49

We now state a theorem that guarantees the existence of a unique solution to an initial value problem. The proof of this theorem is usually studied in a more advanced course in differential equations; we do not give a proof here. Theorem 2.2

Let R be the open rectangle deﬁned by a < t < b, α < y < β. Let f (t, y) be a function of two variables deﬁned on R, where f (t, y) and the partial derivative ∂f/∂y are continuous on R. Suppose (t0 , y0 ) is a point in R. Then there is an open t-interval (c, d), contained in (a, b) and containing t0 , in which there exists a unique solution of the initial value problem y = f (t, y),

y(t0 ) = y0 .

A typical open rectangle R, initial point (t0 , y0 ), and interval (c, d) are shown in Figure 2.8. (The rectangle R is called an open rectangle because it does not contain the four line segments forming its boundary.) y  y0 y(t) ␣ ac

t0

d b

t

FIGURE 2.8

The open rectangle R, deﬁned by a < t < b, α < y < β, contains the initial point (t0 , y0 ). If the hypotheses of Theorem 2.2 hold on R, we are guaranteed a unique solution to the initial value problem on some open interval (c, d).

Although presented in the context of nonlinear differential equations, Theorem 2.2 makes no distinction between linear and nonlinear differential equations. It applies to linear ﬁrst order equations where f (t, y) = −p(t)y + g(t), as well as to nonlinear ﬁrst order equations. Two important observations can be made about Theorem 2.2. 1. The hypotheses of Theorem 2.2 are a natural generalization of those made in Theorem 2.1 for linear differential equations. That is, if f (t, y) = −p(t)y + g(t), then ∂f/∂y = −p(t). Therefore, requiring f (t, y) and ∂f/∂y to be continuous on the rectangle R means that any linear differential equation satisfying the hypotheses of Theorem 2.2 also satisﬁes the hypotheses of Theorem 2.1. Conversely, a linear differential equation satisfying the hypotheses of Theorem 2.1 also satisﬁes the hypotheses of Theorem 2.2.

50

CHAPTER 2

First Order Differential Equations

2. The conclusions of Theorem 2.2, however, differ substantially from those of Theorem 2.1. Since we have broadened our perspective to encompass nonlinear differential equations, the corresponding conclusions of Theorem 2.2 are weaker than those of Theorem 2.1. Theorem 2.1 guarantees existence and uniqueness on the entire (a, b) interval. Theorem 2.2 guarantees existence and uniqueness only on some subinterval (c, d) of (a, b) containing t0 ; it does not guarantee existence and uniqueness on the entire (a, b) interval. Moreover, Theorem 2.2 gives no insight into how large (c, d) is or how we might go about estimating it. Although Theorem 2.2 leaves many questions unanswered, it does provide us with the framework we need to study solution techniques for certain classes of nonlinear differential equations. Examining these special cases will give us valuable insight into the behavior of solutions of nonlinear equations.

Autonomous Differential Equations First order autonomous equations have the form y = f ( y). The right-hand side of the differential equation does not explicitly depend on the independent variable t. Solution curves for an autonomous differential equation have the important geometric property that they can be translated parallel to the t-axis. As an example, consider the autonomous equation y = y(2 − y). The direction ﬁeld for this equation, along with portions of some solution curves, is shown in Figure 2.9. As observed in Section 1.3, the slopes of the direction ﬁeld ﬁlaments for an autonomous equation remain constant along horizontal lines. For instance (see Figure 2.9), at every point along the line y = 1, the direction ﬁeld ﬁlaments have slope equal to 1. Similarly, at every point along the line y = 3, the direction ﬁeld ﬁlaments have slope equal to −3. y y1 (t)

y 2 (t)

3 2.5 2 1.5 1 0.5

t –0.5

1

2

3

4

5

6

–1

FIGURE 2.9

The direction ﬁeld for the autonomous equation y = y(2 − y), together with portions of some typical solutions. Notice that the graph of y1 (t), when translated to the right, looks as though it coincides with the graph of y2 (t).

2.5

First Order Nonlinear Differential Equations

51

Besides illustrating that horizontal lines are isoclines for the autonomous equation, Figure 2.9 illustrates an important property of solutions to autonomous differential equations. That is, it looks as though the graph of y1 (t), when translated about 2 units to the right, will fall exactly on the graph of y2 (t). This is indeed the case, and we show in Theorem 2.3 that the solution y2 (t) is related to the solution y1 (t) by y2 (t) = y1 (t − c), where c is a constant.

Theorem 2.3

Let the initial value problem y = f ( y),

y(0) = y0

satisfy the conditions of Theorem 2.2, and let y1 (t) be the unique solution, where the interval of existence for y1 (t) is a < t < b, with a < 0 < b. Consider the initial value problem y = f ( y),

y(t0 ) = y0 .

(2)

Then the function y2 (t) deﬁned by y2 (t) = y1 (t − t0 ) is the unique solution of initial value problem (2) and has an interval of existence t0 + a < t < t0 + b.

PROOF: Since y1 (t) is deﬁned for a < t < b, we know that y2 (t) = y1 (t − t0 ) is deﬁned for a < t − t0 < b and hence for t0 + a < t < t0 + b. We next observe that y2 (t) satisﬁes the initial condition of (2), since y2 (t0 ) = y1 (t0 − t0 ) = y1 (0) = y0 . Therefore, to complete the proof of Theorem 2.3, we need to show that y2 (t) is a solution of the differential equation y = f ( y). Using the deﬁnition of y2 (t), the chain rule, and the fact that y1 (t) solves the differential equation y = f ( y), we have

●

d d y1 (t − t0 ) = y1 (t − t0 ) (t − t0 ) = y1 (t − t0 ) = f ( y1 (t − t0 )) = f ( y2 (t)). dt dt Therefore, the function y2 (t) = y1 (t − t0 ) is a solution of the initial value problem. ● y2 (t) =

The important conclusion to be reached from Theorem 2.3 is that the solution of the autonomous initial value problem y = f ( y), y(t0 ) = y0 depends on the independent variable t and the initial condition time t0 as a function of the combination t − t0 . What matters is time t measured relative to the initial time t0 . As a simple example, recall that the solution of the linear autonomous equation y = ky, y(t0 ) = y0 is y(t) = y0 ek(t−t0 ) .

52

CHAPTER 2

First Order Differential Equations

Bernoulli Equations We conclude this section by studying a class of nonlinear differential equations known as Bernoulli equations. By making an appropriate change of dependent variable, these nonlinear equations can be transformed into ﬁrst order linear equations and solved using the techniques described in Section 2.2. Bernoulli differential equations4 are ﬁrst order differential equations having the special structure dy + p(t)y = q(t)yn , dt where n is an integer. We do not consider n = 0 and n = 1 since, in those cases, the Bernoulli equation is a ﬁrst order linear equation. A simple example of a Bernoulli equation is dy + e2t y = y3 sin t. dt Bernoulli equations arise in applications such as population models and models of one-dimensional motion inﬂuenced by drag forces. Consider a Bernoulli equation dy + p(t)y = q(t)yn , (3) dt where n is a given integer (n = 0 and n = 1). We look for a change of dependent variable of the form v(t) = y(t)m , where m is a constant to be determined. Using the chain rule, we have dv dy = mym−1 dt dt and therefore dv dv dy = m−1 y1−m = m−1 v(1−m)/m . dt dt dt Equation (3) transforms into the following differential equation for v(t): dv + mp(t)v = mq(t)v(m+n−1)/m . (4) dt At ﬁrst glance, it may seem that our change of variables has accomplished little. The structure of equation (4) seems similar to what we started with. However, we are free to choose the constant m. In particular, if we select m = 1 − n, then equation (4) reduces to the ﬁrst order linear equation dv + (1 − n)p(t)v = (1 − n)q(t). (5) dt We can solve this equation for v(t) and then obtain the desired solution, y(t) = v(t)1/(1−n) . 4

Jacob Bernoulli (1654 –1705) is one of eight members of the extended Bernoulli family remembered for their contributions to mathematics and science. While at the University of Basel, Jacob made important contributions to such areas as inﬁnite series, probability theory, geometry, and differential equations. In 1696 he solved the differential equation that now bears his name. Jacob always had a particular fascination for the logarithmic spiral and requested that this curve be carved on his tombstone.

2.5

First Order Nonlinear Differential Equations

53

E X A M P L E

1

Solve the initial value problem y + y = ty3 ,

y(0) = 2.

Solution: The differential equation is a Bernoulli equation with n = 3. We make the change of dependent variable v = y1−n or, since n = 3, v = y−2 . The initial value problem for v(t) then becomes [recall equation (5)] v − 2v = −2t, The general solution is

v(0) = y(0)−2 = 14 .

v = Ce2t + t + 12 .

Imposing the initial condition, we have

v = − 14 e2t + t + 12 .

Finally, since y = v−1/2 , we arrive at the desired solution −1/2 y = − 14 e2t + t + 12 . ❖

EXERCISES Exercises 1–8: For the given initial value problem, (a) Rewrite the differential equation, if necessary, to obtain the form y = f (t, y),

y(t0 ) = y0 .

Identify the function f (t, y). (b) Compute ∂f/∂y. Determine where in the ty-plane both f (t, y) and ∂f/∂y are continuous. (c) Determine the largest open rectangle in the ty-plane that contains the point (t0 , y0 ) and in which the hypotheses of Theorem 2.2 are satisﬁed. 1. 3y + 2t cos y = 1, 2

3. 2t + (1 + y )y = 0,

1/3

5. y + ty

= tan t,

2. 3ty + 2 cos y = 1,

y(π/2) = −1

2

y(−1) = 1

7. (cos y)y = 2 + tan t,

y(π/2) = −1

4. 2t + (1 + y3 )y = 0,

y(1) = 1

−y

6. ( y − 9)y + e

y(0) = 0

y(1) = 1 2

=t ,

8. (cos 2t)y = 2 + tan y, 2

y(2) = 2 y(π) = 0

2

9. Consider the initial value problem t y − y = 0, y(1) = 1. (a) Determine the largest open rectangle in the ty-plane, containing the point (t0 , y0 ) = (1, 1), in which the hypotheses of Theorem 2.2 are satisﬁed. (b) A solution of the initial value problem is y(t) = t. This solution exists on −∞ < t < ∞. Does this fact contradict Theorem 2.2? Explain your answer. 10. The solution of the initial value problem y = f ( y), y(0) = 8 is known to be y(t) = (4 + t)3/2 . Let y(t) represent the solution of the initial value problem y = f ( y), y(t0 ) = 8. Suppose we know that y(0) = 1. What is t0 ? 11. The solution of the initial value problem y = f ( y), y(0) = 2 is known to be √ y(t) = 2/ 1 − t. Let y(t) represent the solution of the initial value problem y = f ( y), y(1) = 2. What is the value of y(0)?

54

CHAPTER 2

First Order Differential Equations 12. The graph shows the solution of y = −1/(2y), y(0) = 1. Use the graph to answer questions (a) and (b). (a) If z1 (t) is the solution of z1 = −1/(2z1 ), z1 (−2) = 1, what is z1 (−5)? (b) If z2 (t) is the solution of z2 = −1/(2z2 ), z2 (2) = 1, what is z2 (3)? y

2.5 2 1.5 1 0.5 –3.5 –3 –2.5 –2 –1.5 –1 –0.5 –0.5

t

0.5

Figure for Exercise 12

Exercises 13–19: (a) Solve the initial value problem by (i)

transforming the given Bernoulli differential equation and initial condition into a ﬁrst order linear differential equation with its corresponding initial condition,

(ii) solving the new initial value problem, (iii) transforming back to the dependent variable of interest. (b) Determine the interval of existence. 13. y = y(2 − y),

t 2

15. y = −y + e y ,

3 −2

17. ty + y = t y ,

14. y = 2ty(1 − y),

y(0) = 1

y(−1) = −1

16. y = y + y ,

y(1) = 1 −2

19. y = −( y + 1) + t( y + 1) ,

−1

1/3

18. y − y = ty y(0) = 1

y(0) = −1 y(0) = −1

,

y(0) = −8

[Hint: Let z = y + 1.]

20. The initial value problem y + y = q(t)y2 , y(0) = y0 is known to have solution y(t) =

3 (1 − 3t)et

on the interval −∞ < t < 13 . Determine the coefﬁcient function q(t) and the initial value y0 .

2.6

Separable First Order Equations In Section 2.2, we obtained an explicit representation for the solution of a ﬁrst order linear differential equation; recall equation (10) in Section 2.2. By contrast, there is no all-encompassing technique that leads to an explicit representation for the solution of a ﬁrst order nonlinear differential equation. For certain types of nonlinear equations, however, techniques have been discovered that give us some information about the solution. We have already

2.6

Separable First Order Equations

55

seen one type, the Bernoulli equations, in the previous section. In this section we study another type, called separable differential equations.

Separable Equations The term separable differential equation is used to describe any ﬁrst order differential equation that can be put into the form n( y)

dy + m(t) = 0. dt

For example, the differential equations (a)

dy + t2 sin y = 0 dt

and

(b) y + et+y = ey sin t

are separable since they can be rewritten (respectively) as (a ) csc y

dy + t2 = 0 dt

and

(b ) e−y y + [et − sin t] = 0.

A simple example of a nonseparable differential equation is y = 2ty2 + 1. The structure of a separable differential equation, n( y)

dy + m(t) = 0, dt

gives the equation its name. The ﬁrst term is the product of dy/dt and a term n( y) that involves only the dependent variable y. The second term, m(t), involves only the independent variable t. In this sense, the variables “separate.”

Solving a Separable Differential Equation We can get some information about the solution of a separable equation by “reversing the chain rule.” We illustrate this technique in Example 1 and then describe the general procedure. E X A M P L E

1

Solve the initial value problem y = 2ty2 ,

y(0) = 1.

Solution: First, notice that f (t, y) = 2ty2 is continuous on the entire ty-plane, as is the partial derivative ∂f/∂y = 4ty. Therefore, the conditions of Theorem 2.2 are satisﬁed on the open rectangle R deﬁned by −∞ < t < ∞, −∞ < y < ∞. Theorem 2.2 guarantees the existence of a unique solution of the initial value problem, but it provides no insight into the interval of existence of the solution. The differential equation is separable. It can be rewritten as y y2

− 2t = 0. (continued)

56

CHAPTER 2

First Order Differential Equations (continued)

To emphasize the fact that y is the dependent variable, we express the equation as y (t) y(t)2 Taking antiderivatives, we ﬁnd y (t)

− 2t = 0.

dt −

y(t)2

2t dt = C.

Evaluating the integrals on the left-hand side yields −1 − t2 = C. y(t) Solving for y(t), we obtain a family of solutions y(t) =

−1 2

t +C

.

Imposing the initial condition, y(0) = 1, yields the unique solution of the initial value problem: y(t) =

1 1 − t2

.

Having determined the solution, we are now able to see that the interval of existence is −1 < t < 1. ❖ The solution process of Example 1 can be viewed as reversing the chain rule. To explain, we return to the general separable differential equation, n( y)

dy + m(t) = 0. dt

(1)

Let y be a differentiable function of t, and let N(y) be any antiderivative of n( y). By the chain rule, dy d N( y) = n( y) . dt dt

(2a)

Similarly, let M(t) be any antiderivative of m(t), d M(t) = m(t). dt

(2b)

Combining (2a) and (2b), we can rewrite the left-hand side of equation (1) as n( y)

d d d dy + m(t) = N( y) + M(t) = [N( y) + M(t)]. dt dt dt dt

Therefore, equation (1) reduces to d [N( y) + M(t)] = 0. dt

2.6

Separable First Order Equations

57

Since the term N( y) + M(t) is a function of t whose derivative vanishes identically, we have N(y) + M(t) = C,

(3)

where C is an arbitrary constant. Equation (3) provides us with information about the solution y(t). It is not an explicit expression for the solution; rather, it is an equation that the solution must satisfy. An equation in y and t, such as (3), is called an implicit solution. Sometimes (as in Example 1 and in Example 2 below) we can “unravel” the implicit solution and solve for y(t) as an explicit function of the independent variable t. In other cases, we must be content with the implicit solution. Whether or not we can unravel the implicit solution given by equation (3), we can always determine the constant C by imposing the initial condition y(t0 ) = y0 , ﬁnding C = N( y0 ) + M(t0 ). E X A M P L E

2

Solve the initial value problem t dy =− , dt y Solution:

y(0) = −2.

Separating the variables, we obtain y

dy + t = 0. dt

Integrating, we ﬁnd an implicit solution t2 y2 + = C. 2 2 Imposing the initial condition, we ﬁnd C = 2. Thus, an implicit solution of the initial value problem is given by y2 + t2 = 4. Suppose we want an explicit solution. Solving the equation above, we ﬁnd y = ± 4 − t2 . Which root should we take? To satisfy the initial condition, y(0) = −2, we must take the negative root. Thus, the solution is (4) y = − 4 − t2 . This choice of roots is also obvious geometrically, since the graph of y2 + t2 = 4 is a circle of radius 2 in the ty-plane, as shown in Figure 2.10. The solution of the initial value problem has the lower semicircle as its graph. The function given by equation (4) is deﬁned and continuous on [−2, 2]. It is differentiable and satisﬁes the differential equation on the open interval (−2, 2). (continued)

58

CHAPTER 2

First Order Differential Equations (continued)

y

t

y = –√4 – t 2

(0, –2)

FIGURE 2.10

The implicit solution of the initial value problem in Example 2 is y2 + t2 = 4; its graph is a circle of radius 2. The explicit solution of the

initial value problem is y = − 4 − t2 ; its graph is the lower semicircle.

❖

E X A M P L E

3

Solve the initial value problem (2 + sin y)y + t = 0, Solution:

y(2) = 0.

Computing the antiderivatives yields

t2 = C. 2 Imposing the initial condition, we obtain the implicit solution 2y − cos y +

t2 = 1. (5) 2 Although we cannot unravel this equation and determine an explicit solution, we can plot the graph of equation (5); see Figure 2.11. Observe that if the cosine term in (5) were absent, the graph would be that of a concave-down parabola. Loosely speaking, therefore, the cosine term creates the ripples displayed by the graph in Figure 2.11. ❖ 2y − cos y +

Differences between Linear and Nonlinear Differential Equations We can use Examples 1–3 to make several points about Theorem 2.2 and to illustrate some of the differences between nonlinear and linear differential equations. 1. The Interval of Existence May Not Be Obvious If the coefﬁcient functions for a linear differential equation are continuous on an interval (a, b), where (a, b) contains the initial point t0 , then a unique solution

2.6

Separable First Order Equations

59

y

–10

–5

5

10

t

–5

–10

–15

–20

FIGURE 2.11

The initial value problem posed in Example 3 has an implicitly deﬁned solution given by equation (5). The graph of equation (5) is shown above.

of the initial value problem y + p(t)y = g(t), y(t0 ) = y0 exists and is deﬁned on all of (a, b). By way of contrast, consider the initial value problem in Example 1. The function f (t, y) = 2ty2 is continuous everywhere and is about as nice a nonlinear function as we can expect. However, the solution of the initial value problem has vertical asymptotes at t = −1 and at t = 1. Note that we cannot predict that the interval of existence is (−1, 1) by simply looking at the equation y = 2ty2 . In fact, if the initial condition is changed from y(0) = 1 to y(0) = −1, then the interval of existence changes to (−∞, ∞). (See Exercise 4.) Example 2 provides another illustration. The interval of existence in Example 2 is −2 < t < 2. Suppose we leave the initial condition alone and simply change the sign on the right-hand side of the differential equation. This change produces a new initial value problem t dy = , dt y

y(0) = −2.

In this case, the solution is deﬁned for all t. (See Exercise 12.) In each of these examples, a harmless-looking change in the differential equation or in the initial condition leads to a pronounced change in the nature of the solution and in the interval of existence. 2. There May Not Be a Single Formula That Gives All Solutions the family of solutions found in Example 1, y(t) =

−1 2

t +C

,

Note that

(6)

does not include the zero function. The zero function is, however, a solution of y = 2ty2 . In particular, given the initial value problem y = 2ty2 ,

60

CHAPTER 2

First Order Differential Equations

y(0) = 0, there is no choice for C in equation (6) that yields the unique solution, y(t) = 0, that is guaranteed by Theorem 2.2. 3. We May Have to Be Content with Implicitly Deﬁned Solutions In Examples 1 and 2 we were able to ﬁnd an explicit formula for the solution, y(t), of the initial value problem. However, as Example 3 illustrates, it may not be possible to obtain an explicit formula for the solution. By contrast [see equation (10) in Section 2.2], there is an explicit formula for the solution of any ﬁrst order linear differential equation.

EXERCISES Exercises 1–17: (a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can ﬁnd an explicit solution of the problem, determine the t-interval of existence. 1. y

dy − sin t = 0, dt

3. y +

1 = 0, y+1

5. y − ty3 = 0, dy = 1 + y2 , dt dy = t − ty2 , 9. dt

7.

11.

dy = et−y , dt

y(π/2) = −2

2.

dy 1 = 2, dt y

y(1) = 2

4. y − 2ty2 = 0,

y(1) = 0

y(0) = −1

dy + ey t = ey sin t, dt

y(0) = 2

6.

y(π/4) = −1

8. t2 y + sec y = 0,

y(0) =

1 2

10. 3y2

y(0) = 1

13. et y + (cos y)2 = 0,

12. y(0) = π/4

2 t = , y(1) = 2 15. ey y + y+1 y+1 17. ey y = 1 + ey , y(2) = 0

dy + 2t = 1, dt

t dy = , dt y

y(0) = 0 y(−1) = 0 y(−1) = −1

y(0) = −2

14. (2y − sin y)y + t = sin t,

16. (ln y)y + t = 1,

y(0) = 0

y(3) = e

18. For what values of the constants α, y0 and integer n is the function y(t) = (4 + t)−1/2 a solution of the initial value problem y + αyn = 0,

y(0) = y0 ?

19. For what values of the constants α, y0 and integer n is the function y(t) = 6/(5 + t4 ) a solution of the initial value problem y + αtn y2 = 0,

y(1) = y0 ?

20. State an initial value problem, with initial condition imposed at t0 = 2, having implicit solution y3 + t2 + sin y = 4. 21. State an initial value problem, with initial condition imposed at t0 = 0, having implicit solution yey + t2 = sin t.

2.6

Separable First Order Equations

61

22. Consider the initial value problem y = 2y2 ,

y(0) = y0 .

For what value(s) y0 will the solution have a vertical asymptote at t = 4 and a t-interval of existence −∞ < t < 4? 23. (a) A ﬁrst order autonomous differential equation has the form y = f ( y). Show that such an equation is separable. (b) Solve y = y(2 − y),

y(2) = 1.

Exercises 24–26: A differential equation of the form y = p1 (t) + p2 (t)y + p3 (t)y2 is known as a Riccati equation.5 Equations of this form arise when we model onedimensional motion with air resistance; see Section 2.9. In general, this equation is not separable. In certain cases, however (such as in Exercises 24–26), the equation does assume a separable form. Solve the given initial value problem and determine the t-interval of existence. 24. y = 2 + 2y + y2 ,

2

y(0) = 0

26. y = ( y + 2y + 1) sin t,

25. y = t(5 + 4y + y2 ),

y(0) = −3

y(0) = 0

27. Let Q(t) represent the amount of a certain reactant present at time t. Suppose that the rate of decrease of Q(t) is proportional to Q3 (t). That is, Q = −kQ3 , where k is a positive constant of proportionality. How long will it take for the reactant to be reduced to one half of its original amount? Recall that, in problems of radioactive decay where the differential equation has the form Q = −kQ, the half-life was independent of the amount of material initially present. What happens in this case? Does half-life depend on Q(0), the amount initially present? 28. The rate of decrease of a reactant is proportional to the square of the amount present. During a particular reaction, 40% of the initial amount of this chemical remained after 10 sec. How long will it take before only 25% of the initial amount remains? 29. Consider the differential equation y = | y|. (a) Is this differential equation linear or nonlinear? Is the differential equation separable? (b) A student solves the two initial value problems y = | y|, y(0) = 1 and y = y, y(0) = 1 and then graphs the two solution curves on the interval −1 ≤ t ≤ 1. Sketch what she observes. (c) She next solves both problems with initial condition y(0) = −1. Sketch what she observes in this case. 30. Consider the following autonomous ﬁrst order differential equations: y = −y2 ,

y = y3 ,

y = y(4 − y).

Match each of these equations with one of the solution graphs shown. Note that each solution satisﬁes the initial condition y(0) = 1. Can you match them without solving the differential equations?

5

Jacopo Riccati (1676–1754) worked on many differential equations, including the one that now bears his name. His work in hydraulics proved useful to his native city of Venice.

62

CHAPTER 2

First Order Differential Equations y

y

4

7 6

3 5 2

4 3

1 2

–2 –0.4

–0.2

0.2

2

t

t

0.4

Graph A

Graph B y

9 8 7 6 5 4 3 2 1 1

2

3

4

t

Graph C Figure for Exercise 30

31. Let S(t) represent the amount of a chemical reactant present at time t, t ≥ 0. Assume that S(t) can be determined by solving the initial value problem S = −

αS , K +S

S(0) = S0 ,

where α, K, and S0 are positive constants. Obtain an implicit solution of the initial value problem. (The differential equation, often referred to as the Michaelis-Menten equation, arises in the study of biochemical reactions.) 32. Change of Dependent Variable Sometimes a change of variable can be used to convert a differential equation y = f (t, y) into a separable equation. (a) Consider a differential equation of the form y = f (αt + βy + γ ), where α, β, and γ are constants. Use the change of variable z = αt + βy + γ to rewrite the differential equation as a separable equation of the form z = g(z). List the function g(z). (b) A differential equation that can be written in the form y = f ( y/t) is called an equidimensional differential equation. Use the change of variable z = y/t to rewrite the equation as a separable equation of the form tz = g(z). List the function g(z).

2.7

Exact Differential Equations

63

Exercises 33–38: Use the ideas of Exercise 32 to solve the given initial value problem. Obtain an explicit solution if possible. y−t y+t 33. y = , y(2) = 2 34. y = , y(−1) = 0 y+t y+t+1 35. y = (t + y)2 − 1, y(1) = 2 37. y = 2t + y +

1 , 2t + y

y(1) = 1

36. y =

1 , 2t + 3y + 1

38. t2 y = y2 − ty,

39. Consider the initial value problem y = 1 − y2 ,

y(1) = 0

y(−2) = 2

y(0) = 0.

(a) Show that y = sin t is an explicit solution on the t-interval −π/2 ≤ t ≤ π/2. (b) Show that y = sin t is not a solution on either of the intervals −3π/2 < t < −π/2 or π/2 < t < 3π/2. (c) What are the equilibrium solutions of y = 1 − y2 ? Suppose a solution y(t) reaches an equilibrium value at t = t∗ . What happens to the graph of y(t) for t > t∗ ? (d) Show that the solution of the initial value problem is given by ⎧ −∞ < t < −π/2 ⎨−1, −π/2 ≤ t ≤ π/2 y = sin t, ⎩ 1, π/2 < t < ∞.

2.7

Exact Diﬀerential Equations The class of differential equations referred to as exact includes separable equations as a special case. As with separable equations, the solution procedure for this new class consists of reversing the chain rule. This time, however, we use a chain rule that involves a function of two variables.

The Extended Chain Rule Suppose H(t, y) is a function of two independent variables t and y, where H(t, y) has continuous partial derivatives with respect to t and y. If the second independent variable y is replaced with a differentiable function of t, call it y(t), we obtain a composition H(t, y(t)) which is now a function of t only. What is dH/dt? The appropriate chain rule is d ∂H(t, y(t)) ∂H(t, y(t)) dy(t) H(t, y(t)) = + . dt ∂t ∂y dt

(1)

To understand this equation, note that the partial derivatives on the right-hand side refer to H viewed as a function of the two independent variables t and y. Once these partial derivatives are computed, the variable y is replaced by the function y(t). Formula (1) is an extension of the chain rule for functions of a single variable. If the function H has the form H( y) so that it is only a function of the

64

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First Order Differential Equations

single variable y, then the ﬁrst term on the right-hand side of (1) vanishes and the formula reverts to the usual chain rule for the composite function H(y(t)).

Solving Exact Differential Equations The basic idea underlying the solving of exact differential equations is to reverse the extended chain rule when possible. To that end, consider a differential equation of the form dy = 0. (2) dt Notice the similarity between the form of differential equation (2) and that of the chain rule (1). Suppose there exists some function, call it H(t, y), that satisﬁes the following two conditions: M(t, y) + N(t, y)

∂H = M(t, y) and ∂t

∂H = N(t, y). ∂y

(3)

Because of (3), we can rewrite differential equation (2) as ∂H dy ∂H + = 0. ∂t ∂y dt

(4)

By the chain rule (1), equation (4) is the same as d H(t, y) = 0. dt Therefore, we obtain an implicitly deﬁned solution of equation (4), H(t, y) = C.

(5)

If there is a function H(t, y) satisfying the conditions in (3), then differential equation (2) is called an exact differential equation. If we can identify the function H(t, y), then an implicitly deﬁned solution is given by (5).

Recognizing an Exact Differential Equation Once we know that a given differential equation is exact and once we identify a function H(t, y) satisfying the conditions in (3), then we can write down an implicit solution, H(t, y) = C, for the differential equation. Two basic questions therefore need to be answered. 1. Given a differential equation of the form M(t, y) + N(t, y)y = 0, how do we know whether or not it is exact? That is, how do we determine whether there is a function H(t, y) satisfying the conditions ∂H = M(t, y) and ∂t

∂H = N(t, y)? ∂y

2. Suppose we are somehow assured that such a function exists. How do we go about ﬁnding H(t, y)? The answer to the ﬁrst question is given in Theorem 2.4, which is stated without proof. To answer the second question, we will use a process of “antipartial-differentiation.”

2.7

Theorem 2.4

Exact Differential Equations

65

Consider the differential equation M(t, y) + N(t, y)y = 0. Let the functions M, N, ∂M/∂y, and ∂N/∂t be continuous in an open rectangle R of the ty-plane. Then the differential equation is exact in R if and only if ∂N ∂M = ∂y ∂t

(6)

for all points (t, y) in R.

Theorem 2.4 provides an easy test for whether or not a given differential equation is exact. The theorem does not, however, tell how to construct the implicitly deﬁned solution H(t, y) = C. E X A M P L E

1

Which of the following differential equations is (are) exact? (a) y + t + ty = 0

(b) y + sin t + ( y cos t)y = 0

(c) sin y + (2y + t cos y)y = 0

Solution: (a) Using the notation of Theorem 2.4, we have M(t, y) = y + t

N(t, y) = t.

and

Calculating the partial derivatives, we ﬁnd ∂M =1 ∂y

∂N = 1. ∂t

and

Therefore, by Theorem 2.4, the differential equation is exact. (b) For this equation, M(t, y) = y + sin t

and

N(t, y) = y cos t.

Calculating the partial derivatives yields ∂M =1 ∂y

and

∂N = −y sin t. ∂t

Since the partial derivatives are not equal, the differential equation is not exact. (c) Calculating the partial derivatives, we have ∂M = cos y and ∂y

∂N = cos y. ∂t

Since the partial derivatives are equal, the differential equation is exact. ❖ REMARK: Recall that a separable differential equation is one that can be written in the form n( y) dy/dt + m(t) = 0. Notice that ∂m =0 ∂y

and

∂n = 0. ∂t

Thus, by Theorem 2.4, any separable differential equation is also exact.

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First Order Differential Equations

Anti-Partial-Differentiation We can use Theorem 2.4 to determine whether the differential equation M(t, y) + N(t, y)y = 0 is exact. If it is exact, then we know there must be a function H(t, y) such that ∂H ∂H = M(t, y) and = N(t, y). ∂t ∂y Once we determine H(t, y), we have an implicitly deﬁned solution, H(t, y) = C. A process of anti-partial-differentiation can be used to construct H. As an illustration, recall from Example 1 that the following differential equation is exact: sin y + (2y + t cos y)y = 0.

(7)

Thus, there is a function H(t, y) such that ∂H = sin y ∂t

and

∂H = 2y + t cos y. ∂y

(8)

Choose one of these equalities, say ∂H/∂y = 2y + t cos y, and compute an “antipartial-derivative.” Antidifferentiating 2y + t cos y with respect to y, we obtain H(t, y) = y2 + t sin y + g(t),

(9)

where g(t) is an arbitrary function of t. [Note: The “constant of integration” in equation (9) is an arbitrary function of t since t is treated as a constant when the partial derivative with respect to y is computed.] We now determine g(t) so that the representation (9) for H satisﬁes the ﬁrst equality in (8). Taking the partial derivative of H with respect to t, we ﬁnd ∂ 2 dg ∂H = y + t sin y + g(t) = sin y + . ∂t ∂t dt Comparing the preceding result with the ﬁrst condition of equation (8), it follows that we need dg =0 dt or g(t) = C1 , where C1 is an arbitrary constant. Thus, from equation (9), we know H(t, y) = y2 + t sin y + C1 .

(10)

We can drop the arbitrary constant C1 , since we will eventually set H(t, y) equal to an arbitrary constant in the implicit solution. Therefore, H(t, y) = y2 + t sin y. In this illustration, we started with the second equation in (8), ∂H/∂y = 2y + t cos y. We could just as well have started with the ﬁrst equation, ∂H/∂t = sin y. If we had done so, we would have arrived at the same function H; see Exercise 21. E X A M P L E

2

Consider the initial value problem dy = 0, y(0) = 1. dt Verify that the differential equation is exact and solve the initial value problem. 1 + y2 + 2(t + 1)y

2.7

Exact Differential Equations

67

Solution: To verify that the differential equation is exact, we appeal to Theorem 2.4, using M(t, y) = 1 + y2 and N(t, y) = 2(t + 1)y. The functions M, N, ∂M/∂y, and ∂N/∂t are continuous in the entire ty-plane. Since ∂M = 2y and ∂y

∂N = 2y, ∂t

the differential equation is exact. We now ﬁnd H(t, y). Since the equation is exact, ∂H = 1 + y2 . ∂t Antidifferentiating with respect to t yields H(t, y) = t(1 + y2 ) + g( y).

(11)

To determine g( y), we differentiate (11) with respect to y, ﬁnding ∂H dg = 2ty + . ∂y dy Since the differential equation is exact, 2ty +

dg = 2(t + 1)y. dy

Therefore, g( y) = y2 + C1 . Without loss of generality, we let C1 = 0 to obtain H(t, y) = t(1 + y2 ) + y2 = (t + 1)y2 + t. Thus, we have the following implicitly deﬁned solution of the differential equation: (t + 1)y2 + t = C. Imposing the initial condition, we obtain an implicit solution of the initial value problem: (t + 1)y2 + t = 1. We can solve for y: 1−t y = , 1+t 2

or y = ±

1−t . 1+t

Choosing the positive sign so as to satisfy the initial condition y(0) = 1, we arrive at an explicit solution of the initial value problem: 1−t y= . ❖ 1+t √ The explicit solution y = (1 − t)/(1 + t) has −1 < t < 1 as an interval of existence. This should not be surprising in light of the existence-uniqueness

68

CHAPTER 2

First Order Differential Equations

statement in Theorem 2.2. The differential equation in Example 2 has the form y = f (t, y), where f (t, y) = −

1 + y2 . 2(t + 1)y

Since the initial condition point is (t0 , y0 ) = (0, 1), the hypotheses of Theorem 2.2 are satisﬁed in the inﬁnite rectangle deﬁned by −1 < t < ∞, 0 < y < ∞. Along the lines t = −1 and y = 0, both f and ∂f/∂y are undeﬁned. As Figure 2.12 shows, the solution curve has a vertical asymptote at the boundary line t = −1 and approaches the boundary line y = 0 as t → 1 from the left.

y 3 2.5 2 1.5 1 0.5

t –2

–1

1

2

–0.5 –1 –1.5 –2

FIGURE 2.12

A portion of the direction ﬁeld for the differential equation in Example 2. √ The solution y = (1 − t)/(1 + t) is shown as a solid curve.

EXERCISES Exercises 1–10: Show that the given nonlinear differential equation is exact. (Some algebraic manipulation may be required. Also, recall the remark that follows Example 1.) Find an implicit solution of the initial value problem and (where possible) an explicit solution. 1. (2y − t)y − y + 2t = 0,

2

2. (t + y3 )y + y + t3 = 0,

y(1) = 0

2

3

3. y = (3t + 1)( y + 1), y(0) = 1 t+y

5. (e

t+y

+ 2y)y + (e

3

3

2y

2

2

+ 3t ) = 0,

6. ( y − t )y = 3t y + 1,

2

y(0) = −2

4. ( y + cos t)y = 2 + y sin t,

y(0) = −1

y(0) = 0

y(−2) = −1

2

7. (e + t y)y + ty + cos t = 0, y(π/2) = 0 y cos (ty) + 1 1 y(π) = 0 8. y = − y + y2 = 1, 9. 2ty + 2 , y t cos (ty) + 2yey 10. (2y ln t − t sin y)y + t−1 y2 + cos y = 0,

y(2) = 0

y(1) = 1

2.7

Exact Differential Equations

69

Exercises 11–16: Determine the general form of the function M(t, y) or N(t, y) that will make the given differential equation exact. 11. (2t + y)y + M(t, y) = 0

12. (t2 + y2 sin t)y + M(t, y) = 0

13. (tey + t + 2y)y + M(t, y) = 0

14. N(t, y)y + 2t + y = 0

2

2

2

15. N(t, y)y + t + y sin t = 0

16. N(t, y)y + ey + 2ty = 1

Exercises 17–20: The given equation is an implicit solution of N(t, y)y + M(t, y) = 0, satisfying the given initial condition. Assuming the equation N(t, y)y + M(t, y) = 0 is exact, determine the functions M(t, y) and N(t, y), as well as the possible value(s) of y0 . 17. t3 y + et + y2 = 5, 2

18. 2ty + cos (ty) + y2 = 2,

y(0) = y0 yt

19. ln (2t + y) + t + e = 1,

y(0) = y0

3

4

20. y + 4ty + t + 1 = 0,

y(0) = y0 y(0) = y0

21. The differential equation (2y + t cos y)y + sin y = 0 is exact, and thus there exists a function H(t, y) such that ∂H/∂t = sin y and ∂H/∂y = 2y + t cos y. Antidifferentiating the second equation with respect to y, we ultimately arrived at H(t, y) = y2 + t sin y + C1 ; see equation (10). Show that the same result would be obtained if we began the solution process by antidifferentiating the ﬁrst equation, ∂H/∂t = sin y, with respect to t. 22. Making a Differential Equation Exact Suppose the differential equation N(t, y)y + M(t, y) = 0 is not exact; that is, Nt (t, y) = My (t, y). Is it possible to multiply the equation by a function, call it μ(t, y), so that the resulting equation is exact? (a) Show that if μ(t, y)N(t, y)y + μ(t, y)M(t, y) = 0 is exact, the function μ must be a solution of the partial differential equation N(t, y)μt − M(t, y)μy = [My (t, y) − Nt (t, y)]μ. Parts (b) and (c) of this exercise discuss special cases where the function μ can be chosen to be a function of a single variable. In these special cases, the partial differential equation in part (a) reduces to a ﬁrst order linear ordinary differential equation and can be solved using the techniques of Section 2.2. (b) Suppose the quotient [My (t, y) − Nt (t, y)]/N(t, y) is just a function of t, call it p(t). Let P(t) be an antiderivative of p(t). Show that μ can be chosen as μ(t) = eP(t) . (c) Suppose the quotient [Nt (t, y) − My (t, y)]/M(t, y) is just a function of y, call it q( y). Let Q( y) be an antiderivative of q( y). Show that μ can be chosen as μ( y) = eQ( y) .

Exercises 23–28: In each exercise, (a) Show that the given differential equation is not exact. (b) Multiply the equation by the function μ, if it is provided, and show that the resulting differential equation is exact. If the function μ is not given, use the ideas of Exercise 22 to determine μ. (c) Solve the given problem, obtaining an explicit solution if possible. 23. 4tyy + y2 − t = 0,

2 2

3

y(1) = 1,

24. (t y + 1)y + ty = 0, 25. (2t + y)y + y = 0,

2

t

27. tyy + y + e = 0,

μ(t, y) = t−1/2

y(0) = 1,

y(2) = −3 y(1) = −2

μ(t, y) = y−1 26. ty2 y + 2y3 = 1,

2

y(1) = −1

28. (3ty + 2)y + y = 0,

y(−1) = −1

70

CHAPTER 2

2.8

First Order Differential Equations

The Logistic Population Model The art of mathematical modeling involves a trade-off between realism and complexity. A model should incorporate enough reality to make it useful as a predictive tool. However, the model must also be tractable mathematically; if it’s too complex to analyze and if we cannot deduce any useful information from it, then the model is worthless. A key assumption of the population model described in Section 2.4, often referred to as the Malthusian6 model, is that relative birth rate is independent of population. (Recall that the relative birth rate, rb − rd , is the difference between birth and death rates per unit population.) The assumption that rb − rd is independent of population leads to uninhibited exponential growth or decay of solutions. Such behavior is often a poor approximation of reality. For a colony possessing limited resources, a more realistic model is one that accounts for the impact of population on the relative birth rate. When population size is small, resources are relatively plentiful and the population should thrive and grow. When the population becomes larger, however, we expect that resources become scarcer, the population becomes stressed, and the relative birth rate begins to decline (eventually becoming negative). In this section we consider a model that attempts to account for this effect. This model leads to a nonlinear differential equation.

The Verhulst, or Logistic, Model The Verhulst population model7 assumes that the population P(t) evolves according to the differential equation P dP =r 1− (1) P. dt Pe In equation (1), r and Pe are positive constants. Equation (1) is also known as the logistic equation. Comparing equation (1) with the Malthus equation dP = (rb − rd )P, (2) dt we see that the constant relative birth rate rb − rd of equation (2) has been replaced by the population-dependent relative birth rate P(t) r 1− , Pe where r is a positive constant. If P(t) > Pe , then dP/dt < 0 and the population is decreasing. Conversely, if P(t) < Pe , then dP/dt > 0 and the population is increasing. The qualitative behavior of solutions of the logistic equation can also be deduced from the direction ﬁeld. Figure 2.13, for example, shows the direction ﬁeld for the logistic equation for the parameters r = 3 and Pe = 1. 6

Thomas Malthus (1766–1834) was an English political economist whose Essay on the Principle of Population had an important inﬂuence on Charles Darwin’s thinking. Malthus believed that human population growth, if left unchecked, would inevitably lead to poverty and famine. 7 Pierre Verhulst (1804–1849) was a Belgian scientist remembered chieﬂy for his research on population growth. He deduced and studied the nonlinear differential equation named after him.

2.8

The Logistic Population Model

71

y 1.5 1 0.5

t –1

–0.5

0.5

1

1.5

2

–0.5

FIGURE 2.13

The direction ﬁeld for the logistic equation P = 3(1 − P)P. The equilibrium solutions are P = 0 and P = 1.

The logistic equation has two equilibrium solutions, the trivial solution P(t) = 0 and the nontrivial equilibrium solution P(t) = Pe . If P(t0 ) = Pe , then the population remains equal to this value as time evolves. The constant population, Pe , is called an equilibrium population. In terms of its structure, the logistic equation is a nonlinear separable differential equation. It is also a Bernoulli equation. We solve it below as a separable equation. Its treatment as a Bernoulli equation is left to the exercises.

Solution of the Logistic Equation To solve equation (1), we separate variables, obtaining

1 dP = r. P dt 1− P Pe

An implicit solution is therefore

dP = r t + K, P 1− P Pe

(3)

where K is an arbitrary constant. Recalling the method of partial fractions from calculus (also see Section 5.3), we obtain P 1 dP 1 . = dP = ln − P P P − Pe P − Pe 1− P Pe

Therefore,

Hence,

ln

P = r t + K. P − Pe

P(t) P(t) − P

e

= Cer t ,

(4)

72

CHAPTER 2

First Order Differential Equations

where C = eK is an arbitrary positive constant. We can remove the absolute value signs by arguing as follows. The exponential function er t is never zero. Therefore, the quotient P(t)/[P(t) − Pe ] is never zero and, being a continuous function of t, never changes sign; it is either positive or negative for all t. Therefore, we can remove the absolute value signs if we allow the constant C to be either positive or negative. Assume that the initial population P(0) = P0 is positive and P0 = Pe . Then C=

P0 . P 0 − Pe

After some algebraic manipulation, we obtain the explicit solution P(t) =

P0 Pe . P0 − (P0 − Pe )e−r t

(5)

The derivation leading to equation (5) tacitly assumes that P(t) never takes on the value 0 or Pe . The ﬁnal expression (5), however, is valid for any value of P0 (including the equilibrium values 0 and Pe ). What behavior is predicted by the solution (5)? Since r > 0, we see that lim P(t) = Pe

t→∞

for all positive values of P0 . Therefore, any given nonzero population will tend to the equilibrium population value, Pe , as time increases. Figure 2.13 illustrates this behavior, showing the direction ﬁeld for the special case P = 3(1 − P)P. We can ask additional questions. What happens if we allow temporal variations in the relative birth rate by allowing r = r(t) in equation (1)? What happens if migration is introduced into the logistic model? When migration is introduced, the population model has the form P dP =r 1− P + M. dt P∗

(6)

If r, P∗ , and M are constants, the equation is separable. Moreover, the question of equilibrium populations is relevant. What happens to the equilibria as the migration level M changes? Some of these questions are addressed in the exercises. Example 1 illustrates how sketching a diagram similar to a direction ﬁeld enables us to predict the behavior of solutions of equation (6) without actually solving the equation. The basis for such predictions is found in Exercises 14–15. These exercises set r = 1, P∗ = 1 in equation (6), considering the equation P = −(P − P1 )(P − P2 ) with equilibrium populations P1 ≥ P2 > 0. These exercises show the following: 1. If P(0) > P2 , then P(t) exists for 0 ≤ t < ∞ and P(t) → P1 as t → ∞. 2. If 0 ≤ P(0) < P2 , then lim t→t∗ P(t) = −∞ for some ﬁnite t∗ > 0.

2.8

The Logistic Population Model

73

E X A M P L E

1

Let P(t) denote the population of a colony, where P is measured in units of 100,000 individuals and time t is in years. Assume that population is modeled by the logistic model with constant out-migration, dP 2 = (1 − P)P − , dt 9

P(0) = 2.

(a) Determine all the equilibrium populations (that is, the nonnegative equilibrium solutions). Sketch a diagram indicating those regions in the ﬁrst quadrant of the tP-plane where the population is increasing [P (t) > 0] and those regions where the population is decreasing [P (t) < 0]. This diagram gives a rough indication of how the solutions behave. (b) Without solving the initial value problem, use the results of Exercise 14 (as summarized above) to determine lim t→∞ P(t). Solution: (a) The equilibrium solutions are the constant solutions P(t) = 23 and P(t) = 13 . Since these equilibrium solutions are nonnegative, they correspond to equilibrium populations. The diagram sketched in Figure 2.14 indicates the regions in the ﬁrst quadrant where P (t) > 0 and where P (t) < 0. (b) From Figure 2.14, it follows that the solution of the initial value problem . This soP = (1 − P)P − 29 , P(0) = 2 is decreasing at t = 0, since P (0) = − 20 9 lution curve cannot cross the line P = 23 , since P = 23 is also a solution of P = (1 − P)P − 29 . By Exercise 14, either solutions of equation (6) are unbounded or they tend to an equilibrium value as t → ∞. The diagram in Figure 2.14 indicates that lim P(t) = 23 .

t→∞

P P'(t) < 0 2 3

P'(t) = 0 P'(t) > 0

1 3

P'(t) = 0 P'(t) < 0 t

FIGURE 2.14

Consider the population model P = (1 − P)P − 29 . The equilibrium populations are P = 23 and P = 13 ; see Example 1. The diagram shows the regions where population is increasing [P (t) > 0] and where it is decreasing [P (t) < 0]. The solution satisfying the initial condition P(0) = 2 is decreasing and bounded below by the line P = 23 . Therefore, by Exercise 14, P(t) → 23 as t → ∞. (continued)

74

CHAPTER 2

First Order Differential Equations (continued)

[Similarly, if P(0) = P0 where 13 < P0 < 23 , we again obtain lim t→∞ P(t) = 23 . However, if P(0) = P0 where P0 < 13 , then the population becomes extinct when the graph of P(t) intersects the t-axis. At that point, the population model ceases to be meaningful.] ❖

The Spread of an Infectious Disease The logistic differential equation also arises in modeling the spread of an infectious disease. Suppose we have a constant population of N individuals and at time t the number of infected members is P(t). The corresponding number of uninfected individuals is then N − P(t). A reasonable assumption is that the rate of spread of the disease at time t is proportional to the product of noninfected and infected individuals. This assumption leads to the differential equation dP = k(N − P)P, dt where k is the constant of proportionality. If the equation is rewritten as P dP = kN 1 − P, dt N

(7)

we can see that it is similar to the logistic equation (1). The corresponding initial value problem is completed by specifying the number of infected individuals at some initial time, P(t0 ) = P0 . Note that the differential equation (7) has two equilibrium solutions, P = 0 and P = N. This certainly makes sense. If no one is infected or if everyone is infected, the disease will not spread. We consider aspects of this infectious disease model in the exercises.

EXERCISES Exercises 1–3: Let P(t) represent the population of a colony, in millions of individuals. Assume that the population evolves according to the equation dP P P, = 0.1 1 − 3 dt with time t measured in years. Use the explicit solution given in equation (5) to answer the questions. 1. Suppose the colony starts with 100,000 individuals. How long will it take the population to reach 90% of its equilibrium value? 2. Suppose the colony is initially overpopulated, starting with 5,000,000 individuals. How long will it take for the population to decrease to 110% of its equilibrium value?

2.8

The Logistic Population Model

75

3. Suppose, after 3 years of existence, the colony population is found to be 2,000,000 individuals. What was the initial population?

Exercises 4–10: Constant Migration Consider a population modeled by the initial value problem dP = (1 − P)P + M, dt

P(0) = P0 ,

(8)

where the migration rate M is constant. [The model (8) is derived from equation (6) by setting the constants r and P∗ to unity. We did this so that we can focus on the effect M has on the solutions.] For the given values of M and P(0), (a) Determine all the equilibrium populations (the nonnegative equilibrium solutions) of differential equation (8). As in Example 1, sketch a diagram showing those regions in the ﬁrst quadrant of the tP-plane where the population is increasing [P (t) > 0] and those regions where the population is decreasing [P (t) < 0]. (b) Describe the qualitative behavior of the solution as time increases. Use the information obtained in (a) as well as the insights provided by the ﬁgures in Exercises 11–13 (these ﬁgures provide speciﬁc but representative examples of the possibilities). 3 4. M = − 16 ,

P(0) =

7. M = − 14 ,

P(0) = 1

10. M = 2,

3 5. M = − 16 ,

3 2

8. M = − 14 ,

P(0) = P(0) =

6. M = − 14 ,

1 2

9. M = 2,

1 2

P(0) =

1 4

P(0) = 0

P(0) = 4

Exercises 11–13: The direction ﬁelds shown correspond to the differential equation dP = dt

1−

P P∗

P + M.

Determine the constants P∗ and M. 11. The equilibrium solutions are P(t) = 2 and P(t) = 1.

12. The equilibrium solution is P(t) = 2. P

P 3

3.5

2.5

3

2

2.5

1.5

2

1

1.5 1

0.5

t 0.5

1

1.5

2

2.5

3

0.5

3.5

t

–0.5 0.5

1

1.5

2

2.5

3

3.5

76

CHAPTER 2

First Order Differential Equations

13. The equilibrium solutions are P(t) = 2 and P(t) = −1.

P 3 2 1

t 0.5

1

1.5

2

2.5

3

3.5

–1 –2 –3

14. Consider the initial value problem (8). Let P1 and P2 denote the roots of P2 − P − M = 0, and assume that P1 > P2 > 0. [Since P1 and P2 are positive constant solutions of P = (1 − P)P + M, they correspond to equilibrium populations.] (a) What does the assumption P1 > P2 > 0 imply about the sign of M? Is migration occurring at a constant rate into or out of the colony? (b) Solve initial value problem (8), assuming that P(0) = P1 and P(0) = P2 . Note that the differential equation is separable. Your solution will have the form |P(t) − P1 | = Ke−λt , |P(t) − P2 | where K and λ are positive constants. Unravel this implicit solution. [Hint: The graph of P(t) cannot cross the lines P = P1 and P = P2 . Therefore, the terms P(t) − P1 and P(t) − P2 have the same signs as P0 − P1 and P0 − P2 , respectively. In order to unravel the solution, consider the separate cases P0 > P1 , P1 > P0 > P2 , and P2 > P0 and remove the absolute values.] (c) Use the explicit solutions found in part (b) to show that if P0 > P1 , then P(t) → P1 as t → ∞. Similarly, show that if P1 > P0 > P2 , then P(t) → P1 as t → ∞. Finally, show that if P2 > P0 , then lim t→t∗ P(t) = −∞ at some ﬁnite time t∗ . Since populations are nonnegative, what actually happens to the population in this last case? 15. Repeat the analysis of Exercise 14 for the case of a positive repeated root of P2 − P − M = 0 (that is, for the case of P1 = P2 > 0). (a) What does the assumption P1 = P2 > 0 imply about the sign of M? Is migration occurring at a constant rate into or out of the colony? (b) Solve initial value problem (8) for the case in which P(0) = P1 . Argue that P1 = P2 can happen only if P1 = P2 = 12 . (c) By taking limits of the explicit solution found in part (b), show that P(t) → P1 as t → ∞ if P0 ≥ P1 . What happens if P0 < P1 ? Variable Birth Rates In certain situations, a population’s relative birth rate may vary with time. For example, environmental conditions that change over a period of time may affect the birth rate. When food is stored, variations in temperature may affect the growth of harmful bacteria. It is of interest, therefore, to understand the behavior of the logistic equation when the relative birth rate r(t) is variable. 16. Consider the initial value problem dP P P, = r(t) 1 − dt Pe

P(0) = P0 .

t Observe that the differential equation is separable. Let R(t) = 0 r(s) ds. Solve the initial value problem. Note that your solution will involve the function R(t).

2.9

Applications to Mechanics

77

17. Solve the initial value problem in Exercise 16 for the particular case of r(t) = 1 + sin 2πt, Pe = 1, and P0 = 14 . Here, time is measured in years and population in millions of individuals. (The varying relative birth rate might reﬂect the impact of seasonal changes on the population’s ability to reproduce over the course of a year.) How does the population behave as time increases? In particular, does lim t→∞ P(t) exist, and if so, what is this limit? 18. Let P(t) represent the number of individuals who, at time t, are infected with a certain disease. Let N denote the total number of individuals in the population. Assume that the spread of the disease can be modeled by the initial value problem dP = k(N − P)P, dt

P(0) = P0 ,

where k is a constant. Obtain an explicit solution of this initial value problem. 19. Consider the special case of the infectious disease model in Exercise 18, where N = 2,000,000 and P0 = 100,000. Suppose that after 1 year, the number of infected individuals had increased to 200,000. How many members of the population will be infected after 5 years? 20. Consider a chemical reaction of the form A + B → C, in which the rates of change of the two chemical reactants, A and B, are described by the two differential equations A = −kAB,

B = −kAB,

where k is a positive constant. Assume that 5 moles of reactant A and 2 moles of reactant B are present at the beginning of the reaction. (a) Show that the difference A(t) − B(t) remains constant in time. What is the value of this constant? (b) Use the observation made in (a) to derive an initial value problem for reactant A. (c) It was observed, after the reaction had progressed for 1 sec, that 4 moles of reactant A remained. How much of reactants A and B will be left after 4 sec of reaction time? 21. Solve the initial value problem P dP P, =r 1− dt Pe

P(0) = P0

by viewing the differential equation as a Bernoulli equation.

2.9

Applications to Mechanics In Chapter 1, we considered a falling object acted upon only by gravity. For that case, with y(t) used to represent the vertical position of the body at time t and v(t) to represent its velocity, Newton’s second law of motion, ma = F, reduces to m

dv = −mg. dt

Here, m is the mass of the object and g the acceleration due to gravity, nominally 32 ft/sec2 or 9.8 m/s2 . The minus sign on the right-hand side is present because we measure y(t) and v(t) as positive upward while the force of gravity acts downward. For the simple model above, we solved for v(t) and y(t) by computing successive antiderivatives.

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A more realistic model of one-dimensional motion is one that incorporates the effects of air resistance. As an object moves through air, a retarding aerodynamic force is created by the combination of pressure and frictional forces. The air close to the object exerts a normal pressure force upon it. Likewise, friction creates a tangential force that opposes the motion of air past the object. The combination of these effects creates a drag force that acts to reduce the speed of the moving object. The drag force depends on the velocity v(t) of the object and acts on it in such a way as to reduce its speed, |v(t)|. We consider two idealized models of drag force that are consistent with these ideas.

Case 1: Drag Force Proportional to Velocity Assume that velocity v(t) is positive in the upward direction and that the drag force is proportional to velocity. If k is the positive constant of proportionality, Newton’s second law of motion leads to m

dv = −mg − kv. dt

(1)

Does the model of drag that we have postulated act as we want it to? If the object is moving upward [that is, if v(t) > 0], then the drag force −kv(t) is negative and thus acts downward. Conversely, if v(t) < 0, then the object is moving downward. In this case, the drag force −kv(t) = k|v(t)| is a positive (upward) force, as it should be. Therefore, whether the object is moving upward or downward, drag acts to slow the object; this drag model is consistent with our ideas of how drag should act. See Figure 2.15. y

y

drag force

v0

–mg

drag force

–mg y=0

y=0

(a)

(b) FIGURE 2.15

Assume a drag force of the form kv(t), k > 0; see equation (1). (a) When the object is moving downward, drag acts upward and tends to slow the object. (b) When the object is moving upward, drag acts downward and likewise tends to slow the object.

This drag model leads to equation (1), a ﬁrst order linear constant coefﬁcient equation; we can solve it using the ideas of Section 2.2. Before we do, however, let’s consider the question of equilibrium solutions to see if the model makes sense in that regard. The only constant solution of equation (1) is v(t) = −mg/k. At the velocity v(t) = −mg/k, the drag force and gravitational force acting on the object are equal and opposite. This equilibrium velocity is often referred to as the terminal velocity of the object.

2.9

Applications to Mechanics

79

The initial value problem corresponding to equation (1) is m

dv = −mg − kv, dt

v(0) = v 0 .

(2)

Solving initial value problem (2), we obtain mg mg −(k/m)t e v(t) = − . + v0 + k k From this explicit solution it is clear that, for any initial velocity v 0 , v(t) tends to the terminal velocity mg . lim v(t) = − t→∞ k The concept of terminal velocity is a mathematical abstraction since, in any application, the time interval of interest is ﬁnite. However, whether the object is initially moving up or down, it eventually begins to fall and its velocity approaches terminal velocity as the time of falling increases. The velocity that the object actually has the instant before it strikes the ground is called the impact velocity. Once impact occurs, the mathematical model (2) is no longer applicable.

Case 2: Drag Force Proportional to the Square of Velocity A drag force having magnitude κv2 (t) (where the constant of proportionality κ is assumed to be positive) has been found in many cases to be a reasonably good approximation to reality over a range of velocities. However, since it involves an even power of velocity, incorporating this model of drag into the equations of motion requires more care. Drag must act to reduce the speed of the moving object. Therefore, if the object is moving upward, the drag force acts downward and should be −κv2 (t). If the object is moving downward, the drag force acts upward and should be κv2 (t). In other words, when we use this model for drag, Newton’s second law leads to dv = −mg − κv2 , m v(t) ≥ 0, dt dv = −mg + κv2 , m v(t) ≤ 0. dt Here again we can ask about equilibrium solutions and see if the answer makes sense physically. Note that no equilibrium solution exists if v(t) > 0 since −mg − κv2 is never zero. When v(t) < 0, however, there is an equilibrium solution: mg . v(t) = − κ This equilibrium solution is again a terminal velocity corresponding to downward motion; at this velocity, drag and gravity exert equal and opposite forces. Each of the preceding equations is a ﬁrst order separable equation. If the problem involves a falling (rising) object, then velocity is always nonpositive (nonnegative) and a single equation is valid for the entire problem (that is, over the entire t-interval of interest). If, however, the problem involves both upward and downward motion, then both equations will ultimately be needed. The ﬁrst equation must be used to model the upward dynamics, the behavior

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First Order Differential Equations

of the projectile from launch until the time tm at which it reaches its highest point [that is, when v(tm ) = 0]. After time tm , the projectile begins to fall and the second differential equation [with initial condition v(tm ) = 0] is needed to model the descending dynamics. E X A M P L E

1

Assume a projectile having mass m is launched vertically upward from ground level at time t = 0 with initial velocity v(0) = v 0 . Further assume that the drag force experienced by the projectile is proportional to the square of its velocity, with drag coefﬁcient κ. Determine the maximum height reached by the projectile and the time at which this maximum height is reached. Solution: The projectile motion being considered involves only upward motion. Therefore, mv = −mg − κv2 ,

v(0) = v 0 .

Separating variables yields v = −g. κ 2 v 1+ mg Integrating, we ﬁnd

dv = κ 2 v 1+ mg

mg tan−1 κ

κ v = −gt + C. mg

Imposing the initial condition gives mg κ C= tan−1 v0 . κ mg Finally, after some algebra, we obtain the explicit solution mg κ κg −1 tan tan v − t . v(t) = κ mg 0 m

(3)

As a check on (3), note that v(0) does reduce to the given initial velocity v 0 . As a further check, one can show (using L’Hôpital’s rule8 ) that for every ﬁxed time t, lim v(t) = v 0 − gt,

κ→0

which is the velocity in the absence of drag. Let tm denote the time when the maximum height is reached; that is, v(tm ) = 0. From equation (3), we see that the maximum height is attained at the ﬁrst positive value t where the argument of the tangent function is zero. Thus, κ κg v0 − t = 0, tan−1 mg m m 8

Guillaume de L’Hôpital (1661–1704) wrote the ﬁrst textbook on differential calculus, Analyse des inﬁniment petits pour l’intelligence des lignes courbes, which appeared in 1692. The book contains the rule that bears his name.

or

tm =

2.9

Applications to Mechanics

κ v0 . mg

m tan−1 κg

81

(4)

We next want to determine the maximum height, y(tm ), reached by the projectile. To ﬁnd y(tm ), we need an expression for position, y(t). To determine position, y(t), we integrate velocity: t y(t) − y(0) = v(s) ds. 0

For our problem, y(0) = 0. Therefore, y(tm ) =

tm

v(t) dt,

(5)

0

where v(t) is given by equation (3). To carry out the integration in equation (5), we use the fact that tan u du = −ln | cos u| + C, ﬁnding κv 20 m ln 1 + . y(tm ) = 2κ mg Here again, as a check, one can show using L’Hôpital’s rule that lim y(tm ) =

κ→0

v 20 2g

(6)

is the maximum height reached in the absence of drag. Figure 2.16 shows how the size of the drag constant affects the quantities tm and y(tm ) for the case of a 2-lb object launched upward from ground level with an initial velocity of 60 mph (v 0 = 88 ft/sec). y (feet)

t (seconds)

150

3 2.5

100

2 1.5

50

1 0.5 .005

.01

(a)

.005

.01

(b) FIGURE 2.16

A 2-lb object is launched upward from ground level with an initial velocity of 88 ft/sec. The graph in (a) shows the maximum altitude as a function of the drag constant κ; see equation (5). The graph in (b) shows the time when the maximum altitude is reached as a function of the drag constant κ; see equation (4).

❖

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First Order Differential Equations

The computations in Example 1 are relatively complicated. The task of ﬁnding maximum projectile height can be simpliﬁed by transforming the problem to one in which height rather than time is the independent variable. This transformation is discussed in the next subsection.

One-Dimensional Dynamics with Distance as the Independent Variable In many applications, velocity is always nonnegative or always nonpositive over the entire time interval of interest. For the present discussion, assume that motion occurs in a straight line, along the x-axis. Then position x(t) is a monotonic function of time. If velocity is nonnegative, the position of the object is an increasing function of time; if velocity is nonpositive, then position is a decreasing function of time. In these cases, we can simplify the differential equation describing the object’s motion if we use position x as the independent variable rather than time t. Suppose (for deﬁniteness) that position x(t) is an increasing function of time t [a similar analysis is valid if x(t) is decreasing]. Then an inverse function exists and we can express time as a function of position. We can ultimately view velocity as a function of position and use the chain rule to relate dv/dt to dv/dx: dv dx dv dv = =v . (7) dt dx dt dx This change of independent variable is useful when the net force acting on the body is a function of velocity or position or both but does not explicitly depend dv = F transforms on time. In such a case, when F = F(x, v), the equation m dt into dv mv = F(x, v). (8) dx We adopt the customary notation and write v = v(x) when referring to velocity as a function of position and v = v(t) when referring to velocity as a function of time. Equation (8) is typically supplemented by an initial condition that prescribes velocity at some initial position v(x0 ) = v 0 . Equation (8) and the initial condition deﬁne an initial value problem on the underlying x-interval of interest. To illustrate the simpliﬁcations that can be realized by adopting distance as the independent variable, we reconsider the problem solved in Example 1, that of determining the maximum height reached by a projectile when subjected to a drag force proportional to the square of the velocity. E X A M P L E

2

Consider the problem treated in Example 1. The initial value problem describing the projectile’s upward motion is dv = −mg − κv2 , v(0) = v 0 , dt where v = dy/dt. Determine the maximum height reached by the projectile. m

2.9

Applications to Mechanics

83

Solution: We view velocity as a function of position y. Since y = 0 when t = 0, the initial value problem satisﬁed by v( y) becomes mv Therefore, m κ

dv = −mg − κv2 , dy

v v2 + (mg/κ)

dv =

v(0) = v 0 .

m ln [v2 + (mg/κ)] = −y + C. 2κ

Imposing the initial condition yields m m ln [v2 + (mg/κ)] = −y + ln [v 20 + (mg/κ)]. 2κ 2κ

(9)

If we would like to have velocity expressed as a function of height y, we can “unravel” the implicit solution (9). However, to determine ym , the maximum height reached by the projectile, we need only note that the velocity is zero at the maximum height. Setting v = 0 in (9) leads to 2 κv m m m 0 ln [mg/κ] = −ym + ln [v 20 + (mg/κ)], or ym = ln 1 + . 2κ 2κ 2κ mg This is the same expression we found in Example 1. ❖

Impact Velocity The next example, concerning an object falling through the atmosphere, shows that using position as the independent variable may convert a problem we cannot solve into one that we can solve. The force of Earth’s gravitational attraction diminishes as a body moves higher above the surface. For objects near the surface, the usual assumption of constant gravity is fairly reasonable. However, for a body falling from a great height (such as a satellite reentering the atmosphere), a constant gravity assumption is not accurate and may lead to erroneous predictions of the reentry trajectory. While most of the realistic gravity models used in aerospace applications are quite complicated, they are all based on Newton’s law of gravitation, which states that the force of mutual attraction between two bodies is F=

GmM r2

.

(10)

In equation (10), G = 6.673 × 10−11 m3 /(kg · s2 ) is the universal gravitational constant, m and M are the masses of the two bodies, and r is the distance between the centers of mass of the two bodies. [If the bodies are moving relative to each other, then r varies with time and therefore r = r(t).] As a ﬁrst approximation to the force of gravitational attraction for an object falling to the surface of Earth, we assume Earth is a sphere of homogeneous material. Under this assumption, we can use equation (10) to model gravity. The mass of Earth is Me = 5.976 × 1024 kg, and its radius is Re = 6371 km or 6.371 × 106 m.

84

CHAPTER 2 E X A M P L E

3

First Order Differential Equations

Consider an object having mass m = 100 kg which is released from rest at an altitude of h = 200 km above the surface of Earth. Neglecting drag and considering only the force of gravitational attraction, calculate the impact velocity of the object at Earth’s surface. Solution: In this problem we do not assume that the force of gravitational attraction is constant. Rather, we take into account the variation of this force with the separation distance between the two bodies. See Figure 2.17.

Re

r

m h

FIGURE 2.17

An object of mass m is released at an altitude of 200 km above the surface of Earth. As the object falls, its distance from the center of Earth is r. Earth has radius Re , and therefore r = Re + h deﬁnes the object’s altitude, h, above Earth. The quantity r is positive in the direction of increasing altitude.

As shown in Figure 2.17, separation distance r is measured positively from Earth’s center to the 100-kg body. If v = dr/dt, then the application of Newton’s second law of motion leads to the differential equation m

GmMe dv =− . dt r2

(11)

If time is retained as the independent variable, we obtain a differential equation for the dependent variable r(t) by using the fact that v = dr/dt. The resulting differential equation, however, is second order and nonlinear: m

d2 r dt

2

=−

GmMe r2

.

Note that the separation distance r(t) is a decreasing function of time. If we transform differential equation (11) into one in which distance r is the independent variable and use the fact that v = 0 when r = Re + h, we obtain the following initial value problem: GmMe dv =− , dr r2 v(Re + h) = 0. mv

Re < r < Re + h,

(12)

The differential equation in (12), while nonlinear, is a ﬁrst order separable equation for the quantity of interest, v. Solving, we obtain GMe v2 = + C. 2 r

2.9

Applications to Mechanics

85

Imposing the initial condition, we ﬁnd the implicit solution 1 v2 1 = GMe − . 2 r Re + h Since separation distance is decreasing with time, velocity is negative. Therefore, an explicit solution of the problem is 1 1 − v = − 2GMe . r Re + h The impact velocity is found by evaluating velocity at r = Re : 1 1 vimpact = − 2GMe − . Re Re + h Using the values given, we obtain vimpact = −1952 m/s (a speed of about 4350 mph). [Note: The impact velocity does not depend on the mass of the object, but only on its distance from Earth when it is released.] ❖ In the Exercises, we ask you to consider the same problem in the presence of a drag force that is proportional to the square of the velocity. In particular, the governing differential equation becomes m

GmMe dv =− + κv2 . dt r2

(13)

After the independent variable is changed from time to distance, the resulting differential equation can be recast as a Bernoulli equation and then solved; see Exercise 16.

EXERCISES 1. An object of mass m is dropped from a high altitude. How long will it take the object to achieve a velocity equal to one half of its terminal velocity if the drag force is assumed to be proportional to the velocity? 2. A drag chute must be designed to reduce the speed of a 3000-lb dragster from 220 mph to 50 mph in 4 sec. Assume that the drag force is proportional to the velocity. (a) What value of the drag coefﬁcient k is needed to accomplish this? (b) How far will the dragster travel in the 4-sec interval? 3. Repeat Exercise 2 for the case in which the drag force is proportional to the square of the velocity. Determine both the drag coefﬁcient κ and the distance traveled. 4. A projectile of mass m is launched vertically upward from ground level at time t = 0 with initial velocity v 0 and is acted upon by gravity and air resistance. Assume the drag force is proportional to velocity, with drag coefﬁcient k. Derive an expression for the time, tm , when the projectile achieves its maximum height. 5. Derive an expression for the maximum height, ym = y(tm ), achieved in Exercise 4. 6. An object of mass m is dropped from a high altitude and is subjected to a drag force proportional to the square of its velocity. How far must the object fall before its velocity reaches one half its terminal velocity? 7. An object is dropped from altitude y0 . Determine the impact velocity if air resistance is neglected—that is, if we assume no drag force.

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Exercises 8–11: An object undergoes one-dimensional motion along the x-axis subject to the given decelerating forces. At time t = 0, the object’s position is x = 0 and its velocity is v = v 0 . In each case, the decelerating force is a function of the object’s position x(t) or its velocity v(t) or both. Transform the problem into one having distance x as the independent variable. Determine the position xf at which the object comes to rest. (If the object does not come to rest, xf = ∞.) 8. m

dv = −kx2 v dt

9. m

dv = −kxv2 dt

10. m

dv = −ke−x dt

11. m

kv dv =− dt 1+x

12. A boat having mass m is pushed away from a dock with an initial velocity v 0 . The water exerts on the boat a drag force that is proportional to the square of its velocity. Determine the velocity of the boat when it is a distance d from the dock. 13. An object is dropped from altitude y0 . (a) Determine the impact velocity if the drag force is proportional to the square of velocity, with drag coefﬁcient κ. (b) If the terminal velocity is known to be −120 mph and the impact velocity was −90 mph, what was the initial altitude y0 ? (Recall that we take velocity to be negative when the object is moving downward.) 14. An object is dropped from altitude y0 . (a) Assume that the drag force is proportional to velocity, with drag coefﬁcient k. Obtain an implicit solution relating velocity and altitude. (b) If the terminal velocity is known to be −120 mph and the impact velocity was −90 mph, what was the initial altitude y0 ? 15. We need to design a ballistics chamber to decelerate test projectiles ﬁred into it. Assume the resistive force encountered by the projectile is proportional to the square of its velocity and neglect gravity. As the ﬁgure indicates, the chamber is to be constructed so that the coefﬁcient κ associated with this resistive force is not constant but is, in fact, a linearly increasing function of distance into the chamber. Let κ(x) = κ0 x, where κ0 is a constant; the resistive force then has the form κ(x)v2 = κ0 xv2 . If we use time t as the independent variable, Newton’s law of motion leads us to the differential equation m

dv + κ0 xv2 = 0 dt

v=

with

dx . dt

(14)

(a) Adopt distance x into the chamber as the new independent variable and rewrite differential equation (14) as a ﬁrst order equation in terms of the new independent variable. (b) Determine the value κ0 needed if the chamber is to reduce projectile velocity to 1% of its incoming value within d units of distance. Ballistics chamber

Projectile

x Figure for Exercise 15

(x) = 0 x

2.9

Applications to Mechanics

87

16. The motion of a body of mass m, gravitationally attracted to Earth in the presence of a resisting drag force proportional to the square of its velocity, is given by m

dv GmMe =− + κv2 dt r2

[recall equation (13)]. In this equation, r is the radial distance of the body from the center of Earth, G is the universal gravitational constant, Me is the mass of Earth, and v = dr/dt. Note that the drag force is positive, since it acts in the positive r direction. (a) Assume that the body is released from rest at an altitude h above the surface of Earth. Recast the differential equation so that distance r is the independent variable. State an appropriate initial condition for the new problem. (b) Show that the impact velocity can be expressed as 1/2 h −2(κ/m)s e vimpact = − 2GMe ds , 2 0 (Re + s) where Re represents the radius of Earth. (The minus sign reﬂects the fact that v = dr/dt < 0.) 17. On August 24, 1894, Pop Shriver of the Chicago White Stockings caught a baseball dropped (by Clark Grifﬁth) from the top of the Washington Monument. The Washington Monument is 555 ft tall and a baseball weighs 5 18 oz. (a) If we ignore air resistance and assume the baseball was acted upon only by gravity, how fast would the baseball have been traveling when it was 7 ft above the ground? (b) Suppose we now include air resistance in our model, assuming that the drag force is proportional to velocity with a drag coefﬁcient k = 0.0018 lb-sec/ft. How fast is the baseball traveling in this case when it is 7 ft above the ground? 18. A 180-lb skydiver drops from a hot-air balloon. After 10 sec of free fall, a parachute is opened. The parachute immediately introduces a drag force proportional to velocity. After an additional 4 sec, the parachutist reaches the ground. Assume that air resistance is negligible during free fall and that the parachute is designed so that a 200-lb person will reach a terminal velocity of −10 mph. (a) What is the speed of the skydiver immediately before the parachute is opened? (b) What is the parachutist’s impact velocity? (c) At what altitude was the parachute opened? (d) What is the balloon’s altitude? 19. When modeling the action of drag chutes and parachutes, we have assumed that the chute opens instantaneously. Real devices take a short amount of time to fully open and deploy. In this exercise, we try to assess the importance of this distinction. Consider again the assumptions of Exercise 2. A 3000-lb dragster is moving on a straight track at a speed of 220 mph when, at time t = 0, the drag chute is opened. If we assume that the drag force is proportional to velocity and that the chute opens instantaneously, the differential equation to solve is mv = −kv. If we assume a short deployment time to open the chute, a reasonable differential equation might be mv = −k(tanh t)v. Since tanh(0) = 0 and tanh(1) ≈ 0.76, it will take about 1 sec for the chute to become 76% deployed in this model. Assume k = 25 lb-sec/ft. Solve the two differential equations and determine in each case how long it takes the vehicle to slow to 50 mph. Which time do you anticipate will be larger? (Explain.) Is the idealization of instantaneous chute deployment realistic?

88

CHAPTER 2

First Order Differential Equations 20. An object of mass m is dropped from a high platform at time t = 0. Assume the drag force is proportional to the square of the velocity, with drag coefﬁcient κ. As in Example 1, derive an expression for the velocity v(t). 21. Assume the action of a parachute can be modeled as a drag force proportional to the square of the velocity. Determine the drag coefﬁcient κ of the parachute needed so that a 200-lb person using the parachute will have a terminal velocity of −10 mph. Pendulum Motion: Conservation of Energy In Exercises 22 and 23, our goal is to describe the rotational motion of the pendulum shown in the ﬁgure. We neglect the weight of the rod when modeling the motion of this pendulum. m

l O g

Figure for Exercises 22 and 23

Applying the rotational version of Newton’s laws to the pendulum leads to the second order differential equation ml2 θ = −mgl sin θ.

(15)

In equation (15), the right-hand side is negative because it acts to cause clockwise rotation—that is, rotation in the negative θ direction. 22. Suppose that at some initial time the pendulum is located at angle θ0 with an angular velocity dθ/dt = ω0 radians/sec. (a) Equation (15) is a second order differential equation. Rewrite it as a ﬁrst order separable equation by adopting angle θ as the independent variable, using the fact that dω dω dθ dω d dθ θ = = = =ω . dt dt dt dθ dt dθ Complete the speciﬁcation of the initial value problem by specifying an appropriate initial condition. (b) Obtain the implicit solution ml2

ω2 ω2 − mgl cos θ = ml2 0 − mgl cos θ0 . 2 2

(16)

The pendulum is a conservative system; that is, energy is neither created nor destroyed. Equation (16) is a statement of conservation of energy. At a position deﬁned by the angle θ, the quantity ml2 ω2 /2 is the kinetic energy of the pendulum while the term −mgl cos θ is the potential energy, referenced to the horizontal position θ = π/2. The constant right-hand side is the total initial energy. (c) Determine the angular velocity at the instant the pendulum reaches the vertically downward position, θ = 0. Express your answer in terms of the constants ω0 and θ0 . 23. A pendulum, 2 ft in length and initially in the downward position, is launched with an initial angular velocity ω0 . If it achieves a maximum angular displacement of 135 degrees, what is ω0 ?

2.10

Euler’s Method

89

2.10 Euler’s Method Up to this point, we have focused on analytical techniques for solving the initial value problem y = f (t, y),

y(t0 ) = y0 .

(1)

In Section 2.2, we saw that there is an explicit representation, in terms of integrals, for the solution of a linear differential equation. However, we also saw that it might be difﬁcult to work with this representation if the integrand does not have an elementary antiderivative. Sections 2.5–2.7 discussed analytical techniques for solving certain special types of nonlinear differential equations (Bernoulli equations, separable equations, exact equations, etc.). These techniques, however, often lead to implicit solutions, and it may be difﬁcult to “unravel” the implicit solution in order to obtain an explicit solution. In addition, there are many nonlinear differential equations that do not belong to the special categories for which analytical techniques have been developed. Therefore, it’s clear that the analytical methods we’ve discussed, while important and useful, are not totally adequate. We need tools that enable us to obtain quantitative information about the solution of (1) in the general case. Numerical methods are one such tool. We introduce numerical techniques by considering Euler’s method,9 perhaps the simplest and most intuitive numerical method. Besides serving as an introduction to numerical techniques, Euler’s method also provides us with a means (albeit a relatively crude one) to analyze initial value problems for which analytical methods are not applicable. Numerical methods are discussed in greater detail in Chapter 7. We begin by asking the most basic question: “What is a numerical solution of initial value problem (1)?” As we will see, a numerical method for (1) typically generates values y1 ,

y2 ,

...,

yn

that approximate corresponding solution values y(t1 ),

9

y(t2 ),

...,

y(tn )

Leonhard Euler (1707–1783) was one of the most gifted individuals in the history of mathematics and science; his 866 publications (books and articles) make him arguably the most proliﬁc as well. He established the foundations of mathematical analysis, the theory of special functions, and analytical mechanics. Complex analysis, number theory, differential equations, differential geometry, lunar theory, elasticity, acoustics, ﬂuid mechanics, and the wave theory of light are some of the other areas to which Euler made important contributions. A backlog of his work continued to be published for nearly 50 years after his death. Euler’s achievements become even more remarkable when one appreciates the circumstances surrounding his work. He was involved in the world about him. In Russia, he worked on state projects involving cartography, science education, magnetism, ﬁre engines, machines, and shipbuilding. Later, in Berlin, he served as an advisor to the government on state lotteries, insurance, annuities, and pensions. He fathered thirteen children, although only ﬁve survived infancy. Euler claimed that he made some of his greatest discoveries while holding an infant in his arms with other children playing at his feet. In 1771 Euler became totally blind, but he was able to maintain a prodigious output of work until his death.

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at designated abscissa values t0 < t1 < t2 < · · · < tn . Figure 2.18 illustrates the output of a numerical method, where the solid curve indicates the actual (unknown) solution of initial value problem (1). y

y1 y0

y2

y3 y 4

y0

t0 t1 t2 t3 t4

yn y(t)

tn

t

FIGURE 2.18

The solid curve indicates the actual (unknown) solution of initial value problem (1). For i = 1, 2, . . . , n, the points (ti , yi ) are approximations to corresponding points (ti , y(ti )) on the actual solution curve. We say that yi is a numerical approximation to the unknown value y(ti ).

In Figure 2.18, the values y1 , y2 , y3 , . . . are found sequentially—ﬁrst y1 , then y2 , and so forth. Note that the starting value y0 = y(t0 ) is known exactly, since it is speciﬁed in the initial value problem (1). The numerical problem therefore reduces to this question: Given (t0 , y0 ), how do we determine (t1 , y1 )? And in general, given an approximation (tk , yk ), how do we determine the next approximation, (tk+1 , yk+1 )? The simplest answer, the simplest numerical algorithm, is Euler’s method (also known as the tangent line method).

Derivation of Euler’s Method The geometric ideas underlying Euler’s method can be understood in terms of direction ﬁelds. At the initial condition point (t0 , y0 ), the differential equation speciﬁes the slope of the solution curve at (t0 , y0 ); it is equal to f (t0 , y0 ). Therefore, the line tangent to the solution curve y(t) at the point (t0 , y0 ) has equation y = y0 + f (t0 , y0 )(t − t0 ).

(2)

We follow this tangent line over a short time interval, to t = t1 , where t0 < t1 . At t = t1 , we reach the point (t1 , y1 ), where y1 = y0 + f (t0 , y0 )(t1 − t0 ). The value found above, y1 , is the Euler’s method approximation to the solution value y(t1 ). (See Figure 2.19.) While the new point (t1 , y1 ) does not exactly coincide with the point (t1 , y(t1 )), it is generally close to that point (assuming t1 is sufﬁciently close to t0 ). Moreover, since f (t, y) is continuous, the direction ﬁeld ﬁlament at (t1 , y1 )

2.10

Euler’s Method

91

y y1 y(t) y0

y0 t0

y(t 1)

t

t1 FIGURE 2.19

The line tangent to y(t) at the initial point (t0 , y0 ) has slope f (t0 , y0 ). Following the tangent line to time t1 , we arrive at the point (t1 , y1 ) and have an approximation, y1 , to the solution value, y(t1 ).

has nearly the same slope as the ﬁlament at (t1 , y(t1 )). Hence, although we do not know the exact value of y(t) when t = t1 , we are close to it and we do have a good idea of which direction the graph of y(t) is heading. At t = t1 , the graph of y(t) has slope f (t1 , y(t1 )), which is nearly equal to f (t1 , y1 ). So, in an attempt to follow the solution curve y(t), we proceed from (t1 , y1 ) along a line having slope f (t1 , y1 ): y = y1 + f (t1 , y1 )(t − t1 ). Following this line until t = t2 , we reach a point (t2 , y2 ), where y2 = y1 + f (t1 , y1 )(t2 − t1 ). This process is repeated, leading to the algorithm yk+1 = yk + f (tk , yk )(tk+1 − tk ),

k = 0, 1, 2, . . . .

(3)

Iteration (3) is known as Euler’s method and is illustrated in Figure 2.20. Euler’s method amounts to tracing a polygonal path through the direction ﬁeld. y 3 2.5 2 1.5 1 0.5

t –1 –0.5

1

2

3

4

5

6

–1

FIGURE 2.20

Starting on the solution curve at (t0 , y0 ), Euler’s method attempts to track the solution y(t), tracing a polygonal path through the direction ﬁeld. As the path proceeds, its direction is constantly corrected by sampling the direction ﬁeld; see equation (3).

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E X A M P L E

1

Apply Euler’s method to the initial value problem y = t2 + y,

y(2) = 1.

Use t1 = 2.1, t2 = 2.2, and t3 = 2.3. Generate approximations y1 to y(2.1), y2 to y(2.2), and y3 to y(2.3). Solution: The actual (unknown) solution starts at (t0 , y0 ) = (2, 1) and has a starting slope of f (t0 , y0 ) = f (2, 1) = 5. Following the line of slope 5 passing through (2, 1), we obtain, by (3), y1 = y0 + f (t0 , y0 )(t1 − t0 ) = 1 + 5(0.1) = 1.5. Having (t1 , y1 ) = (2.1, 1.5), we take the next step to (t2 , y2 ): y2 = y1 + f (t1 , y1 )(t2 − t1 ) = 1.5 + 5.91(0.1) = 2.091. Having (t2 , y2 ) = (2.2, 2.091), we take the next step to (t3 , y3 ): y3 = y2 + f (t2 , y2 )(t3 − t2 ) = 2.091 + 6.931(0.1) = 2.7841. Note that the differential equation in this example is linear, and hence a formula for the solution can be found. The exact solution is y(t) = 11et−2 − (t2 + 2t + 2). Figure 2.21 compares the Euler’s method approximations with the exact solution at the points t = 2, 2.1, 2.2, and 2.3. y 5 4.5 4 3.5

y(t) t0 t1 t2 t3

3 y3

2.5 2

y2

1.5

= = = =

2.0 2.1 2.2 2.3

y0 y1 y2 y3

= = = =

1.0000 1.5000 2.0910 2.7841

y1

1

y0

0.5 1.9

2

2.1 2.2 2.3 2.4 2.5

t

FIGURE 2.21

The values y1 , y2 , and y3 are the Euler’s method approximations found in Example 1.

❖

Implementing Euler’s Method The simplest way to organize an Euler’s method calculation is to choose an appropriate step size, h, and then use h to deﬁne the equally spaced sample points: t1 = t0 + h, t2 = t1 + h, and, in general, tk+1 = tk + h,

k = 0, 1, . . . , n − 1.

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Euler’s Method

93

In Example 1 we used a step size of h = 0.1 to deﬁne sample points t0 = 2.0, t1 = 2.1, t2 = 2.2, and t3 = 2.3. For a constant step size h, the term tk+1 − tk in (3) is equal to h. Thus, Euler’s method takes the form yk+1 = yk + hf (tk , yk ),

k = 0, 1, . . . , n − 1.

(4)

We anticipate that Euler’s method should become more accurate when we take smaller steps, sampling the direction ﬁeld more often and using this information to correct the “Euler path” that is tracking the solution y(t) (see Exercise 15). Using a small step size h, however, may lead to a signiﬁcant amount of computation. Therefore, numerical methods are usually programmed and run on a computer or programmable calculator. In this section, we have assumed a constant step size h in order to simplify the discussion. Many implementations of numerical methods, however, use variable-step algorithms rather than a ﬁxed-step algorithm. Such variable-step algorithms use error estimates that monitor errors as the algorithm proceeds. When errors are increasing, the steplength is reduced; when errors are decreasing, the steplength is increased. E X A M P L E

2

Apply Euler’s method to the initial value problem y = y(2 − y),

y(0) = 0.1.

(5)

Use h = 0.2 and approximate the solution on the interval 0 ≤ t ≤ 4. Solution: Using a ﬁxed step size of h = 0.2, Euler’s method samples the direction ﬁeld at time values t1 = 0.2, t2 = 0.4, . . . , t19 = 3.8, and t20 = 4.0. For this test case, initial value problem (5) can be solved, since the differential equation is separable. Figure 2.22 shows a portion of the direction ﬁeld for y = y(2 − y). The solid curve in Figure 2.22 is the graph of the actual solution of (5), and the dots show the Euler path through the direction ﬁeld.

y 3 2.5 2 1.5 1 0.5

t –1 –0.5

1

2

3

4

5

6

–1

FIGURE 2.22

The direction ﬁeld for y = y(2 − y). The curve denotes the solution of y = y(2 − y), y(0) = 0.1, the initial value problem posed in Example 2. The dots are the points generated by Euler’s method, using a step size of h = 0.2.

❖

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Runge-Kutta Methods Euler’s method is conceptually important, but it is a relatively crude numerical algorithm. The question then becomes “How do we systematically develop algorithms that are more accurate?” A key to systematically achieving greater accuracy is to view Euler’s method from another, slightly different perspective. Consider the initial value problem (1). Suppose we assume not only that the solution y(t) is differentiable, but also that it has a Taylor series expansion that converges in the interval of interest, a ≤ t ≤ b. We then have, for t and t + h in [a, b], y(t + h) = y(t) + y (t)h +

y (t) 2 y (t) 3 h + h + ···. 2! 3!

(6)

If we set t = tk , t + h = tk+1 and use the fact that y (t) = f (t, y(t)), we obtain y(tk+1 ) = y(tk ) + f (tk , y(tk ))h +

y (tk ) 2 y (tk ) 3 h + h + ···. 2! 3!

(7)

As shown in (7), Euler’s method can be viewed as truncating the Taylor series after the linear term. Equation (7) also shows that the error made in taking one step of Euler’s method is the inﬁnite sum that begins with the term y (tk )h2 /2!. In most applications, it can be shown (using Taylor’s theorem) that this error is bounded by a constant multiple of h2 . Equation (7) provides a blueprint for improving accuracy; an improved algorithm should somehow incorporate more terms of the Taylor series. For example, if the algorithm incorporated the term y (tk )h2 /2!, then it would be including concavity information in addition to the slope information given by the term y (tk )h. Moreover, such an algorithm would have an error bounded by a constant multiple of h3 instead of a multiple of h2 . Since h is small, we would expect increased accuracy. It is shown in Chapter 7 that the terms y(n) (tk )hn /n! can be obtained directly from the differential equation y = f (t, y). In principle, therefore, retaining additional terms in (7) in order to create a more accurate algorithm is straightforward. There are two practical difﬁculties, however. The computations necessary to determine d(n−1) (n) (8) y (tk ) = (n−1) f (t, y(t)) dt t=tk

very quickly become unwieldy as n increases. Moreover, the resulting algorithm is problem speciﬁc, since the calculations in (8) have to be redone every time the method is applied to a new differential equation. The challenge facing the numerical analyst is to retain more terms of the Taylor series (7), but in a way that is both problem independent and computationally friendly. One operation that computers perform very easily is the evaluation of functions. Knowing this, numerical analysts have achieved the desired objectives by creating algorithms that use nested compositions of functions to approximate the higher derivatives y(n) (tk ), n = 2, 3, . . . . Several such algorithms are discussed in Chapter 7. One popular example is the fourth-order

2.10

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95

Runge-Kutta10 method listed in (9). For initial value problem (1), this method has the form h (9a) yk+1 = yk + [K1 + 2K2 + 2K3 + K4 ], 6 where K1 = f (tk , yk ) K2 = f (tk + h/2, yk + (h/2)K1 ) K3 = f (tk + h/2, yk + (h/2)K2 )

(9b)

K4 = f (tk + h, yk + hK3 ). Note that the terms Kj in (9b) are formed by sequentially evaluating compositions of the function f ; for example, K3 = f (tk + h/2, yk + (h/2)f (tk + h/2, yk + (h/2)f (tk , yk ))). When the compositions in (9) are unraveled, it can be seen that the algorithm correctly replicates the Taylor series expansion (7) up to and including the term y(4) (tk )h4 /4!. E X A M P L E

3

As a test case to illustrate how the Runge-Kutta philosophy can improve accuracy, consider the initial value problem y = 2ty + 1,

y(0) = 2.

(a) Solve this initial value problem mathematically. (b) Solve this initial value problem numerically on the interval 0 ≤ t ≤ 2, using Euler’s method and the Runge-Kutta method (9). Use a constant step size of h = 0.05. (c) Tabulate the exact solution values and both sets of numerical approximations from t = 0 to t = 2 in steps of t = 0.25. Is the Runge-Kutta method more accurate than Euler’s method for this test case? Solution: (a) This initial value problem was given in Exercise 45 in Section 2.2. The exact solution is √ 2 π erf(t) , y(t) = et 2 + 2 t 2 2 where erf(t) denotes the error function, erf(t) = √ e−s ds. π 0 (continued) 10

Carle David Tolmé Runge (1856–1927) was a German scientist whose initial interest in pure mathematics was eventually supplanted by an interest in spectroscopy and applied mathematics. During his career, he held positions at universities in Hanover and Gottingen. Runge was a particularly ﬁt and active man; it is reported that he entertained grandchildren on his seventieth birthday by doing handstands. Martin Wilhelm Kutta (1867–1944) held positions at Munich, Jena, Aachen, and Stuttgart. In addition to the Runge-Kutta method (1901), he is remembered for his work in the study of airfoils.

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(b) We coded the Runge-Kutta method (9) for this initial value problem, using MATLAB as a programming environment; a listing, together with a discussion of the practical aspects of coding a Runge-Kutta, is given in the next subsection. (We also provide brief tutorials for MATLAB and Mathematica in our Technical Resource Manual. You may view these tutorials at http://www.aw-bc.com/kohler/.) (c) The results are shown in Table 2.1. As is shown in the table, the Runge-Kutta results for the constant step size h = 0.05 are considerably more accurate than those of Euler’s method using h = 0.05.

TA B L E 2 . 1 The Results of Example 3 Here, y E denotes the Euler’s method estimates of y(t), y RK denotes the Runge-Kutta method estimates, and yT the true values. t

yE

y RK

yT

0.0000 0.2500 0.5000 0.7500 1.0000 1.2500 1.5000 1.7500 2.0000

2.0000 2.3594 3.0726 4.3960 6.9084 11.9543 22.8447 48.3243 113.2709

2.0000 2.3897 3.1603 4.6162 7.4666 13.4434 27.0987 61.4573 157.3539

2.0000 2.3897 3.1603 4.6162 7.4666 13.4434 27.0988 61.4577 157.3563

❖

Coding a Runge-Kutta Method We conclude with a short discussion of the practical aspects of writing a program to implement a Runge-Kutta method. Figures 2.23 and 2.24 list the programs used to generate the numerical solution in Example 3. This particular code was written in MATLAB, but the principles are the same for any programming language. We note ﬁrst that no matter what numerical method we decide to use for the initial value problem y = f (t, y),

y(t0 ) = y0 ,

we need to write a subprogram (or module) that evaluates f (t, y). Such a module is listed in Figure 2.24 for the initial value problem of Example 3. Figure 2.23 lists a MATLAB program that executes 40 steps of the fourth-order Runge-Kutta method (9) for the initial value problem of Example 3. The program listed in Figure 2.23 is not as general as it could be. Normally a Runge-Kutta code is written as a subprogram or module that we can call

2.10

Euler’s Method

97

% % Set the initial conditions for the % initial value problem of Example 3 % t=0; y=2; h=0.05; output=[t,y]; % % Execute the fourth-order Runge-Kutta method % on the interval [0,2] % for i=1:40 ttemp=t; ytemp=y; k1=f(ttemp,ytemp); ttemp=t+h/2; ytemp=y+(h/2)*k1; k2=f(ttemp,ytemp); ttemp=t+h/2; ytemp=y+(h/2)*k2; k3=f(ttemp,ytemp); ttemp=t+h; ytemp=y+h*k3; k4=f(ttemp,ytemp); y=y+(h/6)*(k1+2*k2+2*k3+k4); t=t+h; output=[output;t,y]; end FIGURE 2.23

A Runge-Kutta code for the initial value problem in Example 3.

function yp=f(t,y) yp=2*t*y+1; FIGURE 2.24

A function subprogram that evaluates f (t, y) for the differential equation of Example 3.

whenever we have an initial value problem to solve numerically. Such subprograms allow the user to input a step size h, the number of steps to execute, and the initial conditions. They can be used over and over again and do not have to be modiﬁed whenever the initial value problem changes. We did not list a general module for Figure 2.23 because we wanted the basic steps of a Runge-Kutta program to be obvious.

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Observe that the code listed in Figure 2.23 stays as close as possible to the notation and format of the fourth-order Runge-Kutta method (9). In general, it is a good idea to use variable names (such as K1 and K2 ) that match the names in the algorithm. Beyond the choice of variable names, the code in Figure 2.23 also mimics the steps of algorithm (9) as closely as possible. Adhering to such conventions makes programs much easier to read and debug.

EXERCISES Exercises 1–6: In each exercise, (a) Write the Euler’s method iteration yk+1 = yk + hf (tk , yk ) for the given problem. Also, identify the values t0 and y0 . (b) Using step size h = 0.1, compute the approximations y1 , y2 , and y3 . (c) Solve the given problem analytically. (d) Using the results from (b) and (c), tabulate the errors ek = y(tk ) − yk for k = 1, 2, 3. 1. y = 2t − 1,

3. y = −ty,

2

5. y = y ,

2. y = −y, y(0) = 1

y(1) = 0

4. y = −y + t,

y(0) = 1

y(0) = 1

6. y = y,

y(0) = 0

y(−1) = −1

Exercises 7–10: Reducing the Step Size These exercises examine graphically the effects of reducing step size on the accuracy of the numerical solution. A computer or programmable calculator is needed. (a) Use Euler’s method to obtain numerical solutions on the speciﬁed time interval for step sizes h = 0.1, h = 0.05, and h = 0.025. (b) Solve the problem analytically and plot the exact solution and the three numerical solutions on a single graph. Does the error appear to be getting smaller as h is reduced? 7. y = 2y − 1,

−1

9. y = y ,

y(0) = 1,

y(0) = 1,

0 ≤ t ≤ 0.5

0≤t≤1

8. y = y + e−t ,

2

10. y = −y ,

y(0) = 0,

y(−1) = 2,

0≤t≤1 −1 ≤ t ≤ 0

11. Assume we are considering the direction ﬁeld of an autonomous ﬁrst order differential equation. (a) Suppose we can qualitatively establish, by examining this direction ﬁeld, that all solution curves y(t) in a given region of the ty-plane have one of the following four types of behavior: (i) increasing, concave up (iii) decreasing, concave up

(ii) increasing, concave down (iv) decreasing, concave down.

Suppose we implement an Euler’s method approximation to one of the solution curves in the region, using some reasonable step size h. Consider each of the four cases. In each case, will the values yk underestimate the exact values or overestimate the exact values or is it impossible to reach a deﬁnite conclusion? (b) What do you think will happen if Euler’s method is used to approximate an “S-shaped solution curve” similar to the logistic curve shown in Figure 2.22 on page 93. In that case, a solution curve changes from increasing and concave up to increasing and concave down. Are your answers to part (a) consistent with the behavior exhibited by the Euler approximation shown in the ﬁgure?

2.10

Euler’s Method

99

Exercises 12–15: A programmable calculator or computer is needed for these exercises. 12. Use Euler’s method with step size h = 0.01 to numerically solve the initial value problem y − ty = sin 2πt,

y(0) = 1,

0 ≤ t ≤ 1.

Graph the numerical solution. [Note: Although the differential equation in this problem is a ﬁrst order linear equation and we can get an explicit representation for the exact solution, the representation involves antiderivatives that we cannot express in terms of known functions. From a quantitative point of view, the representation itself is of little use.] 13. Assume a tank having a capacity of 200 gal initially contains 90 gal of fresh water. At time t = 0, a salt solution begins ﬂowing into the tank at a rate of 6 gal/min and the well-stirred mixture ﬂows out at a rate of 1 gal/min. Assume that the inﬂow concentration is given by c(t) = 2 − cos πt oz/gal, where time t is in minutes. Formulate the appropriate initial value problem for Q(t), the amount of salt (in ounces) in the tank at time t. Use Euler’s method to approximately determine the amount of salt in the tank when the tank contains 100 gal of liquid. Use a step size of h = 0.01. 14. Let P(t) denote the population of a certain colony, measured in millions of members. The colony is modeled by P P = 0.1 1 − P + M(t), P(0) = P0 , 3 where time t is measured in years. Assume that the colony experiences a migration inﬂux that is initially strong but that soon tapers off. Speciﬁcally assume that M(t) = e−t . Suppose the colony had 500,000 members initially. Use Euler’s method to estimate its size after 2 years. 15. In Chapter 7, we will examine how the error in numerical algorithms, such as Euler’s method, depends on step size h. In this exercise, we further examine the dependence of errors on step size by studying a particular example, y = y + 1,

y(0) = 0.

(a) Use Euler’s method to obtain approximate solutions to this initial value problem on the interval 0 ≤ t ≤ 1, using step sizes h1 = 0.02 and h2 = 0.01. You will therefore obtain two sets of points, (tk(1) , y(1) k ),

k = 0, . . . , 50

(tk(2) , y(2) k ),

k = 0, . . . , 100

where tk(1) = 0.02k, k = 0, 1, . . . , 50 and tk(2) = 0.01k, k = 0, 1, . . . , 100. (b) Determine the exact solution, y(t). (c) Print a table of the errors at the common points, tk(1) , k = 0, 1, . . . , 50: e(1) (tk(1) ) = y(tk(1) ) − y(1) k

and

e(2) (tk(1) ) = y(tk(1) ) − y(2) 2k .

(d) Note that the approximations y(2) 2k were found using a step size equal to one half of the step size used to obtain the approximations y(1) k ; that is, h2 = h1 /2. Compute the corresponding error ratios. In particular, compute e(2) (t(1) ) k (1) (1) , k = 1, . . . , 50. e (tk ) On the basis of these computations, conjecture how halving the step size affects the error of Euler’s method.

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First Order Differential Equations 16. This exercise treats the simple initial value problem y = λy, y(0) = y0 , where we can see the behavior of the numerical solution as the step size h approaches zero. (a) Show that the solution of the initial value problem is y = eλt y0 . (b) Apply Euler’s method to the initial value problem, using a step size h. Show that yn is related to the initial value by the formula yn = (1 + hλ)n y0 . (c) Consider any ﬁxed time t∗ , where t∗ > 0. Let h = t∗ /n so that t∗ = nh. (The exact solution is y(t) = y0 eλt .) Show that letting h → 0 and n → ∞ in such a way that t∗ remains ﬁxed leads to lim yn = y(t∗ ).

h→0 t ∗ =nh

17. Consider the initial value problem y = 2t − 1, y(1) = 0. (a) Solve this initial value problem. (b) Suppose Euler’s method is used to solve this problem numerically on the interval 1 ≤ t ≤ 5, using step size h = 0.1. Will the numerical solution values be the same as the exact solution values found in part (a)? That is, will yk = y(tk ), k = 1, 2, . . . , 40? Explain. (c) What will be the answer to the question posed in part (b) if the Runge-Kutta method (9) is used instead of Euler’s method?

Exercises 18–22: In each exercise, (a) Using step size h = 01, compute the ﬁrst estimate y1 using Euler’s method and the Runge-Kutta method (9). Let these estimates be denoted by yE1 and yRK 1 , respectively. (b) Solve the problem analytically. (c) Compute the errors | y(t1 ) − y1 | for the two estimates obtained in (a). 18. y = −y, 21. y = y2 ,

y(0) = 1

19. y = −ty,

y(0) = 1

22. y = y,

y(0) = 1

20. y = −y + t,

y(0) = 0

y(−1) = −1

Exercises 23–27: A computer or programmable calculator is needed for these exercises. For the given initial value problem, use the Runge-Kutta method (9) with a step size of h = 0.1 to obtain a numerical solution on the speciﬁed interval. 23. y = −ty + 1, 25. y = −y + t, 27.

y

=

y2 ,

y(0) = 0, y(1) = 0,

y(0) = 1,

0≤t≤2 1≤t≤5

24. y = y3 ,

y(1) = 0.5,

26. y + 2ty = sin t,

1≤t≤2

y(0) = 0,

0≤t≤3

0 ≤ t ≤ 0.9

CHAPTER 2 REVIEW EXERCISES These review exercises provide you with an opportunity to test your understanding of the concepts and solution techniques developed in this chapter. The end-of-section exercises deal with the topics discussed in the section. These review exercises, however, require you to identify an appropriate solution technique before solving the problem.

Exercises 1–30: If the differential equation is linear, determine the general solution. If the differential equation is nonlinear, obtain an implicit solution (and an explicit solution if possible). If an initial condition is given, solve the initial value problem.

Projects y = 3y +4

1. y + 2y = 6

2.

3. y − 3t2 y−1 = 0

4. y2 y = 1 + y3 , y(0) = 0 y 6. y − √ = 0, y(1) = 3e2 t

5. y + 2ty = 2t 7. (t + 3y2 )y + y + 2t = 0 y = 3y, y(0) = 4 9. 2 t +1 √ √ 11. t y − y t = 0

8. y = sin t,

16. y − y = g(t), 17. y = y3 ,

y(0) = 4

10. y − 2y = 0 12. y − 4y = 6e2t

13. (t sin y)y − cos y = 0 15. y + p(t)y = 0,

t2

y(0) = 1; y(0) = 0;

p(t) = g(t) =

2, 1,

3et , 0,

y(0) = 1

14. y = e3t+2y , 1≤t≤2 0≤t0

= 2t(1 + y3 ) 2

y(0) = 1

+ (sin 2t)y = sin 2t,

y2 y

22. (2y + 3t3 )y + 9yt2 = 0

y(0) = 5

27. 2yy + 3t2 = 4,

y(0) = 0

0≤t 0 19.

101

28. t2 (1 + 9y2 )y + 2ty(1 + 3y2 ) = 0

y(π/4) = 4

30. t2 y + csc y = 0,

t>0

PROJECTS Project 1: Flushing Out a Radioactive Spill A lake holding 5,000,000 gallons of water is fed by a stream. Assume that fresh water ﬂows into the lake at a rate of 1000 gal/min and that water ﬂows out at the same rate. At a certain instant, an accidental spill introduces 5 lb of soluble radioactive pollutant into the lake. Assume that the radioactive substance has a half-life of 2 days and dissolves in the lake water to form a well-stirred mixture. 1. Let Q(t) denote the amount of radioactive material present within the lake at time t, measured in minutes from the instant of the spill. Use the conservation principle [rate of change of Q(t) equals rate in minus rate out] to derive a differential equation describing how Q(t) changes with time. Note that the solute is removed by both outﬂow and radioactive decay. Add the appropriate initial condition to obtain the initial value problem of interest. 2. Solve the initial value problem formulated in part 1. 3. How long will it take for the concentration of the radioactive pollutant to be reduced to 0.01% of its original value?

102

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First Order Differential Equations

Project 2: Processing Seafood Many foods, such as crabmeat, are sterilized by cooking. Harvested crabs are laden with bacteria, and the crabmeat must be steamed to reduce the bacteria population to an acceptable level. The longer the crabmeat is steamed, the lower the ﬁnal bacteria count. But steaming forces moisture out of the meat, reducing the amount of crabmeat for sale. Excessive cooking also destroys taste and texture. The processor is therefore faced with a tradeoff when choosing an appropriate steaming time. The basis for a choice of steaming time is the concept of “shelf life.” After the steaming treatment is completed, the product is placed in a sterile package and refrigerated. Under refrigeration, the bacterial content in the meat slowly increases and eventually reaches a size where the crabmeat is no longer suitable for consumption. The time span during which packaged crabmeat is suitable for sale is called the shelf life of the product. We study the following problem: How long must the crabmeat be steamed to achieve a desired shelf life? The ﬁrst step in modeling shelf life is to choose a model that describes the population dynamics of the bacteria. For simplicity, assume dP = k(T)P, dt

(1)

where P(t) denotes the bacteria population at time t. In equation (1), k(T) represents the difference between birth and death rates per unit population per unit time. In this model, k is not constant; it is a function of T, where T denotes the temperature of the crabmeat. [Note that k(T) is ultimately a function of time, since the temperature T of the crabmeat varies with time in the steamer and in the refrigeration case.] We need to choose a reasonable model for the bacteria growth rate, k(T). We do so by reasoning as follows. At low temperatures (near freezing), the rate of growth of the bacteria population is slow; that’s why we refrigerate foods. Mathematically, k(T) is a relatively small positive quantity at those temperatures. As temperature increases, the bacterial growth rate, k(T), ﬁrst increases, with the most rapid rate of growth occurring near 90◦ F. Beyond this temperature, the growth rate begins to decrease. Beyond about 145◦ F, the death rate exceeds the birth rate and the bacteria population begins to decline. A simple model that captures this qualitative behavior is the quadratic function k(T) = k0 + k1 (T − 34)(140 − T),

(2)

where k0 and k1 are positive constants that are typically determined experimentally. We also need a model that describes the thermal behavior of the crabmeat—how the crabmeat temperature T varies in response to the temperature of the surroundings. Assuming Newton’s law of cooling, we have dT = η[S(t) − T]. dt

(3)

In equation (3), η is a positive constant and S(t) is the temperature of the surroundings. Note that the surrounding temperature is not constant, since the crabmeat is initially in the steamer and then in the refrigeration case. We now apply this model to a speciﬁc set of circumstances. Assume the following: (i) Initially the crabmeat is at room temperature (75◦ F) and contains about 107 bacteria per cubic centimeter. (ii) The steam bath is maintained at a constant 250◦ F temperature. (iii) When the crabmeat is placed in the steam bath, it is observed that its temperature rises from 75◦ F to 200◦ F in 5 min. (iv) When the crabmeat is kept at a constant 34◦ F temperature, the bacterial count in the crabmeat doubles in 60 hr.

Projects

103

(v) The bacterial count in the crabmeat begins to decline once the temperature exceeds 145◦ F; that is [see equation (2)], k(145) = 0. (vi) A bacterial count of 105 bacteria per cubic centimeter governs shelf life. Once this bacterial count is reached, the crabmeat can no longer be offered for sale. Determine how long the crabmeat must be steamed to achieve a shelf life of 16 days. Assume that the crabmeat goes directly from the 250◦ F steam bath to the 34◦ F refrigeration case. Assume the 16-day shelf life requirement includes transit time to the point of sale; that is, assume that the measurement of shelf life begins the moment the crabmeat is removed from the steam bath.

Project 3: Belt Friction The slippage of ﬂexible belts, cables, and ropes over shafts or pulleys of circular crosssection is an important consideration in the design of belt drives. When the frictional contact between the belt and the shaft is about to be broken (that is, when slippage is imminent), a belt drive is acting under the most demanding of conditions. The belt tension (force) is not constant along the contact region. Rather, it increases along the contact region between belt and shaft in the direction of the impending slippage. Consider the belt drive shown in Figure 2.25. Suppose we ask the following question: “How much greater can we make tension T2 relative to the opposing tension T1 before the belt slips over the pulley in the direction of T2 ?” The answer obviously depends in part on friction—that is, on the roughness of the belt-shaft contact surface.

2

1

T2

T1 FIGURE 2.25

Consider the belt drive. How much greater can we make tension T2 relative to the opposing tension T1 before the belt slips over the pulley in the direction of T2 ? When slippage is imminent, the tension in the belt has been found to satisfy the differential equation dT(θ) = μT(θ), dθ where the angle θ (in radians) is measured in the direction of the impending slippage over the belt-pulley contact region. The parameter μ is an empirically determined constant known as the coefﬁcient of friction. The larger the value of μ, the rougher the contact surface. In Figure 2.25, with slippage impending in the direction of T2 , the value of T2 is determined relative to T1 by solving the initial value problem dT = μT, dθ

T(θ1 ) = T1 ,

θ1 ≤ θ ≤ θ2 .

Consider the belt drive conﬁgurations shown in Figures 2.26 and 2.27. Assume that belt slippage is impending in the direction shown by the dashed arrow. Compute the belt tensions at the locations shown for the geometries and coefﬁcients of friction given. Which conﬁguration can support the greater load Tl before slipping?

104

CHAPTER 2

First Order Differential Equations

T1 = 100 lb = 0.2

a

T2 = ?

4a

= 0.2

a Tl = ? FIGURE 2.26

5a

5a Tl = ?

a

a

2a T2 = ?

T3 = ? = 0.2

T1 = 100 lb FIGURE 2.27

Project 4: The Baranyi Population Model Milk and milk products depend on refrigeration for storage after pasteurization. Most microorganisms do not grow at refrigeration temperatures (0◦ C–7◦ C). One exception is Listeria monocytogenes. This anaerobic pathogen can multiply at refrigeration temperatures, and the microbe has been responsible for some recent outbreaks of listeriosis, caused by human consumption of contaminated milk products. Food scientists are interested in developing predictive mathematical models that can accurately model the growth of harmful organisms. These models should be able to relate environmental conditions (such as temperature and pH) to the growth rate of a microbial population. In this regard, the modeler walks a ﬁne line. On the one hand, there is an ongoing need to “build more reality” into the model. On the other hand, the model must be kept simple enough to be mathematically tractable and useful. A population model currently being studied and used in food science research is the Baranyi population model.11 It attempts to account for the way certain critical substances affect bacterial cell growth. The essence of the model is a pair of initial value problems, q(t) dP(t) P(t) m P(t), P(0) = P0 , =μ 1− dt 1 + q(t) Pe (4) dq(t) = νq(t), q(0) = q0 . dt 11

J. Baranyi, T. A. Roberts, and P. McClure, “A Non-autonomous Differential Equation to Model Bacterial Growth,” Food Microbiology, Vol. 10, 1993, pp. 43–59.

Projects

105

In this model, P(t) represents the population of bacteria at time t, while q(t) represents the concentration of critical substance present. The positive constant ν represents the growth rate of the critical substance. The parameter μ accounts for the effects of environmental conditions, such as temperature, on the growth rate of the bacteria. We shall, for simplicity, assume both μ and ν to be constants. In (4), the initial values P0 and q0 represent the bacterial population and the amount of critical substance present at time t = 0, respectively. The integer exponent m is introduced into the relative birth rate to allow for greater modeling ﬂexibility. (When m = 1, the differential equation reduces to the logistic equation.) 1. Solve the initial value problem for q(t). For brevity, let q(t) . 1 + q(t)

α(t) =

The differential equation for P(t) now takes the form dP(t) P(t) m P(t). = μα(t) 1 − dt Pe

(5)

2. Show that lim α(t) = 1.

t→∞

Therefore, in the model (5), the critical substance exerts only a certain limited effect on bacterial growth. 3. Make a “change-of-clock” change of independent variable by introducing the new independent variable τ , where dτ = μα(t), dt Therefore,

τ (0) = 0.

t

τ (t) = μ

α(s) ds.

(6)

0

Using the chain rule, show that if we introduce the normalized dependent variable p = P/Pe and view p to be a function of τ , then the initial value problem for p becomes dp(τ ) = [1 − pm (τ )]p(τ ), dτ

p(0) =

P0 . Pe

(7)

4. Solve initial value problem (7) for p(τ ). Note that this differential equation is separable (as well as being a Bernoulli equation). For a general integer m, the equation is most easily solved as a Bernoulli equation. In particular, show that − m1 −m P0 −mτ p(τ ) = 1 + −1 e . Pe 5. Noting that P(t) = Pe p(τ (t)), determine P(t).

(8)

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C H A P T E R

Second and Higher Order Linear Diﬀerential Equations CHAPTER OVERVIEW 3.1

Introduction

3.2

The General Solution of Homogeneous Equations

3.3

Constant Coefﬁcient Homogeneous Equations

3.4

Real Repeated Roots; Reduction of Order

3.5

Complex Roots

3.6

Unforced Mechanical Vibrations

3.7

The General Solution of a Linear Nonhomogeneous Equation

3.8

The Method of Undetermined Coefﬁcients

3.9

The Method of Variation of Parameters

3.10

Forced Mechanical Vibrations, Electrical Networks, and Resonance

3.11

Higher Order Linear Homogeneous Differential Equations

3.12

Higher Order Homogeneous Constant Coefﬁcient Differential Equations

3.13

Higher Order Linear Nonhomogeneous Differential Equations Review Exercises

107

3

108

CHAPTER 3

3.1

Second and Higher Order Linear Differential Equations

Introduction Most of this chapter discusses initial value problems of the form y + p(t)y + q(t)y = g(t),

y(t0 ) = y0 ,

y (t0 ) = y0 ,

a < t < b.

(1)

In equation (1), p(t), q(t), and g(t) are continuous functions on the interval a < t < b, and t0 is some point in this t-interval of interest. The differential equation in problem (1) is called a second order linear differential equation. If g(t) is the zero function, then the differential equation is a homogeneous differential equation; otherwise the equation is a nonhomogeneous differential equation. An initial value problem for a second order equation involves two supplementary or initial conditions, y(t0 ) = y0 and y (t0 ) = y0 . Second order differential equations owe much of their relevance and importance to Newton’s laws of motion. Since acceleration is the second derivative of position, modeling one-dimensional dynamics often leads to a differential equation of the form my = F(t, y, y ). When the applied force F is a linear function of position and velocity, we obtain a second order linear differential equation. The two supplementary conditions in (1) specify position and velocity at time t0 . The concepts and techniques used to analyze the second order initial value problem (1) extend naturally to analogous problems involving higher order linear equations. We consider these extensions in Sections 3.11–3.13.

An Example: The Bobbing Motion of a Floating Object We have all observed a cork, a block of wood, or some other object bobbing up and down in a liquid such as water. How do we mathematically model this bobbing motion? In its rest or equilibrium state, a ﬂoating object is subjected to two equal and opposite forces—the weight of the object is counteracted by an upward buoyant force equal to the weight of the displaced liquid. (This is the law of buoyancy discovered by Archimedes.1 ) If we disturb this equilibrium state by pushing down or pulling up on the object and then releasing it, the object will begin to bob up and down. The physical principle governing the object’s movement is Newton’s second law of motion: ma = F, the product of the mass and acceleration of an object is equal to the sum of the forces acting on it. For example, consider a cylindrical object having uniform mass density ρ, constant cross-sectional area A, and height L (see Figure 3.1). We assume the object is ﬂoating in a liquid having density ρl , where ρ < ρl . In its rest or equilibrium state, the object sinks into the liquid until the weight of the liquid 1

Archimedes of Syracuse (287–212 B.C.) was a remarkable mathematician and scientist, contributing important results in geometry, mechanics, and hydrostatics. Archimedes developed an early form of integration that empowered his work. He was also an inventor, developing the compound pulley, a pump known as Archimedes’ screw, and military machines used to defend his native Syracuse in Sicily from attack by the Romans. Archimedes was killed when Syracuse was captured by the Romans during the Second Punic War.

3.1

Introduction

109

Perturbed state

Equilibrium state Cross-sectional area A Density

y(t)

L Y Y Density l

(a)

(b) FIGURE 3.1

(a) The ﬂoating object is in its equilibrium or rest state when the weight of the displaced liquid is equal to the weight of the object. (b) The object is in a perturbed state when it is displaced from its equilibrium position. At any time t, the quantity y(t) measures how far the object is from its equilibrium position.

displaced equals the weight of the object; we denote the depth to which the object sinks as Y . It can be shown (see Exercise 12) that ρ Y = L. (2) ρl Suppose now that the object is displaced from its equilibrium state, as illustrated in Figure 3.1(b). Let the depth to which the body is immersed in the liquid at time t be denoted by Y + y(t). Thus, y(t) represents the time-varying displacement of the object from its equilibrium state. For deﬁniteness, we assume y(t) to be positive in the downward direction. In the perturbed state, the net force acting upon the object is typically nonzero. In Exercise 12, you are asked to show that Newton’s law, ma = F, leads to the equation ρg ω2 = l , (3) y (t) + ω2 y(t) = 0, ρL where g is the acceleration due to gravity. The perturbation depth, y(t), is thus described by a second order linear homogeneous differential equation. We need more than just the differential equation to uniquely characterize the motion of the bobbing object. Specifying the object’s depth and velocity at some particular instant of time by specifying y(t0 ) = y0 and y (t0 ) = y0 would seem (on physical grounds) to uniquely characterize the motion. The discussion of existence and uniqueness issues given later shows that this physical intuition is, in fact, correct. Consider the differential equation in (3). Assume for the present discussion that the initial value problem y (t) + ω2 y(t) = 0,

y(0) = y0 ,

y (0) = y0

(4)

has a unique solution on any time interval of interest. For simplicity, we’ve

110

CHAPTER 3

Second and Higher Order Linear Differential Equations

chosen the initial time to be t0 = 0. Does the solution of initial value problem (4) describe a bobbing or oscillating motion that is consistent with our experience? For some insight and a preview of what’s to come, note that the functions y(t) = sin ωt and y(t) = cos ωt are each solutions of the differential equation in (4). Section 3.5 shows how to obtain these solutions. For now, the assertion can be veriﬁed by direct substitution. In fact, for any choice of constants C1 and C2 , the function y(t) = C1 sin ωt + C2 cos ωt

(5)

is a solution of y + ω2 y = 0. You will see later that y(t) = C1 sin ωt + C2 cos ωt is, in fact, the general solution of the differential equation; that is, any solution of the differential equation can be constructed by making an appropriate choice of the constants C1 and C2 . For initial value problem (4), imposing the initial conditions upon the general solution (5) leads to the set of equations y(0) = C1 sin(0) +

C2 cos(0) = y0 y (0) = C1 ω cos(0) − C2 ω sin(0) = y0 .

Solving this system of equations, we ﬁnd C1 = y0 /ω and C2 = y0 . The unique solution of initial value problem (4) is therefore y(t) =

y0 sin ωt + y0 cos ωt. ω

(6)

If either y0 = 0 or y0 = 0, it’s obvious that the solution represents the type of sinusoidal oscillating behavior that is consistent with our experience. In general, as you will see later in this chapter, the solution (6) can always be written as a sinusoid. Figure 3.2, for example, shows the behavior of (6) for the special case y0 = y0 = 1, ω = 2. Thus, the mathematical model (4) and its solution (6) do, in fact, predict an oscillatory behavior that is consistent with physical intuition about the motion of a bobbing body. y 1.5 1 0.5 2

4

6

8 10 12 14 16 18 20

t

–0.5 –1 –1.5

FIGURE 3.2

The graph of the solution of y + 4y = 0, y(0) = 1, y (0) = 1. The solution is given by equation (6), using values ω = 2, y0 = 1, y0 = 1.

3.1

Introduction

111

Existence and Uniqueness We begin to develop the necessary mathematical underpinnings by stating an existence-uniqueness theorem proved in advanced texts. Theorem 3.1

Let p(t), q(t), and g(t) be continuous functions on the interval (a, b), and let t0 be in (a, b). Then the initial value problem y + p(t)y + q(t)y = g(t),

y(t0 ) = y0 ,

y (t0 ) = y0

has a unique solution deﬁned on the entire interval (a, b). Compare this theorem with Theorem 2.1, which states an analogous existence-uniqueness result for ﬁrst order linear initial value problems. Both theorems assume that the coefﬁcient functions and the nonhomogeneous term on the right-hand side are continuous on the interval of interest. Both theorems reach the same three conclusions: 1. The solution exists. 2. The solution is unique. 3. The solution exists on the entire interval (a, b). Theorem 3.1 deﬁnes the framework within which we will work. It assures us that, given an initial value problem of the type described, there is one and only one solution. Our job is to ﬁnd it. The similarity of Theorems 2.1 and 3.1 is no accident. You will see in Chapter 4 that these two theorems, as well as an analogous theorem stated for higher order linear initial value problems, can be viewed as special cases of a single, all-encompassing existence-uniqueness theorem for ﬁrst order linear systems. E X A M P L E

1

Determine the largest t-interval on which we can guarantee the existence of a solution of the initial value problem ty + (cos t)y + t2 y = t,

y(−1) = −1,

y (−1) = 2.

Solution: Before we apply Theorem 3.1, we need to write the differential equation in standard form: cos t y + ty = 1. y + t With the equation in this form, we can identify the coefﬁcient functions in the hypotheses of Theorem 3.1. One of the coefﬁcient functions is not continuous at t = 0, but there are no other points of discontinuity on the t-axis. Since the initial conditions are posed at the point t0 = −1, it follows from Theorem 3.1 that the given initial value problem is guaranteed to have a unique solution on the interval −∞ < t < 0. Figure 3.3 shows a numerical solution for this initial value problem on the interval [−1, −0.002]. As you can see, it appears that the solution is not deﬁned at t = 0. (continued)

112

CHAPTER 3

Second and Higher Order Linear Differential Equations y

(continued)

10 8 6 4 2 –1

–0.8

–0.6

–0.4

t

–0.2 –2

FIGURE 3.3

The graph of a numerical solution of the initial value problem in Example 1. The solution appears to have a vertical asymptote at t = 0.

❖

EXERCISES Exercises 1– 4: For each initial value problem, determine the largest t-interval on which Theorem 3.1 guarantees the existence of a unique solution. 1. y + 3t2 y + 2y = sin t, 2. y + y + 3ty = tan t, 3. e t y + 4. ty +

4 1 y= , t t2 − 1

y(1) = 1,

y (1) = −1

y(π) = 1,

y (π) = −1

y(−2) = 1,

y (−2) = 2

sin 2t y + 2y = 0, t2 − 9

y(1) = 0,

y (1) = 1

5. Consider the initial value problem t2 y − ty + y = 0, y(1) = 1, y (1) = 1. (a) What is the largest interval on which Theorem 3.1 guarantees the existence of a unique solution? (b) Show by direct substitution that the function y(t) = t is the unique solution of this initial value problem. What is the interval on which this solution actually exists? (c) Does this example contradict the assertion of Theorem 3.1? Explain.

Exercises 6–7: Let y(t) denote the solution of the given initial value problem. Is it possible for the corresponding limit to hold? Explain your answer. 6. y +

1 y = 0, t − 16 2

7. y + 2y +

y(0) = 1,

1 y = 0, t−3

y (0) = 1,

lim− y(t) = +∞

t→3

y(1) = 1,

y (1) = 2,

lim+ y(t) = +∞

t→0

Exercises 8–10: In each exercise, assume that y(t) = C1 sin ω t + C2 cos ω t is the general solution of y + ω2 y = 0. Find the unique solution of the given initial value problem.

3.1 8. y + y = 0,

y (π) = 2

y(π) = 0,

10. y + 16y = 0,

y(π/4) = 1,

9. y + 4y = 0,

Introduction

y(0) = −2,

113

y (0) = 0

y (π/4) = −4

11. Concavity of the Solution Curve In the discussion of direction ﬁelds in Section 1.3, you saw how the differential equation deﬁnes the slope of the solution curve at a point in the ty-plane. In particular, given the initial value problem y = f (t, y), y(t0 ) = y0 , the slope of the solution curve at initial condition point (t0 , y0 ) is y (t0 ) = f (t0 , y0 ). In like manner, a second order equation provides direct information about the concavity of the solution curve. Given the initial value problem y = f (t, y, y ), y(t0 ) = y0 , y (t0 ) = y0 , it follows that the concavity of the solution curve at the initial condition point (t0 , y0 ) is y (t0 ) = f (t0 , y0 , y0 ). (What is the slope of the solution curve at that point?) Consider the four graphs shown. Each graph displays a portion of the solution of one of the four initial value problems given. Match each graph with the appropriate initial value problem. (a) y + y = 2 − sin t, (b) y + y = −2t, (c) y − y = t2 ,

y(0) = 1,

y(0) = 1, y(0) = 1,

(d) y − y = −2 cos t,

y (0) = −1

y (0) = −1 y (0) = 1

y(0) = 1,

y (0) = 1

y

y

2.5

2

2 1.5 1.5 1 1 0.5

0.5

–0.8

–0.4

0.4

0.8

t

–0.8

–0.4

0.4

Graph A

Graph B

y

y

1.5

1.5

1

1

0.5

0.5

–0.4

0.4

0.8

t

–0.8

–0.5

–0.4

0.4 –0.5

Graph C

Graph D Figure for Exercise 11

0.8

t

t

114

CHAPTER 3

Second and Higher Order Linear Differential Equations

12. The Bobbing Cylinder Model Using Figure 3.1 for reference, carry out the following derivations. (a) Derive expressions for the mass of the cylinder and the displaced liquid, in terms of the mass densities and cylinder geometry. Recall that weight W is given by W = mg. Apply the law of buoyancy to the equilibrium state shown in Figure 3.1(a) and establish equation (2), Y = (ρ/ρl )L. (b) Apply Newton’s law ma = F to the cylinder shown in its perturbed state in Figure 3.1(b). Since y is positive downward, the net force F equals the cylinder weight minus the buoyant force. Show that ρg y + l y = 0. ρL [Hint: The equilibrium equality of part (a) can be used to simplify the differential equation obtained from ma = F.] 13. Since sin(ωt + 2π) = sin ωt and cos(ωt + 2π) = cos ωt, the amount of time T it takes a bobbing object to go through one cycle of its motion is determined by the relation ωT = 2π , or T = 2π/ω. This time T is called the period of the motion (see Section 3.6). As the period decreases, the bobbing motion of the ﬂoating object becomes more rapid. (a) Two identically shaped cylindrical drums, made of different material, are ﬂoating at rest as shown in part (a) of the ﬁgure. (b) Two cylindrical drums, made of identical material, are ﬂoating at rest as shown in part (b). For each case, when the drums are put into motion, is it possible to identify the drum that will bob up and down more rapidly? Explain.

1

2 1

2

(a)

(b) Figure for Exercise 13

14. A buoy having the shape of a right circular cylinder 3 ft in diameter and 5 ft in height is initially ﬂoating upright in water. When it was put into motion at time t = 0, the following 10-sec record of its displacement from equilibrium, measured in inches positive in the downward direction, was obtained. (a) Determine the initial displacement y0 and the period T of the motion (see Exercise 13). (b) Determine the constant ω and the initial velocity y0 of the buoy.

3.2

The General Solution of Homogeneous Equations

115

y

6 4 2

2

4

6

8

t

–2 –4 –6

Figure for Exercise 14

3.2

The General Solution of Homogeneous Equations Consider the second order linear homogeneous differential equation y + p(t)y + q(t)y = 0,

a < t < b,

(1)

where p(t) and q(t) are are continuous on (a, b). We begin with this homogeneous equation because understanding its solution structure is basic to developing methods for solving linear differential equations, whether they are homogeneous or nonhomogeneous. The general solution of equation (1) is often described as a “linear combination” of functions. In particular, let f1 (t) and f2 (t) be any two functions having a common domain, and let c1 and c2 be any two constants. A function of the form f (t) = c1 f1 (t) + c2 f2 (t) is called a linear combination of the functions f1 and f2 . For example, the function f (t) = 3 sin t + 8 cos t is a linear combination of the functions sin t and cos t.

The Principle of Superposition The ﬁrst result we establish for the homogeneous equation (1) is a superposition principle. It shows that a linear combination of two solutions is also a solution. An analogous superposition principle is also valid for higher order linear homogeneous equations; see Section 3.11.

116

CHAPTER 3

Theorem 3.2

Second and Higher Order Linear Differential Equations

Let y1 (t) and y2 (t) be any two solutions of y + p(t)y + q(t)y = 0 deﬁned on the interval (a, b). Then, for any constants c1 and c2 , the linear combination y(t) = c1 y1 (t) + c2 y2 (t) is also a solution on (a, b).

● PROOF: The hypotheses state that y1 (t) and y2 (t) are both solutions of the homogeneous equation. Therefore,

y1 + p(t)y1 + q(t)y1 = 0

and y2 + p(t)y2 + q(t)y2 = 0.

Substituting y(t) = c1 y1 (t) + c2 y2 (t) into the differential equation, we obtain y + p(t)y + q(t)y = (c1 y1 + c2 y2 ) + p(t)(c1 y1 + c2 y2 ) + q(t)(c1 y1 + c2 y2 ). (2) Using basic properties from calculus, we can write the right-hand side of (2) as c1 [ y1 + p(t)y1 + q(t)y1 ] + c2 [ y2 + p(t)y2 + q(t)y2 ] = c1 [0] + c2 [0] = 0. Therefore, the linear combination y(t) = c1 y1 (t) + c2 y2 (t) is also a solution.

●

It is important to understand that the superposition principle of Theorem 3.2 is valid for homogeneous linear equations. In general, a linear combination of solutions of a linear nonhomogeneous equation is not a solution of the nonhomogeneous equation. Similarly, a linear combination of solutions of a nonlinear differential equation is normally not a solution of the nonlinear equation.

Fundamental Sets of Solutions Theorem 3.2 shows that we can form a linear combination of two solutions of equation (1) and create a new solution. We now turn this idea around and ask, “Is it possible to ﬁnd two solutions, y1 (t) and y2 (t), such that every solution of y + p(t)y + q(t)y = 0,

a 0. In this case, the two roots λ1 and λ2 are real and distinct. We obtain two solutions, y1 (t) = eλ1 t and y2 (t) = eλ2 t . We show later in this section that these two solutions form a fundamental set of solutions for equation (1). (b) Suppose b2 − 4ac = 0. In this case, the two roots are equal, λ 1 = λ2 =

−b . 2a

Our computation, based on the trial form y(t) = eλt , therefore yields only one solution, namely y1 (t) = e−(b/2a)t . Since a fundamental set of solutions consists of two solutions having a nonvanishing Wronskian, we must ﬁnd another solution having a different functional form. In Section 3.4, we will discuss this real “repeated root” case and show how to obtain the second function needed for a fundamental set. (c) Suppose b2 − 4ac < 0. In this case, the roots are complex-valued and we have 4ac − b2 b ±i . λ1,2 = − 2a 2a Since a, b, and c are real constants, the roots λ1,2 form a complex conjugate pair. For brevity, let 4ac − b2 b β= . α=− , 2a 2a Then λ1,2 = α ± iβ, where β is nonzero. Several questions arise. What mathematical interpretation do we give to expressions of the form e(α±iβ)t ? Once we make sense of such expressions mathematically, how do we obtain realvalued, physically meaningful solutions of equation (1)? These issues will be addressed in Section 3.5.

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The General Solution When the Characteristic Equation Has Real Distinct Roots We now consider case (a), where the discriminant b2 − 4ac is positive. In this case, the two roots λ1 and λ2 are real and distinct and y1 (t) = eλ1 t and y2 (t) = eλ2 t are two solutions of ay + by + cy = 0. To determine whether { y1 , y2 } forms a fundamental set of solutions, we calculate the Wronskian: eλ 1 t eλ2 t W(t) = λ t = (λ2 − λ1 )e(λ1 +λ2 )t . λ1 e 1 λ2 eλ2 t The factor (λ2 − λ1 ) is nonzero since the roots are distinct. In addition, the exponential function e(λ1 +λ2 )t is nonzero for all t. This calculation establishes, once and for all, that the two exponential solutions obtained in the real, distinct root case form a fundamental set of solutions. There is no need to reestablish this fact for every particular example. The corresponding general solution of ay + by + cy = 0 is y(t) = c1 eλ1 t + c2 eλ2 t .

(6)

E X A M P L E

2

Solve the initial value problem y + 4y + 3y = 0, Solution:

y(0) = 7,

y (0) = −17.

The characteristic equation is λ2 + 4λ + 3 = 0,

or (λ + 1)(λ + 3) = 0. Therefore, the general solution is y(t) = c1 e−t + c2 e−3t , and its derivative is y (t) = −c1 e−t − 3c2 e−3t . To satisfy the initial conditions, c1 and c2 must satisfy c1 + c2 = 7 −c1 − 3c2 = −17. We ﬁnd c1 = 2 and c2 = 5. The unique solution of the initial value problem is y(t) = 2e−t + 5e−3t . ❖ E X A M P L E

3

Solve the initial value problem y + y − 2y = 0,

y(0) = y0 ,

y (0) = y0 .

For what values of the constants y0 and y0 can we guarantee that lim t→∞ y(t) = 0?

3.3

Constant Coefﬁcient Homogeneous Equations

125

The characteristic polynomial for y + y − 2y = 0 is

Solution:

λ2 + λ − 2 = (λ + 2)(λ − 1). Thus, the general solution is y(t) = c1 e−2t + c2 e t . Imposing the initial conditions, we have c1 + c2 = y0 −2c1 + c2 = y0 . The solution of this system of equations is c1 = ( y0 − y0 )/3, c2 = (2y0 + y0 )/3. The solution of the initial value problem is therefore

y0 − y0 2y0 + y0 y(t) = e−2t + e t. 3 3 Since lim t→∞ e−2t = 0 and lim t→∞ e t = +∞, the solution of the initial value problem will tend to zero as t increases if the coefﬁcient of e t in the solution is zero. Therefore, lim t→∞ y(t) = 0 if y0 = −2y0 . ❖

EXERCISES Exercises 1–15: (a) Find the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value problem. (c) Describe the behavior of the solution y(t) as t → −∞ and as t → ∞. Does y(t) approach −∞, +∞, or a ﬁnite limit? 1. y + y − 2y = 0, 2. y − 14 y = 0,

4. 2y − 5y + 2y = 0,

5. y − y = 0,

y (2) = 0

y(2) = 1,

3. y − 4y + 3y = 0,

y(0) = −1,

y (0) = 1

y(0) = −1, y (0) = −5

y(0) = 1,

6. y + 2y = 0,

y (0) = −3

y(0) = 3,

y (0) = −1 y (−1) = 2

y(−1) = 0,

7. y + 5y + 6y = 0,

y(0) = 1,

y (0) = −1

8. y − 5y + 6y = 0,

y(0) = 1,

y (0) = −1

9. y − 4y = 0,

y(3) = 0,

10. 8y − 6y + y = 0, 11. 2y − 3y = 0,

13. y + 4y + 2y = 0, 14. y − 4y − y = 0, 15. 2y − y = 0,

y (3) = 0

y(1) = 4,

y (1) =

3 2

y (−2) = 0

y(−2) = 3,

12. y − 6y + 8y = 0,

y(1) = 2,

y (1) = −8

y(0) = 0,

y (0) = 4

y (0) = 2 + √ y (0) = 2

y(0) = 1,

y(0) = −2,

√ 5

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Second and Higher Order Linear Differential Equations 16. Consider the initial value problem y + αy + βy = 0, y(0) = 1, y (0) = y0 , where α, β, and y0 are constants. It is known that one solution of the differential equation is y1 (t) = e−3t and that the solution of the initial value problem satisﬁes lim t→∞ y(t) = 2. Determine the constants α, β, and y0 . 17. Consider the initial value problem y + αy + βy = 0, y(0) = 3, y (0) = 5. The differential equation has a fundamental set of solutions, { y1 (t), y2 (t)}. It is known that y1 (t) = e−t and that the Wronskian formed by the two members of the fundamental set is W(t) = 4e2t . (a) Determine the second member of the fundamental set, y2 (t). (b) Determine the constants α and β. (c) Solve the initial value problem. 18. The three graphs display solutions of initial value problems on the interval 0 ≤ t ≤ 3. Each solution satisﬁes the initial conditions y(0) = 1, y (0) = −1. Match the differential equation with the graph of its solution. (a) y + 2y = 0

(b) 6y − 5y + y = 0

(c) y − y = 0

y y

1 0.9 0.8

0.5

0.7

1

1.5

–2

0.6 –4

0.5 0.4

–6

0.3 –8

0.2 0.1

–10 0.5

1

1.5

2

t

2.5

Graph A

Graph B y

1

0.9

0.8

0.7

0.6

0.5 0.5

1

1.5

2

Graph C Figure for Exercise 18

2.5

t

2

2.5

t

3.4

Real Repeated Roots; Reduction of Order

127

19. Obtain the general solution of y − 5y + 6y = 0. [Hint: Make the change of dependent variable u(t) = y (t), determine u(t), and then antidifferentiate to obtain y(t).] 20. Rectilinear Motion with a Drag Force In Chapter 2, we considered rectilinear motion in the presence of a drag force proportional to velocity. We solved the ﬁrst order linear equation for velocity and antidifferentiated the solution to obtain distance as a function of time. We now consider directly the second order linear differential equation for the distance function. A particle of mass m moves along the x-axis and is acted upon by a drag force proportional to its velocity. The drag constant is denoted by k. If x(t) represents the particle position at time t, Newton’s law of motion leads to the differential equation mx (t) = −kx (t). (a) Obtain the general solution of this second order linear differential equation. (b) Solve the initial value problem if x(0) = x0 and x (0) = v 0 . (c) What is lim t→∞ x(t)?

3.4

Real Repeated Roots; Reduction of Order In Section 3.3, it was shown that the constant coefﬁcient differential equation ay + by + cy = 0 has the general solution y(t) = c1 eλ1 t + c2 eλ2 t whenever the characteristic equation aλ2 + bλ + c = 0 has distinct real roots λ1 and λ2 . In this section, we consider the case where the characteristic equation has a repeated real root (that is, when the discriminant b2 − 4ac = 0). In this event, looking for solutions of the form y(t) = eλt leads to only one solution, since the characteristic equation has just one distinct root. We must somehow ﬁnd a second solution in order to form a fundamental set of solutions.

The Method of Reduction of Order To obtain a second solution for ay + by + cy = 0 in the repeated root case, we use a method called reduction of order. We apply the method ﬁrst to the problem at hand, ay + by + cy = 0.

(1)

Then, at the end of this section, we’ll discuss reduction of order as a technique for ﬁnding a second solution of the general homogeneous linear equation, y + p(t)y + q(t)y = 0, given that we have somehow found one solution, y1 (t), of the equation. Assume, without loss of generality, that a = 1 in equation (1). Then, since b2 − 4c = 0, we know that c is positive. We can represent c as c = α 2 and choose b = −2α. With these simpliﬁcations in notation, differential equation (1) becomes y − 2αy + α 2 y = 0. The characteristic polynomial for equation (2) is λ2 − 2αλ + α 2 = (λ − α)2 , and therefore one solution is y1 (t) = eαt .

(2)

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Second and Higher Order Linear Differential Equations

To ﬁnd a second solution, y2 (t), we use the method of reduction of order. The basic idea underlying the method is to look for a second solution, y2 (t), of the form y2 (t) = y1 (t)u(t) = eαt u(t).

(3)

The function u(t) in (3) must be chosen so that y2 (t) is also a solution of equation (2). At this point, there’s no obvious reason to believe that this assumed form of the solution provides any simpliﬁcation. We must substitute (3) into differential equation (2) and see what happens. Substituting, we obtain y2 − 2αy2 + α 2 y2 = (eαt u) − 2α(eαt u) + α 2 (eαt u),

(4)

which simpliﬁes to y2 − 2αy2 + α 2 y2 = eαt u .

(5) αt

Since the exponential function is nonzero everywhere, y2 (t) = e u(t) is a solution of y − 2αy + α 2 y = 0 if and only if u = 0. The equation u = 0 can be solved by antidifferentiation to obtain u(t) = a1 t + a2 , where a1 and a2 are arbitrary constants. Thus, the method of reduction of order has led us to a second solution, y2 (t) = eαt (a1 t + a2 ) = a1 teαt + a2 eαt . Notice that the term a2 eαt is simply a constant multiple of y1 (t). Since the general solution of the differential equation contains y1 (t) multiplied by an arbitrary constant, we lose no generality by setting a2 = 0. We can likewise take a1 = 1 since y2 (t) will also be multiplied by an arbitrary constant in the general solution. With these simpliﬁcations, the second solution is y2 (t) = teαt . To verify that { y1 , y2 } = {eαt , teαt } forms a fundamental set, we compute the Wronskian: eαt teαt W(t) = αt = e2αt . αe (αt + 1)eαt Since the Wronskian is nonzero, we have shown that the general solution is y(t) = c1 eαt + c2 teαt .

(6)

E X A M P L E

1

Solve the initial value problem 4y + 4y + y = 0,

y(2) = 1,

y (2) = 0.

Solution: Looking for solutions of the form y(t) = eλt leads to the characteristic equation 4λ2 + 4λ + 1 = (2λ + 1)2 = 0. Therefore, the characteristic equation has real repeated roots λ1 = λ2 = − 12 . By (6), the general solution is y(t) = c1 e−t/2 + c2 te−t/2 .

3.4

Real Repeated Roots; Reduction of Order

129

Imposing the initial conditions leads to c1 e−1 + c2 2e−1 = 1 c = 0. − 1 e−1 2 The solution is c1 = 0, c2 = e/2. The solution of the initial value problem is y(t) =

e −t/2 t = e1−t/2 . ❖ te 2 2

Method of Reduction of Order (General Case) The method of reduction of order is not restricted to constant coefﬁcient equations. It can be applied to the general second order linear homogeneous equation y + p(t)y + q(t)y = 0,

(7)

where p(t) and q(t) are continuous functions on the t-interval of interest. Suppose we know one solution of equation (7); call it y1 (t). We again assume that there is another solution, y2 (t), of the form y2 (t) = y1 (t)u(t). Substituting the assumed form into (7) leads (after some rearranging of terms) to y1 u + [2y1 + p(t)y1 ]u + [ y1 + p(t)y1 + q(t)y1 ]u = 0. At ﬁrst it seems as though this equation offers little improvement. Recall, however, that y1 is not an arbitrary function; it is a solution of differential equation (7). Therefore, the factor multiplying u in the preceding equation vanishes, and we obtain a considerable simpliﬁcation: y1 u + [2y1 + p(t)y1 ] u = 0.

(8)

The structure of equation (8) is what gives the method its name. Although equation (8) is a second order linear differential equation for u, we can deﬁne a new dependent variable v(t) = u (t). Under this change of variables, equation (8) reduces to a ﬁrst order linear differential equation for v, y1 (t)v + [2y1 (t) + p(t)y1 (t)]v = 0.

(9)

Thus, the task of solving a second order linear differential equation has been replaced by that of solving a ﬁrst order linear differential equation. Once we have v = u , we obtain u (and ultimately y2 ) by antidifferentiation. E X A M P L E

2

Observe that y1 (t) = t is a solution of the homogeneous linear differential equation t2 y − ty + y = 0,

0 < t < ∞.

(10)

(a) Use reduction of order to obtain a second solution, y2 (t). Does the pair { y1 , y2 } form a fundamental set of solutions for this differential equation? (continued)

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Second and Higher Order Linear Differential Equations (continued)

(b) If { y1 , y2 } is a fundamental set of solutions, solve the initial value problem t2 y − ty + y = 0, Solution: comes

y (1) = 8.

y(1) = 3,

Note that the initial value problem, written in standard form, bey −

1 1 y + 2 y = 0, t t

y (1) = 8.

y(1) = 3,

(11)

The coefﬁcient functions are not continuous at t = 0. Our t-interval of interest, 0 < t < ∞, is the largest interval containing t0 = 1 on which we are guaranteed the existence of a unique solution of the initial value problem. (a) Since one solution is known, we apply reduction of order. Assuming y2 (t) = tu(t), we have y2 = u + tu

and

Substituting these expressions t2 y − ty + y = 0, we ﬁnd

y2 = 2u + tu . into

the

differential

equation,

t2 (2u + tu ) − t(u + tu ) + tu = t2 (tu + u ) = 0. Therefore, tu + u = 0. Setting v = u leads to the ﬁrst order linear equation t v + v = 0.

(12)

The general solution of equation (12) is c v(t) = . t Since v(t) = u (t), it follows that u(t) = c ln t + d, and we obtain a second solution, y2 (t) = tu(t) = t(c ln t + d).

(13)

[Note that ln |t| = ln t since t > 0.] Using the same rationale as before, we can take c = 1, d = 0 and let y2 (t) = t ln t. The Wronskian of y1 and y2 is W(t) = t, which is nonzero on the interval 0 < t < ∞. Therefore, the general solution is y(t) = c1 y1 (t) + c2 y2 (t) = c1 t + c2 t ln t,

0 < t < ∞.

(14)

(b) For y(t) = c1 t + c2 t ln t, we have y (t) = c1 + c2 (1 + ln t). Imposing the initial conditions y(1) = 3 and y (1) = 8, we obtain =3 c1 c1 + c2 = 8. The solution of the initial value problem is y(t) = 3t + 5t ln t. ❖

3.4

Real Repeated Roots; Reduction of Order

131

EXERCISES Exercises 1–9: (a) Obtain the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value problem. (c) Describe the behavior of the solution as t → −∞ and t → ∞. In each case, does y(t) approach −∞, +∞, or a ﬁnite limit? 1. y + 2y + y = 0,

y (1) = 0

y(1) = 1,

2. 9y − 6y + y = 0,

y(3) = −2,

3. y + 6y + 9y = 0,

y(0) = 2,

4. 25y + 20y + 4y = 0,

y (3) = − 53 y (0) = −2

y(5) = 4e−2 ,

y (5) = − 35 e−2

5. 4y − 4y + y = 0,

y(1) = −4,

y (1) = 0

6. y − 4y + 4y = 0,

y(−1) = 2,

y (−1) = 1

7. 16y − 8y + y = 0, y(0) = −4, √ 8. y + 2 2y + 2y = 0, y(0) = 1, 9. y − 5y + 6.25y = 0,

y(−2) = 0,

y (0) = 3 y (0) = 0 y (−2) = 1

10. Consider the simple differential equation y = 0. (a) Obtain the general solution by successive antidifferentiation. (b) View the equation y = 0 as a second order linear homogeneous equation with constant coefﬁcients, where the characteristic equation has a repeated real root. Obtain the general solution using this viewpoint. Is it the same as the solution found in part (a)?

Exercises 11–12: In each exercise, the graph shown is the solution of y − 2αy + α 2 y = 0, y(0) = y0 , y (0) = y0 . Determine the constants α, y0 , and y0 as well as the solution y(t). In Exercise 11, the maximum point shown on the graph has coordinates (2, 8e−1 ). y

11.

y

12.

(2, 8e–1)

1.5 2.5

1 0.5

2

1 1.5

2

3

4

5

6

7

8

9

t

–0.5 –1

1

–1.5 –2

0.5

–2.5 1

2

3

4

5

6

7

8

9

t

13. The graph of a solution y(t) of the differential equation 4y + 4y + y = 0 passes through the points (1, e−1/2 ) and (2, 0). Determine y(0) and y (0).

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Exercises 14 –20: One solution, y1 (t), of the differential equation is given. (a) Use the method of reduction of order to obtain a second solution, y2 (t). (b) Compute the Wronskian formed by the solutions y1 (t) and y2 (t). 14. ty − (2t + 1)y + (t + 1)y = 0, 2

15. t y − ty + y = 0,

y1 (t) = e t

y1 (t) = t

16. y − (2 cot t)y + (1 + 2 cot2 t)y = 0, 2

17. (t + 1) y − 4(t + 1)y + 6y = 0, 18. y + 4ty + (2 + 4t2 )y = 0,

y1 (t) = sin t

y1 (t) = (t + 1)2

y1 (t) = e−t

2

19. (t − 2)2 y + (t − 2)y − 4y = 0, y1 (t) = (t − 2)2

n−1 n−1 y + 1 + y = 0, where n is a positive integer, y1 (t) = e t 20. y − 2 + t t

3.5

Complex Roots We now complete the discussion of ﬁnding the general solution for the differential equation ay + by + cy = 0, by obtaining the general solution when the discriminant is negative. Looking for solutions of the form y(t) = eλt when the discriminant is negative leads to a characteristic equation aλ2 + bλ + c = 0 having complex conjugate roots, 4ac − b2 b ±i . λ1,2 = − 2a 2a Using α = −b/2a and β = 4ac − b2 /2a for simplicity, we have for the roots λ1,2 = α ± iβ.

(1)

The Complex Exponential Function The approach to solving ay + by + cy = 0 has been based on looking for solutions of the form y(t) = eλt . When the roots (1) of the characteristic equation are complex, we are led to consider exponential functions with complex arguments: y1 (t) = e(α+iβ)t

and

y2 (t) = e(α−iβ)t .

(2)

We need to clarify the mathematical meaning of these two expressions. How is the deﬁnition of the exponential function extended or broadened to accommodate complex as well as real arguments? Once such a generalization is understood mathematically, we then need to demonstrate for complex λ that the function eλt is, in fact, a differentiable function of t satisfying the fundamental differentiation formula d λt e = λeλt . dt

3.5

Complex Roots

133

This differentiation formula was tacitly assumed in Section 3.3, and it is needed now if we want to show that y = e(α+iβ)t and y = e(α−iβ)t are, in fact, solutions of ay + by + cy = 0. A second issue that needs to be addressed is that of physical relevance. We will be able to show that the functions {e(α+iβ)t , e(α−iβ)t } form a fundamental set of solutions. How do we use them to obtain real-valued solutions? As a case in point, recall the buoyancy example discussed in Section 3.1. When we modeled the object’s position, y(t), we arrived at a differential equation of the form y + ω2 y = 0. The characteristic equation for this differential equation is λ2 + ω2 = 0 and has roots λ1 = iω and λ2 = −iω. How do we relate the two solutions y1 (t) = eiωt

and

y2 (t) = e−iωt

to the real-valued general solution y = A sin ωt + B cos ωt used to describe the bobbing motion of the object?

The Deﬁnition of the Complex Exponential Function In calculus, the exponential function y = e t is often introduced as the inverse of the natural logarithm function. For our present purposes, however, we want a representation of the exponential function that permits us to generalize from a real argument to a complex argument in a straightforward and natural way. In this regard, the power series representation of the function e z is very convenient. From calculus, the Maclaurin4 series expansion for e z is the inﬁnite series ∞

zn z3 z2 + + ··· = , e =1+z+ 2! 3! n! n=0 z

(3)

where z0 and 0! are understood to be equal to 1. For a given value of z, the Maclaurin series (3) is interpreted as the limit of the sequence of partial sums, e z = lim

M→∞

M zn n=0

n!

.

(4)

When limit (4) exists, we say that the series converges. It is shown in calculus that the Maclaurin series (3) converges to e z for every real value of z. Although we do not do so here, it is possible to derive a number of familiar properties of the exponential function from the power series (3). For example, it can be shown that e z is differentiable and that e z1 +z2 = e z1 e z2 . 4

Colin Maclaurin (1698–1746) was a professor of mathematics at the University of Aberdeen and later at the University of Edinburgh. In a two-volume exposition of Newton’s calculus, titled the Treatise of Fluxions (published in 1742), Maclaurin used the special form of Taylor series now bearing his name. Maclaurin also is credited with introducing the integral test for the convergence of an inﬁnite series.

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Second and Higher Order Linear Differential Equations

The importance of the power series representation (3) for e z is that the representation is not limited to real values of z. It can be shown that power series (3) converges for all complex values of z. In this manner, the function eλt is given meaning even for a complex value of λ. The power series representation can also be used to show that eλt = e(α+iβ)t = eαt+iβt = eαt eiβt .

(5)

Euler’s Formula From equation (5), eλt = eαt+iβt = eαt eiβt . This result simpliﬁes our task of understanding the function y = eλt , since we already know the behavior of the factor eαt when α is real. Thus, we focus on the other factor, eiβt . Using z = iβt in power series (3), we obtain (iβt)3 (iβt)4 (iβt)5 (iβt)2 + + + + ··· 2! 3! 4! 5! (βt)4 (βt)3 (βt)5 (βt)2 + − · · · + i βt − + − ··· = 1− 2! 4! 3! 5!

eiβt = 1 + (iβt) +

=

∞ (−1)n (βt)2n n=0

(2n)!

+i

∞ (−1)n (βt)2n+1 n=0

(2n + 1)!

(6)

.

In (6), we used the fact that i2 = −1 and regrouped the terms into real and imaginary parts. The two series on the right-hand side of (6) are Maclaurin series representations of familiar functions: ∞

cos βt = 1 −

(−1)n (βt)2n (βt)2 (βt)4 + − ··· = , 2! 4! (2n)! n=0 ∞

sin βt = βt −

(−1)n (βt)2n+1 (βt)5 (βt)3 + − ··· = . 3! 5! (2n + 1)! n=0

Using these results in equation (6), we obtain Euler’s formula, eiβt = cos βt + i sin βt.

(7)

The symmetry properties of the sine and cosine functions [cos(−θ) = cos θ and sin(−θ ) = − sin θ ], together with Euler’s formula (7), lead to an analogous expression for e−iβt , e−iβt = ei(−βt) = cos(−βt) + i sin(−βt) = cos βt − i sin βt.

(8)

Equations (5), (7), and (8) can be used to express e(α+iβ)t and e(α−iβ)t in terms of familiar functions: e(α+iβ)t = eαt eiβt = eαt (cos βt + i sin βt), e(α−iβ)t = eαt e−iβt = eαt (cos βt − i sin βt).

(9)

We can use equation (9) to show that the Wronskian of y1 (t) = e(α+iβ)t and y2 (t) = e(α−iβ)t is W(t) = −i2βe2αt , which is nonzero for all t since β = 0. The two solutions, y1 and y2 , therefore form a fundamental set of solutions.

3.5

Complex Roots

135

From a mathematical point of view, we are done with the complex roots case since we have found a fundamental set. From a physical point of view, however, if the mathematical problem arises from a (real-valued) physical process and we seek real-valued, physically meaningful answers, the two solutions in (9) are not satisfactory. We want a fundamental set consisting of real-valued solutions.

The Real and Imaginary Parts of a Complex-Valued Solution Are Also Solutions We now state and prove a theorem that shows how to obtain a fundamental set of real-valued solutions when the characteristic equation has complex roots. Theorem 3.3

Let y(t) = u(t) + iv(t) be a solution of the differential equation y + p(t)y + q(t)y = 0, where p(t) and q(t) are real-valued and continuous on a < t < b and where u(t) and v(t) are real-valued functions deﬁned on (a, b). Then u(t) and v(t) are also solutions of the differential equation on this interval.

●

PROOF:

Since y(t) is known to be a solution, we have (u + iv) + p(t)(u + iv) + q(t)(u + iv) = 0.

Therefore, u + iv + p(t)(u + iv ) + q(t)(u + iv) = 0. Collecting real and imaginary parts, we have [u + p(t)u + q(t)u] + i[v + p(t)v + q(t)v] = 0,

a < t < b.

Since a complex quantity vanishes if and only if its real and imaginary parts both vanish, we know that u + p(t)u + q(t)u = 0,

a 0, y(t) is zero at t = 3π/8, 7π/8, 11π/8, . . . . ❖

E X A M P L E

4

Solve the initial value problem y + y = 0,

y(0) = −1,

√ y (0) = − 3

and put the solution in the form Reαt cos(βt − δ). Solution: The characteristic equation is λ2 + 1 = 0 and has roots λ1,2 = ±i. The general solution is y(t) = A cos t + B sin t. Imposing the initial conditions, we obtain the solution √ y(t) = − cos t − 3 sin t.

3.5

√ Using R cos δ = −1 and R sin δ = − 3, we have R=2

and

tan δ =

Complex Roots

139

√ 3.

Since cos δ and sin δ are both negative, the phase angle δ must lie in the third quadrant, √ 4π . δ = tan−1 3 + π = 3 Hence [see (12)],

4π y(t) = 2 cos t − . 3

The graph of y(t) is shown in Figure 3.6. It is the graph of 2 cos t shifted to the right by 4π/3 ≈ 4.19 radians. y 2 1.5 1 0.5 5

10

15

20

25

t

–0.5 –1

4 3

–1.5 –2

FIGURE 3.6

The graph of the solution found in Example 4, y(t) = 2 cos[t − (4π/3)].

❖

EXERCISES The identity e z1 +z2 = e z1 e z2 , from which we obtain (e z )n = enz , is useful in some of the exercises. 1. Write each of the complex numbers in the form α + iβ, where α and β are real numbers. √ (a) 2eiπ/3 (b) −2 2 e−iπ/4 (c) (2 − i)ei3π/2 √ 4 1 (d) √ ei7π/6 (e) 2 eiπ/6 2 2 2. Write each of the functions in the form Aeαt cos βt + iBeαt sin βt, where α, β, A, and B are real numbers. √ 2 1 (a) 2ei 2 t (b) e−(2+3i)t (c) − e2t+i(t+π) π 2

√ (1+i)t 3 1 iπt 3 (d) 3e (e) − √ e 2

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Second and Higher Order Linear Differential Equations

Exercises 3–12: For the given differential equation, (a) Determine the roots of the characteristic equation. (b) Obtain the general solution as a linear combination of real-valued solutions. (c) Impose the initial conditions and solve the initial value problem. 3. y + 4y = 0,

4. y + 2y + 2y = 0, 5. 9y + y = 0,

y (π/4) = 1

y(π/4) = −2,

y (0) = −1

y(0) = 3,

y(π/2) = 4, y (π/2) = 0

6. 2y − 2y + y = 0, 7. y + y + y = 0,

y (−π) = −1

y(−π) = 1,

y (0) = −2

y(0) = −2,

8. y + 4y + 5y = 0,

y(π/2) = 1/2,

9. 9y + 6y + 2y = 0,

y(3π) = 0,

y (π/2) = −2 y (3π) = 1/3

10. y + 4π 2 y = 0, y(1) = 2, y (1) = 1 √ √ 11. y − 2 2 y + 3y = 0, y(0) = −1/2, y (0) = 2 12. 9y + π 2 y = 0,

y (3) = −π

y(3) = 2,

Exercises 13–21: The function y(t) is a solution of the initial value problem y + ay + by = 0, y(t0 ) = y0 , y (t0 ) = y0 , where the point t0 is speciﬁed. Determine the constants a, b, y0 , and y0 . √ 13. y(t) = sin t − 2 cos t, t0 = π/4 14. y(t) = 2 sin 2t + cos 2t, −2t

15. y(t) = e

−2t

cos t − e

t−π/6

t0 = π/4 t0 = 0

sin t, t−π/6

cos 2t − e sin 2t, t0 = π/6 16. y(t) = e √ 17. y(t) = 3 cos πt − sin π t, t0 = 1/2 √ 18. y(t) = 2 cos(2t − π/4), t0 = 0 19. y(t) = 2e t cos(π t − π),

t0 = 1

20. y(t) = e−t cos(π t − π),

t0 = 0

21. y(t) = 3e−2t cos(t − π/2),

t0 = 0

Exercises 22–26: Rewrite the function y(t) in the form y(t) = Reαt cos(βt − δ), where 0 ≤ δ < 2π . Use this representation to sketch a graph of the given function, on a domain sufﬁciently large to display its main features. 22. y(t) = sin t + cos t √ 24. y(t) = e t cos t + 3 e t sin t

23. y(t) = cos π t − sin πt √ 25. y(t) = −e−t cos t + 3 e−t sin t

26. y(t) = e−2t cos 2t − e−2t sin 2t

Exercises 27–29: In each exercise, the ﬁgure shows the graph of the solution of an initial value problem y + ay + by = 0, y(0) = y0 , y (0) = y0 . Use the information given to express the solution in the form y(t) = R cos(βt − δ), where 0 ≤ δ < 2π. Determine the constants a, b, y0 , and y0 .

3.5

27. The graph has a maximum value at (0, 2) and a t-intercept at (1, 0).

Complex Roots

141

28. The graph has a maximum value at (π/12, 1) and a t-intercept at (5π/12, 0). y

y

0.8 1 0.4

–4

–2

2

4

t

–4

–2

2

4

t

–0.4 –1 –0.8

29. The graph has a maximum value at (5π/12, 1/2) and a t-intercept at (π/6, 0). y

0.4

0.2

–4

–2

2

4

t

–0.2

–0.4

30. Consider the differential equation y + ay + by = 0, where a and b are positive real constants. Show that lim t→∞ y(t) = 0 for every solution of this equation. 31. Consider the differential equation y + ay + 9y = 0, where a is a real constant. Suppose we know that the Wronskian of a fundamental set of solutions for this equation is constant. What is the general solution for this equation? 32. Buoyancy Problems with Drag Force We discussed modeling the bobbing motion of ﬂoating cylindrical objects in Section 3.1. Bobbing motion will not persist indefinitely, however. One reason is the drag resistance a ﬂoating object experiences as it moves up and down in the liquid. If we assume a drag force proportional to velocity, an application of Newton’s second law of motion leads to the differential equation y + μy + ω2 y = 0, where y(t) is the downward displacement of the object from its static equilibrium position, μ is a positive constant describing the drag force, and ω2 is a positive constant determined by the mass densities of liquid and object and the vertical extent of the cylindrical object. (See Figure 3.1.) (a) Obtain the general solution of this differential equation, assuming that μ2 < 4ω2 . (b) Assume that a cylindrical ﬂoating object is initially displaced downward a distance y0 and released from rest [so the initial conditions are y(0) = y0 , y (0) = 0].

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CHAPTER 3

Second and Higher Order Linear Differential Equations Obtain, in terms of μ, ω, and y0 , the solution y(t) of the initial value problem. Show that lim t→∞ y(t) = 0. 33. This section has focused on the differential equation ay + by + cy = 0, where a, b, and c are real constants. The fact that the roots of the characteristic equation occur in conjugate pairs when they are complex is due to the fact that the coefﬁcients a, b, and c are real numbers. To see that this may not be true if the coefﬁcients are allowed to be complex, determine the roots of the characteristic equation for the differential equation y + 4iy + 5y = 0. Find two complex-valued solutions of this equation.

3.6

Unforced Mechanical Vibrations In this section, we model the motion of a simple mechanical system—that of a mass suspended from the end of a hanging spring and subjected to some initial disturbance. The resulting up-and-down motion of the mass will be similar to the bobbing motion of a ﬂoating cylindrical object.

Hooke’s Law A spring hangs vertically from a ceiling. We assume that the weight of the spring is negligibly small. The natural or unstretched length of the spring is denoted by l, as in Figure 3.7. Suppose we now apply a vertical force to the end of the spring. If the force is directed downward, the spring will stretch. As it stretches, the spring develops an upward restoring force that resists this stretching or elongation. Conversely, if the applied force is directed upward, the spring compresses or shortens in length. In this case, the spring develops a counteracting downward restoring force that tends to resist compression. (a)

(b)

(c)

l Δy1 < 0 FR = 0

FR = –kΔ y1 = k⏐Δ y1⏐ > 0

Δy 2 > 0 FR = –kΔy2 < 0

Distance y ( positive downward) FIGURE 3.7

(a) A spring with natural length l. (b) The restoring force, FR = −k y1 , is positive (directed downward) when the spring is compressed. (c) The restoring force, FR = −k y2 , is negative (directed upward) when the spring is stretched.

3.6

Unforced Mechanical Vibrations

143

We need an equation that relates the restoring force developed by the spring to the amount of elongation or compression that has occurred. The relation we use is Hooke’s law,5 which assumes the restoring force is proportional to the amount of stretching or compression that the spring has undergone. We assume that the displacement y is positive when the spring is stretched and negative when it is compressed. Whether the spring is stretched or compressed, its length is given by the quantity l + y. Hooke’s law states that the restoring force is FR = −k y,

(1)

where y is the displacement and k is a positive constant of proportionality, called the spring constant. The negative sign in equation (1) arises because the restoring force acts to counteract the displacement of the spring end. The spring constant k in equation (1) has the dimensions of force per unit length and represents a measure of spring stiffness. A stiffer spring has a larger value of k, and the same restoring force arises from a smaller displacement. Hooke’s law is a useful description of reality when the displacement magnitude, | y|, is reasonably small. It cannot remain valid for arbitrarily large | y| since one cannot stretch or compress a spring indeﬁnitely. We assume in all our modeling and computations that displacement magnitudes are small enough to permit the use of Hooke’s law.

A Mathematical Model for the Spring-Mass System An object having mass m is attached to the end of the unstretched spring, as in Figure 3.8. The weight of the object is W = mg,

Y

m Equilibrium or rest position

y(t)

m Perturbed state

Distance (positive downward) FIGURE 3.8

As the mass moves, the quantity y(t) measures its displacement from the equilibrium position.

5

Robert Hooke (1635–1703) served as professor of geometry at Gresham College, London, for 30 years. He worked on problems in elasticity, optics, and simple harmonic motion. Hooke invented the conical pendulum and was the ﬁrst to build a Gregorian reﬂecting telescope. He was a competent architect and helped Christopher Wren rebuild London after the Great Fire of 1666.

144

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Second and Higher Order Linear Differential Equations

where g is the acceleration due to gravity. The spring stretches until it achieves a new rest or equilibrium conﬁguration. Let Y represent the distance the spring stretches to achieve this new equilibrium position. The displacement, Y , is determined by Hooke’s law—the spring stretches until the restoring force exactly counteracts the object’s weight: W + FR = mg − kY = 0.

(2)

It follows that Y = mg/k. When the spring-mass system in Figure 3.8 is perturbed from its equilibrium position, we have from Newton’s second law of motion m

d2 dt2

(Y + y) = W + FR = mg − k(Y + y).

Using (2) and noting that Y is a constant, we can reduce this equation to my + ky = 0.

(3)

2

2

If we deﬁne ω = k/m, we obtain the differential equation y + ω y = 0, which characterizes the bobbing motion of a ﬂoating object. Suppose we now assume a damping mechanism, a dashpot, is attached and suppresses the vibrating motion of the spring-mass system. It is shown schematically in Figure 3.9. We assume the damping force is proportional to velocity, FD = −γ

dy(t) . dt

(4)

In (4), γ is a positive constant of proportionality, referred to as the damping coefﬁcient. The negative sign is present because the damping force acts to oppose the motion. A similar model of velocity damping was assumed in the linear model of projectile motion with air resistance discussed in Section 2.9.

m

FIGURE 3.9

A spring-mass-dashpot system.

The differential equation describing the motion of the mass, again obtained from Newton’s second law of motion, is m

d2 y dt

2

= W + FR + FD = mg − k(Y + y) − γ

dy . dt

(5)

3.6

Unforced Mechanical Vibrations

145

Since mg − kY = 0, equation (5) simpliﬁes to m

d2 y dt

2

+γ

dy + ky = 0. dt

(6)

Equation (6) is a second order linear constant coefﬁcient differential equation. It is homogeneous because we are considering only the unforced or free vibration of the system. That is, there are no external forces applied to the system. In Section 3.10, we will consider forced vibrations, where equation (6) is modiﬁed to include a time-varying applied force, F(t). Inclusion of an applied force leads to a nonhomogeneous equation of the form m

d2 y dt

2

+γ

dy + ky = F(t). dt

(7)

Behavior of the Model In this subsection, we discuss the solutions of equation (6), my + γ y + ky = 0. These solutions describe how the mass-spring-dashpot system behaves—predicting the position, y(t), and the velocity, y (t), of the moving mass at any time t. The characteristic equation, mλ2 + γ λ + k = 0, has roots λ1,2 =

−γ ±

γ 2 − 4mk

2m

.

(8)

The corresponding mass-spring-dashpot system exhibits different behavior, depending on the roots of the characteristic equation. The roots, in turn, are determined by the relative values of the mass, spring constant, and damping coefﬁcient. Case 1 If γ 2 > 4km (if damping is relatively strong), the characteristic equation has two negative real roots

−γ ± γ 2 − 4mk . λ1,2 = 2m

(Both roots are negative since γ 2 − 4mk is less than γ .) As shown in Section 3.3, the general solution is given by y(t) = c1 eλ1 t + c2 eλ2 t .

(9a)

Therefore, the strong damping suppresses any vibratory motion of the mass. The general solution is a linear combination of two decreasing exponential functions. This case is referred to as the overdamped case. Case 2 If γ 2 = 4km, then the roots are real and repeated. As we saw in Section 3.4, the general solution is given by y(t) = c1 eλ1 t + c2 teλ1 t ,

(9b)

146

CHAPTER 3

Second and Higher Order Linear Differential Equations

where λ1 = −γ /2m. In this case, known as the critically damped case, damping is also sufﬁciently strong to suppress oscillatory vibrations of the mass. Case 3 If γ 2 < 4km, then the roots are complex conjugates, 4mk − γ 2 −γ ±i = α ± iβ. λ1,2 = 2m 2m As we saw in Section 3.5, the general solution is given by y(t) = eαt (c1 cos βt + c2 sin βt).

(9c)

In this case, known as the underdamped case, damping is too weak to totally suppress the vibrations of the mass. Note that the underdamped case also includes the case where there is no damping whatsoever—that is, the case where γ = 0. Later, we refer to this as the undamped case. Finally, recall from Section 3.5 that solution (9c) can be restated in amplitude-phase form as y(t) = Reαt cos(βt − δ),

(10)

where R = c21 + c22 , R cos δ = c1 , and R sin δ = c2 . If damping is present (that is, if γ > 0), then α = −γ /2m < 0 and the motion of the mass described by equation (10) consists of damped vibrations (oscillations that decrease in amplitude as time progresses). If there is no damping, then α = 0 and the oscillations do not decrease in magnitude. Representative examples of the motion that occurs in these three cases are shown in Figure 3.10, parts (a)– (c). When damping is present, it follows from equations (9a)– (9c) that lim t→∞ y(t) = 0 for any choices of c1 and c2 . This is to be expected, since energy is dissipated and any initial disturbance will diminish in strength as time increases.

Vibrations and Periodic Functions As a special case, assume that damping is absent; that is, γ = 0. In this case, α = 0 and solution (10) reduces to k t−δ . (11) y(t) = R cos m If we set ω = k/m, equation (11) becomes y(t) = R cos(ωt − δ).

(12)

In this case, the amplitude of the vibrations remains constant and equal to R. The function y(t) in (12) is an example of a periodic function. In general, let f (t) be deﬁned on −∞ < t < ∞ or a ≤ t < ∞ for some a. The function f (t) is called a periodic function if there exists a positive constant T, called the period, such that f (t + T) = f (t)

(13)

for all values of t in the domain. The smallest value of the constant T satisfying (13) is called the fundamental period of the function.

3.6

Unforced Mechanical Vibrations

147

y 2 1.5 1 0.5 –1

1

2

3

4

5

t

–0.5

Equation: y ″ + 4y′ + 3y = 0 y(0) = 1, y′(0) = – 9 Solution: y(t) = 4e–3t – 3e –t

–1 –1.5 –2

(a) Overdamped motion y 2 1.5 1 0.5 –1

1

2

3

4

5

t

–0.5 –1

Equation: y ″ + 2y′ + 17y = 0 y(0) = 1, y′(0) = –5 Solution: y(t) = e–t (cos 4t – sin 4t) = √2 e–t cos(4t + /4)

–1.5 –2

(b) Underdamped motion y 2 1.5 1 0.5 –1

1

2

3

4

5

t

–0.5 –1

Equation: y ″ + 16y = 0 y(0) = 1, y′(0) = –4 Solution: y(t) = cos 4t – sin 4t = √2 cos(4t + /4)

–1.5 –2

(c) Undamped motion FIGURE 3.10

Examples of (a) overdamped motion, (b) underdamped motion with nonzero damping, and (c) undamped motion.

148

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Second and Higher Order Linear Differential Equations

The basic qualitative feature of a periodic function is that its graph repeats itself. If we know what the graph looks like on any time segment of duration T, we can obtain the graph on the entire domain simply by replicating this segment. Figure 3.11 provides an illustration. f(t)

t T

FIGURE 3.11

The graph of a periodic function repeats itself over any time period of duration T, where T is the fundamental period of the function.

Consider again the solution y(t) = R cos(ωt − δ) in equation (12). Since the cosine function repeats itself whenever its argument changes by 2π , y(t) is a periodic function. To ﬁnd the period for y(t), we set y(t + T) = y(t): y(t + T) = R cos[ω(t + T) − δ] = R cos[ωt + ωT − δ] = y(t), if ωT = 2nπ,

n = 1, 2, . . . .

Therefore, the function y(t) has fundamental period T = 2π/ω. In terms of the spring-mass system, T = 2π/ω is referred to as the fundamental period of the motion or simply the period. The period represents the time required for the mass to execute one cycle of its oscillatory motion. The motion itself is often referred to as periodic motion. The reciprocal of the period, f = 1/T, is called the frequency of the oscillations. The frequency represents the number of cycles of the periodic motion executed per unit time. For example, if T = 0.01 sec, the system completes 100 cycles of its motion per second. In current terminology, one cycle per second is referred to as one Hertz.6 Therefore, we would say that the system oscillations have a frequency of 100 Hertz (100 Hz). From the relations T = 2π/ω and f = 1/T, it follows that ω = 2π f . The constant ω is called the angular frequency or the radian frequency. It represents the change, in radians, that cos(ωt − δ) undergoes in one period. It’s worthwhile to check that the model predictions are consistent with our everyday experience. In the absence of damping, angular frequency is ω = k/m and frequency is f = (1/2π ) k/m. Frequency therefore increases as either k increases or m decreases. Thus, when a given mass is attached in turn to two springs of differing stiffness, the model predicts it will vibrate more rapidly when suspended from the stiffer spring. Likewise, if two bodies of dif6

Heinrich Hertz (1857–1894) was a German physicist who conﬁrmed Maxwell’s theory of electromagnetism by producing and studying radio waves. He demonstrated that these waves travel at the velocity of light and can be reﬂected, refracted, and polarized. The unit of frequency was named in his honor.

3.6

Unforced Mechanical Vibrations

149

fering mass (and therefore weight) are suspended from the same spring, the smaller mass will vibrate more rapidly than the larger mass. Now consider what happens when damping is added. The solution y(t) in equation (10) has the form y(t) = Reαt cos(βt − δ), where −γ α= 2m

and

(14)

4km − γ 2 β= . 2m

Equation (14) predicts that when two different masses are attached to a springdashpot system having the same damping coefﬁcient γ and spring constant k, the solution envelope of the larger mass (the heavier body) will decrease more slowly with time because the associated value α is smaller. Note also that the introduction of damping changes the cosine term in equation (12) from cos[( k/m)t − δ], when damping is absent, to 4km − γ 2 t−δ , cos 2m when damping is present. Since k 4km − γ 2 > , m 2m the introduction of damping causes the vibrations to “slow down” while simultaneously being reduced in amplitude. Are these model predictions consistent with your everyday experience? What experiments might test these predictions, both qualitatively and quantitatively? We conclude this section with two examples illustrating the motion of a spring-mass-dashpot system. E X A M P L E

1

A block weighing 8 lb is attached to the end of a spring, causing the spring to stretch 6 in. beyond its natural length. The block is then pulled down 3 in. and released. Determine the motion of the block, assuming there are no damping forces or external applied forces. Solution:

The motion of the block is governed by equation (3), my + ky = 0,

along with the initial conditions of the problem. Assuming the gravitational constant to be g = 32 ft /sec2 and noting that the weight is given by W = mg, we ﬁnd the block has mass m=

1 lb-sec2 . 4 ft

[Note that 1 lb-sec2 /ft = 1 slug.] The spring constant k can be determined by the fact that an 8-lb force (the weight of the block) causes the spring to stretch (continued)

150

CHAPTER 3

Second and Higher Order Linear Differential Equations (continued)

6 in.; Hooke’s law implies k = 86 lb/in. = 16 lb/ft. Since 3 in. = value problem governing the motion is 1 y 4

+ 16y = 0,

1 4

ft, the initial

y (0) = 0.

y(0) = 14 ,

(15)

The units of y(t) are feet and of y (t) are feet per second. The solution of the initial value problem is y(t) = 14 cos 8t. A graph of the block’s position, y(t), is shown in Figure 3.12(a). ❖ y

y

0.4

0.4

0.2

0.2

–0.5

0.5

1

1.5

t

2

–0.5

0.5

–0.2

–0.2

–0.4

–0.4

(a)

1

1.5

2

t

(b) FIGURE 3.12

(a) The position, y(t), of the mass in Example 1 (no damping). (b) The position, y(t), of the mass in Example 2 (includes damping). E X A M P L E

2

Consider the spring-mass system in Example 1. Assume that damping is present and that the damping coefﬁcient is given by γ = 1 lb-sec/ft. Determine the motion of the block. Solution: To account for the assumed damping force, the equation of Example 1 will be modiﬁed to include a damping term γ y , where γ = 1: 1 y 4

+ y + 16y = 0.

Therefore, the initial value problem governing the motion of the block is y + 4y + 64y = 0,

y(0) = 14 ,

y (0) = 0.

The general solution is

√ √ y(t) = e−2t c1 cos(2 15 t) + c2 sin(2 15 t) .

Imposing the initial conditions y(0) = 14 and y (0) = 0, we obtain √ √ 1 −2t 1 y(t) = e cos(2 15 t) + √ sin(2 15 t) . 4 15 The graph of y(t) is shown in Figure 3.12(b). The block still oscillates about its equilibrium position, but the envelope of the oscillations decreases with time. ❖

3.6

Unforced Mechanical Vibrations

151

EXERCISES 1. The given function f (t) is periodic with fundamental period T; therefore, f (t + T) = f (t). Use the information given to sketch the graph of f (t) over the time interval 0 ≤ t ≤ 4T. 1, 0 ≤ t ≤ 32 , T=2 (a) f (t) = t(2 − t), 0 ≤ t < 2, T = 2 (b) f (t) = 0, 32 < t < 2 t, 0≤t≤2 (c) f (t) = , T=4 (d) f (t) = −1 + t, 0 ≤ t < 2, T = 2 4 − t, 2 < t < 4 (e) f (t) = 2e−t , 0 ≤ t < 1, T = 1 2 sin πt, 0 ≤ t ≤ 1 (g) f (t) = , T=2 0, 1 0, think of the solution y(t) as being a function of the damping constant γ , and determine lim γ →∞ y(t). Do you have any intuitive insight as to what the answer should be? Develop the answer with the following steps. (a) Show that the roots of the characteristic polynomial are γ − λ1 = − 2m

γ 2 − 4mk , 2m

γ λ2 = − + 2m

γ 2 − 4mk . 2m

Solve the initial value problem, assuming the system to be overdamped. Express the solution in terms of the two roots λ1 and λ2 . (Since we are interested in the system’s behavior for large values of γ , the overdamped assumption is appropriate.) (b) Show that lim γ →∞ λ1 = −∞ and lim γ →∞ λ2 = 0. (c) Use the results of parts (a) and (b) to determine lim γ →∞ y(t) (with k, m, and t > 0 ﬁxed). What is the physical meaning of your answer? Does it agree with your intuition? Does it make physical sense in retrospect? 13. In this problem, we explore computationally the question posed in Exercise 12. Consider the initial value problem y + γ y + y = 0,

y(0) = 1,

y (0) = 0,

where, for simplicity, we have given the mass, spring constant, and initial displacement all a numerical value of unity. (a) Determine γcrit , the damping constant value that makes the given spring-massdashpot system critically damped. (b) Use computational software to plot the solution of the initial value problem for γ = γcrit , 2γcrit , and 20γcrit over a common time interval sufﬁciently large to display the main features of each solution. What trend do you observe in the behavior of the solutions as γ increases? Is it consistent with the conclusions reached in Exercise 12?

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Second and Higher Order Linear Differential Equations

The General Solution of a Linear Nonhomogeneous Equation We consider the linear second order nonhomogeneous differential equation y + p(t)y + q(t)y = g(t),

a0 t > −1

yP (t) = t + 1

Exercises 22–26: The general solution of the nonhomogeneous differential equation y + αy + βy = g(t) is given, where c1 and c2 are arbitrary constants. Determine the constants α and β and the function g(t). 22. y(t) = c1 e t + c2 e2t + 2e−2t

23. y(t) = c1 + c2 e−t + t2

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25. y(t) = c1 e t cos t + c2 e t sin t + e t + sin t

26. y(t) = c1 sin 2t + c2 cos 2t − 1 + sin t

3.8

The Method of Undetermined Coeﬃcients In Section 3.7, we discussed the structure of the general solution for the nonhomogeneous equation y + p(t)y + q(t)y = g(t),

a < t < b.

We saw that the general solution has the form y(t) = yC (t) + yP (t),

(1)

where yC is the general solution of the homogeneous equation y + p(t)y + q(t)y = 0 and yP is a particular solution of the nonhomogeneous equation y + p(t)y + q(t)y = g(t). In this section, we describe a technique that often can be used to ﬁnd a particular solution, yP . The technique is known as the method of undetermined coefﬁcients. We ﬁrst illustrate the method through a series of examples. Later, we summarize the method in tabular form. In Section 3.9, we describe a different technique for ﬁnding a particular solution, the method of variation of parameters. Before we discuss these two methods for obtaining a particular solution, it’s worth stating the procedure that should be followed to obtain the general solution (1): 1. The ﬁrst step is to ﬁnd the complementary solution, yC . As you will see, knowledge of the complementary solution is a prerequisite for using either the method of undetermined coefﬁcients or the method of variation of parameters. 2. Next, use undetermined coefﬁcients or variation of parameters (or anything else that works) to ﬁnd a particular solution, yP . 3. Finally, obtain the general solution by forming yC + yP . If you are solving an initial value problem, the initial conditions are imposed as a last step, step 4.

Introduction to the Method of Undetermined Coefﬁcients Consider the nonhomogeneous differential equation ay + by + cy = g(t),

(2)

where a, b, and c are constants. You will see that we can guess the form of a particular solution for certain types of functions g(t). Example 1 introduces some of the main ideas.

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159

E X A M P L E

1

Find the general solution of the nonhomogeneous equation y − y − 2y = 8e3t . Solution: We always begin by ﬁnding the complementary solution, yC . The characteristic polynomial for y − y − 2y = 0 is λ2 − λ − 2 = (λ + 1)(λ − 2). Therefore, the complementary solution is yC (t) = c1 e−t + c2 e2t . We now look for a particular solution, a function yP (t) such that yP − yP − 2yP = 8e3t .

(3)

Since all derivatives of y = e3t are again multiples of e3t , it seems reasonable that a particular solution might have the form yP = Ae3t , where A is a coefﬁcient to be determined. Inserting yP = Ae3t into equation (3), we obtain 9Ae3t − 3Ae3t − 2(Ae3t ) = 8e3t . Collecting terms on the left-hand side yields 4Ae3t = 8e3t . Therefore, A = 2, and yP (t) = 2e3t is a particular solution of the nonhomogeneous equation. Having the complementary solution yC and a particular solution yP , we form the general solution of the nonhomogeneous equation y(t) = yC (t) + yP (t) = c1 e−t + c2 e2t + 2e3t . ❖

Example 1 suggests a reasonable approach to ﬁnding a particular solution, yP . If the right-hand side of nonhomogeneous equation (2) is of a certain special type, then it might be possible to guess an appropriate form for yP . The method of undetermined coefﬁcients amounts to a recipe for choosing the form of yP . This recipe involves unknown (or undetermined) coefﬁcients that must be evaluated by inserting the form, yP , into the differential equation. (The role that the complementary solution plays in this process will be clariﬁed shortly.)

Trial Forms for the Particular Solution The method of undetermined coefﬁcients can be applied to differential equations of the form ay + by + cy = g(t), where a, b, and c are constants and where the nonhomogeneous term g(t) is one of several possible types. It’s important to understand what types of functions g(t) are suitable and why. To gain insight, we start with some examples. The ﬁrst few examples will treat, for various right sides g(t), the differential equation y − y − 2y = g(t).

(4)

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Equation (4) with g(t) = 8e3t was discussed in Example 1; the complementary solution of (4) is yC (t) = c1 e−t + c2 e2t . E X A M P L E

2

Find the general solution of y − y − 2y = 4t2 . Solution: We already know the complementary solution from Example 1. Therefore, we can form the general solution after ﬁnding a particular solution. Following the approach taken in Example 1, we are tempted to look for a solution of the form yP (t) = At2 , where A is an unknown (undetermined) coefﬁcient. The guess yP (t) = At2 , however, does not work because forming the ﬁrst and second derivatives of t2 generates multiples of some new functions, t and 1. In particular, this guess leads to a contradiction when the trial form yP (t) = At2 is inserted into the differential equation. Instead, we assume a particular solution of the form yP (t) = At2 + Bt + C, where the constants A, B, and C must be chosen so that yP − yP − 2yP = 4t2 . Substituting yP , we obtain the condition (2A) − (2At + B) − 2(At2 + Bt + C) = 4t2 or, after collecting terms, −2At2 − (2A + 2B)t + (2A − B − 2C) = 4t2 .

(5)

This equality must hold for all t in the interval of interest. Therefore, the coefﬁcients of t2 , t, and 1 on the left-hand side of this equation must equal their counterparts on the right-hand side. We obtain the following three equations for the three unknown coefﬁcients A, B, and C. −2A =4 −2A − 2B =0 2A − B − 2C = 0.

(6)

The solution of this system is A = −2, B = 2, C = −3. Therefore, a particular solution is given by yP (t) = −2t2 + 2t − 3. The desired general solution is y(t) = yC (t) + yP (t), or y(t) = c1 e−t + c2 e2t − 2t2 + 2t − 3. ❖

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The Method of Undetermined Coefﬁcients

161

REMARKS ABOUT EXAMPLE 2: 1. In retrospect, it is clear why our ﬁrst guess, yP (t) = At2 , failed. Substitution into the nonhomogeneous equation leads [see system (6)] to the contradictory constraints −2A = 4, −2A = 0, 2A = 0. The emergence of such contradictions is a clear indicator that the assumed form for the particular solution is incorrect. 2. The assumed form yP (t) = At2 + Bt + C of the particular solution is appropriate because of another fact that we did not mention—none of the functions 1, t, or t2 is a part of the complementary solution, yC . If one of these functions had been part of the complementary solution, then (see Examples 4 and 5) our guess would have failed. E X A M P L E

3

Find the general solution of y − y − 2y = −20 sin 2t. Solution:

The complementary solution is yC (t) = c1 e−t + c2 e2t .

In choosing a guess for the particular solution, we observe that differentiation of the right-hand side, g(t) = −20 sin 2t, produces a multiple of cos 2t but that continued differentiation of the set of functions {sin 2t, cos 2t} simply produces multiples of the functions in the set. In addition, neither of the functions sin 2t or cos 2t appears as part of the complementary solution. Therefore, we choose the following trial form for a particular solution: yP (t) = A sin 2t + B cos 2t. Substituting the trial form yP (t) = A sin 2t + B cos 2t, we obtain yP − yP − 2yP = −20 sin 2t, or (−4A sin 2t− 4B cos 2t)− (2A cos 2t − 2B sin 2t)−2(A sin 2t+B cos 2t) = −20 sin 2t. Collecting like terms reduces this equation to (−6A + 2B) sin 2t − (2A + 6B) cos 2t = −20 sin 2t. Since this equation must hold for all t in the interval of interest, it follows that −6A + 2B = −20 −2A − 6B = 0. The solution of this system is A = 3 and B = −1, leading to a particular solution yP (t) = 3 sin 2t − cos 2t and the general solution y(t) = c1 e−t + c2 e2t + 3 sin 2t − cos 2t. ❖

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Our next two examples illustrate how the trial form for yP must be modiﬁed if portions of g(t) or derivatives of g(t) are present in the complementary solution. E X A M P L E

4

Find the general solution of y − y − 2y = 4e−t . Solution: In this case, we observe that the function e−t is a solution of the homogeneous equation. To illustrate that the trial form yP (t) = Ae−t is not correct, we substitute it into the differential equation, obtaining Ae−t − (−Ae−t ) − 2Ae−t = 4e−t , or 0Ae−t = 4e−t . Since the condition 0A = 4 cannot hold for any value A, the assumed form of the trial solution is not correct. We obtain the correct form for a particular solution of y − y − 2y = 4e−t if we multiply e−t by t—that is, if we assume a trial solution of the form yP (t) = Ate−t .

(7)

At ﬁrst glance, it may seem surprising that this form is correct. Nevertheless, when we substitute yP (t) = Ate−t , we obtain (Ate−t − 2Ae−t ) − (−Ate−t + Ae−t ) − 2Ate−t = 4e−t ,

(8)

or −3Ae−t = 4e−t . This equation is satisﬁed by choosing A = − 43 , leading to a particular solution yP (t) = − 43 te−t and the general solution y(t) = yC (t) + yP (t) = c1 e−t + c2 e2t − 43 te−t . ❖ REMARK: In retrospect, it should be clear why the te−t terms vanish on the left-hand side of equation (8). After yP (t) = Ate−t is substituted into the differential equation, the terms that survive as te−t terms are precisely those in which differentiation under the product rule has acted on the e−t factor and not on the t factor. Such terms ultimately vanish because e−t is a solution of the homogeneous equation. E X A M P L E

5

Find the general solution of y + 2y + y = 2e−t . Solution: For this problem, the homogeneous equation has characteristic equation λ2 + 2λ + 1 = (λ + 1)2 = 0. We obtain two real repeated roots, λ1 = λ2 = −1, and the complementary solution is yC (t) = c1 e−t + c2 te−t .

3.8

The Method of Undetermined Coefﬁcients

163

It’s clear that a guess of the form yP (t) = Ae−t will not be the appropriate form for a particular solution, because e−t is a solution of the homogeneous equation. A guess of the form yP (t) = Ate−t will fail for the same reason. It’s perhaps not surprising that a guess of the form yP (t) = At2 e−t does work. Substituting this form of the trial solution leads to (At2 e−t − 4Ate−t + 2Ae−t ) + 2(−At2 e−t + 2Ate−t ) + At2 e−t = 2e−t . Simplifying, we obtain 2Ae−t = 2e−t , so 2A = 2 and hence yP (t) = t2 e−t . The general solution is y(t) = c1 e−t + c2 te−t + t2 e−t . ❖

A Table Summarizing the Method of Undetermined Coefﬁcients We summarize the method of undetermined coefﬁcients in Table 3.1. The method applies to the nonhomogeneous linear differential equation ay + by + cy = g(t), where a, b, and c are constants and g(t) has one of the forms listed on the lefthand side of the table. The corresponding form to assume for the particular solution is listed on the right-hand side of the table. The forms listed in Table 3.1 will work; that is, they will always yield a particular solution. In the Exercises, we ask you to solve problems using Table 3.1. The role of the factor t r in the TA B L E 3 . 1 The right-hand column gives the proper form to assume for a particular solution of ay + by + cy = g(t). In the right-hand column, choose r to be the smallest nonnegative integer such that no term in the assumed form is a solution of the homogeneous equation ay + by + cy = 0. The value of r will be 0, 1, or 2. Form of g(t)

Form to Assume for a Particular Solution yP (t)

an tn + · · · + a1 t + a0

tr [An tn + · · · + A1 t + A0 ]

[an tn + · · · + a1 t + a0 ]eαt

tr [An tn + · · · + A1 t + A0 ]eαt

⎫ [an tn + · · · + a1 t + a0 ] sin βt ⎪ ⎬ or n

[an t + · · · + a1 t + a0 ] cos βt eαt sin βt or eαt cos βt

⎫ eαt [an tn + · · · + a0 ] sin βt ⎪ ⎬ or ⎪ ⎭ eαt [an tn + · · · + a0 ] cos βt

⎪ ⎭

tr [(An tn + · · · + A1 t + A0 ) sin βt + (Bn tn + · · · + B1 t + B0 ) cos βt] tr [Aeαt sin βt + Beαt cos βt] tr [(An tn + · · · + A0 )eαt sin βt + (Bn tn + · · · + B0 )eαt cos βt]

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right-hand column of Table 3.1 is to ensure that no term in the assumed form for yP is present in the complementary solution. You need to choose the proper value for r; the procedure for doing so is described in the table.

E X A M P L E

6

Using Table 3.1, choose an appropriate form for a particular solution of (a) y + 4y = t2 e3t

(b) y + 4y = te2t cos t

(c) y + 4y = 2t2 + 5 sin 2t + e3t

(d) y + 4y = t2 cos 2t

Solution: We note ﬁrst that the complementary solution for each of parts (a)–(d) is yC (t) = c1 sin 2t + c2 cos 2t. (a) For g(t) = t2 e3t , Table 3.1 speciﬁes yP (t) = tr [A2 t2 + A1 t + A0 ]e3t . If r = 0, no term in the assumed form for yP is present in the complementary solution. So the appropriate form for a trial particular solution is yP (t) = [A2 t2 + A1 t + A0 ]e3t . (b) For g(t) = te2t cos t, the speciﬁed form is yP (t) = tr [(A1 t + A0 )e2t sin t + (B1 t + B0 )e2t cos t]. If r = 0, no term in the assumed form for yP is present in the complementary solution. So the appropriate form for a trial particular solution is yP (t) = (A1 t + A0 )e2t sin t + (B1 t + B0 )e2t cos t. (c) Note that the nonhomogeneous term g(t) = 2t2 + 5 sin 2t + e3t does not match any of the forms listed in Table 3.1. However, we can use the superposition principle described by Theorem 3.4. Suppose u(t) is a particular solution of y + 4y = 2t2 , v(t) is a particular solution of y + 4y = 5 sin 2t, and w(t) is a particular solution of y + 4y = e3t . By Theorem 3.4, yP (t) = u(t) + v(t) + w(t) is a particular solution of y + 4y = 2t2 + 5 sin 2t + e3t .

(9)

To determine the individual particular solutions u(t), v(t), and w(t), we turn to Table 3.1 to ﬁnd suitable trial forms. In particular, an appropriate trial form for y + 4y = 2t2 is u(t) = A2 t2 + A1 t + A0 . A suitable trial form for y + 4y = 5 sin 2t is the function v(t) = B0 t cos 2t + C0 t sin 2t and a suitable trial form for y + 4y = e3t is w(t) = D0 e3t . (In the ﬁrst and last cases, r = 0.

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165

In the second case, r = 1.) Therefore, yP (t) = A2 t2 + A1 t + A0 + B0 t cos 2t + C0 t sin 2t + D0 e3t . (d) For g(t) = t2 cos 2t, Table 3.1 prescribes the form yP (t) = tr [(A2 t2 + A1 t + A0 ) sin 2t + (B2 t2 + B1 t + B0 ) cos 2t]. If we set r = 0, the assumed form for yP (t) will contain two terms, A0 sin 2t and B0 cos 2t, that are solutions of the homogeneous equation. Therefore, r cannot be zero. With r = 1, we see that no term in the assumed form is a solution of the homogeneous equation. Therefore, the appropriate form is yP (t) = t[(A2 t2 + A1 t + A0 ) sin 2t + (B2 t2 + B1 t + B0 ) cos 2t] = (A2 t3 + A1 t2 + A0 t) sin 2t + (B2 t3 + B1 t2 + B0 t) cos 2t. ❖ Although we presented Table 3.1 in the context of discussing nonhomogeneous constant coefﬁcient second order linear differential equations, the method of undetermined coefﬁcients is not restricted to second order equations; the ideas extend naturally to nonhomogeneous constant coefﬁcient linear equations of order higher than two.

EXERCISES Exercises 1–15: For the given differential equation, (a) Determine the complementary solution. (b) Use the method of undetermined coefﬁcients to ﬁnd a particular solution. (c) Form the general solution. 1. y − 4y = 4t2

2. y − 4y = sin 2t

3. y + y = 8e t

4. y + y = e t sin t

5. y − 4y + 4y = e2t

6. y − 4y + 4y = 8 + sin 2t

7. y + 2y + 2y = t3

8. 2y − 5y + 2y = te t

10. y + y = 6t

2

9. y + 2y + 2y = cos t + e−t

11. 2y − 5y + 2y = −6e t/2

12. y + y = cos t

13. 9y − 6y + y = 9te t/3

14. y + 4y + 5y = 5t + e−t

15. y + 4y + 5y = 2e−2t + cos t

Exercises 16–22: For the given differential equation, (a) Determine the complementary solution. (b) List the form of particular solution prescribed by the method of undetermined coefﬁcients; you need not evaluate the constants in the assumed form. [Hint: In Exercises 20 and 22, rewrite the hyperbolic functions in terms of exponential functions. In Exercise 21, use trigonometric identities.] 16. y − 2y − 3y = 2e−t cos t + t2 + te3t

17. y + 9y = t2 cos 3t + 4 sin t

18. y − y = t2 (2 + e t )

19. y − 2y + 2y = e−t sin 2t + 2t + te−t sin t

20. y − y = cosh t + sinh 2t

21. y + 4y = sin t cos t + cos2 2t

22. y + 4y = 2 sinh t cosh t + cosh t 2

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Exercises 23–27: Consider the differential equation y + αy + βy = g(t). In each exercise, the complementary solution, yC (t), and nonhomogeneous term, g(t), are given. Determine α and β and then ﬁnd the general solution of the differential equation. 23. yC (t) = c1 e−t + c2 e2t , −t

24. yC (t) = c1 + c2 e , −2t

25. yC (t) = c1 e

+ c2 te

g(t) = 4t g(t) = t

−2t

,

g(t) = 5 sin t

26. yC (t) = c1 cos t + c2 sin t, g(t) = t + sin 2t 27. yC (t) = c1 e−t cos 2t + c2 e−t sin 2t,

g(t) = 8e−t

Exercises 28–30: Consider the differential equation y + αy + βy = g(t). In each exercise, the nonhomogeneous term, g(t), and the form of the particular solution prescribed by the method of undetermined coefﬁcients are given. Determine the constants α and β. 28. g(t) = t + e3t ,

yP (t) = A1 t2 + A0 t + B0 te3t

29. g(t) = 3e2t − e−2t + t,

yP (t) = A0 te2t + B0 te−2t + C1 t + C0

30. g(t) = −e t + sin 2t + e t sin 2t, yP (t) = A0 e t + B0 t cos 2t + C0 t sin 2t + D0 e t cos 2t + E0 e t sin 2t 31. Consider the initial value problem y + 4y = e−t , y(0) = y0 , y (0) = y0 . Suppose we know that y(t) → 0 as t → ∞. Determine the initial conditions y0 and y0 as well as the solution y(t). 32. Consider the initial value problem y − 4y = e−t , y(0) = 1, y (0) = y0 . Suppose we know that y(t) → 0 as t → ∞. Determine the initial condition y0 as well as the solution y(t). 33. Consider the initial value problem y − y + 2y = 3, y(0) = y0 , y (0) = y0 . Suppose we know that | y(t) | ≤ 2 for all t ≥ 0. Determine the initial conditions y0 and y0 as well as the solution y(t).

Exercises 34–36: Consider the initial value problem y + ω2 y = g(t), y(0) = 0, y (0) = 0, where ω is a real nonnegative constant. For the given function g(t), determine the values of ω, if any, for which the solution satisﬁes the constraint | y(t) | ≤ 2, 0 ≤ t < ∞. 34. g(t) = 1

35. g(t) = cos 2ωt

36. g(t) = sin ωt

37. Each of the ﬁve graphs on the next page is the solution of one of the ﬁve differential equations listed. Each solution satisﬁes the initial conditions y(0) = 1, y (0) = 0. Match each graph with one of the differential equations. 3t (a) y + 3y + 2y = sin t (b) y + y = sin t (c) y + y = sin 2 t t/10 (e) y + y + y = sin (d) y + 3y + 2y = e 3 Complex-Valued Solutions Although we have emphasized the need to obtain real-valued, physically relevant solutions to problems of interest, it is sometimes computationally convenient to consider differential equations with complex-valued nonhomogeneous terms. The corresponding particular solutions will then likewise be complex-valued functions. Exercises 38 and 39 illustrate some aspects of this type of calculation. Exercises 40–44 provide some additional examples. 38. Consider the differential equation y + p(t)y + q(t)y = g1 (t) + ig2 (t), where p(t), q(t), g1 (t), and g2 (t) are all real-valued functions continuous on some t-interval of interest. Assume that yP (t) is a particular solution of this equation. Generally, yP (t)

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The Method of Undetermined Coefﬁcients

167

y y 8

2

7

1.5

6

1

5

0.5 5

10

15

20

t

25

4

–0.5

3

–1

2

–1.5

1

–2 5

10

15

Graph A

20

t

25

Graph B

y

y

0.8

1

0.6

0.5

0.4 5

0.2

5

10

15

20

15

20

25

t

–0.5

t

25

10

–1

–0.2

Graph C

Graph D y

10 5

5

10

15

25

t

–5 –10

Graph E Figure for Exercise 37

will be a complex-valued function. Let yP (t) = u(t) + iv(t), where u(t) and v(t) are real-valued functions. Show that u + p(t)u + q(t)u = g1 (t)

and

v + p(t)v + q(t)v = g2 (t).

That is, show that the real and imaginary parts of the complex-valued particular solution, yP (t), are themselves particular solutions corresponding to the real and imaginary parts of the complex-valued nonhomogeneous term, g(t).

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Second and Higher Order Linear Differential Equations 39. Consider the nonhomogeneous differential equation y − y = ei2t . The complementary solution is yC = c1 e t + c2 e−t . Recall from Euler’s formula that ei2t = cos 2t + i sin 2t. Therefore, the right-hand side is a (complex-valued) linear combination of functions for which the method of undetermined coefﬁcients is applicable. (a) Assume a particular solution of the form yP = Aei2t , where A is an undetermined (generally complex) coefﬁcient. Substitute this trial form into the differential equation and determine the constant A. (b) With the constant A as determined in part (a), write yP (t) = Aei2t in the form yP (t) = u(t) + iv(t), where u(t) and v(t) are real-valued functions. (c) Show that u(t) and v(t) are themselves particular solutions of the following differential equations: u − u = Re[ei2t ] = cos 2t

and

v − v = Im[ei2t ] = sin 2t.

Therefore, the single computation with the complex-valued nonhomogeneous term yields particular solutions of the differential equation for the two real-valued nonhomogeneous terms forming its real and imaginary parts.

Exercises 40–44: For each exercise, (a) Use the indicated trial form for yP (t) to obtain a (complex-valued) particular solution for the given differential equation with complex-valued nonhomogeneous term g(t). (b) Write yP (t) as yP (t) = u(t) + iv(t), where u(t) and v(t) are real-valued functions. Show that u(t) and v(t) are particular solutions of the given differential equation with nonhomogeneous terms Re[g(t)] and Im[g(t)], respectively. 40. y + 2y + y = eit , 42. y + 4y = ei2t ,

44. y + y = e

3.9

(1+i)t

yP (t) = Aeit

yP (t) = Atei2t ,

41. y + 4y = eit , 43. y + y = e−i2t ,

yP (t) = Aeit yP (t) = Ae−i2t

(1+i)t

yP (t) = Ae

The Method of Variation of Parameters Section 3.8 discussed the method of undetermined coefﬁcients as a technique for ﬁnding a particular solution of the constant coefﬁcient equation ay + by + cy = g(t).

(1)

The method of undetermined coefﬁcients can be applied to equation (1) as long as the nonhomogeneous term, g(t), is one of the types listed in Table 3.1 of Section 3.8 or is a linear combination of types listed in the table. In this section, we consider the general linear second order nonhomogeneous differential equation y + p(t)y + q(t)y = g(t).

(2)

Unlike in Section 3.8, we do not insist that this differential equation have constant coefﬁcients or that g(t) belong to some special class of functions. The only restriction we place on differential equation (2) is that the functions p(t), q(t), and g(t) be continuous on the t-interval of interest. The technique we discuss, the method of variation of parameters, is one that uses a knowledge of the complementary solution of (2) to construct a corresponding particular solution.

3.9

The Method of Variation of Parameters

169

Discussion of the Method Assume that { y1 (t), y2 (t)} is a fundamental set of solutions of the homogeneous equation y + p(t)y + q(t)y = 0. In other words, the complementary solution is given by yC (t) = c1 y1 (t) + c2 y2 (t). To obtain a particular solution of (2), we “vary the parameters.” That is, we replace the constants c1 and c2 by functions u1 (t) and u2 (t) and look for a particular solution of (2) having the form yP (t) = y1 (t)u1 (t) + y2 (t)u2 (t).

(3)

In (3), there are two functions, u1 (t) and u2 (t), that we are free to specify. One obvious constraint on u1 (t) and u2 (t) is that they must be chosen so that yP is a solution of nonhomogeneous equation (2). Generally speaking, however, two constraints can be imposed when we want to determine two functions. We will impose a second constraint as we proceed, choosing it so as to simplify the calculations. Substituting (3) into the left-hand side of equation (2) requires us to compute the ﬁrst and second derivatives of (3). Computing the ﬁrst derivative leads to yP = [ y1 u1 + y2 u2 ] + [ y1 u1 + y2 u2 ]. The grouping of terms in this equation is motivated by the fact that we now impose a constraint on the functions u1 (t) and u2 (t). We require y1 u1 + y2 u2 = 0.

(4)

With this constraint, the derivative of yP becomes while

yP

yP = y1 u1 + y2 u2 ,

(5)

yP = y1 u1 + y2 u2 + y1 u1 + y2 u2 .

(6)

is

Notice that the ﬁrst and second derivatives of yP have been simpliﬁed and yP does not involve u1 or u2 . Inserting yP = y1 u1 + y2 u2 into the differential equation y + p(t)y + q(t)y = g(t) and using (5) and (6), we ﬁnd [y1 u1 + y2 u2 + y1 u1 + y2 u2 ] + p(t)[ y1 u1 + y2 u2 ] + q(t)[ y1 u1 + y2 u2 ] = g(t). Rearranging terms yields [y1 + p(t)y1 + q(t)y1 ]u1 + [ y2 + p(t)y2 + q(t)y2 ]u2 + [ y1 u1 + y2 u2 ] = g(t).

(7)

Since y1 and y2 are solutions of y + p(t)y + q(t)y = 0, equation (7) reduces to y1 u1 + y2 u2 = g(t).

(8)

We therefore obtain two constraints, equations (4) and (8), for the two unknown functions u1 (t) and u2 (t). We can combine these two equations into the matrix equation 0 y1 (t) y2 (t) u1 (t) = . (9) y1 (t) y2 (t) u2 (t) g(t)

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If the (2 × 2) coefﬁcient matrix has an inverse, then we can solve equation (9) for the unknowns u1 (t) and u2 (t). Once they are determined, we can ﬁnd u1 (t) and u2 (t) by computing antiderivatives. The coefﬁcient matrix in equation (9) is invertible if and only if its determinant is nonzero. Note, however, that the determinant of this matrix is the Wronskian of the functions y1 and y2 , W(t) = y1 (t)y2 (t) − y1 (t)y2 (t). Since { y1 (t), y2 (t)} is a fundamental set of solutions, we are assured that the Wronskian is nonzero for all values of t in our interval of interest. Solving equation (9) for u1 and u2 gives 0 u1 (t) y2 (t) −y2 (t) 1 = , W(t) −y1 (t) u2 (t) y1 (t) g(t) or u1 (t) =

−y2 (t)g(t) W(t)

and

u2 (t) =

y1 (t)g(t) . W(t)

Antidifferentiating to obtain u1 (t) and u2 (t), we have a particular solution, yP (t) = y1 (t)u1 (t) + y2 (t)u2 (t). Explicitly, the particular solution we have obtained is y2 (s)g(s) y1 (s)g(s) yP (t) = −y1 (t) ds + y2 (t) ds. W(s) W(s)

(10)

Once we calculate the two antiderivatives in equation (10), we will have determined a particular function that solves the nonhomogeneous equation (2).

E X A M P L E

1

Find the general solution of the differential equation y − 2y + y = e t ln t,

t > 0.

(11)

Solution: Note that the nonhomogeneous term, g(t) = e t ln t, does not appear in Table 3.1 as a candidate for the method of undetermined coefﬁcients. Since the method of undetermined coefﬁcients is not applicable, we turn to the method of variation of parameters. For equation (11), the complementary solution is yC (t) = c1 y1 (t) + c2 y2 (t) = c1 e t + c2 te t . Using variation of parameters to ﬁnd a particular solution, we assume yP (t) = e t u1 (t) + te t u2 (t). Substituting this expression into the nonhomogeneous differential equation (11) and applying the constraint from equation (4), e t u1 (t) + te t u2 (t) = 0, we obtain 0 u1 (t) et te t = t . e ln t e t e t + te t u2 (t)

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The Method of Variation of Parameters

171

The determinant of the coefﬁcient matrix is W(t) = e2t . Solving this matrix equation gives −t ln t u1 (t) = . u2 (t) ln t Computing antiderivatives yields t2 t2 u1 (t) = u1 (t) dt = − t ln t dt = − ln t + + K1 2 4 and

u2 (t) =

u2 (t) dt =

ln t dt = t ln t − t + K2 .

We can choose the constants of integration to suit our convenience because all we need is some particular solution, yP = y1 u1 + y2 u2 . For convenience, we set both K1 and K2 equal to zero and obtain t2 t2 t yP (t) = e − ln t + + te t [t ln t − t] 2 4 t2 e t 3 = ln t − . 2 2 The general solution of equation (11) is, therefore, t2 e t 3 ln t − . ❖ y(t) = c1 e t + c2 te t + 2 2 E X A M P L E

2

Observe that y1 (t) = t is a solution of the homogeneous equation t2 y − ty + y = 0,

t > 0.

Use this observation to solve the nonhomogeneous initial value problem t2 y − ty + y = t,

y(1) = 1,

y (1) = 4.

Solution: The ﬁrst step in ﬁnding the general solution of the nonhomogeneous equation is determining a fundamental set of solutions {y1 , y2 }. Thus, we need to ﬁnd a second solution, y2 (t), to go along with the given solution, y1 (t) = t. The method of reduction of order, described in Section 3.4, can be used. It leads to a second solution, y2 (t) = t ln t. The functions t and t ln t can be shown to form a fundamental set of solutions for the homogeneous equation. Therefore, the complementary solution of the nonhomogeneous equation is yC (t) = c1 t + c2 t ln t,

t > 0.

Since the differential equation has variable coefﬁcients, we cannot use the method of undetermined coefﬁcients to ﬁnd a particular solution. Instead, we (continued)

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use the method of variation of parameters. Assume a particular solution of the form yP (t) = tu1 (t) + [t ln t]u2 (t). The simplest approach is to substitute this form into the given nonhomogeneous differential equation to determine one equation for u1 and u2 , then use the constraint tu1 (t) + [t ln t]u2 (t) = 0 to form a second equation for u1 and u2 . As an alternative, we could go directly to equation (9) or equation (10) to determine a particular solution. If we proceed in this fashion, however, we have to make certain that the nonhomogeneous equation under consideration is in the standard form given by equation (2). Since the differential equation in this example does not have this form, we need to rewrite it as y −

1 1 1 y + 2y = . t t t

Doing so, we can identify the term g(t) in equations (9) and (10), g(t) = 1/t. Both approaches lead to the following system of equations for u1 and u2 : ⎡0⎤ t t ln t u1 = ⎣1⎦ . 1 1 + ln t u2 t Solving this system, we obtain

u1

u2

=

−t−1 ln t t−1

.

Computing antiderivatives yields (ln t)2 1 ln t dt = − + K1 , u1 (t) = − t 2 1 dt = ln t + K2 . u2 (t) = t We can set both of the arbitrary constants equal to zero, obtaining a particular solution t (ln t)2 + [t ln t] ln t = (ln t)2 . yP (t) = t − 2 2 The general solution of the nonhomogeneous equation is therefore y(t) = c1 t + c2 t ln t +

t (ln t)2 , 2

t > 0.

Imposing the initial conditions shows that the solution of the initial value problem is t y(t) = t + 3t ln t + (ln t)2 , t > 0. ❖ 2

3.9

The Method of Variation of Parameters

173

Figure 3.13 displays the graph of the solution y(t). Note that lim t→0+ y(t) = 0. The solution is well behaved near t = 0, even though the differential equation has coefﬁcient functions that are not deﬁned at t = 0. y 16 14 12 10 8 6 4 2 0.5

1

1.5

2

2.5

t

3

–2

FIGURE 3.13

The solution of the initial value problem in Example 2. Even though some of the coefﬁcient functions are not deﬁned at t = 0, the solution y(t) has a limit as t approaches 0 from the right.

EXERCISES Exercises 1–14: For the given differential equation, (a) Determine the complementary solution, yC (t) = c1 y1 (t) + c2 y2 (t). (b) Use the method of variation of parameters to construct a particular solution. Then form the general solution. 1. y + 4y = 4

2. y + y = sec t,

−π/2 < t < π/2

t

3. y − [2 + (1/t)]y + [1 + (1/t)]y = te , 0 < t < ∞. [The functions y1 (t) = e t and y2 (t) = t2 e t are solutions of the homogeneous equation.] 1 5. y − y = e t 4. y − y = 1 + et 6. y − (2/t)y + (2/t2 )y = t/(1 + t2 ), 0 < t < ∞. [The function y1 (t) = t2 is a solution of the homogeneous equation.] 7. y − 2y + y = e t

8. y + 36y = csc3 (6t)

9. y − (2 cot t)y + (2 csc2 t − 1)y = t sin t, 0 < t < π . [The functions y1 (t) = sin t and y2 (t) = t sin t are both solutions of the homogeneous equation.] 10. t2 y − ty + y = t ln t, 0 < t < ∞. [The functions y1 (t) = t and y2 (t) = t ln t are both solutions of the homogeneous equation.] 11. y + [t/(1 − t)]y − [1/(1 − t)]y = (t − 1)e t , 1 < t < ∞. [The functions y1 (t) = t and y2 (t) = e t are both solutions of the homogeneous equation.] 2

2

2

12. y + 4ty + (2 + 4t2 )y = t2 e−t . [The functions y1 (t) = e−t and y2 (t) = te−t are both solutions of the homogeneous equation.]

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Second and Higher Order Linear Differential Equations 13. (t − 1)2 y − 4(t − 1)y + 6y = t, 1 < t < ∞. [The function y1 (t) = (t − 1)2 is a solution of the homogeneous equation.] 14. y − [2 + (2/t)]y + [1 + (2/t)]y = e t , tion of the homogeneous equation.]

0 < t < ∞. [The function y1 (t) = e t is a solu-

15. Consider the homogeneous differential equation y + p(t)y + q(t)y = g(t). Let { y1 , y2 } be a fundamental set of solutions for the corresponding homogeneous equation, and let W(t) denote the Wronskian of this fundamental set. Show that the particular solution that vanishes at t = t0 is given by t g(λ) dλ. yP (t) = [ y1 (t)y2 (λ) − y2 (t)y1 (λ)] W(λ) t0

Exercises 16–18: The given expression is the solution of the initial value problem y + αy + βy = g(t),

y(0) = y0 ,

y (0) = y0 .

Determine the constants α, β, y0 , and y0 . 1 t 16. y(t) = sin[2(t − λ)]g(λ) dλ 2 0 t t−λ t e − e−(t−λ) 17. y(t) = e−t + g(λ) dλ = e−t + sinh(t − λ)g(λ) dλ 2 0 0 t (t − λ)g(λ) dλ 18. y(t) = t + 0

3.10 Forced Mechanical Vibrations, Electrical Networks, and Resonance

In this section, we use what we know about solving nonhomogeneous second order linear differential equations to study the behavior of mechanical systems (such as ﬂoating objects and spring-mass-dashpot systems) that are subjected to externally applied forces. We also consider simple electrical networks containing resistors, inductors, and capacitors (called RLC networks) that are driven by voltage and current sources. All of these applications ultimately give rise to the same mathematical problem. And, as we shall see, the physical phenomenon of resonance is an important consideration common to all of these applications.

Forced Mechanical Vibrations A Buoyant Body Consider again the problem of the buoyant body discussed in Section 3.1. (See Figure 3.1.) A cylindrical block of cross-sectional area A, height L, and mass density ρ is placed in a liquid having mass density ρl . Since we assume ρ < ρl , the block ﬂoats in the liquid. In equilibrium, it sinks a depth Y into the liquid; at this depth, the weight of the block equals the weight of the liquid displaced. The quantity y(t) represents the instantaneous vertical displacement of the block from its equilibrium position, measured positive downward. Suppose now that an externally applied vertical force, Fa (t), acts

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175

on the buoyant body. Newton’s second law of motion ultimately leads to a nonhomogeneous differential equation for the displacement y(t): d2 y dt

2

+

ρl g 1 y= F (t). ρL ρAL a

(1)

In equation (1), g denotes gravitational acceleration and the term ρAL is the block’s mass. Deﬁning a radian frequency, ω0 = ρl g/ρL, and a force per unit mass, fa (t) = (1/ρAL)Fa (t), we can rewrite equation (1) as y + ω02 y = fa (t).

(2)

Equation (2) is a second order constant coefﬁcient linear nonhomogeneous differential equation. To uniquely prescribe the bobbing motion of the ﬂoating body, we would add initial conditions that specify the initial displacement, y(t0 ) = y0 , and the initial velocity, y (t0 ) = y0 , of the block. A Spring-Mass-Dashpot System As in Section 3.6, we can use Newton’s second law of motion to derive a nonhomogenous differential equation for the displacement of a mass suspended from a spring-dashpot connection and acted upon by an applied force. The resulting differential equation is my + γ y + ky = Fa (t),

(3)

where Fa (t) denotes an applied vertical force. The positive constants m, γ , and k represent the mass, damping coefﬁcient, and spring constant of the system, respectively. The dependent variable y(t) measures downward displacement from the equilibrium rest position. (See Figure 3.8.) If there is no damping (that is, if γ = 0) and if we deﬁne radian frequency ω0 = k/m and force per unit mass fa (t) = (1/m)Fa (t), equation (3) can be rewritten as y + ω02 y = fa (t).

(4)

Note that equation (4), describing a spring-mass system, is identical in structure to equation (2), which describes the bobbing motion of a buoyant body. Both applications lead to the same mathematical problem. As we shall see shortly, other applications (such as RLC networks) also lead to differential equations having exactly the same structure as equations (2) and (3). Since these applications all lead to the same mathematical problem, we will discuss equations (2) and (3) in their own right rather than investigate each application separately. The Exercises focus on speciﬁc applications.

Oscillatory Applied Forces and Resonance Examples 1 and 2 concern an important special case of equation (2) where the applied force is a sinusoidally varying force, fa (t) = F cos ω1 t, where F is a constant. For simplicity, we assume the system is initially at rest in equilibrium. Thus, for the special case under consideration, the corresponding

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initial value problem is y + ω02 y = F cos ω1 t, y(0) = 0, y (0) = 0.

t > 0,

(5)

The complementary solution of (5) is yC (t) = c1 sin ω0 t + c2 cos ω0 t.

(6)

In order to ﬁnd a particular solution, we need to consider two separate cases, ω1 = ω0 and ω1 = ω0 . From a mathematical perspective, the method of undetermined coefﬁcients discussed in Section 3.8 leads us to expect two different types of particular solutions. If ω1 = ω0 , the nonhomogeneous term, g(t) = F cos ω1 t, is not a solution of the homogeneous equation; when ω1 = ω0 , g(t) is a solution of the homogeneous equation. This mathematical perspective is consistent with the physics of the problem. We should expect different behavior on purely physical grounds. Radian frequency ω0 (or, more properly, f0 = ω0 /2π ) is called the natural frequency of the vibrating system. It represents the frequency at which the system would vibrate if no applied force were present and the system were merely responding to some initial disturbance. The applied force acts on the system with its own applied frequency ω1 . In the special case where the natural and applied frequencies are equal, the applied force pushes and pulls on the system with a frequency precisely equal to that at which the system tends naturally to vibrate. This precise reinforcement leads to the phenomenon of resonance. For this reason, the natural frequency of the system is also referred to as its resonant frequency. E X A M P L E

1

Assume ω1 = ω0 . Solve the initial value problem (5). Solution: Since ω1 = ω0 , Table 3.1 in Section 3.8 suggests a particular solution of the form yP (t) = A cos ω1 t + B sin ω1 t. Substituting this form into the nonhomogeneous equation, we obtain (see Exercise 1) yP (t) = −

F ω12

− ω02

cos ω1 t.

Imposing the initial conditions on the general solution y(t) = c1 sin ω0 t + c2 cos ω0 t −

F ω12

− ω02

cos ω1 t,

we obtain the solution of initial value problem (5), y(t) =

F ω12

− ω02

[cos ω0 t − cos ω1 t].

(7)

Figure 3.14 shows solution (7) for the special case where ω1 = 12π s−1 , ω0 = 10π s−1 , and F is chosen so that F/(ω12 − ω02 ) = 2 cm. The example therefore assumes that the natural frequency of the system is 5 Hz while the applied frequency is 6 Hz. (For deﬁniteness, the unit of length is chosen to be the centimeter.)

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Forced Mechanical Vibrations, Electrical Networks, and Resonance

177

y 6 4 2 1

2

3

t

–2 –4 –6

FIGURE 3.14

The solution of initial value problem (5), as given by equation (7). For the case shown, ω1 = 12π, ω0 = 10π, and F = 2(ω12 − ω02 ).

We can better understand Figure 3.14, including the dashed envelope, if we use trigonometric identities to recast equation (7). Suppose we deﬁne an average radian frequency, ω, and a difference (or beat) radian frequency, β, by ω=

ω1 + ω0 , 2

β=

ω1 − ω0 . 2

With these deﬁnitions, we have ω1 = ω + β and ω0 = ω − β. Equation (7) becomes F [cos(ωt − βt) − cos(ωt + βt)]. y(t) = 4ωβ Using the trigonometric identity cos(θ1 ± θ2 ) = cos θ1 cos θ2 ∓ sin θ1 sin θ2 , we note that cos(ωt − βt) − cos(ωt + βt) = 2 sin ωt sin βt. Therefore, we obtain an alternative representation for the solution y(t): y(t) = A(t) sin ωt,

(8)

where A(t) =

F sin βt . 2ωβ

Using equation (8), we can interpret solution (7) as the product of a variable amplitude term, A(t), and a sinusoidal term, sin ωt. In cases where ω1 and ω0 are nearly equal, the amplitude term, whose behavior is governed by the factor sin βt, is slowly varying relative to the sinusoidal term, sin ωt (because |β| ω). The combination of these disparate rates of variation gives rise to the phenomenon of “beats” seen in Figure 3.14. Since ω = 11π s−1 and β = π s−1 , sin βt = sin πt varies much more slowly than sin ωt = sin 11π t. The multiplicative factor A(t) =

F sin βt = 4 sin π t 2ωβ (continued)

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Second and Higher Order Linear Differential Equations (continued)

deﬁnes a slowly varying sinusoidal envelope for the more rapidly varying sin ωt = sin 11π t. The dashed envelope shown in Figure 3.14 is deﬁned by the graphs of y = ±4 sin πt. ❖ E X A M P L E

2

Assume ω1 = ω0 . Solve initial value problem (5). Solution:

In this case, y + ω02 y = F cos ω0 t.

Since the functions cos ω0 t and sin ω0 t are both solutions of the homogeneous equation, Table 3.1 of Section 3.8 prescribes a trial solution of the form yP (t) = At cos ω0 t + Bt sin ω0 t. Substituting this form leads (see Exercise 1) to the particular solution yP (t) =

F t sin ω0 t 2ω0

and the general solution y(t) = c1 sin ω0 t + c2 cos ω0 t +

F t sin ω0 t. 2ω0

The initial conditions imply that c1 and c2 are zero. The solution of initial value problem (5) is therefore y(t) =

F t sin ω0 t. 2ω0

(9)

Figure 3.15 shows solution (9) for the case ω0 = 10π s−1

F = 4π cm/s. 2ω0

and

y 50 40 30 20 10 –10

1

2

3

t

–20 –30 –40 –50

FIGURE 3.15

The solution of initial value problem (5) as given by equation (9). For the case shown, ω0 = 10π and F = 8πω0 . The solution envelope grows linearly with time, illustrating the phenomenon of resonance.

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179

In this case, the applied force reinforces the natural frequency vibrations of the mechanical system and the envelope of the solution (shown by the dashed lines in Figure 3.15) grows linearly with time. This is the phenomenon of resonance. Does the solution (9) make sense? The vibration amplitude of a real physical system certainly does not continue to grow indeﬁnitely. Therefore, one would expect equation (9) to describe the behavior of a real system for a limited time at best. Once the vibration amplitude becomes sufﬁciently large, the assumptions made in deriving the mathematical models cease to be valid. For example, equation (3) and special case (4) for the spring-mass system assume the validity of Hooke’s law. When mass displacement amplitude becomes too large, however, the force-displacement relation becomes more complicated than the simple linear relation embodied in Hooke’s law. ❖ REMARK: One property of a well-posed problem is continuous dependence upon the data. Roughly speaking, if an initial value problem is to be a reasonable model of reality, its solution should not change uncontrollably when a parameter (such as a coefﬁcient in the differential equation or an initial condition) is changed slightly. Therefore, we might reasonably ask whether it is possible to see resonant solution (9) emerge from nonresonant solution (7) or (8) in the limit as ω1 → ω0 . Note that (8) can be rewritten as

sin βt F t (10) y(t) = sin ωt. 2ω βt Suppose we ﬁx t and let ω1 → ω0 . Then, from their deﬁnitions, ω → ω0 and β → 0. We know from calculus that sin x = 1. x Therefore, for any ﬁxed value of t, we do indeed obtain resonant solution (9) from nonresonant expression (8) in the limit as ω1 → ω0 . lim x→0

The Effect of Damping on Resonance There are no perpetual motion machines. All physical systems have at least some small loss or damping present. Therefore, it is of interest to see what happens if we add damping to an otherwise resonant system. Suppose we consider the initial value problem y + 2δy + ω02 y = F cos ω0 t,

y(0) = 0,

y (0) = 0,

(11a)

where δ is a positive constant (the factor 2 is added for convenience). What does the solution look like? In Exercise 11(a), you are asked to show that the solution of this initial value problem is

⎤ ⎡ −δt 2 2 ω0 − δ t ⎥ e sin F ⎢ ⎥. ⎢ sin ω0 t −

y(t) = (11b) ⎦ 2δ ⎣ ω0 ω02 − δ 2 [In (11), we tacitly assume that ω02 > δ 2 .] As a check, you can show [see Exercise 11(b)] that for any ﬁxed t and ω0 , expression (11b) reduces to (9) as δ → 0. Equation (11b) shows the effect of damping on the otherwise resonant system. If we ﬁx δ at some positive value and let t → ∞, the second term in equa-

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Second and Higher Order Linear Differential Equations

tion (11b) tends to zero. Therefore, with the addition of damping, displacement does not grow indeﬁnitely, as it appears to in Figure 3.15. As time increases, displacement approaches a steady-state behavior given by the ﬁrst term, F sin ω0 t. 2δω0 As δ becomes smaller, the amplitude of these steady-state oscillations becomes correspondingly larger. Figure 3.16 shows the variation of displacement for the case F = 4π cm/s, δ = 0.5 s−1 . ω0 = 10π s−1 , 2ω0 y 8

4

1

2

3

4

5

6

7

8

9

10

t

–4

–8

FIGURE 3.16

ω02 y

The solution of y + 2δy + = F cos ω0 t for the case ω0 = 10π, F = 8πω0 , and δ = 0.5. As you can see by comparing this graph with the resonant solution graphed in Figure 3.15, the inclusion of damping eliminates the unbounded linear growth in the solution envelope that characterizes the resonant case. As noted, however, the steady-state oscillations, (F/2δω0 ) sin ω0 t, have an amplitude proportional to δ −1 .

Nonresonant Excitation with Damping Present Suppose we now change the radian frequency of the applied force in the previous problem to ω1 = ω0 . In that case, the problem becomes y + 2δy + ω02 y = F cos ω1 t,

y(0) = 0,

y (0) = 0.

(12a)

This amounts to the addition of a damping force to the problem deﬁned by equation (5). Again assuming that ω02 > δ 2 , the solution of problem (12a) is y(t) =

F (ω02 −

−

ω12 )2

+ (2δω1 )2

[(ω02 − ω12 ) cos ω1 t + 2δω1 sin ω1 t]

Fe−δ t (ω02 − ω12 )2 + (2δω1 )2

2 2 2 2 ω0 − δ t (ω0 − ω1 ) cos δ(ω2 + ω12 ) + 0 sin ω02 − δ 2

⎤

ω02 − δ 2 t ⎦.

(12b)

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181

This solution seems relatively complicated. However, two checks can be made. At a ﬁxed value of t, the solution should reduce to (11b) in the limit as ω1 → ω0 . Also, the solution should reduce to (7) if we ﬁx ω1 but let δ → 0. Exercise 12 asks you not only to derive this solution, but also to make these checks upon its correctness. The second term in the solution of equation (12a) represents a transient term, one that tends to zero as time increases. The ﬁrst term is the steady-state portion of the solution. Figure 3.17 shows the behavior of the solution for the case where F ω0 = 10π s−1 , ω1 = 12π s−1 , = 4π cm/s, δ = 0.5 s−1 . 2ω0 Compare this behavior with that exhibited in Figures 3.14 and 3.16. When time t is relatively small, before the effects of damping become pronounced, the solution exhibits a difference frequency modulation envelope that is qualitatively similar to the behavior shown in Figure 3.14. As time progresses, however, damping eventually diminishes the second term in the solution to a negligibly small contribution and the solution becomes essentially the steady-state portion given by the ﬁrst term. In this respect, the long-term behavior qualitatively resembles the damping-perturbed resonant case exhibited in Figure 3.16. y 4 3 2 1 1

2

3

4

5

6

7

8

9

10

t

–1 –2 –3 –4

FIGURE 3.17

The solution of equation (12) for representative values of ω0 , ω1 , F, and δ. Initially, for small values of t, damping is not signiﬁcant and the motion is similar to that shown in Figure 3.14, exhibiting beats. As t grows, damping diminishes the second term in equation (12) and the motion has a period similar to the applied force.

RLC Networks We now consider networks containing resistors, inductors, and capacitors. The application of Kirchhoff ’s7 voltage law and Kirchhoff ’s current law to these networks leads us to second order differential equations. 7

Gustav Robert Kirchhoff (1824–1887) was a German physicist who made important contributions to network theory, elasticity, and our understanding of blackbody radiation. A lunar crater is named in his honor.

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Consider ﬁrst the series RLC network shown in Figure 3.18. A voltage source VS (t) having the polarity shown is connected in series with circuit elements having resistance R, inductance L, and capacitance C. A current I(t), assumed positive in the sense shown, ﬂows in the loop. In essence, Kirchhoff’s voltage law asserts that the voltage at each point in the network is a well-deﬁned singlevalued quantity. Therefore, as we make an excursion around the loop, the sum of the voltage rises must equal the sum of the voltage drops. If we proceed around the loop in Figure 3.18 in a clockwise manner, the voltage rise is the source voltage VS (t), while the voltage drops are the drops across the three circuit elements. The voltage drop across the resistor is I(t)R, the drop across the inductor is L (dI/dt), and the drop across the capacitor is (1/C)Q(t), where Q(t) represents the electric charge on the capacitor. R I(t)

+

Voltage source

VS (t)

L

~

C

–

FIGURE 3.18

A series RLC network, with voltage source VS (t) and loop current I(t).

An application of Kirchhoff’s voltage law therefore leads to the equation VS (t) = RI + L

1 dI + Q(t). dt C

(13a)

To obtain a differential equation for a single dependent variable, we use the fact that electric current is the rate of change of electric charge with respect to time, I(t) =

dQ . dt

One approach is to rewrite equation (13a) as a second order differential equation for the electric charge, obtaining L

d2 Q dt

2

+R

1 dQ + Q = VS (t). dt C

This equation, supplemented by initial conditions specifying the charge Q(t0 ) and current Q (t0 ) = I(t0 ) at some initial time t0 , can be solved for the charge Q(t). Differentiating this solution yields the desired current, I(t). A second approach is simply to differentiate equation (13a), obtaining a second order differential equation for the current I(t). In that case, d2 I dt

2

+

1 1 dVS (t) R dI + I= . L dt LC L dt

(13b)

Equation (13b) is a nonhomogeneous second order linear differential equation for the unknown loop current I(t). To uniquely prescribe circuit performance, we must add initial conditions I(t0 ) = I0 and I (t0 ) = I0 at some initial time

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183

t0 . [Specifying I (t0 ) is tantamount to specifying the voltage drop across the inductor at time t0 .] As a second example of an RLC network, consider the network shown in Figure 3.19. In this case, the three circuit elements are connected in parallel with a current source IS (t), whose current output is assumed to ﬂow in the direction shown. This time, the dependent variable of interest is nodal voltage V(t), assumed to have the polarity shown.

+ Current source

IS (t)

V(t)

L R

C

– FIGURE 3.19

A parallel RLC network, with current source IS (t) and nodal voltage V(t).

The governing physical principle, Kirchhoff’s current law, states that electric current does not accumulate at a circuit node. Therefore, the total current ﬂowing into a node must equal the total current ﬂowing out. Consider the upper node. The current ﬂowing in is the source current, while the current ﬂowing out is the current ﬂowing “down” through each of the circuit elements. The current through the resistor is (1/R)V(t), the current through the capacitor is C(dV/dt), ! and the current through the inductor is (1/L) V(s) ds (an antiderivative of the nodal voltage). Applying Kirchhoff’s current law to the network in Figure 3.19 leads us to the equation dV 1 1 + V(s) ds. IS (t) = V + C R dt L Upon differentiating and rearranging terms, the equation becomes d2 V dt2

+

1 1 dIS (t) 1 dV + V= . RC dt LC C dt

(14)

Specifying V and V (that is, the currents through the resistor and capacitor) at some initial time t0 will uniquely determine circuit performance.

REMARK: If we short circuit the resistor (that is, set R = 0) in the series circuit (Figure 3.18) or if we open circuit the resistor (that is, let R → ∞) in the parallel circuit (Figure 3.19), we remove the dissipative (damping) element in each case and obtain a lossless LC circuit. Such circuits can exhibit resonance. Note that equations (13b) and (14) become identical in structure to equations (2) and (4) with a resonant radian frequency deﬁned by ω02 =

1 . LC

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EXERCISES 1. Consider the differential equation y + ω02 y = F cos ωt. (a) Determine the complementary solution of this differential equation. (b) Use the method of undetermined coefﬁcients to ﬁnd a particular solution in each of the cases: (i) ω = ω1 = ω0 , (ii) ω = ω0 .

Exercises 2–5: A 10-kg object suspended from the end of a vertically hanging spring stretches the spring 9.8 cm. At time t = 0, the resulting spring-mass system is disturbed from its rest state by the given applied force, F(t). The force F(t) is expressed in newtons and is positive in the downward direction; time is measured in seconds. (a) Determine the spring constant, k. (b) Formulate and solve the initial value problem for y(t), where y(t) is the displacement of the object from its equilibrium rest state, measured positive in the downward direction. (c) Plot the solution and determine the maximum excursion from equilibrium made by the object on the t-interval 0 ≤ t < ∞ or state that there is no such maximum. 3. F(t) = 20e−t 20, 5. F(t) = 0,

2. F(t) = 20 cos 10t 4. F(t) = 20 cos 8t

0≤t≤π π δ 2 and ω1 = ω0 . The radian frequency of the applied force is therefore not equal to ω0 . (a) Solve the initial value problem for y(t) and verify that equation (12b) represents the solution. (b) Assume that t > 0 and δ > 0 are ﬁxed. Show that ⎤ ⎡ −δt ω02 − δ 2 t F ⎣ sin(ω0 t) e sin ⎦.

− lim y(t) = ω1 →ω0 2δ ω0 ω2 − δ 2 0

(c) Assume now that t > 0 and ω1 are ﬁxed. Show that lim+ y(t) =

δ→0

ω12

% & F cos(ω0 t) − cos(ω1 t) . 2 − ω0

13. The Great Zacchini, daredevil extraordinaire, is a circus performer whose act consists of being “shot from a cannon” to a safety net some distance away. The “cannon”

Second and Higher Order Linear Differential Equations

is a frictionless tube containing a large spring, as shown in the ﬁgure. The spring constant is k = 150 lb/ft, and the spring is precompressed 10 ft prior to launching the acrobat. Assume that the spring obeys Hooke’s law and that Zacchini weighs 150 lb. Neglect the weight of the spring. (a) Let x(t) represent spring displacement along the tube axis, measured positive in the upward direction. Show that Newton’s second law of motion leads to the differential equation mx = −kx − mg cos(π/4), x < 0, where m is the mass of the daredevil. Specify appropriate initial conditions. (b) With what speed does he emerge from the tube when the spring is released? (c) If the safety net is to be placed at the same height as the mouth of the “cannon,” how far downrange from the cannon’s mouth should the center of the net be placed? g

y'(t) Uncompressed spring

45°

2 -ft unstretched rubber band

45°

Compressed spring prior to launch

Figure for Exercise 13

y' = 0

3 ft

g ft

CHAPTER 3

10

186

1-oz ball y(t)

Figure for Exercise 14

14. A Change of Independent Variable (see Section 2.9) A popular child’s toy consists of a small rubber ball attached to a wooden paddle by a rubber band; see the ﬁgure. Assume a 1-oz ball is connected to the paddle by a rubber band having an unstretched length of 2 ft. When the ball is launched vertically upward by the paddle with an initial speed of 30 ft/sec, the rubber band is observed to stretch 1 ft (to a total length of 3 ft) when the ball has reached its highest point. Assume the rubber band behaves like a spring and obeys Hooke’s law for this amount of stretching. Our objective is to determine the spring constant k. (Neglect the weight of the rubber band.) The motion occurs in two phases. Until the ball has risen 2 ft above the paddle, it acts like a projectile inﬂuenced only by gravity. Once the height of the ball exceeds 2 ft, the ball acts like a mass on a spring, acted upon by the elastic restoring force of the rubber band and gravity. (a) Assume the ball leaves the paddle at time t = 0. Let t2 and t3 represent the times at which the height of the ball is 2 ft and 3 ft, respectively, and let m denote the mass of the rubber ball. Show that an application of Newton’s second law of motion leads to the following two-part description of the problem: (i)

my = −mg,

0 < t < t2 , y(0) = 0, y (0) = 30

(ii) my = −k( y − 2) − mg,

t2 < t < t3 .

Here, y and y are assumed to be continuous at t = t2 . We also know that y(t2 ) = 2, y(t3 ) = 3, and y (t3 ) = 0. If we attempt to solve the problem “directly” with time as the independent variable, it is relatively difﬁcult, since the times t2 and t3 must be determined as part of the problem. Since height y(t) is an increasing function of time t over the interval of interest, however, we can view time t as a function of height, y, and use y as the independent variable.

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187

(b) Let v = y = dy/dt. If we adopt y as the independent variable, the acceleration becomes y = dv/dt = (dv/dy) (dy/dt) = v(dv/dy). Therefore, dv (i) mv = −mg, 0 < y < 2, v(0) = 30 dy dv = −k( y − 2) − mg, 2 < y < 3. dy Here, v is continuous at y = 2 and v|y=3 = 0. (ii) mv

Solve these two separable differential equations, impose the accompanying supplementary conditions, and determine the spring constant k. Network Problems Use the following consistent set of scaled units (referred to as the Scaled SI Unit System) in Exercises 15–18. Quantity

Unit

Symbol

Voltage Current Time Resistance Inductance Capacitance

volt milliampere (mA) millisecond (ms) kilohm (k ) henry (H) microfarad (μF)

V I t R L C

Exercises 15–16: Consider the series LC network shown in the ﬁgure. Assume that at time t = 0, the current and its time rate of change are both zero. For the given source voltage, determine the current I(t).

L = 1H + VS (t)

I(t) C = 4 F

~ –

Figure for Exercises 15–16

16. VS (t) = 10te−t volts

15. VS (t) = 5 sin 3t volts

Exercises 17–18: Consider the parallel RLC network shown in the ﬁgure. Assume that at time t = 0, the voltage V(t) and its time rate of change are both zero. For the given source current, determine the voltage V(t).

+

IS (t)

V(t)

R

C L

R = 1 kΩ L = 1H C = 12 F

– Figure for Exercises 17–18

17. IS (t) = 1 − e−t mA

18. IS (t) = 5 sin t mA

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3.11 Higher Order Linear Homogeneous Diﬀerential Equations

So far in this chapter, we have studied second order linear differential equations y + p1 (t)y + p0 (t)y = g(t), with emphasis on the important special case of constant coefﬁcient equations. We now consider higher order linear equations. An nth order linear differential equation has the form y(n) + pn−1 (t)y(n−1) + · · · + p2 (t)y + p1 (t)y + p0 (t)y = g(t).

(1)

Our study of equation (1) and the associated initial value problem follows a familiar pattern. We ﬁrst generalize the existence and uniqueness results of Section 3.1, then examine the homogeneous case of equation (1), and ﬁnally discuss nonhomogeneous equations. The basic theory and solution techniques for second order linear equations extend naturally to the higher order case. Our motivation for studying higher order equations is twofold. The theory of higher order linear differential equations is important in certain applications, and the way it generalizes second order linear theory is aesthetically appealing. Fourth order linear equations arise, for example, in modeling the loading and bending of beams (see the Projects at the end of this chapter).

Existence and Uniqueness Theorem 3.5 generalizes the existence and uniqueness results presented earlier for initial value problems involving linear differential equations; see Theorem 2.1 (ﬁrst order problems) and Theorem 3.1 (second order problems). Theorem 3.5

Let p0 (t), p1 (t), . . . , pn−1 (t) and g(t) be continuous functions deﬁned on the interval a < t < b, and let t0 be in (a, b). Then the initial value problem y(n) + pn−1 (t)y(n−1) + · · · + p2 (t)y + p1 (t)y + p0 (t)y = g(t), y(t0 ) = y0 ,

y (t0 ) = y0 ,

y (t0 ) = y0 ,

...,

y(n−1) (t0 ) = y(n−1) 0

has a unique solution deﬁned on the entire interval (a, b).

Comparing Theorems 2.1, 3.1, and 3.5, we see that the language and conclusions of the three theorems are virtually identical. In fact, we shall see in Chapter 4 that Theorems 2.1, 3.1, and 3.5 can all be viewed as special cases of an overarching existence-uniqueness theorem for systems of ﬁrst order linear equations (see Theorem 4.1 in Section 4.2).

The Principle of Superposition and Fundamental Sets of Solutions Consider the nth order linear homogeneous equation y(n) + pn−1 (t)y(n−1) + · · · + p2 (t)y + p1 (t)y + p0 (t)y = 0,

a < t < b.

(2)

3.11

Higher Order Linear Homogeneous Differential Equations

189

The principle of superposition, stated in Theorem 3.2 for second order linear equations, applies to higher order linear equations as well. In particular, if y1 (t), y2 (t), . . . , yr (t) are solutions of equation (2), then the linear combination y(t) = c1 y1 (t) + c2 y2 (t) + · · · + cr yr (t) is also a solution of equation (2). The idea of a fundamental set of solutions for a second order linear homogeneous differential equation also extends to nth order linear homogeneous equations. Let { y1 (t), y2 (t), . . . , yn (t)} be a set of n solutions of the homogeneous differential equation (2). This set is a fundamental set of solutions if every solution of (2) can be represented as a linear combination of the form y(t) = c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t),

a < t < b.

(3)

Constructing Fundamental Sets Consider the initial value problem y(n) + pn−1 (t)y(n−1) + · · · + p2 (t)y + p1 (t)y + p0 (t)y = 0, y(t0 ) = y0 ,

y (t0 ) = y0 ,

y (t0 ) = y0 ,

...,

y(n−1) (t0 ) = y0(n−1) .

(4)

Note that every solution of homogeneous equation (2) can be viewed as the unique solution of some initial value problem represented by (4). Simply ﬁx a point t0 in (a, b), and use the values of the function and its ﬁrst n − 1 derivatives at t0 as initial conditions. Let y0 , y0 , . . . , y0(n−1) be an arbitrary set of n constants, and let u(t) be the corresponding unique solution of initial value problem (4). Assume that { y1 , y2 , . . . , yn } is a fundamental set of solutions of (2). Since {y1 , y2 , . . . , yn } is a fundamental set of solutions, there are constants c1 , c2 , . . . , cn such that u(t) = c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t), a < t < b. The initial conditions in (4) lead to a system of equations that can be written in matrix form as ⎤ ⎡ ⎤⎡ ⎤ ⎡ y0 y1 (t0 ) y2 (t0 ) ··· yn (t0 ) c1 ⎢ y (t ) ⎢ ⎥ ⎢ ⎥ y2 (t0 ) yn (t0 ) ⎥ ⎢ 1 0 ⎥ ⎢c2 ⎥ ⎢ y0 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ (5) .. .. ⎢ ⎥ ⎢ .. ⎥ = ⎢ ... ⎥. . . ⎦ ⎣ ⎦⎣ . ⎦ ⎣ y(n−1) (t0 ) cn y1(n−1) (t0 ) y2(n−1) (t0 ) · · · y(n−1) n 0 By Theorem 3.5, initial value problem (4) has a unique solution for any choice of the initial conditions. Therefore, matrix equation (5) has a solution for any , and this means that the (n × n) coefﬁcient matrix choice of y0 , y0 , . . . , y(n−1) 0 has a nonzero determinant. [From linear algebra, if an (n × n) matrix equation Ax = b is consistent for every right-hand side b, then the matrix A is invertible. Equivalently, the determinant of A is nonzero.] As in Section 3.2, the determinant of the coefﬁcient matrix in equation (5) is called the Wronskian and is denoted as W(t): y1 (t) y2 (t) ··· yn (t) y (t) y2 (t) yn (t) 1 . W(t) = (6) .. .. . . (n−1) y (t) y2(n−1) (t) · · · y(n−1) (t) n 1

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We have just seen that if { y1 (t), y2 (t), . . . , yn (t)} is a fundamental set of solutions for (2), then the corresponding Wronskian (6) is nonzero for all t, a < t < b. Conversely, if { y1 (t), y2 (t), . . . , yn (t)} is a set of n solutions of (2) and if the corresponding Wronskian is nonzero on (a, b), then {y1 (t), y2 (t), . . . , yn (t)} is a fundamental set of solutions. To prove this, let u(t) be any solution of (2), and let y0 = u(t0 ), y0 = u (t0 ), . . . , y0(n−1) = u(n−1) (t0 ). Solve equation (5) for c1 , c2 , . . . , cn , and deﬁne yˆ (t) = c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t). Note that u(t) and yˆ (t) are both solutions of initial value problem (4). Therefore, by Theorem 3.5, u(t) = yˆ (t) = c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t), a < t < b. E X A M P L E

1

Consider the fourth order differential equation d4 y dt4

− y = 0,

−∞ < t < ∞.

It can be veriﬁed that the functions y1 (t) = e t , y2 (t) = e−t , y3 (t) = cos t, and y4 (t) = sin t are solutions of this equation. (a) Show that these four solutions form a fundamental set of solutions. (b) Represent the solution y(t) = sinh t + sin(t + π/3) in terms of this fundamental set. Solution: (a) Computing the Wronskian, we obtain t e e−t cos t sin t t −t − sin t cos t e −e = −8(cos 2 t + sin 2 t) = −8. W(t) = t e−t − cos t − sin t e t sin t − cos t e −e−t Since the Wronskian is nonzero on (−∞, ∞), it follows that { y1 , y2 , y3 , y4 } is a fundamental set of solutions. (b) We use the fact that sinh t = (e t − e−t )/2 and the trigonometric identity sin(A + B) = sin A cos B + sin B cos A to obtain √ 3 π 1 t 1 −t 1 cos t. ❖ y(t) = sinh t + sin t + = e − e + sin t + 3 2 2 2 2

Abel’s Theorem Let {y1 (t), y2 (t), . . . , yn (t)} be a set of solutions of the linear homogeneous equation (2). We have seen that { y1 (t), y2 (t), . . . , yn (t)} is a fundamental set of solutions if and only if the corresponding Wronskian, W(t), is nonzero on (a, b). Abel’s theorem shows that the Wronskian is either zero throughout (a, b) or is never zero in (a, b). This fact allows us to choose a single convenient test point, t0 , and use the value W(t0 ) to decide whether {y1 (t), y2 (t), . . . , yn (t)} is a fundamental set of solutions. We state Abel’s theorem for the general nth order case, but prove it only for n = 2.

3.11

Theorem 3.6

Higher Order Linear Homogeneous Differential Equations

191

Let y1 (t), y2 (t), . . . , yn (t) denote n solutions of the differential equation y(n) + pn−1 (t)y(n−1) + · · · + p1 (t)y + p0 (t)y = 0,

a < t < b,

where p0 (t), p1 (t), . . . , pn−1 (t) are continuous on (a, b). Let W(t) be the Wronskian of y1 (t), y2 (t), . . . , yn (t). Then the function W(t) is a solution of the ﬁrst order linear differential equation W = −pn−1 (t)W. Therefore, if t0 is any point in the interval (a, b), it follows that −

W(t) = W(t0 )e

!t t 0

pn−1 (s) ds

,

a < t < b.

(7)

PROOF: We prove Theorem 3.6 for the case n = 2. Let y1 and y2 be solutions of the second order linear equation

●

y + p1 (t)y + p0 (t)y = 0. The Wronskian is W = y1 y2 − y2 y1 . Differentiating and simplifying, we obtain W = y1 y2 − y2 y1

= y1 [−p1 (t)y2 − p0 (t)y2 ] − y2 [−p1 (t)y1 − p0 (t)y1 ] = −p1 (t)[ y1 y2 − y2 y1 ]

= −p1 (t)W. Solving the resulting ﬁrst order linear equation W = −p1 (t)W leads to (7).

●

The proof of Theorem 3.6 for general n can be found in most advanced texts on differential equations. The basic argument is similar to that used for the case n = 2. The computations are more involved, however, since one needs to compute the derivative of an (n × n) determinant of functions. By equation (7), if W(t0 ) = 0, then W(t) = 0 for all t in (a, b). On the other hand, if W(t0 ) = 0 then W(t) is also zero for all t in (a, b). The point t0 is arbitrary; Abel’s theorem implies that the Wronskian of a set of solutions of (2) either is zero throughout (a, b) or is never zero in (a, b).

Additional Observations We conclude by making some additional observations about fundamental sets of solutions of the homogeneous equation (2).

Fundamental Sets Always Exist When differential equation (2) has constant coefﬁcients, we can explicitly construct fundamental sets of solutions. For the general case of variable coefﬁcients, however, we are usually unable to explicitly construct solutions of (2). In such cases, it is logical to ask whether fundamental sets of solutions do, in fact, exist. The following theorem provides an afﬁrmative answer.

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Theorem 3.7

Consider the nth order linear homogeneous differential equation y(n) + pn−1 (t)y(n−1) + · · · + p1 (t)y + p0 (t)y = 0,

a < t < b,

where p0 (t), p1 (t), . . . , pn−1 (t) are continuous on (a, b). A fundamental set of solutions exists for this equation.

PROOF: We prove Theorem 3.7 for the case n = 2. Let t0 be any point in (a, b), and let y1 and y2 be solutions of the initial value problems

●

y1 + p1 (t)y1 + p0 (t)y1 = 0, y2 + p1 (t)y2 + p0 (t)y2 = 0,

y1 (t0 ) = 0

y1 (t0 ) = 1,

y2 (t0 ) = 1.

y2 (t0 ) = 0,

Existence-uniqueness Theorem 3.5 assures us that each of these initial value problems has a unique solution on (a, b). The fact that {y1 , y2 } forms a fundamental set of solutions follows immediately from the observation that y (t ) y (t ) 1 0 1 0 2 0 W(t0 ) = = 1. ● = y1 (t0 ) y2 (t0 ) 0 1

Fundamental Sets of Solutions Are Linearly Independent A set of functions deﬁned on a common domain, say f1 (t), f2 (t), . . . , fr (t) deﬁned on the interval a < t < b, is called a linearly dependent set if there exist constants k1 , k2 , . . . , kr , not all zero, such that k1 f1 (t) + k2 f2 (t) + · · · + kr fr (t) = 0,

a < t < b.

(8)

A set of functions that is not linearly dependent is called linearly independent. Thus, a set of functions is linearly independent if equation (8) implies k1 = k2 = · · · = kr = 0. If a set of functions is linearly dependent, then at least one of the functions can be expressed as a linear combination of the others. For example, if k1 = 0 in (8), then f1 (t) = −(k2 /k1 )f2 (t) − (k3 /k1 )f3 (t) − · · · − (kr /k1 )fr (t),

a < t < b.

Loosely speaking, the functions comprising a linearly independent set are all “basically different,” while those forming a linearly dependent set are not. The following theorem, whose proof is left to the exercises, shows that a fundamental set of solutions is a linearly independent set.

Theorem 3.8

Consider the nth order linear homogeneous differential equation y(n) + pn−1 (t)y(n−1) + · · · + p1 (t)y + p0 (t)y = 0,

a < t < b,

where p0 (t), p1 (t), . . . , pn−1 (t) are continuous on (a, b). A fundamental set of solutions for this equation is necessarily a linearly independent set of functions.

3.11

Higher Order Linear Homogeneous Differential Equations

193

How Fundamental Sets Are Related Fundamental sets of solutions for equation (2) always exist. In fact, this linear homogeneous equation has inﬁnitely many fundamental sets. These fundamental sets must be related to each other, since any given fundamental set can be used to construct all solutions of (2). The following theorem shows how these fundamental sets are related. Theorem 3.9

Let {y1 (t), y2 (t), . . . , yn (t)} be a fundamental set of solutions of the nth order linear homogeneous differential equation y(n) + pn−1 (t)y(n−1) + · · · + p1 (t)y + p0 (t)y = 0,

a < t < b,

where p0 (t), p1 (t), . . . , pn−1 (t) are continuous on (a, b). Let { y1 (t), y2 (t), . . . , yn (t)} be a set of solutions of the differential equation. Then there exists an (n × n) matrix A such that [ y1 (t), y2 (t), . . . , yn (t) ] = [y1 (t), y2 (t), . . . , yn (t)]A. Moreover, {y1 (t), y2 (t), . . . , yn (t)} is a fundamental set of solutions if and only if the determinant of A is nonzero.

PROOF: We prove Theorem 3.9 for the case n = 2. Since {y1 , y2 } is a fundamental set, we can express y1 and y2 as linear combinations of y1 and y2 : y1 = a11 y1 + a21 y2 a11 a12 = [ y1 , y2 ]A. or [y1 , y2 ] = [ y1 , y2 ] y2 = a12 y1 + a22 y2 a21 a22

●

Since the equation [ y1 , y2 ] = [ y1 , y2 ]A holds for all t in (a, b), we can differentiate it and obtain the matrix equation y1 y2 y1 y2 = A. y1 y2 y1 y2 Using the fact that the determinant of the product of two matrices is the product of their determinants, we have W(t) = W(t) det(A), where W(t) and W(t) denote the Wronskians of { y1 , y2 } and {y1 , y2 }, respectively. Since W(t) = 0 on (a, b), it follows that W(t) = 0 if and only if det(A) = 0. ●

EXERCISES Exercises 1–6: In each exercise, (a) Verify that the given functions form a fundamental set of solutions. (b) Solve the initial value problem. 1. y = 0; y1 (t) = 2,

y(1) = 4,

y (1) = 2,

y2 (t) = t − 1,

y (1) = 0

y3 (t) = t2 − 1

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Second and Higher Order Linear Differential Equations 2. y − y = 0; y1 (t) = 1, 3. y

(4)

y2 (t) = e ,

+ 4y = 0;

y2 (t) = t,

4. y + 2y = 0; y1 (t) = 1,

y3 (t) = cos 2t,

t > 0;

y4 (t) = sin 2t

y (0) = −8

y (0) = 3,

y (2) = − 54 ,

y(2) = 12 , y3 (t) = t

6. t y + ty − y = 0, y1 (t) = 1,

y (0) = 8

y3 (t) = e

y2 (t) = t,

y (0) = −4,

−2t

y2 (t) = t,

5. ty + 3y = 0, 2

y (0) = −1,

y(0) = 0,

y1 (t) = 1,

y3 (t) = e

y(0) = 0,

y (0) = 3

−t

t

y1 (t) = 1,

y (0) = 1,

y(0) = 4,

t < 0;

y2 (t) = ln(−t),

y (2) =

1 4

−1

y (−1) = −1,

y(−1) = 1,

y (−1) = −1

y3 (t) = t2

Exercises 7–10: Consider the given differential equation on the interval −∞ < t < ∞. Assume that the members of a solution set satisfy the initial conditions. Do the solutions form a fundamental set? 7. y + 2ty + t2 y = 0,

8. y + ty = 0,

y1 (1) = 2, y1 (0)

y1 (0) = 0,

9. y + (sin t)y = 0,

y1 (0) = 1,

y2 (0) = 2, 10. y + e t y + y = 0,

= 0,

y2 (1) = −4, y2 (0)

= 2,

y2 (0) = −1,

y1 (0)

y1 (0)

= −1,

= 0,

y3 (0) = −2,

y3 (0) = 2,

y1 (1) = 0,

y2 (1)

y1 (1) = −1,

y1 (1) = 1,

y3 (1) = −1,

y1 (1) = y3 (1) = 0,

y2 (1) = 2

=0 y2 (0) = 0,

y2 (0) = 0,

y3 (0) = 1 y2 (1) = y3 (1) = 0

1,

1,

y2 (1) = −1,

Exercises 11–15: The given differential equation has a fundamental set of solutions whose Wronskian W(t) is such that W(0) = 1. What is W(4)? t t 13. y + y + ty = 0 11. y + y + y = 0 12. y + y + y = 0 2 2 15. (t2 + 1)y − 2ty + y = 0 14. y(4) − y + y = 0

Exercises 16–19: Find a fundamental set { y1 , y2 } satisfying the given initial conditions. 16. y − y = 0,

y1 (0) = 1,

y1 (0) = 0,

y2 (0) = 0,

y2 (0) = 1

17. y + y = 0,

y1 (0) = 1,

y1 (0) = 1,

y2 (0) = 1,

y2 (0) = −1

18. y + 4y + 5y = 0,

19. y + 4y + 4y = 0,

y1 (0) = 1, y1 (0) = −1,

y1 (0) = −1, y1 (0)

= 2,

y2 (0) = 0,

y2 (0) = 1

y2 (0) = 1,

y2 (0) = −1

Exercises 20–22: In each exercise, { y1 , y2 , y3 } is a fundamental set of solutions and { y1 , y2 , y3 } is a set of solutions. (a) Find a (3 × 3) constant matrix A such that [ y1 (t), y2 (t), y3 (t)] = [ y1 (t), y2 (t), y3 (t)]A. (b) Determine whether { y1 , y2 , y3 } is also a fundamental set by calculating det(A). 20. y − y = 0,

{ y1 (t), y2 (t), y3 (t)} = {1, e t , e−t }, { y1 (t), y2 (t), y3 (t)} = {cosh t, 1 − sinh t, 2 + sinh t}

21. y − y = 0,

{ y1 (t), y2 (t), y3 (t)} = {1, t, e−t }, { y1 (t), y2 (t), y3 (t)} = {1 − 2t, t + 2, e−(t+2) }

3.12

Higher Order Homogeneous Constant Coefﬁcient Differential Equations

22. t2 y + ty − y = 0, t > 0,

195

{ y1 (t), y2 (t), y3 (t)} = {1, ln t, t2 }, { y1 (t), y2 (t), y3 (t)} = {2t2 − 1, 3, ln t3 }

Exercises 23–26: The Wronskian formed from a solution set of the given differential equation has the speciﬁed value at t = 1. Determine W(t). 23. y − 3y + 3y − y = 0;

W(1) = 1

24. y + (sin t)y + (cos t)y + 2y = 0; 3

2

25. t y + t y − 2y = 0, 3

26. t y − 2y = 0,

t > 0;

t > 0;

W(1) = 0

W(1) = 3

W(1) = 3

27. Linear Independence For deﬁniteness, consider Theorem 3.8 in the case n = 3. Let { y1 , y2 , y3 } be a fundamental set of solutions for y + p2 (t)y + p1 (t)y + p0 (t)y = 0, where p0 , p1 , p2 are continuous on a < t < b. Show that the fundamental set is linearly independent. [Hint: Consider the equation k1 y1 (t) + k2 y2 (t) + k3 y3 (t) = 0 along with the ﬁrst and second derivatives of this equation, evaluated at some point t0 in a < t < b.]

3.12 Higher Order Homogeneous Constant Coeﬃcient Diﬀerential Equations

Consider the nth order linear homogeneous differential equation y(n) + an−1 y(n−1) + · · · + a1 y + a0 y = 0,

−∞ < t < ∞,

(1)

where a0 , a1 , . . . , an−1 are real constants. The general solution is given by y(t) = c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t), where { y1 (t), y2 (t), . . . , yn (t)} is a fundamental set of solutions. How do we determine a fundamental set of solutions for equation (1)? As in Section 3.3, we look for solutions of the form y(t) = eλt , where λ is a constant to be determined. Substituting this form into equation (1) leads to (λn + an−1 λn−1 + · · · + a1 λ + a0 )eλt = 0,

−∞ < t < ∞.

The exponential function eλt is nonzero for all values of λ (whether real or complex). Therefore, in order for y(t) = eλt to be a solution, we need λn + an−1 λn−1 + · · · + a1 λ + a0 = 0.

(2)

The nth degree polynomial in equation (2) is called the characteristic polynomial, while equation (2) itself is called the characteristic equation. The roots of the characteristic equation deﬁne solutions of (1) having the form y(t) = eλt .

Roots of the Characteristic Equation When we considered the case n = 2, we were able to list three possibilities for the roots of the characteristic polynomial (two distinct real roots, one repeated real root, or two distinct complex roots). While we cannot list all the possibilities for the general case, we can make some useful observations.

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Complex Roots Occur in Complex Conjugate Pairs Since the coefﬁcients a0 , a1 , . . . , an−1 are real constants, the complex-valued roots of the characteristic equation always occur in complex conjugate pairs. One simple consequence of this observation is the fact that every characteristic polynomial of odd degree has at least one real root. We noted in Section 3.5 that if λ = α + iβ and λ = α − iβ are a complex conjugate pair of roots of the characteristic equation, then real-valued solutions corresponding to these two roots are y1 (t) = eαt cos βt

and

y2 (t) = eαt sin βt.

(3)

This result is true for the general nth order linear homogeneous equation as well. E X A M P L E

1

Find the general solution of the third order differential equation y + 4y = 0,

−∞ < t < ∞.

Solution: Looking for solutions of the form eλt leads to the characteristic equation λ3 + 4λ = 0, or λ(λ2 + 4) = λ(λ + 2i)(λ − 2i) = 0. The three roots are therefore λ1 = 0, λ2 = 2i, and λ3 = λ2 = −2i. The corresponding real-valued solutions are y1 (t) = 1,

y2 (t) = cos 2t,

y3 (t) = sin 2t.

To show these three solutions constitute a fundamental set, we calculate the Wronskian and ﬁnd 1 cos 2t sin 2t W(t) = 0 −2 sin 2t 2 cos 2t = 8(sin 2 2t + cos 2 2t) = 8. 0 −4 cos 2t −4 sin 2t Since the Wronskian is nonzero, the three solutions form a fundamental set of solutions. The general solution of the differential equation is y(t) = c1 + c2 cos 2t + c3 sin 2t,

−∞ < t < ∞.

In this example, we computed the determinant W(t). But, recalling Abel’s theorem, we could have simply evaluated W(t) at a convenient point, say t = 0. Note, for this equation, that Abel’s theorem would also have anticipated that W(t) is a constant function. ❖

If the Characteristic Polynomial Has Distinct Roots, the Corresponding Solutions Form a Fundamental Set In Example 1, the characteristic equation had three distinct roots: the real number λ1 = 0 and the complex conjugate pair λ2 = 2i and λ3 = −2i. The corresponding set of solutions formed a fundamental set of solutions. In the exercises, we ask you to show that this result holds in general. That is, if the characteristic equation has n distinct roots λ1 , λ2 , . . . , λn , then the set of solutions {eλ1 t , eλ2 t , . . . , eλn t } forms a fundamental set of solutions. (If some of these

3.12

Higher Order Homogeneous Constant Coefﬁcient Differential Equations

197

distinct roots are complex-valued, we can replace the conjugate pair of complex exponentials with the corresponding real-valued pair of solutions and the conclusion remains the same.)

Roots of the Characteristic Equation May Have Multiplicity Greater than 2 and May Be Complex For the second order linear homogeneous equations discussed earlier, a repeated root of the corresponding quadratic characteristic equation is, of necessity, real-valued. If λ1 is a repeated root of the quadratic characteristic equation, then the corresponding solutions y1 (t) = eλ1 t

and

y2 (t) = teλ1 t

form a fundamental set of solutions. For higher order differential equations, the situation is more complicated. For example, (a) A root can be repeated more than once. polynomial has the form

In particular, if the characteristic

p(λ) = λn + an−1 λn−1 + · · · + a1 λ + a0 = (λ − λ1 )r q(λ),

(4)

where q is a polynomial of degree n − r and q(λ1 ) = 0, then we say λ1 is a root of multiplicity r. If λ1 is a root of multiplicity r, then the functions eλ1 t ,

teλ1 t ,

t2 eλ1 t ,

t r−1 eλ1 t

...,

(5)

form a set of r solutions of the differential equations. The remaining n − r solutions needed to form a fundamental set of solutions can be determined by examining the roots of q(λ) = 0. (b) A repeated root might be complex. We recall, however, that complex roots arise in complex conjugate pairs. Therefore, if λ1 = α + iβ is a root of multiplicity r, then λ1 = α − iβ is also a root of multiplicity r. In such cases, the characteristic polynomial has the form ˆ λn + an−1 λn−1 + · · · + a1 λ + a0 = (λ − λ1 )r (λ − λ1 )r q(λ), ˆ ˆ 1 ) = 0 and where q(λ) is a polynomial of degree n − 2r and where q(λ ˆ 1 ) = 0. In this case, the functions q(λ eαt cos βt,

teαt cos βt,

t2 eαt cos βt,

eαt sin βt,

teαt sin βt,

t2 eαt sin βt,

..., ...,

t r−1 eαt cos βt t r−1 eαt sin βt

(6)

form a set of 2r real-valued solutions. E X A M P L E

2

Find the general solution of (a) y(6) + 3y(5) + 3y(4) + y = 0

(b) y(5) − y(4) + 2y − 2y + y − y = 0

Solution: (a) The characteristic polynomial is p(λ) = λ6 + 3λ5 + 3λ4 + λ3 = λ3 (λ3 + 3λ2 + 3λ + 1) = λ3 (λ + 1)3 . (continued)

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CHAPTER 3

Second and Higher Order Linear Differential Equations (continued)

Therefore, λ = 0 and λ = −1 are each roots of multiplicity 3. By (5), the functions {1, t, t2 , e−t , te−t , t2 e−t } are solutions. The Wronskian can be shown to be nonzero, and hence the solutions form a fundamental set; the general solution is y(t) = c1 + c2 t + c3 t2 + c4 e−t + c5 te−t + c6 t2 e−t . (b) The characteristic polynomial is p(λ) = λ5 − λ4 + 2λ3 − 2λ2 + λ − 1 = (λ − 1)(λ2 + 1)2 . The characteristic equation has one real root, λ = 1, and a repeated pair of complex conjugate roots, ±i. The set of solutions {e t , cos t, sin t, t cos t, t sin t} forms a fundamental set of solutions. The general solution is y(t) = c1 e t + c2 cos t + c3 sin t + c2 t cos t + c3 t sin t. ❖ It can be shown that solutions (5) and (6), arising from repeated roots, will always form a fundamental set of solutions when they are combined with the solutions that arise from any remaining roots of the characteristic equation.

Solving the Differential Equation y (n) – ay = 0 Let a be a real number. The characteristic equation for y(n) − ay = 0 is λn − a = 0. Finding the roots of this equation amounts to ﬁnding the n different nth roots of the real number a. The ﬁrst step is to write a in polar form as a = Reiα , where R = |a| and where α = 0 when a > 0 and α = π when a < 0. Recall Euler’s formula from Section 3.5, eiθ = cos θ + i sin θ. For any integer k, it follows that ei2kπ = 1. Therefore, we can write a = Rei(α+2kπ ) , k = 0, ±1, ±2, . . . , and hence a1/n = R1/n ei(α+2kπ)/n ,

k = 0, ±1, ±2, . . . .

(7)

In equation (7), R1/n is the positive real nth root of R = |a|. We generate the n distinct roots of λn − a = 0 by setting k = 0, 1, . . . , n − 1 in equation (7). (Other integer values of k simply replicate these values.) Once we determine the n roots, we can use Euler’s formula to rewrite them in the form x + iy. E X A M P L E

3

Find the general solution of y(4) + 16y = 0,

−∞ < t < ∞.

Solution: The characteristic equation is λ4 + 16 = 0. Therefore, in this example, a = −16 and n = 4. Using equation (7) with R = 16 and α = π , we ﬁnd the four roots are λk = 2ei(π+2kπ )/4 ,

k = 0, 1, 2, 3,

3.12

Higher Order Homogeneous Constant Coefﬁcient Differential Equations

or

199

√ √ 2+i 2 √ √ =− 2+i 2 √ √ =− 2−i 2 √ √ = 2 − i 2.

λ0 = 2e iπ/4 = λ1 = 2e i3π/4 λ2 = 2e i5π/4 λ3 = 2e i7π/4

The general solution, expressed in terms of real-valued solutions, is √ √ √ √ √ √ y(t) = e 2 t c1 cos 2 t + c2 sin 2 t + e− 2 t c3 cos 2 t + c4 sin 2 t . ❖ When a is a real number, the complex roots of λn − a = 0 occur in complex conjugate pairs. If we plot these n roots in the complex plane, we see they all lie on a circle of radius |a|1/n and the angular separation between roots is 2π/n. Figure 3.20 shows the four roots of λ4 + 16 = 0 found in Example 3. Im

√2 + i√2

–√2 + i√2

Re

√2 – i√2

–√2 – i√2

FIGURE 3.20 4

The four roots of λ + 16 = 0 lie on a circle of radius 161/4 = 2 and have an angular separation of 2π/4 = π/2 radians. The roots occur in two complex conjugate pairs.

EXERCISES Exercises 1–18: In each exercise, (a) Find the general solution of the differential equation. (b) If initial conditions are speciﬁed, solve the initial value problem. 1. y − 4y = 0

2. y + y − y − y = 0

3. y + y + 4y + 4y = 0

4. 16y(4) − 8y + y = 0

5. 16y(4) + 8y + y = 0

6. y − y = 0

7. y − 2y − y + 2y = 0

8. y(4) − y = 0

9. y + 8y = 0

10. 2y − y = 0

11. y + y = 0

13. y(6) − y = 0

14. y(4) − y + y − y = 0

15. y + 2y + y = 0,

y(0) = 0, y (0) = 0, y (0) = 1

12. y(4) + 2y + y = 0

200

CHAPTER 3

Second and Higher Order Linear Differential Equations 16. y + 4y = 0,

y(0) = 1, y (0) = 6, y (0) = 4

17. y + 3y + 3y + y = 0, (4)

18. y

− y = 0,

y(0) = 0, y (0) = 1, y (0) = 0

y(0) = 0, y (0) = 0, y (0) = 0, y (0) = 1

Exercises 19–20: The given differential equations have repeated roots; α and β are real constants. (a) Use equation (5) or (6), as appropriate, to determine a solution set. (b) Show that the solution set found in part (a) is a fundamental set of solutions by evaluating the Wronskian at t = 0. 19. y − 3αy + 3α 2 y − α 3 y = 0

20. y(4) + 2β 2 y + β 4 y = 0

Exercises 21–25: In each exercise, you are given the general solution of y(4) + a3 y + a2 y + a1 y + a0 y = 0, where a3 , a2 , a1 , and a0 are real constants. Use the general solution to determine the constants a3 , a2 , a1 , and a0 . [Hint: Construct the characteristic equation from the given general solution.] 21. y(t) = c1 + c2 t + c3 cos 3t + c4 sin 3t 22. y(t) = c1 cos t + c2 sin t + c3 cos 2t + c4 sin 2t 23. y(t) = c1 e t + c2 te t + c3 e−t + c4 te−t 24. y(t) = c1 e−t sin t + c2 e−t cos t + c3 e t sin t + c4 e t cos t 25. y(t) = c1 e t + c2 te t + c3 t2 e t + c4 t3 e t

Exercises 26–30: Consider the nth order homogeneous linear differential equation y(n) + an−1 y(n−1) + · · · + a3 y + a2 y + a1 y + a0 y = 0, where a0 , a1 , a2 , . . . , an−1 are real constants. In each exercise, several functions belonging to a fundamental set of solutions for this equation are given. (a) What is the smallest value n for which the given functions can belong to such a fundamental set? (b) What is the fundamental set? 26. y1 (t) = t,

y2 (t) = e t ,

t

t

27. y1 (t) = e ,

y2 (t) = e cos 2t,

2

28. y1 (t) = t sin t, 29. y1 (t) = t sin t, 30. y1 (t) = t2 ,

y3 (t) = cos t y3 (t) = e−t cos 2t

t

y2 (t) = e sin t y2 (t) = t2 e t

y2 (t) = e2t

Exercises 31–35: Consider the nth order differential equation y(n) − ay = 0, where a is a real number. In each exercise, some information is presented about the solutions of this equation. Use the given information to deduce both the order n (n ≥ 1) of the differential equation and the value of the constant a. (If more than one answer is possible, determine the smallest possible order n and the corresponding value of a.) 31. |a| = 1 and lim t→∞ y(t) = 0 for all solutions y(t) of the equation.

3.13

Higher Order Linear Nonhomogeneous Differential Equations

201

32. |a| = 2 and all nonzero solutions of the differential equation are exponential functions. 33. y(t) = t3 is a solution of the differential equation. 34. |a| = 4 and all solutions of the differential equation are bounded functions on the interval −∞ < t < ∞. √ 35. Two solutions are y1 (t) = e−t and y2 (t) = e t/2 sin( 3 t/2). 36. Assume the characteristic equation of y(n) + an−1 y(n−1) + · · · + a1 y + a0 y = 0 has distinct roots λ1 , λ2 , . . . , λn . It can be shown that the Vandermonde8 determinant has the value 1 1 ··· 1 λ λn λ2 1 n ' 2 2 2 λ1 λ λ = (λi − λj ). 2 n . .. i, j=1 .. . i>j λn−1 λn−1 · · · λn−1 1

Use this fact to show that {e

2

λ1 t

n

λ2 t

,e

,...,e

λn t

} is a fundamental set of solutions.

3.13 Higher Order Linear Nonhomogeneous Diﬀerential Equations

We now consider the nth order linear nonhomogeneous differential equation y(n) + pn−1 (t)y(n−1) + · · · + p1 (t)y + p0 (t)y = g(t),

a < t < b.

(1)

The arguments made in Section 3.7, establishing the solution structure for second order equations, apply as well in the nth order case. The arguments rely on the fact that the differential equation is linear; they do not depend on the order of the equation. The solution structure of equation (1) can be represented schematically as The general solution The general solution A particular solution of the nonhomogeneous = of the homogeneous + of the nonhomogeneous equation equation equation.

The general solution of the homogeneous equation is the complementary solution, denoted by yC (t). The one solution of the nonhomogeneous equation that we have somehow found is the particular solution, denoted by yP (t). The general solution of nonhomogeneous equation (1) has the form y(t) = yC (t) + yP (t).

Finding a Particular Solution In Sections 3.8 and 3.9, we discussed the method of undetermined coefﬁcients and the method of variation of parameters for constructing particular solutions.

8

Alexandre-Theophile Vandermonde (1735–1796) was a violinist who turned to mathematics when he was 35 years old. His four published mathematical papers made noteworthy contributions to the study of algebraic equations, topology, combinatorics, and determinants. Surprisingly, the determinant that now bears his name appears nowhere in his published works.

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Second and Higher Order Linear Differential Equations

Both methods have straightforward generalizations to nth order equations. With respect to undetermined coefﬁcients, Table 3.1 in Section 3.8 applies to higher order equations as well as second order equations. The only change is that the integer r in Table 3.1 can now range as high as the order of the equation, n. We illustrate the method of undetermined coefﬁcients in Example 1.

E X A M P L E

1

Choose an appropriate form for a particular solution of y(6) + 3y(5) + 3y(4) + y = t + 2te−t + sin t,

−∞ < t < ∞.

Solution: The ﬁrst step is to ﬁnd the complementary solution. The characteristic polynomial is λ6 + 3λ5 + 3λ4 + λ3 = λ3 (λ3 + 3λ2 + 3λ + 1) = λ3 (λ + 1)3 . Since λ = 0 and λ = −1 are repeated roots, the general solution of the homogeneous equation is yC (t) = c1 + c2 t + c3 t2 + c4 e−t + c5 te−t + c6 t2 e−t . Therefore, the method of undetermined coefﬁcients suggests that we look for a particular solution having the form yP (t) = t3 (A1 t + A0 ) + t3 (B1 te− t + B0 e− t ) + C cos t + D sin t. The t3 multipliers ensure that no term in the assumed form is a solution of the homogeneous equation. ❖

Variation of Parameters The method of variation of parameters, discussed in Section 3.9, can be extended to ﬁnd a particular solution of a linear nonhomogeneous nth order equation. As with second order equations, we assume we know a fundamental set of solutions of the homogeneous equation, { y1 (t), y2 (t), . . . , yn (t)}. The complementary solution is therefore yC (t) = c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t),

a < t < b.

We now “vary the parameters,” by replacing the constants c1 , c2 , . . . , cn with functions u1 (t), u2 (t), . . . , un (t), and assume a particular solution of the form yP (t) = y1 (t)u1 (t) + y2 (t)u2 (t) + · · · + yn (t)un (t),

a < t < b.

(2)

The functions u1 (t), u2 (t), . . . , un (t) must be chosen so that (2) is a solution of nonhomogeneous equation (1). However, since there are n functions in (2), we are free to impose n − 1 additional constraints on the n functions. Speciﬁcally, we impose the following n − 1 constraints: y1 u1 + y2 u2 + · · · + yn un = 0 y1 u1 + y2 u2 + · · · + yn un = 0 y1 u1 + y2 u2 + · · · + yn un = 0 .. . u1 + y2(n−2) u2 + · · · + yn(n−2) un = 0. y(n−2) 1

(3)

3.13

Higher Order Linear Nonhomogeneous Differential Equations

203

The purpose of (3) is to make successive derivatives of yP (t) [where yP (t) is deﬁned by equation (2)] have the following simple forms: yP = y1 u1 + y2 u2 + · · · + yn un yP = y1 u1 + y2 u2 + · · · + yn un .. .

(4)

y(n−1) = y1(n−1) u1 + y(n−1) u2 + · · · + yn(n−1) un . P 2 When we substitute representation (2) for yP into differential equation (1), use (3), and also use the fact that each of the functions y1 , y2 , . . . , yn is a solution of the homogeneous equation, we obtain u2 + · · · + y(n−1) un = g. y1(n−1) u1 + y(n−1) n 2

(5)

Taken together, equations (3) and (5) form a set of n linear equations for the n unknowns, u1 , u2 , . . . , un . In matrix form, this system of equations is ⎡ ⎤⎡ ⎤ ⎡ ⎤ y1 y2 ··· yn u1 0 ⎥⎢ ⎥ ⎢ ⎢ y y2 ··· yn ⎥ ⎢u2 ⎥ ⎢0⎥ ⎥ ⎢ 1 ⎢ ⎥ ⎢ ⎥ ⎢ . a < t < b. (6) .. ⎥ ⎢ . ⎥ ⎢ .. ⎥ = ⎢ .. ⎥ , . ⎦⎣ . ⎦ ⎣.⎦ ⎣ . g un y1(n−1) y2(n−1) · · · yn(n−1) The determinant of the (n × n) coefﬁcient matrix is the Wronskian of a fundamental set of solutions, { y1 , y2 , . . . , yn }. Therefore, since the Wronskian determinant is nonzero for all values t in the interval (a, b), the system (6) has a unique solution for the unknowns u1 , u2 , . . . , un . Once these n functions are determined, we compute u1 , u2 , . . . , un by antidifferentiation and form yP as prescribed by equation (2). In principle, the method of variation of parameters is very general. However, the practical limitations noted in Section 3.9 for the second order case also apply to the nth order case. If the differential equation coefﬁcients are not constants, it may be very difﬁcult to determine a fundamental set of solutions of the homogeneous equation. Even if we know a fundamental set, it may be impossible to express the antiderivatives of u1 , u2 , . . . , un in terms of known functions. The following example, however, is one in which the entire computational program of variation of parameters can be performed explicitly. E X A M P L E

2

Consider the nonhomogeneous differential equation t3 y − 3t2 y + 6ty − 6y = t,

0 < t < ∞.

(7)

(a) Verify that the functions y1 (t) = t, y2 (t) = t2 , y3 (t) = t3 form a fundamental set of solutions for the associated homogeneous equation. (b) Use variation of parameters to ﬁnd a particular solution of the nonhomogeneous equation. Solution: (a) In Section 8.3, we discuss methods of ﬁnding solutions for homogeneous equations such as t3 y − 3t2 y + 6ty − 6y = 0. For now, the fact that these (continued)

204

CHAPTER 3

Second and Higher Order Linear Differential Equations (continued)

functions are solutions can be veriﬁed by direct substitution. The Wronskian of the solutions y1 (t) = t, y2 (t) = t2 , y3 (t) = t3 is nonzero on 0 < t < ∞: t t2 t3 W(t) = 1 2t 3t2 = 2t3 . 0 2 6t Therefore, the given functions form a fundamental set of solutions. (b) Assume a particular solution of the form yP = tu1 (t) + t2 u2 (t) + t3 u3 (t). If we want to use equations (5) and (6), we must ﬁrst divide equation (7) by t3 to put it in the standard form (1); this step is necessary in order to properly identify the nonhomogeneous term, g(t). Here, the function g(t) is given by g(t) = t−2 . Using equation (6), we arrive at ⎤⎡ ⎤ ⎡ ⎤ ⎡ 0 u1 t t2 t3 ⎢ ⎢ ⎥ 2⎥ ⎢ ⎥ (8) ⎣1 2t 3t ⎦ ⎣u2 ⎦ = ⎣ 0 ⎦ . 0 Solving system (8), we ﬁnd

2

6t

u3

t−2

⎤ ⎡ ⎤ ⎡ 1/(2t) u1 ⎢ ⎥ ⎢ 2 ⎥ ⎣u2 ⎦ = ⎣ −1/t ⎦ . u3

1/(2t3 )

We obtain the functions u1 , u2 , u3 by computing convenient antiderivatives: u1 (t) =

1 1 ln |t| = ln t, 2 2

u2 (t) =

1 , t

u3 (t) = −

where 0 < t < ∞. Thus, one particular solution is

3t t 1 −1 t 2 3 = ln t + , yP (t) = ln t + t +t 2 2 t 2 4 4t

1 , 4t2

0 < t < ∞.

Since the term 3t/4 is a solution of the homogeneous equation, we can dispense with it and use a simpler particular solution, yP (t) =

t ln t, 2

0 < t < ∞.

The general solution is y(t) = yC (t) + yP (t) = c1 t + c2 t2 + c3 t3 +

EXERCISES Exercises 1–14: For each differential equation, (a) Find the complementary solution. (b) Find a particular solution. (c) Formulate the general solution.

t ln t, 2

0 < t < ∞. ❖

3.13

Higher Order Linear Nonhomogeneous Differential Equations

1. y − y = e2t

2. y − y = 4 + 2 cos 2t

4. y − y = −4e t

7. y − 2y + y = t + 4e

8. y − 3y + 3y − y = 12e t

10. y + y = e t + cos t

11. y(4) − y = t + 1 13. y + y = t

6. y − y = 4e−2t

t

9. y − y = e t

3. y − y = 4t

5. y + y = 6e−t

205

12. y(4) − y = cos 2t 14. y + y = 4

3

Exercises 15–21: For each differential equation, (a) Find the complementary solution. (b) Formulate the appropriate form for the particular solution suggested by the method of undetermined coefﬁcients. You need not evaluate the undetermined coefﬁcients. 15. y − 4y + 4y = t3 + 4t2 e2t (4)

17. y

16. y − 3y + 3y − y = e t + 4e t cos 3t + 4 18. y(4) + 8y + 16y = t cos 2t

− 16y = 4t sin 2t

19. y(4) − y = te−t + (3t + 4) cos t

20. y − y = t2 + cos t

21. y(4) + 4y = e t sin t

Exercises 22–24: Consider the nonhomogeneous differential equation y + ay + by + cy = g(t). In each exercise, the general solution of the differential equation is given, where c1 , c2 , and c3 represent arbitrary constants. Use this information to determine the constants a, b, c and the function g(t). 22. y = c1 + c2 t + c3 e2t + 4 sin 2t 2

24. y = c1 + c2 t + c3 t − 2t

23. y = c1 sin 2t + c2 cos 2t + c3 e t + t2

3

Exercises 25–26: Consider the nonhomogeneous differential equation t3 y + at2 y + bty + cy = g(t),

t > 0.

In each exercise, the general solution of the differential equation is given, where c1 , c2 , and c3 represent arbitrary constants. Use this information to determine the constants a, b, c and the function g(t). 25. y = c1 + c2 t + c3 t3 + t4

26. y = c1 t + c2 t2 + c3 t4 + 2 ln t

27. (a) Verify that {t, t2 , t4 } is a fundamental set of solutions of the differential equation t3 y − 4t2 y + 8ty − 8y = 0. (b) Find the general solution of

√ t3 y − 4t2 y + 8ty − 8y = 2 t,

t > 0.

[Hint: Cramer’s rule can be used to solve the system of equations arising in the method of variation of parameters.] 28. Using the information of Exercise 27(a), ﬁnd the general solution of t3 y − 4t2 y + 8ty − 8y = 2t,

t > 0.

29. Using the information of Exercise 27(a), ﬁnd the general solution of t3 y − 4t2 y + 8ty − 8y = 6t 3 ,

t > 0.

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CHAPTER 3

Second and Higher Order Linear Differential Equations

Exercises 30–33: Find the solution of the differential equation that satisﬁes the given conditions. 30. y(4) − y = e−t ,

y(0) = 0,

31. y + y = cos t,

y(0) = 1,

32. y + y = 4e−2t , 33. y + y = e−t ,

lim t→∞ y(t) = 0 | y(t)| ≤ 2 for all t,

y(0) = 2,

y(0) = 1,

0≤t 0 c2 > 0 (b) c1 < 0 c2 = 0 (c) c1 > 0 c2 < 0

(c) FIGURE 4.6

When A = αI, phase plane trajectories move along rays emanating from the origin. Here, where α < 0, motion is toward the origin.

If A is defective, the general solution has the form y(t) = c1 eαt v1 + c2 eαt (tv1 + v2 ). In this case, as in equation (11), motion is toward the origin when α < 0 and away from the origin when α > 0. Unlike the situation illustrated in Figure 4.6, however, the solution points need not move along rays. Figure 4.7 shows the phase plane direction ﬁeld for the linear system treated in Example 1. In this example, the general solution was found to be

3t −1 3t −t + 1 y(t) = c1 e (12) + c2 e . 1 t As can be seen from Figure 4.7, all trajectories move outward from the origin. If c2 = 0 in (12), the solution points move outward along the line y2 = −y1 . If c2 = 0, then y2 (t) ≈ −y1 (t) for large values of t. y2 2 1.5 1 0.5

y1 –2 –1.5 –1 –0.5 –0.5

0.5

1

1.5

2

–1 –1.5 –2

FIGURE 4.7

The phase plane direction ﬁeld for Example 1. From the general solution in equation (12), we see that phase plane solution trajectories approach the line y2 = −y1 as t increases.

274

CHAPTER 4

First Order Linear Systems

EXERCISES Exercises 1–12: Consider the given initial value problem y = Ay, y(t0 ) = y0 . (a) Find the eigenvalues and eigenvectors of the coefﬁcient matrix A. (b) Construct a fundamental set of solutions. (c) Solve the initial value problem. 1. y1 = 2y1 + y2 , y2 3.

y

5.

y

7.

y

9. y

y1 (0) = 3

= −y1 , y2 (0) = −1 −2 1 1 y, y(0) = = 0 −2 −1 6 0 −2 = y, y(0) = 2 6 0

−4 1 1 y, y(−1) = = −1 −2 0 ⎡ ⎤ ⎡ ⎤ 2 1 0 1 ⎢ ⎥ ⎢ ⎥ = ⎣0 2 0⎦ y, y(0) = ⎣ 3⎦ 0 0 1 −2

11. y1 = 0, y2 = 5y2 − y3 ,

2. y1 =

y2 ,

y1 (0) = 0

y2 = −y1 − 2y2 , y2 (0) = 2 4. 1 −1 1 y = y, y(0) = 4 5 1 6. y1 =

y2 , y1 (1) = −1

y2 = −y1 + 2y2 , y2 (1) = 2

8. 6 1 4 y, y(0) = y = −1 4 −4 ⎡ ⎤ ⎡ ⎤ 10. 1 −1 0 −3 ⎢ ⎥ ⎢ ⎥ 3 0⎦ y, y(0) = ⎣−2⎦ y = ⎣1 0 0 2 0

y1 (0) = 4 y2 (0) = 1

y3 = 4y2 + y3 , y3 (0) = 1 ⎡ ⎤ ⎡ ⎤ 12. 0 −3 0 −36 ⎢ ⎥ ⎢ ⎥ 0⎦ y, y(0) = ⎣−1⎦ y = ⎣ 0 1 2 1 0 9 13. Find the eigenvalues and eigenvectors of ⎡ ⎡ ⎤ ⎤ 2 1 0 2 1 0 ⎢ ⎢ ⎥ ⎥ (a) A1 = ⎣0 2 1⎦ (b) A2 = ⎣0 2 0⎦ 0 0 2 0 0 2 14. Consider the homogeneous linear system y = A1 y, where A1 is given in Exercise 13. (a) Write the three component differential equations of y = A1 y, and solve these equations sequentially, ﬁnding ﬁrst y3 (t), then y2 (t), and then y1 (t). [For example, the third component equation is y3 = 2y3 . Therefore, y3 (t) = c3 e2t .] (b) Rewrite the component solutions obtained in part (a) as a single matrix equation of the form y(t) = (t)c, where (t) is a (3 × 3) solution matrix and ⎡ ⎤ c1 ⎢ ⎥ c = ⎣c2 ⎦ . c3 Show that (t) is, in fact, a fundamental matrix. [Note that this observation is consistent with the fact that the component solutions obtained in part (a) form the general solution.]

4.7

Repeated Eigenvalues

275

15. Repeat Exercise 14 for the homogeneous linear system y = A2 y. 16. The Scalar Repeated-Root Equation Revisited Consider the homogeneous scalar equation y − 2αy + α 2 y = 0, where α is a real constant. Recall from Section 3.4 that the general solution is y(t) = c1 eαt + c2 teαt . (a) Recast y − 2αy + α 2 y = 0 as a ﬁrst order linear system z = Az. (b) Show that the (2 × 2) matrix A has eigenvalue α with algebraic multiplicity 2 and geometric multiplicity 1. (c) Obtain the general solution of z = Az. As a check, does z1 (t) equal the general solution y(t) = c1 eαt + c2 teαt ? Is z2 (t) equal to z1 (t)?

Exercises 17–23: For each matrix A, ﬁnd the eigenvalues and eigenvectors. Give the geometric and algebraic multiplicity of each eigenvalue. Does A have a full set of eigenvectors? ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 5 0 0 5 0 0 5 0 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 17. A = ⎣1 5 0⎦ 18. A = ⎣1 5 0⎦ 19. A = ⎣0 5 0⎦ 1 1 5 1 0 5 0 0 5 ⎡ ⎤ 2 0 0 0 ⎡ ⎤ 5 0 0 ⎢1 2 0 0⎥ ⎢ ⎥ ⎢ ⎥ 20. A = ⎣0 5 0⎦ 21. A = ⎢ ⎥ ⎣0 0 3 0⎦ 0 0 5 0 0 1 3 ⎤ ⎡ ⎡ ⎤ 2 0 0 0 2 0 0 0 ⎢0 2 0 0⎥ ⎢0 2 0 0⎥ ⎥ ⎢ ⎢ ⎥ 23. A = ⎢ 22. A = ⎢ ⎥ ⎥ ⎣0 0 2 0⎦ ⎣0 0 2 0⎦ 0

0

1

2

0

0

1

3

24. Let A be a real (2 × 2) matrix having repeated eigenvalue λ1 = λ2 = α and a full set of eigenvectors, x1 and x2 . Show that A = αI. [Hint: Let T = [x1 , x2 ] be the invertible (2 × 2) matrix whose columns are the eigenvectors. Show that AT = αT.]

Exercises 25–28: In each exercise, the general solution of the linear system y = Ay is given. Determine the coefﬁcient matrix A.

−2t c1 e 26. 25. y1 (t) = c1 e−t (1 + 2t) + 4c2 te−t y(t) = −2t −t −t c y2 (t) = −c1 te + c2 e (1 − 2t) 2e

t

e (1 + t) −te t 27. + c2 t y(t) = c1 e (1 − t) te t

t/2 c1 e 0 28. y(t) = t/2 c2 0 e 29. Match the linear system with one of the phase plane direction ﬁelds on the next page.

2 0 (a) y1 = −y1 (b) y = y 0 2 y2 = 2y1 − y2

−0.5 0 (c) y1 = y1 + y2 (d) y = y 0 −0.5 y2 = y2

276

CHAPTER 4

First Order Linear Systems y2

y2

2

2

1.5

1.5

1

1

0.5

0.5

y1 –2 –1.5 –1 –0.5 –0.5

0.5

1

1.5

y1 –2 –1.5 –1 –0.5 –0.5

2

–1

–1

–1.5

–1.5

–2

–2

0.5

1

Direction Field 1

Direction Field 2

y2

y2

2

2

1.5

1.5

1

1

0.5

0.5

1.5

2

1.5

2

y1

y1 –2 –1.5 –1 –0.5 –0.5

0.5

1

1.5

2

–2 –1.5 –1 –0.5 – 0.5

–1

–1

–1.5

–1.5

–2

–2

Direction Field 3

0.5

1

Direction Field 4 Figure for Exercises 29

Exercises 30–33: Consider the linear system y = Ay, where A is a real (2 × 2) constant matrix with repeated eigenvalues. Use the given information to determine the matrix A. 30. Phase plane solution trajectories have horizontal tangents on the line y2 = 2y1 and vertical tangents on the line y1 = 0. The matrix A has a nonzero repeated eigenvalue and a21 = −1. 31. All nonzero phase plane solution points move away from the origin on straight line paths as t increases. In addition, a22 = 32 . 32. Phase plane solution trajectories have horizontal tangents on the line y2 = 0 and vertical tangents on the line y2 = 2y1 . All nonzero phase plane solution points move away from the origin as t increases. In addition, a12 = −1. 33. All phase plane solution points remain stationary as t increases. Generalized Eigenvectors Let A be an (n × n) matrix. The ideas introduced in equation (8) can be extended. Let v1 = 0 be an eigenvector of A corresponding to the eigenvalue λ, and suppose we can generate the following “chain” of nonzero vectors: (A − λI)v1 = 0 (A − λI)v2 = v1 (A − λI)v3 = v2 .. . (A − λI)vr = vr−1

(13)

4.8

Nonhomogeneous Linear Systems

In (13), the vector vj is called a generalized eigenvector of order j. Deﬁne t k−1 v1 . yk (t) = eλt vk + t vk−1 + · · · + (k − 1)!

277

(14)

Exercise 36 asks you to show, for r = 3 and k = 1, 2, 3, that yk (t) is a solution of y = Ay. If λ has algebraic multiplicity m and geometric multiplicity 1, then it can be shown that there is a chain (13) consisting of m different generalized eigenvectors and that these generalized eigenvectors are linearly independent (see Exercise 37). Thus, equation (14) deﬁnes a set of m linearly independent solutions of y = Ay—as many solutions as the multiplicity of the eigenvalue. (If λ has geometric multiplicity 2 or larger, the situation is more complicated.)

Exercises 34 –35: Using equations (13) and (14), ﬁnd a fundamental set of solutions for the linear system y = Ay. In each exercise, you are given an eigenvalue λ, where λ has algebraic multiplicity 3 and geometric multiplicity 1 and an eigenvector v1 . ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 4 0 0 1 0 2 1 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 35. A = ⎣2 4 0⎦ , λ = 4, v1 = ⎣0⎦ 34. A = ⎣0 2 1⎦ , λ = 2, v1 = ⎣0⎦ 1 3 4 0 1 0 0 2 36. Let v1 , v2 , and v3 be a chain of nonzero vectors, as in equation (13). Show that the vector function y3 (t) deﬁned in equation (14) is a solution of y = Ay. 37. Let v1 , v2 , and v3 be a chain of nonzero vectors, as in equation (13). Show that these vectors form a linearly independent set of vectors. [Hint: Begin with the dependence equation c1 v1 + c2 v2 + c3 v3 = 0 and multiply both sides by A − λI.]

4.8

Nonhomogeneous Linear Systems We now address the problem of ﬁnding the general solution of a nonhomogeneous ﬁrst order linear system, y = P(t)y + g(t),

a < t < b.

(1)

In (1), y(t) is an (n × 1) vector function, P(t) is an (n × n) matrix function, and the nonhomogeneous term, g(t), is an (n × 1) vector function. The component functions of P(t) and g(t) are assumed to be continuous on a < t < b.

The Structure of the General Solution In analyzing the structure of the general solution of (1), we return once more to a theme that has permeated all our discussions of nonhomogeneous linear equations. If y1 (t) and y2 (t) represent any two solutions of y = P(t)y + g(t), we ask, “How do they differ?” To answer this question, we form the difference function, w(t) = y1 (t) − y2 (t). Differentiating w(t) yields w = (y1 − y2 ) = y 1 − y 2

= [P(t)y1 + g(t)] − [P(t)y2 + g(t)] = P(t)(y1 − y2 ) = P(t)w.

278

CHAPTER 4

First Order Linear Systems

Thus, the difference between any two solutions of the nonhomogeneous linear equation is a solution of the homogeneous linear equation. This leads to the familiar decomposition The general solution of The general solution of A particular solution of the nonhomogeneous the homogeneous the nonhomogeneous = + linear system linear system linear system y = P(t)y + g(t) y = P(t)y y = P(t)y + g(t).

As before, we refer to the general solution of the homogeneous system, y = P(t)y, as the complementary solution and denote it by yC (t). A solution of the nonhomogeneous system that we have somehow found is called a particular solution and is denoted by yP (t). The following theorem, an analog of the superposition principle given in Theorem 3.4, holds for nonhomogeneous linear systems. We leave the proof as an exercise.

Theorem 4.8

Let u(t) be a solution of y = P(t)y + g1 (t),

a < t < b,

and let v(t) be a solution of y = P(t)y + g2 (t),

a < t < b.

Let a1 and a2 be any constants. Then the vector function yP (t) = a1 u(t) + a2 v(t) is a particular solution of y = P(t)y + a1 g1 (t) + a2 g2 (t),

a < t < b.

The following example illustrates Theorem 4.8. E X A M P L E

1

Find the general solution of

1 2 e2t y+ y = , 2 1 −2t

−∞ < t < ∞.

Solution: We saw earlier (in Example 1 of Section 4.4) that the general solution of the homogeneous equation,

1 2 y = y, −∞ < t < ∞, 2 1 is

yC (t) = c1

e−t −e−t

+ c2

e3t e3t

=

e−t −e−t

c1 . 3t c2 e e3t

4.8

Nonhomogeneous Linear Systems

279

Having the complementary solution, yC (t), we turn our attention to the task of somehow ﬁnding a particular solution, yP (t), of the nonhomogeneous equation. Note that the nonhomogeneous term,

e2t , g(t) = −2t can be decomposed as follows:

e2t

1 0 =e +t . 0 −2 −2t 2t

Using the superposition principle in Theorem 4.8, we decompose the differential equation and separately ﬁnd particular solutions of

1 2 1 2 0 2t 1 u = u+e v+t . (2) and v = 2 1 2 1 −2 0 Consider the ﬁrst equation in (2). Remembering the method of undetermined coefﬁcients, we look for a solution of the form uP (t) = e2t a, where a is a constant (2 × 1) vector to be determined. Substituting uP (t) = e2t a into the ﬁrst differential equation in (2) leads to

1 2 2t 2t 2t 1 2e a = (e a) + e , −∞ < t < ∞. 2 1 0 Canceling the common e2t factor and rearranging terms, we see that a must satisfy the condition ⎡ ⎤

− 13 1 −2 1 ⎣ ⎦. a= , or a = 2 −2 1 0 − 3

Thus, a particular solution is

⎤ ⎡ − 13 e2t ⎦. uP (t) = e2t a = ⎣ 2 2t −3e

To ﬁnd a particular solution of the second equation in (2), we look for a solution having the form vP (t) = tb + c, where b and c are constant (2 × 1) vectors to be determined. Substituting this guess into the differential equation leads to

0 1 2 (tb + c) + t b= −2 2 1 or, after collecting like powers of t,

1 2 0 1 t b+ + 2 1 −2 2

2 c − b = 0, 1

−∞ < t < ∞.

Since the set of functions {t, 1} is linearly independent on any t-interval, this equation holds only if the coefﬁcients of t and 1 are 0—that is,

1 2 b1 0 1 2 c1 b (3) = and = 1 . 2 1 b2 b2 2 2 1 c2 (continued)

280

CHAPTER 4

First Order Linear Systems (continued)

The two matrix equations in (3) can be solved sequentially, yielding ⎡ ⎤ ⎡ ⎤ 4 − 89 3 ⎣ ⎦ ⎣ ⎦. b= and c = 10 − 23 9 Therefore, the solution v(t) = tb + c is ⎡ v(t) = ⎣

⎤

4 t 3

−

8 9

− 23 t

+

10 9

⎦.

Applying Theorem 4.8, we have for a particular solution of the given differential equation ⎤ ⎡ − 13 e2t + 43 t − 89 ⎦. yP (t) = u(t) + v(t) = ⎣ − 23 e2t − 23 t + 10 9 The general solution is therefore

y(t) = yC (t) + yP (t) = c1

e−t

−e−t

+ c2

e3t e3t

⎡

− 13 e2t + 43 t −

+⎣ − 23 e2t − 23 t +

8 9 10 9

⎤ ⎦. ❖

Comparing Solution Methods In Chapter 3, the method of undetermined coefﬁcients was seen to be an effective way to ﬁnd particular solutions when the differential equation had constant coefﬁcients and when the nonhomogeneous term was of a certain form; see Table 3.1. Example 1 provides a simple illustration of how these ideas can be extended and applied to the constant coefﬁcient linear system y = Ay + g(t) when the nonhomogeneous vector function g(t) has components of a certain form. The Exercises give additional illustrations. However, in contrast to the scalar problem, the complexity of the matrix problem makes the “educated guesswork” at the core of this method difﬁcult to implement systematically. We shall not discuss the method of undetermined coefﬁcients any further. The method of variation of parameters (considered in Sections 3.9 and 3.13 for scalar linear equations) extends to linear systems. Therefore, we shall concentrate on the method of variation of parameters. In Chapter 5, we show how Laplace transforms also can be used to solve constant coefﬁcient nonhomogeneous linear ﬁrst order systems. As the ﬁrst step in applying the method of variation of parameters to a nonhomogeneous system of differential equations, we revisit the concept of a fundamental matrix.

Fundamental Matrices Section 4.3 introduced the concepts of a solution matrix and a fundamental matrix. A solution matrix, (t), is an (n × n) matrix whose n columns are each solutions of the homogeneous linear ﬁrst order system y = P(t)y, a < t < b.

4.8

Nonhomogeneous Linear Systems

281

Thus, if y1 (t), y2 (t), . . . , yn (t) are solutions of y = P(t)y, then (t) = [y1 (t), y2 (t), . . . , yn (t)] is a solution matrix. In addition, if these solutions form a fundamental set of solutions, then the solution matrix (t) is called a fundamental matrix. When we introduced the concepts of a solution matrix and a fundamental matrix in Section 4.3, our primary focus was on the solutions {y1 , y2 , . . . , yn }. We used solution matrices and fundamental matrices as a way to organize solutions into an array. Initial conditions are conveniently imposed using such arrays. We also use solution matrices to deﬁne the Wronskian of the solution set. At this point, we begin a subtle but important shift of emphasis. We now view solution matrices and fundamental matrices as (n × n) matrix functions that are mathematical entities in their own right. In particular, solution matrices and fundamental matrices for y = P(t)y can themselves be viewed as solutions of the matrix differential equation = P(t). Some important properties of solution matrices and fundamental matrices are summarized in Theorem 4.9. Theorem 4.9

Consider the homogeneous linear ﬁrst order system y = P(t)y,

a < t < b,

where y(t) is an (n × 1) vector function and P(t) is an (n × n) coefﬁcient matrix, continuous on (a, b). (a) Let (t) be any solution matrix of y = P(t)y, a < t < b. Then (t) satisﬁes the matrix differential equation = P(t),

a < t < b.

(b) Let 0 represent any given constant (n × n) matrix, and let t0 be any ﬁxed point in the interval a < t < b. Then there is a unique (n × n) matrix (t) that solves the initial value problem = P(t),

(t0 ) = 0 ,

a < t < b.

Moreover, if the constant matrix 0 is invertible, then the matrix (t) is a fundamental matrix of y = P(t)y, a < t < b. ˆ (c) If (t) is any fundamental matrix and (t) is any solution matrix of y = P(t)y, a < t < b, then there exists an (n × n) constant matrix C such that ˆ (t) = (t)C, a < t < b. ˆ Moreover, the matrix (t) is also a fundamental matrix if and only if det[C] = 0.

●

PROOF:

(a) Express the solution matrix in column form as (t) = [y1 (t), y2 (t), . . . , yn (t)].

282

CHAPTER 4

First Order Linear Systems

Recall, from Section 4.1, that (t) = [ y 1 (t), y 2 (t), . . . , y n (t)]. Therefore, (t) = [ y 1 (t), y 2 (t), . . . , y n (t)] = [P(t)y1 (t), P(t)y2 (t), . . . , P(t)yn (t)] = P(t)[y1 (t), y2 (t), . . . , yn (t)] = P(t)(t),

a < t < b.

(b) Let the constant matrix 0 be represented in terms of its columns as 0 = [1 , 2 , . . . , n ]. The initial value problem = P(t),

(t0 ) = 0 ,

a 0 be a given step size. From calculus, we know that the derivative, y (t), is deﬁned by y(t + t) − y(t) = y (t). t→0 t Therefore, if y(t) is the solution of initial value problem (1) and if the step size h is small, we expect the following approximation to be good: lim

y(t + h) − y(t) ≈ y (t) = f (t, y(t)). h Evaluating approximation (3) at a grid point t = tk , we obtain y(tk + h) ≈ y(tk ) + hf (tk , y(tk )),

(3)

0 ≤ k ≤ N − 1.

Therefore, once we have an estimate yk of y(tk ), this approximation leads us to an estimate yk+1 of y(tk+1 ): yk+1 = yk + hf (tk , yk ),

0 ≤ k ≤ N − 1,

y0 = y(t0 ).

(4)

Equation (4) is Euler’s method applied to the scalar initial value problem y = f (t, y),

y(t0 ) = y0 .

The approximation (3) is called a ﬁnite difference approximation to y (t). Methods derived from ﬁnite difference approximations, such as Euler’s method, 1

We assume a constant step size h in order to simplify the discussion. As noted in Section 2.10, many implementations of numerical methods use variable-step algorithms rather than a ﬁxedstep algorithm. Such variable-step methods use error estimates to monitor errors as the algorithm proceeds; when errors exceed a prescribed upper level, the steplength is reduced, and when errors are below a prescribed lower level, the steplength is lengthened.

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are known as ﬁnite difference methods. Finite difference methods can be generalized in a natural way to systems of ﬁrst order differential equations.

Extending Euler’s Method to First Order Linear Systems Consider the initial value problem y = P(t)y + g(t),

y(t0 ) = y0 ,

a ≤ t ≤ b,

(5)

where the (n × n) matrix function P(t) and the (n × 1) vector function g(t) are continuous on [a, b]. Euler’s method for initial value problem (5) begins with a generalization of the ﬁnite difference approximation (3). In particular, let y(t) be the unique solution of initial value problem (5), where ⎡ ⎤ y1 (t) ⎢ . ⎥ . ⎥ y(t) = ⎢ ⎣ . ⎦. yn (t) As we know from Section 4.1, ⎡

⎤ y1 (t + t) − y1 (t) lim ⎢t→0 ⎥ y1 (t) t ⎢ ⎥ ⎢ ⎥ . . ⎥ ⎢ .. .. ⎥ = ⎢ y (t) = ⎢ ⎥ ⎣ ⎦ ⎢ ⎥ ⎦ ⎣ y (t + t) − y (t) yn (t) n n lim t→0 t ⎡ ⎤ y1 (t + t) − y1 (t) ⎥ 1 ⎢ .. ⎢ ⎥ = lim . ⎣ ⎦ t→0 t yn (t + t) − yn (t) ⎡

= lim

t→0

⎤

1 [ y(t + t) − y(t)]. t

As before, let h > 0 be a given step size, where h = (b − a)/N, and let tk = t0 + kh,

0 ≤ k ≤ N,

t0 = a.

Since y (t) = P(t)y(t) + g(t), we expect (for small h) that 1 [y(t + h) − y(t)] ≈ P(t)y(t) + g(t). h Evaluating this approximation at the grid point t = tk , we obtain y(tk + h) ≈ y(tk ) + h[P(tk )y(tk ) + g(tk )]. Therefore, once we have an estimate yk of y(tk ), this approximation gives us an estimate yk+1 of y(tk+1 ). Deﬁne yk+1 = yk + h[P(tk )yk + g(tk )],

0 ≤ k ≤ N − 1,

y0 = y(t0 ).

Iteration (6) is Euler’s method for the initial value problem (5).

(6)

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291

There are obvious mathematical questions about the algorithm (similar to those raised in Section 2.10 for the scalar problem) that need to be answered. These will be addressed in Chapter 7.

E X A M P L E

1

Consider the two-tank mixing problem formulated in Section 4.1. The ﬂow schematic is shown in Figure 4.8. The corresponding initial value problem is ⎡ ⎤ 6 6

⎢− 200 + 10t Q1 (0) 300 ⎥ 0 5 d Q1 ⎢ ⎥ Q1 . , 0 ≤ t ≤ 30, = =⎢ + ⎥ 50 ⎣ 0 dt Q2 Q2 (0) 2 14 ⎦ Q2 − 200 + 10t 300

(7)

10 gal/min

12 gal/min

1 lb/gal 2

Fresh water Q1 lb/gal V1

Tank 1

Tank 2 2 gal/min

Q1(t) lb V1(t) gal

Q2(t) lb Q2 lb/gal V2

V2 (t) gal

6 gal/min Q2 lb/gal V2

Q1 lb/gal V1 4 gal/min

8 gal/min FIGURE 4.8

The two-tank mixing problem discussed in Example 1.

In (7), Q1 (t) and Q2 (t) represent the amounts of salt (in pounds) in Tanks 1 and 2, respectively, at time t (in minutes). Recall that salt solutions enter and leave Tank 1 at different rates, leading to the variable coefﬁcient matrix in (7). At time t = 30 min, Tank 1 is ﬁlled to capacity and the ﬂow stops. Use Euler’s method (6) to estimate Q1 (t) and Q2 (t) on this time interval; use a step size of h = 0.01. Plot Q1 (t) and Q2 (t) as functions of t, and also plot the concentration of salt in each tank as a function of t. Solution:

The ﬁrst order system has the form Q = P(t)Q + g,

0 ≤ t ≤ 30,

where [see equation (7)] P(t) is a (2 × 2) matrix function and g is a (2 × 1) constant vector. Euler’s method, applied to this problem, is the iteration 0 , 0 ≤ k ≤ N − 1, Q0 = Qk+1 = Qk + h[P(tk )Qk + g], 50 (continued)

First Order Linear Systems (continued)

where tk = kh, h = 30/N. In component form, the algorithm is 6 6 Q1,k + Q1,k+1 = Q1,k + h − Q2,k + 5 200 + 10tk 300 2 14 Q Q − , 0 ≤ k ≤ N − 1, Q2,k+1 = Q2,k + h 200 + 10tk 1,k 300 2,k

(8)

where Q1,0 = 0, Q2,0 = 50. The vector Qj is an approximation to the exact solution, Q(tj ), at time tj = jh. Since h = 0.01, the number of steps is N = 3000. Figure 4.9(a) shows the result of implementing Euler’s method with h = 0.01. Figure 4.9(b) displays the Euler’s method approximations to the concentrations, Q (t) cm (t) = m , m = 1, 2. Vm (t) As graphed in Figure 4.9, the answers seem reasonable. We expect the salt concentration in Tank 1 to increase, but not to exceed the maximum inﬂow concentration of 0.5 lb/gal. Likewise, the 2 gal/min inﬂow from Tank 1 into Tank 2 mitigates the “ﬂushing out” of Tank 2 that would otherwise occur. The concentration in Tank 2 therefore decreases somewhat slowly over the 30-min interval. Q

c

140

0.35

120

0.3 Salt concentration

CHAPTER 4

Amount of salt

292

Q1 (t)

100 80 60

0.25

c1(t)

0.2 0.15

40

0.1 Q2(t)

20

c2 (t)

0.05 5

10

15 20 Time

25

30

t

(a)

5

10

15 20 Time

25

30

t

(b) FIGURE 4.9

The results of applying Euler’s method to the initial value problem in Example 1.

❖

Runge-Kutta Methods for Systems Euler’s method is a conceptually important but relatively crude numerical algorithm. In order to obtain a high level of accuracy, Euler’s method usually

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293

demands a very small step size. And, of course, small steps require longer runtimes as well as close attention to the accumulation of roundoff errors. One alternative described in Section 2.10, higher order Runge-Kutta methods, usually provides more accuracy for a given step size h and is generally more efﬁcient than Euler’s method. Recall the fourth order Runge-Kutta method described in Section 2.10. For the scalar initial value problem (1), this method has the form yk+1 = yk +

h [K + 2K2 + 2K3 + K4 ], 6 1

(9)

where K1 = f (tk , yk ) K2 = f (tk + h/2, yk + (h/2)K1 ) K3 = f (tk + h/2, yk + (h/2)K2 )

(10)

K4 = f (tk + h, yk + hK3 ). Just as Euler’s method can be extended to systems, most Runge-Kutta methods can be extended to systems without any loss of accuracy. In particular, consider the initial value problem y = f(t, y),

y(t0 ) = y0 .

(11)

The extension of the Runge-Kutta method (9)–(10) takes the form yk+1 = yk +

h K1 + 2K2 + 2K3 + K4 , 6

(12)

where K1 = f(tk , yk ) K2 = f(tk + h/2, yk + (h/2)K1 ) K3 = f(tk + h/2, yk + (h/2)K2 )

(13)

K4 = f(tk + h, yk + hK3 ).

E X A M P L E

2

As a test case to illustrate how the Runge-Kutta philosophy can improve accuracy, consider the initial value problem y1 = y1 − y2 + 3, y2 = −y1 + y2 + 2t,

y1 (0) = 1 y2 (0) = 3.

(a) Solve this initial value problem mathematically. (b) Solve this initial value problem numerically on the interval 0 ≤ t ≤ 2, using Euler’s method and the Runge-Kutta method (12). Use a constant step size of h = 0.1. (c) Tabulate the exact solution values and both sets of numerical approximations at t = 0.5, t = 1.0, t = 1.5, and t = 2.0. Is the Runge-Kutta method more accurate than Euler’s method for this test case? (continued)

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Solution: (a) The exact (mathematical) solution is

1 −e2t + t2 + 4t + 3 . y(t) = 2 e2t + t2 + 2t + 5 (b) For this problem, we have

f(t, y) =

y1 − y2 + 3 −y1 + y2 + 2t

.

We used MATLAB as a programming environment for this example. The MATLAB code that evaluates f(t, y) is shown in Figure 4.10. MATLAB codes implementing Euler’s method and the Runge-Kutta method are shown in Figures 4.11 and 4.12, respectively.

function yp=f(t,y) yp=zeros(2,1); yp(1)=y(1)-y(2)+3; yp(2)=-y(1)+y(2)+2*t;

FIGURE 4.10

The MATLAB function subprogram to evaluate f(t, y) in Example 2.

% % Set the initial conditions for the % initial value problem in Example 2 % t=0; y=[1,3]’; h=0.1; output=[t,y(1),y(2)]; % % % Execute Euler’s method % on the interval [0,2] % for i=1:20 y=y+h*f(t,y); t=t+h; output=[output;t,y(1),y(2)]; end FIGURE 4.11

A MATLAB script that carries out Euler’s method for the initial value problem in Example 2.

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Numerical Methods for Systems of Linear Differential Equations

295

% % Set the initial conditions for the % initial value problem in Example 2 % t=0; y=[1,3]’; h=0.1; output=[t,y(1),y(2)]; % % Execute the fourth-order Runge-Kutta method % on the interval [0,2] % for i=1:20 ttemp=t; ytemp=y; k1=f(ttemp,ytemp); ttemp=t+h/2; ytemp=y+(h/2)*k1; k2=f(ttemp,ytemp); ttemp=t+h/2; ytemp=y+(h/2)*k2; k3=f(ttemp,ytemp); ttemp=t+h; ytemp=y+h*k3; k4=f(ttemp,ytemp); y=y+(h/6)*(k1+2*k2+2*k3+k4); t=t+h; output=[output;t,y(1),y(2)]; end FIGURE 4.12

A Runge-Kutta code for the initial value problem in Example 2.

(c) Table 4.1 compares the values.

TA B L E 4 . 1 The results of Example 2. Here, y E1 denotes Euler’s method estimates of y1 (t), y RK 1 denotes the Runge-Kutta method estimates of y1 (t), and yT1 denotes the true value of y1 (t). Similar notation is used in the last three columns. Note that the Runge-Kutta estimates are more accurate than the Euler’s method estimates. t

y E1

y RK 1

y T1

y E2

y RK 2

y T2

0.5000 1.0000 1.5000 2.0000

1.3558 0.8541 –2.1535 –11.7688

1.2659 0.3056 –4.4174 –19.7978

1.2659 0.3055 –4.4178 –19.7991

4.3442 7.0459 12.7535 25.5688

4.4841 7.6944 15.1674 33.7978

4.4841 7.6945 15.1678 33.7991

❖

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E X A M P L E

3

Use the fourth order Runge-Kutta method (12) to estimate the solution of y − 2ty + y = et/2 ,

y (1) = 0,

y(1) = 1,

1 ≤ t ≤ 2.

Use a uniform step size of h = 0.01. Solution: The differential equation is scalar, second order, linear, and nonhomogeneous, with variable coefﬁcients. None of the analytic techniques described in Chapter 3 are applicable. In order to use the Runge-Kutta method (12), we have to reformulate the second order equation as a system of two ﬁrst order equations: y1 = y2 , y2

y1 (1) = 1

= −y1 + 2ty2 + e

t/2

,

y2 (1) = 0.

We can use the Runge-Kutta code listed in Figure 4.12, changing the ﬁrst three lines to read t=1; y=[1,0]’; h=0.01; We also have to change the for loop to read “for i=1:100” and modify the last two lines of the function m-ﬁle in Figure 4.10 to read yp(1)=y(2); yp(2)=-y(1)+2*t*y(2)+exp(t/2); After making these changes and executing the program, we obtain the results shown in Figure 4.13.

y 10 9 8 7 6 5 4 3 2 1

y2(t)

y1(t)

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2

t

FIGURE 4.13

The results of Example 3. The solid curve is the numerical estimate of the graph of y1 (t) = y(t); the dashed curve is the estimate of the graph of y2 (t) = y (t).

❖

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Numerical Methods for Systems of Linear Differential Equations

297

EXERCISES Exercises 1–5: In each exercise, assume that a numerical solution is desired on the interval t0 ≤ t ≤ t0 + T, using a uniform step size h. (a) As in equation (8), write the Euler’s method algorithm in explicit form for the given initial value problem. Specify the starting values t0 and y0 . (b) Give a formula for the kth t-value, tk . What is the range of the index k if we choose h = 0.01? (c) Use a calculator to carry out two steps of Euler’s method, ﬁnding y1 and y2 . Use a step size of h = 0.01 for the given initial value problem. Hand calculations such as these are used to check the coding of a numerical algorithm.

1. y =

2. y =

3. y =

1

2

2

3

y+

1

1

t

2+t

2 t

−t2

1

y+

,

y(0) =

1 t

,

−1

y(1) =

, 1

2

1

1

2

0≤t≤1 ,

1 ≤ t ≤ 1.5

, y(1) = y+ , 1≤t≤4 t 0 2−t 0 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ 0 0 1 0 1 ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ −1 ≤ t ≤ 0 4. y = ⎣3 2 1⎦ y + ⎣2⎦ , y(−1) = ⎣0⎦ , 1 t 1 2 0 ⎤ ⎡

1 sin t⎥ 0 0 ⎢ t 5. y = ⎣ , 1≤t≤6 ⎦ y + 2 , y(1) = 0 t 1−t 1

Exercises 6–9: In each exercise, (a) As in Example 3, rewrite the given scalar initial value problem as an equivalent initial value problem for a ﬁrst order system. (b) Write the Euler’s method algorithm, y k+1 = y k + h[P(tk )y k + g(tk )], in explicit form for the given problem. Specify the starting values t0 and y 0 . (c) Using a calculator and a uniform step size of h = 0.01, carry out two steps of Euler’s method, ﬁnding y1 and y2 . What are the corresponding numerical approximations to the solution y(t) at times t = 0.01 and t = 0.02? 6. y + y = t3/2 ,

y(0) = 1, y (0) = 0

7. y + y + t2 y = 2,

y(0) = 1, y (0) = 1

8. y + 2y + ty = t + 1, y(0) = 1, y (0) = −1, y (0) = 0 dy d 9. et + y = 2e t , y(0) = −1, y (0) = 1 dt dt

Exercises 10–18: For the problem in the exercise speciﬁed, (a) Write a program that carries out Euler’s method. Use a step size of h = 0.01. (b) Run your program on the interval given.

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(c) Check your numerical solution by comparing the ﬁrst two values, y1 and y2 , with the hand calculations. (d) Plot the components of the numerical solution on a common graph over the time interval of interest. 10. Exercise 1

11. Exercise 2

12. Exercise 3

13. Exercise 4

14. Exercise 5

15. Exercise 6

16. Exercise 7

17. Exercise 8

18. Exercise 9

Estimating the Numerical Error of Euler’s Method When solving problems, we should apply all available tests or checks before accepting an answer. In addition to the checks provided by common sense, the physics of the problem being modeled, and the mathematical theory of differential equations, there are additional checks available for testing the accuracy of numerical algorithms. We now describe such a test. Suppose we apply Euler’s method to the initial value problem y = P(t)y + g(t), y(t0 ) = y0 , a ≤ t ≤ b. We observed in Section 2.10 that the error in Euler’s method is reduced approximately in half when the step size h is halved (this result will be justiﬁed in Chapter 7). This approximate halving of the error leads to a process for estimating the error. In particular, let t∗ be a point in [a, b] and choose a step size h by deﬁning h = (t∗ − t0 )/n, where n is a positive integer. Let yn denote the Euler’s method estimate to y(t∗ ) obtained using a step size h. Let y2n denote the Euler’s method estimate to y(t∗ ) obtained using a step size of h/2. We anticipate, by halving the step size, that the error will be (approximately) halved as well: y(t∗ ) − y2n ≈

y(t∗ ) − yn . 2

Some algebraic manipulation leads to the following estimate of the error, y(t∗ ) − y2n : y(t∗ ) − y2n ≈ y2n − yn .

(14)

Exercises 19–22: (a) Compute the error estimate (14) by using your Euler’s method program to solve the given initial value problem. In each case, let t∗ = 1. Use h = 0.01 and h = 0.005. (b) Solve the initial value problem mathematically, and determine the exact solution at t = t∗ . (c) Compare the actual error, y(t∗ ) − y2n , with the estimate of the error y2n − yn . [Note that estimate (14) is also applicable at any of the intermediate points 0.01, 0.02, . . . , 0.99.] 19. y − y = t,

y(0) = 2,

y (0) = −1

20. y + 4y = 3 cos t + 3 sin t, y(0) = 4,

2 −1 0 21. y = y, y(0) = 1 2 2

y (0) = 5

22. y =

−1

1

1

−1

y,

y(0) =

3

−1

23. Draining a Two-Tank System Consider the ﬂow system shown in the ﬁgure. Tank 1 initially contains 40 lb of salt dissolved in 200 gal of water, while Tank 2 initially contains 40 lb of salt dissolved in 500 gal of water. At time t = 0, the draining process is initiated, with well-stirred mixtures ﬂowing at the rates and directions shown. The volumes of ﬂuid in each tank change with time. Note, in particular, that Tank 1 empties in 20 min. Therefore, the ﬂow processes shown in the ﬁgure cease to be valid after 20 min. (Tank 2 will still contain 100 gal of ﬂuid after 20 min.) (a) Let Qj (t) represent the amount of salt in tank j at time t, j = 1, 2. Formulate the

4.9

Numerical Methods for Systems of Linear Differential Equations

initial value problem for

Q(t) =

Q1 (t) Q2 (t)

299

,

valid for 0 < t < 20 min. (b) Solve the initial value problem on the interval 0 ≤ t ≤ 19.9 min, using Euler’s method. Use a uniform step size with h = 0.01. (c) Plot the amounts of salt (in pounds) in each tank, Q1 (t) and Q2 (t), on the same graph for 0 ≤ t ≤ 19.9 min. (d) On what positive t-interval does Theorem 4.1 guarantee a unique solution of the initial value problem formulated in (a)? Should we be on the alert for possible numerical problems as time t increases to 20 min? Explain. Q1 lb/gal V1 Tank 1

Tank 2 15 gal/min Q2 lb/gal V2

Q1(t) lb V1(t) gal

Q2(t) lb V2 (t) gal

5 gal/min 30 gal/min

Q2 lb/gal V2

Figure for Exercise 23

24. A Spring-Mass-Dashpot System with Variable Damping As we saw in Section 3.6, the differential equation modeling unforced damped motion of a mass suspended from a spring is my + γ y + ky = 0, where y(t) represents the downward displacement of the mass from its equilibrium position. Assume a mass m = 1 kg and a spring constant k = 4π 2 N/m. Also assume the damping coefﬁcient γ is varying with time: γ (t) = 2te−t/2 kg/sec. Assume, at time t = 0, the mass is pulled down 20 cm and released. (a) Formulate the appropriate initial value problem for the second order scalar differential equation, and rewrite it as an equivalent initial value problem for a ﬁrst order linear system. (b) Applying Euler’s method, numerically solve this problem on the interval 0 ≤ t ≤ 10 min. Use a step size of h = 0.005. (c) Plot the numerical solution on the time interval 0 ≤ t ≤ 10 min. Explain, in qualitative terms, the effect of the variable damping upon the solution.

Exercises 25–27: Write a program that applies the Runge-Kutta algorithm (12) to the given problem. Run the program on the interval given, with a constant step size of h = 0.01. Plot the components of the solution. 25. y1 = −y1 + y2 + 2, 26.

y1 (0) = 1,

y2

= −y1 − y2 ,

y1 y2

= −y1 + y2 ,

y1 (0) = 1,

=

y2 (0) = 0

y2 + t,

0≤t≤2

y2 (0) = 0 0≤t≤1

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27. y =

1

t

0

1

y,

y(1) =

0 1

,

1≤t≤2

28. Suppose the Runge-Kutta method (12) is applied to the initial value problem y = Ay, y(0) = y0 , where A is a constant square matrix [thus, f(t, y) = Ay]. (a) Express each of the vectors Kj in terms of h, A, and yk , j = 1, 2, 3, 4. (b) Show that the Runge-Kutta method, when applied to this initial value problem, can be unraveled to obtain h2 2 h3 3 h4 4 A + A + A yk . (15) yk+1 = I + hA + 2! 3! 4! (c) Use the differential equation y = Ay to express the nth derivative, y(n) (t), in terms of A and y(t). Express the Taylor series expansion y(t + h) =

∞

y(n) (t)

n=0

hn n!

in terms of h, A, and y(t). Compare the Taylor series with the right-hand side of (15), with t = tk and y(tk ) = yk . How well does (15) replicate the Taylor series? 29. The exact solution of the initial value problem

0.5 1 1 y, y(0) = y = is given by 1 0.5 0

−t/2 + e3t/2 1 e . y(t) = 2 −e−t/2 + e3t/2

(a) Write a program that applies the Runge-Kutta method (12) to this problem. (b) Run your program on the interval 0 ≤ t ≤ 1, using step size h = 0.01. (c) Run your program on the interval 0 ≤ t ≤ 1, using step size h = 0.005. (d) Let y100 and y200 denote the numerical approximations to y(1) computed in parts (b) and (c), respectively. Compute the error vectors y(1) − y100 and y(1) − y200 . By roughly what fractional amount is the error reduced when the step size is halved?

4.10 The Exponential Matrix and Diagonalization The solution of the scalar initial value problem y = ay, y(0) = y0 is y(t) = eat y0 . If A is a constant (n × n) matrix, the solution of y = Ay, y(0) = y0 is y(t) = (t)y0 ,

(1)

where (t) is the fundamental matrix that reduces to the identity at t = 0. Is it possible to represent the solution (1) in the form y(t) = eAt y0 ?

(2)

There are two issues to be resolved. First, how do we give meaning to the exponential of a square matrix? Second, if we can give meaning to eAt , is expression (2) the unique solution of y = Ay, y(0) = y0 ? In other words, is eAt = (t)?

The Exponential Matrix To see how we might deﬁne the exponential matrix eAt , assume for the present discussion that the solution y(t) of the initial value problem y = Ay, y(0) = y0

4.10

The Exponential Matrix and Diagonalization

301

has the series expansion ∞ tm (m) y (0). m! m=0

y(t) =

(3)

We can use the initial value problem itself to evaluate the vectors y(m) (0): y(0) (0) = y(0) = y0 y(1) (0) = y (0) = Ay(0) = Ay0 y(2) (t) = [Ay(t)] = Ay (t) = A2 y(t);

therefore,

y(2) (0) = A2 y0 .

In general, we obtain y(m) (0) = Am y0 , and hence the series (3) becomes ∞ tm m t2 2 t3 3 A y0 = I + tA + A + A + · · · y0 . y(t) = m! 2! 3! m=0

(4)

The partial sums of the series of matrix powers in equation (4) have the form t2 2 t3 3 tk k Sk (t) = I + tA + A + A + · · · + A . 2! 3! k! It can be shown, for any constant (n × n) matrix A, that lim Sk (t)

k→∞

exists for all values of t. We deﬁne the (n × n) limit matrix to be the exponential matrix and denote it as eAt . Therefore, e

At

∞ tm m A . = m! m=0

(5)

It can be shown that the matrix exponential is differentiable and that its derivative can be calculated by differentiating the series (5) term by term. Assuming the validity of termwise differentiation, it follows that d At e = AeAt . dt

(6)

By (6), y(t) = eAt y0 is a solution of y = Ay for any (n × 1) vector y0 . Using this fact, along with the observation that eAt reduces to the identity when t = 0 and the fact that the solution of

(t) = A (t),

(0) = I

is unique, it follows that eAt = (t).

Properties of the Exponential Matrix In view of the close correspondence of the series deﬁning eAt when A is a matrix and eαt when α is a scalar, it is not surprising that the exponential matrix and the scalar exponential function possess similar properties. For instance, eAt1 eAt2 = eA(t1 +t2 )

(7a)

(eAt )−1 = e−At .

(7b)

and

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First Order Linear Systems

While properties (7a) and (7b) resemble properties of the scalar exponential function, they also have interpretations in terms of differential equations. In particular, consider the initial value problem y = Ay, y(0) = y0 . We can think of eAt as a matrix that propagates solutions forward by t units in time; that is, multiplying the initial state y(0) = y0 by eAt moves the solution forward in time to y(t) = eAt y0 . Taking this propagator viewpoint, property (7a) says either we can move the solution forward in one step or we can move it in stages. We can evolve the solution forward in one step from time t = 0 to time t = t1 + t2 by forming y(t1 + t2 ) = eA(t1 +t2 ) y0 . On the other hand, we can achieve the same result by ﬁrst moving forward t1 units in time using y(t1 ) = eAt1 y0 and then moving forward an additional t2 units by forming y(t1 + t2 ) = eAt2 y(t1 ) = eAt2 eAt1 y0 . In general, it can be shown (see Exercise 20) that y(t + t) = eAt y(t).

(8)

In a similar vein, suppose we are given y(t) = eAt y0 . We can recover the initial state y0 either by multiplying by the inverse matrix, obtaining y0 = (eAt )−1 y(t), or by propagating the solution backwards t units in time by forming y0 = e−At y(t).

E X A M P L E

1

Use the series (5) to calculate the exponential matrix eAt for

λ1 0 A= . 0 λ2 Solution:

Since A is a diagonal matrix,

λm m A = 1 0

0 . λm 2

Therefore, the matrix series (5) becomes

2

3

λ1 0 t 2 λ1 0 t 3 λ1 0 tA e = I+t + + + ··· 0 λ2 2! 0 λ22 3! 0 λ32 ⎡ ∞ ⎤ tm m

m λ 0 ∞ tm λ1 0 ⎢m=0 m! 1 ⎥ ⎢ ⎥ = m =⎣ ∞ m 0 λ2 m⎦ m! t λ 0 m=0 m! 2

=

E X A M P L E

2

eλ1 t

0

0

eλ 2 t

m=0

. ❖

Use the series (5) to calculate the exponential matrix eAt for

λ 1 A= . 0 λ

4.10

The Exponential Matrix and Diagonalization

303

Solution: It can be shown (see Exercise 21) that Am = λm I + mλm−1 E, where E denotes the matrix

0 1 E= . 0 0 Therefore,

⎡

etA =

m

t λm m!

∞ ⎢ tm m m=0 (λ I + mλm−1 E) = ⎢ ⎣ m! 0 m=0

=

∞

eλt

teλt

0

eλt

∞

⎤

m

m=0

t λm−1 ⎥ (m−1)!

∞

m=0

m

t λm m!

⎥ ⎦

. ❖

Similar Matrices We say that an (n × n) matrix A is similar to an (n × n) matrix B if there exists an (n × n) invertible matrix T such that T −1 AT = B. If A is similar to B, it follows that B is similar to A since T −1 AT = B implies A = TBT −1 = (T −1 )−1 B(T −1 ). Therefore, it is appropriate to refer to the matrices A and B as a pair of similar matrices. The act of forming the matrix product T −1 AT is often referred to as a similarity transformation. Among other things, the concept of similarity is important because (a) If A and B are similar (n × n) matrices, then A and B have the same characteristic polynomial and, hence, the same eigenvalues (see Theorem 4.10). (b) If A and B are similar (n × n) matrices, then solutions of w = Bw are related to solutions of y = Ay by the transformation y(t) = Tw(t).

Theorem 4.10

Let A and B be similar (n × n) matrices. Then A and B have the same characteristic polynomial.

● PROOF: Since A and B are similar, there is an invertible (n × n) matrix T such that T −1 AT = B. Let pA (λ) and pB (λ) denote the characteristic polynomials of matrices A and B, respectively. Observe that

pB (λ) = det[B − λI] = det[T −1 AT − λI] = det[T −1 AT − λT −1 T] = det[T −1 (A − λI)T] = det[T −1 ]det[A − λI]det[T] = pA (λ). (To obtain the last equality, we used the fact that det [T −1 ] det [T] = det [I] = 1.)

●

304

CHAPTER 4

First Order Linear Systems

Diagonalization Consider a similarity transformation of the form T −1 AT = D, where the matrix D is a diagonal matrix, ⎡ ⎤ 0 ··· 0 d11 0 ⎢ ⎥ ⎢ 0 d22 0 · · · 0 ⎥ ⎢ ⎥ ⎢ 0 d33 · · · 0 ⎥ D=⎢ 0 ⎥. ⎢ . .. ⎥ ⎢ .. . ⎥ ⎦ ⎣ 0 0 0 · · · dnn In such cases, we say that matrix A is similar to a diagonal matrix or that A is diagonalizable. Two questions arise with regard to diagonalization: 1. When is it possible to diagonalize a square matrix A? 2. If it is possible to diagonalize a given matrix A, how do we ﬁnd the matrix T that accomplishes the diagonalization? Theorem 4.11 and its corollary address these two questions.

Theorem 4.11

Let A be an (n × n) matrix similar to a diagonal matrix D. Let T be an invertible matrix such that T −1 AT = D. Then the diagonal elements of matrix D are the eigenvalues of matrix A, and the columns of matrix T are corresponding eigenvectors of matrix A.

●

PROOF: It follows from Theorem 4.10 that matrices A and D have the same eigenvalues. The eigenvalues of diagonal matrix D, however, are its diagonal elements. Therefore, the diagonal elements of D are the eigenvalues of A. To ﬁnish the proof, let D be the diagonal matrix ⎤ ⎡ λ1 0 0 · · · 0 ⎢0 λ 0 ··· 0 ⎥ ⎥ ⎢ 2 ⎥ ⎢ ⎥, ⎢ 0 0 λ · · · 0 D=⎢ 3 ⎥ .. ⎥ ⎢ .. ⎣ . . ⎦ 0

0

0

· · · λn

and let t1 , t2 , . . . , tn denote the n columns of the matrix [t1 , t2 , . . . , tn ]. Because T −1 AT = D, it follows that AT = TD: ⎡ λ1 0 0 · · · ⎢0 λ 0 ··· ⎢ 2 ⎢ A[t1 , t2 , . . . , tn ] = [t1 , t2 , . . . , tn ] ⎢ ⎢ 0 0 λ3 · · · ⎢ .. ⎣ . 0

0

0

T so that T = 0

⎤

0 ⎥ ⎥ ⎥ 0 ⎥ ⎥. .. ⎥ . ⎦ · · · λn

(9)

4.10

The Exponential Matrix and Diagonalization

305

The matrix product on the left-hand side of equation (9) can be rewritten as [At1 , At2 , . . . , Atn ]. Similarly, the matrix product on the right-hand side of equation (9) can be rewritten as [λ1 t1 , λ2 t2 , . . . , λ n t n ]. Since matrix equality implies that corresponding columns are equal, we obtain At j = λj t j ,

j = 1, 2, . . . , n.

(10)

To complete the argument, note that invertibility of the matrix T implies that none of its columns can be the (n × 1) zero vector. Therefore, t j = 0, 1 ≤ j ≤ n, and this fact, in conjunction with (10), shows that the columns of T are eigenvectors of A. ● A corollary of Theorem 4.11 characterizes diagonalizable matrices. Corollary

An (n × n) matrix A is diagonalizable if and only if it has a set of n linearly independent eigenvectors.

From what we know already, matrices with distinct eigenvalues as well as real symmetric matrices and Hermitian matrices are diagonalizable. In general, A is diagonalizable if and only if A has a full set of eigenvectors. E X A M P L E

3

As noted, real symmetric matrices are diagonalizable. Let A be the matrix

1 2 A= . 2 1 Find an invertible (2 × 2) matrix T such that T −1 AT = D. Solution:

We saw in Example 1 of Section 4.4 that eigenpairs of A are

1 1 and λ2 = −1, x2 = . λ1 = 3, x1 = 1 −1

Therefore, a matrix T made from the eigenvectors will diagonalize A. So, let T be

1 1 T= . 1 −1 A direct calculation shows ⎡ T

−1

AT = ⎣

1 2

1 2

1 2

⎤

1 ⎦ 2 − 12

2 1 1 3 0 = . ❖ 1 1 −1 0 −1

The Exponential Matrix Revisited We again consider the exponential matrix eAt , assuming now that A is an (n × n) diagonalizable matrix. Therefore, A = TDT −1 ,

(11)

306

CHAPTER 4

First Order Linear Systems

where D is the (n × n) diagonal matrix consisting of the eigenvalues of A and T is the (n × n) matrix formed from the corresponding eigenvectors; recall equations (9) and (10). We now see that equation (11) simpliﬁes the inﬁnite series (5) deﬁning eAt . Observe, by (11), that A2 = (TDT −1 )(TDT −1 ) = TD(T −1 T)DT −1 = TD2 T −1 . In general, it can be shown that Am = TDm T −1 . Therefore, At

e

∞ ∞ tm m tm A = TDm T −1 = T = m! m! m=0 m=0

∞ tm m D m! m=0

T −1 = TeDt T −1 .

Since D is a diagonal matrix, it follows (as in Example 1) that ⎤ ⎡ λt e1 0 ··· 0 ⎥ ⎢ ∞ 0 e λ2 t 0 ⎥ tm m ⎢ D =⎢ .. ⎥ ⎢ .. ⎥. m! . ⎦ ⎣ .

(12a)

(12b)

m=0

0 Using (12b) in (12a), we obtain ⎡ λt e1 ⎢ ⎢ 0 eAt = T ⎢ ⎢ .. ⎣ . 0

0 e λ2 t 0

0 ···

· · · e λn t 0 0 .. .

⎤ ⎥ ⎥ −1 ⎥T . ⎥ ⎦

(12c)

· · · eλn t

Decoupling Transformations There is an alternative approach to solving y = Ay when A is diagonalizable. This alternative involves making an appropriate change of dependent variable that transforms y = Ay into a collection of decoupled scalar problems. This decoupled system can then be solved using the techniques of Chapters 2 and 3. Example 4 illustrates the ideas, and additional examples are considered in the Exercises. E X A M P L E

4

Solve the initial value problem y1 = y1 + 2y2 + 1, y2 = 2y1 + y2 + t, Solution:

y1 (0) = 1 y2 (0) = −1.

This problem has the form y = Ay + g(t), y(0) = y0 , where

1 2 1 1 A= , g(t) = , y0 = . 2 1 t −1

(13)

4.10

The Exponential Matrix and Diagonalization

307

From Example 3, we know that A is diagonalizable. In particular, T −1 AT = D, where

1 1 1 3 0 1 1 −1 T = , T= , D= . 2 1 −1 1 −1 0 −1 Let us make the change of variable z(t) = T −1 y(t) or, equivalently, y(t) = Tz(t). Since y (t) = Tz (t), the equation y = Ay + g(t) transforms into Tz = ATz + g(t), or z = T −1 ATz + T −1 g(t). The initial condition for the transformed system is z(0) = T −1 y(0) = T −1 y0 . Since T −1 AT = D, the original problem (13) has become z = Dz + T −1 g(t), z(0) = T −1 y0 , or

1 1 1 1 3 0 1 1 1 1 , z(0) = . z = z+ 2 1 −1 t 2 1 −1 −1 0 −1 In terms of components, this system has the form z1 = 3z1 + (1 + t)/2, z2 = − z2 + (1 − t)/2,

z1 (0) = 0 z2 (0) = 1.

(14)

As can be seen, the system (14) is an uncoupled system of ﬁrst order linear equations of the type studied in Chapter 2. The solutions are z1 (t) = (4e3t − 3t − 4)/18 z2 (t) = (2 − t)/2. In terms of the original variables, y(t) = Tz(t), or

1 1 1 4e3t − 12t + 14 y(t) = z(t) = . ❖ 18 4e3t + 6t − 22 1 −1

A Warning Some properties of the scalar exponential function eat generalize to the exponential matrix eAt ; properties (7a) and (7b) are two such examples. There are limits, however, to the extent to which the scalar properties generalize to eAt . If A and B are (n × n) matrices, then it is generally the case that eAt eBt = eBt eAt

and

eAt eBt = e(A+B)t .

(15)

The reason the expected results do not materialize is that matrix multiplication is not commutative. That is, it is usually the case that AB = BA.

308

CHAPTER 4

First Order Linear Systems

EXERCISES Exercises 1–10: The given matrix A is diagonalizable. (a) Find T and D such that T −1 AT = D. (b) Using (12c), determine the exponential matrix eAt .

5 −6 3 4 1 1. A = 3. A = 2. A = 3 −4 −2 −3 1

2 3 1 −2 2 5. A = 7. A = 6. A = 3 2 −2 1 1

0 2 1 9. A = 10. A = −2 0 −2

1

4. A =

1

0

8. A =

1 2

2

3

2

3

−2

2

0

3

1

Exercises 11–16: In each exercise, the coefﬁcient matrix A of the given linear system has a full set of eigenvectors and is therefore diagonalizable. (a) As in Example 4, make the change of variables z(t) = T −1 y(t), where T −1 AT = D. Reformulate the given problem as a set of uncoupled problems. (b) Solve the uncoupled system in part (a) for z(t), and then form y(t) = Tz(t) to obtain the solution of the original problem.

11. 12. 6 −6 1 1 1 0 y, y(0) = y+ y = y = 2 −1 2 2 −3 3

2t 13. 0 −4 −6 e , y(0) = y+ y = 2t 0 3 5 −e 14. y1 = 3y1 + 2y2 + 4 y2 16.

15. y1 = −9y1 − 5y2 ,

= y1 + 4y2 + 1

3

2

−4

−3

y + y =

1

y2 y3

= =

8y1 + 4y2

y1 (0) = 1 + 1,

y2 (0) = 0

y3 + 2,

y3 (0) = 0

1 2 1 y. Example 2 shows that the corre17. Consider the differential equation y = 0 2 2t 2t te 1 e sponding exponential matrix is eAt = . Use the 2t . Suppose that y(1) = 2 0 e propagator property (8) to determine y(4) and y(−1). 18. Determine by inspection whether or not the matrix is diagonalizable. Give a reason that supports your conclusion.

1 1 1 1 1 1 (a) A1 = (b) A2 = (c) A3 = 0 1 0 −1 1 1 19. Let A be a constant (n × n) diagonalizable matrix. Use the representation (12c) to establish properties (7a) and (7b). That is, show that eAt1 eAt2 = eA(t1 +t2 ) and (eAt )−1 = e−At . 20. Use property (7a) to establish the propagator property (8). That is, show that y(t + t) = eAt y(t). 0 1 λ 1 , and let E = . Use mathematical induction or the binomial 21. Let A = 0 0 0 λ formula to show that Am = λm I + mλm−1 E.

4.10

The Exponential Matrix and Diagonalization ⎡

λ 22. Consider the differential equation y = Ay, where A = ⎣0 0 ⎤ ⎡ λt teλt 12 t2 eλt e ⎢ eλt teλt ⎥ series (5), show that eAt = ⎣ 0 ⎦. 0 0 eλt

1 λ 0

309

⎤ 0 1⎦. Using the inﬁnite λ

Exercises 23–24: In each exercise, (a) Does AB = BA? (b) Calculate the exponential matrices eAt , eBt , and e(A+B)t . Does eAt eBt = e(A+B)t ?

23. A =

1

1

0

−1

,

B=

1

0

0

−1

24. A =

2

−1

−1

2

,

B=

1

1

1

1

Exercises 25–30: Second Order Linear Systems We consider systems of second order linear equations. Such systems arise, for instance, when Newton’s laws are used to model the motion of coupled spring-mass systems, such as those in Exercises 31–32. In each of Exercises 2 1 1 25–30, let A = . Note that the eigenpairs of A are λ1 = 3, x1 = and 1 2 1 1 λ2 = 1, x2 = . −1 (a) Let T = [x1 , x2 ] denote the matrix of eigenvectors that diagonalizes A. Make the change of variable z(t) = T −1 y(t), and reformulate the given problem as a set of uncoupled second order linear problems. (b) Solve the uncoupled problem for z(t), and then form y(t) = Tz(t) to solve the original problem. 25. y + Ay = 0 26. y1 − 2y1 − y2 = 0,

y1 (0) = 0,

y1 (0) = 1

y2 − y1 − 2y2 = 0, y2 (0) = 0, y2 (0) = −1

1 0 1 27. y + Ay = , y(0) = , y (0) = 0 1 0 28. y + y + Ay = 0

29. y1 + 4y1 + 2y2 = 0 y2 + 2y1 + 4y2 = 1

30. Ay + 4y = 0

Exercises 31–32: Consider the spring-mass system shown in the ﬁgure on the next page. The system can execute one-dimensional motion on the frictionless horizontal surface. The unperturbed and perturbed systems are labeled (a) and (b) respectively. 31. (a) Show that an application of Newton’s second law of motion leads the second order system Mx + Kx = 0, where ⎡ m1 ⎢ M=⎣ 0 0

0 m2 0

⎤

⎡

⎥ 0 ⎦,

⎢ K = ⎣−k1

0

m3

k1 0

−k1 k1 + k2 −k2

0

⎤

⎥ −k2 ⎦ , k2

⎡ ⎤ x1 (t) ⎢ ⎥ x(t) = ⎣x2 (t)⎦ . x3 (t)

310

CHAPTER 4

First Order Linear Systems k1

(a)

m1

x1(t)

(b)

k2 m2

m3

x2(t) m1

x3(t) m2

m3

Figure for Exercises 31–32

(b) Let m1 = m2 = m3 = m and k1 = k2 = k. Determine the eigenpairs of A = M−1 K. (c) Obtain the general solution of x + Ax = 0. 32. Consider the second order linear system derived in part (a) of Exercise 31. (a) Show that the matrix K has an eigenvalue λ = 0. Determine a corresponding eigenvector and denote it as v0 . (b) Explain the physical signiﬁcance of the eigenpair (0, v0 ). In particular, what motion does the system execute if the initial conditions are x(0) = 0, x (0) = v0 ? [Hint: Look for a solution of the form x(t) = f (t)v0 , where f (t) is a scalar function to be determined.] Describe, in words, how the system is behaving. 33. We know that similar matrices have the same eigenvalues (in fact, they have the same characteristic polynomial). There are many examples that show the converse is not true; that is, there are examples of matrices A and B that have the same characteristic polynomial but are not similar. Show that the following matrices A and B cannot be similar:

1 0 1 0 A= and B = . 3 1 0 1 34. Drawing on the ideas involved in working Exercise 33, show that if an (n × n) real matrix A is similar to the (n × n) identity I, then A = I. 35. Give an example that shows that while similar matrices have the same eigenvalues, they may not have the same eigenvectors. 36. Deﬁne matrices P(t) and Q(t) as follows:

1 cos t P(t) = , 2t 0

Q(t) =

t

P(s) ds. 0

Show that P(t) and its derivative Q(t) do not commute. That is, P(t)Q(t) = Q(t)P(t).

CHAPTER 4 REVIEW EXERCISES These review exercises provide you with an opportunity to test your understanding of the concepts and solution techniques developed in this chapter. The end-of-section exercises deal with the topics discussed in the section. These review exercises, however, require you to identify an appropriate solution technique before solving the problem.

Exercises 1–22: In each exercise, determine the general solution. If initial conditions are given, solve the initial value problem.

Projects

1.

4.

7.

9.

11.

y =

y = y = y1 y2 y3 y1 y2

−4

2

−3

−2 1

3

0 −1

2. y

−1

y =

−7

6

−9

8

y

5. y1 = 3y1 − y2

y

y2

−4

2 6 y+ 3 7

8.

10.

= y1 + 2y2 + y3 = y1 + y2 + 2y3 y1 (0) = 7

12.

= − 3y1 + 5y2 , y2 (0) = 5

13. −6 10 5 y, y(0) = y = −4 6 5 ⎡ ⎤ 2 −1 1 15. ⎢ ⎥ 2 1⎦ y y = ⎣−1 1 1 2

17. 1 4 7 y = y+ −1 5 11

19. 0 4 y y = −1 4

21. −2 1 y y = −1 0

14.

16.

18.

20.

22.

y =

0

−4

4

0

y

6. y1 = − 2y1 + y2 y2 = − y1 − 2y2

= y1 + y 2

= 2y1 + y2 + y3

= − 4y1 + 6y2 ,

3.

311

y = y1 y2 y3

3

−2

2

−1

y, y(0) =

4 2

= 2y1 + 2y2 = 5y1 + 5y2 =

2y3

3 −5 1 y+ y = 1 −1 1

−2 −1 y y = 2 −4 ⎡ ⎤ 1 1 0 ⎢ ⎥ y = ⎣0 1 0⎦ y 0 0 1

7 −5 y y = 10 −7

9 20 y = y −4 −9

11 −3 y = y 30 −8

PROJECTS Consider the conﬁguration shown in Figure 4.14. The three identical springs of unstretched length l are assumed to be weightless, and the two identical masses are asl

l m

l m

(a) x1(t)

x2(t) m

m

(b) FIGURE 4.14

A coupled spring-mass system. (a) The equilibrium state. (b) The perturbed state.

312

CHAPTER 4

First Order Linear Systems

sumed to slide on a frictionless surface. In the vertical direction, the surface exerts a normal force upon each mass equal and opposite to its weight. Therefore, we need only consider motion in the horizontal direction.

Project 1: Derivation of the Equations of Motion Assume the system is set in motion at time t = 0 and there are no externally applied forces. Let x1 (t) and x2 (t) represent the respective horizontal displacements of the two masses from their equilibrium positions, measured positive to the right as shown. Show that the application of Newton’s second law of motion to each mass leads to the system of equations mx1 = k x2 − x1 − kx1 mx2 = −k x2 − x1 − kx2 , t > 0. These equations can be rewritten as a second order linear system

−2 1 x1 (t) . mx = k x, t > 0, where x(t) = x2 (t) 1 −2

(1)

In addition to the equations of motion (1), we specify the initial position and velocity of each mass by x(0) = x 0 ,

x (0) = x 0 .

(2)

Project 2: Numerical Solution Using the Exponential Matrix The initial value problem deﬁned by equation (1) and initial condition (2) can be solved by using the diagonalization techniques of Section 4.10 to transform the problem into two decoupled scalar problems. In this project, however, we will solve the problem by using the exponential matrix to propagate the solution forward in time. Our solution will be numerical in the sense that we will tabulate the solution at discrete times. The ﬁrst task is to recast the problem as a ﬁrst order system. Deﬁne y1 = x1 ,

y2 = x1 ,

y3 = x2 ,

y4 = x2 .

With this, the initial value problem can be rewritten as y = Ay,

y(0) = y0 ,

(3)

where A is a (4 × 4) constant matrix and y0 is a (4 × 1) vector. 1. Determine A and y0 . 2. Suppose we want to solve (3) on the interval 0 ≤ t ≤ T. Choose a time step t = T/N, where N is a positive integer. The solution of (3) can be tabulated in 0 ≤ t ≤ T at the time steps tj = j t, j = 0, 1, 2, . . . , N, using the iteration y(tj ) = eAt y(tj−1 ),

j = 1, 2, . . . , N.

(4)

[The iteration (4) comes from the propagator property of the exponential matrix, y(t + t) = eAt y(t); recall equation (8) in Section 4.10.] Assume m = 2 kg, k = 8 N/m, l = 1 m, T = 10 sec, and t = 0.05 sec. For the initial conditions

0 −0.25 m, x0 = x0 = m/sec, 0.25 0 ﬁnd y(tj ) using iteration (4), and plot the spring displacements at times t0 , t1 , . . . , tN . Interpret your results physically, describing how the two masses are moving in the time interval 0 ≤ t ≤ T. [Software such as MATLAB has built-in exponential matrix

Projects

313

routines you can call to form eAt . Or you can simply solve the initial value problem

= A , (0) = I and use the fact that (t) = eAt .] 3. Repeat part (2), but with the initial conditions

0 0.25 x0 = m, x0 = m/sec. 0 0.25

Project 3: Resonant Behavior of a Coupled Spring-Mass System Consider the spring-mass system shown in Figure 4.15. The ﬁgure shows two springs of negligible weight. The springs have unstretched lengths l1 and l2 and have spring constants k1 and k2 . Masses m1 and m2 are attached, and the springs stretch appropriately to achieve the rest conﬁguration shown. The amount of stretching done by each spring is determined by imposing conditions of static equilibrium on each mass. Since the sum of the vertical forces acting on each mass must vanish, we obtain k1 Y1 = m1 g + k2 Y2 and m2 g = k2 Y2 . Therefore, k1 Y1 = m1 g + m2 g, where g represents acceleration due to gravity. When the system is disturbed from its equilibrium state, both masses move vertically. Let y1 (t) and y2 (t) represent the displacements of the masses from their equilibrium positions at time t, as shown in Figure 4.15. We assume the system is initially at rest and is set into motion by a force f (t) = F sin ωt, applied vertically to the mass m2 ; see Figure 4.15(c).

k1

l1 l 1 + Y1

k2

l2

l 1 + Y1 + y1(t) l 1 + Y1 + l 2 + Y2 + y2(t)

m1 m1 l 2 + Y2 m2 m2 f(t)

(a) Unstretched springs

(b) System at rest with masses attached

(c) System disturbed from rest state

FIGURE 4.15

1. Show that Newton’s second law of motion leads to the following nonhomogeneous system of second order differential equations: m1 y1 = −(k1 + k2 )y1 + k2 y2 m2 y2 = k2 y1 − k2 y2 + F sin ωt.

(5)

Since the system is initially at rest, the initial value problem is y + Ay = b sin ωt,

y(0) = 0,

y (0) = 0.

(6)

Determine the constant matrix A and the constant vector b. 2. Assume that m1 = m2 = m and that k1 = 2k and k2 = k. Denote the eigenpairs of the real symmetric matrix A by (λ1 , x1 ) and (λ2 , x2 ). Make the change of variable

314

CHAPTER 4

First Order Linear Systems z(t) = T −1 y(t), where T = [x1 , x2 ], and reformulate problem (6) as a pair of decoupled second order initial value problems. 3. Determine the resonant frequencies of the pair of initial value problems derived in part (2). That is, for what values of ω will at least one component of z(t) have an amplitude that grows linearly with time? 4. Solve the initial value problem derived in part (2) in the case where m = 1 kg, k = 1 N/m, l = 1 m, F = 0.1 N, and ω is equal to the largest of the values determined in part (3). Form y(t) = Tz(t) to obtain the solution of equation (6). Discuss the physical relevance of the solution. Is it physically meaningful for all t, 0 ≤ t < ∞?

Project 4: A Control Problem in Charged Particle Ballistics Consider a particle, having mass m and electric charge q, moving in a magnetic ﬁeld. The magnetic ﬁeld is a vector ﬁeld B. The motion of the particle in this magnetic ﬁeld is most conveniently described in terms of vectors. Let i, j, and k represent unit vectors in the x, y, and z directions, respectively. The position of the particle is deﬁned by the position vector, r(t) = x(t)i + y(t)j + z(t)k, and its velocity by the corresponding velocity vector, v(t) = vx (t)i + vy (t)j + vz (t)k. Since v(t) = dr/dt, the velocity components are vx (t) =

d x(t), dt

vy (t) =

d y(t), dt

vz (t) =

d z(t). dt

Figure 4.16 illustrates the problem.

B z m, q v(t )

r(t)

y

x FIGURE 4.16

A charged particle having mass m and electric charge q moves in the magnetic ﬁeld B. Its motion is described by its position vector r(t) and velocity vector v(t).

Projects

315

If the weight of the charged particle is neglected, the only force acting on it is the Lorentz2 force (the force the magnetic ﬁeld exerts on the moving charge). The Lorentz force, described using vector notation, is qv(t) × B. An application of Newton’s second law of motion leads to the vector equation d v(t) = qv(t) × B. (7) dt Suppose the charged particle is launched from the origin at time t = 0 with initial velocity v(0) = v0 . Is it possible to select v0 so that the particle will be located at a desired location R at a speciﬁc later time t = τ > 0? In other words, can we choose v(0) = v0 so that r(τ ) = R? m

1. Let B = Bk and deﬁne ωc = qB/m. The constant ωc is a radian frequency known as the cyclotron frequency. Rewrite vector equation (7) in the form v = Av,

v(0) = v0 .

2. Solve the initial value problem, expressing the solution in the form v(t) = eAt v0 . t 3. Form r(t) = 0 v(s) τ ds and determine conditions under which we can choose v0 such that r(τ ) = R = 0 v(s) ds.

2

Hendrik Lorentz (1853–1928) was professor of mathematical physics at Leiden University from 1878 until 1912; he thereafter directed research at the Teyler Institute in Haarlem. Lorentz is noted for his studies of atomic structure and of the mathematical transformations (called Lorentz transformations) that form the basis of Einstein’s theory of special relativity. Along with his student Pieter Zeeman, Lorentz was awarded the Nobel Prize in 1902.

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C H A P T E R

5

Laplace Transforms CHAPTER OVERVIEW 5.1

Introduction

5.2

Laplace Transform Pairs

5.3

The Method of Partial Fractions

5.4

Laplace Transforms of Periodic Functions and System Transfer Functions

5.5

Solving Systems of Differential Equations

5.6

Convolution

5.7

The Delta Function and Impulse Response

5.1

Introduction When you begin to study a new topic such as the Laplace transform, two questions arise: “What is it?” and “Why is it important?” A scientist often uses Laplace1 transforms to solve a mathematical problem for the same reason that a motorist leaves a congested highway and travels a network of back roads to reach his or her destination. The most easily traveled path between two points is not always the most direct one. One of the important applications of Laplace transforms is solving constant coefﬁcient linear differential equations that have discontinuous righthand sides. In particular, many mechanical and electrical systems are driven by sources that switch on and off. Such systems are often modeled by an initial value problem of the form y + ay + by = g(t), 1

y(t0 ) = y0 ,

y (t0 ) = y0 ,

Pierre-Simon Laplace (1749–1827) was a French scientist renowned for his work in mathematics, celestial mechanics, probability theory, and physics. The Laplace transform, the Laplace probability density function, and Laplace’s equation (arising in the study of potential theory) are some mathematical entities named in his honor.

317

318

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where the right-hand side g(t) has discontinuities at those times when the source changes abruptly. Laplace transform techniques are a convenient tool for analyzing such initial value problems. The philosophy underlying the use of Laplace transforms is illustrated in Figure 5.1. We have a problem to solve—for example, determining the behavior of some mechanical or electrical system. Instead of attacking the problem directly, we transform (or map) the original problem into a new problem. This mapping is accomplished by means of the mathematical operation known as the Laplace transform. The original problem is often referred to as the time domain problem since the independent variable for the original problem is usually time. The new problem, resulting from the Laplace transformation, is referred to as the transform domain problem. After obtaining this new transform domain problem, we solve it and then transform the solution back into the time domain by performing another mapping, known as the inverse Laplace transform. The inverse mapping thus gives us the solution of the original time domain problem, the problem of interest. Time domain Original problem of interest

Transform domain

Laplace transform

Transform domain problem

Solution

Desired solution

Inverse Laplace transform

Solution of transform domain problem

FIGURE 5.1

The philosophy underlying the use of Laplace transforms.

In order for the problem-solving approach diagrammed in Figure 5.1 to be attractive, the following three steps must be easier to implement than the direct solution approach: 1. calculating the Laplace transform, 2. solving the transformed problem, and 3. calculating the inverse Laplace transform. For many of the problems we treat, this will be the case. Constant coefﬁcient linear differential equations will be transformed into algebraic equations. The resulting transform domain problem typically entails solving a single linear algebraic equation or a system of linear algebraic equations. We will consider a variety of constant coefﬁcient linear differential equations (both scalar equations and systems) and show how these problems can be solved using Laplace transforms. We will also consider several problems, such

5.1

Introduction

319

as the parameter identiﬁcation problem in the following example, that are not so straightforwardly solved using the techniques developed so far. E X A M P L E

1

Consider a vibrating mechanical system that exists in a “black box,” as in Figure 5.2. Assume you are conﬁdent that the vibrating system can be modeled as the spring-mass-dashpot arrangement shown, but you do not have the internal access needed to directly measure the spring constant k, the mass m, or the damping constant γ . You can only apply a force at the external access point and measure the resulting displacement.

k

m

␥

Applied force f (t)

FIGURE 5.2

A cutaway schematic of a “black box” vibrating system.

Mathematically (as we saw in Section 3.10), the mechanical system in Figure 5.2 is described by the initial value problem, my + γ y + ky = f (t),

t ≥ 0,

y(0) = 0,

y (0) = 0.

The system is at rest at time t = 0 when force f (t) is applied. The applied force f (t) is known for t ≥ 0; the displacement y(t) is monitored and is thus known for t ≥ 0. The parameters m, γ , and k, however, are unknown. Assuming we know the input-output relation [that is, f (t) and y(t)] for one given applied force and the corresponding measured displacement, we ask the following two questions: 1. Is it possible to predict what the output will be if another input is applied to the system? In other words, if we apply a different external force, f˜ (t), is it possible to predict the resulting displacement, y˜ (t)? 2. Is it possible to determine the constants m, γ , and k from a knowledge of the single input-output history given by f (t) and y(t)? We will see in Section 5.4 that the use of Laplace transforms provides a relatively easy way to answer both questions. ❖

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The Laplace Transform The ﬁrst use of Laplace transforms as an operational tool for solving constant coefﬁcient linear differential equations is often attributed to the British physicist Oliver Heaviside.2 As noted earlier, the Laplace transform maps a function of t, say f (t), into a new function, F(s), of a new transform variable s. [In terms of notation, we generally use lowercase letters to designate time domain functions, such as f (t), and capital letters to denote corresponding transform domain functions, F(s).] Let f (t) be a function deﬁned on the interval 0 ≤ t < ∞. The Laplace transform of f (t) is deﬁned by the improper integral ∞ L{ f (t)} = F(s) = f (t)e−st dt, (1) 0

provided the integral exists. As the notation of equation (1) indicates, we often denote F(s), the Laplace transform of f (t), by the symbol L{ f (t)}. The new transform variable s is assumed to be a real variable. (In more advanced treatments of the Laplace transform, the variable s is allowed to be a complex variable.) As we look at equation (1), the ﬁrst issue to settle is that of identifying the properties f (t) must possess in order for its Laplace transform to exist—that is, in order for the improper integral in equation (1) to converge. Recall from calculus that the improper integral in (1) is deﬁned by ∞ T f (t)e−st dt = lim f (t)e−st dt, (2) T→∞

0

0

provided the limit exists. When the limit (2) exists, we say that the improper integral converges, and we deﬁne the improper integral to be this limit value. If the limit in (2) does not exist, we say that the improper integral diverges. Whether or not limit (2) exists depends on the nature of f (t) and on the value of the transform variable s; note that s plays the role of a parameter in the integral. In this section, we identify a large class of functions that possess Laplace transforms. It is important to realize, however, that not every function has a Laplace transform. The third function considered in Example 2 illustrates this fact. E X A M P L E

2

Find the Laplace transform, if it exists, of 2 (a) f (t) = eat (b) f (t) = t (c) f (t) = et Solution: (a) Applying the deﬁnition, we see that ∞ at at −st e e dt = lim L{e } = 0

2

T→∞

T

e−(s−a)t dt,

(3)

0

Oliver Heaviside (1850–1925) studied electricity and magnetism while employed as a telegrapher. He is remembered for his great simpliﬁcation of Maxwell’s equations, his contributions to vector analysis, and his development of operational calculus. In 1902, Heaviside conjectured that a conducting layer exists in the atmosphere which allows radio waves to follow the curvature of Earth. This layer, now called the Heaviside layer, was detected in 1923.

5.1

provided the limit exists. Since

T

e−(s−a)t dt =

⎧ ⎨T,

Introduction

321

s=a −(s−a)T

⎩1 − e , s = a, s−a the improper integral deﬁned by the limit (3) exists if and only if s > a. Therefore, 1 F(s) = L{eat } = , s > a. (4) s−a (b) Similarly, ∞ T te−st dt = lim te−st dt, (5) L{t} = 0

0

T→∞

0

provided the limit exists. Since ⎧ 2 T ⎪ ⎪ s=0 ⎪ T ⎨ 2 , te−st dt = ⎪ −sT 0 ⎪ 1 ⎪ e ⎩ − 2 (1 + sT) + 2 , s = 0, s s the improper integral deﬁned by the limit (5) exists if and only if s > 0. Therefore, 1 F(s) = L{t} = 2 , s > 0. (6) s (c) Applying the deﬁnition gives ∞ T t2 t2 −st L{e } = e e dt = lim et(t−s) dt, 0

T→∞

0

provided the limit exists. For any ﬁxed value of s, however, the integrand, the limit does not exet(t−s) , is greater than 1 whenever t > s. Therefore, 2 ist for any value s and the function f (t) = et does not possess a Laplace transform. ❖

Piecewise Continuous Functions and Exponentially Bounded Functions We now identify a class of functions that possess Laplace transforms. If f (t) is a member of this class, then its Laplace transform exists for all s > a, where a is a constant that depends on the particular function f . We begin with two deﬁnitions. A function f (t) is called piecewise continuous on the interval 0 ≤ t ≤ T if (a) There are at most ﬁnitely many points, 0 ≤ t1 < t2 < · · · < tN ≤ T, at which T f (t) is not continuous, and (b) At any point of discontinuity, tj , the one-sided limits lim f (t) and

t→t−j

lim f (t)

t→ t+j

both exist. (If a discontinuity occurs at an endpoint, 0 or T, then only the interior one-sided limits need exist.) These discontinuities are called jump discontinuities.

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Condition (b) says the only discontinuities allowed for a piecewise continuous function are jump discontinuities. Condition (a) says a function that is piecewise continuous on the interval [0, T] never has more than a ﬁnite number of jump discontinuities in [0, T]. Note that the number of discontinuity points is a nondecreasing function of the interval length T; that is, if T2 > T1 , then NT ≥ NT . 2 1 A function deﬁned on 0 ≤ t < ∞ is called piecewise continuous on the interval 0 ≤ t < ∞ if it is piecewise continuous on 0 ≤ t ≤ T for all T > 0. An example of a function f (t) that is piecewise continuous on 0 ≤ t < ∞ is the “square wave” shown in Figure 5.3, where 1, 0 ≤ t ≤ 1, f (t) = f (t + 2) = f (t). 0, 1 < t < 2, Note that f (t) is a periodic function with period 2. Every discontinuity of f (t) is a jump discontinuity. While f (t) has inﬁnitely many discontinuities in 0 ≤ t < ∞, the function never has more than a ﬁnite number of discontinuities in a ﬁnite interval 0 ≤ t ≤ T.

f

f (t ) =

1, 0 ≤ t ≤ 1 0, 1 < t < 2 f (t + 2) = f (t )

1

1

2

3

4

5

6

t

FIGURE 5.3

The function f (t) is called a “square wave.” It has inﬁnitely many discontinuities in 0 ≤ t < ∞. They are all jump discontinuities, and there are never more than a ﬁnite number in any ﬁnite subinterval of 0 ≤ t < ∞. Therefore, f (t) is piecewise continuous on 0 ≤ t < ∞.

Our next deﬁnition is concerned with measuring how fast | f (t)| grows as t → ∞. A function f (t) deﬁned on 0 ≤ t < ∞ is called exponentially bounded on 0 ≤ t < ∞ if there are constants M and a, with M ≥ 0, such that | f (t)| ≤ Meat ,

0 ≤ t < ∞.

Figure 5.4 illustrates the nature of this deﬁnition. A function f (t) is exponentially bounded if we can ﬁnd constants M and a such that the graph of f (t) is contained in the region R, where R is bounded above by y = Meat and below by y = −Meat . If a function f (t) is bounded on 0 ≤ t < ∞, then it is also exponentially bounded; that is, if | f (t)| ≤ M, 0 ≤ t < ∞, then | f (t)| ≤ Meat , 0 ≤ t < ∞, with a = 0.

Existence of the Laplace Transform Theorem 5.1 establishes the existence of the Laplace transform for all functions that are piecewise continuous and exponentially bounded on 0 ≤ t < ∞. The proof is given in advanced texts.

5.1

Introduction

323

y

y = Me at

y = f (t) t

y = –Me at

FIGURE 5.4

The function f (t) is exponentially bounded because its graph is bounded above by y = Meat and below by y = −Meat . Hence, | f (t)| ≤ Meat , 0 ≤ t < ∞.

Theorem 5.1

Let f (t) be piecewise continuous and exponentially bounded on 0 ≤ t < ∞, where | f (t)| ≤ Meat , 0 ≤ t < ∞. Then the Laplace transform, ∞ F(s) = f (t)e−st dt, 0

exists for all s > a. In this chapter, we restrict our attention to functions that are piecewise continuous and exponentially bounded on 0 ≤ t < ∞. The next theorem, stated without proof, gives some closure properties for this special class of functions, asserting that we can form linear combinations and products of functions in the class and that any new functions produced will also belong to the class (and thus have Laplace transforms). Theorem 5.2

Let f1 (t) and f2 (t) be piecewise continuous and exponentially bounded on 0 ≤ t < ∞, where | f1 (t)| ≤ M1 ea1 t

and | f2 (t)| ≤ M2 ea2 t .

(a) Let f (t) = c1 f1 (t) + c2 f2 (t), where c1 and c2 are arbitrary constants. Then f (t) is also piecewise continuous and exponentially bounded on 0 ≤ t < ∞. In fact, | f (t)| ≤ Meat , where M = |c1 |M1 + |c2 |M2 and a = max {a1 , a2 }. Moreover, F(s) = L{ f (t)} is given by F(s) = L{c1 f1 (t) + c2 f2 (t)} = c1 L{ f1 (t)} + c2 L{ f2 (t)} = c1 F1 (s) + c2 F2 (s),

s > a.

(b) Let g(t) = f1 (t)f2 (t). Then g(t) is also piecewise continuous and exponentially bounded on 0 ≤ t < ∞. In fact, | g(t)| ≤ Meat , where M = M1 M2 and a = a1 + a2 . It follows that G(s) = L{ g(t)} exists for s > a.

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Since the Laplace transform satisﬁes the formula L{c1 f1 (t) + c2 f2 (t)} = c1 L{ f1 (t)} + c2 L{ f2 (t)},

(7)

we say that the Laplace transform is a linear transformation on the set of piecewise continuous, exponentially bounded functions. E X A M P L E

3

Determine whether the functions are exponentially bounded and piecewise continuous on 0 ≤ t < ∞: 1, 0 ≤ t ≤ 1, (a) f (t) = 0, 1 < t < 2, (b) g(t) = te t , 2

(c) k(t) = et ,

f (t) = f (t − 2) for

t≥2

0≤t 1. (c) If k(t) were exponentially bounded, then there would be constants M and 2 a such that et ≤ Meat for all nonnegative values t. In turn, this inequality would imply et(t−a) ≤ M,

0 ≤ t < ∞. 2

But, as t grows, the inequality has to fail eventually. Thus, k(t) = et is not exponentially bounded. The function is, however, piecewise continuous since it is continuous. ❖

5.1

Introduction

325

The Inverse Laplace Transform and Uniqueness Using the Laplace transform to solve problems involves three separate steps: (1) applying the transform to obtain a new transform domain problem, (2) solving the new transform domain problem, and (3) applying the inverse transform that maps the transform domain solution back to the time domain, resulting in the solution of the problem of interest. In order to deﬁne the inverse mapping (that is, the inverse Laplace transform), we need to know that the Laplace transform operation, when applied to functions that are piecewise continuous and exponentially bounded, possesses an underlying uniqueness property. In particular, given a transform domain function F(s), we want unambiguously to identify a function f (t) that has F(s) as its transform. The following theorem, which we present without proof, addresses the uniqueness question.

Theorem 5.3

Let f1 (t) and f2 (t) be piecewise continuous and exponentially bounded on 0 ≤ t < ∞. Let F1 (s) and F2 (s) represent their respective Laplace transforms. Suppose, for some constant a, that F1 (s) = F2 (s),

s > a.

Then f1 (t) = f2 (t) at all points t ≥ 0 where both functions are continuous.

This theorem gives about the best result we can hope for. As an illustration, consider the function f1 (t) = e−t , t ≥ 0. We saw in Example 2 that L{eat } =

1 , s−a

s > a.

Therefore, L{e−t } = F(s) =

1 , s+1

s > −1.

Suppose we create a new function f2 (t) by simply redeﬁning f1 (t) to be zero at each of the positive integers: e−t , t not an integer f2 (t) = (8) 0, t an integer. The graph of the function f2 (t) is shown in Figure 5.5. Observe, even though f1 (t) and f2 (t) are different functions, that T T −st f1 (t)e dt = f2 (t)e−st dt (9) 0

0

for every T > 0. Therefore, L{ f1 (t)} = L{ f2 (t)} = F(s), s > −1. As we see from equation (9), the improper integral deﬁning the Laplace transform is insensitive to changes in the value of a function at a ﬁnite number of points in 0 ≤ t ≤ T. This insensitivity, however, does not pose a serious

326

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Laplace Transforms f2 1.5

1

0.5

1

2

t

3

–0.5

FIGURE 5.5

The graph of the function f2 (t) deﬁned by equation (8). Note that the graph of f2 (t) is identical to the graph of f1 (t) = e−t except at t = 0, 1, 2, . . . . Even though the functions f1 (t) and f2 (t) are different, their Laplace transforms are the same [see equation (9)].

practical problem since we are interested in physically relevant functions. For example, in deﬁning the inverse Laplace transform of F(s) =

1 , s−a

we will choose it to be the continuous function f (t) = eat , t ≥ 0. Our approach to determining inverse Laplace transforms will be a tabular one. In the next several sections, we will compute Laplace transforms of functions and build up a library of Laplace transform pairs, such as the pair f (t) = eat ,

t≥0

F(s) =

and

1 , s−a

s > a.

(10)

Determining an inverse Laplace transform will essentially consist of a simple “table look-up” process. That is, we ﬁnd the appropriate transform domain function F(s) in the table and then take the corresponding time domain function f (t) in the table to be the inverse transform of F(s). The next example illustrates this approach. (In more advanced treatments, within the theory of complex variables, a more fundamental approach to computing inverse Laplace transforms is developed.) [Note: We will use the symbol L−1 { } to denote the operation of taking the inverse Laplace transform.]

E X A M P L E

4

What is the inverse Laplace transform of F(s) =

2s 2

s −1

,

s > 1?

That is, for what function f (t) do we have L{ f (t)} = F(s) = 2s/(s2 − 1)?

5.1

Introduction

327

Solution: We ﬁrst observe that the rational function F(s) has the following partial fraction expansion: 2s 2

s −1

=

1 1 + . s−1 s+1

(The topic of partial fractions is reviewed in Section 5.3. For now you can verify this claim by simply recombining the right-hand side.) Since the Laplace transform is a linear transformation, the inverse Laplace transform is likewise a linear transformation. In particular, 2s 1 1 1 1 −1 −1 −1 −1 L =L + =L +L . s−1 s+1 s−1 s+1 s2 − 1 Recalling the Laplace transform pair listed earlier in equation (10), we obtain 1 1 −1 t −1 L = e and L = e−t . s−1 s+1 Therefore, −1

L

2s s2 − 1

= e t + e−t = 2 cosh t,

t ≥ 0. ❖

EXERCISES Exercises 1–12: As in Example 2, use the deﬁnition to ﬁnd the Laplace transform for f (t), if it exists. In each exercise, the given function f (t) is deﬁned on the interval 0 ≤ t < ∞. If the Laplace transform exists, give the domain of F(s). In Exercises 9–12, also sketch the graph of f (t). 1. f (t) = 1

2. f (t) = e3t

√

(t−1)2

t t

5. f (t) = te 0, 9. f (t) = 1, ⎧ ⎪ ⎨0, 11. f (t) = 1, ⎪ ⎩ 0,

6. f (t) = e 0≤t 0,

to calculate the Laplace transform R(s) = L{r(t)} of the given function r(t). For what values s does the Laplace transform exist? 22. r(t) = 2e−5t + 6t

23. r(t) = 5e−7t + t + 2e2t

Exercises 24 –31: In each exercise, a function f (t) is given. In Exercises 28 and 29, the symbol [[ u ]] denotes the greatest integer function, [[ u ]] = n when n ≤ u < n + 1, n an integer, n = . . . , −2, −1, 0, 1, 2, . . . . (a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on 0 ≤ t < ∞, or neither? (b) Is f (t) exponentially bounded on 0 ≤ t < ∞? If so, determine values of M and a such that | f (t)| ≤ Meat , 0 ≤ t < ∞. 24. f (t) = tan t 28. f (t) = [[ t ]]

Exercises 32–35:

25. f (t) = e t sin t

29. f (t) = e2t

26. f (t) = t 2 e−t 2 et 30. f (t) = 2t e +1

27. f (t) = cosh 2t 1 31. f (t) = t

Determine whether the given improper integral converges. If the integral converges, give its value. ∞ ∞ 1 t 32. dt 33. dt 2 1 + t 1 + t2 0 0 ∞ ∞ 2 e−t cos(e−t ) dt 35. te−t dt 34. 0

0

Exercises 36–39: Suppose that L{ f1 (t)} = F1 (s) and L{ f2 (t)} = F2 (s), s > a. Use the fact that

L−1 {c1 F1 (s) + c2 F2 (s)} = c1 L−1 {F1 (s)} + c2 L−1 {F2 (s)},

a< s

5.2

Laplace Transform Pairs

329

to determine the inverse Laplace transform of the given function. Refer to the examples in this section and equation (11) in Exercise 15.

5.2

36. F(s) =

3 s−2

38. F(s) =

4s 2 2 = + s+2 s−2 s2 − 4

2 1 + s+1 s2 1 2 1 = 39. F(s) = 2 − s−1 s+1 s −1

37. F(s) = −

Laplace Transform Pairs This section develops a library of Laplace transform pairs that we will use to solve problems. We begin by deﬁning a function known as the unit step function or the Heaviside step function.

The Unit Step Function The unit step function or Heaviside step function, h(t), is the piecewise continuous function deﬁned by 1, t≥0 h(t) = 0, t < 0. Figure 5.6 displays graphs of h(t) and its “shifted argument” counterpart, h(t − α), α > 0.

1

h (t)

t

h (t – ␣)

1 ␣

t

FIGURE 5.6

The graphs of the unit step function, h(t), and the shifted step function, h(t − α).

The Laplace transform of the unit step function, h(t), is given by ∞ ∞ ∞ e−st 1 −st −st L{h(t)} = s > 0. h(t)e dt = e dt = − = s, s 0 0 0 In equation (1a), we use a common notation: ∞ f (t) a = lim f (t) − f (a). t→∞

(1a)

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For the shifted step function, h(t − α), we obtain the Laplace transform

∞

L{h(t − α)} =

h(t − α)e−st dt =

0

α

∞

e−st dt = −

∞ e−st e−sα , = s α s s>0

and

(1b)

α ≥ 0.

Note that the unit step function h(t) and the constant function f (t) = 1 are identical on 0 ≤ t < ∞. Therefore, they have the same Laplace transform, L{1} =

1 , s

s > 0.

(1c)

Transforms of Polynomial, Exponential, and Trigonometric Functions In this subsection, we develop some common transform pairs, starting with the polynomial function f (t) = t n . Also see Exercise 15 in Section 5.1. The Laplace Transform of f(t) = t n For n = 1, we use integration by parts to obtain

∞ ∞ e−st te−st 1 −st − 2 = 2, te dt = − s > 0. (2a) L{t} = s s s 0 0 In general, for any positive integer n, integration by parts yields n

L{t } =

∞

n −st

t e 0

∞ t n e−st n ∞ n−1 −st dt = − + t e dt. s 0 s 0

You can use L’Hôpital’s rule to show that lim t→∞ t n e−st = 0, s > 0. Therefore, we obtain the following reduction formula for L{t n }: L{t n } =

n L{t n−1 }, s

s > 0.

(2b)

Applying reduction formula (2b) recursively, we ﬁnd, for s > 0, L{t 2 } =

2 2 L{t} = 3 , s s

L{t3 } =

3 3·2 L{t 2 } = 4 s s

and, in general, L{t n } =

n! sn+1

,

n = 1, 2, 3, . . . ,

s > 0.

(3)

The Laplace Transform of f(t) = e αt We saw in Section 5.1 that L{eαt } =

1 , s−α

s > α.

(4)

5.2

Laplace Transform Pairs

331

The Laplace Transforms of f(t) = sin ωt and f(t) = cos ωt Using integration by parts twice yields ∞ L{sin ωt} = e−st sin ωt dt 0

−st ∞ e sin ωt ωe−st cos ωt ω2 ∞ −st − = − − e sin ωt dt s s2 s2 0 0 =

ω s2

−

ω2 s2

L{sin ωt},

s > 0.

Solving for L{sin ωt}, we ﬁnd L{sin ωt} =

ω 2

s + ω2

,

s > 0.

(5a)

,

s > 0.

(5b)

Similarly, L{cos ωt} =

s s2 + ω 2

We know from Section 5.1 that the Laplace transform deﬁnes a linear transformation on the set of piecewise continuous and exponentially bounded functions; that is, if f (t) and g(t) are piecewise continuous and exponentially bounded, then L{c1 f (t) + c2 g(t)} = c1 L{ f (t)} + c2 L{ g(t)}. We can use this linearity property to extend our library of transforms. For example, combining linearity with the transform for L{t n } listed in equation (3), we obtain the Laplace transform of any polynomial. The next example provides an illustration. E X A M P L E

1

Use the transform pairs developed above to ﬁnd the Laplace transform of (a) p(t) = 2t3 + 5t − 3, t≥0 (b) f (t) = 4 cos2 3t, t≥0 Solution: (a) Using linearity and equation (3), we have L{ p(t)} = L{2t3 + 5t − 3} = 2L{t3 } + 5L{t} − 3L{1} =2

3! s4

+5

1 s2

−3

12 + 5s2 − 3s3 1 = , s s4

s > 0.

(b) We have no transform pair directly involving cos2 ωt. However, we can use a trigonometric identity to rewrite f (t) = 4 cos2 3t as f (t) = 2 + 2 cos 6t. Using linearity and equation (5b) yields L{ f (t)} = 2L{1} + 2L{cos 6t} =2

s 4s2 + 72 1 +2 2 = , s s + 36 s(s2 + 36)

s > 0. ❖

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Two Shift Theorems The next two results, established in Theorem 5.4, are often referred to as the ﬁrst and second shift theorems. Like the linearity property illustrated in Example 1, the shift theorems increase the number of functions for which we can easily ﬁnd Laplace transforms.

Theorem 5.4

Let f (t) be piecewise continuous and exponentially bounded on 0 ≤ t < ∞, where |f (t)| ≤ Meat , 0 ≤ t < ∞. Let F(s) = L{ f (t)}, and let h(t) denote the unit step function. Then (a) L{eαt f (t)} = F(s − α), −αs

(b) L{ f (t − α)h(t − α)} = e

s>a+α F(s), α ≥ 0,

s > a.

Since h(t − α) = 0 when t < α (see Figure 5.6), f (t − α)h(t − α) = 0,

t < α.

The graph of f (t − α)h(t − α) looks just like the graph of f (t) except for the fact that it is shifted to the right and remains zero until t = α. Figure 5.7 provides an example.

f (t) t

f (t – ␣)h (t – ␣) t

␣ FIGURE 5.7

The graphs of f (t) and f (t − α)h(t − α). Note that the graph of f (t − α)h(t − α) looks like the graph of f (t) except that it is shifted α units to the right.

●

PROOF (of Theorem 5.4):

(a) The following calculation establishes part (a): ∞ ∞ L{eαt f (t)} = eαt f (t)e−st dt = f (t)e−(s−α)t dt = F(s − α), 0

0

s > a + α.

5.2

Laplace Transform Pairs

(b) To establish the second shift theorem, we begin with ∞ −st L{ f (t − α)h(t − α)} = f (t − α)h(t − α)e dt =

∞

α

0

333

f (t − α)e−st dt.

Making the change of variable τ = t − α, we have ∞ ∞ f (t − α)e−st dt = f (τ )e−s(τ +α) dτ L{ f (t − α)h(t − α)} = α

= e−sα

0 ∞

f (τ )e−sτ dτ = e−sα F(s),

s > a.

●

0

Note that parts (a) and (b) of Theorem 5.4 possess a certain duality. Roughly speaking, multiplying a function by an exponential function in the time domain shifts the argument of its Laplace transform. Likewise, shifting the argument in the time domain leads to an exponential multiplicative factor in the transform domain. E X A M P L E

2

Find

αt

2t 4

(a) L{e t }

(b) L{e cos ωt}

−1

(c) L

e−5s

s2

−1

(d) L

e−αs

s2 + 1

Solution: (a) By Theorem 5.4, multiplying f (t) by eαt shifts the argument of its transform, F(s). That is, 4! L{e2t t4 } = L{t4 } = , s > 2. s→s−2 (s − 2)5 (b) As in part (a), L{eαt cos ωt} = L{cos ωt}|s→s−α =

s−α (s − α)2 + ω2

,

s > α.

(c) We know that L{t} = 1/s2 , t ≥ 0. By the second shift theorem, e−5s s2 Therefore,

−1

L

e−5s s2

= L{(t − 5)h(t − 5)}.

0, 0≤t max{a, 0}.

(c) Let f (t) be piecewise continuous and exponentially bounded on 0 ≤ t < ∞, where | f (t)| ≤ Meat , 0 ≤ t < ∞. Then t F(s) L{ f (t)} L = , s > max{a, 0}. f (u) du = s s 0

●

PROOF: The proof of part (a) is presented to illustrate the relevant ideas. By hypothesis, the function f (t) is continuous. We now show it is also exponen-

5.2

Laplace Transform Pairs

335

tially bounded, and thus f (t) has a Laplace transform. We have t t t f (u) du ≤ | f (0)| + f (u) du ≤ | f (0)| + Meau du, | f (t)| = f (0) + 0

0

0

t ≥ 0.

Therefore, for 0 ≤ t < ∞, ⎧

M at M at ⎪ ⎪ | f (0)| + (e − 1) ≤ | f (0)| + e , a>0 ⎪ ⎪ ⎪ a a ⎪ ⎨ | f (t)| ≤ | f (0)| + Mt, a=0 ⎪ ⎪ ⎪ ⎪ M M ⎪ ⎪ (1 − eat ) ≤ | f (0)| + , a < 0. ⎩| f (0)| + |a| |a|

(6)

From these inequalities, we are able to conclude that L{f (t)} = F(s) exists for s > max{a, 0}. To obtain (a), consider the interval 0 ≤ t ≤ T for some arbitrary T > 0. Let t1 < t2 < · · · < tN represent the points of discontinuity of f (t) on this interval. Then t T 1 f (t)e−st dt = f (t)e−st dt 0

0

+

t2

f (t)e−st dt + · · · +

t1

tN

f (t)e−st dt +

T

f (t)e−st dt.

tN

tN−1

Performing integration by parts on each of these integrals yields T t t t T f (t)e−st dt = f (t)e−st 01 + f (t)e−st t2 + · · · + f (t)e−st tN + f (t)e−st t 0

+s

1

t1

f (t)e

0

−st

dt +

t2

−st

f (t)e

N−1

dt + · · · +

t1

tN

f (t)e

−st

N

dt +

T

−st

f (t)e

dt .

tN

tN−1

Since f (t) is continuous, the sum of the endpoint evaluations reduces to f (T)e−sT − f (0). Similarly, the sum of integrals on the right-hand side can be expressed as a single integral from 0 to T. Therefore, we obtain T T f (t)e−st dt = f (T)e−sT − f (0) + s f (t)e−st dt. 0

0

Now let T → ∞, while assuming that s > max{a, 0}. For these values of s, T lim T→∞ f (T)e−sT = 0 and lim T→∞ 0 f (t)e−st dt = L{ f (t)} = F(s). Therefore, the result follows. ● Note that differentiation in the time domain corresponds, roughly speaking, to multiplication by s in the transform domain, while antidifferentiation in the time domain corresponds to division by s in the transform domain.

Solving Initial Value Problems The next example, while quite simple, illustrates how we can use Laplace transforms to solve initial value problems. Following a review of the method of partial fractions in Section 5.3, we give a more detailed discussion.

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E X A M P L E

3

Consider the initial value problem y − 3y = g(t),

y(0) = 1,

where g(t) is the step function given by 0, 0 ≤ t < 2 g(t) = 6, 2 ≤ t < ∞. Let Y (s) denote the Laplace transform of y(t), where y(t) is the unique solution of this initial value problem. Using Theorem 5.5, derive an equation for Y (s) and, taking the inverse Laplace transform, ﬁnd y(t). Solution: The nonhomogeneous term g(t) can be represented as g(t) = 6h(t − 2), where h(t − 2) denotes the shifted Heaviside step function. (See Figure 5.6.) Taking Laplace transforms of y (t) − 3y(t) = 6h(t − 2),

0 ≤ t < ∞,

we have L{ y (t)} − 3L{ y(t)} = 6L{h(t − 2)}. Using part (a) of Theorem 5.5 to evaluate L{ y (t)} and part (b) of Theorem 5.4 to evaluate L{ h(t − 2)}, we ﬁnd [sY (s) − y(0)] − 3Y (s) =

6e−2s . s

Solving for Y (s) and using the fact that y(0) = 1, we obtain Y (s) =

1 6e−2s + , s − 3 s(s − 3)

s > 3.

Using a partial fraction expansion for the second term on the right-hand side gives 2 1 2 −2s Y (s) = +e − , s > 3. s−3 s−3 s Therefore, y(t) = L−1

1 s−3

+ 2L−1 e−2s

1 s−3

1 − 2L−1 e−2s . s

Using the second shifting theorem, we see that y(t) = e3t + 2h(t − 2)[e3t−6 − 1], t ≥ 0, is the unique solution of the initial value problem. In particular, y(t) is the piecewise-deﬁned function e3t , 0≤t a for some value a. After we formally execute the three steps—Laplace transformation, solution of the transformed problem, and inverse Laplace transformation—we will obtain what we can regard as a candidate for the solution of our initial value problem. If we can directly verify that the “candidate solution” obtained by the use of transforms is, in fact, the unique solution of the original problem of interest, then we are done. Example 4 illustrates these points and the three-step Laplace transform solution procedure, using it to analyze a network problem.

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Laplace Transforms

E X A M P L E

4

The series RLC network shown in Figure 5.9 is assumed to be initially quiescent; that is, the current and the charge on the capacitor are both zero for t ≤ 0. At time t = 0, a voltage source v(t) = v0 te−αt , having the polarity shown, is turned on. Determine the current i(t) for t ≥ 0. R

L i(t)

+

~

v(t)

C

–

FIGURE 5.9

The RLC network analyzed in Example 4.

Solution: Recall that the underlying principle for describing our problem mathematically is Kirchhoff’s voltage law (see Section 3.10). As we traverse the network in a clockwise manner, the voltage rise through the source must equal the sum of the voltage drops through the resistor R, inductor L, and capacitor C. The resulting equation, along with the accompanying supplementary conditions, is 1 t di(t) v(t) = Ri(t) + L + i(u) du, i(0) = 0, t ≥ 0, dt C 0 or 1 di(t) R + i(t) + L LC dt

t

i(u) du =

0

1 v(t), L

i(0) = 0,

t ≥ 0.

(7)

When we considered this problem in Section 3.10, we differentiated equation (7) to obtain a second order differential equation for the current i(t). Now, however, we will work directly with equation (7), which is an integrodifferential equation for the unknown current, i(t). The ﬁrst step is to compute the Laplace transform of both sides of equation (7), obtaining t 1 1 R di L i(u) du = L{v(t)}. + L{i} + L dt L LC L 0 This equation can be written as sI(s) − i(0) +

R 1 I(s) 1 I(s) + = V(s). L LC s L

(8)

Notice that the supplementary condition involving i(0) enters directly into the transformed equation (8). In our case, i(0) = 0. Since v(t) = v0 te−αt , we have V(s) = v0 L{te−αt } = v0 L{t} s→s+α = v0

1 (s + α)2

.

5.2

Laplace Transform Pairs

339

The transform domain problem is therefore entirely deﬁned by the algebraic equation: sI(s) +

1 I(s) v 1 R I(s) + = 0 . L LC s L (s + α)2

(9)

The second step is to solve transform domain problem (9). We ﬁnd I(s) =

v0 L

s

(s + α)2 s2 +

R 1 s+ L LC

.

(10)

The third step is to ﬁnd the inverse Laplace transform of I(s). To accomplish this, we use a partial fraction expansion to decompose rational function (10) into a sum of terms, each of whose inverse Laplace transform is known. For an illustration of this third step with a speciﬁc case, suppose that I(s) in equation (10) is given by I(s) =

50s 2

(s + 1) (s2 + 4s + 13)

.

(11)

Expression (11) has the partial fraction expansion 5 6s + 13 6 − − 2 s + 1 (s + 1) (s + 2)2 + 9 5 3 s+2 6 1 − −6 = − , 2 s + 1 (s + 1)2 3 (s + 2) + 9 (s + 2)2 + 9

I(s) =

(12)

where we have used the fact that s2 + 4s + 13 = (s + 2)2 + 9. The algebraic manipulations leading to the last expression in (12) were done in anticipation of the inverse Laplace transform computation. Applying the inverse Laplace transform to I(s) yields i(t) = L−1 {I(s)} 1 1 = 6L−1 − 5L−1 s+1 (s + 1)2 s+2 3 1 −1 −1 − 6L − L . 3 (s + 2)2 + 9 (s + 2)2 + 9

(13)

The required inverse transforms can be obtained from Table 5.1 at the end of this section. When these inverse transforms are used in equation (13), it follows that i(t) = 6e−t − 5te−t − 6e−2t cos 3t − 13 e−2t sin 3t,

t ≥ 0.

As a ﬁnal check, one should verify that this expression for the network current is, in fact, the desired solution. ❖ The network current is plotted in Figure 5.10; its behavior seems reasonable. Since the source voltage is proportional to te−t , one would expect the current to exhibit a transient behavior followed by an approach to zero as t → ∞.

340

CHAPTER 5

Laplace Transforms i 2 1.5

i(t)

1 0.5 1

2

3

4

5

6

t

–0.5 –1

FIGURE 5.10

The network current found in Example 4 for the RLC network of Figure 5.9. Since the source voltage is proportional to te−t , we expect the current to consist of an initial transient variation followed by an approach to zero as time increases.

TA B L E 5 . 1 A Table of Laplace Transform Pairs Time Domain Function f (t), t ≥ 0 1, t ≥ 0

1.

h(t) =

2.

t n,

3.

eαt

4.

sin ωt

5.

cos ωt

6.

sinh αt

7.

cosh αt

8.

eαt f (t), with |f (t)| ≤ Meat

0, t < 0 n = 1, 2, 3, . . .

Laplace Transform F (s) 1 , s>0 s n! , s>0 sn+1 1 , s>α s−α ω , s>0 s2 + ω2 s , s>0 s2 + ω2 α , s > |α| s2 − α 2 s , s > |α| s2 − α 2 F(s − α), s>α+a

(9)–(12) are four special cases of (8): 9.

eαt h(t)

10. eαt t n , 11. eαt sin ωt 12. eαt cos ωt

n = 1, 2, 3, . . .

1 , s>α s−α n! , s>α (s − α)n+1 ω , s>α (s − α)2 + ω2 (s − α) , s>α (s − α)2 + ω2

5.2

TA B L E 5 . 1 13. f (t − α)h(t − α),

Laplace Transform Pairs

341

(continued) e−αs F(s),

α ≥ 0,

s>a

at

with | f (t)| ≤ Me

(14) is a special case of (13): 14. h(t − α),

e−αs , s

α≥0

s>0

15. f (t), with f (t) continuous and | f (t)| ≤ Meat

sF(s) − f (0), s > max{a, 0}

16. f (t), with f (t) continuous and | f (t)| ≤ Meat

s2 F(s) − sf (0) − f (0), s > max{a, 0}

17. f (n) (t), with f (n−1) (t) continuous and | f (n) (t)| ≤ Meat

sn F(s) − sn−1 f (0) − · · · − sf (n−2) (0) − f (n−1) (0), s > max{a, 0}, n = 1, 2, 3, . . .

t

F(s) , s

f (u) du, with | f (t)| ≤ Meat

18. 0

s > max{a, 0}

19.

1 (sin ωt − ωt cos ωt) 2ω3

1 , (s2 + ω2 )2

s>0

20.

t sin ωt 2ω

s , (s + ω2 )2

s>0

2

−F (s)

21. tf (t)

EXERCISES Exercises 1–12: Use Table 5.1 to ﬁnd L{ f (t)} for the given function f (t) deﬁned on the interval t ≥ 0. 1. f (t) = 3t 2 + 2t + 1 3t−3

4. f (t) = e

2. f (t) = 2e t + 5 2

h(t − 1)

5. f (t) = (t − 1) h(t − 1)

7. f (t) = 2te−2t

8. f (t) = sin 3t cos 3t

10. f (t) = e2t cos 3t

11. f (t) = e3t h(t − 1)

3. f (t) = 1 + sin 3t 6. f (t) = sin2 ωt 9. f (t) = 2th(t − 2) 12. f (t) = e4t (t 2 + 3t + 5)

Exercises 13–21: −1

Use Table 5.1 to ﬁnd L {F(s)} for the given F(s). 3 24 + 4 s s 2s − 4 15. F(s) = (s − 2)2 + 9 3 17. F(s) = e−2s 2 s +9 4s − 6 19. F(s) = 2 s − 2s + 10 13. F(s) =

21. F(s) =

48(e−3 s + 2e−5 s ) s5

4 10 + s2 + 25 s − 3 5 16. F(s) = (s − 3)4 e−2 s 18. F(s) = s−9 e−3 s (2s + 7) 20. F(s) = s2 + 16

14. F(s) =

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Laplace Transforms

Exercises 22–33: Combinations of Shifted Heaviside Step Functions Exercises 22–33 deal with combinations of Heaviside step functions. As the two examples below show, we can use combinations of shifted Heaviside step functions to represent pulses.

f

f

1

1

1

2

3

t

4

1

(a) f (t) = h(t – 1) – h(t – 3)

2

3

t

4

(b) f (t) = (t – 1)[h(t – 1) – h(t – 2)]

In each exercise, graph the function f (t) for 0 ≤ t < ∞, and use Table 5.1 to ﬁnd the Laplace transform of f (t). 22. f (t) = h(t − 1) + h(t − 3)

23. f (t) = sin(t − 2π)h(t − 2π)

24. f (t) = t[h(t − 1) − h(t − 3)]

25. f (t) = h(t) − h(t − 3)

26. f (t) = 3[h(t − 1) − h(t − 4)]

27. f (t) = (2 − t)[h(t − 1) − h(t − 3)]

28. f (t) = |2 − t|[h(t − 1) − h(t − 3)] 29. f (t) = [h(t − 1) − h(t − 2)] − [h(t − 2) − h(t − 3)] 30. h(2 − t)

31. e−2t h(1 − t)

32. h(t − 1) + h(4 − t)

33. h(t − 2) − h(3 − t)

Exercises 34 –37: In each exercise, the graph of f (t) is given. Represent f (t) using a combination of Heaviside step functions, and use Table 5.1 to calculate the Laplace transform of f (t). 34. f

f

35.

1

2

1

2

3

4

5

t

1

1

2

3

4

1

2

3

4

t

37. f

36. f

1

1

1

2

3

4

t

t

An Introduction to the Method of Partial Fractions We will present a review of the method of partial fractions in Section 5.3. By way of introduction, however, we consider here a special case of the method and use it to solve some initial value problems, as in Examples 3 and 4. Suppose F(s) = 1/Q(s), where Q(s) = (s − r1 )(s − r2 ) · · · (s − rn ) and where the roots r1 , r2 , . . . , rn are real and distinct. In this case, there are constants A1 , A2 , . . . , An

5.2

Laplace Transform Pairs

343

such that F(s) =

A1 A2 An 1 = + + ··· + . (s − r1 )(s − r2 ) · · · (s − rn ) s − r1 s − r2 s − rn

(14)

One way to determine the constants A1 , A2 , . . . , An is to recombine the right-hand side into a single rational function and equate the resulting numerator to 1.

Exercises 38– 41: −1

Using a partial fraction expansion, ﬁnd L {F(s)}. In Exercise 40, compare your answer with (6) in Table 5.1. 12 (s − 3)(s + 1) 24e−5s 40. F(s) = 2 s −9

38. F(s) =

4 s(s + 2) 10e−s 41. F(s) = 2 s − 5s + 6 39. F(s) =

Exercises 42– 45: As in Examples 3 and 4, use Laplace transform techniques to solve the initial value problem. ⎧ 0≤t α 2 , complete the square and rewrite this factor in the form (s + α)2 + ω2 . 1. F(s) =

2s + 3 (s − 1)(s − 2)2

s3 + 3s + 1 (s − 1)3 (s − 2)2 s2 + 5s − 3 4. F(s) = 2 (s + 16)(s − 2) s3 − 1 6. F(s) = 2 (s + 1)2 (s + 4)2 s4 + 5s2 + 2s − 9 8. F(s) = 2 (s + 8s + 17)2 (s − 2)2 2. F(s) =

s2 + 1 s (s + 2s + 10) s2 − 1 5. F(s) = 2 (s − 9)2 s2 + s + 2 7. F(s) = 2 (s + 8s + 17)(s2 + 6s + 13) 3. F(s) =

2

2

Exercises 9–17: Find the inverse Laplace transform. 2 s−3 2s − 3 12. F(s) = 2 s − 3s + 2 9. F(s) =

15. F(s) =

3s2 + s + 8 s3 + 4s

1 (s + 1)3 3s + 7 13. F(s) = 2 s + 4s + 3

10. F(s) =

16. F(s) =

s2 + 6s + 8 s + 8s2 + 16 4

4s + 5 s2 + 9 4s2 + s + 1 14. F(s) = s3 + s s 17. F(s) = 3 s − 3s2 + 3s − 1 11. F(s) =

Exercises 18–29: Use the Laplace transform to solve the initial value problem. 18. y + 2y = 26 sin 3t, y(0) = 3

19. y − 3y = 13 cos 2t, y(0) = 1

20. y + 2y = 4t, y(0) = 3

21. y − 3y = e3t , y(0) = 1

350

CHAPTER 5

Laplace Transforms 22. y + 3y + 2y = 6e−t , y(0) = 1, y (0) = 2 23. y + 4y = 8t, y(0) = 2, y (0) = 6

24. y + 4y = cos 2t, y(0) = 1, y (0) = 1

25. y + 4y = sin 2t, y(0) = 1, y (0) = 0 26. y − 2y + y = e2t , y(0) = 0, y (0) = 0 27. y + 2y + y = e−t , y(0) = 0, y (0) = 1 28. y + 9y = g(t), y(0) = 1, y (0) = 3, 29. y + y = g(t), y(0) = 1, y (0) = 0,

g(t) = g(t) =

6,

0≤t 0.

Evaluating this last integral, we ﬁnd 1 − e−sT/2 s−1 1 . ❖ = L{ f (t)} = −sT s 1 + e−sT/2 1−e

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Laplace Transforms

E X A M P L E

2

Find the inverse transform of F(s) =

1 s2

−

e−s , s(1 − e−s )

s > 0.

Applying the inverse transform operation yields 1 e−s e−s −1 −1 −1 −1 L {F(s)} = L −L =t−L . s(1 − e−s ) s(1 − e−s ) s2

Solution:

Expanding the factor (1 − e−s )−1 as a geometric series and applying the inverse operation termwise to the convergent series, we obtain −s e−s e −1 −1 −s −2s −3s [1 + e + e L +e + · · ·] =L s s(1 − e−s ) 1 −s = L−1 [e + e−2s + e−3s + · · ·] s −s e−2s e−3s e −1 −1 −1 =L +L +L + ··· s s s = h(t − 1) + h(t − 2) + h(t − 3) + · · · . The function g(t) = h(t − 1) + h(t − 2) + h(t − 3) + · · · has the staircase-like graph shown in Figure 5.12(a). Thus, the inverse transform of F(s) is the sawtooth wave function f (t) = t − g(t) whose graph is shown in Figure 5.12(b). [Note: For any ﬁxed value of t, g(t) = h(t − 1) + h(t − 2) + h(t − 3) + · · · is actually a ﬁnite sum, since h(t − α) = 0 when t < α. For instance, if t = 2.5, then g(2.5) = 2.] g 2

f

1

1

1

2

3

t

(a)

1

2

3

t

(b) FIGURE 5.12

(a) The graph of g(t) = h(t − 1) + h(t − 2) + h(t − 3) + · · · resembles a staircase. (b) The inverse transform of F(s) in Example 2 is f (t) = t − g(t); this graph is often called a sawtooth wave.

❖

Solution of Parameter Identiﬁcation Problems and the System Transfer Function Example 1 of Section 5.1 posed the problem of studying a “black box” that housed a spring-mass-dashpot mechanical system (see Figure 5.2). Two speciﬁc questions are 1. If we subject the initially quiescent system to a known force f (t), starting at t = 0, and measure the subsequent displacement y(t) for t ≥ 0, can we

5.4

Laplace Transforms of Periodic Functions and System Transfer Functions

353

use our measurements to predict what the displacement y˜ (t) would be if a different force f˜ (t) were applied? 2. Can we use our knowledge of the input-output relation [that is, our knowledge of f (t) and y(t)] to determine the mass m, the spring constant k, and the damping coefﬁcient γ of the unknown mechanical system? The relevant mathematical problem is my + γ y + ky = f (t), y(0) = 0, y (0) = 0.

t>0

(3)

We now use Laplace transforms to provide afﬁrmative answers to these two questions. Taking Laplace transforms of both sides of equation (3) and noting the zero initial conditions, we have ms2 Y (s) + γ sY (s) + kY (s) = F(s), or

Y (s) =

1 ms2 + γ s + k

F(s).

(4)

Although the computations in (4) are simple, the result is important. In the time domain, we obtain the output y(t) from the input f (t) by solving an initial value problem. In the transform domain, however, we obtain the output Y (s) from the input F(s) by multiplying F(s) by the function 1 (5) (s) = . ms2 + γ s + k Note that the function (s) in (5) depends only on the mechanical system; it is sometimes referred to as the system transfer function. If we know (s), we can use multiplication to determine the output Y (s) arising from a given input F(s). Conversely, if we know some input-output pair F(s) and Y (s), we can determine the system transfer function (s) by forming the quotient Y (s) . (s) = F(s) The role of the system transfer function is shown schematically in Figure 5.13. Time domain f (t)

Transform domain Input

Solve the initial value problem my ″ + ␥ y ′ + ky = f (t) y (0) = 0, y′(0) = 0

y(t)

F(s)

Form the product Φ(s) F(s) where Φ is the system transfer function

Output

Y(s)

FIGURE 5.13

There are two ways to analyze the mechanical spring-mass-dashpot system. In the time domain, solve the initial value problem given by equation (3). In the transform domain, form the product of the system transfer function (s) and the input F(s), as in equation (4).

354

CHAPTER 5

Laplace Transforms

Example 3 illustrates how these transform domain ideas can be used to answer the two questions posed above. E X A M P L E

3

Suppose we know that the response of an initially quiescent mechanical system to an applied force can be modeled as the solution of the spring-mass-dashpot initial value problem my + γ y + y = f (t), y(0) = 0, y (0) = 0.

t>0

Assume, for a known applied force f (t), we can measure the resulting displacement y(t) for t ≥ 0. We are, however, unable to directly determine the parameters m, γ , and k. In particular, suppose when we apply a unit step force f (t) = h(t), the displacement is y(t) = − 12 e−t cos t − 12 e−t sin t + 12 ,

t ≥ 0.

Use this information to (a) Predict the displacement should the force f˜ (t) = e−2t , t ≥ 0 be applied. (b) Determine the parameters m, γ , and k. Solution: To solve the problem, we ﬁrst compute Laplace transforms of the applied force f (t) = h(t) and the ensuing response y(t) = − 12 e−t cos t − 1 −t e sin t + 12 . We obtain 2 F(s) =

1 s

Y (s) = −

1

s+1 2

2 (s + 1) + 1

−

1

1 2

2 (s + 1) + 1

+

1 2s

=

1 2

s(s + 2s + 2)

.

In the transform domain, Y (s) = (s)F(s), where (s) is the system transfer function. Therefore, the system transfer function is given by (s) =

Y (s) 1 = 2 . F(s) s + 2s + 2

(6)

Once we know the system transfer function, we can readily predict the output corresponding to any input. (a) Suppose the applied force is f˜ (t) = e−2t . The Laplace transform of the ap˜ plied force is F(s) = 1/(s + 2). We can ﬁnd the transform of the displace˜ ment from the relationship Y˜ (s) = (s)F(s):

1 1 1 = 2 Y˜ (s) = 2 s + 2s + 2 s + 2 (s + 2s + 2)(s + 2) s+1 1 1 1 1 1 + . = − 2 2 s + 2 2 (s + 1) + 1 2 (s + 1)2 + 1 The corresponding time domain output is thus y˜ (t) = L−1 {Y˜ (s)} = 12 e−2t − 12 e−t cos t + 12 e−t sin t,

t ≥ 0.

5.4

Laplace Transforms of Periodic Functions and System Transfer Functions

355

(b) The problem posed in question (b) can be solved by comparing the transfer function (s) =

1 2

s + 2s + 2

with the previously determined form of the transfer function in equation (5), (s) =

1 2

ms + γ s + k

.

Comparing coefﬁcients, we conclude that m = 1,

γ = 2,

k = 2. ❖

In the preceding discussion, we assumed that the mechanical system was initially at rest and an applied force then activated the system at time t = 0. If the initial conditions are nonzero but known, the same general approach can be used. These ideas are developed in the Exercises.

EXERCISES Exercises 1–8: Find the Laplace transform of the periodic function whose graph is shown. y

1.

y

2.

3

3 1 2

4

t

6

t

1

2

3

4

5

1

2

3

4

5

6

7

1

2

3

4

5

6

7

–3 y

3.

y

4. 2

2

1 2

4

t

6

t

8

–3 y

5.

y

6.

1

1 1

2

3

4

5

6

t

t

356

CHAPTER 5

Laplace Transforms y

7. 1

y

8. 1 – t2 1

2 2

3

4

5

t

6

1 1

2

3

4

5

6

t

Exercises 9–12: Sketch the graph of f (t), state the period of f (t), and ﬁnd L{ f (t)}. 9. f (t) = |sin 2t| sin t, 10. f (t) = 0,

0 ≤ t < π, π ≤ t < 2π,

f (t + 2π) = f (t)

11. f (t) = e−t , 0 ≤ t < 1, f (t + 1) = f (t) 12. f (t) = 1 − e−t , 0 ≤ t < 2, f (t + 2) = f (t) 13. Let α be a positive constant. As in Example 2, show that −1

L

e−αs s(1 − e−αs )

= h(t − α) + h(t − 2α) + h(t − 3α) + · · · .

Sketch the graph of g(t) = h(t − α) + h(t − 2α) + h(t − 3α) + · · · for α = 1 and 0 ≤ t < 5.

Exercises 14 –15: In each exercise, use linearity of the inverse transformation and Exercise 13 to ﬁnd −1 f (t) = L {F(s)} for the given transform F(s). Sketch the graph of f (t) for 0 ≤ t < 5 in Exercise 14 and 0 ≤ t < 10 in Exercise 15. 14. F(s) =

e−s s−1 + 2 s s(1 − e−s )

15. F(s) =

3 3e−2s − 2 s s(1 − e−2s )

−1

16. As in Example 2, ﬁnd f (t) = L {F(s)} for F(s) = 1/2s2 − (1/s2 )[e−2s /(1 + e−2s )]. Sketch the graph of f (t) for 0 ≤ t < 12.

Exercises 17–19: One-Dimensional Motion with Drag and Periodic Thrust Assume a body of mass m moves along a horizontal surface in a straight line with velocity v(t). The body is subject to a frictional force proportional to velocity and is propelled forward with a periodic propulsive force f (t). Applying Newton’s second law, we obtain the following initial value problem: mv + kv = f (t),

t ≥ 0,

v(0) = v0 .

Assume that m = 1 kg, k = 1 kg/s, and v0 = 1 m/s. (a) Use Laplace transform methods to determine v(t) for the propulsive force f (t), where f (t) is given in newtons. (b) Plot v(t) for 0 ≤ t ≤ 10 [this time interval spans the ﬁrst ﬁve periods of f (t)]. In Exercise 17, explain why v(t) is constant on the interval 0 ≤ t ≤ 1. 1, 0 ≤ t ≤ 1, 17. f (t) = f (t + 2) = f (t) 0, 1 < t < 2, 0, 0 ≤ t ≤ 1, 18. f (t) = f (t + 2) = f (t) 1, 1 < t < 2, 19. f (t) = t/2,

0 ≤ t < 2,

f (t + 2) = f (t)

5.4

Laplace Transforms of Periodic Functions and System Transfer Functions

357

20. An object having mass m is initially at rest on a frictionless horizontal surface. At time t = 0, a periodic force is applied horizontally to the object, causing it to move in the positive x-direction. The force, in newtons, is given by f0 , 0 ≤ t ≤ T/2, f (t) = f (t + T) = f (t). 0, T/2 < t < T, The initial value problem for the horizontal position, x(t), of the object is mx (t) = f (t), x(0) = 0, x (0) = 0. (a) Use Laplace transforms to determine the velocity, v(t) = x (t), and the position, x(t), of the object. (b) Let m = 1 kg, f0 = 1 N, and T = 1 s. What are the velocity v and position x of the object at t = 1.25 s? 21. A lake containing 50 million gal of fresh water has a stream ﬂowing through it. Water enters the lake at a constant rate of 5 million gal/day and leaves at the same rate. At some initial time, an upstream manufacturer begins to discharge pollutants into the feeder stream. Each day, during the hours from 8 A.M. to 8 P.M., the stream has a pollutant concentration of 1 mg/gal (10−6 kg/gal); at other times, the stream feeds in fresh water. Assume that a well-stirred mixture leaves the lake and that the manufacturer operates seven days per week. (a) Let t = 0 denote the instant that pollutants ﬁrst enter the lake. Let q(t) denote the amount of pollutant (in kilograms) present in the lake at time t (in days). Use a “conservation of pollutant” principle (rate of change = rate in − rate out) to formulate the initial value problem satisﬁed by q(t). (b) Apply Laplace transforms to the problem formulated in (a) and determine Q(s) = L{q(t)}. −1

(c) Determine q(t) = L {Q(s)}, using the ideas of Example 2. In particular, what is q(t) for 1 ≤ t < 2, the second day of manfacturing? 22. Consider the RL and RC networks shown, with the associated equations for the current i(t).

R

R

L(t) e(t)

L(t)

~

e(t)

L

~

C

Ri + 1 C

Ri + Li' = e(t), i(0) = 0

t 0

i ()d = e(t)

Figure for Exercise 22

Assume that the network element values are R = 1 k , L = 1 H, C = 1μF and that e(t), given in volts, is 0, 0 ≤ t ≤ 0.5, e(t) = e(t + 1) = e(t). 1, 0.5 < t < 1, The associated units of current and time are milliamperes and milliseconds, respectively. (a) Determine i(t) for the RL network. (b) Determine i(t) for the RC network. Transfer Function Problems Consider the initial value problem ay + by + cy = f (t), y(0) = 0,

y (0) = 0,

0 0.

Given this model of the impulsive force, Newton’s laws of motion lead to the following initial value problem: mv + kv = F0 δ(t − t0 ), v(0) = v0 .

t>0

(9)

Once we know v(t), the position of the body is given by t x(t) = v(λ) dλ.

(10)

0

We will use Laplace transforms to solve the problem. Let V(s) = L{v(t)}

and

X(s) = L{x(t)}.

Noting equation (8), we have for the Laplace transform of equation (9) m[sV(s) − v0 ] + kV(s) = F0 e−st0 . Therefore, v0

V(s) =

s+

k m

+

F0 e−st0 , k m s+ m

and hence v(t) = v0 e−(k/m)t +

F0 −(k/m)(t−t ) 0 h(t − t ), e 0 m

t ≥ 0.

(11)

We can ﬁnd position x(t) by computing the antiderivative of velocity v(t), as in equation (10). Alternatively, we can use the fact that X(s) = to obtain

⎡ X(s) = v0

m⎢ ⎢1 − k ⎣s

1 V(s) s

⎤

⎡

⎤

⎥ F0 −st ⎢ 1 1 ⎥ ⎥+ ⎥. e 0⎢ − ⎣ ⎦ k k ⎦ k s s+ s+ m m 1

Taking inverse transforms, we ﬁnd ' F $ ' m$ x(t) = v0 1 − e−(k/m)t + 0 1 − e−(k/m)(t−t0 ) h(t − t0 ), k k

t ≥ 0. ❖

(12)

The solid curves in Figure 5.22 are graphs of velocity and position of the body for the parameter values m = 5 kg,

k = 0.5 kg/s,

v0 = 20 m/s,

F0 = 500 N,

t0 = 3 s. (13)

382

CHAPTER 5

Laplace Transforms v

x 600

120 100

500

v(t)

80

400

60

300

40

200

20

100 1

2

3

4

5

6

7

8

x (t)

t

1

2

3

(a)

4

5

6

7

8

t

(b) FIGURE 5.22

The results from Example 1. (a) The graph of velocity, v(t), as given in equation (11). The discontinuity is the result of an impulsive force applied at t = 3. (b) The graph of position, x(t), as given in equation (12).

As the graph illustrates, application of the impulsive force creates a jump discontinuity in the velocity. This jump is the idealization of the very rapid velocity transition that would occur if the applied force were a very narrow pulse of integrated strength 500 newton-seconds. The dotted curves in Figure 5.23 show the velocity and position that would result from a force of 5000 N being applied during the interval 3 ≤ t ≤ 3.1 sec. x

v 120

600

100

500

80

400

60

300

40

200

20

100 1

2

3

4

5

6

7

8

t

(a)

1

2

3

4

5

6

7

8

(b) FIGURE 5.23

The graphs of (a) velocity and (b) position for the problems described by equations (9) and (14). Equation (9) models the idealized problem, using the delta function. Equation (14) models the problem using a large (but ﬁnite) pulse applied over a t-interval of small (but nonzero) duration. As you can see, the graphs are qualitatively similar.

t

5.7

The Delta Function and Impulse Response

383

In other words, the dotted curves arise from solving the initial value problem mv + kv = F0 pε (t − t0 ),

t>0

(14)

v(0) = v0 ,

with ε = 0.1 sec and all other parameter values as given by (13). The comparison of these graphs illustrates the “idealizing nature” of using the delta function in modeling applications. [The solution of problem (14) is outlined in the Exercises.]

The Impulse Response and the System Transfer Function The formal use of the delta function as an impulsive source leads to the fact that the impulse response and the system transfer function form a Laplace transform pair. For example, consider the initial value problem an

dn y dn−1 y dn−2 y dy + a0 y = δ(t) + a + · · · + a1 n + an−1 n−2 n−1 n−2 dt dt dt dt

(n−1)

y

(n−2)

(0) = 0,

y

(0) = 0,

...,

y (0) = 0,

(15)

y(0) = 0,

where we use the delta function to model an impulsive nonhomogeneous term. The solution of (15) is the impulse response of an nth order linear system. Taking Laplace transforms of (15) and using equation (8), we ﬁnd (an sn + an−1 sn−1 + an−2 sn−2 + · · · + a1 s + a0 )Y (s) = 1, and therefore Y (s) =

1 an sn + an−1 sn−1 + an−2 sn−2 + · · · + a1 s + a0

.

(16)

The right-hand side of equation (16) is the Laplace transform of the impulse response and is equal to the system transfer function.

EXERCISES 1. Evaluate 3 (1 + e−t )δ(t − 2) dt (a)

−2

0

2

(c) −1

cos 2t te−t

1

(b)

⎡ ⎤ δ(t + 2) ⎢ ⎥ (e2t + t) ⎣δ(t − 1)⎦ dt −3 δ(t − 3)

δ(t) dt

(d)

(1 + e−t )δ(t − 2) dt

2

2. Let f (t) be a function deﬁned and continuous on 0 ≤ t < ∞. Determine t f ∗δ = f (t − λ)δ(λ) dλ. 0

3. Determine a value of the constant t0 such that 5 4. If 1 t n δ(t − 2) dt = 8, what is the exponent n?

1 0

sin2 [π(t − t0 )]δ(t − 12 ) dt = 34 .

t 5. Sketch the graph of the function f (t) deﬁned by f (t) = 0 δ(λ − 1) dλ, 0 ≤ t < ∞. Can the graph obtained be characterized in terms of a Heaviside step function?

384

CHAPTER 5

Laplace Transforms tλ 6. Sketch the graph of the function g(t) that is deﬁned by g(t) = 0 0 δ(σ − 1) dσ dλ, 0 ≤ t < ∞. t 7. Sketch the graph of the function k(t) = 0 [δ(λ − 1) − δ(λ − 2)] dλ, 0 ≤ t < ∞. Can the graph be characterized in terms of a Heaviside step function or Heaviside step functions? t 8. The graph of the function g(t) = 0 eαt δ(t − t0 ) dt, 0 ≤ t < ∞ is shown. Determine the constants α and t0 . g(t) e –2

t

2 Figure for Exercise 8

Exercises 9–11: In each exercise, a function g(t) is given. (a) Solve the initial value problem y − y = g(t), y(0) = 0, using the techniques developed in Chapter 2. (b) Use Laplace transforms to determine the transfer function φ(t), φ − φ = δ(t),

φ(0) = 0.

t (c) Evaluate the convolution integral φ ∗ g = 0 φ(t − λ)g(λ) dλ, and compare the resulting function with the solution obtained in part (a). 9. g(t) = h(t)

10. g(t) = e t

11. g(t) = t

Exercises 12–20: Solve the given initial value problem, in which inputs of large amplitude and short duration have been idealized as delta functions. Graph the solution that you obtain on the indicated interval. (In Exercises 19 and 20, plot the two components of the solution on the same graph.) 12. y + y = 2 + δ(t − 1), y(0) = 0, 0 ≤ t ≤ 6 13. y + y = δ(t − 1) − δ(t − 2), y(0) = 0, 0 ≤ t ≤ 6 14. y = δ(t − 1) − δ(t − 3), y(0) = 0, y (0) = 0, 0 ≤ t ≤ 6 15. y + 4π 2 y = 2πδ(t − 2), y(0) = 0, y (0) = 0, 0 ≤ t ≤ 6 16. y − 2y = δ(t − 1), y(0) = 1, y (0) = 0, 0 ≤ t ≤ 2 17. y + 2y + 2y = δ(t − 1), y(0) = 0, y (0) = 0, 0 ≤ t ≤ 6 18. y + 2y + y = δ(t − 2), y(0) = 0, y (0) = 1, 0 ≤ t ≤ 6

y1 (0) 0 1 1 y1 d y1 1 , = = + δ(t − 1) 19. , 0≤t≤2 0 dt y2 y2 (0) 0 1 1 y2 0 2 1 y1 0 1 y1 (0) d y1 − δ(t − 1) 20. = = + , , 0≤t≤2 dt y2 0 y2 (0) 0 1 y2 1 0

Projects

385

PROJECTS Project 1: Periodic Pinging of a Spring-Mass System In Chapter 3, we considered the response of a spring-mass system to a periodic applied force. In particular, we saw that the solution of the initial value problem y + ω02 y = F cos ω1 t,

y(0) = 0,

y (0) = 0

has an envelope that grows linearly with time when ω1 = ω0 . When ω1 = ω0 , on the other hand, the solution envelope remains bounded with time. Suppose that, instead of applying a continuous force, we ping the spring-mass system periodically. In other words, we apply a force of very short duration at equally spaced time intervals. Can we achieve an analogous resonant growth in the solution envelope if the time interval between pings is properly chosen? Consider the initial value problem y + ω02 y = F

∞

δ(t − mT),

y(0) = 0,

y (0) = 0,

m=1

where we use the delta function to model applied pings of short duration. The positive constant T represents the time interval between successive pings. 1. Solve the initial value problem using Laplace transforms. Assume that formal manipulations, such as interchanging the order of inverse Laplace transformation and inﬁnite summation, are valid. 2. Consider the case where the time interval is T = 2π/ω0 . In this case, the interval between pings equals the resonant period of the vibrating system. Discuss the qualitative behavior of the solution. Does the solution envelope exhibit some form of resonant growth? As a speciﬁc case, assume ω0 = 2π and F = 2π . Plot the solution over the time interval 0 ≤ t < 10. 3. Now consider the case where T = π/ω0 . (This interval between pings is half the resonant period.) Again, assume that ω0 = 2π and F = 2π. Plot the solution over the time interval 0 ≤ t < 10. Does the solution envelope remain bounded or grow with time? Provide a physical rationale for the observed behavior of the solution.

Project 2: Curing Sick Fish Assume that the tropical ﬁsh in a 100-gal aquarium have contracted an ailment and that a soluble medication must be administered to combat the illness. The medicine is packaged in 800-mg doses, and one dose is to be administered daily. Assume that the following facts are known: (i) A “well-stirred” approximation is valid; that is, the medicine dissolves and disperses itself throughout the tank very rapidly. (ii) The medicine loses potency at a rate proportional to the amount of medicine present. In fact, the half-life of the medicine (the time span over which the potency is reduced to one-half its initial strength) is one day. (iii) In order to effectively combat the illness, the concentration of medicine in the tank must be maintained at a level greater than or equal to 5 mg per gallon for a period of 7 days.

386

CHAPTER 5

Laplace Transforms

Let q(t) denote the amount of potent medicine (in milligrams) in the tank at time t (in days). Assume that the illness is detected at time t = 0 and that the ﬁrst dose of medicine is administered at time t = 1. If N doses are administered on consecutive days, the problem to be solved is q (t) + kq(t) = 800

N

δ(t − n),

q(0) = 0.

n=1

Because of the well-stirred assumption, we can use the delta function to model the administration of each dose. Moreover, the concentration is c(t) = q(t)/100 mg/gal. 1. Determine the constant k and solve the initial value problem using Laplace transforms. 2. Determine the minimum number of doses needed to effectively combat the ailment. 3. What would happen if we continued to administer the medicine (that is, if N became arbitrarily large)? Would the maximum amount of medicine in the tank continue to grow, or would q(t) undergo an initial transient phase and then settle into a periodic, steady-state behavior as time increased? 4. Suppose we deﬁne N q(N) =

q(λ) dλ,

N = 1, 2, 3, . . . .

N−1

Thus, q(N) is the average amount of medicine present in the tank during the Nth day. Show, from the differential equation itself, that if q(t) does settle into a periodic behavior as time increases, then lim q(N) =

N→∞

800 . k

Project 3: Locating a Transmission Line Fault Laplace transformation is an operational tool that can be used to map a given problem into a simpler “transformed problem.” We have seen how problems involving ordinary differential equations can be transformed into problems involving simpler algebraic equations. We now consider a problem where Laplace transforms can be used to transform a problem involving partial differential equations into a simpler problem involving ordinary differential equations. The steps outlined in Figure 5.1 remain the same; we ﬁrst solve this simpler problem and then use the inverse Laplace transform to ﬁnd the desired solution. The problem considered is a simple application of the idea of echo location. Knowing how fast sound travels in air, we can determine the distance to a reﬂection point by measuring the time separation between when a sound is emitted and when its echo is heard. This basic idea can be used to determine where a transmission line fault or disruption is located. A transmission line is an example of a distributed network. A transmission line is unlike the networks considered earlier in that the voltage and current are functions of both space and time. Consider Figure 5.24, where the transmission line is represented by two parallel cables. The variable x measures distance along the line, with a voltage source or generator positioned at x = 0 and the fault (assumed to be an open circuit) located at x = l. We assume that the location of the fault is unknown; our goal is to locate it by sending a short pulse down the line and measuring the two-way transit time—the time it takes for the pulse reﬂected by the fault to return to the source. As shown in Figure 5.24, we represent the voltage across the line and the current along the line at position x and time t by v(x, t) and i(x, t), respectively. The voltage

Projects

i(x, t)

Rg eg(t)

387

+ v(x, t) –

~ x=0

x=l FIGURE 5.24

A transmission line network. A voltage generator is connected at x = 0, and an open circuit is assumed to exist at the unknown fault location, x = l. generator is assumed to have an internal resistance Rg . Figure 5.25 depicts a snapshot of a differential transmission line segment taken at some time t. As the ﬁgure indicates, the transmission line itself is characterized by a series inductance L per unit length and a shunt capacitance C per unit length. To determine how the transmission line voltage and current behave as functions of space and time, we apply Kirchhoff’s voltage and current laws to this differential segment of line. The voltage drop across the inductance is L

∂i(x, t) dx, ∂t

while the current ﬂow through the capacitance is C

i(x, t)

+

∂v(x, t) dx. ∂t

i(x + dx, t)

–

L dx

+ v(x, t)

+ Cdx

v(x + dx, t)

–

–

x

x + dx FIGURE 5.25

Differential transmission line segment equivalent circuit.

If we apply Kirchhoff’s voltage law to the circuit in Figure 5.25, we obtain v(x, t) − L

∂i(x, t) dx − v(x + dx, t) = 0. ∂t

Similarly, applying Kirchhoff’s current law leads to i(x, t) − C

∂v(x, t) dx − i(x + dx, t) = 0. ∂t

If we divide by dx and let dx → 0, we obtain a pair of partial differential equations ∂v(x, t) ∂i(x, t) = −L ∂x ∂t

(1)

∂v(x, t) ∂i(x, t) = −C . ∂x ∂t We assume that the transmission line is quiescent for t ≤ 0. That is, we assume i(x, 0) = 0,

v(x, 0) = 0,

0 ≤ x ≤ l.

(2)

388

CHAPTER 5

Laplace Transforms At time t = 0, the voltage generator is turned on, emitting a signal eg (t), t > 0. Applying Kirchhoff’s voltage law at the generator leads to eg (t) − i(0, t)Rg − v(0, t) = 0,

t > 0.

(3)

Lastly, the assumption that an open circuit exists at fault location x = l leads us to the constraint i( l, t) = 0,

t > 0.

(4)

Equations (1)–(4) constitute the mathematical problem of interest. We are free to select the generator voltage eg (t). Our goal is to determine a formula for v(0, t), t > 0. As we will see, this formula contains l as a parameter. Since v(0, t) is a quantity that can be measured, we will use a voltage measurement to determine the distance l to the fault. Once the location of the fault is known, appropriate repairs can be made. Since 0 < t < ∞ is the time interval of interest, we can deﬁne the Laplace transforms: ∞ ∞ V(x, s) ≡ v(x, t)e−st dt, I(x, s) ≡ i(x, t)e−st dt. (5) 0

0

In (5), the variable x is treated as a parameter. 1. Apply the Laplace transform (5) to both sides of equations (1)–(4). Assume that the order of operations can be interchanged. For example,

∞ ∞ ∂ ∂V(x, s) ∂v(x, t) −st e dt = . v(x, t)e−st dt = ∂x ∂x 0 ∂x 0 Show that an application of the Laplace transform leads to the following transformed problem: ∂V(x, s) = −sLI(x, s) ∂x ∂I(x, s) = −sCV(x, s) ∂x Eg (s) − I(0, s)Rg = V(0, s) I(l, s) = 0,

(6a) (6b) (6c) (6d)

where Eg (s) denotes the Laplace transform of eg (t). Note that problem (6) is, in fact, simpler. The only differentiation performed in (6) is with respect to the spatial variable x. If we view the transform variable s as a parameter, then equations (6a)–(6b) are essentially a linear system of ordinary differential equations. A problem involving partial differential equations has been transformed into one that de facto involves only ordinary differential equations. Note that problem (6) is not an initial value problem. It is a two-point boundary value problem; the spatial domain is 0 ≤ x ≤ l, and the supplementary conditions (6c)–(6d) are prescribed at the two endpoints. 2. Obtain the general solution of equations (6a)–(6b), viewed as a linear system of ordinary differential equations. Note that since transform variable s is being viewed as a parameter, the two arbitrary constants appearing in the general solution will generally be functions √ of s. The quantity L/C has the dimensions of resistance; it is called the characteristic impedance of the transmission line and is often denoted by the symbol Z0 . Assume that Rg = Z0 . When this condition holds, the voltage generator is said to be “matched to the transmission line.” Impose constraints (6c)–(6d) and show that ' √ Eg (s) $ V(0, s) = (7) 1 + e−s(2l LC) . 2

Projects

389

3. Determine v(0, t), t > 0 by computing the inverse Laplace transform of (7). The prod√ uct l LC has the dimensions of time. Assume that the generator voltage eg (t) is a very short pulse, say 10, 0 < t ≤ 0.1 eg (t) = 0, 0.1 < t < ∞, √ and that l LC = 5. Graph v(0, t) as a function of time t for t > 0. Explain the physical signiﬁcance of the two terms comprising v(0, t). Suppose, for example, we know the properties √ of the transmission line; speciﬁcally, suppose we know L and C and, therefore, LC. Explain how your solution can be used to determine the unknown distance l.

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6

C H A P T E R

Nonlinear Systems CHAPTER OVERVIEW 6.1

Introduction

6.2

Equilibrium Solutions and Direction Fields

6.3

Conservative Systems

6.4

Stability

6.5

Linearization and the Local Picture

6.6

Two-Dimensional Linear Systems

6.7

Predator-Prey Population Models

6.1

Introduction In this chapter, we consider systems of nonlinear differential equations y1 = f1 (t, y1 , y2 , . . . , yn ) y2 = f2 (t, y1 , y2 , . . . , yn ) .. .

yn = fn (t, y1 , y2 , . . . , yn ),

(1)

a < t < b.

To formulate an initial value problem, we specify n initial conditions y1 (t0 ) = y01 ,

y2 (t0 ) = y02 ,

...,

yn (t0 ) = y0n ,

(2)

where t0 is some point belonging to the interval a < t < b. The special case of n = 1 reduces to the scalar nonlinear problem y = f (t, y), y(t0 ) = y0 , treated in Chapter 2.

391

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The Vector Form for a Nonlinear System We can express the nonlinear system (1) in a compact fashion using vector notation. In particular, deﬁne the vector functions ⎡ ⎤ ⎡ ⎡ ⎤ ⎤ y01 f y1 (t) 1 (t, y1 , y2 , . . . , yn ) ⎥ ⎢ ⎢ ⎢ ⎥ ⎥ ⎢ 0⎥ ⎢f2 (t, y1 , y2 , . . . , yn )⎥ ⎢y2 (t)⎥ y ⎥ ⎢ 2 ⎥ ⎥, f(t, y) = ⎢ y0 = ⎢ . ⎥ . y(t) = ⎢ .. ⎢ ⎥ ⎢ .. ⎥ , . ⎢ ⎥ . ⎣ ⎦ ⎣ . ⎦ ⎣.⎦ 0 yn (t) fn (t, y1 , y2 , . . . , yn ) yn With this notation, we can write the initial value problem as y (t) = f(t, y(t)),

a 0. Therefore, all the direction ﬁeld arrows in region 1 point upward and to the right. Figure 6.6 shows, in schematic form, the general orientation of direction ﬁeld arrows in each of the four open regions. It also shows the orientation of the direction ﬁeld arrows on the nullclines; these arrows are either horizontal (if f = 0 and g = 0) or vertical (if f = 0 and g = 0). The information in Figure 6.6 is sufﬁcient to deduce the general qualitative behavior of phase-plane trajectories. For example, if x and y are both initially nonzero, then solutions will tend toward the equilibrium state x = y = 1 as time increases. While quite useful, a rough graph such as the one in Figure 6.6 may not be detailed enough to yield a good qualitative picture of the phase-plane trajectories. The next example illustrates this point.

6.2

Equilibrium Solutions and Direction Fields

y

407

y

1: 2: 3: 4:

f=0 2

(0, 32 )

3

f f f f

> > <

< < >

0 0 0 0

(1, 1) 1

g=0 (0, 0)

4

x

(2, 0)

x

(a)

(b) FIGURE 6.5

(a) The lines denote the nullclines for system (10). (b) The nullclines divide the direction ﬁeld of Figure 6.4 into regions where arrows all have the same general orientation. y

x

FIGURE 6.6

Along a nullcline, the arrows are either vertical or horizontal. In an open region bounded by nullclines, the single arrow indicates the general orientation of the solution curves in that region. This ﬁgure suggests that any solution curve starting in one of the open regions will move toward the equilibrium point (1, 1). E X A M P L E

5

Consider the pendulum equation treated in Example 1, where g/l = 1: x = y y = − sin x. (a) Sketch the nullclines, as in Figure 6.5(a), marking the equilibrium points with a heavy dot. Then, as in Figure 6.6, add arrows to indicate the ﬂow of solution curves. Does this sketch have enough detail to predict the qualitative nature of phase-plane trajectories? (continued)

408

CHAPTER 6

Nonlinear Systems (continued)

(b) Using a computer, sketch a portion of the direction ﬁeld for −8 ≤ x ≤ 8 and −6 ≤ y ≤ 6. Using your sketch, describe the two different types of phaseplane trajectories and give a physical interpretation for each type. (Recall that x denotes angular displacement from the pendulum’s downward-hanging equilibrium position and y denotes angular velocity.) Solution: (a) The nullclines consist of the x-axis and the inﬁnite set of vertical lines x = mπ, m = 0, ±1, ±2, . . . . A portion of the phase plane is shown in Figure 6.7. This ﬁgure also shows the direction ﬁeld arrows on the nullclines as well as the general “sense of direction” arrows within the vertical phaseplane strips. The arrows are vertical on the nullcline y = 0 (the x-axis) and are horizontal on the nullclines x = mπ , m = 0, ±1, ±2, . . . . For gaining a good qualitative picture of the phase-plane trajectories, however, this level of description is inadequate. y

f = 0: g = 0:

(1, 1) –2

–

x 2

FIGURE 6.7

A sketch showing the main features of the direction ﬁeld for the pendulum equation in Example 5. Note that we cannot tell from this sketch whether a solution curve passing through the point (1, 1) will continue down until it is below the x-axis or whether it will remain above the x-axis.

For example, consider a solution curve passing through the point (1, 1). As we see from Figure 6.7, the solution point is moving downward and to the right when it passes through (1, 1). But is the curve falling fast enough that it will cross the x-axis and continue to move down but now to the left? Or is its rate of descent slowing enough that it will intersect the line x = π in the upper half of the phase plane and then move up and to the right? We need a more detailed direction ﬁeld in order to give a reasonable assessment. (b) Figure 6.8 presents a more detailed direction ﬁeld plot. This plot indicates that there are two basic types of trajectories. Near the x-axis, there appear to be closed phase-plane trajectories; as time increases, the solution point seems to make clockwise orbits around these closed curves. Far from the x-axis, the trajectories no longer appear to be closed curves. Rather, they appear to be undulating curves that are basically horizontally oriented and that tend to become ﬂatter as distance from the x-axis increases.

6.2

Equilibrium Solutions and Direction Fields

409

y 6

4

2

x –8

–6

–4

–2

2

4

6

8

–2

–4

–6

FIGURE 6.8

A portion of the direction ﬁeld for the pendulum equation in Example 5. Some phase-plane trajectories appear to be closed curves, centered at equilibrium points on the x-axis. Other trajectories appear to be undulating curves where the solution point moves basically in one direction (to the right above the x-axis and to the left below the x-axis). The typical closed trajectory corresponds to a pendulum swinging back and forth, motion in which the pendulum never reaches the vertically upward position. The undulating trajectories correspond to a pendulum that continues to rotate in one direction (counterclockwise if y = θ is positive, clockwise if y = θ is negative).

❖

We can understand Figure 6.8 in terms of its physical interpretation. Recall that x = θ represents the angular displacement of the pendulum from its downward hanging equilibrium position and y = θ represents the instantaneous angular velocity of the pendulum. The closed orbits close to the x-axis therefore correspond to motion in which the pendulum swings back and forth. For example, consider a closed trajectory centered at the equilibrium point (0, 0). On such a trajectory, the maximum excursion of x = θ from zero is less than π . This maximum displacement is reached when the trajectory intersects the xaxis—that is, when angular velocity y is zero. In such a motion, the pendulum never reaches the vertically upward position. It swings up to some maximum angular displacement less than π and then swings back the same amount in the other direction. The continual orbiting of the solution point around a closed trajectory corresponds to this continual back-and-forth swing of the pendulum. The horizontally conﬁgured undulating trajectories correspond to motion in which the pendulum continually rotates about its pivot. In the upper halfplane, y = θ is positive and the pendulum is always rotating in the counterclockwise direction. In the lower half-plane, y = θ is negative and the pendulum is always rotating in the clockwise direction. Since y = θ is never zero, the pendulum never stops and consequently never changes direction.

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Nonlinear Systems

The pendulum is an example of a conservative system (that is, a system in which energy is conserved). For such systems, and for a more general twodimensional class of autonomous systems known as Hamiltonian systems, we can derive equations for the phase-plane trajectories. We consider such systems in the next section.

EXERCISES Exercises 1–10: Find all equilibrium points of the autonomous system. 1. x = −x + xy

y = y − xy 3. x = (x − 2)( y + 1) y = x2 − 4x + 3 5. x = x2 − 2xy

2. x = y(x + 3) y = (x − 1)( y − 2) 4. x = xy − y + x − 1 y = xy − 2y 6. x = y2 − xy

y = 3xy − y2

y = 2xy + x2

7. x = x2 + y2 − 8

8. x = x2 + 2y2 − 3

y = x2 − y2 9. x = y − 1

y = 2x2 + y2 − 3 10. x = z2 − 1

y = xy + x2

y = z − 2xz + yz

z = 2y − yz

z = −(1 − x − y)2

Exercises 11–15: Rewrite the given scalar differential equation as a ﬁrst order system, and ﬁnd all equilibrium points of the resulting system. 11. y + y + y3 = 0 2y + y2 = 1 1 + y4 4 − y2 =0 15. y − ( y )2 + 2 + ( y )2 13. y +

12. y + ey y + sin2 π y = 1 14. y − y + 2 sin y = 1

Exercises 16– 19: Use the information provided to determine the unspeciﬁed constants. 16. The system x = x + αxy + β y = γ y − 3xy + δ has equilibrium points at (x, y) = (0, 0) and (2, 1). Is (−2, −2) also an equilibrium point? 17. The system x = αx + βxy + 2 y = γ x + δy2 − 1 has equilibrium points at (x, y) = (1, 1) and (2, 0).

6.2

Equilibrium Solutions and Direction Fields

411

18. Consider the system x = x + αy3 y = x + βy + y4 . The slopes of the phase-plane trajectories passing through the points (x, y) = (2, 1) and (1, −1) are 1 and 0, respectively. 19. Consider the system x = αx2 + βy + 1 y = x + γ y + y2 . The slopes of the phase-plane trajectories passing through the points (x, y) = (1, 1) and (1, −1) are 0 and 4, respectively. The phase-plane trajectory passing through the point (x, y) = (0, −1) has a vertical tangent. 20. Consider the system x = x + y2 − xyn y = −x + y−1 . The slope of the phase-plane trajectory passing through the point (x, y) = (1, 2) is 1 . Determine the exponent n. 6 21. The scalar differential equation y − y + 2y2 = α, when rewritten as a ﬁrst order system, results in a system having an equilibrium point at ( x, y) = (2, 0). Determine the constant α. 22. For the given system, compute the velocity vector v(t) = x (t)i + y (t)j at the point (x, y) = (2, 3). (a) x = −x + xy

y = y − xy

(b) x = y(x + 3)

y = (x − 1)( y − 2)

(c) x = (x − 2)( y + 1) y = x2 − 4x + 3

23. Let A be a (2 × 2) constant matrix, and let (λ, u) be an eigenpair for A. Assume that λ is real, λ = 0, and

u1 u= . u2 Consider the phase plane for the autonomous linear system y = Ay. We can deﬁne a phase-plane line through the origin by the parametric equations x = τ u1 , y = τ u2 , −∞ < τ < ∞. Let P be any point on this line, say P = (τ0 u1 , τ0 u2 ) for some τ0 = 0. (a) Show that at the point P, x = τ0 λu1 and y = τ0 λu2 . (b) How is the velocity vector v(t) = x (t)i + y (t)j at point P oriented relative to the line?

Exercises 24–27: In each exercise, a matrix A is given. For each matrix, the vectors 1 1 and u2 = u1 = 1 −1 are eigenvectors of A. As discussed in Exercise 23, these eigenvectors are associated with the phase-plane lines y = x and y = −x. Solution points of y = Ay originating on these two lines remain on these lines as time evolves. Match the given matrix A to one of the four direction ﬁelds shown (on the next page) for y = Ay.

−9 1 −1 −3 −4 6 4 2 24. A = 25. A = 26. A = 27. A = 1 −9 −3 −1 6 −4 2 4

412

CHAPTER 6

Nonlinear Systems y

y

3

3

2

2

1

1

x

x –3

–2

–1

1

2

–3

3

–2

–1

1

–1

–1

–2

–2

–3

–3

Direction Field A

Direction Field B

y

y

3

3

2

2

1

1

2

3

2

3

x –3

–2

–1

1

2

3

x –3

–2

–1

1

–1

–1

–2

–2

–3

–3

Direction Field C

Direction Field D Figure for Exercises 24 –27

28. Suppose that the nonlinear second order equation y + f ( y) = 0 is recast as an autonomous ﬁrst order system. Show that the nullclines for the resulting system are the horizontal line y = 0 and vertical lines of the form x = α, where α is a root of f ( y) = 0. For each such root, what is the nature of the phase-plane point ( x, y) = (α, 0)?

Exercises 29–31: (a) Rewrite the given second order equation as an equivalent ﬁrst order system. (b) Graph the nullclines of the autonomous system and locate all equilibrium points. (c) As in Figure 6.6, sketch direction ﬁeld arrows on the nullclines. Also, sketch an arrow in each open region that suggests the direction in which a solution point is moving when it is in that region. 29. y + y + y3 = 0

30. y + y(1 − y2 ) = 0

31. y + 2 sin2 y = 1

6.3

Conservative Systems

413

Exercises 32–36: In each exercise, (a) Graph the nullclines of the autonomous system and locate all equilibrium points. (b) As in Figure 6.6, sketch direction ﬁeld arrows on the nullclines. Also, sketch an arrow in each open region that suggests the direction in which a solution point is moving when it is in that region. 32. x = 3x − y − 2

y = x−y 35. x = (2x − y − 6)(x − y) y = x + y

6.3

33. x = −x − y + 2

34. x = (2x − y − 2)(4 − x − y) y = x − 2y

y = x−y

36. x = x2 + y − 1 y = −x2 + y + 1

Conservative Systems A mathematical model of a physical system often neglects effects such as friction or electrical resistance if they are small enough. We have already encountered several of these idealized mathematical models in our discussion of spring-mass systems, buoyant bodies, and pendulums. As a consequence of such assumptions, these idealized models obey what is usually called a conservation law. In particular, a conservation law means that a quantity, such as energy, remains constant. For example, consider an idealized pendulum. On its upswing, as the bob elevates and simultaneously slows down, energy is converted from kinetic energy to potential energy. On the downswing, potential energy is, in turn, transformed back into kinetic energy. We will show, for this idealized pendulum model, that total pendulum energy (the sum of kinetic and potential energy) remains constant in time. Thus, total pendulum energy is a conserved quantity in the idealized pendulum model. In general, consider a second order scalar differential equation y = f (t, y, y ), and let y(t) be a solution of this differential equation. If there is a function of two variables H(u, v) such that H( y(t), y (t)) remains constant in time, then we call H a conserved quantity and say that the differential equation y = f (t, y, y ) possesses a conservation law. We use the same terms to describe the general case of an n-dimensional system with solution components y1 (t), y2 (t), . . . , yn (t) for which some function H( y1 (t), y2 (t), . . . , yn (t)) remains constant. In this section, we are interested in the following questions: 1. Given a mathematical model, how can we determine (from the structure of the differential equation itself) whether or not it satisﬁes a conservation law? 2. If the model does possess a conserved quantity, how can we explicitly describe this quantity and use its mathematical description to better understand the system’s dynamics?

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An Important Class of Second Order Scalar Equations Consider the differential equation y + f ( y) = 0.

(1)

Such differential equations often arise when we apply Newton’s laws to a body in one-dimensional motion. In such applications, y corresponds to acceleration and the term −f ( y) corresponds to the applied force per unit mass. In fact, three of the mathematical models we have discussed—the undamped massspring system, the buoyant body, and the pendulum—all have the structure of equation (1). We now show that equation (1) possesses a conservation law. To see why, and to obtain a description of this law, we ﬁrst multiply the equation by y , obtaining y y + f ( y)y = 0.

(2)

Consider the terms in (2). Recalling the chain rule of calculus, we see that the ﬁrst term on the left-hand side can be written as d 1 2 . ( y (t)) (3) y (t)y (t) = dt 2 If F( y) denotes an antiderivative of f ( y), the chain rule allows us to express the second term in (2) as d F( y(t)). dt Using (3) and (4), we can rewrite (2) in the form d 1 ( y (t))2 + F( y(t)) = 0. 2 dt Therefore, f ( y(t))y (t) =

1 y (t)2 2

+ F( y(t)) = C.

(4)

(5)

(6)

Equation (6) is the underlying conservation law. For instance, if y(t) represents displacement, then the term 12 y (t)2 is kinetic energy per unit mass. The other term, F( y(t)), is potential energy per unit mass (measured relative to some reference value that depends on the particular antiderivative F chosen). The constant C can be interpreted as the (constant) total energy per unit mass.

Phase-Plane Interpretation Differential equation (1) can be recast as the ﬁrst order autonomous system x = y y = −f (x), where x and y now play the roles of y(t) and y (t), respectively. Thus, the conservation law (6) takes the form 1 ( y)2 2

+ F(x) = C.

(7)

The family of curves obtained by graphing this equation for different values of C is a set of phase-plane trajectories describing the motion. The next example develops these ideas for the pendulum.

6.3

Conservative Systems

415

E X A M P L E

1

Consider the pendulum equation (recall Example 5, Section 6.2) x = y y = − sin x. From (7), the corresponding conservation law is 1 2 y 2

− cos x = C.

If we revert to the original variables, θ and θ , and use E to denote the constant energy of the system, the conservation law has the form1 1 2 (θ ) 2

− cos θ = E.

(8)

The term −cos θ represents the potential energy of the pendulum bob, measured relative to a zero reference at the horizontal position. Equation (8) is graphed in Figure 6.9 for various energy levels. The direction of solution point motion on these trajectories is indicated with arrows. (The direction ﬁeld arrows in Figure 6.8 of Section 6.2 are tangent to these phase-plane trajectories.) The entire phase plane is simply a periodic repetition of the portion shown; every 2π increment in x = θ brings the pendulum back to the same physical conﬁguration. To understand Figure 6.9, observe from equation (8) that the equilibrium point (θ, θ ) = (0, 0) (with the bob at rest hanging vertically downward) corresponds to the energy level E = −1. In a similar fashion, the equilibrium points (θ, θ ) = (π, 0) and (θ, θ ) = (−π, 0) (with the bob at rest and positioned vertically upward) correspond to the energy level E = 1. The energy value E = 1 is a delineating, or “separating,” value. Phase-plane trajectories for −1 < E < 1 are the closed curves in Figure 6.9 centered at (0, 0). These trajectories correspond to motion in which the pendulum continuously swings back and forth; it does not have enough energy to reach the vertical upward position. The pendulum swings upward to some θmax < π , stops, and then swings downward, achieving the same maximum elevation on its backswing. The two closed curves, labeled (b) and (c) in Figure 6.9, correspond to energy levels E = − 12 and E = 12 . The maximum angular displacements achieved by the pendulum in these two cases are |θmax | = π/3 and |θmax | = 2π/3, respectively. These values correspond to the θ-axis intercepts of the curves. Energy levels E > 1 correspond to motion in which the system possesses enough energy to permit the pendulum to reach the vertical upward position and continue to rotate. Since energy is conserved in this idealized model, the pendulum continues to rotate forever. For each energy level greater than 1, two trajectories are possible. These are not closed trajectories, since total angular displacement increases or decreases monotonically. The trajectories in the upper half-plane (where θ > 0) correspond to counterclockwise pendulum rotation, while the counterpart trajectories in the lower half-plane (where θ < 0) represent clockwise pendulum rotation. The eight such trajectories shown in Figure 6.9 correspond to E = 2, E = 3, E = 4, and E = 5. (continued) 1

In Section 2.9, Exercise 22, we derived this conservation law in a different way. A change of independent variable was used to transform θ + (g/l ) sin θ = 0 into a ﬁrst order separable differential equation. The implicit solution of this equation yielded the conservation law (8).

416

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Nonlinear Systems (continued)

′ (a) E = (b) E = (c) E = (d) E = (e) E = (f ) E = (g) E = (h) E =

6

4

–1 –1/2 1/2 1 2 3 4 5

2

(h) (g)

(f) (e) (d) (a)

–2

–

(b)

(c)

(b)

(c)

2

(d) (e)

–2

(f)

(g) (h) –4

FIGURE 6.9

Some of the phase-plane trajectories for the pendulum equation discussed in Example 1. These curves are graphs of equation (8) for various energy levels E.

The two trajectories corresponding to E = 1 appear to connect the equilibrium points (−π, 0) and (π, 0). These trajectories are called separatrices; they separate the closed trajectory curves from the open ones. On the upper separatrix, the solution point approaches the equilibrium point (π, 0) as t → ∞. The pendulum swings upward in the counterclockwise direction, slowing down as it approaches the vertical upward position. The pendulum bob approaches this inverted position in the limit as t → ∞. On the lower separatrix, the solution point approaches the equilibrium point (−π, 0) as t → ∞. In this case, the pendulum swings upward in the clockwise direction, again slowing down and approaching the inverted position in the limit as t → ∞. ❖

Hamiltonian Systems We now discuss a class of autonomous ﬁrst order systems, called Hamiltonian systems, that always satisfy a conservation law. We restrict our attention to two-dimensional systems; the Exercises show how the underlying principle

6.3

Conservative Systems

417

can be extended to higher dimensional systems. Hamiltonian systems include the second order scalar equation (1) as a special case. As a ﬁrst step, recall the following chain rule from calculus. Assume that a function H(x, y), viewed as a function of two independent variables x and y, is continuous and has continuous ﬁrst and second partial derivatives (it will be apparent later why continuous second partial derivatives are required). We now form a composition, replacing the variables x and y with differentiable functions of t; we refer to these two functions as x(t) and y(t). The resulting composite function, H(x(t), y(t)), is a differentiable function of t, and its derivative can be found by the chain rule: d ∂H(x(t), y(t)) dx ∂H(x(t), y(t)) dy H(x(t), y(t)) = + . dt ∂x dt ∂y dt

(9)

Consider now the two-dimensional autonomous system x (t) = f (x(t), y(t)) y (t) = g(x(t), y(t)).

(10)

System (10) is called a Hamiltonian system2 if there exists a function of two variables H(x, y) that is continuous, with continuous ﬁrst and second partial derivatives, and such that ∂H(x, y) = −g(x, y) ∂x ∂H(x, y) = f (x, y). ∂y

(11)

The function H(x, y) is called the Hamiltonian function (or simply the Hamiltonian) of the system. If system (10) is a Hamiltonian system, then the composition H(x(t), y(t)) is a conserved quantity of the system. To see why, note that ∂H(x(t), y(t)) dx(t) ∂H(x(t), y(t)) dy(t) d H(x(t), y(t)) = + dt ∂x dt ∂y dt = [−g(x(t), y(t))]f (x(t), y(t)) + f (x(t), y(t)) g(x(t), y(t)) = 0. It follows, therefore, that H(x(t), y(t)) = C. Two important questions about Hamiltonian systems are “How can we determine whether a given autonomous system is a Hamiltonian system?” and

2

Sir William Rowan Hamilton (1805–1865) was an Irish mathematician noted for his contributions to optics and dynamics and for the development of the theory of quaternions. Shortly before his death, he was elected the ﬁrst foreign member of the United States National Academy of Sciences.

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“If a system is known to be a Hamiltonian system, how do we determine the conserved quantity H?” We will address both of these questions after the next example, which shows that the second order scalar equation (1), when recast as an autonomous ﬁrst order system, is a Hamiltonian system.

E X A M P L E

2

If the second order scalar equation y + v( y) = 0 is rewritten as a ﬁrst order system, we obtain the autonomous system x = y y = −v(x). Show that this system is a Hamiltonian system with H(x, y) = 12 y2 + V(x), where V(x) is any antiderivative of v(x). Solution: With the notation of (10) and (11), f (x, y) = y and g(x, y) = −v(x). Calculating the partial derivatives of H(x, y) = 12 y2 + V(x), we ﬁnd ∂H(x, y) dV(x) = = v(x) = −g(x, y) ∂x dx ∂H(x, y) = y = f (x, y). ∂y Thus, from equation (11), the system is a Hamiltonian system. ❖

Recognizing a Hamiltonian System The following discussion about identifying Hamiltonian systems and constructing Hamiltonians closely parallels the discussion in Section 2.7 about identifying exact differential equations and constructing solutions of exact equations. In particular, suppose H(x, y) is a Hamiltonian for system (10). Then, from equation (11), ∂H(x, y) = −g(x, y) and ∂x

∂H(x, y) = f (x, y). ∂y

(12)

From calculus, if the second partial derivatives of H(x, y) exist and are continuous, then the second mixed partial derivatives are equal; that is, ∂ 2 H(x, y) ∂ 2 H(x, y) = . ∂x∂y ∂y∂x Therefore, if system (10) is a Hamiltonian system, it necessarily follows that ∂f (x, y) ∂g(x, y) =− . ∂x ∂y The following theorem, stated without proof, asserts that this condition is both necessary and sufﬁcient.

6.3

Theorem 6.2

Conservative Systems

419

Consider the two-dimensional autonomous system x = f (x, y) y = g(x, y). Assume that f (x, y) and g(x, y) are continuous in the xy-plane. Assume as well that the partial derivatives ∂f , ∂x

∂f , ∂y

∂g , ∂x

∂g ∂y

exist and are continuous in the xy-plane. Then the system is a Hamiltonian system if and only if ∂f (x, y) ∂g(x, y) =− ∂x ∂y

(13)

for all (x, y).

Constructing Hamiltonians Once a system is known to be a Hamiltonian system, we can construct the Hamiltonian function by the same process of anti-partial-differentiation we used to solve exact differential equations in Section 2.7. We illustrate the ideas in the next example by constructing a Hamiltonian function for a Hamiltonian system. E X A M P L E

3

Consider the autonomous system x = y2 + cos x y = 2x + 1 + y sin x. (a) Use Theorem 6.2 to show that this system is a Hamiltonian system. (b) Find a Hamiltonian function for the system. Solution: (a) Calculating the partial derivatives required by test (13), we ﬁnd ∂f = − sin x ∂x

and

∂g = sin x. ∂y

Since ∂f /∂x = −∂g/∂y, we know the system is a Hamiltonian system. (b) Since the given system is a Hamiltonian system, there must be a function H(x, y) such that ∂H(x, y) = −g(x, y) = −2x − 1 − y sin x ∂x ∂H(x, y) = f (x, y) = y2 + cos x. ∂y

(14)

(continued)

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Nonlinear Systems (continued)

Choose one of these equations, say the ﬁrst, and compute an anti-partialderivative, obtaining H(x, y) = −x2 − x + y cos x + q( y),

(15)

where q( y) is an arbitrary differentiable function of y. [Note: Since the variable y is held ﬁxed when partial differentiation is performed with respect to x, we must allow for this arbitrary function of y when reversing the operation.] From equations (14) and (15), we ﬁnd ∂H(x, y) dq( y) = y2 + cos x = cos x + . ∂y dy Therefore, dq( y) = y2 , dy and q( y) = y3 /3 + C. We can drop the additive arbitrary constant and obtain a Hamiltonian y3 . (16) 3 Figure 6.10 shows some phase-plane trajectories [that is, graphs of H(x, y) = C] for a few representative values of the constant C. H(x, y) = −x2 − x + y cos x +

y 3 2 1 –3

–2

–1

1

2

3

x

–1 –2 –3

FIGURE 6.10

Some phase-plane trajectories for the autonomous system in Example 3. These curves are level curves of the Hamiltonian (16).

❖

6.3

Conservative Systems

421

EXERCISES Exercises 1–6: In each exercise, (a) As in Example 1, derive a conservation law for the given autonomous equation x + u(x) = 0. (Your answer should contain an arbitrary constant and therefore deﬁne a one-parameter family of conserved quantities.) (b) Rewrite the given autonomous equation as a ﬁrst order system of the form x = f (x, y) y = g(x, y) by setting y(t) = x (t). The phase plane is then the xy-plane. Express the family of conserved quantities found in (a) in terms of x and y. Determine the equation of the particular conserved quantity whose graph passes through the phase-plane point (x, y) = (1, 1). (c) Plot the phase-plane graph of the conserved quantity found in part (b), using a computer if necessary. Determine the velocity vector v = f (x, y)i + g(x, y)j at the phaseplane point (1, 1). Add this vector to your graph with the initial point of the vector at (1, 1). What is the geometric relation of this velocity vector to the graph? What information does the velocity vector give about the direction in which the solution point traverses the graph as time increases? (d) For the solution whose phase-plane trajectory passes through (1, 1), determine whether the solution x(t) is bounded. If the solution is bounded, use the phaseplane plot to estimate the maximum value attained by |x(t)|. 1. x + 4x = 0

2. x − x = 1

4. x − x3 = π sin(π x)

5. x + x2 = 0

3. x + x3 = 0 x 6. x + =0 1 + x2

Exercises 7–8: The conservation law for an autonomous second order scalar differential equation having the form x + f (x) = 0 is given (where y corresponds to x ). Determine the differential equation. 7. y2 + x2 cos x = C

2

8. y2 − e−x = C

9. Consider the differential equation x + x + x3 = 0. It has the same structure as the equation used to model the cubically nonlinear spring. (a) Rewrite the differential equation as a ﬁrst order system. On the xy-phase plane, sketch the nullclines and locate any equilibrium point(s). Place direction ﬁeld arrows on the nullclines, indicating the direction in which the solution point traverses the nullclines. (b) Compute the velocity vector v = x i + y j at the four phase-plane points (x, y) = (±1, ±1). Locate these points, and draw the velocity vectors on your phaseplane sketch. Use this information, together with the information obtained in part (a), to draw a rough sketch of some phase-plane solution trajectories. Indicate the direction in which the solution point moves on these trajectories. (c) Determine the conservation law satisﬁed by solutions of the given differential equation. Determine the equation of the conserved quantity whose graph passes through the phase-plane point (x, y) = (1, 1). Plot the graph of this equation on your phase plane, using computational software as appropriate. Does the graph pass through the other three points, (−1, 1), (−1, −1), and (1, −1), as well? Is the sketch made in part (b) consistent with this graph of the conserved quantity?

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10. Each ﬁgure shows a phase-plane graph of a conserved quantity for the autonomous differential equation x + αx = 0, where α is a real constant. (a) Determine the value of the constant α in each case. What is the equation whose phase-plane graph is shown? (b) Indicate the direction in which the solution point travels along these phaseplane curves as time increases.

y

y 2 1 –2

–1

y

3 2 1 1

2

–1 –2

x

–3 –2 –1 –1 –2 –3

(a)

1 2 3

2

x –4

–2

2

4

x

–2

(b)

(c)

Figure for Exercise 10 In (b), the asymptotes are y = ±x.

11. Consider the autonomous third order scalar equation y + f ( y ) = 0, where f is a continuous function. Does this differential equation have a conservation law? If so, obtain the equation of the family of conserved quantities. 12. Consider the equation mx + kx = 0. We saw in Chapter 3 that this equation models the vibrations of a spring-mass system. The conserved quantity 12 m(x )2 + 12 kx2 = E is the (constant) total energy of the system. The ﬁrst term, 12 m(x )2 , is the kinetic energy, while the second term, 12 kx2 , is the elastic potential energy. Suppose that damping is now added to the system. The differential equation mx + γ x + kx = 0 now models the motion (with γ a positive constant). Deﬁne E(t) = 12 m(x )2 + 12 kx2 . (a) Show, in the case of damping, that E(t) is no longer constant. Show, rather, that dE(t)/dt ≤ 0. (b) Discuss the physical relevance of the observation made in part (a).

Exercises 13–20: For the given system, (a) Use Theorem 6.2 to show that the system is a Hamiltonian system. (b) Find a Hamiltonian function for the system. (c) Use computational software to graph the phase-plane trajectory passing through (1, 1). Also, indicate the direction of motion for the solution point. 13. x = 2x y = −2y 16. x = −8y

y = 2x 19. x = −2y y = 3x2

14. x = 2xy y = −y2 17. x = 2y cos x y = y2 sin x 20. x = xexy y = −2x − yexy

15. x = x − x2 + 1 y = −y + 2xy + 4x 18. x = 2y − x + 3 y = y + 4x3 − 2x

6.3

Conservative Systems

423

Exercises 21–26: Use Theorem 6.2 to decide whether the given system is a Hamiltonian system. If it is, ﬁnd a Hamiltonian function for the system. 21. x = x3 + 3 sin(2x + 3y)

22. x = exy + y3 y = −exy − x3

2

y = −3x y − 2 sin(2x + 3y) 23. x = − sin(2xy) − x

24. x = −3x2 + xey

y = sin(2xy) + y

y = 6xy + 3x − ey

25. x = y

26. x = x + 2y

y = x − x2

y = x3 − 2x + y

Exercises 27–30: Let f (u) and g(u) be deﬁned and continuously differentiable on the interval −∞ < u < ∞, and let F(u) and G(u) be antiderivatives for f (u) and g(u), respectively. In each exercise, use Theorem 6.2 to show that the given system is Hamiltonian. Determine a Hamiltonian function for the system, expressed in terms of F and/or G. 27. x = f ( y)

28. x = f ( y) + 2y y = g(x) + 6x

y = g(x) 29. x = 3f ( y) − 2xy

30. x = f (x − y) + 2y y = f (x − y)

y = g(x) + y2 + 1

31. A Generalized Hamiltonian System Consider the two-dimensional autonomous system x = f (x, y) y = g(x, y). Suppose there exist two functions K(x, y) and μ(x, y) satisfying ∂K(x, y) = −μ(x, y)g(x, y) ∂x ∂K(x, y) = μ(x, y)f (x, y). ∂y Does the given autonomous system have a conserved quantity? If so, what is the conserved quantity? 32. Higher-Dimensional Autonomous Systems The ideas underlying Hamiltonian systems extend to higher-dimensional systems. For example, consider the threedimensional autonomous system x = f (x, y, z) y = g(x, y, z)

(17)

z = h(x, y, z). (a) Use the chain rule to show that autonomous system (17) has a conserved quantity if there exists a function H(x, y, z) for which ∂ ∂ ∂ H(x, y, z)f (x, y, z) + H(x, y, z)g(x, y, z) + H(x, y, z)h(x, y, z) = 0. ∂x ∂y ∂z (b) Show that H(x, y, z) = cos2 (x) + ye z is a conserved quantity for the system x = ye z y = y cos x sin x z = cos x sin x.

424

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6.4

Nonlinear Systems

Stability Differential equations can model different physical behavior at different equilibrium points. For instance, consider the pendulum. The equilibrium points are (θ, θ ) = (mπ, 0), where m is an integer. Equilibrium points with m an even integer correspond to the pendulum bob hanging vertically downward, while equilibrium points for m an odd integer correspond to the pendulum bob resting in the inverted position. Suppose a pendulum bob, initially in an equilibrium state, is subjected to a slight perturbation; in other words, it is given a slight displacement and/or a very small angular velocity. If the pendulum is initially hanging downward, we expect the perturbation to remain small— the bob will swing back and forth, making small excursions from the vertical. If the pendulum is initially in the inverted position, however, we expect dramatic changes. The pendulum bob, displaced from its precarious equilibrium state, will fall, ultimately making large departures from its initial equilibrium position. In everyday language, we might describe the pendulum’s downward rest position as a “stable” conﬁguration and the inverted rest position as an “unstable” conﬁguration. Mathematicians have taken these everyday terms and given them precise deﬁnitions consistent with our intuitive notion of stable and unstable. In this section, we present and discuss these mathematical definitions. The next section introduces the technique of linearization, which, in many cases, enables us to study and characterize equilibrium point stability by analyzing a simpler associated linear system. The pendulum example illustrates the question of primary concern: If perturbed slightly from an equilibrium state, will a system exhibit a markedly different behavior? In the case of mechanical systems, instability often means vibrations that grow in amplitude, leading to possible system failure.

Stable and Unstable Equilibrium Points Consider the autonomous system y1 = f1 ( y1 , y2 , . . . , yn ) y2 = f2 ( y1 , y2 , . . . , yn ) .. . yn = fn ( y1 , y2 , . . . , yn ), which we write in vector form as y = f(y). Assume that the constant vector function

⎡

y1e

(1)

⎤

⎢ ⎥ ⎢y2e ⎥ ⎥ y(t) = ye = ⎢ ⎢ .. ⎥ ⎣ . ⎦ yne is an equilibrium solution of the system; that is, f(ye ) = 0.

6.4

Stability

425

In order to deﬁne precisely what it means for the equilibrium point ye to be stable or unstable, we need to be able to compute the distance between points in n-dimensional space. Let u and v denote two points in n-dimensional space, ⎡ ⎤ ⎡ ⎤ v1 u1 ⎢ ⎥ ⎢ ⎥ ⎢ v2 ⎥ ⎢u2 ⎥ ⎢ ⎥ ⎥ u=⎢ ⎢ .. ⎥ and v = ⎢ .. ⎥ . ⎣ . ⎦ ⎣.⎦ un

vn

We deﬁne the norm of u, denoted by u, by u = u21 + u22 + · · · + u2n . The distance between u and v, denoted by u − v, is the size (the norm) of the difference u − v: (2) u − v = (u1 − v1 )2 + (u2 − v2 )2 + · · · + (un − vn )2 . Now, let ye be an equilibrium point of the autonomous system y = f(y). We say that the equilibrium point ye is stable if Given any ε > 0, there exists a corresponding δ > 0 such that every solution satisfying y(0) − ye < δ also satisﬁes y(t) − ye < ε for all t ≥ 0. If an equilibrium point of y = f(y) is not stable, it is called unstable.

Interpreting Stability in the Phase Plane When n = 2, we can use the phase plane to give a graphical interpretation of stability. Consider the autonomous system x = f (x, y) y = g(x, y) having equilibrium solution ye =

xe

We have adopted the notation

x(t) y(t) = and y(t)

ye

.

f(y) =

f (x, y)

g(x, y)

so that we can speak of the phase plane as the xy-plane. We can identify ye as the point in the phase plane having coordinates (xe , ye ). The set of all phase-plane points y satisfying y − ye < r is the set of all points lying within a circle of radius r centered at (xe , ye ). Consider now the deﬁnition of stability. It involves two circles centered at (xe , ye ), one of radius ε and the other of radius δ (see Figure 6.11). The stability criterion requires that all solutions lying within the circle of radius δ at the initial time t = 0 remain within the circle of radius ε for all subsequent time.

426

CHAPTER 6

Nonlinear Systems y

y

y(0) y(t) y(0)

ye

␦

ye

␦

x

(a)

x

(b) FIGURE 6.11

Two examples of behavior near a stable equilibrium point. In each case, y(t) = (x(t), y(t)) represents a typical solution trajectory near an equilibrium point of the autonomous system y = f(y). (a) When the initial point y(0) is sufﬁciently close to ye , the solution trajectory is a closed curve surrounding ye . (b) When y(0) is sufﬁciently close to ye , the solution trajectory spirals in toward ye .

This situation must hold for all possible choices of ε > 0; whenever we are given an ε > 0, we must be able to ﬁnd a corresponding δ > 0 that “works.” The real test of the deﬁnition occurs as we consider smaller and smaller ε > 0. Can we continue to ﬁnd corresponding values δ > 0 that work? If so, the equilibrium point is stable; if not, it is unstable. For higher order systems, the same geometrical ideas hold. However, instead of circles in the phase plane, we must consider n-dimensional spheres. We illustrate the concept of stability with two examples involving autonomous linear systems for which explicit general solutions are known. E X A M P L E

1

Consider the two-dimensional autonomous linear system y = Ay, where

−2 1 A= . 1 −2

(3)

Show that ye = 0 is the only equilibrium point, and determine whether it is stable or unstable. Solution: In this case, f(y) = Ay. The matrix A is invertible, and therefore solving Ay = 0 leads to a single equilibrium point, ye = 0. To determine the stability properties of this equilibrium point, we apply the stability deﬁnition directly to the general solution of this ﬁrst order linear system. Using the methods of Chapter 4, we ﬁnd the general solution is

−t

−3t 1 1 e + c e c 1 2 (4) . y(t) = c1 e−t + c2 e−3t = 1 −1 c1 e−t − c2 e−3t

6.4

Stability

427

In order to apply the deﬁnition of stability, we need to relate the following two quantities: 1. the distance y(0) − ye , between the initial point and the equilibrium point, and 2. the distance y(t) − ye , between y(t) and ye for t ≥ 0. Since ye = 0, it follows from (4) that y(0) − ye = y(0) = (c1 + c2 )2 + (c1 − c2 )2 = 2(c21 + c22 ). Similarly, for t ≥ 0,

(c1 e−t + c2 e−3t )2 + (c1 e−t − c2 e−3t )2 = 2(c21 e−2t + c22 e−6t ) = 2(c21 + c22 e−4t )e−2t ≤ 2(c21 + c22 )e−2t = 2(c21 + c22 )e−t

y(t) − ye = y(t) =

(5)

= y(0)e−t . Now consider the deﬁnition of stability where some value ε > 0 is given. We need to determine a corresponding value δ > 0 such that If y(t) is a solution of y = Ay and if y(0) < δ, then y(t) < ε for all t ≥ 0. By (5), we know that y(t) ≤ y(0)e−t ≤ y(0),

t ≥ 0.

Therefore, we can guarantee that y(t) < ε by taking δ ≤ ε. This shows the equilibrium point ye = 0 is a stable equilibrium point of the system y = Ay. ❖ E X A M P L E

2

Consider the two-dimensional autonomous linear system y = Ay, where

−1 −2 A= . −2 −1

(6)

Show that ye = 0 is the only equilibrium point, and determine whether it is stable or unstable. Solution: As in Example 1, f(y) = Ay, and we see that the matrix A is invertible. Therefore, solving Ay = 0 leads to a single equilibrium point, ye = 0. The general solution of y = Ay is

1 1 c1 et + c2 e−3t t −3t + c2 e = . y(t) = c1 e (7) −1 1 −c1 et + c2 e−3t In Example 1, the coefﬁcient matrix had two negative eigenvalues. In this example, however, the coefﬁcient matrix has a positive eigenvalue, λ = 1. Since (continued)

428

CHAPTER 6

Nonlinear Systems (continued)

lim t→∞ et = ∞, we anticipate that any solution (7) with c1 = 0 will become unboundedly large in norm as t increases. Therefore, we anticipate that ye = 0 is an unstable equilibrium point of this system. To prove that ye = 0 is an unstable equilibrium point, we show that, for some ε > 0, there is no δ that works. That is, for every δ > 0, there is at least one solution y(t) that originates in the circle of radius δ but that eventually gets outside the circle of radius ε. In particular, solutions given by (7) with c2 = 0 and c1 = 0 have the form

1 t y(t) = c1 e , t ≥ 0. −1 Moreover, y(0) =

√ 2c1

and

y(t) =

√ t 2c1e .

This particular family of solutions has phase-plane trajectories that lie on the line √ y=t −x (see Figure 6.12 and Exercise 23 in Section 6.2). Since y(t) = 2 c1 e , the solution moves away from the origin along this line, growing in norm as t increases. No matter what value δ > 0 we take, we can always choose |c1| = 0 but sufﬁciently small that y(0) is within the circle of radius δ. But, as long as |c1| = 0, the solution y(t) eventually exits the circle of radius ε. Therefore, ye = 0 is an unstable equilibrium point of the autonomous system y = Ay. y 3 2 1

x –3

–2

–1

1

2

3

–1 –2 –3

FIGURE 6.12

The direction ﬁeld for the system in Example 2. The arrows indicate the direction the solution point moves as time increases.

❖

In Examples 1 and 2, it was relatively straightforward to determine the stability properties of the equilibrium point ye = 0 because we had an explicit representation of the general solution. The obvious question is “How do we analyze the stability properties of equilibrium points of nonlinear problems, such as the pendulum and competing species problems, when explicit solutions are not attainable?” We address this issue in the next section.

6.4

Stability

429

Asymptotic Stability Example 1 has an interesting aspect. The equilibrium point ye = 0 not only is stable but also has the feature that all solutions approach it in the limit as t → ∞. This additional feature, wherein all solutions originating sufﬁciently close to a stable equilibrium point actually approach the equilibrium point as t → ∞, is important enough to warrant its own deﬁnition. Let ye be an equilibrium point of the autonomous system y = f(y). We say that ye is an asymptotically stable equilibrium point if (a) it is a stable equilibrium point and (b) there exists a δ0 > 0 such that lim t→∞ y(t) = ye for all solutions initially satisfying y(0) − ye < δ0 . Roughly speaking, all solutions starting close enough to a stable equilibrium point remain close to it for all subsequent time. All solutions starting sufﬁciently close to an asymptotically stable equilibrium point not only remain close to it for all subsequent time but, in fact, approach it in the limit as t → ∞. Note that asymptotic stability implies stability. However, as the next example shows, an equilibrium point can be stable but not asymptotically stable. E X A M P L E

3

Consider the two-dimensional autonomous linear system y = Ay, where

0 1 A= . −1 0 Show that ye = 0 is the only equilibrium point, and determine whether it is asymptotically stable. Solution: As in Examples 1 and 2, f(y) = Ay, where the matrix A is invertible. Therefore, solving Ay = 0 leads to a single equilibrium point, ye = 0. Eigenpairs of the matrix A are

1 1 λ1 = i, u1 = and λ2 = λ1 = −i, u2 = u1 = . i −i Using the ideas of Section 4.6, we ﬁnd a real-valued general solution of this system to be

cos t sin t y(t) = c1 (8) + c2 . − sin t cos t It follows from (8) that

y(t) − ye = y(t) = (c1 cos t + c2 sin t)2 + (−c1 sin t + c2 cos t)2 = c21 + c22 = y(0), t ≥ 0.

(9)

Therefore, the distance of a solution from the phase-plane origin remains constant in time—the phase-plane trajectories are circles centered at the origin. (continued)

430

CHAPTER 6

Nonlinear Systems (continued)

Thus, the equilibrium point ye = 0 is stable (given ε > 0, we simply take δ = ε). The equilibrium point is not asymptotically stable, however. In particular, no nonzero solution approaches the equilibrium point as t → ∞. ❖

Stability Characteristics of y = Ay The three examples previously considered illustrate the following general stability result, which we present without proof.

Theorem 6.3

Let A be a real invertible (2 × 2) matrix. Then the autonomous linear system y = Ay has a unique equilibrium point ye = 0. This equilibrium point is (a) asymptotically stable if all eigenvalues of A have negative real parts. (In other words, the eigenvalues can be real and negative, or they can be a complex conjugate pair with negative real parts.) (b) stable but not asymptotically stable if the eigenvalues of A are purely imaginary. (c) unstable if at least one eigenvalue of A has a positive real part. (In other words, the eigenvalues can be real with at least one being positive, or they can be a complex conjugate pair with positive real parts.)

Equilibrium points are sometimes characterized as being isolated or not isolated. A phase-plane equilibrium point is called an isolated point if it is the center of some small disk whose interior contains no other equilibrium points. The equilibrium point in Theorem 6.3 (as in all the examples in this section) is the only equilibrium point. It is, therefore, an isolated equilibrium point. We also note, with respect to Theorem 6.3, that the assumption that A is invertible implies det(A) = 0. Therefore, λ = 0 is not an eigenvalue of A. However, if A is not invertible, then ye = 0 is not the only equilibrium point (nor is it isolated). For example, suppose

α β A= , cα cβ with α and β not both zero. Thus, det(A) = 0, but A is not the zero matrix. One can show (see Exercise 33) that every point on the phase-plane line αx + βy = 0 is an equilibrium point. Theorem 6.3 applies equally well whether the (2 × 2) matrix A has distinct or repeated eigenvalues. Recall that if A has a repeated (real) eigenvalue λ1 = λ2 = λ, then solutions involving the function teλt , as well as eλt , are possible. If λ < 0, however, we know from calculus that teλt is bounded for t ≥ 0 and that lim t→∞ teλt = 0. It follows that ye = 0 will always be an asymptotically stable equilibrium point when λ < 0.

6.4

Stability

431

In the higher-dimensional case, stability characterization is somewhat more complicated. If A is a real invertible (n × n) matrix, then the unique equilibrium point ye = 0 of y = Ay is asymptotically stable if all eigenvalues of A have negative real parts. The equilibrium point is unstable if at least one eigenvalue of A has a positive real part. For n ≥ 4, repeated complex conjugate pairs of purely imaginary eigenvalues, say λ = ±iω, are possible. In this case, solutions of the form t cos ωt and t sin ωt are possible when the matrix A does not have n linearly independent eigenvectors; in that event, ye = 0 is an unstable equilibrium point.

EXERCISES 1. Assume that a two-dimensional autonomous system has an isolated equilibrium point at the origin and that the phase-plane solution curves consist of the family of concentric ellipses x2 /4 + y2 = C, C ≥ 0. (a) Apply the deﬁnition to show that the origin is a stable equilibrium point. In particular, given an ε > 0, determine a corresponding δ > 0 so that all solutions starting within a circle of radius δ centered at the origin stay within the circle of radius ε centered at the origin for all t ≥ 0. (The δ you determine should be expressed in terms of ε.) (b) Is the origin an asymptotically stable equilibrium point? Explain. 2. Assume that a two-dimensional autonomous system has an isolated equilibrium point at the origin and that the phase-plane solution curves consist of the family of hyperbolas −x2 + y2 = C, C ≥ 0. Is the equilibrium point stable or unstable? Explain. 3. Consider the differential equation x + γ x + x = 0, where γ is a real constant. (a) Rewrite the given scalar equation as a ﬁrst order system, deﬁning y = x . (b) Determine the values of γ for which the system is (i) asymptotically stable, (ii) stable but not asymptotically stable, (iii) unstable.

Exercises 4–15: Each exercise lists a linear system y = Ay, where A is a real constant invertible (2 × 2) matrix. Use Theorem 6.3 to determine whether the equilibrium point ye = 0 is asymptotically stable, stable but not asymptotically stable, or unstable.

−3 −2 5 −14 0 −2 4. y = y 5. y = y 6. y = y 3 −8 2 0 4 3

1 4 9. x = 9x + 5y 7. y = y 8. x = −7x − 3y −1 1 y = 5x + y y = −7x − 3y 10. x = −3x − 5y

y = 2x − y

11. x = 9x − 4y

y = 15x − 7y

13. x = 3x − 2y

14. x = x − 5y

y = 5x − 3y

y = x − 3y

12. x = −13x − 8y y = 15x + 9y 15. x = −3x + 3y y = x − 5y

Exercises 16–23: Each exercise lists the general solution of a linear system of the form x = a11 x + a12 y y = a21 x + a22 y,

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CHAPTER 6

Nonlinear Systems where a11 a22 − a12 a21 = 0. Determine whether the equilibrium point ye = 0 is asymptotically stable, stable but not asymptotically stable, or unstable. 16. x = c1 e−2t + c2 e3t

17. x = c1 e2t + c2 e3t

y = c1 e−2t + 2c2 e3t

y = c1 e2t + 2c2 e3t

19. x = c1 et cos 2t + c2 et sin 2t 21. x = c1 e

−2t

cos 2t + c2 e

y = c1 e−2t + 2c2 e−4t

20. x = c1 cos 2t + c2 sin 2t

y = −c1 et sin 2t + c2 et cos 2t −2t

18. x = c1 e−2t + c2 e−4t

y = −c1 sin 2t + c2 cos 2t 22. x = c1 e−2t + c2 e3t

sin 2t

y = −c1 e−2t sin 2t + c2 e−2t cos 2t

y = c1 e−2t − c2 e3t

23. x = c1 e−2t + c2 e−3t y = c1 e−2t − c2 e−3t 24. Consider the nonhomogeneous linear system y = Ay + g0 , where A is a real invertible (2 × 2) matrix and g0 is a real (2 × 1) constant vector. (a) Determine the unique equilibrium point, ye , of this system. (b) Show how Theorem 6.3 can be used to determine the stability properties of this equilibrium point. [Hint: Adopt the change of dependent variable z(t) = y(t) − ye .]

Exercises 25–28: Locate the unique equilibrium point of the given nonhomogeneous system, and determine the stability properties of this equilibrium point. Is it asymptotically stable, stable but not asymptotically stable,

or unstable? −2 1 −4 25. y = y+ 26. x = y + 2 1 −2 2 y = −x + 1

3 2 −2 y+ 27. y = 28. x = −x + y + 1 −4 −3 2 y = −10x + 5y + 2

Exercises 29–32:

Higher Dimensional Systems In each exercise, locate all equilibrium points for the given autonomous system. Determine whether the equilibrium point or points are asymptotically stable, stable but not asymptotically stable, or unstable. ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ y 1 −1 0 2 y1 ⎥ ⎢ 1⎥ ⎢ ⎥ ⎥ ⎢ d ⎢ ⎢ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥ 2⎦ ⎣y2 ⎦ + ⎣0⎥ y = 0 −1 29. y1 = 2y1 + y2 + y3 30. ⎦ dt ⎣ 2 ⎦ ⎣ 0 0 −1 3 y3 y3 y2 = y1 + y2 + 2y3 y3 = y1 + 2y2 + y3 ⎡ ⎤ ⎡ y −3 −5 ⎢ 1⎥ ⎢ ⎢ ⎥ ⎢ d ⎢y2 ⎥ ⎢ 2 −1 31. ⎢ ⎥=⎢ dt ⎢y3 ⎥ ⎢ 0 0 ⎣ ⎦ ⎣ y4 0 0 32.

y1 y2 y3 y4

0 0 0 −2

⎤⎡ ⎤ 0 y1 ⎥⎢ ⎥ ⎢ ⎥ 0⎥ ⎥ ⎢y2 ⎥ ⎥⎢ ⎥ 2⎥ ⎢y3 ⎥ ⎦⎣ ⎦ y4 0

= y2 − 1 = y1 + 2 = −y3 + 1 = −y4

33. Let A be a real (2 × 2) matrix. Assume that A has eigenvalues λ1 and λ2 , and consider the linear homogeneous system y = Ay. (a) Prove that if λ1 and λ2 are both nonzero, then ye = 0 is an isolated equilibrium point.

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433

(b) Suppose that eigenvalue λ1 = 0 but that λ2 = 0 with corresponding eigenvector β . Show that all points on the phase-plane line αx + βy = 0 are equilibrium −α points. (In this case, ye = 0 is not an isolated equilibrium point.) 34. Consider the linear system y =

−1

α

α

−1

y,

where α is a real constant. (a) What information can be obtained about the eigenvalues of the coefﬁcient matrix simply by examining its structure? (b) For what value(s) of the constant α is the equilibrium point ye = 0 an isolated equilibrium point? For what value(s) of the constant α is the equilibrium point ye = 0 not isolated? (c) In the case where ye = 0 is not an isolated equilibrium point, what is the equation of the phase-plane line of equilibrium points? (d) Is it possible in this example for ye = 0 to be an isolated equilibrium point that is stable but not asymptotically stable? Explain. (e) For what values of the constant α, if any, is the equilibrium point ye = 0 an isolated asymptotically stable equilibrium point? For what values of the constant α, if any, is the equilibrium point ye = 0 an unstable equilibrium point?

1 a12 be a real (2 × 2) matrix. Assume that 35. Let A = a21 a22 A

1 2

=

1

a12

a21

a22

1 2

=2

1 2

and that the origin is not an isolated equilibrium point of the system y = Ay. Determine the constants a12 , a21 , and a22 .

6.5

Linearization and the Local Picture We now consider nonlinear autonomous systems and a technique, known as linearization, for investigating the stability properties of such systems. The stability results cited in Theorem 6.3 for the linear system y = Ay will be useful for the nonlinear equations treated in this section. In Section 6.6, we will examine the phase-plane geometry of the linear two-dimensional system y = Ay in more detail.

Nonlinear Systems Although we are going to focus on the case n = 2, the ideas are applicable to general n-dimensional autonomous systems. Let

xe ye = ye

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be an equilibrium solution of y = f(y). In component form, y = f(y) is given by x = f (x, y) y = g(x, y).

(1)

For a nonlinear system such as (1), it is usually impossible to obtain explicit solutions. In the absence of explicit solutions, we look for approximations or simpliﬁcations that provide qualitative insight into the stability properties of the equilibrium point ye .

Linearization From the deﬁnition, we know that the issue of equilibrium point stability is ultimately determined by the behavior of solutions very close to the equilibrium point. If any nearby solutions diverge from the equilibrium point, it is an unstable equilibrium point. If all nearby solutions can be suitably conﬁned, the equilibrium point is stable (and perhaps asymptotically stable). To address the issue of equilibrium point stability, we begin with the observation that if the point (x, y) is near the equilibrium point (xe , ye ), then the ﬁrst few terms of the Taylor series expansions of f (x, y) and g(x, y) will yield good approximations to their values near the equilibrium point: f (x, y) = f (xe , ye ) +

∂f (xe , ye ) ∂f (xe , ye ) (x − xe ) + ( y − ye ) + · · · ∂x ∂y

∂g(xe , ye ) ∂g(xe , ye ) (x − xe ) + ( y − ye ) + · · · . g(x, y) = g(xe , ye ) + ∂x ∂y

(2)

We make the following observations: 1. Since (xe , ye ) is an equilibrium point, f (xe , ye ) = g(xe , ye ) = 0. Thus, the ﬁrst term on the right-hand side of each equation in (2) vanishes. 2. The error made in truncating the series [retaining only the linear terms shown on the right-hand sides of (2)] can usually be bounded by a multiple of y − ye 2 = (x − xe )2 + ( y − ye )2 . If the Taylor expansion (2) is used in differential equation (1), we can write the system in matrix form as ⎡ ⎤ ∂f (xe , ye ) ∂f (xe , ye ) ⎢ ⎥ ⎢ ⎥ ∂x ∂y ⎢ ⎥ (y(t) − y ) + · · · . y (t) = ⎢ (3) e ⎥ ⎣ ∂g(xe , ye ) ∂g(xe , ye ) ⎦ ∂x ∂y Note that the (2 × 2) coefﬁcient matrix in (3) is a constant matrix since the partial derivatives are evaluated at the equilibrium point. [In vector calculus, the matrix of ﬁrst order partial derivatives in (3) is called the Jacobian matrix of f(y).] Linearization is based on the following two ideas:

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435

1. Since ye is a constant vector, the term y (t) on the left-hand side of (3) can be replaced by [y(t) − ye ] . 2. If we consider solutions close enough to the equilibrium point for the purpose of determining its stability characteristics, the higher order terms in (3) are typically small relative to the linear term, y(t) − ye . We will neglect these higher order terms. Introduce a new dependent variable, z(t) = y(t) − ye . Retaining only the linear term in (3) leads to the corresponding linearized system, ⎤ ⎡ ∂f (xe , ye ) ∂f (xe , ye ) ⎥ ⎢ ⎥ ⎢ ∂x ∂y ⎥ z(t). ⎢ (4) z (t) = ⎢ ⎥ ⎣ ∂g(xe , ye ) ∂g(xe , ye ) ⎦ ∂x ∂y Note that equation (4) is a homogeneous constant coefﬁcient linear system and that z = 0 is an equilibrium point of the linear system. The stability properties of z = 0 are easy to analyze since we can explicitly ﬁnd the general solution of equation (4); these properties are summarized in Theorem 6.3. The underlying premise of linearization is that the stability properties of z = 0 for the linear system (4) should be the same as the stability properties of y = ye for the original nonlinear system (1). We illustrate linearization in the next example. Then we address the question “When does linearization work?” E X A M P L E

1

Develop the linearized-system approximation for each of the equilibrium points of the nonlinear autonomous system x = 12 1 − 12 x − 12 y x y = 14 1 − 13 x − 23 y y. Also, determine the stability characteristics of the linearized system in each case. Solution: This is the competing species model considered in Example 4 of Section 6.2. Recall that this system has four equilibrium solutions,

0 0 2 1 (1) (2) (3) (4) ye = , ye = 3 , , ye = . ye = 0 0 1 2 The Jacobian matrix is given by ⎤ ⎡ ∂f (x, y) ∂f (x, y) ⎡ 1 ⎥ ⎢ − 1 x − 14 y ⎢ ∂x ∂y ⎥ ⎢ ⎥ = ⎣2 2 ⎢ ∂g(x, y) ∂g(x, y) ⎥ 1 ⎦ ⎣ − 12 y ∂x ∂y

⎤

− 14 x 1 4

−

1 x 12

−

⎦.

1 y 3

(5)

The Jacobian matrix (5) must be evaluated at each of the equilibrium points in order to obtain the coefﬁcient matrix of the appropriate linearized system. At (continued)

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the equilibrium point (0, 0), the coefﬁcient matrix is ⎡ ⎤ ⎡ 1 1 1 1 1 − x − y − x 4 4 ⎣2 2 ⎦ ⎣2 = 1 1 1 − 12 0 y − 12 x − 13 y x= y=0 4

0 1 4

⎤ ⎦.

The linearized system is therefore ⎡ z = ⎣

1 2

0

0

1 4

⎤ ⎦ z.

[Observe, for this case, that z(t) = y(t) − y(1) e = y(t).] Since the eigenvalues of the coefﬁcient matrix (λ = 12 and λ = 14 ) are positive, we conclude from Theorem 6.3 that z = 0 is an unstable equilibrium solution of the linearized system. Therefore, we anticipate that y(1) e will also be an unstable equilibrium point of the nonlinear system. The remaining three equilibrium points can be analyzed in the same manner. The results for all four equilibrium points are summarized in Table 6.1.

TA B L E 6 . 1

Equilibrium Point

z(t)

(0, 0)

0,

3 2

⎡1

x(t)

y(t) ⎡

⎣ ⎤

⎢ x(t) ⎥ ⎢ ⎥ ⎣ ⎦ y(t) − 32

(2, 0)

x(t) − 2 y(t)

(1, 1)

Linearized System Coefﬁcient Matrix

x(t) − 1 y(t) − 1

⎡

2

0

0

1 4

1 8

⎣ − 18

Eigenvalues of the Coefﬁcient Matrix

Stability Properties of the Linearized System

1 1 , 2 4

Unstable

1 1 ,− 8 4

Unstable

1 1 − , 2 12

Unstable

⎤ ⎦

0

⎤ ⎦

− 14

⎡ 1 ⎤ − 2 − 12 ⎣ ⎦ 1 0 12 ⎡ 1 ⎤ − 4 − 14 ⎣ ⎦ 1 − 12 − 16

√ √ −5 − 13 −5 + 13 , 24 24

Asymptotically stable

❖ The stability properties obtained by studying the linearized systems in Table 6.1 are consistent with the phase-plane direction ﬁeld portrait of Figure 6.4 in Section 6.2. The three equilibria designated as unstable in Table 6.1 are the ones that appear to have direction ﬁeld arrows pointing away from them in Figure 6.4. The fourth equilibrium point, designated as asymptotically stable in Table 6.1, appears to have all trajectories moving toward it in Figure 6.4.

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437

When Does Linearization Work? We restrict our attention to the two-dimensional autonomous system y = f(y). Let ye be an equilibrium solution,

xe . ye = ye We’ll proceed as before by introducing a new dependent variable that shifts the equilibrium point to 0 and by rewriting y = f(y) in a form that explicitly exhibits the linearized part. Let z(t) = y(t) − ye , and let A denote the Jacobian matrix evaluated at ye , ⎡ ⎤ ∂f (xe , ye ) ∂f (xe , ye ) ⎢ ⎥ ⎢ ⎥ ∂x ∂y ⎥ A=⎢ (6) ⎢ ∂g(x , y ) ∂g(x , y ) ⎥ . ⎣ e e e e ⎦ ∂x ∂y The system y = f(y) can be rewritten (without any approximation) as z (t) = f(z(t) + ye ) = Az(t) + [f(z(t) + ye ) − Az(t)].

(7)

Deﬁne g(z(t)) = f(z(t) + ye ) − Az(t). With this, equation (7) becomes z (t) = Az(t) + g(z(t)).

(8)

Written in this form, we see that g(z(t)) represents a nonlinear perturbation of the linearized system z (t) = Az(t). The linearization approximation amounts to discarding the nonlinear term, g(z(t)). If the behavior of the linearized system z = Az is going to be qualitatively similar to that of the nonlinear system (8) near the equilibrium point z = 0, it seems clear that g(z) must be suitably “small” near z = 0. In other words, the linear part of (8), Az, must control the basic behavior of solutions near the equilibrium point. We will now describe such a class of nonlinear systems. A two-dimensional autonomous system y = f(y) is called an almost linear system at an equilibrium point ye if (a) f(y) is a continuous vector-valued function whose component functions possess continuous partial derivatives in an open region of the phase plane containing the equilibrium point, ye . (b) The matrix ⎡ ⎤ ∂f (xe , ye ) ∂f (xe , ye ) ⎢ ⎥ ⎢ ⎥ ∂x ∂y ⎥ A=⎢ ⎢ ∂g(x , y ) ∂g(x , y ) ⎥ ⎣ e e e e ⎦ ∂x ∂y is invertible. (c) The perturbation function g(z) = f(z + ye ) − Az is such that lim

z→0

g(z) = 0. z

(9)

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REMARKS: 1. The perturbation function g(z) = f(z + ye ) − Az inherits the continuity and differentiability properties assumed for f. Thus, g(z) is continuous with continuous partial derivatives in an open region of the z-plane containing the origin, z = 0. 2. Limit (9) establishes how “small” the nonlinear perturbation must be near the equilibrium point; the norm of the perturbation must tend to zero faster than z as z approaches the origin. 3. Since matrix A is invertible, z = 0 is the only equilibrium point of the linearized problem z = Az. It is also clear that g(0) = 0 [since f(ye ) = 0]. Therefore, z = 0 is an equilibrium point of the nonlinear system z = Az + g(z). In addition, it can be shown that the assumptions made in (a)–(c) imply that z = 0 is an isolated equilibrium point of z = Az + g(z). Therefore, y = ye is an isolated equilibrium point of the original system, y = f(y). Theorem 6.4

Let y = f(y) be a two-dimensional autonomous system that is almost linear at an equilibrium point y = ye . Let z = Az be the corresponding linearized system. (a) If z = 0 is an asymptotically stable equilibrium point of z = Az, then y = ye is an asymptotically stable equilibrium point of y = f( y). (b) If z = 0 is an unstable equilibrium point of z = Az, then y = ye is an unstable equilibrium point of y = f( y). (c) If z = 0 is a stable (but not asymptotically stable) equilibrium point of z = Az, then no conclusions can be drawn about the stability properties of equilibrium point y = ye .

The proof of Theorem 6.4 can be found in more advanced treatments of differential equations, such as Coddington and Levinson.3 However, the assertions made in Theorem 6.4 should strike you as reasonable. When the linearized system is asymptotically stable, both eigenvalues of the coefﬁcient matrix A have negative real parts. We know, therefore, that the norm of the solution of the linearized system, z(t), is exponentially decreasing to zero. If the nonlinear perturbation is sufﬁciently weak, we might expect this qualitative behavior to persist in the nonlinear system (8). Similarly, if the linearized system is unstable, at least one of the two eigenvalues of A has a positive real part. The linearized system, therefore, has some solutions that grow exponentially in norm. In this case, we might expect instability to persist when the nonlinear perturbation g is sufﬁciently weak. Finally, if the linearized system is stable but not asymptotically stable, then the eigenvalues of A form a purely imaginary complex conjugate pair. In this case, the linearized system is sitting on the fence between stability and instability. It is 3

Earl A. Coddington and Norman Levinson, Theory of Ordinary Differential Equations (Malabar, FL: R. E. Krieger, 1984).

6.5

Linearization and the Local Picture

439

possible for the nonlinear perturbation (however small) to tip the balance either way—causing the nonlinear system to be stable or causing it to be unstable. Example 3, found later in this section, illustrates this last point [and thereby proves condition (c) of Theorem 6.4]. E X A M P L E

2

Consider again the nonlinear system discussed in Example 1, x = 12 1 − 12 x − 12 y x y = 14 1 − 13 x − 23 y y.

(10)

Use Theorem 6.4 to determine the stability properties of equilibrium point (1, 1) for the nonlinear system (10). Solution: We begin by making the change of dependent variable z(t) = y(t)−ye : z1 (t) = x(t) − 1

and

z2 (t) = y(t) − 1.

With this change of variables, we can rewrite system (10) as ⎤ ⎡ ⎡ ⎤ − 14 (z21 + z1 z2 ) − 14 − 14 ⎦z + ⎣ ⎦. z = ⎣ 1 1 − 12 − 16 (z1 z2 + 2z22 ) − 12

(11)

We also know (see Table 6.1) that the linearized system ⎤ ⎡ − 14 − 14 ⎦z z = ⎣ 1 − 12 − 16 has an asymptotically stable equilibrium point at z = 0. Theorem 6.4 asserts that z = 0 will be an asymptotically stable equilibrium point of nonlinear system (11) (and therefore ye is an asymptotically stable equilibrium point of the original system) if we can show that the system is almost linear at the equilibrium point ye . Note that the ﬁrst two conditions of the deﬁnition are clearly satisﬁed. So, to apply Theorem 6.4, all we need to do is establish the limit (9): g(z) →0 z where [see equation (11)]

⎡

as z → 0,

− 14 (z21 + z1 z2 )

⎤

⎦. g(z) = ⎣ 1 (z1 z2 + 2z22 ) − 12 In order to calculate the quotient g(z)/z, it is convenient to introduce polar coordinates z1 = r cos θ and z2 = r sin θ . Under this change of variables, we see that z = r and ⎤ ⎡ r2 2 ⎢ − (cos θ + sin θ cos θ) ⎥ ⎥ ⎢ 4 ⎥. g(z) = ⎢ ⎥ ⎢ 2 ⎦ ⎣ r − (sin θ cos θ + 2 sin2 θ) 12 (continued)

440

CHAPTER 6

Nonlinear Systems (continued)

Thus, g(z) g(z) = z r 2 2 1 1 2 2 =r − (cos θ + sin θ cos θ) + − (sin θ cos θ + 2 sin θ) 4 12 √ 2 2 5 1 1 (1 + 1) + (1 + 2) = r. 0, construct a circle of radius ε centered at the origin. For a value of energy E sufﬁciently close to −1, we can ﬁnd a closed trajectory lying entirely within this circle. Let this value of energy be Eε . Now we choose δ > 0 sufﬁciently small that a circle of radius δ lies within this closed trajectory. This choice of δ will work as far as satisfying the stability deﬁnition is concerned—all solutions originating within the circle of radius δ will remain within the closed trajectory of energy Eε since solutions

444

CHAPTER 6

Nonlinear Systems ′ (a) E = (b) E = (c) E = (d) E = (e) E = (f ) E = (g) E = (h) E =

6

4

–1 –1/2 1/2 1 2 3 4 5

2

(h) (g)

(f) (e) (d) (a)

–2

–

(b)

(c)

(b)

(c)

2

(d) (e) (f)

–2

(g) (h) –4

(a)

␦

(b) Equilibrium point (0, 0)

(c) Equilibrium point (, 0)

FIGURE 6.14

(a) Some of the phase-plane trajectories for the pendulum equation (19). Note that the separatrices approach the unstable equilibrium point (π, 0). (b) Any trajectory originating within the circle of radius δ must stay within the trajectory having energy Eε and must therefore remain within the circle of radius ε. (c) Every small circle centered at (π, 0) has at least one trajectory that eventually exits the circle.

cannot intersect. Consequently, the trajectories will remain within the circle of radius ε. Figure 6.14(b) illustrates these ideas. These geometrical ideas also provide another way of seeing that (π, 0) is an unstable equilibrium point of the nonlinear system. In particular [see Figure 6.14(c)], any circle of radius δ > 0 centered at (π, 0) must contain portions of

6.5

Linearization and the Local Picture

445

the separatrices and portions of some of the other trajectories that correspond to large-scale motions of the pendulum. Solution points originating on the latter trajectories eventually exit any small circle of radius ε centered at (π, 0). Therefore, the deﬁnition fails, and the equilibrium point (π, 0) is unstable.

EXERCISES Exercises 1–9: In each exercise, the given system is an almost linear system at each of its equilibrium points. (a) Find the (real) equilibrium points of the given system. (b) As in Example 2, ﬁnd the corresponding linearized system z = Az at each equilibrium point. (c) What, if anything, can be inferred about the stability properties of the equilibrium point(s) by using Theorem 6.4? 1. x = x2 + y2 − 32 y = y − x 2

2

y = x −y +1 7. x = (x − 2y)( y + 4) y = 2x − y

3. x = 1 − x2

y = x

4. x = x − y − 1

2. x = x2 + 9y2 − 9

y = x 2 + y2 − 2

5. x = (x − 2)( y − 3)

6. x = (x − y)( y + 1)

y = (x + 2)( y − 4)

y = (x + 2y)( y − 1) 8. x = xy − 1

9. x = y2 − x

y = (x + 4y)(x − 1)

y = x2 − y

10. Perform a stability analysis of the competing species model at the equilibrium point (0, 0): x = 12 1 − 12 x − 12 y x y = 14 1 − 13 x − 23 y y. Speciﬁcally, repeat the analysis of Example 2 to determine the stability properties of the nonlinear system at this point. 11. Consider the system encountered in the study of pendulum motion, x = y y = − sin x, at its equilibrium points (0, 0) and (π, 0). (a) Let z1 = x, z2 = y. Show that the system becomes

0 1 0 . z+ z = z1 − sin z1 −1 0 (b) Let z1 = x − π, z2 = y. Show that the system becomes

0 1 0 z = . z− z1 − sin z1 1 0 (c) Show that the system is almost linear at both equilibrium points. [Hint: One approach is to use Taylor’s theorem and polar coordinates.]

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Exercises 12–20: Each exercise lists a nonlinear system z = Az + g(z), where A is a constant (2 × 2) invertible matrix and g(z) is a (2 × 1) vector function. In each of the exercises, z = 0 is an equilibrium point of the nonlinear system. (a) Identify A and g(z). (b) Calculate g(z). (c) Is lim z→0 g(z)/z = 0? Is z = Az + g(z) an almost linear system at z = 0? (d) If the system is almost linear, use Theorem 6.4 to choose one of the three statements: (i) z = 0 is an asymptotically stable equilibrium point. (ii) z = 0 is an unstable equilibrium point. (iii) No conclusion can be drawn by using Theorem 6.4. 12. z1 = 9z1 − 4z2 + z22

13. z1 = 5z1 − 14z2 + z1 z2

z2 = 15z1 − 7z2

z2 = 3z1 − 8z2 + z21 + z22

z21 + z22 z2 = −z1 − 5z2 + z1 cos z21 + z22

14. z1 = −3z1 + z2 + z21 + z22

15. z1 = −z1 + 3z2 + z2 cos

z2 = 2z1 − 2z2 + (z21 + z22 )1/3 16. z1 = −2z1 + 2z2 + z1 z2 cos z2 z2 = z1 − 3z2 + z1 z2 sin√ z2

17. z1 = 2z2 + z22 z2 = −2z1 + z1 z2

2

2

18. z1 = −3z1 − 5z2 + z1 e− z1 +z2 √ 2 2 z2 = 2z1 − z2 + z2 e− z1 +z2 20. z1 = 2z1 + 2z2

19. z1 = 9z1 + 5z2 + z1 z2 z2 = −7z1 − 3z2 + z21

z2 = −5z1 − 2z2 + z21 21. Consider the autonomous system x = −x + xy + y y = x − xy − 2y. This is the reduced system for the chemical reaction discussed in Exercise 19 of Section 6.1 with a(t) = x(t), c(t) = y(t), e0 = 1, and all rate constants set equal to 1. (a) Show that this system has a single equilibrium point, (xe , ye ) = (0, 0). (b) Determine the linearized system z = Az, and analyze its stability properties. (c) Show that the system is an almost linear system at equilibrium point (0, 0). (d) Use Theorem 6.4 to determine the equilibrium properties of the given nonlinear system at (0, 0). 22. Consider the nonlinear scalar differential equation x = 1 − (1 + x)3/2 . An equation having this structure arises in modeling the bobbing motion of a ﬂoating parabolic trough. (a) Let y = x and rewrite the given scalar equation as an equivalent ﬁrst order system. (b) Show that the system has a single equilibrium point at (xe , ye ) = (0, 0). (c) Determine the linearized system z = Az, and analyze its stability properties. (d) Assume that the system is an almost linear system at equilibrium point (0, 0). Does Theorem 6.4 provide any information about the stability properties of the nonlinear system obtained in part (a)? Explain.

6.5

Linearization and the Local Picture

447

23. Consider again the differential equation of Exercise 22, x = 1 − (1 + x)3/2 . (a) In Section 6.3, equations having this structure were shown to have a conservation law. Derive this conservation law; it will have the form y2 /2 + F(x) = C, where y = x . (b) Let C = 12 , 34 , and 1 in the conservation law derived in part (a). Plot the corresponding phase-plane solution trajectories, using computational software. Are these phase-plane trajectories consistent with a bobbing motion? Assuming these plots typify the behavior of solution trajectories near the origin, do they suggest that the origin is a stable or unstable equilibrium point for the nonlinear system? Explain. 24. Each of the autonomous nonlinear systems fails to satisfy the hypotheses of Theorem 6.4 at the equilibrium point (0, 0). Explain why. (a) x = x − y + xy

(b) x = x − 2y − x2/3

y = −x + y + 2x2 y2

y = x + y + 2y1/3

25. Polar Coordinates Consider the system z = Az + g(z), where A=

a11

a12

a21

a22

and

g(z) =

g1 (z) g2 (z)

.

Show that adopting polar coordinates z1 (t) = r(t) cos[θ(t)] and z2 (t) = r(t) sin[θ (t)] transforms the system z = Az + g(z) into r = r[a11 cos2 θ + a22 sin2 θ + (a12 + a21 ) sin θ cos θ] + [g1 cos θ + g2 sin θ] θ = [a21 cos2 θ − a12 sin2 θ + (a22 − a11 ) sin θ cos θ] + r −1 [−g1 sin θ + g2 cos θ]. 26. Use the polar equations derived in Exercise 25 to show that if a11 = a22 ,

a21 = −a12 ,

g1 (z) = z1 h

z21 + z22 ,

g2 (z) = z2 h

z21 + z22

for some function h, then the polar equations uncouple into r = a11 r + rh(r) θ = a21 . Note that the radial equation is a separable differential equation and the angle equation can be solved by antidifferentiation. 27. Consider the system x = y + αx(x2 + y2 ), y = −x + αy(x2 + y2 ). Introduce polar coordinates and use the results of Exercises 25 and 26 to derive differential equations for r(t) and θ (t). Solve these differential equations, and then form x(t) and y(t).

Exercises 28–29: Introduce polar coordinates and transform the given initial value problem into an equivalent initial value problem for the polar variables. Solve the polar initial value problem, and use the polar solution to obtain the solution of the original initial value problem. If the solution exists at time t = 1, evaluate it. If not, explain why. √ 29. x = y − x ln[x2 + y2 ], x(0) = e/ 2 28. x = x + x x2 + y2 , x(0) = 1 √ √ y = −x − y ln[x2 + y2 ], y(0) = e/ 2 y = y + y x2 + y2 , y(0) = 3

448

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6.6

Nonlinear Systems

Two-Dimensional Linear Systems We continue studying the phase-plane behavior of solutions of the linear system y = Ay, where A is a (2 × 2) real invertible matrix. Since A is invertible, y = 0 is the only equilibrium solution of y = Ay. As we have seen, there are two principal reasons for studying this phaseplane behavior. First, y = Ay is an important and intrinsically interesting system. Second, such systems arise whenever we linearize about an equilibrium point, zooming in to study the behavior of a nonlinear autonomous system close to an equilibrium point. By studying the eigenvalues and the phase-plane geometry of the associated eigenvectors at an equilibrium point, we can often sketch a good local picture— one that gives a qualitative description of the nonlinear system behavior near the equilibrium point. Such local pictures complement the large-scale overview provided by the direction ﬁeld. Taken together, they provide a good overall view of system behavior. To illustrate the ideas, we consider the competing species problem that has served as a vehicle for discussion throughout this chapter.

E X A M P L E

1

We use linearized system approximations to develop local pictures of system behavior near each of the equilibrium points of the nonlinear system x = 12 1 − 12 x − 12 y x (1) y = 14 1 − 13 x − 23 y y. System (1) has four equilibrium points, (0, 0), (0, 32 ), (2, 0), and (1, 1). We focus on the equilibrium point (0, 32 ) in order to illustrate the basic ideas. Using

x(t) z(t) = , y(t) − 32 we have for the linearized system at (0, 32 ) ⎡ 1 8

z = ⎣ − 18

⎤ 0 ⎦ z.

− 14

The general solution of this linear system is

3 0 z(t) = c1 et/8 + c2 e−t/4 . −1 1

(2)

We can use the eigenpair information in (2) to sketch the qualitative behavior of solution trajectories of z = Az. In turn, these sketches provide qualitative information about solutions of the original nonlinear system near the equilibrium point (0, 32 ). To begin, consider the special case where c2 = 0 and c1 = 0. In this case, z1 (t) = 3c1 et/8

and

z2 (t) = −c1 et/8 .

(3)

6.6

Two-Dimensional Linear Systems

449

From (3), these solutions lie on the z-plane line z2 = − 13 z1 . Since et/8 increases as t increases, we also see from (3) that both z1 (t) and z2 (t) increase in magnitude as t increases. Therefore, solution points originating on this line remain on this line and move away from the origin as t increases, as shown in the direction ﬁeld plot in Figure 6.15(a), on the next page. Similarly, consider the companion case where c1 = 0 and c2 = 0 in equation (2). In this case, solution points lie on the phase-plane line z1 = 0 and approach the origin as t increases, since e−t/4 decreases as t increases [see Figure 6.15(a)]. Solution point behavior in these two special cases enables us to determine qualitatively the general phase-plane characteristics of (2). Consider the general case where c1 = 0, c2 = 0. For sufﬁciently small values of t, both exponential functions are roughly comparable in size and both terms in the general solution inﬂuence solution point behavior. However, as t increases, the term

3 t/8 c1 e −1 becomes increasingly dominant and all solution trajectories approach the phase-plane line z2 = − 13 z1 as an asymptote. Therefore, we obtain the phaseplane behavior shown in Figure 6.15(a). A similar analysis can be used to study behavior near the other equilibrium points, (0, 0), (2, 0), and (1, 1). In all cases, the eigenvalues of the linearized system coefﬁcient matrix are real and distinct. The eigenvectors determine lines through the z-plane origin on which solution points travel either toward the origin if the corresponding eigenvalue is negative or away from the origin if the eigenvalue is positive. Using this behavior as a guide, we can sketch in the qualitative behavior of solution points originating elsewhere in the plane. This qualitative behavior is shown in Figures 6.15(b)–(d). ❖

The four z-plane-phase portraits in Figure 6.15, when positioned at the corresponding y-plane equilibrium points, provide local pictures that are complementary to and consistent with the large-scale overview developed in Section 6.2. This is illustrated in Figure 6.16, where the local equilibrium pictures from Figure 6.15 have been superimposed on Figure 6.6 from Section 6.2. Attention is restricted to the ﬁrst quadrant, since the dependent variable y(t) has components that represent (nonnegative) populations.

Classifying Equilibrium Points In Example 1, the coefﬁcient matrix of the linearized system at each of the four equilibrium points had real, distinct eigenvalues. In two cases, the eigenvalues had the same sign; in the other two, the eigenvalues had opposite signs. If A is an invertible (2 × 2) matrix, other possibilities exist for the (nonzero) pair of eigenvalues. The eigenvalues might be real and repeated. They might be a complex conjugate pair with nonzero real parts, or they might be a purely imaginary complex conjugate pair. Table 6.2 enumerates the various possibilities and the names assigned to them.

450

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Nonlinear Systems z2

–3

–2

z2

3

3

2

2

1

1

–1

1

2

3

z1

–3

–2

–1

–1

1

2

z1

3

–1

z2 = – 13 z1

–3

–2

–2

–2

–3

–3

(a)

(b)

z2

z2

3

3

2

2

1

1

–1

1

2

3

z1

–3

–1

1

2

(

2 1 + √13

)z

1

)z

1

z1

3

–1

–1

–2

–2

z2 =

z2 = – 76 z1

–2

–3

–3

(c)

(d) FIGURE 6.15

Direction ﬁelds for the various linearizations z = Az associated with the nonlinear system (2). Each z-plane direction ﬁeld corresponds to an equilibrium point of the nonlinear system. The equilibrium points are (a) (0, 32 ), (b) (0, 0), (c) (2, 0), and (d) (1, 1).

z2 =

(

2 1 – √13

6.6

Two-Dimensional Linear Systems

451

y 1.5

1 g=0 0.5 f=0 0.5

1

1.5

2

x

FIGURE 6.16

The local equilibrium pictures from Figure 6.15, superimposed on the qualitative picture developed in Section 6.2 (see Figure 6.6 of Section 6.2).

TA B L E 6 . 2 Classiﬁcation of the Equilibrium Point at the Origin for y = Ay

Eigenvalues of A

Type of Equilibrium Point

Stability Characteristics of the Linear System

Real eigenvalues λ1 , λ2 where λ1 ≤ λ2 < 0 0 < λ1 ≤ λ2

Node Node

Asymptotically stable Unstable

Real eigenvalues λ1 , λ2 where λ1 < 0 < λ2

Saddle point

Unstable

Complex eigenvalues where λ1,2 = α ± iβ, α0

Spiral point Spiral point

Asymptotically stable Unstable

Complex eigenvalues where λ1,2 = ±iβ, β = 0

Center

Stable but not asymptotically stable

The “node” designation is often divided into two subcategories. If matrix A has two equal (real) eigenvalues and is a scalar multiple of the (2 × 2) identity matrix, the equilibrium point is called a proper node. In all other cases (when A has equal real eigenvalues but only one linearly independent eigenvector or when A has unequal real eigenvalues of the same sign), the equilibrium point is called an improper node. Figures 6.15 and 6.17–6.19 provide some examples of phase-plane behavior at nodes and saddle points. The following three examples illustrate typical behavior at a proper node, a spiral point, and a center.

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E X A M P L E

2

Proper Node Consider the linear system

y =

α

0

0

α

α = 0.

y,

The origin is a proper node since the eigenvalues are real and equal (λ1 = λ2 = α), and the coefﬁcient matrix is a nonzero multiple of the identity matrix. The phase-plane behavior of trajectories is easily recognized if we adopt polar coordinates. Let x = r cos θ

and

y = r sin θ.

With this change of variables, the component equations transform into the differential equations for the polar variables, r = αr θ = 0. As time increases, the solution points move on rays, since the polar angle θ remains constant. If α < 0, solutions approach the origin and the origin is an asymptotically stable equilibrium point. If α > 0, solution points move outward along the rays and the origin is an unstable equilibrium point. Note that the rays (the trajectories themselves) are independent of the value of α. The parameter α governs only how quickly solution points move inward or outward along the rays. Figure 6.17 depicts behavior for the case α > 0. y 3 2 1 x –3

–2

–1

1

2

3

–1 –2 –3

FIGURE 6.17

The origin is a proper node for the system in Example 2. Since α > 0 in this example, the origin is an unstable equilibrium point. E X A M P L E

3

Spiral Point Consider the linear system

−1 −1 y = y. 1 −1

❖

6.6

Two-Dimensional Linear Systems

453

The eigenvalues of the coefﬁcient matrix are the complex conjugate pair λ1 = −1 + i, λ2 = −1 − i. According to Table 6.2, the origin is an asymptotically stable spiral point. The behavior of solutions, as well as the reason for the terminology “spiral point,” can be clearly seen when we change to polar coordinates. For this system, we obtain the following pair of differential equations for the polar variables: r = −r θ = 1. Let the initial conditions be r(0) = r0 and θ(0) = θ0 . Then the solutions are r(t) = r0 e−t ,

θ (t) = t + θ0 .

Thus, as time increases, a solution point spirals inward toward the origin. Its distance from the origin decreases at an exponential rate while it moves counterclockwise about the origin. This behavior is shown in Figure 6.18. y 3 2 1

x –3

–2

–1

1

2

3

–1 –2 –3

FIGURE 6.18

The origin is a spiral point for the system in Example 3. Since the eigenvalues of A have real part −1, the origin is asymptotically stable. A solution point follows a trajectory that spirals in toward the origin.

❖

E X A M P L E

4

Center Consider the linear system

−4 y = −5

5 y. 4

The eigenvalues of the coefﬁcient matrix are the purely imaginary complex conjugate pair λ1 = 3i, λ2 = −3i. According to Table 6.2, the origin is classiﬁed as a stable center. Phase-plane behavior is shown in Figure 6.19. One way to derive the equations for the elliptical trajectories in Figure 6.19 is to change to polar coordinates. For this linear system, the differential equations for the polar variables are r = −4r cos 2θ θ = −5 + 4 sin 2θ.

(continued)

454

CHAPTER 6

Nonlinear Systems y

(continued) 3 2 1

x –3

–2

–1

1

2

3

–1 –2 –3

FIGURE 6.19

The origin is a center for the system in Example 4 since the eigenvalues of A are purely imaginary. The origin is a stable equilibrium point, but it is not asymptotically stable. The solution points follow elliptical trajectories about the origin.

Notice that θ is a decreasing function of t. Therefore, an inverse function exists, and we can view r as a function of θ. Using the chain rule, we have dr dt 1 dr = = −4r cos 2θ , dθ dt dθ −5 + 4 sin 2θ or dr 4 cos 2θ = r. dθ 5 − 4 sin 2θ This equation is a ﬁrst order linear differential equation. Assuming an initial condition of r = r0 when θ = 0, we ﬁnd the solution r0 . r=√ 1 − 0.8 sin 2θ Note that r is a periodic function of θ , with period π. Since θ is a decreasing function of t, the solution points (r, θ ) move clockwise around the closed elliptical trajectories, as shown in Figure 6.19. ❖ An alternative derivation of the trajectory equations is outlined in Exercise 31. This approach leads to equations in terms of the original x, y phaseplane variables.

EXERCISES Exercises 1–5: In each exercise, the eigenpairs of a (2 × 2) matrix A are given where both eigenvalues are real. Consider the phase-plane solution trajectories of the linear system y = Ay, where

x(t) y(t) = . y(t)

6.6

Two-Dimensional Linear Systems

455

(a) Use Table 6.2 to classify the type and stability characteristics of the equilibrium point at y = 0. (b) Sketch the two phase-plane lines deﬁned by the eigenvectors. If an eigenvector is u1 , the line of interest is u2 x − u1 y = 0. Solution trajectories originating on such u2 a line stay on the line; they move toward the origin as time increases if the corresponding eigenvalue is negative or away from the origin if the eigenvalue is positive. (c) Sketch appropriate direction ﬁeld arrows on both lines. Use this information to sketch a representative trajectory in each of the four phase-plane regions having these lines as boundaries. Indicate the direction of motion of the solution point on each trajectory.

1 1 1. λ1 = 2, x1 = ; λ2 = −1, x2 = 1 −1

1 2 ; λ2 = 2, x2 = 2. λ1 = 1, x1 = 2 −1

2 0 3. λ1 = 2, x1 = ; λ2 = 1, x2 = 0 2

1 1 4. λ1 = −2, x1 = ; λ2 = −1, x2 = 0 1

1 2 ; λ2 = −1, x2 = 5. λ1 = 1, x1 = 0 1

Exercises 6–20: In each exercise, consider the linear system y = Ay. Since A is a constant invertible (2 × 2) matrix, y = 0 is the unique (isolated) equilibrium point. (a) Determine the eigenvalues of the coefﬁcient matrix A. (b) Use Table 6.2 to classify the type and stability characteristics of the equilibrium point at the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node.

6. y =

1

−6

1

−4

12. y =

15. y = 18. y =

7. y =

y

1

2

−5

−1

9. y =

2

−3

3

2

−1

8

−1

5

y

10. y =

y

13. y =

3

0

0

3

y

16. y =

y

19. y =

6

−10

2

−3

y

8. y =

1

−1

−1

−2

−4

5

2

−2

1

−1

−2

−1

y

11. y =

1

2

−8

1

y

14. y =

y

17. y =

y

20. y =

−6

14

−2

5

y

1

−6

2

−6

7

−24

2

−7

y

y

2

4

−4

−6

−1

−2

2

3

y

y

21. Consider the linear system y = Ay. Four direction ﬁelds are shown. Determine which of the four coefﬁcient matrices listed corresponds to each of the direction ﬁelds shown.

456

CHAPTER 6

Nonlinear Systems (a) A1 = (c) A3 =

−2

1

1

−2

2

1

−1

−2

(b) A2 =

(d) A4 =

1

2

−2

−1

1

2

−2

1

y

y

2

2

1.5

1.5

1

1

0.5

0.5

x –2 –1.5 –1 –0.5 –0.5

0.5

1

1.5

2

x –2 –1.5 –1 –0.5 –0.5

–1

–1

–1.5

–1.5

–2

–2

0.5

1

Direction Field 1

Direction Field 2

y

y

2

2

1.5

1.5

1

1

0.5

1.5

2

1.5

2

0.5

x –2 –1.5 –1 –0.5 –0.5

0.5

1

1.5

2

x –2 –1.5 –1 –0.5 –0.5

–1

–1

–1.5

–1.5

–2

–2

Direction Field 3

0.5

1

Direction Field 4 Figure for Exercise 21

Exercises 22–25: Use the information given about the nature of the equilibrium point at the origin to determine the value or range of permissible values for the unspeciﬁed entry in the coefﬁcient matrix. 2 3 y; determine α. 22. The origin is a center for the linear system y = −3 α −4 α y, for what values of α (if any) can the origin be an asymptoti23. Given y = −2 2 cally stable spiral point? −2 0 24. The origin is an asymptotically stable proper node of y = y; determine α −2 the value(s) of α.

6.6

457

−2 y, for what values of α (if any) can the origin be an (unstable) −4

4 25. Given y = α saddle point?

Two-Dimensional Linear Systems

Exercises 26–29: Locate the equilibrium point of the given nonhomogeneous linear system y = Ay + g0 . [Hint: Introduce the change of dependent variable z(t) = y(t) − y0 , where y0 is chosen so that the equation can be rewritten as z = Az.] Use Table 6.2 to classify the type and stability characteristics of the equilibrium point.

3 4 1 4 6 5 26. y = y+ 27. y = y+ −1 1 −7 −6 2 −6 28. x = 5x − 14y + 2

29. x = −x + 2

y = 3x − 8y + 1

y = 2y − 4

30. Let

A=

a11

a12

a21

a22

be a real invertible matrix, and consider the system y = Ay. (a) What conditions must the matrix entries ai j satisfy to make the equilibrium point ye = 0 a center? (b) Assume that the equilibrium point at the origin is a center. Show that the system y = Ay is a Hamiltonian system. (c) Is the converse of the statement in part (b) true? In other words, if the system y = Ay is a Hamiltonian system, does it necessarily follow that ye = 0 is a center? Explain. 31. Consider the linear system of Example 4, −4 y = −5

5 4

y.

The coefﬁcient matrix has eigenvalues λ1 = 3i, λ2 = −3i; the equilibrium point at the origin is a center. (a) Show that the linear system is a Hamiltonian system. Either use the results of Exercise 30 or apply the criterion directly to this example. (b) Derive the conservation law for this system. The result, 52 x2 − 4xy + 52 y2 = C > 0, deﬁnes a family of ellipses. These ellipses are the trajectories on which the solution point moves as time changes. (c) Plot the ellipses found in part (b) for C = 14 , 12 , and 1. Indicate the direction in which the solution point moves on these ellipses.

Exercises 32–34: A linear system is given in each exercise. (a) Determine the eigenvalues of the coefﬁcient matrix A. (b) Use Table 6.2 to classify the type and stability characteristics of the equilibrium point at y = 0. (c) The given linear system is a Hamiltonian system. Derive the conservation law for this system. 32.

y =

−2

1

5

2

y

33. x = x + 3y y = −3x − y

34.

y =

2 0

1 −2

y

458

CHAPTER 6

6.7

Nonlinear Systems

Predator-Prey Population Models By way of introduction, we pose a question. This question involves the familiar problem of a predatory population being introduced into a colony, either by accident or by design. If the predator is undesirable, our goal may be to remove it from the colony. If the predator is desirable, our goal may be to establish a coexisting ecological balance between the predators and their prey. Which scenarios lead to predator eradication, and which scenarios lead to predatorprey coexistence? And what are the factors responsible for the desired outcome?

Mathematical Modeling We develop a mathematical model of two-species predator-prey interaction to gain insight into the question posed above.4 Let P1 (t) and P2 (t) represent the populations of predators and prey (respectively) in a colony at time t. Both populations change with time because of births, deaths, and harvesting. A “conservation of population” principle of the type discussed in Section 2.4 leads to differential equations having the following general structure: dP1 = R1 P1 − μ1 P1 dt

(1)

dP2 = R2 P2 − μ2 P2 . dt In equation (1), the term R j represents the net birth rate per unit population of the jth species. We assume that R1 and R2 are functions of the populations P1 and P2 (but not explicit functions of time t). The nonnegative constants μ1 and μ2 represent harvesting rates per unit population. Applying the ideas underlying the logistic population model developed in Section 2.8, we assume the rate functions R1 and R2 have the form R1 = r1 (−1 − α1 P1 + β1 P2 ) R2 = r2 (1 − β2 P1 − α2 P2 ),

(2)

where the parameters rj , αj , and βj are nonnegative constants, j = 1, 2. What are we actually assuming in (2)? In the absence of prey for food (that is, if β1 = 0), the predator rate function would be R1 = −r1 (1 + α1 P1 ) < 0 and the predator population would continually decrease. The β1 P2 term embodies the beneﬁcial aspects of the prey food supply on the predator growth rate. The −α1 P1 term allows for competition among the predators for the available food. Consider now the prey rate function, R2 . In the absence of predators and limitations on the prey’s food supply (that is, if β2 = 0 and α2 = 0), the rate function would be R2 = r2 > 0. In that case, the prey population would grow exponentially whenever r2 > μ2 . The terms −β2 P1 and −α2 P2 account for the 4

Important early work in developing and applying such models was done by Volterra. Vito Volterra (1860–1940) was an Italian mathematician and scientist noted for his work on functional calculus, partial differential equations, integral equations and mathematical biology. During his career, he held distinguished positions at the universities of Pisa, Turin, and Rome. In 1931, he was forced to leave the University of Rome after refusing to take an oath of allegiance to the Fascist government. He left Italy the following year and spent the rest of his life abroad.

6.7

Predator-Prey Population Models

459

negative effects of predation and limits on the prey’s food supply, respectively. The predator-prey equations we use as our basic model are therefore dP1 = r1 (−1 − α1 P1 + β1 P2 )P1 − μ1 P1 dt dP2 = r2 (1 − β2 P1 − α2 P2 )P2 − μ2 P2 . dt

(3)

Managing a Predator Population We now take up the question posed at the beginning of this section: How do we manage a predator population that has been introduced into a colony, whether by accident or by design? We assume the colony has resource limitations that exert a constraining inﬂuence on each of the predator and prey populations. We allow for the possibility of harvesting the predator population but not the prey population. Under these assumptions, system (3) becomes μ dP1 = r1 −1 − − α1 P1 + β1 P2 P1 dt r1 (4) dP2 = r2 1 − β2 P1 − α2 P2 P2 , dt where all the constants on the right-hand side of (4) are assumed positive with the possible exception of the harvesting rate μ, which we may allow to be zero. From an ecological point of view, we want to know what combination of harvesting strategies and environmental factors will cause the predator population to (a) die out or (b) achieve a coexisting balance with the prey population as time evolves. Rephrasing in mathematical terms, we want to discover which relations among the constants in (4) will cause all solutions to (a) converge to an equilibrium value (0, P2e ), where P2e > 0, or (b) converge to an equilibrium value (P1e , P2e ), where P1e > 0, P2e > 0. Autonomous system (4) has at most three equilibrium points in the ﬁrst quadrant of the phase plane: ⎡ ⎤ μ + β1 ⎥ ⎢ −α2 1 + r ⎡ ⎤ ⎥ ⎢ 1

0 ⎢ ⎥ α α + β β ⎥ ⎢ 0 1 2 1 2 ⎢ ⎥ (2) (3) ⎥. ⎢ = , P = = , P P(1) ⎦ ⎣ 1 e e e ⎢ ⎥ 0 μ ⎥ ⎢ ⎢ β2 1 + α2 + α1 ⎥ ⎦ ⎣ r1 α 1 α 2 + β 1 β2

460

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Nonlinear Systems

The equilibrium point Pe(1) , where P1e = P2e = 0, corresponds to neither species being present. The equilibrium point P(2) e , where P1e = 0,

P2e =

1 , α2

(5)

corresponds to a complete absence of predators. The third equilibrium point, P(3) e , is found by solving the system of equations −1 −

μ − α1 P1e + β1 P2e = 0 r1

1 − β2 P1e − α2 P2e = 0. This equilibrium point is given by μ −α2 1 + + β1 r1 , P1e = α1 α2 + β1 β2

P2e

μ + α1 β2 1 + r1 = . α1 α2 + β1 β2

(6)

By deﬁnition, the two species coexist when both P1e and P2e are positive. Thus, model (4) predicts that both species can coexist in equilibrium only if μ (7) . β1 > α2 1 + r1 If inequality (7) does not hold, then the two species cannot coexist in equilibrium and the only nontrivial equilibrium solution in the ﬁrst quadrant is (5), wherein predators are absent. Suppose our goal is to eradicate the predators. We see from (7) that we can eliminate the possibility of equilibrium predator-prey coexistence by sufﬁciently increasing α2 [1 + (μ/r1 )] and/or by decreasing β1 . Does this make sense? Increasing μ corresponds to increasing the harvesting rate of predators, while decreasing β1 corresponds to somehow reducing the beneﬁcial effects of the prey as food for the predators. It seems reasonable that either of these two strategies would be harmful to the predators. What about increasing α2 , however? Recall that α2 is the parameter modeling the constraining effects of the available colony resources on the prey population. In the absence of predators, the equilibrium prey population P2e = 1/α2 decreases as α2 increases. Is it reasonable to conclude that we can adversely impact the predator population by indirectly constraining its food supply? We want to focus on the role of the parameter α2 in controlling the population. Toward that end, the parameters r1 , r2 , α1 , β1 , β2 , and μ will all be assigned the value 1, leading to the system dP1 = (−2 − P1 + P2 )P1 dt dP2 = (1 − P1 − α2 P2 )P2 . dt

(8)

Table 6.3 summarizes the information we can deduce from linearizing system (8). In either case (whether equilibrium coexistence is possible or impossible), the origin is an unstable saddle point. If equilibrium coexistence is possible, then the equilibrium point (0, 1/α2 ) is an unstable saddle point. If coexistence is

6.7

Predator-Prey Population Models

461

TA B L E 6 . 3

Equilibrium Point P1e = 0 P2e = 0

Linearized System z = Az [z(t) = P(t) − Pe ] ⎡ ⎢−2 A=⎣ 0

Eigenvalues

⎤ 0⎥ ⎦ 1

λ1 = −2

P2e = 1/α2

⎤ 0 ⎦ −1

1 − 2α2 α2 + 1 3 P2e = α2 + 1

⎡ 1 − 2α2 ⎢− α + 1 ⎢ 2 A=⎢ 3 ⎣ − α2 + 1

⎤ 1 − 2α2 α2 + 1 ⎥ ⎥ ⎥ 3α2 ⎦ − α2 + 1

P1e =

Unstable

λ2 = 1

⎡ −2 + 1/α2 A=⎣ −1/α2

P1e = 0

Stability Properties of System (4)

Unstable if α2 < 12 ,

λ1 = −2 + 1/α2

asymptotically stable if α2 > 12

λ2 = −1

λ1,2

1 1 =− ± 2 2

1−

12(1 − α2 − 2α22 ) 2

(1 + α2 )

Asymptotically stable if α2 < 12

impossible, however, this equilibrium point becomes an asymptotically stable node. The third equilibrium point in Table 6.3, corresponding to the coexistence of predators and prey, requires [see equation (7)] that the numerator of P1e be positive; that is, α2 < 12 . When α2 < 12 , this third equilibrium point is either an asymptotically stable spiral point or a node. The three phase-plane plots in Figure 6.20 correspond to different values of α2 in system (8). In Figure 6.20(a), α2 = 0. In this case, no resource limitations constrain prey growth, and predator-prey coexistence is possible. All solution trajectories having both species initially present spiral in toward the asymptotically stable equilibrium point (1, 3). In Figure 6.20(b), α2 has been 9 . In this case, the equilibrium point of the linearized system increased to α2 = 20 is a stable node. Here again, all solution trajectories of the nonlinear system having both species initially present approach the asymptotically stable equi2 60 librium point at ( 29 , 29 ). Lastly [see Figure 6.20(c)], when α2 = 1, coexistence is not possible. All solution trajectories having both species initially present approach the asymptotically stable equilibrium point (0, 1) as t → ∞. In this last case, the predator population tends toward extinction as time progresses. These direction ﬁeld observations support our previous conclusions. On one hand, the parametric study illustrated in Figure 6.20 indicates that our interpretation of the model’s behavior seems correct. On the other hand, a model such as (4) is at best a gross simpliﬁcation of reality. The trade-off in

462

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Nonlinear Systems P2 2.2

P2

2.15 4

2.1

3.5

2.05

3

2

2.5 1.95

2

1.9

1.5 1

1.85

0.5

1.8 0.5

1

1.5

2

2.5

3

3.5

4

P1

0.05

0.1

0.15

(a)

0.2

0.25

0.3

0.35

0.4

P1

(b)

P2 4 3.5 3 2.5 2 1.5 1 0.5 0.5

1

1.5

2

2.5

3

3.5

4

P1

(c) FIGURE 6.20

Portions of the direction ﬁeld for the predator-prey equations (8), with 9 different choices of α2 : (a) α2 = 0, (b) α2 = 20 , (c) α2 = 1.

modeling is always one of reducing a problem to its essential features without “throwing away” reality. In particular, when model predictions seem counterintuitive, we need to proceed with a healthy skepticism—both scrutinizing and reﬁning the model to gain further conﬁdence and insight.

EXERCISES Exercises 1– 4: Assume the given autonomous system models the population dynamics of two species, x and y, within a colony. (a) For each of the two species, answer the following questions.

6.7

Predator-Prey Population Models

463

(i) In the absence of the other species, does the remaining population continuously grow, decline toward extinction, or approach a nonzero equilibrium value as time evolves? (ii) Is the presence of the other species in the colony beneﬁcial, harmful, or a matter of indifference? (b) Determine all equilibrium points lying in the ﬁrst quadrant of the phase plane (including any lying on the coordinate axes). (c) The given system is an almost linear system at the equilibrium point (x, y) = (0, 0). Determine the stability properties of the system at (0, 0). 1. x = x − x2 − xy

2. x = −x − x2 y = −y + xy

y = y − 3y2 − 12 xy 3. x = x − x2 − xy

4. x = x − x2 + xy

y = −y − y2 + xy

y = y − y2 + xy

5. A scientist adopted the following mathematical model for a colony containing two species, x and y: x = r1 (1 + α1 x + β1 y)x y = r2 (1 + β2 x + α2 y)y. The following information is known: (i) If only species x is present in the colony, any initial amount will vary with time as shown in graph (a). If only species y is present, any initial amount will vary as shown in graph (b). (ii) If both species are initially present, (xe , ye ) = (2, 3) is an equilibrium point. ln[x(t)]

ln[y(t)] 1

4

1

2

4

6

t

2 1

1

2 2

4

–1

1

–2

6

t –4

(a)

(b) Figure for Exercise 5

(a) Determine the six constants r1 , α1 , β1 , r2 , α2 , and β2 . (b) How do the two populations relate to each other? Is population x beneﬁcial, harmful, or indifferent to population y? Is population y beneﬁcial, harmful, or indifferent to population x?

Exercises 6–7: Two Competing Species These exercises explore the question “When one of two species in a colony is desirable and the other is undesirable, is it better to use resources to nurture the growth of the desirable species or to harvest the undesirable one?”

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Let x(t) and y(t) represent the populations of two competing species, with x(t) the desirable species. Assume that if resources are invested in promoting the growth of the desirable species, the population dynamics are given by x = r(1 − αx − βy)x + μx y = r(1 − αy − βx)y.

(9)

If resources are invested in harvesting the undesirable species, the dynamics are x = r(1 − αx − βy)x y = r(1 − αy − βx)y − μy.

(10)

In (10), r, α, β, and μ are positive constants. For simplicity, we assume the same parameter values for both species. For deﬁniteness, assume that α > β > 0. 6. Consider system (9), which describes the strategy in which resources are invested into nurturing the desirable species. (a) Determine the four equilibrium points for the system. (b) Show that it is possible, by investing sufﬁcient resources (that is, by making μ large enough), to prevent equilibrium coexistence of the two species. (c) Assume that μ is large enough to preclude equilibrium coexistence of the two species. Compute the linearized system at each of the three physically relevant equilibrium points. Determine the stability characteristics of the linearized system at each of these equilibrium points. (d) System (9) can be shown to be an almost linear system at each of the equilibrium points. Use this fact and the results of part (c) to infer the stability properties of system (9) at each of the three equilibrium points of interest. (e) Sketch the direction ﬁeld. Will a sufﬁciently aggressive nurturing of species x ultimately drive undesirable species y to extinction? If so, what is the limiting population of species x? 7. Consider system (10), which describes the strategy in which resources are invested in harvesting the undesirable species. Again assume that α > β > 0. (a) Determine the four equilibrium points for the system. (b) Show that it is possible, by investing sufﬁcient resources (that is, by making μ large enough), to prevent equilibrium coexistence of the two species. In fact, if μ > r, show that there are only two physically relevant equilibrium points. (c) Assume μ > r. Compute the linearized system at each of the two physically relevant equilibrium points. Determine the stability characteristics of the linearized system at each of these equilibrium points. (d) System (10) can be shown to be an almost linear system at each of the equilibrium points. Use this fact and the results of part (c) to infer the stability properties of system (10) at each of the two equilibrium points of interest. (e) Sketch the direction ﬁeld. Will sufﬁciently aggressive harvesting of species y ultimately drive undesirable species y to extinction? If so, what is the limiting population of species x? 8. Compare the conclusions reached in Exercises 6 and 7. Assume we have sufﬁcient resources to implement either strategy. Which strategy will result in the larger number of desirable species x eventually being present: promoting the desirable species or harvesting the undesirable one? Could this answer have been anticipated by assuming that both strategies will lead to the eventual extinction of species y? Will comparing the resulting one-species equilibrium values for x provide the answer?

6.7

Predator-Prey Population Models

465

9. Three species, designated as x, y, and z, inhabit a colony. Species x and y are two mutually competitive varieties of prey, while z is a predator that depends on x and y for sustenance. In the absence of the other two species, both x and y are known to evolve toward a nonzero equilibrium value as time increases. Species z, however, decreases exponentially toward extinction when both species of prey are absent. Assume that a mathematical model having the following structure is adopted to describe the population dynamics: x = ±a1 x ± b1 x2 ± c1 xy ± d1 xz y = ±a2 y ± b2 y2 ± c2 xy ± d2 yz z = ±a3 z ± c3 xz ± d3 yz. If we want the constants a1 , b1 , c1 , a2 , . . . , d3 to be nonnegative, use the information given to select the correct (plus or minus) sign in the model.

Exercises 10–11: Infectious Disease Dynamics Consider a colony in which an infectious disease (such as the common cold) is present. The population consists of three “species” of individuals. Let s represent the susceptibles—healthy individuals capable of contracting the illness. Let i denote the infected individuals, and let r represent those who have recovered from the illness. Assume that those who have recovered from the illness are not permanently immunized but can become susceptible again. Also assume that the rate of infection is proportional to si, the product of the susceptible and infected populations. We obtain the model s = −αsi + γ r i = αsi − βi

(11)

r = βi − γ r, where α, β, and γ are positive constants. 10. (a) Show that the system of equations (11) describes a population whose size remains constant in time. In particular, show that s(t) + i(t) + r(t) = N, a constant. (b) Modify (11) to model a situation where those who recover from the disease are permanently immunized. Is s(t) + i(t) + r(t) constant in this case? (c) Suppose that those who recover from the disease are permanently immunized but that the disease is a serious one and some of the infected individuals perish. How does the system of equations you formulated in part (b) have to be further modiﬁed? Is s(t) + i(t) + r(t) constant in this case? 11. (a) Consider system (11). Use the fact that s(t) + i(t) + r(t) = N to obtain a reduced system of two differential equations for the two dependent variables s(t) and i(t). (b) For simplicity, set α = β = γ = 1 and N = 9. Determine the equilibrium points of the reduced two-dimensional system. (c) Determine the linearized system at each of the equilibrium points found in part (b). Use Table 6.2 to analyze the stability characteristics of each of these linearized systems. (d) Show that the nonlinear system is an almost linear system at each of the equilibrium points found in part (b). What are the stability characteristics of the nonlinear system at these points?

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PROJECTS Project 1: A Bobbing Sphere Consider Figure 6.21. Assume a sphere of radius R weighs half as much as an equivalent volume of water. In its equilibrium state, the sphere ﬂoats half-submerged, as shown in Figure 6.21(a). The sphere is disturbed from equilibrium at some instant. Its position is as shown in Figure 6.21(b), with displacement y(t) measured positive downward.

R R + y(t)

FIGURE 6.21

(a) The equilibrium state of a ﬂoating sphere whose weight is one half the weight of an equal volume of water. (b) The perturbed state of the sphere, with y(t) > 0 as shown.

1. Compute the volume of the submerged portion of the sphere at the instant when the displacement from equilibrium is y(t). (Archimedes’ law of buoyancy states that the upward force acting on the sphere is the weight of the water displaced at that instant.) 2. Apply Newton’s second law of motion to obtain the differential equation governing the bobbing motion of the sphere. Considering only the weight and buoyant force, equate my to the net downward force (sphere weight minus upward buoyant force). Show that the resulting equation is

g y + 2

y3 3y − 3 R R

= 0.

(1)

In (1), g denotes the acceleration due to gravity. For what range of values of y(t) is the differential equation (1) physically relevant? 3. Equation (1) deﬁnes a conservative system. Derive the corresponding conservation law, and use it to answer the following questions. Assume R = 0.5 m and g = 9.8 m/s2 . (a) If the sphere is raised 10 cm above its equilibrium position and released from rest, what is the maximum vertical speed attained by the sphere in its bobbing motion? (b) If the sphere is set into motion with initial conditions y(0) = 0, y (0) = 1 m/s, what is the maximum depth that the sphere center will reach as it bobs? 4. Rewrite (1) as an equivalent two-dimensional system of ﬁrst order equations, where z1 = y and z2 = y . Show that the nonlinear system is an almost linear system at its only equilibrium point, ze = 0. 5. Perform a stability analysis of the linearized system in part 4 at z = 0. Can we use this analysis to infer the stability properties of the corresponding nonlinear system? Explain.

Projects

467

Project 2: Introduction of an Infectious Disease into a Predator-Prey Ecosystem In this chapter, we have discussed a model of a predator-prey ecosystem, x = −ax + bxy y = cy − dxy.

(2)

In (2), x(t) and y(t) represent the populations of predators and prey, respectively, at time t. The terms a, b, c, and d are positive constants. The terms bxy and −dxy account for the beneﬁcial and detrimental impacts of predation on the predator and prey populations. We also discussed a model for the dynamics of an infectious disease within a population (see Exercise 10 of Section 6.7). In the present discussion, we will assume that infected individuals, when they have recovered, immediately become susceptible again. Therefore, we need not consider a “recovered” population as a separate entity. With this assumption, the infectious disease model considered in Section 6.7 simpliﬁes to s = −αsi + βi i = αsi − βi.

(3)

In (3), s(t) and i(t) are the populations of susceptible and infected individuals at time t, while α and β are positive constants. In the model (3), the total population, s(t) + i(t), remains constant in time. We will combine the ideas embodied in these two problems to model a situation where an infectious disease has been introduced into a predator-prey colony. The goal is to determine the behavior of the colony. Assume the following additional facts. (i) (ii)

(iii) (iv) (v)

The disease is benign to the prey; that is, the prey are “carriers.” The relative birth rate for infected prey remains the same as that for healthy, susceptible prey. The disease is debilitating and ultimately fatal for predators. Once a predator is infected, it can basically be assumed to be deceased. Therefore, we need only consider one population of predators—those that are susceptible. The disease is spread among the prey by contact. We assume the rate of infection to be proportional to the product of susceptible and infected populations. The predators make no distinction between susceptible and infected individuals in their consumption of prey. The predators contract the disease only by consumption of prey. The rate of predator infection is proportional to the product of infected prey and susceptible predators.

We obtain a model by dividing the prey population into two subgroups: healthy, susceptible prey and infected prey. Let s(t) and i(t) represent the populations of susceptible and infected prey, respectively, at time t. Let x(t) denote the population of healthy, susceptible predators. The autonomous system that will model the ecosystem is x = −ax + bxs − δxi s = cs − dsx − αsi + βi

(4)

i = ci − dxi + αsi − βi, where the constants a, b, c, d, α, β, and δ are all positive. 1. Explain the modeling role played by each term in the three differential equations. (For example, the term −ax accounts for the fact that, in the absence of prey, the predator population would decrease and exponentially approach extinction.) 2. As usual, we assume that the variables x, s, and i have been scaled so that one unit of population corresponds to a large number of actual individuals. Assume the following

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values for the constants in equation (4): a = 1,

b = 1,

c = 1,

d = 1,

α = 12 ,

β = 1,

δ = 1.

With this, equation (4) becomes x = −x + xs − xi s = s − sx − 12 si + i

i = i − xi +

1 si 2

− i = −xi +

(5) 1 si. 2

Show that autonomous system (5) has just one equilibrium point in the ﬁrst octant of xsi-space, where all three components are strictly positive. What is this equilibrium point? 3. Linearize system (5) about the equilibrium point found in part 2. Let A denote the (3 × 3) constant coefﬁcient matrix of the linearized system. Without performing any further calculations, answer the following questions: (a) Must the matrix A have at least one real eigenvalue? (b) Is it possible for A to have exactly two real eigenvalues? (c) Is the matrix A real and symmetric? Does it possess any special structure to suggest that it must have three real eigenvalues? Now use computer software to determine the three eigenvalues of A. It can be shown (see Coddington and Levinson5 ) that the (isolated) equilibrium point (xe , se , ie ) = (1, 2, 1) of nonlinear system (5) is (i)

asymptotically stable if all three eigenvalues have negative real parts.

(ii) unstable if at least one eigenvalue has a positive real part. Can either of these results be applied in this case to determine the stability properties of the equilibrium point? If so, describe the stability properties of this equilibrium point.

Project 3: Chaos and the Lorenz Equations In the early 1960s, Edward N. Lorenz,6 an MIT meteorologist, formulated and studied a system of three nonlinear differential equations that today bear his name. These equations, arising from a model of ﬂuid convection, were analyzed by Lorenz to gain insight into the feasibility of long-range weather forecasting. His ﬁndings, published in a classic 1963 paper,7 illustrate what is now called deterministic chaos, a phenomenon wherein even a small number of nonlinear differential equations can exhibit behavior that is highly complicated and extremely sensitive to perturbations in the initial conditions. This sensitivity is sometimes called the “butterﬂy effect,” an allusion to the notion that the ﬂapping of a butterﬂy’s wings on one continent can, after a time, inﬂuence the weather on another continent. This project uses a Runge-Kutta method to solve the Lorenz equations numerically and illustrate solution complexity and the “butterﬂy effect.” 5

Earl A. Coddington and Norman Levinson, Theory of Ordinary Differential Equations (Malabar, FL: R. E. Krieger, 1984). 6 Edward N. Lorenz, professor emeritus, Department of Earth, Atmospheric and Planetary Sciences, MIT, received the 1983 Crafoord Prize “for fundamental contributions to the ﬁeld of geophysical hydrodynamics, which in a unique way have contributed to a deeper understanding of the large-scale motions of the atmosphere and the sea.” 7 Edward N. Lorenz, “Deterministic Nonperiodic Flow,” Journal of the Atmospheric Sciences, Vol. 20, March 1963, pp. 130–141. An interesting account of the research activity culminating in these results appears in the book Chaos by James Gleick (Viking Press, 1987).

Projects

469

The Lorenz equations are X = −σ X + σ Y Y = −XZ + rX − Y Z = XY − bZ, where σ, r, and b are positive constants. We will refer to the three dependent variables X(t), Y (t), and Z(t) as “coordinates.” In reality, however, they are variables that characterize the intensity and temperature variations of the ﬂuid convective motion. 1. Write a computer program to solve an initial value problem for the Lorenz equations using the fourth order Runge-Kutta method given in equations (11)–(13) of Section 4.9. (Section 4.9 also provides an example of a MATLAB code for the algorithm.) 2. Use your program to solve the Lorenz equations numerically on the time interval 0 ≤ t ≤ 50. Use a step size of h = 0.01, and use the following parameter values and initial conditions: σ = 10, X(0) = 0,

b = 83 ,

r = 28

Y (0) = 1,

Z(0) = 0.

3. We now introduce a very small perturbation of the initial conditions. Repeat the computations of step 2 using the same parameter values and step size, but with the following (different) initial conditions: X(0) = 0.0005,

Y (0) = 0.9999,

Z(0) = 0.0001.

Let the “unperturbed” numerical solution obtained in step 2 be denoted by X(t), Y (t), Z(t) and the “perturbed” solution of step 3 by Xp (t), Yp (t), Zp (t). 4. Since there are three dependent variables, parametric solution curves are threedimensional space curves. Such curves are said to exist in phase space (in contrast to the two-dimensional phase plane). Instead of plotting such space curves, we will display solution curve complexity by plotting their projections onto each of the three coordinate planes. We will use computer software to create three separate parametric plots, X(t) vs. Y (t),

X(t) vs. Z(t),

Y (t) vs. Z(t),

0 ≤ t ≤ 50.

Illustrate the “butterﬂy effect” as follows: (a) Create three separate graphs, displaying each of the “unperturbed” and corresponding “perturbed” pairs of coordinates on the same graph as a function of time. What happens as time progresses? Is there a time beyond which the corresponding coordinate curves bear virtually no resemblance to each other? (b) To see this chaotic behavior from a different point of view, select one pair of “unperturbed” and “perturbed” coordinates, say X(t) and Xp (t). Now create the parametric plot X(t) vs. Xp (t),

0 ≤ t ≤ 50.

If X(t) and Xp (t) were to remain close in value to each other for all time, what would this parametric plot look like? Does the plot actually obtained look anything like this?

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7

C H A P T E R

Numerical Methods CHAPTER OVERVIEW 7.1

Introduction

7.2

Euler’s Method, Heun’s Method, and the Modiﬁed Euler’s Method

7.3

Taylor Series Methods

7.4

Runge-Kutta Methods

7.1

Introduction A simple numerical method, Euler’s method, was introduced in Section 2.10. In Section 4.9, we extended Euler’s method to linear systems. We also described a fourth order Runge-Kutta method that served as the basis for an improved, more accurate algorithm. In this chapter, we discuss the ideas underlying a systematic development of more accurate algorithms. We begin with the ﬁrst order scalar initial value problem y = f (t, y),

y(t0 ) = y0 ,

t0 ≤ t ≤ t0 + T.

We assume that this problem has a unique solution on the given t-interval. Our goal is to develop algorithms that generate accurate approximations to the solution y(t). A numerical method frequently begins by imposing a partition of the form t0 < t1 < t2 < · · · < tN−1 < tN = t0 + T on the t-interval [t0 , t0 + T]. Often this partition is uniformly spaced—that is, the partition points are deﬁned by tn = t0 + nh,

n = 0, 1, 2, . . . , N,

where

h=

T . N

The partition spacing, h = T/N, is called the step length or the step size. At each partition point, tn , the numerical algorithm generates an approximation, yn , to the exact solution value, y(tn ). A numerical solution of the differential 471

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Numerical Methods

equation consists of the points {(t0 , y0 ), (t1 , y1 ), . . . , (tN , yN )}, where yn ≈ y(tn ),

n = 0, 1, . . . , N.

Note that the initial condition provides us with an exact starting point (t0 , y0 ). A “good” numerical algorithm is one that generates points (tn , yn ) that lie “close” to their exact solution counterparts, (tn , y(tn )) for n = 1, 2, . . . , N. The terms “good” and “close,” while intuitively clear, will be made precise later. Figure 7.1 displays the exact solution of the initial value problem y = y2 , y(0) = 1 on the interval 0 ≤ t ≤ 0.95 and a pair of numerical approximations corresponding to different step lengths h. [The exact solution, y(t) = (1 − t)−1 , does not exist for t ≥ 1.] y 20 16 12 8 4

0.2

0.4

0.6

0.8

1

t

FIGURE 7.1

2

The initial value problem y = y , y(0) = 1 has solution y(t) = (1 − t)−1 , t < 1. The solid curve is the graph of y(t) for 0 ≤ t ≤ 0.95. The points marked by an ◦ represent a numerical solution with step length h = 0.05, and the points marked by an × represent a numerical solution with h = 0.1. The numerical solutions were generated by Euler’s method. As is usually the case, the smaller step length generates approximations that are more accurate.

Numerical Solutions for Systems of Differential Equations Focusing our attention on scalar ﬁrst order initial value problems may seem to be overly restrictive, but that is not the case. The algorithms we develop for ﬁrst order scalar problems extend directly to ﬁrst order systems. And, as you have seen, ﬁrst order systems basically encompass all the differential equations we have considered so far. We concentrate on ﬁrst order scalar problems because they possess the virtues of relative simplicity and ease of visualization. In particular, we can graph and compare the exact solution and the numerical solution. To further simplify the development, we restrict our discussions to uniformly spaced partitions of step size h. The computational aspects of Euler’s method were treated earlier. The algorithm was introduced and applied to ﬁrst order scalar problems in Section 2.10 and extended to ﬁrst order linear systems in Section 4.9. Euler’s method

7.2

Euler’s Method, Heun’s Method, and the Modiﬁed Euler’s Method

473

serves as our starting point in the next section, where we brieﬂy review the method and explore ways of improving it.

7.2

Euler’s Method, Heun’s Method, and the Modiﬁed Euler’s Method Euler’s Method As we saw in Section 2.10, Euler’s method develops a numerical solution of the initial value problem y = f (t, y),

y(t0 ) = y0

(1)

using the algorithm yn+1 = yn + hf (tn , yn ),

n = 0, 1, 2, . . . , N − 1.

(2)

There are several different ways to derive Euler’s method. In Section 2.10, we used a geometric approach based on direction ﬁelds. We now discuss two other ways of looking at Euler’s method. While they are variations on the basic theme, they provide useful insights as we look for ways to improve Euler’s method.

Approximating the Integral Equation Let y(t) denote the exact solution of initial value problem (1). For now, we restrict our attention to the interval tn ≤ t ≤ tn+1 . Assume that we do not know the exact solution, y(t), but that we have already calculated approximations, yk , of y(tk ) for k = 0, 1, . . . , n (see Figure 7.2). Our goal is to ﬁnd the next approximation, yn+1 , of y(tn+1 ). y yn – 2 y0

y1 y2

yn – 1 yn t

t0 t1 t2

tn – 1 tn tn + 1 FIGURE 7.2

Let y(t) denote the exact solution of initial value problem (1). While we do not know y(t), we assume we have calculated approximations, yk , to y(tk ) for k = 0, 1, . . . , n. Our goal is to ﬁnd the next approximation, yn+1 , to y(tn+1 ).

Consider differential equation (1) and its exact solution, y(t). Integrating both sides of equation (1) over the interval [tn , tn+1 ], we obtain tn+1 tn+1 y (s) ds = f (s, y(s)) ds. tn

tn

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By the fundamental theorem of calculus, the left-hand integral is y(tn+1 ) − y(tn ). Therefore, we obtain an equation for y(tn+1 ): tn+1 f (s, y(s)) ds. (3) y(tn+1 ) = y(tn ) + tn

We cannot use equation (3) computationally because we do not know y(s), tn ≤ s ≤ tn+1 . Suppose, however, that the step length h is small enough that f (s, y(s)) is nearly constant over the interval tn ≤ s ≤ tn+1 . In this case, we can approximate the integral by the left Riemann sum, tn+1 f (s, y(s)) ds ≈ hf (tn , y(tn )). (4) tn

Using approximation (4) in equation (3), we obtain y(tn+1 ) ≈ y(tn ) + hf (tn , y(tn )). Replacing y(tn ) by the previously calculated estimate yn , we are led to Euler’s method, y(tn+1 ) ≈ yn + hf (tn , yn ) = yn+1 . In other words, we can view Euler’s method as a left Riemann sum approximation of integral equation (3).

Heun’s Method Looked at in this light, Euler’s method might be improved by asking “Are there better numerical integration schemes than approximation (4)?” The trapezoidal rule is one such numerical integration scheme. Using the trapezoidal rule, we can approximate the integral in (3) by tn+1 h f (tn , y(tn )) + f (tn+1 , y(tn+1 )) . f (s, y(s)) ds ≈ 2 tn Using this integral approximation in (3), we obtain y(tn+1 ) ≈ y(tn ) +

h f (tn , y(tn )) + f (tn+1 , y(tn+1 )) . 2

Replacing y(tn ) by its estimate yn leads to y(tn+1 ) ≈ yn +

h f (tn , yn ) + f (tn+1 , y(tn+1 )) . 2

(5)

At ﬁrst glance, it appears we have made matters worse since the unknown y(tn+1 ) appears on the right-hand side of (5), in the term f (tn+1 , y(tn+1 )). Approximation (5), if used as it stands, leads to an implicit algorithm with a nonlinear equation that has to be solved for y(tn+1 ). Suppose, however, that we use Euler’s method to approximate the unknown value y(tn+1 ) on the right-hand side of (5): y(tn+1 ) ≈ yn +

h f (tn , yn ) + f (tn+1 , yn + hf (tn , yn )) . 2

7.2

Euler’s Method, Heun’s Method, and the Modiﬁed Euler’s Method

475

This yields the explicit iteration h f (tn , yn ) + f (tn+1 , yn + hf (tn , yn )) , n = 0, 1, . . . , N − 1. (6) 2 Algorithm (6) is often called Heun’s method or the improved Euler’s method. yn+1 = yn +

The Modiﬁed Euler’s Method Another simple numerical integration scheme is the modiﬁed Euler’s method, in which the integrand is approximated over the interval tn ≤ t ≤ tn+1 by its midpoint value. If we use the midpoint rule to approximate the integral in (3) and again use Euler’s method to approximate the unknown value y(t) at the midpoint t = tn + h/2, we obtain the algorithm h h n = 0, 1, . . . , N − 1. (7) yn+1 = yn + hf tn + , yn + f (tn , yn ) , 2 2 Algorithm (7) is known as the modiﬁed Euler’s method. [There is no universal agreement on the names of algorithms (6) and (7).1 ] Although algorithms (6) and (7) appear somewhat complicated, they are relatively easy to implement, since computers can readily evaluate compositions of functions. However, you may rightly ask whether algorithm (6) is, as one of its names implies, an improvement on Euler’s method. If so, how do we quantitatively describe this improvement? The same question applies to (7), and we address it in Section 7.3. For now, we content ourselves with an example that compares Euler’s method with algorithms (6) and (7) for a particular initial value problem. E X A M P L E

1

Consider the initial value problem y = y2 ,

y(0) = 1,

0 ≤ t ≤ 0.95.

Using a step length of h = 0.05, compare the results of Euler’s method (2), Heun’s method (6), and the modiﬁed Euler’s method (7). [The exact solution is y(t) = 1/(1 − t), t < 1.] Solution: For this example, t0 = 0, T = 0.95, N = T/h = 19, and f (t, y) = y2 . Table 7.1 on the next page lists the results. Note that algorithms (6) and (7) do, in fact, represent an improvement over Euler’s method. ❖ The relationship between numerical methods for solving differential equations and numerical integration schemes is a reciprocal one. Every numerical integration technique suggests an algorithm for the initial value problem—this is the approach we used in obtaining algorithms (6) and (7) from equation (3). Conversely, an algorithm for the initial value problem gives rise to a corresponding numerical integration scheme. To see why, consider t the initial value problem y = f (t), y(t0 ) = 0. The solution is simply y(t) = t f (s) ds. Therefore, 0 any numerical method used to solve this initial value problem gives rise to an 1

We are using names found in Peter Henrici, Discrete Variable Methods in Ordinary Differential Equations (New York: Wiley, 1962).

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TA B L E 7 . 1 The Results of Example 1 As is usually the case, algorithms (6) and (7) give better approximations to y(tn ) than does Euler’s method. As is also typical, algorithms (6) and (7) have comparable accuracy. [Note: For this particular initial value problem, Heun’s method (6) yields slightly better approximations than the modiﬁed Euler’s method (7). For other examples, (7) may give slightly better approximations than (6).]

tn

Euler’s Method

Heun’s Method

Modiﬁed Euler’s Method

Exact Solution

0.0000 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500

1.0000 1.0500 1.1051 1.1662 1.2342 1.3104 1.3962 1.4937 1.6052 1.7341 1.8844 2.0620 2.2745 2.5332 2.8541 3.2614 3.7932 4.5126 5.5308 7.0603

1.0000 1.0526 1.1109 1.1762 1.2495 1.3326 1.4275 1.5370 1.6645 1.8151 1.9954 2.2153 2.4894 2.8402 3.3049 3.9488 4.8975 6.4264 9.2615 15.9962

1.0000 1.0525 1.1109 1.1761 1.2493 1.3323 1.4271 1.5363 1.6636 1.8137 1.9934 2.2124 2.4850 2.8333 3.2935 3.9289 4.8597 6.3449 9.0471 15.2001

1.0000 1.0526 1.1111 1.1765 1.2500 1.3333 1.4286 1.5385 1.6667 1.8182 2.0000 2.2222 2.5000 2.8571 3.3333 4.0000 5.0000 6.6667 10.0000 20.0000

approximation of the integral. In particular, Euler’s method, Heun’s method, and the modiﬁed Euler’s method, when applied to the initial value problem y = f (t), y(t0 ) = 0, reduce to a left Riemann sum, the trapezoidal rule, and the midpoint rule, respectively.

Approximating the Taylor Series Expansion This subsection presents another derivation of Euler’s method. Since y(t) is the solution of initial value problem (1) and is assumed to exist on the interval t0 ≤ t ≤ t0 + T, we know y(t) is differentiable on that interval. Assume for now not only that the solution is differentiable, but that it can be expanded in a Taylor series at t = tn , where the Taylor series converges in an interval containing [tn , tn + h]. Therefore, we can express y(tn+1 ) as y(tn+1 ) = y(tn + h) = y(tn ) + y (tn )h +

y (tn ) 2 y (tn ) 3 h + h + ···. 2! 3!

(8)

7.2

Euler’s Method, Heun’s Method, and the Modiﬁed Euler’s Method

477

Truncating the series (8) after two terms, we obtain the approximation y(tn+1 ) ≈ y(tn ) + y (tn )h.

(9)

Since y (tn ) = f (tn , y(tn )), we can rewrite approximation (9) as y(tn+1 ) ≈ y(tn ) + f (tn , y(tn ))h. Replacing y(tn ) in this approximation by its estimate yn , we are once more led to Euler’s method: y(tn+1 ) ≈ yn + f (tn , yn )h = yn+1 . Thus, we obtain Euler’s method by truncating the Taylor series (8) after two terms. Viewed in this light, Euler’s method might be improved by retaining more terms of the Taylor series—truncating after three, four, or more terms. We investigate this possibility in Section 7.3.

EXERCISES Most exercises in this chapter require a computer or programmable calculator.

Exercises 1–5: In each exercise, (a) Solve the initial value problem analytically, using an appropriate solution technique. (b) For the given initial value problem, write the Heun’s method algorithm, yn+1 = yn +

h [ f (tn , yn ) + f (tn+1 , yn + hf (tn , yn ))]. 2

(c) For the given initial value problem, write the modiﬁed Euler’s method algorithm, h h yn+1 = yn + hf tn + , yn + f (tn , yn ) . 2 2 (d) Use a step size h = 0.1. Compute the ﬁrst three approximations, y1 , y2 , y3 , using the method in part (b). (e) Use a step size h = 0.1. Compute the ﬁrst three approximations, y1 , y2 , y3 , using the method in part (c). (f ) For comparison, calculate and list the exact solution values, y(t1 ), y(t2 ), y(t3 ). 1. y = 2t − 1,

4. y = −y + t,

y(1) = 0 y(0) = 0

2. y = −y, 2

y(0) = 1

5. y y + t = 0,

3. y = −ty,

y(0) = 1

y(0) = 1

Exercises 6–9: In each exercise, (a) Find the exact solution of the given initial value problem. (b) As in Example 1, use a step size of h = 0.05 for the given initial value problem. Compute 20 steps of Euler’s method, Heun’s method, and the modiﬁed Euler’s method. Compare the numerical values obtained at t = 1 by calculating the error |y(1) − y20 |. t 6. y = 1 + y2 , y(0) = −1 7. y = − , y(0) = 3 y 8. y + 2y = 4, y(0) = 3

9. y + 2ty = 0, y(0) = 2

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Exercises 10–14: In each exercise, the given iteration is the result of applying Euler’s method, Heun’s method, or the modiﬁed Euler’s method to an initial value problem of the form y = f (t, y),

y(t0 ) = y0 ,

t0 ≤ t ≤ t0 + T.

Identify the numerical method, and determine t0 , T, and f (t, y). 10.

11.

12.

13.

14.

yn+1 = yn + h( yn + tn2 y3n ), y0 = 1 tn = 2 + nh, h = 0.02, n = 0, 1, 2, . . . , 49 h [t y2 + 2 + (tn + h)( yn + h(tn y2n + 1))2 ], y0 = 2 2 n n tn = 1 + nh, h = 0.05, n = 0, 1, 2, . . . , 99 h h 2 2 sin yn + tn sin ( yn ) , y0 = 1 yn+1 = yn + h tn + 2 2 yn+1 = yn +

tn = nh, h = 0.01, n = 0, 1, 2, . . . , 199 h , y0 = −1 yn+1 = yn 1 + 2 tn + y2n tn = 2 + nh, h = 0.01, n = 0, 1, 2, . . . , 99 h h yn+1 = yn + h sin tn + + yn + sin(tn + yn ) , y0 = 1 2 2 tn = −1 + nh, h = 0.05, n = 0, 1, 2, . . . , 199

15. Let h be a ﬁxed positive step size, and let λ be a nonzero constant. Suppose we apply Heun’s method or the modiﬁed Euler’s method to the initial value problem y = λy, y(t0 ) = y0 , using this step size h. Show, in either case, that k

(hλ)2 (hλ)2 yk−1 and hence yk = 1 + hλ + y0 , k = 1, 2, . . . . yk = 1 + hλ + 2! 2!

Exercises 16–17: Assume a tank having a capacity of 200 gal initially contains 90 gal of fresh water. At time t = 0, a salt solution begins to ﬂow into the tank at a rate of 6 gal/min and the wellstirred mixture ﬂows out at a rate of 1 gal/min. Assume that the inﬂow concentration ﬂuctuates in time, with the inﬂow concentration given by c(t) = 2 − cos(πt) lb/gal, where t is in minutes. Formulate the appropriate initial value problem for Q(t), the amount of salt (in pounds) in the tank at time t. Our objective is to approximately determine the amount of salt in the tank when the tank contains 100 gal of liquid. 16. (a) Formulate the initial value problem. (b) Obtain a numerical solution, using Heun’s method with a step size h = 0.05. (c) What is your estimate of Q(t) when the tank contains 100 gal? 17. (a) Formulate the initial value problem. (b) Obtain a numerical solution, using the modiﬁed Euler’s method with a step size h = 0.05. (c) What is your estimate of Q(t) when the tank contains 100 gal?

Exercises 18–19: Let P(t) denote the population of a certain colony, measured in millions of members. Assume that P(t) is the solution of the initial value problem P P = 0.1 1 − P + M(t), P(0) = P0 , 3

7.3

Taylor Series Methods

479

where time t is measured in years. Let M(t) = e−t . Therefore, the colony experiences a migration inﬂux that is initially strong but soon tapers off. Let P0 = 12 ; that is, the colony had 500,000 members at time t = 0. Our objective is to estimate the colony size after two years. 18. Obtain a numerical solution of this problem, using Heun’s method with a step size h = 0.05. What is your estimate of colony size at the end of two years? 19. Obtain a numerical solution of this problem, using the modiﬁed Euler’s method with a step size h = 0.05. What is your estimate of colony size at the end of two years? 20. Error Estimation In most applications of numerical methods, as in Exercises 16– 19, an exact solution is unavailable to use as a benchmark. Therefore, it is natural to ask, “How accurate is our numerical solution?” For example, how accurate are the solutions obtained in Exercises 16–19 using the step size h = 0.05? This exercise provides some insight. Suppose we apply Heun’s method or the modiﬁed Euler’s method to the initial value problem y = f (t, y), y(t0 ) = y0 and we use a step size h. It can be shown, for most initial value problems and for h sufﬁciently small, that the error at a ﬁxed point t = t∗ is proportional to h2 . That is, let n be a positive integer, let h = (t∗ − t0 )/n, and let yn denote the method’s approximation to y(t∗ ) using step size h. Then lim

h→0 t ∗ ﬁxed

y(t∗ ) − yn = C, h2

C = 0.

As a consequence of this limit, reducing a sufﬁciently small step size by 12 will reduce the error by approximately 14 . In particular, let yˆ 2n denote the method’s approximation to y(t∗ ) using step size h/2. Then, for most initial value problems, we expect that y(t∗ ) − yˆ 2n ≈ [ y(t∗ ) − yn ]/4. Rework Example 1, using Heun’s method and step sizes of h = 0.05, h = 0.025, and h = 0.0125. (a) Compare the three numerical solutions at t = 0.05, 0.10, 0.15, . . . , 0.95. Are the errors reduced by about 14 when the step size is reduced by 12 ? (Since the solution becomes unbounded as t approaches 1 from the left, the expected error reduction may not materialize near t = 1.) (b) Suppose the exact solution is not available. How can the Heun’s method solutions obtained using different step sizes be used to estimate the error? [Hint: Assuming that y(t∗ ) − yˆ 2n ≈

[ y(t∗ ) − yn ] , 4

derive an expression for y(t∗ ) − yˆ 2n that involves only yˆ 2n and yn .] (c) Test the error monitor derived in part (b) on the initial value problem in Example 1.

7.3

Taylor Series Methods In Section 7.2, we saw that we could obtain Euler’s method by truncating the Taylor series for the solution y(t) after the ﬁrst two terms of the expansion. We therefore anticipate that Euler’s method can be improved by retaining more terms of the Taylor series. In this section, we describe how such an improvement of Euler’s method is carried out. In addition, we use the Taylor series expansion as a basis for

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quantifying the accuracy of numerical algorithms. We begin with some preliminaries:

• First, we state Theorem 7.1. This theorem gives conditions guaranteeing that the solution of an initial value problem has a convergent Taylor series expansion. • We then present Theorem 7.2, Taylor’s theorem. This theorem from calculus enables us to measure the error that arises when we truncate a Taylor series. Once these preliminary results are in place, we can use Taylor series as a basis for systematically developing algorithms of increasing accuracy. These Taylor series algorithms can, in principle, be made as accurate as we wish. They are not, however, computationally friendly. We combine accuracy with ease of implementation in Section 7.4, when we discuss Runge-Kutta methods.

Preliminaries We begin with two deﬁnitions and then present a theorem guaranteeing that the solution of initial value problem (1), y = f (t, y),

y(t0 ) = y0 ,

(1)

can be expanded in a Taylor series that converges in a neighborhood of the point t0 . A function y(t), deﬁned on an open interval containing the point t, is said to be analytic at t = t if it has a Taylor series expansion y(t) =

∞

an (t − t )n

(2a)

n=0

that converges in an interval t − δ < t < t + δ, where δ > 0. It is shown in calculus that if y(t) is analytic at t = t, then y(t) has derivatives of all orders in the interval (t − δ, t + δ). Moreover, the coefﬁcients of the Taylor series are given by an =

y(n) (t ) , n!

n = 0, 1, 2, . . . .

(2b)

In general, a function y(t) is said to be analytic in the interval a < t < b if it is analytic at every point t in this interval. Consider the function f (t, y) appearing on the right-hand side of differential equation (1). In the context of differential equation (1), f (t, y) is understood to represent f (t, y(t)), where y(t) is the unknown solution of interest. In the next deﬁnition, however, we view f as a function of two independent variables, t and y. Let f (t, y) be a function deﬁned in an open region R of the ty-plane containing the point (t, y ). The function f (t, y) is said to be analytic at (t, y ) if it has a two-variable Taylor series expansion f (t, y) =

∞ ∞ m=0 n=0

bmn (t − t )m ( y − y )n

(3a)

7.3

Taylor Series Methods

481

that converges in a neighborhood Nρ of (t, y ),

Nρ = (t, y) : (t − t )2 + ( y − y )2 < ρ . We say that f (t, y) is analytic in a region R if it is analytic at every point (t, y ) in R. The coefﬁcients bmn can be evaluated in terms of the function f and its partial derivatives, evaluated at (t, y ); the two-variable Taylor series expansion has the form f (t, y) = f (t, y ) + ft (t, y )(t − t ) + fy (t, y )( y − y ) + 12 ftt (t, y )(t − t )2 + 2fty (t, y )(t − t )( y − y ) + fyy (t, y )( y − y )2 + · · · . (3b)

The Existence of Analytic Solutions It is natural to ask whether analyticity of f (t, y) guarantees analyticity of the solution of initial value problem (1). An afﬁrmative answer is contained in Theorem 7.1, which can be regarded as a reﬁnement of Theorem 2.2. A proof of Theorem 7.1 can be found in Birkhoff and Rota.2

Theorem 7.1

Let R denote the rectangle deﬁned by a < t < b, α < y < β. Let f (t, y) be a function deﬁned and analytic in R, and suppose that (t0 , y0 ) is a point in R. Then there is a t-interval (c, d) containing t0 in which there exists a unique analytic solution of the initial value problem y = f (t, y),

y(t0 ) = y0 .

E X A M P L E

1

Consider the initial value problem y = y2 + t 2 , 2

y(t0 ) = y0 .

2

Here, the function f (t, y) = y + t is a polynomial in the variables t and y and is therefore analytic in the entire ty-plane. Hence, the region R can be assumed to be the entire ty-plane. Theorem 7.1 guarantees the existence of a unique analytic solution y(t) in an interval of the form t0 − δ < t < t0 + δ for some δ > 0. Note that the theorem does not tell us the value of δ, only that such a positive δ exists. Since the solution is analytic in t0 − δ < t < t0 + δ, we know y(t) has the form ∞ y(n) (t0 ) y(t) = (t − t0 )n , t0 − δ < t < t0 + δ. ❖ n! n=0 We assume throughout this chapter that the hypotheses of Theorem 7.1 are satisﬁed. This theorem assures us that an analytic solution y(t) exists on some interval of the form t0 − δ < t < t0 + δ. As noted in Example 1, however, 2

Garrett Birkhoff and Gian-Carlo Rota, Ordinary Differential Equations, 4th ed. (New York: Wiley, 1989).

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Theorem 7.1 does not tell us the size of δ. Since we are interested in generating a numerical solution on an interval of the form t0 ≤ t ≤ t0 + T, we shall also assume that the interval of interest, [t0 , t0 + T], lies within the existence interval, (t0 − δ, t0 + δ). Given this assumption, we can expand solution y(t) in a Taylor series about any point t lying in the interval of interest. It is important to realize, however, that in practical computations involving nonlinear differential equations there is no a priori guarantee that the solution exists on a designated interval of interest, [t0 , t0 + T].

Using the Differential Equation to Compute the Taylor Series Coefﬁcients When Euler’s method was discussed in Section 2.10, we based the development on the fact that the differential equation determines the direction ﬁeld. In particular, if we evaluate f at a point (t, y ) in the ty-plane, then the value f (t, y ) tells us the slope of the solution curve passing through (t, y ). We now show that the differential equation determines much more. In particular, suppose that a solution curve y(t) passes through the point (t, y ). We will see that f (t, y) and its partial derivatives, evaluated at (t, y ), can be used to calculate all the derivatives of y(t). In turn [see equations (2a) and (2b)], these derivative evaluations completely determine the Taylor series expansion of the solution y(t). In particular, we know the identity y (t) = f (t, y(t)) holds for t in a neighborhood of t. Therefore, y (t ) = f (t, y(t )) = f (t, y ). We ﬁnd higher derivatives by differentiating the identity y (t) = f (t, y(t)). For example, y (t) =

d d y (t) = f (t, y(t)). dt dt

(4a)

We use the chain rule to calculate the derivative in equation (4a), ∂f ∂f dy ∂f ∂f df = + = + f = ft + fy f . dt ∂t ∂y dt ∂t ∂y

(4b)

Once the partial derivatives in equation (4b) are computed, we substitute the function y(t) for the second independent variable y, obtaining y (t) = ft (t, y(t)) + fy (t, y(t))f (t, y(t)). Using the fact that y(t ) = y, we have y (t ) = ft (t, y ) + fy (t, y )f (t, y ).

(5)

Equation (5) determines the concavity of the solution curve at the point (t, y ), just as y (t ) = f (t, y ) determines the slope of the solution curve at (t, y ). This differentiation process can be continued to compute higher derivatives of the solution at (t, y ). To simplify the notation, we continue to use subscripts to denote partial derivatives and do not explicitly indicate their ultimate eval-

7.3

Taylor Series Methods

483

uation at (t, y ). Thus, y = ft + fy f (6) d ft + fy f = ftt + fty f + ( fyt + fyy f )f + fy ( ft + fy f ) . y = dt It is possible, in principle, to continue this differentiation process and compute as many derivatives of y(t) at t = t as desired. It is clear from (6), however, that the computations can quickly become cumbersome. The next example illustrates, however, that when the differential equation has a simple structure, it may be relatively easy to calculate higher derivatives. E X A M P L E

2

Consider the initial value problem y = y2 ,

y(0) = 1.

Evaluate the derivatives y (0), y (0), y (0), and y(4) (0). Solution: In this case, f (t, y) = y2 is a polynomial in y. Therefore, Theorem 7.1 applies, and we know the solution y(t) is an analytic function of t in the open interval (−δ, δ) for some δ > 0. Since y (t) = y2 (t), the chain rule yields y (t) = [ y2 (t)] = 2y(t)y (t) = 2y3 (t) y (t) = [2y3 (t)] = 6y2 (t)y (t) = 6y4 (t) y(4) (t) = [6y4 (t)] = 24y3 (t)y (t) = 24y5 (t). Therefore, y(0) = 1, y (0) = 1, y (0) = 2, y (0) = 6 = 3!, and y(4) (0) = 24 = 4!. Given these derivative values, the ﬁrst few terms in the Taylor series expansion of y(t) are y(t) = y(0) + y (0)t +

y (0) 2 y (0) 3 y(4) (0) 4 t + t + t + ··· 2! 3! 4!

= 1 + t + t2 + t3 + t4 + · · · . We recognize this expansion as a geometric series that converges to the exact solution, y(t) =

1 , 1−t

−1 < t < 1.

For this initial value problem, we ﬁnd (after the fact) that δ = 1. ❖

Taylor Series Methods The preceding discussion shows how to calculate higher derivatives of the solution y(t) of the initial value problem y = f (t, y),

y(t0 ) = y0 .

We can use these ideas to improve Euler’s method. Let yn be an approximation to y(tn ), where tn and tn+1 = tn + h are in the interval t0 ≤ t ≤ tN . As in equation (8) of Section 7.2, we have y(tn+1 ) = y(tn ) + y (tn )h +

y (tn ) 2 y (tn ) 3 h + h + ···. 2! 3!

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Truncating this expansion after p terms, we obtain the approximation y(tn+1 ) ≈ y(tn ) + y (tn )h +

y (tn ) 2 y (tn ) 3 y( p) (tn ) p h + h + ··· + h . 2! 3! p!

(7)

As we saw in equation (6), the Taylor series coefﬁcients, y (tn ), y (tn ), y (tn ), . . . , can be expressed in terms of f and its partial derivatives evaluated at (tn , y(tn )). For instance, with p = 1, (7) becomes y(tn+1 ) ≈ y(tn ) + f (tn , y(tn ))h. Similarly, for p = 2, we obtain from (7) h2 y(tn+1 ) ≈ y(tn ) + f (tn , y(tn ))h + ft (tn , y(tn )) + fy (tn , y(tn ))f (tn , y(tn )) . 2! We ﬁnd similar approximations when p ≥ 3. In order to use these approximations for computations, we replace y(tn ) by its estimate, yn . The algorithms we obtain in this manner are collectively referred to as Taylor series methods. We use the term Taylor series method of order p to identify the Taylor series method obtained from approximation (7). The Taylor series methods of orders 1, 2, and 3 are as follows: Taylor Series Method of Order 1 (Euler’s Method) yn+1 = yn + hf (tn , yn ),

n = 0, 1, . . . , N − 1

(8a)

Taylor Series Method of Order 2 yn+1 = yn + hf (tn , yn ) +

h2 ft (tn , yn ) + fy (tn , yn )f (tn , yn ) , 2!

n = 0, 1, . . . , N − 1 (8b)

Taylor Series Method of Order 3 yn+1 = yn + hf (tn , yn ) + +

h2 ft (tn , yn ) + fy (tn , yn )f (tn , yn ) 2!

h3 f (t , y ) + 2fty (tn , yn )f (tn , yn ) + fyy (tn , yn )f 2 (tn , yn ) 3! tt n n + fy (tn , yn )ft (tn , yn ) + fy2 (tn , yn )f (tn , yn ) , n = 0, 1, . . . , N − 1. (8c)

It is cumbersome to write out all the terms of the general pth order Taylor series method. In order to simplify the notation when discussing Taylor series methods, it is common to denote a pth order Taylor series method as yn+1 = yn + hyn +

h2 h3 hp yn + yn + · · · + y(np) , 2! 3! p!

n = 0, 1, . . . , N − 1.

(9)

We are using the name “pth order Taylor series method” to denote method (9). The term order has a precise meaning that is given later in this section. Once we state the formal deﬁnition of order, however, we will see that method (9) is properly named.

7.3

Taylor Series Methods

485

E X A M P L E

3

Consider the initial value problem y = y2 ,

y(0) = 1.

Using h = 0.05, execute 19 steps of the Taylor series method of order p for p = 1, 2, 3, and 4. Do the results improve as p increases? Solution: As we saw in Example 2, y = 2y3 , y = 6y4 , and y(4) = 24y5 . The Taylor series methods of orders 1, 2, 3, and 4 are, respectively, yn+1 = yn + hy2n , yn+1 = yn + hy2n + h2 y3n yn+1 = yn + hy2n + h2 y3n + h3 y4n yn+1 = yn + hy2n + h2 y3n + h3 y4n + h4 y5n . Table 7.2 illustrates how the Taylor series method estimates improve as the order increases. TA B L E 7 . 2 In this table, we designate the results of the pth order Taylor series method as “order p” for p = 1, 2, 3, 4 and the value of the exact solution at t = tn as y(tn ). As anticipated, the results improve when we retain more terms in the Taylor series expansion—that is, as the order p increases. Taylor Series Methods tn

Order 1

Order 2

Order 3

Order 4

y(tn )

0.0000 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500

1.0000 1.0500 1.1051 1.1662 1.2342 1.3104 1.3962 1.4937 1.6052 1.7341 1.8844 2.0620 2.2745 2.5332 2.8541 3.2614 3.7932 4.5126 5.5308 7.0603

1.0000 1.0525 1.1108 1.1759 1.2491 1.3320 1.4266 1.5357 1.6626 1.8123 1.9914 2.2095 2.4805 2.8264 3.2822 3.9093 4.8227 6.2661 8.8443 14.4850

1.0000 1.0526 1.1111 1.1764 1.2500 1.3333 1.4285 1.5383 1.6664 1.8178 1.9994 2.2212 2.4984 2.8543 3.3281 3.9894 4.9756 6.5980 9.7297 17.8859

1.0000 1.0526 1.1111 1.1765 1.2500 1.3333 1.4286 1.5385 1.6666 1.8182 2.0000 2.2221 2.4999 2.8569 3.3328 3.9987 4.9963 6.6536 9.9301 19.1273

1.0000 1.0526 1.1111 1.1765 1.2500 1.3333 1.4286 1.5385 1.6667 1.8182 2.0000 2.2222 2.5000 2.8571 3.3333 4.0000 5.0000 6.6667 10.0000 20.0000

❖

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Example 3 illustrates (for the special case of the differential equation y = y2 ) how the Taylor series method of order p becomes more accurate as p increases. We are now ready to make the concept of order precise and to discuss why we expect that higher order methods are usually more accurate than lower order methods.

Taylor’s Theorem We consider the error made when we truncate a Taylor series. Theorem 7.2, known as Taylor’s theorem, gives a convenient way of estimating the resulting truncation error. A proof of Taylor’s theorem can be found in most calculus books.

Theorem 7.2

Let y(t) be analytic at t = t, where the Taylor series expansion (2) converges in the interval t − δ < t < t + δ. Let m be a positive integer, and let t be in the interval (t − δ, t + δ). Then y(t) = y(t ) + y (t )(t − t ) +

y (t ) (t − t )2 + · · · 2!

y(m) (t ) y(m+1) (ξ ) (t − t )m + (t − t )m+1 , + m! (m + 1)!

(10)

where ξ is some point lying between t and t.

In Theorem 7.2, the polynomial Pm (t) = y(t ) + y (t )(t − t ) +

y (t ) y(m) (t ) (t − t )2 + · · · + (t − t )m 2! m!

is referred to as the Taylor polynomial of degree m. The term Rm (t) =

y(m+1) (ξ ) (t − t )m+1 (m + 1)!

is the remainder, and it measures the error made in approximating y(t) by the Taylor polynomial, Pm (t). When we consider the errors of a numerical method, the role of t is typically played by tn and the generic point t lies in the interval tn ≤ t ≤ tn+1 .

One-Step Methods and the Local Truncation Error The methods we have considered thus far (Euler’s method, Heun’s method, the modiﬁed Euler’s method, and Taylor series methods) are classiﬁed as one-step methods. In general, a one-step method has the form yn+1 = yn + hφ(tn , yn ; h),

n = 0, 1, 2, . . . , N − 1.

(11)

7.3

Taylor Series Methods

487

These methods are called one step because they use only the most recently computed point, (tn , yn ), to compute the next point, (tn+1 , yn+1 ). [By contrast, a multistep method uses multiple back values, (tn , yn ), (tn−1 , yn−1 ), (tn−2 , yn−2 ), . . . , (tn−k , yn−k ), to compute (tn+1 , yn+1 ).3 We restrict our consideration to one-step methods.] In equation (11), the term φ(tn , yn ; h) is called an increment function. Different increment functions deﬁne different one-step methods. For instance, Euler’s method, yn+1 = yn + hf (tn , yn ), is a one-step method with increment function φ(tn , yn ; h) = f (tn , yn ). Heun’s method is a one-step method with increment function φ(tn , yn ; h) = 12 f (tn , yn ) + f (tn + h, yn + hf (tn , yn )) .

E X A M P L E

4

Write the second order Taylor series method in the form of a one-step method, and identify the increment function φ(tn , yn ; h). Solution: form

From equation (8b), the second order Taylor series method has the

yn+1

h = yn + h f (tn , yn ) + f (t , y ) + fy (tn , yn )f (tn , yn ) . 2! t n n

Thus, φ(tn , yn ; h) = f (tn , yn ) +

h ft (tn , yn ) + fy (tn , yn )f (tn , yn ) . ❖ 2!

A quantity known as the local truncation error is one of the keys to understanding and assessing the accuracy of one-step methods. Let y(t) denote the solution of the initial value problem y = f (t, y), y(t0 ) = y0 , and assume y(t) exists on the interval of interest, [t0 , t0 + T]. Let tn and tn+1 = tn + h lie in the interval [t0 , t0 + T]. For a given one-step method (11), we deﬁne the quantity Tn+1 by y(tn+1 ) = y(tn ) + hφ(tn , y(tn ); h) + Tn+1 .

(12)

The quantities Tn+1 , n = 0, 1, . . . , N − 1 are called local truncation errors. A local truncation error 4 measures how much a single step of the numerical method misses the true solution value, y(tn+1 ), given that the numerical method starts on the solution curve at the point (tn , y(tn )). 3

John D. Lambert, Numerical Methods for Ordinary Differential Systems (Chichester, England: Wiley, 1991). 4 There is no universal agreement about the deﬁnition of local truncation errors. Some texts express the quantity Tn+1 in equation (12) as hτn+1 = Tn+1 and refer to τn+1 as a local truncation error. However, no matter how local truncation errors are deﬁned, there is universal agreement on the deﬁnition of the “order” of a one-step method, as given in the next subsection in equation (15).

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E X A M P L E

5

Derive an expression for the local truncation errors of Euler’s method. Solution: Since Euler’s method is given by yn+1 = yn + hf (tn , yn ), the local truncation errors are deﬁned by y(tn+1 ) = y(tn ) + hf (tn , y(tn )) + Tn+1 ,

(13)

where y(t) is the unique solution of the initial value problem y = f (t, y), y(t0 ) = y0 . However, f (tn , y(tn )) = y (tn ), and so equation (13) can be expressed as y(tn+1 ) = y(tn ) + y (tn )h + Tn+1 .

(14a)

By Taylor’s theorem, we can also write y(tn+1 ) = y(tn ) + y (tn )h +

y (ξ ) 2 h , 2!

(14b)

where tn < ξ < tn+1 . Comparing (14a) and (14b), we see that Tn+1 =

y (ξ ) 2 h , 2!

where ξ is some point in the t-interval tn < t < tn+1 . For later use, we note from (14c) that max Tn+1 ≤ Kh2 , 0≤n≤N−1

where K = maxt

0

≤t≤t0 +T

(14c)

(14d)

|y (t)|/2!. ❖

The Order of a Numerical Method We now deﬁne the order of a numerical method and show that the terminology “Taylor series method of order p” is appropriate. We say that a one-step method has order p if there are positive constants K and h0 such that For any point tn in the interval [t0 , t0 + T − h0 ] and any step size h satisfying 0 < h ≤ h0 , we have |Tn+1 | ≤ Khp+1 .

(15)

Note that, in inequality (15), the constant K does not depend on the index n. From inequality (14d) of Example 5, we see that Euler’s method has order p = 1. Similar arguments show that the Taylor series methods (8b) and (8c) have orders 2 and 3, respectively. In general, the Taylor series method yn+1 = yn + hyn +

h2 h3 hp yn + yn + · · · + y(np) 2! 3! p!

has order p; this is consistent with our prior use of the term. The order of a numerical method is a measure of how well the method replicates the Taylor expansion of the solution. A numerical method of order p has local truncation errors that satisfy |Tn+1 | ≤ Khp+1 . From Taylor’s theorem, therefore, it follows that a pth order one-step method correctly replicates the Taylor series up to and including the term of order hp .

7.3

Taylor Series Methods

489

The Global Error The size of the local truncation error for a numerical method tells us how far we would deviate from y(tn+1 ) if we were to take a single step of the method starting on the solution curve at the point (tn , y(tn )). However, except for the ﬁrst step of the method [when we start at the initial point (t0 , y(t0 )) = (t0 , y0 )], we do not expect to take steps that begin on the solution curve. In this sense, the local truncation error is not a quantity that we can calculate without knowing the true solution of the initial value problem. We are using the concept of the local truncation error to deﬁne the order of a numerical method and to establish the convergence of numerical methods. In practical computations, we are primarily interested in the global errors, y(tn ) − yn

for

n = 0, 1, . . . , N,

(16)

where y(t) is the true solution of the initial value problem and yn is the numerical method’s estimate to y(tn ). In discussing local truncation errors and global errors, it is convenient to use the “Big O” order symbol (also known as the Landau symbol). This symbol is frequently used to characterize inequalities such as (15). We use the notation q(h) = O(hr ),

h→0

or simply

q(h) = O(hr )

to mean there exists some positive constant K such that |q(h)| ≤ Khr for all positive, sufﬁciently small h. Thus, inequality (15) can be written as Tn+1 = O(hp+1 ). Note that the order of a numerical method, p, is one integer less than the order of the local truncation error. For example, from equation (14d), the local truncation error of Euler’s method is O(h2 ), and therefore we say that Euler’s method is a ﬁrst order method. In an appendix to Section 7.4, we state a theorem that shows how (for the types of problems and numerical methods we are considering) the order of the numerical method and the size of the global errors are related. In particular, there is a positive constant M such that the global errors for a pth order method satisfy the inequality max |y(tn ) − yn | ≤ Mhp .

0≤n≤N

(17)

Inequality (17) tells us how the global errors are reduced when h is reduced. If we are using a pth order method and if we reduce the step size h by 12 , then we p anticipate that the global errors will be reduced by about 12 . E X A M P L E

6

We again consider the example y = y2 ,

y(0) = 1,

0 ≤ t ≤ 0.95.

Use Euler’s method to generate numerical solutions, ﬁrst using step size h1 = 0.05 and then using step size h2 = 0.025. From (17) with p = 1, we expect the global errors to be reduced by approximately 12 when h is reduced by 12 . Compare the global errors at t = 0.05, 0.10, 0.15, . . . , 0.95. Does it appear (continued)

490

CHAPTER 7

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that the errors resulting from the smaller step size are about half the size of the errors of the larger step? Solution: The results are listed in Table 7.3. The column headed E1 gives the global errors y(tk ) − yk , made using h1 = 0.05. Similarly, the column headed E2 lists the global errors, at the same values of t, made using h2 = 0.025. As predicted by (17), the ratios of E2 to E1 (given in the column headed E2 /E1 ) are close to 0.5 for smaller values of t. The ratios tend to deviate from 0.5 as the values tk approach t = 1, where the exact solution has a vertical asymptote. TA B L E 7 . 3 The Results of Example 6 Note, as predicted by (17), that E2 ≈ E1 /2. tk

E1 (h = 0.05)

E2 (h = 0.025)

E2 / E1

0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500

0.0026 0.0060 0.0103 0.0158 0.0230 0.0324 0.0448 0.0614 0.0841 0.1156 0.1603 0.2255 0.3239 0.4793 0.7386 1.2068 2.1541 4.4692 12.9397

0.0014 0.0031 0.0054 0.0083 0.0121 0.0171 0.0238 0.0329 0.0454 0.0630 0.0883 0.1259 0.1840 0.2782 0.4412 0.7491 1.4111 3.1700 10.4052

0.5191 0.5206 0.5223 0.5242 0.5264 0.5290 0.5320 0.5355 0.5396 0.5446 0.5507 0.5583 0.5680 0.5805 0.5973 0.6207 0.6551 0.7093 0.8041

❖

The Need for Computationally Friendly Algorithms Taylor series expansions provide a clear blueprint for how to improve the accuracy of a numerical algorithm. The Exercises develop such algorithms for a variety of problems. In speciﬁc cases, as in Examples 1 and 2, the computations are not overly difﬁcult. In other cases, as the order of the algorithm increases, the computations rapidly become unwieldy and the possibility of mistakes in programming the numerical method grows as well. Moreover, Taylor series

7.3

Taylor Series Methods

491

methods are problem speciﬁc; the various partial derivatives of f (t, y) must be recomputed every time we are given a new differential equation. For these reasons, a Taylor series method is not very attractive as an allpurpose method for solving initial value problems. The challenge is to develop algorithms that replicate the desired number of terms in the Taylor series expansion (thereby achieving the desired accuracy) but do not require calculation of partial derivatives. In particular, we want algorithms that require only evaluations of the function f . Heun’s method and the modiﬁed Euler’s method, developed in Section 7.2, provide insight into how these goals might be achieved using compositions of functions. Computers can evaluate functions with relative ease, and compositions of functions, while they might look formidable to us, are also evaluated with relative ease on a computer. Nested compositions of functions, such as those used in Heun’s method and the modiﬁed Euler’s method, form the basis of Runge-Kutta methods that are discussed in Section 7.4. Runge-Kutta methods achieve the accuracy of Taylor series methods, but in a computationally friendly way.

EXERCISES Exercises 1–10: Assume, for the given differential equation, that y(0) = 1. (a) Use the differential equation itself to determine the values y (0), y (0), y (0), y(4) (0) and form the Taylor polynomial P4 (t) = y(0) + y (0)t +

y (0) 2 y (0) 3 y(4) (0) 4 t + t + t . 2! 3! 4!

(b) Verify that the given function is the solution of the initial value problem consisting of the differential equation and initial condition y(0) = 1. (c) Evaluate both the exact solution y(t) and P4 (t) at t = 0.1. What is the error E(0.1) = y(0.1) − P4 (0.1)? [Note that E(0.1) is the local truncation error incurred in using a Taylor series method of order 4 to step from t0 = 0 to t1 = 0.1 using step size h = 0.1.] 1. y = −y + 2; y(t) = 2 − e−t −1

t2 2 3. y = ty ; y(t) = 1 − 2 t 2 5. y = y1/2 ; y(t) = 1 + 2

2. y = 2ty; y(t) = et

2

4. y = t2 + y; y(t) = 3e t − (t2 + 2t + 2) 6. y = ty−1 ; y(t) =

1 + t2

3e t − cos t − sin t 7. y = y + sin t; y(t) = 2 t 4 π 3/4 8. y = y ; y(t) = 1 + 9. y = 1 + y2 ; y(t) = tan t + 4 4 10. y = −4t3 y; y(t) = e−t

4

Results analogous to Theorem 7.1 guaranteeing the existence of analytic solutions can be established for higher order scalar problems and ﬁrst order systems. The development

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of higher order numerical methods for such problems will be addressed in Section 7.4. Exercises 11–14 illustrate how a series expansion of the solution of a higher order scalar problem can be obtained from the differential equation itself. For example, consider the initial value problem y = f (t, y, y ), y(t0 ) = y0 , y (t0 ) = y0 . From the equation, we have y (t0 ) = f (t0 , y0 , y0 ). Differentiating the identity y (t) = f (t, y(t), y (t)) allows us to obtain y (t0 ) and then y(4) (t0 ) and so forth.

Exercises 11–14: In each exercise, for the given t0 , (a) Obtain the ﬁfth degree Taylor polynomial approximation of the solution, P5 (t) = y(t0 ) + y (t0 )(t − t0 ) +

y (t0 ) y(5) (t0 ) (t − t0 )2 + · · · + (t − t0 )5 . 2! 5!

(b) If the exact solution is given, calculate the error at t = t0 + 0.1. 11. y − 3y + 2y = 0, y(0) = 1, y (0) = 0; t0 = 0. The exact solution is y(t) = 2e t − e2t . 12. y − y = 0, y(1) = 1, y (1) = 2; t0 = 1. The exact solution is y(t) = −1 + 2e(t−1) . 13. y − y = 0, y(0) = 1, y (0) = 2, y (0) = 0; t0 = 0. The exact solution is y(t) = 1 + e t − e−t . 14. y + y + y3 = 0, y(0) = 1, y (0) = 0; t0 = 0

Exercises 15–18: In each exercise, determine the largest positive integer r such that q(h) = O(hr ). [Hint: Determine the ﬁrst nonvanishing term in the Maclaurin expansion of q.] 15. q(h) = sin 2h

16. q(h) = 2h + h3

17. q(h) = 1 − cos h

18. q(h) = eh − (1 + h)

19. Give an example of functions f and g such that f (h) = O(h), g(h) = O(h) but ( f + g)(h) = O(h2 ).

Exercises 20–23: For the given initial value problem, (a) Execute 20 steps of the Taylor series method of order p for p = 1, 2, 3. Use step size h = 0.05. (b) In each exercise, the exact solution is given. List the errors of the Taylor series method calculations at t = 1. t , y(0) = 1. The exact solution is y(t) = −1 + t2 + 4. 20. y = y+1 −1 The exact solution is y(t) = . 21. y = 2ty2 , y(0) = −1. 1 + t2 √ 1 , y(0) = 1. The exact solution is y(t) = 1 + t. 22. y = 2y 23. y =

1 + y2 , y(0) = 0. 1+t

The exact solution is y(t) = tan [ln(1 + t)].

Exercises 24–27: Assume that a pth order Taylor series method is used to solve an initial value problem. When the step size h is reduced by 12 , we expect the global error to be reduced by about 1 p . Exercises 24–27 investigate this assertion using a third order Taylor series method 2 for the initial value problems of Exercises 20–23.

7.4

Runge-Kutta Methods

493

Use the third order Taylor series method to numerically solve the given initial value problem for 0 ≤ t ≤ 1. Let E1 denote the global error at t = 1 with step size h = 0.05 and E2 the error at t = 1 when h = 0.025. Calculate the error ratio E2 /E1 . Is the ratio close to 1/8? t 25. y = 2ty2 , y(0) = −1 24. y = , y(0) = 1 y+1 1 1 + y2 , y(0) = 1 26. y = 27. y = , y(0) = 0 2y 1+t

7.4

Runge-Kutta Methods In this section, we discuss Runge-Kutta methods as a way of numerically solving the initial value problem y = f (t, y),

y(t0 ) = y0 .

(1)

Runge-Kutta methods are based on Taylor series methods, but they use nested compositions of function evaluations instead of the partial derivatives of f (t, y) required by a Taylor series method. In theory, one can achieve any desired level of accuracy using the Runge-Kutta approach. Heun’s method and the modiﬁed Euler’s method are two familiar algorithms that use the Runge-Kutta philosophy of evaluating compositions of functions. For instance, Heun’s method has the form yn+1 = yn +

h f (tn , yn ) + f (tn + h, yn + hf (tn , yn )) , 2

n = 0, 1, 2, . . . , N − 1.

Heun’s method is easy to implement—in order to take a step, we need only evaluate the function f (t, y) at the current estimate (tn , yn ) and at the point (tn + h, yn + hf (tn , yn )). Moreover, as is shown in Example 1, Heun’s method is a second order method. In contrast, a second order Taylor series method requires the calculation of two partial derivatives, ft (tn , yn ) and fy (tn , yn ), in order to make a step with comparable second order accuracy.

E X A M P L E

1

Calculate the order of Heun’s method, yn+1 = yn +

h f (tn , yn ) + f (tn + h, yn + hf (tn , yn )) . 2

(2)

Solution: Let y(t) denote the unique solution of the initial value problem (1). To determine the order of the one-step method (2), we need to ﬁnd an expression for the local truncation errors, Tn+1 [recall equation (12) in Section 7.3]. Assume that we apply Heun’s method starting on the exact solution curve at tn —that is, with yn = y(tn ). To determine the local truncation error, we must ﬁrst unravel the composition f (tn + h, yn + hf ) [where functions without arguments will be assumed to be evaluated at (tn , yn )]. Expanding f (tn + h, yn + hf ) in a Taylor series about (tn , yn ), we obtain f (tn + h, yn + hf ) = f + ( ft + fy f )h + 12 ( ftt + 2ft y f + fy y f 2 )h2 + O(h3 ).

(3)

(continued)

494

CHAPTER 7

Numerical Methods (continued)

Using this expansion in (2) yields h 1 yn+1 = yn + f + f + ( ft + fy f )h + ( ftt + 2ft y f + fy y f 2 )h2 + O(h3 ) 2 2

(4)

h2 h3 = yn + f h + ( ft + fy f ) + ( ft t + 2ft y f + fy y f 2 ) + O(h4 ). 2 4 We compare this expansion with the Taylor series of the exact solution, y(tn+1 ). Using the fact that y(tn ) = yn and using the expressions for y (tn ), y (tn ), y (tn ) derived in Section 7.3, we have y(tn+1 ) = yn + f h + ( ft + fy f )

h2 h3 + ( ft t + 2ft y f + fy y f 2 + fy ft + f 2y f ) + O(h4 ). 2 6

(5)

Comparing expansions (4) and (5), we see that they agree up to and including the O(h2 ) terms but that the O(h3 ) term in the Heun method expansion does not correctly replicate the O(h3 ) term in the Taylor series of the exact solution. Therefore, the local truncation error of Heun’s method is Tn+1 = O(h3 ), and Heun’s method is second order. ❖

Second Order Runge-Kutta Methods To generalize the approach suggested by Heun’s method, we choose a set of points (θi , γi ), i = 1, 2, . . . , k that lie in the ty-plane, in the vertical strip bounded by the lines t = tn and t = tn+1 . As Figure 7.3 suggests, these points sample the direction ﬁeld in the vicinity of the point (tn , yn ). To formalize this idea of sampling the direction ﬁeld, consider a one-step method yn+1 = yn + hφ(tn , yn ; h),

(6)

where the increment function is deﬁned by φ(tn , yn ; h) = A1 f (θ1 , γ1 ) + A2 f (θ2 , γ2 ) + · · · + Ak f (θk , γk ).

(7)

The constants A1 , A2 , . . . , Ak are the weights of the method (6). Thus, the increment function is a weighted sum of direction ﬁeld slopes. For a ﬁxed integer k, y

(tn, yn )

t tn

tn + 1

FIGURE 7.3

A portion of the direction ﬁeld for y = f (t, y) near our latest estimate (tn , yn ). We use a weighted sum of direction ﬁeld evaluations at the points marked “×” to evolve the numerical solution from tn to tn+1 .

7.4

Runge-Kutta Methods

495

the local truncation error is reduced by selecting weights, {Ai }ki=1 , and direction ﬁeld sampling points, {(θi , γi )}ki=1 , so that the method (6) replicates as many terms in the Taylor series expansion of the local solution as possible. When k = 2, method (6) has the form yn+1 = yn + h[A1 f (θ1 , γ1 ) + A2 f (θ2 , γ2 )].

(8)

We need to choose the sampling points (θ1 , γ1 ) and (θ2 , γ2 ) and weights A1 and A2 . Since we are viewing the term A1 f (θ1 , γ1 ) + A2 f (θ2 , γ2 ) as an average slope, we want the sampling points to be near (tn , yn ) and to be representative of the direction ﬁeld between t = tn and t = tn+1 . A reasonable choice for one of the sampling points is (θ1 , γ1 ) = (tn , yn ). For a second point, our previous study suggests that we might sample somewhere along the “Euler line”—the line of slope f (tn , yn ) that passes through the point (tn , yn ). Thus, as a second sampling point, we choose (θ2 , γ2 ) = (tn + αh, yn + αhf (tn , yn )), where α is a constant, 0 < α ≤ 1. See Figure 7.4.

tn

tn + ␣h

t tn + 1

FIGURE 7.4

Given the two-sample method (8), we generally choose one sample at (tn , yn ) and the second somewhere along the “Euler line,” at (tn + αh, yn + αhf (tn , yn )), where α is a constant, 0 < α ≤ 1.

With the choices shown in Figure 7.4, method (8) has the form yn+1 = yn + h A1 f (tn , yn ) + A2 f (tn + αh, yn + αhf (tn , yn )) .

(9)

We now need to select constants A1 , A2 , and α. Since the right-hand side of equation (9) is a function of h, it makes sense to expand the right-hand side in powers of h, with the objective of choosing the constants so that yn+1 matches a Taylor series method through as many powers of h as possible. Expanding the right-hand side of (9) gives yn+1 = yn + h[A1 f (tn , yn ) + A2 f (tn + αh, yn + αhf (tn , yn ))] = yn + h[A1 f (tn , yn ) + A2 {f (tn , yn ) + ft (tn , yn )αh + fy (tn , yn )αhf (tn , yn ) + O(h2 )}] = yn + h(A1 + A2 )f (tn , yn ) + h2 αA2 ft (tn , yn ) + fy (tn , yn )f (tn , yn ) + O(h3 ).

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We now attempt to match this expansion with the Taylor series method of order 2, h2 ft (tn , yn ) + fy (tn , yn )f (tn , yn ) . yn+1 = yn + hf (tn , yn ) + 2 Our objective is to select the parameters at our disposal, A1 , A2 , and α, so as to maximize the agreement between the expansions. Comparing the two expansions, we see that we can obtain agreement through terms of order h2 if A1 , A2 , and α satisfy the equations A1 + A2 = 1 (10)

αA2 = 12 .

Once we satisfy these constraints, the method (9) matches the second order Taylor series method up through terms of order h2 and therefore, like the second order Taylor series method, has an O(h3 ) local truncation error. [This is the best we can do with method (9). It is impossible to select A1 , A2 , and α to match the terms of the third order Taylor series method.] In (10), we have a system of two (nonlinear) equations in three unknowns. This system has inﬁnitely many solutions, A2 =

1 2α

and

A1 = 1 −

1 , 2α

(11)

with 0 < α ≤ 1. Since α represents the fraction of the step we move along the Euler line to the second sampling point, (θ2 , γ2 ) = (tn + αh, yn + αhf (tn , yn )), there are two “natural” choices for α, namely α = 12 and α = 1. If α = 1 in equation (11), then A1 = 12 and A2 = 12 . With this choice, method (9) reduces to Heun’s method, yn+1 = yn +

h f (tn , yn ) + f (tn + h, yn + hf (tn , yn )) . 2

If α = 12 in equation (11), then A1 = 0 and A2 = 1. With this choice, method (9) reduces to the modiﬁed Euler’s method, h hf (tn , yn ) yn+1 = yn + hf tn + , yn + . 2 2

R-stage Runge-Kutta Methods In general, a Runge-Kutta method has the form yn+1 = yn + hφ(tn , yn ; h),

(12a)

where the increment function, φ(tn , yn ; h), is given by φ(tn , yn ; h) =

R

Aj K j (tn , yn ).

(12b)

j=1

In (12b), the terms Aj are constants (the weights) and the terms K j (tn , yn ) are direction ﬁeld samples, usually called stages. The stages are deﬁned sequentially

7.4

Runge-Kutta Methods

497

as follows: K1 (tn , yn ) = f (tn , yn ) K j (tn , yn ) = f (tn + αj h, yn + h

j−1

βji Ki (tn , yn )),

j = 2, 3, . . . , R,

(12c)

i=1

where 0 < αj ≤ 1 and where βj,1 + βj,2 + · · · + βj, j−1 = αj . Method (12) is called an R-stage Runge-Kutta method. A Runge-Kutta method can be viewed as a “staged” sampling process. That is, for each j, we choose a value αj that determines the t-coordinate of the jth sampling point. Then [see (12c)] the y-coordinate of the jth sampling point is determined using the prior stages. In this sense, the sampling process is recursive. In (12c), the constraint 0 < αj ≤ 1 means that all sampling points lie between t = tn and t = tn+1 . While this description of an R-stage Runge-Kutta method may seem complicated, the format of equation (12) makes programming a Runge-Kutta method very simple (see Figures 7.7 and 7.8 on pages 501 and 502). An example of a three-stage Runge-Kutta method is h (K + 4K2 + K3 ) 6 1 K1 = f (tn , yn ) h h K2 = f tn + , yn + K1 2 2

yn+1 = yn +

(13)

K3 = f (tn + h, yn − hK1 + 2hK2 ). It is not difﬁcult to show that method (13) has order 3; it matches the third order Taylor series method up through terms of order h3 but not the order h4 term. As we saw in equations (9) and (11), there are inﬁnitely many two-stage, second order Runge-Kutta methods. Similarly, there is an inﬁnite two-parameter family of three-stage, third order Runge-Kutta methods (see Exercises 31–34). Likewise, when the parameters in (12) are chosen properly, there are fourstage, fourth order Runge-Kutta methods. One of the most popular fourth order Runge-Kutta methods (recall Sections 2.10 and 4.9) is h (K + 2K2 + 2K3 + K4 ) 6 1 K1 = f (tn , yn ) h h K2 = f tn + , yn + K1 2 2 h h K3 = f tn + , yn + K2 2 2 yn+1 = yn +

(14)

K4 = f (tn + h, yn + hK3 ). Viewing algorithm (14) geometrically, we can envision it as being formed in the following way. First, we calculate K1 , the slope of the tangent line at starting point (tn , yn ). We proceed a half-step along this tangent line to locate the direction ﬁeld point at which slope K2 is evaluated. We use this new slope K2 to deﬁne another line through (tn , yn ). Proceeding, in turn, a half step along this new line locates the point that determines slope K3 . Finally, we proceed a full step from (tn , yn ) along the line having slope K3 to determine the point at which

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slope K4 is evaluated. The appropriately weighted average of these four slopes deﬁnes the algorithm.

Runge-Kutta Methods for Systems The discussion in this chapter has focused on the scalar initial value problem y = f (t, y),

y(t0 ) = y0 .

As mentioned earlier, the ideas developed and the ensuing methods extend naturally to ﬁrst order systems. Consider the initial value problem y = f(t, y), where

⎡

y1 (t)

y(t0 ) = y0 ,

⎥ ⎢ ⎢ y2 (t)⎥ ⎢ y(t) = ⎢ . ⎥ ⎥, ⎣ .. ⎦ ym (t) and

⎡

f1 (t, y)

⎤

⎡

⎤

⎡

y(1) 0

(15)

⎤

⎥ ⎢ ⎢ (2) ⎥ ⎢ y0 ⎥ y0 = ⎢ . ⎥ ⎢ .. ⎥ ⎣ ⎦ y(m) 0 f1 (t, y1 , y2 , . . . , ym )

⎤

⎥ ⎢ ⎥ ⎢ ⎥ ⎢ f2 (t, y)⎥ ⎢ ⎢ f2 (t, y1 , y2 , . . . , ym )⎥ ⎢ ⎥ f(t, y) = ⎢ . ⎥ = ⎢ ⎥. .. ⎥ ⎣ .. ⎦ ⎢ . ⎣ ⎦ fm (t, y) fm (t, y1 , y2 , . . . , ym ) The concept of an analytic function developed in Section 7.3 can be extended to the vector-valued functions y(t) and f(t, y). Theorem 7.1 can be extended to give analogous conditions sufﬁcient for the existence of an analytic solution of (15) on an interval of the form t0 − δ < t < t0 + δ for some δ > 0. We saw in Section 4.9 how Euler’s method and higher order Runge-Kutta methods extend naturally to initial value problems such as (15). For example, the system counterpart of algorithm (14) is h K1 + 2K2 + 2K3 + K4 6 K1 = f(tn , yn ) h h K2 = f tn + , yn + K1 2 2 h h K3 = f tn + , yn + K2 2 2 yn+1 = yn +

(16)

K4 = f(tn + h, yn + hK3 ).

The Damped Pendulum The next example illustrates how we can apply a Runge-Kutta method to a ﬁrst order system of the form y = f(t, y),

y(t0 ) = y0 .

7.4

Runge-Kutta Methods

499

E X A M P L E

2

Consider a pendulum whose motion is inﬂuenced not only by its weight but also by a resistive or damping force. The mathematical formulation of this problem leads to an initial value problem involving a scalar second order nonlinear differential equation. We rewrite this scalar second order problem as an equivalent problem for a ﬁrst order (nonlinear) system and then use the fourth order Runge-Kutta method (16), with a step size of h = 0.05, to obtain a numerical solution. Problem Formulation: The pendulum is formed by a mass m attached to a rod of length l (see Figure 7.5). We neglect the mass of the rod. As the pendulum moves, it is acted on by the force of gravity and also by a damping force, which acts to retard the pendulum motion. We assume this damping force is proportional to the angular velocity of the pendulum and acts in the tangential direction to retard the motion. We obtain κ g θ + sin θ = 0, ml2 θ = −mgl sin θ − κlθ , or θ + ml l where κ is a positive damping constant. We complete the formulation by specifying both θ and θ at the initial time of interest, say t = 0. These two constants give the initial position and initial angular velocity of the pendulum. We adopt the numerical values κ g = 0.2 s−1 , = 1 s−2 , θ (0) = 0 rad, θ (0) = 3 rad/s, ml l and the initial value problem of interest becomes θ + 0.2θ + sin θ = 0,

θ (0) = 0,

θ (0) = 3.

The differential equation is recast as a ﬁrst order system by deﬁning y1 (t) y1 (t) = θ(t), y2 (t) = θ (t), and y(t) = . y2 (t) The initial value problem becomes y2 y = , − sin y1 − 0.2y2

y(0) =

0 3

.

(17)

We will solve initial value problem (17) numerically using algorithm (16). Pivot O

l

mgsin

Mass m Weight mg

FIGURE 7.5

The damped pendulum described in Example 2. (continued)

500

CHAPTER 7

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What Should We Expect? Before embarking on a numerical solution, it’s usually worthwhile to bring to bear any available physical insights that will help determine what to expect. We know that, generally, solutions of nonlinear initial value problems do not exist on arbitrarily large time intervals. However, because of the nature of the pendulum motion it describes, we expect the exact solution of (17) to exist on an arbitrarily large time interval. We saw in Chapter 6 that, in the absence of damping, a pendulum starting at θ (0) = 0 with θ (0) = 2 has just enough energy to reach the inverted position (in the limit as t → ∞). In our case, the initial angular velocity is greater, since θ (0) = 3. Damping, however, retards the motion and causes the pendulum to lose energy. If damping is not too large, we expect the pendulum to go past the inverted position at least once. If damping is large enough, however, the accompanying loss of energy will more than offset the increase in initial energy and the pendulum will not reach the inverted position. It’s not clear at the outset which possibility will occur. In any event, the pendulum eventually will have insufﬁcient energy to reach the inverted position, and it will simply swing back and forth with decreasing amplitude as time increases. Based on these observations, what do you expect the graphs of θ(t) and θ (t) to look like? Interpreting the Results Figure 7.6 shows the results of the numerical computation. Note that the graph of y1 (t) = θ(t) increases from zero to a maximum of about 8.29 rad. Since 2π ≈ 6.28, the graph tells us that the pendulum makes one complete counterclockwise revolution, rotating an additional 2 rad ≈ 115◦ beyond the vertically downward position before falling back, beginning to swing back and forth with decreasing amplitude as time progresses. The graph has a horizontal asymptote of 2π, since the pendulum approaches the vertically downward rest position as t → ∞. ′

9

3

8

2.5

7

2

6

1.5

5

1

4

0.5

3

5

2

–0.5

1

–1 5

10 15 20 25 30 35 40

t

10 15 20 25 30 35 40

t

–1.5

(a)

(b) FIGURE 7.6

(a) The graph of y1 (t) = θ(t). (b) The graph of y2 (t) = θ (t).

Is the graph of y2 (t) = θ (t) consistent with this physical interpretation? What do the initial minimum and maximum (for t > 0) of this graph corre-

7.4

Runge-Kutta Methods

501

spond to? Should they occur while the dependent variable (angular velocity) is positive? Should the zero crossings of this graph occur at the critical points of y1 (t) = θ (t)? Should the maxima of y2 (t) = θ (t) occur when the pendulum is in the vertically downward position? Should the graph of y2 (t) = θ (t) have a horizontal asymptote of zero? Subjecting your numerical solution to simple common-sense checks such as these is an important ﬁnal step. ❖

Coding a Runge-Kutta Method We conclude this section with a short discussion about the practical aspects of writing a program to implement a Runge-Kutta method. Figures 7.7 and 7.8 list the program used to generate the numerical solution of Example 2. This particular code was written in MATLAB, but the principles are the same for any programming language.

% % Set the initial conditions for the % initial value problem of Example 2 % t=0; y=[0,3]’; h=0.05; output=[t,y(1),y(2)]; % % % Execute the fourth order Runge-Kutta method % on the interval [0, 30] % for i=1:600 ttemp=t; ytemp=y; k1=f(ttemp,ytemp); ttemp=t+h/2; ytemp=y+(h/2)*k1; k2=f(ttemp,ytemp); ttemp=t+h/2; ytemp=y+(h/2)*k2; k3=f(ttemp,ytemp); ttemp=t+h; ytemp=y+h*k3; k4=f(ttemp,ytemp); y=y+(h/6)*(k1+2*k2+2*k3+k4); t=t+h; output=[output;t,y(1),y(2)]; end FIGURE 7.7

A Runge-Kutta code for the initial value problem in Example 2.

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function yp=f(t,y) yp=zeros(2,1); yp(1)=y(2); yp(2)=-sin(y(1))-0.2*y(2); FIGURE 7.8

A function subprogram that evaluates f(t, y) for the differential equation of Example 2.

Note ﬁrst that no matter what numerical method we decide to use for the initial value problem y = f(t, y),

y(t0 ) = y0 ,

we need to write a subprogram (or module) that evaluates f(t, y). Such a module is listed in Figure 7.8 for the initial value problem of Example 2. Figure 7.7 lists a MATLAB program that executes 600 steps of the fourth order Runge-Kutta method (16) for the initial value problem of Example 2. The code listed in Figure 7.7 stays as close as possible to the notation and format of the fourth order Runge-Kutta method (16). It is always a good idea to use variable names (such as k1 and k2 ) that match the names in the algorithm. Beyond the choice of variable names, the code in Figure 7.8 also mimics the steps of algorithm (16) as closely as possible. Adhering to such conventions makes programs easier to read and debug.

EXERCISES Exercises 1–10: We reconsider the initial value problems studied in the Exercises of Section 7.3. The solution of the differential equation satisfying initial condition y(0) = 1 is given. (a) Carry out one step of the third order Runge-Kutta method (13) using a step size h = 0.1, obtaining a numerical approximation of the exact solution at t = 0.1. (b) Carry out one step of the fourth order Runge-Kutta method (14) using a step size h = 0.1, obtaining a numerical approximation of the exact solution at t = 0.1. (c) Examine the exact solution. Should either or both of the Runge-Kutta methods, in principle, yield an exact answer for the particular problem being considered? Explain. (d) Compare the numerical values obtained in parts (a) and (b) with the exact solution evaluated at t = 0.1. Are the results consistent with your answer in part (c)? Is the error incurred using the four-stage algorithm less than the error for the three-stage calculation? 2

1. y = −y + 2; y(t) = 2 − e−t

2. y = 2ty; y(t) = et

2 2 − t2 t 2 √ 5. y = y; y(t) = 1 + 2

4. y = t2 + y; y(t) = 3e t − (t2 + 2t + 2)

3. y = ty2 ; y(t) =

7. y = y + sin t; y(t) =

3e t − cos t − sin t 2

6. y =

t ; y(t) = 1 + t2 y

7.4 8. y = y3/4 ; y(t) =

1+

10. y = −4t3 y; y(t) = e−t

t 4

4

Runge-Kutta Methods

503

π 9. y = 1 + y2 ; y(t) = tan t + 4

4

Exercises 11–16: For the given initial value problem, an exact solution in terms of familiar functions is not available for comparison. If necessary, rewrite the problem as an initial value problem for a ﬁrst order system. Implement one step of the fourth order Runge-Kutta method (14), using a step size h = 0.1, to obtain a numerical approximation of the exact solution at t = 0.1. 11. y + ty + y = 0, y(0) = 1, y (0) = −1 d dy et + ty = 1, y(0) = 1, y (0) = 2 12. dt dt 2 1 0 t −1 y+ , y(0) = 14. y = 13. y = t t 1 e 0 2

t 0

y, y(0) =

−1

1

15. y − ty = 0, y(0) = 1, y (0) = 0, y (0) = −1 16. y + z + ty = 0 z − y = t, y(0) = 1, y (0) = 2, z(0) = 0

Exercises 17–18: One differential equation for which we can explicitly demonstrate the order of the Runge-Kutta algorithm is the linear homogeneous equation y = λy, where λ is a constant. 17. (a) Verify that the exact solution of y = λy, y(t0 ) = y0 is y(t) = y0 eλ(t−t0 ) . (b) Show, for the three-stage Runge-Kutta method (13), that

(λh)3 (λh)2 + y(tn ) + hφ(tn , y(tn ); h) = y(tn ) 1 + λh + . 2! 3! (c) Show that y(tn+1 ) = y(tn )eλh . (d) What is the order of the local truncation error? 18. Repeat the calculations of Exercise 17 using the four-stage Runge-Kutta method (14). In this case, show that the local truncation error is O(h5 ).

Exercises 19–22: In these exercises, we ask you to use the fourth order Runge-Kutta method (14) to solve the problems in Exercises 20–23 of Section 7.3. (a) For the given initial value problem, execute 20 steps of the method (14); use step size h = 0.05. (b) The exact solution is given. Compare the numerical approximation y20 and the exact solution y(t20 ) = y(1). t , y(0) = 1. The exact solution is y(t) = −1 + t2 + 4. 19. y = y+1 −1 The exact solution is y(t) = . 20. y = 2ty2 , y(0) = −1. 1 + t2 √ 1 , y(0) = 1. The exact solution is y(t) = 1 + t. 21. y = 2y

504

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22. y =

1 + y2 , y(0) = 0. 1+t

The exact solution is y(t) = tan[ln(1 + t)].

Exercises 23–25: In each exercise, (a) Verify that the given function is the solution of the initial value problem posed. If the initial value problem involves a higher order scalar differential equation, rewrite it as an equivalent initial value problem for a ﬁrst order system. (b) Execute the fourth order Runge-Kutta method (16) over the speciﬁed t-interval, using step size h = 0.1, to obtain a numerical approximation of the exact solution. Tabulate the components of the numerical solution with their exact solution counterparts at the endpoint of the speciﬁed interval. 23. y + 2y + 2y = −2, y(0) = 0, y (0) = 1; y(t) = e−t (cos t + 2 sin t) − 1; 0 ≤ t ≤ 2 ⎤ ⎡ ⎡ ⎤ 1 −1 2 e−t/2 + e−3t/2 2 ⎦; ⎦ y, y(0) = 0≤t≤1 24. y = ⎣ ; y(t) = ⎣ −t/2 1 e − e−3t/2 0 −1 2 25. t2 y − ty + y = t2 , y(1) = 2, y (1) = 2;

y(t) = t(t + 1 − ln t);

1≤t≤2

Exercises 26–30: These exercises ask you to use numerical methods to study the behavior of some scalar second order initial value problems. In each exercise, use the fourth order Runge-Kutta method (16) and step size h = 0.05 to solve the problem over the given interval. 26. y + 4(1 + 3 tanh t)y = 0, y(0) = 1, y (0) = 0;

0 ≤ t ≤ 10.

This problem might model the motion of a spring-mass system in which the mass is released from rest with a unit initial displacement at t = 0 and with the spring stiffening as the motion progresses in time. Plot the numerical solutions for y(t) and y (t). Since tanh t approaches 1 for large values of t, we might expect the solution to approximate a solution of y + 16y = 0 for time t sufﬁciently large. Do your graphs support this conjecture? 27. y + y + y3 = 0, y(0) = 0, y (0) = 1;

0 ≤ t ≤ 10.

A nonlinear differential equation having this structure arose in modeling the motion of a nonlinear spring. We are interested in assessing the impact of the nonlinear y3 term on the motion. Plot the numerical solution for y(t). If the nonlinear term were not present, the initial value problem would have solution y(t) = sin t. On the same graph, plot the function sin t. Does the nonlinearity increase or decrease the period of the motion? How do the amplitudes of the motion differ? 28. θ + sin θ = 0.2 sin t, θ(0) = 0, θ (0) = 0;

0 ≤ t ≤ 50.

This nonlinear differential equation is used to model the forced motion of a pendulum initially at rest in the vertically downward position. For small angular displacements, the approximation sin θ ≈ θ is often used in the differential equation. Note, however, that the solution of the resulting initial value problem θ + θ = 0.2 sin t, θ (0) = 0, θ (0) = 0 is given by θ(t) = −0.1(sin t − t cos t), leading to pendulum oscillations that continue to grow in amplitude as time increases. Our goal is to determine how the nonlinear sin θ term affects the motion. Plot the numerical solutions for θ(t) and θ (t). Describe in simple terms what the pendulum is doing on the time interval considered.

7.4 29. θ + sin θ = 0, θ(0) = 0, θ (0) = 2;

Runge-Kutta Methods

505

0 ≤ t ≤ 20.

This problem models pendulum motion when the pendulum is initially in the vertically downward position with an initial angular velocity of 2 rad/s. For this conservative system, it was shown in Chapter 6 that (θ )2 − 2 cos θ = 2. Therefore, the initial conditions have been chosen so that the pendulum will rotate upward in the positive (counterclockwise) direction, slowing down and approaching the vertically upward position as t → ∞. The phase-plane solution point is moving on the separatrix; thus, loosely speaking, the exact solution is “moving on a knife’s edge.” If the initial velocity is slightly less, the pendulum will not reach the upright position but will reach a maximum value less than π and then proceed to swing back and forth. If the initial velocity is slightly greater, the pendulum will pass through the vertically upright position and continue to rotate counterclockwise. What happens if we solve this problem numerically? Plot the numerical solutions for θ (t) and θ (t). Interpret in simple terms what the numerical solution is saying about the pendulum motion on the time interval considered. Does the numerical solution conserve energy? πx 2kδ 30. mx + tan = F(t), x(0) = 0, x (0) = 0; 0 ≤ t ≤ 15. π 2δ This problem was used to model a nonlinear spring-mass system (see Exercise 18 in Section 6.1). The motion is assumed to occur on a frictionless horizontal surface. In this equation, m is the mass of the object attached to the spring, x(t) is the horizontal displacement of the mass from the unstretched equilibrium position, and δ is the length that the spring can contract or elongate. The spring restoring force has vertical asymptotes at x = ±δ. Time t is in seconds. Let m = 100 kg, δ = 0.15 m, and k = 100 N/m. Assume that the spring-mass system is initially at rest with the spring at its unstretched length. At time t = 0, a force of large amplitude but short duration is applied: F0 sin πt, 0 ≤ t ≤ 1 F(t) = newtons. 0, 1 < t < 15 Solve the problem numerically for the two cases F0 = 4 N and F0 = 40 N. Plot the corresponding displacements on the same graph. How do they differ?

Exercises 31–34: Third Order Runge-Kutta Methods Runge-Kutta method is

As given in equation (12), the form of a three-stage

yn+1 = yn + h[A1 K1 (tn , yn ) + A2 K2 (tn , yn ) + A3 K3 (tn , yn )],

(18a)

where K1 (tn , yn ) = f (tn , yn ) K2 (tn , yn ) = f (tn + α2 h, yn + hβ2,1 K1 (tn , yn ))

(18b)

K3 (tn , yn ) = f (tn + α3 h, yn + h[β3,1 K1 (tn , yn ) + β3,2 K2 (tn , yn )]) and where [see equation (12c)] 0 < α2 ≤ 1, 0 < α3 ≤ 1, β2,1 = α2 , and β3,1 + β3,2 = α3 . It can be shown (see Lambert5 ) that this three-stage Runge-Kutta method has order 3 if

5

John D. Lambert, Numerical Methods for Ordinary Differential Systems (Chichester, England: Wiley, 1991).

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the following four equations are satisﬁed: A1 +

A2 +

A3 = 1

α2 A2 +

α3 A3 =

1 2

α22 A2 +

α32 A3 =

1 3

(19)

α2 β3,2 A3 = 16 . One way to ﬁnd a solution of this system of four nonlinear equations is ﬁrst to select values for α2 and α3 . [Note that α2 and α3 determine the t-coordinate of the sampling points deﬁning K2 (tn , yn ) and K3 (tn , yn ), respectively.] Once α2 and α3 are chosen, the ﬁrst three equations in (19) can be solved for A1 , A2 , and A3 . The parameters α2 and α3 are nonzero; if they are distinct, then there are unique values A1 , A2 , and A3 that satisfy the ﬁrst three equations. Having A1 , A2 , and A3 , you can determine β3,2 from the fourth equation and β3,1 from the condition β3,1 + β3,2 = α3 . 31. Consider a three-stage Runge-Kutta method. Show that if the ﬁrst equation in system (19) holds, then the method has order at least 1. 32. Consider a three-stage Runge-Kutta method. Show that if the ﬁrst and second equations in system (19) hold, then the method has order at least 2. [Note: In order to obtain order 3, the last two equations in (19) must hold as well.] 33. Determine the values of α2 and α3 that give rise to the three-stage third order RungeKutta method (13). Then solve equations (19), and verify that Runge-Kutta method (13) results. 7 2 34. (a) Verify that the choice of α2 = 10 and α3 = 10 leads to another solution of (19) having the same weights A1 , A2 , and A3 as (13).

(b) Use the values from part (a) to form another three-stage, third order RungeKutta method. Test this method on y = t/( y + 1), y(0) = 1, using step size h = 0.05. Compute the error at t = 1 [the exact solution is y(t) = −1 + t2 + 4 ].

Appendix 1 Convergence of One-Step Methods In this appendix, we state a theorem that guarantees convergence of the onestep method, yn+1 = yn + hφ(tn , yn ; h),

n = 0, 1, 2, . . . , N − 1.

(1)

The convergence theorem, Theorem 7.3, applies to an initial value problem y = f (t, y),

y(t0 ) = y0 .

Let f (t, y) be a function deﬁned on the rectangle R given by a < t < b, α < y < β. The function f is said to satisfy a Lipschitz condition in y if there is a positive constant K such that Whenever (t, y1 ) and (t, y2 ) are two points in R, then |f (t, y1 ) − f (t, y2 )| ≤ K|y1 − y2 |.

(2)

The constant K in (2) is called a Lipschitz constant. Note that Lipschitz constants are not unique; if a particular constant K can be used in inequality (2),

7.4

Runge-Kutta Methods

507

then so can K + M for any positive constant M. A Lipschitz condition is not an overly restrictive assumption; if the partial derivative fy (t, y) exists on R, then (by the mean value theorem) f (t, y1 ) − f (t, y2 ) = fy (t, y∗ )( y1 − y2 ), where y∗ is some value between y1 and y2 . Thus, if we know that |fy (t, y)| ≤ K for all (t, y) in R, then the Lipschitz condition (2) holds where the bound on |fy (t, y)| serves as a Lipschitz constant K.

Theorem 7.3

Consider the initial value problem y = f (t, y),

y(t0 ) = y0 ,

(3)

where f (t, y) is analytic and Lipschitz continuous in the vertical inﬁnite strip deﬁned by a < t < b, −∞ < y < ∞. Assume that a < t0 < t0 + T < b. Let yn+1 = yn + hφ(tn , yn ; h) be a pth order one-step method, and let h = T/N. Assume that for all step sizes less than some h0 the increment function φ, when applied to the initial value problem (3), satisﬁes a Lipschitz condition in y with Lipschitz constant L. Then max |y(tn ) − yn | = O(hp ).

0≤n≤N

(4)

In words, conclusion (4) says that the global error can be bounded by some constant multiple of hp as long as 0 < h ≤ h0 . Also note that we are asking for a Lipschitz condition to hold on a vertical inﬁnite strip. This rather restrictive condition simpliﬁes the theorem, since it ensures that initial value problem (3) has a unique solution on [t0 , t0 + T] and that φ(tn , yn ; h) is deﬁned for all points tn in t0 ≤ t ≤ t0 + T. Although Theorem 7.3 was stated for the scalar problem, a similar result can be established for a system of differential equations.

Appendix 2 Stability of One-Step Methods Theorem 7.3 shows that we can, in principle, achieve arbitrarily good accuracy by using a one-step method with a sufﬁciently small step size h. We now consider the opposite situation and show that results sometimes become disastrously bad if we inadvertently use a step size that is just a little too large. In particular, numerical methods for initial value problems are subject to difﬁculties of “stability.” When a numerical method is applied to a given differential equation, it can happen that there is a sharp division between a step size h that is too large and that produces terrible results and a step size h that is small enough to produce acceptable results. Such a stability boundary is illustrated in Figures 7.9 and 7.10. In each case, we used Euler’s method to solve

508

CHAPTER 7

Numerical Methods y

y

10

150

8

100

6

50

4

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

2

t

–50 1

2

3

4

5

6

7

8

9 10

t

–2

–100 –150

FIGURE 7.9

FIGURE 7.10

h = 1/130

h = 1/120

the initial value problem y + 251y + 250 y = 500 cos 5t,

y(0) = 10,

y (0) = 0.

(1)

This differential equation might model the forced vibrations of a spring-massdashpot system. (For the coefﬁcients chosen, the spring constant and the damping coefﬁcient per unit mass are relatively large.) As we know from Section 3.10, once the initial transients die out, the solution, y(t), should tend toward a periodic steady-state solution. This expected behavior is exhibited by the results in Figure 7.9, but not by those in Figure 7.10. The only difference between the two computations is that the results in Figure 7.9 were obtained using a step 1 , whereas the results in Figure 7.10 were obtained using a slightly size of h = 130 1 larger step size of h = 120 . The sharp division between the accurate results of Figure 7.9 and the terrible results of Figure 7.10 can be explained by examining the behavior of Euler’s method when it is applied to the homogeneous initial value problem y = Ay, y(0) = y0 . Assume that A is a (2 × 2) constant matrix, with distinct eigenvalues λ1 and λ2 and corresponding eigenvectors u1 and u2 . Since the eigenvalues are distinct, the eigenvectors are linearly independent. Therefore, the initial condition can be represented as y0 = α1 u1 + α2 u2

(2)

for some constants α1 and α2 . When applied to the initial value problem y = Ay, y(0) = y0 , Euler’s method takes the form yn = yn−1 + hAyn−1 , or yn = (I + hA)yn−1 ,

n = 1, 2, . . . .

(3)

It follows from (3) that yn = (I + hA)n y0 , n = 1, 2, . . . . From (2), it follows that yn = (I + hA)n y0 = (I + hA)n (α1 u1 + α2 u2 ) = α1 (I + hA)n u1 + α2 (I + hA)n u2 . By Exercises 31 and 32 in Section 4.4, (I + hA)n u j = (1 + hλj )n u j , j = 1, 2. Thus, yn = (I + hA)n y0 = α1 (1 + hλ1 )n u1 + α2 (1 + hλ2 )n u2 .

(4)

7.4

Runge-Kutta Methods

509

If the eigenvalues λ1 and λ2 are both negative, then the exact solution y(t) tends to 0 as t increases. Thus, the output from Euler’s method [the sequence yn in equation (4)] should also tend to 0 as t increases. Assume that λ1 and λ2 are both negative and that λ1 < λ2 < 0. If α1 and α2 are both nonzero, having yn → 0 as n → ∞ requires |1 + hλ1 | < 1 and |1 + hλ2 | < 1. These two inequalities reduce to −2 < hλ1 < 0

−2 < hλ2 < 0.

and

Therefore, to obtain yn → 0 as n → ∞, we need to use a step size h that satisﬁes the inequality h < −2/λ1 . When we write the homogeneous second order equation y + 251y + 250 y = 0 as a ﬁrst order system y = Ay, y(0) = y0 , we ﬁnd A=

1

0 −250

−251

and

λ1 = −250,

λ2 = −1.

Therefore, for this problem, the critical step size is h = −2/(−250) = 1/125. If we apply Euler’s method to the homogeneous problem y + 251y + 250 y = 0, we expect to see results qualitatively similar to those shown in Figure 7.10 when we use a step size h, where h > h = 1/125. We now return to initial value problem (1), which we represent as y = Ay + g(t): y =

0

1

−250

−251

y + (500 cos 5t)

0 1

,

y(0) =

10 0

.

Applying Euler’s method to this problem, we obtain yn+1 = yn + h[Ayn + g(tn )] = [I + hA]n+1 y0 + h

n j=0

[I + hA]j g(tn−j ).

If we represent the vectors y0 and g(tn−j ) in terms of the eigenvectors of A, y0 = α1 u1 + α2 u2

and

g(tn−j ) = (500 cos 5tn−j )[β1 u1 + β2 u2 ],

Euler’s method produces yn+1 = α1 (1 + hλ1 )n+1 u1 + α2 (1 + hλ2 )n+1 u2 + 500hβ1

n j=0

cos(tn−j )(1 + hλ1 ) j u1 + 500hβ2

n

cos(tn−j )(1 + hλ2 ) j u2 .

j=0

Thus, as we saw with the homogeneous problem y = Ay, y(0) = y0 , if we do not choose a step size h such that |1 + hλ1 | < 1 and |1 + hλ2 | < 1, Euler’s method will produce results qualitatively similar to those in Figure 7.10. The ideas regarding stability that we have discussed in relation to Euler’s method apply to one-step methods in general.

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PROJECTS Project 1: Projectile Motion At some initial time, a projectile (such as a meteorite) is traveling above Earth; assume that its position and velocity at that instant are known. We consider a model in which the only force acting on the projectile is the gravitational force exerted by Earth. Given this assumption, the projectile’s trajectory lies in the plane determined by the projectile’s initial position vector and initial velocity vector. For simplicity, we assume this plane is the xy-plane. In our model, the projectile eventually strikes the surface of Earth. Our goal is to determine where and when the impact occurs. The dynamics of the projectile can be described by the equations −Gme x(t) x (t) = 3/2 , 2 x (t) + y2 (t)

−Gme y(t) y (t) = 3/2 , 2 x (t) + y2 (t)

(1)

where G is the universal gravitational constant and me is the mass of Earth. The center of Earth is at the origin, and we let Re denote the radius of Earth. The values of these constants are taken to be G = 6.673 × 10−11

m3 , kg · s2

me = 5.976 × 1024 kg,

Re = 6.371 × 106 m.

According to equation (1), the projectile dynamics are governed by a pair of coupled nonlinear second order differential equations. We will solve the problem numerically. 1. The problem geometry and the nature of the force acting on the projectile suggest the use of polar coordinates. Let x(t) = r(t) cos[θ(t)],

y(t) = r(t) sin[θ(t)].

(2)

Show that equation (1) transforms into the following pair of equations for the polar variables: r − (θ )2 r = −

r θ + 2 θ = 0. r

Gme , r2

(3)

2. Assume that the projectile is launched at time t = 0 at a point above Earth’s surface, as shown in Figure 7.11. Thus, r(0) = R0 > Re , θ(0) = 0. Show that r (0) = v0 cos α

and

θ (0) =

v0 sin α, R0

(4)

where initial speed v0 and angle α are as shown in the ﬁgure. 3. When performing the numerical calculations, we want to deal with variables whose magnitudes are comparable to unity. To achieve this, we will adopt Earth’s radius, Re , as the unit of length and the hour as the unit of time. For bookkeeping purposes, let T = 3600 s/hr. Deﬁne the scaled variables ρ(t) =

r(t) Re

and

τ=

t . T

Projects y

511

Projectile at time t > 0 r(t)

Re

␣

v0 x Projectile at time t = 0

(t)

FIGURE 7.11

The initial conditions for the projectile whose trajectory is described by equation (3).

Thus, points on Earth’s surface correspond to ρ = 1, while 3600 seconds corresponds to one unit of time on the τ -scale. Note, from the chain rule, that d dτ d 1 d = = . dt dt dτ T dτ Show that the initial value problem posed by (3) and (4) transforms into the following problem: 2 d2 ρ v T 1 R dρ(0) dθ = 0 cos α − ρ = −19.985 2 , ρ(0) = 0 , dτ Re dτ R0 dτ 2 ρ

(5) dρ dθ(0) v0 T d2 θ dθ dτ = 0, θ(0) = 0, = +2 sin α. ρ dτ dτ R0 dτ 2 The constant Gme T 2 /R3e = 19.985 has units of hr−2 . 4. Assume that the projectile is initially 9000 km above the surface of Earth with a speed v0 = 2000 m/s and angle α = 10◦ . Translate the assumed information into initial conditions for problem (5). 5. Recast initial value problem (5) as an initial value problem for a ﬁrst order system, where y1 (τ ) = ρ,

y2 (τ ) =

dρ , dτ

y3 (τ ) = θ,

y4 (τ ) =

dθ . dτ

6. Solve this problem using a fourth order Runge-Kutta method and a step size h = 0.005. The projectile will strike Earth when ρ = 1. Execute the program on a τ -interval sufﬁciently large to achieve this condition. [Hint: Gradually build up the size of the τ -interval. If too large an interval is used at the outset, the numerical solution will “blow up.”] 7. Determine the polar coordinates of the impact point and the time of impact. 8. Suppose that the point ρ = 1, θ = 0 corresponds to the point where the equator intersects the prime meridian, while the point ρ = 1, θ = π/2 corresponds to the North Pole. Use a globe and determine the approximate location of impact.

512

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Numerical Methods

Project 2: The Double Pendulum The double pendulum shown in Figure 7.12 consists of one pendulum attached to another. Two bobs, having masses m1 and m2 , are connected to frictionless pivots by rigid rods of lengths l1 and l2 .

1

l1

m1 l2 2 m2 FIGURE 7.12

A double pendulum. Assume that both pendulums can rotate freely about their pivots and that the masses of the two rigid rods are negligibly small. With respect to the coordinate system shown, the positions of the two bobs are x1 = l1 sin θ1 ,

y1 = −l1 cos θ1

x2 = l1 sin θ1 + l2 sin θ2 ,

y2 = −l1 cos θ1 − l2 cos θ2 .

(6)

Using g to denote gravitational acceleration, we can show that the angles θ1 and θ2 satisfy the following system of coupled nonlinear second order differential equations: d2 θ1 d2 θ2 dθ2 2 + m l l cos(θ − θ ) + m l l sin(θ1 − θ2 ) + l1 g(m1 + m2 ) sin θ1 = 0 2 1 2 1 2 2 1 2 dt dt2 dt2 dθ1 2 d2 θ d2 θ sin(θ1 − θ2 ) + l2 m2 g sin θ2 = 0. m2 l22 22 + m2 l1 l2 21 cos(θ1 − θ2 ) − m2 l1 l2 dt dt dt (m1 + m2 )l12

Prescribing the initial angular position and velocity of each pendulum will complete the speciﬁcation of the initial value problem. 1. As a check on the differential equations, determine what happens to these equations when (a) m2 = 0

(b) l2 = 0

(c) l1 = 0

In each of these cases, are the equations consistent with what you would expect on purely physical grounds? Note that, in the case of interest, l1 and l2 are both positive. Therefore, we can remove an l1 factor from the ﬁrst equation and an l2 factor from the second. 2. Transform the differential equations into an equivalent pair of equations of the form θ1 = f1 (θ1 , θ2 , θ1 , θ2 ) θ2 = f2 (θ1 , θ2 , θ1 , θ2 ).

(7)

Projects

513

Hint: The original differential equations can be written as m2 l2 cos(θ1 − θ2 ) θ1 (m1 + m2 )l1 m2 l1 cos(θ1 − θ2 )

m2 l2

θ2 =−

m2 l2 (θ2 )2 sin(θ1 − θ2 ) + g(m1 + m2 ) sin θ1

−m2 l1 (θ1 )2 sin(θ1 − θ2 ) + m2 g sin θ2

.

Is the determinant of the (2 × 2) matrix ever zero? 3. Rewrite system (7) as an equivalent four-dimensional ﬁrst order system by deﬁning y 1 = θ1 ,

y2 = θ1 ,

y3 = θ2 ,

y4 = θ2 .

4. Let m1 = m2 = 2 kg, l1 = l2 = 0.5 m, and let g = 9.8 m/s2 . Assume initial conditions θ1 (0) =

π , 2

θ1 (0) = 0,

θ2 (0) =

5π , 6

θ2 (0) = 0.

Solve the initial value problem for y on the time interval 0 ≤ t ≤ 10 using the fourth order Runge-Kutta method. Plot θ1 (t) and θ2 (t) versus t on separate graphs. 5. To obtain a better insight into how the double pendulum actually moves, use the numerical solutions obtained in part 4 and equations (6) to determine the bob coordinates (xi (t), yi (t)), i = 1, 2. Create parametric plots of the two bob trajectories on the same graph over the ten-second interval. On this graph, sketch the double pendulum conﬁgurations at initial and ﬁnal times, t = 0 and t = 10.

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C H A P T E R

Series Solutions of Linear Diﬀerential Equations

8

CHAPTER OVERVIEW 8.1

Introduction

8.2

Series Solutions Near an Ordinary Point

8.3

The Euler Equation

8.4

Solutions Near a Regular Singular Point and the Method of Frobenius

8.5

The Method of Frobenius Continued: Special Cases and a Summary

8.1

Introduction In this chapter, attention is focused mainly on problems involving second order linear differential equations with variable coefﬁcients, y + p(t)y + q(t)y = 0, y(t0 ) = y0 ,

y (t0 ) =

a 1, the inﬁnite series (3) is divergent. If L = 1, the ratio rest is inconclusive. The ratio test can be used to determine the radius of convergence of a power series, as we see in Example 1.

E X A M P L E

1

Use the ratio test to determine the radius of convergence of the power series ∞

nt n = t + 2t 2 + 3t3 + · · · .

n=1

Solution: The power series clearly converges at t = 0. Applying the ratio test at an arbitrary value t, t = 0, we ﬁnd (n + 1)t n+1 1 lim |t| = |t|. = lim 1 + n→∞ n→∞ nt n n Therefore, by the ratio test, the power series converges if |t| < 1 and diverges if |t| > 1. The radius of convergence is R = 1. ❖

Suppose a power series has a ﬁnite and positive radius of convergence R. The preceding discussion says nothing about whether or not the power series converges at the points t = t0 ± R. In fact, no general statements can be made about convergence or divergence at these points, which separate the open interval of absolute convergence from the semi-inﬁnite intervals of divergence. For instance, the power series ∞

nt n

n=0

considered in Example 1 diverges at t = ±1. In general, a power series might converge absolutely, converge conditionally (that is, converge but not converge absolutely), or diverge at the point t = t0 + R. The same statement can be made with regard to the point t = t0 − R.

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Operations with Power Series Every power series deﬁnes a function f (t). The domain of f (t) is the set of t-values for which the series converges. Consider the function f (t) deﬁned by f (t) =

∞

an (t − t0 )n

n=0

= a0 + a1 (t − t0 ) + a2 (t − t0 )2 + a3 (t − t0 )3 + · · · . Assume that the power series deﬁning f (t) has radius of convergence R, where R > 0. The following results are established in calculus and say, roughly, that power series can be treated like polynomials with respect to the operations of addition, subtraction, multiplication, and division.

Power Series Can Be Added and Subtracted If f (t) and g(t) are given by f (t) =

∞

n

an (t − t0 )

and

g(t) =

n=0

∞

bn (t − t0 )n ,

n=0

with both series converging in |t − t0 | < R, then the sum and difference functions are given by ( f ± g)(t) =

∞

(an ± bn )(t − t0 )n ,

n=0

where the sum and difference both converge absolutely in |t − t0 | < R.

Power Series Can Be Multiplied If f (t) and g(t) are given by f (t) =

∞

an (t − t0 )n

and

n=0

g(t) =

∞

bn (t − t0 )n ,

n=0

with both series converging in |t − t0 | < R, then the product function, ( fg)(t), has a power series representation ( fg)(t) =

∞

cn (t − t0 )n ,

n=0

which likewise converges in |t − t0 | < R. Moreover, the coefﬁcients cn can be obtained by formally multiplying the power series for f (t) and g(t) as if they were polynomials and grouping terms. In other words, c0 + c1 (t − t0 ) + c2 (t − t0 )2 + · · · = [a0 + a1 (t − t0 ) + a2 (t − t0 )2 + · · ·][b0 + b1 (t − t0 ) + b2 (t − t0 )2 + · · ·] = a0 b0 + (a0 b1 + a1 b0 )(t − t0 ) + (a0 b2 + a1 b1 + a2 b0 )(t − t0 )2 + · · · . Therefore, c0 = a0 b0 ,

c1 = a0 b1 + a1 b0 ,

c2 = a0 b2 + a1 b1 + a2 b0 ,

8.1

Introduction

519

and, in general, cn =

n

ai bn−i ,

n = 0, 1, 2, . . . .

i=0

The product power series ( fg)(t) is called the Cauchy1 product.

In Some Cases Power Series Can Be Divided If f (t) and g(t) are given by f (t) =

∞

an (t − t0 )n

and

g(t) =

n=0

∞

bn (t − t0 )n ,

n=0

with both converging in |t − t0 | < R, and if g(t0 ) = b0 = 0, then the quotient function ( f /g)(t) has a power series representation ( f /g)(t) =

∞

dn (t − t0 )n ,

n=0

which converges in some neighborhood of t0 . Again, we can determine the coefﬁcients dn by formally manipulating the power series as if they were polynomials. We have a0 + a1 (t − t0 ) + a2 (t − t0 )2 + · · · b0 + b1 (t − t0 ) + b2 (t − t0 )2 + · · ·

= d0 + d1 (t − t0 ) + d2 (t − t0 )2 + · · ·

or, after multiplying by the denominator series, a0 + a1 (t − t0 ) + a2 (t − t0 )2 + · · · = [b0 + b1 (t − t0 ) + b2 (t − t0 )2 + · · ·][d0 + d1 (t − t0 ) + d2 (t − t0 )2 + · · ·]. The coefﬁcients dn can be recursively determined by forming the Cauchy product of the two series on the right and solving the resulting hierarchy of linear equations. We obtain a a0 = b0 d0 and hence d0 = 0 , b0 a1 = b0 d1 + b1 d0

and hence

a2 = b0 d2 + b1 d1 + b2 d0 .. .

d1 =

and hence

a1 − b1 d0 , b0 d2 =

a2 − b1 d1 − b2 d0 , b0

Notice how the statement made in the case of division differs from that made in the previous cases. In particular, even though the series for f (t) and g(t) 1

Augustin Louis Cauchy (1789–1857) was a scientiﬁc giant whose life was enmeshed in the political turmoil of early nineteenth-century France. He contributed to many areas of mathematics and science and is considered to be the founder of the theory of functions of a complex variable. Numerous terms in mathematics bear his name, such as the Cauchy integral theorem, the CauchyRiemann equations, and Cauchy sequences. His collected works, when published, ﬁlled 27 volumes.

520

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Series Solutions of Linear Differential Equations

converge in |t − t0 | < R, it does not necessarily follow that the quotient series also converges in |t − t0 | < R. All we can say in general is that the quotient series converges in some neighborhood of t0 .

A Function Deﬁned by a Power Series Can Be Differentiated Termwise Let f (t) be given by the power series f (t) =

∞

an (t − t0 )n ,

(5)

n=0

which converges in |t − t0 | < R. The function f (t) has derivatives of all orders on the interval t0 − R < t < t0 + R. We can obtain these derivatives by termwise differentiation of the original power series. That is, f (t) = f (t) =

∞

nan (t − t0 )n−1 = a1 + 2a2 (t − t0 ) + 3a3 (t − t0 )2 + · · · ,

n=1 ∞

n(n − 1)an (t − t0 )n−2 = 2a2 + 6a3 (t − t0 ) + · · · ,

n=2

and so forth. Each of these derived series also converges absolutely in the interval t0 − R < t < t0 + R. The derived series can be used to express the coefﬁcient an in terms of the nth derivative of f (t) evaluated at t = t0 . In particular, by evaluating the derived series at t = t0 , we see that f (t0 ) = a0 , f (t0 ) = a1 , f (t0 ) = (2 · 1)a2 , f (t0 ) = (3 · 2 · 1)a3 , . . . . In general, for f (t) given by (5), f (n) (t0 ) = n!an ,

n = 0, 1, 2, . . . .

Some Functions Are Deﬁned by a Taylor Series If f (t) is a function deﬁned by the power series (5), then f (n) (t0 ) = n!an , n = 0, 1, 2, . . . . Conversely, if we are given a function f (t) that is deﬁned and inﬁnitely differentiable on an interval t0 − R < t < t0 + R, then we can associate f (t) with its formal Taylor series ∞ f (n) (t0 ) (t − t0 )n . n! n=0

(6)

Recall from calculus that the Taylor series for f (t) need not necessarily converge to f (t). However, for most of the functions considered in this chapter, the Taylor series converges to f (t), so f (t) =

∞ f (n) (t0 ) (t − t0 )n , n! n=0

t0 − R < t < t0 + R.

If t0 = 0, the Taylor series is usually referred to as a Maclaurin series. For later reference, we list the Maclaurin series for several functions. We also give the

8.1

Introduction

521

interval of absolute convergence for the series. et =

∞ tn n=0

sin t =

=1+t+

n!

∞

∞

−∞ < t < ∞

(−1)n

t3 t5 t7 t2n+1 =t− + − + ···, (2n + 1)! 3! 5! 7!

(−1)n

t2 t4 t6 t2n =1− + − + ···, (2n)! 2! 4! 6!

n=0

cos t =

t2 t3 t4 + + + ···, 2! 3! 4!

n=0

(7a)

−∞ < t < ∞

−∞ < t < ∞

(7b)

(7c)

∞

1 t n = 1 + t + t 2 + t3 + t4 + · · · , = 1 − t n=0 ln(1 + t) =

∞

∞

(−1)n

n=0

(7d)

t2 t3 t4 tn =t− + − + ···, n 2 3 4

−1 < t < 1

(7e)

t3 t5 t7 t2n+1 = t − + − + ···, 2n + 1 3 5 7

−1 < t < 1

(7f )

(−1)n−1

n=1

tan−1 t =

−1 < t < 1

Note that these basic series can be used to ﬁnd the Taylor series of certain simple compositions. For example, by (7d), ∞

1 1 (t − 2)n , = = 3−t 1 − (t − 2) n=0

|t − 2| < 1

and 1 1 + 4t 2

=

∞

(−1)n 4n t2n ,

n=0

|t|

0. For later use, we also recall that If f (t) and g(t) are analytic at t0 , then the functions f (t) ± g(t) and f (t)g(t) are also analytic at t0 . Furthermore, the quotient f (t)/g(t) is analytic at t0 if g(t0 ) = 0. Polynomial functions are analytic at all points. Rational functions are analytic at all points where the denominator polynomial is nonzero. (When discussing rational functions, we assume the denominator and numerator have no factors in common.) If the denominator is nonzero at t0 , the radius of convergence is equal to the distance from t0 to the nearest zero (either real or complex) of the denominator. As we will show, the general solution of y + p(t)y + q(t)y = 0 can be expressed in terms of power series that converge in a neighborhood of t0 whenever both p(t) and q(t) are analytic at t0 . The point t0 is called an ordinary point when both p(t) and q(t) are analytic at t0 . If p(t) and/or q(t) is not analytic at t0 , t0 is called a singular point. As noted in Section 8.1, if a function f (t) is deﬁned by a power series, n f (t) = ∞ n=0 an (t − t0 ) , and if this power series has radius of convergence R > 0, then f (t) has derivatives of all orders at t = t0 . Therefore, if some derivative of f (t) fails to exist at a point t0 , then f (t) cannot be analytic at t0 . To help identify ordinary points, we can use some facts noted earlier: Sums, differences, and products of functions analytic at t0 are again analytic at t0 . Quotients of

528

CHAPTER 8

Series Solutions of Linear Differential Equations

functions analytic at t0 are also analytic at t0 if the denominator function is nonzero at t0 . E X A M P L E

1

Consider the differential equation (1 − t 2 )y + (tan t)y + t5/3 y = 0 in the open interval −2 < t < 2. Classify each point in this interval as an ordinary point or a singular point. Solution: We ﬁrst rewrite the equation in the form (1), y +

tan t 1 − t2

t5/3

y +

1 − t2

y = 0.

Therefore, the coefﬁcient functions p(t) and q(t) are p(t) =

tan t 1−t

2

and

q(t) =

t5/3 1 − t2

.

The function p(t) fails to be analytic at the points t = ±1 (where the denominator vanishes) and t = ±π/2 (where the graph of y = tan t has vertical asymptotes). The function q(t) is not analytic at t = ±1 or at t = 0. (The numerator function t5/3 is not analytic at t = 0; it is continuous and has a continuous ﬁrst derivative at t = 0, but its second derivative does not exist at t = 0.) Thus, in the interval −2 < t < 2, the ﬁve points t = 0, ±1, ± π/2 are singular points and all other points are ordinary points. ❖

Series Solutions Near an Ordinary Point Theorem 8.1 shows that, in a neighborhood of an ordinary point, we can represent the general solution of equation (1) in terms of convergent power series. The proof of Theorem 8.1 is given in more advanced texts, such as Birkhoff and Rota.2 Theorem 8.1

Let p(t) and q(t) be analytic at t0 , and let R denote the smaller of the two radii of convergence of their respective Taylor series representations. Then the initial value problem y + p(t)y + q(t)y = 0,

y(t0 ) = y0 ,

y (t0 ) = y0

(2)

has a unique solution that is analytic in the interval |t − t0 | < R.

According to Theorem 8.1, if t0 is an ordinary point, then initial value problem (2) has a power series solution of the form y(t) =

∞

an (t − t0 )n = a0 + a1 (t − t0 ) + a2 (t − t0 )2 + · · · .

n=0 2

Garrett Birkhoff and Gian-Carlo Rota, Ordinary Differential Equations, 4th ed. (New York: Wiley, 1989).

8.2

Series Solutions Near an Ordinary Point

529

Note that the ﬁrst two coefﬁcients, a0 and a1 , are determined by the initial conditions in (2), since y(t0 ) = a0 and y (t0 ) = a1 . Theorem 8.1 assures us that the power series we obtain by solving the recurrence relation for the remaining coefﬁcients a2 , a3 , . . . converges in the interval |t − t0 | < R, where R is the smaller of the radii of convergence of the coefﬁcient functions p(t) and q(t). Note that Theorem 8.1 does not rule out the possibility that the power series for y(t) may converge on a larger interval. This happens in some cases. The following corollary is a consequence of Theorem 8.1.

Corollary

Let p(t) and q(t) be analytic at t0 , and let R denote the smaller of the two radii of convergence of their respective Taylor series representations. The general solution of the differential equation y + p(t)y + q(t)y = 0

(3)

can be expressed as y(t) =

∞

an (t − t0 )n = a0 y1 (t) + a1 y2 (t),

n=0

where the constants a0 and a1 are arbitrary. The functions y1 (t) and y2 (t) form a fundamental set of solutions, analytic in the interval |t − t0 | < R.

The solutions y1 (t) and y2 (t) forming the fundamental set can be obtained by adopting the particular initial conditions y1 (t0 ) = 1, y1 (t0 ) = 0 and y2 (t0 ) = 0, y2 (t0 ) = 1.

E X A M P L E

2

Consider the initial value problem y +

t+1 t3 + t

y +

1 t 2 − 4t + 5

y = 0,

y(2) = y0 ,

y (2) = y0 .

n If y(t) = ∞ n=0 an (t − 2) is the solution, determine a lower bound for the radius of convergence R of this series. Solution: Since t3 + t = t(t 2 + 1), the coefﬁcient function p(t) has denominator zeros at t = 0, t = −i, and t = i. Likewise, the coefﬁcient function q(t) has denominator zeros at t = 2 ± i. The radius of convergence Rp of the Taylor deseries expansion for p(t) is equal to the distance from t0 = 2 to the nearest √ nominator zero; that is, Rp is the smaller of |2 ± 0| = 2 and |2 − i| = 5. Thus, Rp = 2. (See Figure 8.1.) Similarly, the radius of convergence of the Taylor series for q(t) is Rq = |2 − (2 ± i)| = 1. Thus, by Theorem 8.1, the radius of convergence of the Taylor series for y(t) is guaranteed to be no smaller than R = 1. (continued)

530

CHAPTER 8

Series Solutions of Linear Differential Equations (continued)

2+i

i

0

2

2

–i

2–i

FIGURE 8.1

The radius of convergence of the expansion of p(t) is equal to the distance from t0 = 2 to the nearest of the denominator zeros, 0, i, and −i. For q(t), the two nearest denominator zeros, 2 + i and 2 − i, are equidistant from t0 = 2. The radius of convergence for the series in Example 2 is R = 1.

❖

When the coefﬁcient functions of y + p(t)y + q(t)y = 0 possess certain symmetries, some useful observations can be made; see Theorem 8.2.

Theorem 8.2

Consider the differential equation y + p(t)y + q(t)y = 0. (a) Let p(t) be a continuous odd function deﬁned on the domain (−b, −a) ∪ (a, b), where a ≥ 0. Let q(t) be a continuous even function deﬁned on the same domain. If f (t) is a solution of the differential equation on the interval a < t < b, then f (−t) is a solution on the interval (−b, −a). (b) Let the coefﬁcient functions p(t) and q(t) be analytic at t = 0 with a common radius of convergence R > 0. Let p(t) be an odd function and q(t) an even function. Then the differential equation has even and odd solutions that are analytic at t = 0 with radius of convergence R.

Recall the deﬁnitions of even and odd functions. We are assuming that p(−t) = −p(t) and q(−t) = q(t) for all t in (−b, −a) ∪ (a, b). The proof of Theorem 8.2 is outlined in Exercises 31–32. In Example 2 of Section 8.1, we obtained a power series solution of y + ω2 y = 0 and observed that the ratio test could be used to show that each of the two series forming the general solution has an inﬁnite radius of convergence. This fact is also an easy consequence of Theorem 8.1, since the differential equation y + ω2 y = 0 has coefﬁcient functions p(t) = 0 and q(t) = ω2 that are analytic on −∞ < t < ∞. Moreover, since p(t) is an odd function and q(t) is an even function, it follows from Theorem 8.2 that even and odd solutions of this differential equation exist; they are cos ωt and sin ωt, respectively.

8.2

Series Solutions Near an Ordinary Point

531

Polynomial Solutions Some of the second order linear differential equations that arise in mathematical and scientiﬁc applications (such as the Legendre equation, the Hermite equation, and the Chebyshev equation) have polynomial solutions. The next example treats the Chebyshev equation. Other equations are considered in the Exercises.

E X A M P L E

3

The Chebyshev3 differential equation is (1 − t 2 )y − ty + μ2 y = 0,

(4)

where μ is a constant. Find a Maclaurin series solution of (4). Show that if μ is an integer, the Chebyshev differential equation (4) has a polynomial solution. Solution:

Rewriting the equation as y −

t 1−t

y + 2

μ2 1 − t2

y = 0,

we see that t = ±1 are singular points. All other points are ordinary points. In addition, we deduce from the structure of the differential equation itself that it possesses solutions with even symmetry and solutions with odd symmetry (see Theorem 8.2). By Theorem 8.1, the general solution of the Chebyshev equation can be represented of Maclaurin series that we know will converge in (−1, 1). Let in terms n a t . Substitution into differential equation (4) leads to y(t) = ∞ n=0 n (1 − t 2 )

∞

an n(n − 1)t n−2 − t

n=2

∞ n=1

an nt n−1 + μ2

∞

an t n = 0.

(5)

n=0

Equation (5) can be rewritten as ∞

an n(n − 1)t n−2 −

n=2

∞

[an n(n − 1) + nan − μ2 an ] t n = 0,

n=0

or, after adjusting the index in the ﬁrst summation and collecting terms, ∞

[an+2 (n + 2)(n + 1) − an (n2 − μ2 )] t n = 0.

n=0

The recurrence relation is therefore an+2 =

n2 − μ2 a , (n + 1)(n + 2) n

n = 0, 1, 2, . . . .

(6)

The recurrence relation determines all the even-indexed coefﬁcients to be (continued) 3

Pafnuty Lvovich Chebyshev (1821–1894) was appointed to the University of St. Petersburg in 1847. He contributed to many areas of mathematics and science, including number theory, mechanics, probability theory, special functions, and the calculation of geometric volumes.

532

CHAPTER 8

Series Solutions of Linear Differential Equations (continued)

multiples of a0 and all the odd-indexed coefﬁcients to be multiples of a1 . The general solution is

−μ2 2 −(4 − μ2 )μ2 4 t + t + ··· y(t) = a0 1 + 2 24 + a1

2

t+

2

2

(7)

1 − μ 3 (1 − μ ) (9 − μ ) 5 t + t + ··· . 6 120

The ratio test, in conjunction with recurrence relation (6), can be used to show that the power series in (7) have radius of convergence R = 1 (except in the case when they terminate after a ﬁnite number of terms). If μ is an even integer, we see from (6) that all the even coefﬁcients having index greater than μ will vanish. For example, if μ = 4, then a6 = 0, a8 = 0, . . . . Thus, when μ = 4, recurrence relation (6) leads to a2 = −8a0 ,

a4 = −a2 = 8a0 ,

a6 = a8 = a10 = · · · = 0.

From this, we see that the fourth degree polynomial P(t) = a0 (1 − 8t 2 + 8t4 ) is a solution of the Chebyshev equation. Similarly, if μ is an odd positive integer, then we obtain an odd polynomial solution of degree μ. These polynomial solutions, generated as μ ranges over the nonnegative integers, are known as Chebyshev polynomials of the ﬁrst kind. The Nth degree Chebyshev polynomial of the ﬁrst kind is usually denoted TN (t). The ﬁrst few Chebyshev polynomials are T0 (t) = 1,

T1 (t) = t,

T2 (t) = 2t 2 − 1,

T3 (t) = 4t3 − 3t,

T4 (t) = 8t4 − 8t 2 + 1. The Chebyshev polynomials are normalized; that is, the arbitrary constant is selected so that TN (1) = 1. ❖ REMARKS: 1. Even though the differential equation has singular points at t = ±1, the Chebyshev polynomial solutions are well behaved at these points. The polynomial solutions are analytic with inﬁnite radius of convergence. It is important to remember that solutions need not necessarily behave badly at singular points. 2. Chebyshev polynomials ﬁnd important application in the design of antenna arrays and electrical ﬁlters. Consider, for example, the low-pass ﬁltering problem illustrated in Figure 8.2. We want to build an electrical network having the power transfer function shown in Figure 8.2(a). Energy at all frequencies less than the cutoff frequency fc should pass through the network unscathed, while the passage of energy at all frequencies above fc should be completely blocked. The problem, however, is that the network elements we have available to build the network only allow us to realize power transfer functions that are rational functions of frequency. The design problem is to ﬁnd a rational function that closely approximates the ideal behavior in Figure 8.2(a).

8.2

Series Solutions Near an Ordinary Point

y

533

y 1

1

f/fc

1

0.5

(a)

1

1.5

2

f/fc

(b) FIGURE 8.2

(a) The graph of an ideal power transfer function. (b) The graph of an approximation of the form (8) to the function graphed in (a).

Chebyshev polynomials are particularly suited for such problems because they possess an “equal ripple” property (see Exercise 27). The polynomial TN (t) oscillates between ±1 for t in the range −1 ≤ t ≤ 1 and grows monotonically in magnitude when |t| ≥ 1. Therefore, a power transfer function of the form 1 1+

ε TN2

, f fc

(8)

with positive constant ε sufﬁciently small and integer N sufﬁciently large, can serve as a good rational approximation to the ideal power transfer function. Figure 8.2(b) illustrates the particular choice ε = 0.02, N = 12.

EXERCISES Exercises 1–6: Identify all the singular points of y + p(t)y + q(t)y = 0 in the interval −10 < t < 10. t 2. y + t2/3 y + (sin t)y = 0 y=0 1. y + (sec t)y + 2 t −4 t 3. (1 − t 2 )y + ty + (csc t)y = 0 y=0 4. (sin 2t)y + et y + 25 − t 2 t 5. (1 + ln|t|)y + y + (1 + t 2 )y = 0 y + (tan t)y = 0 6. y + 1 + |t|

Exercises 7–12: In each exercise, t = t0 is an ordinary point of y + p(t)y + q(t)y = 0. Apply Theorem 8.1 to determine a value R > 0 such that an initial value problem, with the initial conditions prescribed at t0 , is guaranteed to have a unique solution that is analytic in the interval t0 − R < t < t0 + R. 7. y +

1 t y + y = 0, t0 = 0 1 + 2t 1 − t2

8. (1 − 9t 2 )y + 4y + ty = 0, t0 = 1 1 3t y + y = 0, t0 = −1 4 − 3t 5 + 30t 1 t y = 0, t0 = 0 y + 10. y + 4+t 1 + 4t 2 9. y +

534

CHAPTER 8

Series Solutions of Linear Differential Equations 1 y + (sin t)y = 0, t0 = 2 1 + 3(t − 2) t+3 y + t 2 y = 0, t0 = 1 12. y + 1 + t2 11. y +

Exercises 13–21: In each exercise, t = 0 is an ordinary point of y + p(t)y + q(t)y = 0. (a) Find the relation that deﬁnes the coefﬁcients of the power series solution recurrence n y(t) = ∞ a t . n=0 n (b) As in equation (7), ﬁnd the ﬁrst three nonzero terms in each of two linearly independent solutions. (c) State the interval −R < t < R on which Theorem 8.1 guarantees convergence. (d) Does Theorem 8.2 indicate that the differential equation has solutions that are even and odd? 13. y + ty + y = 0

14. y + 2ty + 3y = 0

15. (1 + t 2 )y + ty + 2y = 0

16. y − 5y + 6y = 0

17. y − 4y + 4y = 0

18. (1 + t)y + y = 0

2

19. (3 + t)y + 3ty + y = 0 20. (2 + t )y + 4y = 0

21. y + t 2 y = 0

Exercises 22–25: In each exercise, t = 1 is an ordinary point of y + p(t)y + q(t)y = 0. (a) Find the that deﬁnes the coefﬁcients of the power series solution recurrence relation n y(t) = ∞ n=0 an (t − 1) . (b) Find the ﬁrst three nonzero terms in each of two linearly independent solutions. (c) State the interval −R < t − 1 < R on which Theorem 8.1 guarantees convergence. 22. y + (t − 1)y + y = 0

23. y + y = 0

24. (t − 2)y + y + y = 0

25. y + y + (t − 2)y = 0

26. Recall Chebyshev’s equation from Example 3, (1 − t 2 )y − ty + μ2 y = 0. As you saw in Example 3, this equation has a polynomial solution, TN (t), when μ = N is a nonnegative integer. Using recurrence relation (6), ﬁnd T5 (t) and T6 (t). 27. The Equal Ripple Property of Chebyshev Polynomials Consider the Chebyshev differential equation (1 − t 2 )y − ty + N 2 y = 0, where N is a nonnegative integer. (a) Show by substitution that the function y(t) = cos(N arccos t) is a solution for −1 < t < 1. (b) Show, for N = 0, 1, 2, that the function cos(N arccos t) is a polynomial in t and that TN (t) = cos(N arccos t). This result holds, in fact, for all nonnegative integers N. It can be shown that cos(N arccos t), −1 ≤ t ≤ 1 TN (t) = cosh(N arccosh t), 1 < |t|. (c) Use a computer graphics package to plot TN (t) for N = 2, 5, and 8 and for −1.2 ≤ t ≤ 1.2. (d) What serves as a bound for |TN (t)| when −1 ≤ t ≤ 1? What is the behavior of |TN (t)| when 1 < |t|? 28. Legendre’s Equation Legendre’s equation is (1 − t 2 )y − 2ty + μ(μ + 1)y = 0. Byn Theorem 8.1, this equation has a power series solution of the form y(t) = ∞ n=0 an t that is guaranteed to be absolutely convergent in the interval −1 < t < 1. As in Example 3,

8.2

Series Solutions Near an Ordinary Point

535

(a) Find the recurrence relation for the coefﬁcients of the power series. (b) Argue, when μ = N is a nonnegative integer, that Legendre’s equation has a polynomial solution, PN (t). (c) Show, by direct substitution, that the Legendre polynomials P0 (t) = 1 and P1 (t) = t satisfy Legendre’s equation for μ = 0 and μ = 1, respectively. (d) Use the recurrence relation and the requirement that Pn (1) = 1 to determine the next four Legendre polynomials, P2 (t), P3 (t), P4 (t), P5 (t). Theorem 8.1, this 29. Hermite’s Equation Hermite’s equation is y − 2ty + 2μy = 0. By n equation has a power series solution of the form y(t) = ∞ n=0 an t that is guaranteed to be absolutely convergent in the interval −∞ < t < ∞. As in Example 3, (a) Find the recurrence relation for the coefﬁcients of the power series. (b) Argue that, when μ = N is a nonnegative integer, Hermite’s equation has a polynomial solution, HN (t). (c) Show, by direct substitution, that the Hermite polynomials H0 (t) = 1 and H1 (t) = 2t satisfy Hermite’s equation for μ = 0 and μ = 1, respectively. (d) Use the recurrence relation and the requirement that Hn (t) = 2n t n + · · · to determine the next four Hermite polynomials, H2 (t), H3 (t), H4 (t), H5 (t). 0. In some cases, we may be 30. Consider the differential equation y + p(t)y + q(t)y = n able to ﬁnd a power series solution of the form y(t) = ∞ n=0 an (t − t0 ) even when t0 is not an ordinary point. In other cases, there is no power series solution. (a) The point t = 0 is a singular point ∞of ty n+ y − y = 0. Nevertheless, ﬁnd a nontrivial power series solution, y(t) = n=0 an t , of this equation. 2 (b) The point t = 0 is a singular point of t y + ny = 0. Show that the only solution of this equation having the form y(t) = ∞ n=0 an t is the trivial solution.

Exercises 31 and 32 outline the proof of parts (a) and (b) of Theorem 8.2, respectively. In each exercise, consider the differential equation y + p(t)y + q(t)y = 0, where p and q are continuous on the domain (−b, −a) ∪ (a, b), a ≥ 0. 31. Let f (t) be a solution on the interval (a, b). (a) Let t lie in the interval (−b, −a) and set τ = −t, so that a < τ < b. Show that if d2 f (τ ) df (τ ) + q(τ )f (τ ) = 0, + p(τ ) dτ dτ 2

a < τ < b,

then df (−t) d2 f (−t) + q(−t)f (−t) = 0, − p(−t) 2 dt dt

−b < t < −a.

(b) Use the fact that p and q are odd and even functions, respectively, to show that f (−t) is a solution of the given differential equation on the interval −b < t < −a. 32. Now let p and q be analytic at t = 0 with a common radius of convergence R > 0, where p is an odd function and q is an even function. (a) Let f1 (t) and f2 (t) be solutions of the given differential equation, satisfying initial conditions f1 (0) = 1, f1 (0) = 0, f2 (0) = 0, f2 (0) = 1. What does Theorem 8.1 say about the solutions f1 (t) and f2 (t)? (b) Use the results of Exercise 31 to show that f1 (−t) and f2 (−t) are also solutions on the interval −R < t < R. (c) Form the functions fe (t) = [ f1 (t) + f1 (−t)]/2 and fo (t) = [ f2 (t) − f2 (−t)]/2. Show that fe (t) and fo (t) are even and odd analytic solutions, respectively, on the interval −R < t < R.

536

CHAPTER 8

Series Solutions of Linear Differential Equations (d) Show that fe (t) and fo (t) are nontrivial solutions by showing that fe (t) = f1 (t) and fo (t) = f2 (t). [Hint: Use the fact that solutions of initial value problems are unique.] (e) Show that the solutions fe (t) and fo (t) form a fundamental set of solutions.

Exercises 33–38: Suppose a linear differential equation y + p(t)y + q(t)y = 0 satisﬁes the hypotheses of Theorem 8.2(b), on the interval −∞ < t < ∞. Then, by Exercise 32, we can assume the general solution of y + p(t)y + q(t)y = 0 has the form y(t) = c1 ye (t) + c2 yo (t),

(9)

where ye (t) is an even solution of y + p(t)y + q(t)y = 0 and yo (t) is an odd solution. In each of the following exercises, determine whether Theorem 8.2(b) can be used to guarantee that the given differential equation has a general solution of the form (9). If your answer is no, explain why the equation fails to satisfy the hypotheses of Theorem 8.2(b). 33. y + (sin t)y + t 2 y = 0

34. y + (cos t)y + ty = 0

35. y + t 2 y = 0

36. y + y + t 2 y = 0

37. y + ty = 0

38. y + et y + y = 0

39. Consider the differential equation y + ay + by = 0, where a and b are constants. For what values of a and b will the differential equation have nontrivial solutions that are odd and even? 40. Consider the initial value problem (1 + t 2 )y + y = 0, y(0) = 1, y (0) = 0. (a) Show, by Theorem 8.1, that this initialnvalue problem is guaranteed to have a unique solution of the form y(t) = ∞ n=0 an t , where the series converges absolutely in −1 < t < 1. (b) The recurrence relation has the form r(n)an+2 = s(n)an , where r(n) and s(n) are quadratic polynomials. Therefore, an+2 s(n) = a r(n) . n Show that this ratio tends to 1 as n tends to ∞, and conclude, therefore, that the series solution of the initial value problem diverges for 1 < |t|. (c) Note, however, that the coefﬁcient functions p(t) = 0 and q(t) = (1 + t 2 )−1 are continuous for −∞ < t < ∞, and hence, by Theorem 3.1, a unique solution exists for all values t. Do the conclusions of Theorem 8.1 and Theorem 3.1 contradict each other? Explain.

8.3

The Euler Equation The second order linear homogeneous equation t 2 y + αty + βy = 0,

t = 0

(1)

is known as the Euler equation (it is also known as the Cauchy-Euler equation or the equidimensional equation). In equation (1), α and β are real constants. Note the special structure of the differential equation; the power of each monomial coefﬁcient matches the order of the derivative that it multiplies—for example, t 2 multiplies y . This special structure enables us to obtain an explicit representation for the general solution. In this section, we present two related approaches to deriving the general solution.

8.3

The Euler Equation

537

If we rewrite the Euler equation in the form y + p(t)y + q(t)y = 0, then p(t) =

α t

q(t) =

and

β t2

.

The coefﬁcients p(t) and q(t) are analytic at every point except t = 0. Ignoring the trivial case where α and β are both zero, we see that t = 0 is a singular point and all other values of t are ordinary points. Note as well that p(t) and q(t) are not continuous at t = 0. Therefore, the basic existence-uniqueness result, Theorem 3.1, alerts us to possible problems—solutions may or may not exist at t = 0. The Euler equation arises in a variety of applications. An example is the problem of determining the time-independent (or steady-state) temperature distribution within a circular geometry (such as the interior of a circular pipe) from a knowledge of the temperature on the boundary; see the Projects at the end of this chapter. The Euler equation is also of interest because it serves as a prototype or model equation. In Section 8.4, we will deﬁne a special type of singular point, called a regular singular point; our understanding of the Euler equation will serve as the basis for studying regular singular points.

The General Solution of the Euler Equation If f (t) is a solution of the Euler equation (1), then so is f (−t) (see Theorem 8.2 in Section 8.2). Therefore, we assume our interval of interest is t > 0. Once the general solution is obtained for the interval t > 0, we can obtain the general solution in t < 0 by replacing t with −t. We present two approaches to deriving the general solution of (1). In retrospect, you will see that the two approaches are closely related. Nevertheless, both points of view are worthy of consideration.

Solutions of the Form y(t) = t λ The special structure of the Euler equation makes it possible to ﬁnd solutions of the form y(t) = t λ , where λ is a constant to be determined and where t > 0 is assumed. Substitution of this trial form into equation (1) leads to t 2 λ(λ − 1)t λ−2 + αtλt λ−1 + βt λ = [λ(λ − 1) + αλ + β]t λ = 0,

t > 0.

(2)

Since t λ is not identically zero on 0 < t < ∞ for any real or complex value λ, we must have λ2 + (α − 1)λ + β = 0, or λ1,2 =

−(α − 1) ±

(α − 1)2 − 4β

2

(3)

.

(4)

In other words, y(t) = t λ is a solution of the Euler equation whenever λ is a root of equation (3). We refer to equation (3) as the characteristic equation or the indicial equation. There are three possibilities for the roots of the characteristic equation:

538

CHAPTER 8

Series Solutions of Linear Differential Equations

1. There are two real distinct roots if the discriminant, (α − 1)2 − 4β, is positive. In this case, the Wronskian of the solution set is λ t 1 t λ2 W(t) = λ1 −1 = (λ2 − λ1 )t λ1 +λ2 −1 . λ2 −1 λ1 t λ2 t Since W(t) is nonzero for t > 0, the functions y1 (t) = t λ1 and y2 (t) = t λ2 form a fundamental set. The general solution of equation (1) is y(t) = c1 t λ1 + c2 t λ2 , t > 0, where c1 and c2 are arbitrary constants. As previously mentioned, we can obtain the general solution for t < 0 by replacing t by −t; the general solution for t < 0 is y(t) = c1 (−t)λ1 + c2 (−t)λ2 , where c1 and c2 again denote arbitrary constants. Since −t > 0 when t < 0, the general solution can be expressed in the form y(t) = c1 |t|λ1 + c2 |t|λ2 ,

t = 0.

(5)

2. There is one real repeated root if (α − 1)2 − 4β = 0. In this case, one solution is y1 (t) = t λ , t > 0, where λ = −(α − 1)/2. We can use reduction of order (see Section 3.4) to ﬁnd a second linearly independent solution, y2 (t) = t λ ln t, t > 0 (see Exercise 1 of this section). The general solution in the repeated root case is therefore y(t) = c1 |t|λ + c2 |t|λ ln|t|,

t = 0.

(6)

3. There are complex conjugate roots if (α − 1)2 − 4β < 0. For brevity, let −(α − 1) ± i 4β − (α − 1)2 λ1,2 = = γ ± iδ. 2 In this case, tγ +iδ = e(γ +iδ) ln t = eγ ln t eiδ ln t = tγ eiδ ln t ,

t > 0.

Hence, by Euler’s formula (see Section 3.5), tγ +iδ = tγ [cos(δ ln t) + i sin(δ ln t)],

t > 0.

(7)

From equation (7), we obtain the two real-valued solutions y1(t) = tγ cos(δ ln t) and y2 (t) = tγ sin(δ ln t), which can be shown to form a fundamental set on 0 < t < ∞ (see Exercise 2). Therefore, the general solution is y(t) = c1 |t|γ cos(δ ln|t|) + c2 |t|γ sin(δ ln|t|),

t = 0.

(8)

E X A M P L E

1

Find the general solution of each of the Euler equations (a) t 2 y − 2ty + 2y = 0

(b) t 2 y + 5ty + 4y = 0

(c) t 2 y + 3ty + 5y = 0

Solution: (a) For this equation, α = −2 and β = 2. Looking for solutions of the form t λ , t > 0 leads to λ2 − 3λ + 2 = (λ − 1)(λ − 2) = 0. The general solution is therefore y(t) = c1 t + c2 t 2 ,

−∞ < t < ∞.

(9)

Since the general solution involves integer powers of t, we can dispense with the absolute value signs. Note further that solution (9) is deﬁned for

8.3

The Euler Equation

539

all values of t, including the singular point t = 0. For this equation, every solution is well behaved at t = 0. (b) For this differential equation, α = 5 and β = 4. The characteristic equation λ2 + 4λ + 4 = 0 has real repeated roots λ1 = λ2 = −2. The general solution is y(t) = c1 t−2 + c2 t−2 ln|t|,

t = 0.

(c) For this equation, α = 3 and β = 5. The characteristic equation λ2 + 2λ + 5 = 0 has complex conjugate roots λ1,2 = −1 ± 2i. The general solution is y(t) = c1 t−1 cos(2 ln|t|) + c2 t−1 sin(2 ln|t|),

t = 0.

Neither (b) nor (c) has solutions that exist at t = 0. ❖

Change of Independent Variable After a change of independent variable, the Euler equation can be transformed into a new homogeneous constant coefﬁcient equation. We begin with the Euler equation t 2 y + αty + βy = 0 for t > 0 and introduce a new independent variable z = ln t, or (equivalently) t = e z . The t-interval 0 < t < ∞ transforms into the z-interval −∞ < z < ∞. Let y(t) = y(e z ) = Y (z). Then, using the chain rule, we have dy(t) dy(e z ) dz dY (z) dt −1 dY (z) = = . = e−z dt dz dt dz dz dz Therefore, t (dy(t)/dt) transforms into e z e−z

dY (z) dY (z) = , dz dz

(10)

while t 2 (d2 y(t)/dt 2 ) transforms into d2 Y (z) dz

2

−

dY (z) . dz

(11)

Under this change of independent variable, the Euler equation (1) transforms into the constant coefﬁcient equation Y + (α − 1)Y + βY = 0.

(12)

Therefore, we can solve the Euler equation (1) for t > 0 by 1. making the change of independent variable t = e z , 2. solving the constant coefﬁcient equation (12) using solution procedures developed in Chapter 3, and then 3. using the inverse map to obtain the desired solution, y(t) = Y (ln t). Looking for solutions of t 2 y + αty + βy = 0 having the form y(t) = t λ is equivalent to looking for solutions of Y + (α − 1)Y + βY = 0 having the form Y (z) = eλz . The characteristic equation (3) results in either case. Each of the two solution approaches has its utility. The ﬁrst approach, looking for solutions of the form |t|λ , serves as a guide in the next section when

540

CHAPTER 8

Series Solutions of Linear Differential Equations

we study the behavior of solutions of differential equations in a neighborhood of certain types of singular points. The change of independent variable approach is useful because it permits us to use the techniques developed in Chapter 3. As the next example shows, the change of variables approach can be applied to nonhomogeneous Euler equations.

E X A M P L E

2

Obtain the general solution of 4t 2 y + 8ty + y = t + ln t,

t>0

by making the change of independent variable t = e z . Solution: Under the change of variable t = e z , the nonhomogeneous term transforms into e z + z. Thus, using (10) and (11) yields for the given differential equation 4(Y − Y ) + 8Y + Y = e z + z, or 4Y + 4Y + Y = e z + z,

(13)

where Y = dY /dz. As seen in Chapter 3, the solution of (13) is the sum of a complementary solution YC (z) and a particular solution YP (z). The complementary solution is YC (z) = c1 e−z/2 + c2 ze−z/2 .

(14)

A particular solution YP (z) can be obtained using the method of undetermined coefﬁcients. We ﬁnd YP (z) = 19 e z + z − 4. The general solution is therefore Y (z) = c1 e−z/2 + c2 ze−z/2 + 19 e z + z − 4. Using z = ln t to convert to the original independent variable t, we have y(t) = c1 t−1/2 + c2 t−1/2 ln t + 19 t + ln t − 4,

t > 0. ❖

Generalizations of the Euler Equation There are two natural ways to generalize the Euler equation. One such generalization is given by (t − t0 )2 y + (t − t0 )αy + βy = 0.

(15)

In equation (15), t0 is a point in the interval −∞ < t < ∞. To solve equation (15) for t > t0 , we assume a solution of the form y(t) = (t − t0 )λ . (We could also adopt the change of independent variable t − t0 = e z .) Another natural generalization is a higher order version of the Euler equation, such as t3 y + αt 2 y + βty + γ y = 0. Examples of each are given in the exercises.

(16)

8.3

The Euler Equation

541

EXERCISES 1. Consider the Euler equation t 2 y − (2α − 1)ty + α 2 y = 0. (a) Show that the characteristic equation has a repeated root λ1 = λ2 = α. One solution is therefore y(t) = t α , t > 0. (b) Use the method of reduction of order (Section 3.4) to obtain a second linearly independent solution for the interval 0 < t < ∞. (c) Compute the Wronskian of the set of solutions {tα , tα ln t}, and show that it is a fundamental set of solutions on 0 < t < ∞. 2. Let y1 (t) = tγ cos(δ ln t) and y2 (t) = tγ sin(δ ln t), t > 0, where δ and γ are real constants with δ = 0. Solutions of this form arise when the characteristic equation has complex roots. Compute the Wronskian of this pair of solutions, and show that they form a fundamental set of solutions on 0 < t < ∞.

Exercises 3–18: Identify the singular point. Find the general solution that is valid for values of t on either side of the singular point. 3. t 2 y − 4ty + 6y = 0

4. t 2 y − 6y = 0

5. t 2 y − 3ty + 4y = 0

6. t 2 y − ty + 5y = 0

7. t 2 y − 3ty + 29y = 0

8. t 2 y − 5ty + 9y = 0

9. t 2 y + ty + 9y = 0 12. y +

11 25 y + 2y=0 t t

10. t 2 y + 3ty + y = 0 13. y +

5 40 y + 2y=0 t t

11. t 2 y + 3ty + 17y = 0 14. t 2 y − 2ty = 0

15. (t − 1)2 y − (t − 1)y − 3y = 0

16. (t − 1)2 y + 3(t − 1)y + 17y = 0

17. (t + 2)2 y + 6(t + 2)y + 6y = 0

18. (t − 2)2 y + (t − 2)y + 4y = 0

Exercises 19–21: A Euler equation (t − t0 )2 y + α(t − t0 )y + βy = 0 is known to have the given general solution. What are the constants t0 , α, and β? 1 , t = −2 19. y(t) = c1 (t + 2) + c2 (t + 2)2 20. y(t) = c1 + c2 ln|t − 1|, t = 1 21. y(t) = c1 t 2 cos( ln|t|) + c2 t 2 sin( ln|t|), t = 0

Exercises 22–23: A nonhomogeneous Euler equation t 2 y + αty + βy = g(t) is known to have the given general solution. Determine the constants α and β and the function g(t). 22. y(t) = c1 t 2 + c2 t−1 + 2t + 1, t > 0

23. y(t) = c1 t 2 + c2 t3 + ln t, t > 0

Exercises 24–29: Find the general solution of the given equation for 0 < t < ∞. [Hint: You can, as in Example 2, use the change of variable t = e z .] 24. t 2 y − 2y = 2

25. t 2 y − ty + y = t−1

26. t 2 y + ty + 9y = 10t

27. t 2 y − 6y = 10t−2 − 6

2

28. t y − 4ty + 6y = 3ln t

29. t 2 y + 8ty + 10y = 36(t + t−1 )

Exercises 30–33: Solve the given initial value problem. What is the interval of existence of the solution? 30. t 2 y − ty − 3y = 8t + 6, y(1) = 1, y (1) = 3

542

CHAPTER 8

Series Solutions of Linear Differential Equations 31. t 2 y − 5ty + 5y = 10, y(1) = 4, y (1) = 6 32. t 2 y + 3ty + y = 8t + 9, y(−1) = 1, y (−1) = 0 33. t 2 y + 3ty + y = 2t−1 , y(1) = −2, y (1) = 1 34. Consider the third order equation t3 y + αt 2 y + βty + γ y = 0, t > 0. Make the change of independent variable t = e z and let Y (z) = y(e z ). Derive the third order constant coefﬁcient equation satisﬁed by Y .

Exercises 35–38: Obtain the general solution of the given differential equation for 0 < t < ∞.

8.4

35. t3 y + 3t 2 y − 3ty = 0

36. t3 y + ty − y = 0

37. t3 y + 3t 2 y + ty = 8t 2 + 12

38. t3 y + 6t 2 y + 7ty + y = 2 + ln t

Solutions Near a Regular Singular Point and the Method of Frobenius We introduced the term singular point in Section 8.2 to denote a point t where at least one of the coefﬁcient functions of y + p(t)y + q(t)y = 0 fails to be analytic. Near a singular point, the possible types of solution behavior are diverse and complicated. In this section, we restrict our attention to a particular type of singular point, one known as a regular singular point. Deﬁnitive statements can be made about the behavior of solutions near regular singular points. Many of the important equations of mathematical physics, such as the Euler equation, Bessel’s equation, and Legendre’s equation, possess regular singular points.

Regular Singular Points We begin with the Euler equation t 2 y + αty + βy = 0, where α and β are constants, not both zero. Since the coefﬁcient functions are p(t) =

α t

q(t) =

β

, t2 t = 0 is a singular point and all other points are ordinary points. The Euler equation serves as a model for a differential equation having a regular singular point. In general, let t0 be a singular point of the differential equation and

y + p(t)y + q(t)y = 0. The point t0 is a regular singular point if both of the functions (t − t0 )p(t)

and

(t − t0 )2 q(t)

are analytic at t0 . A singular point that is not a regular singular point is called an irregular singular point. If (t − t0 )p(t) and/or (t − t0 )2 q(t) is indeterminate at t = t0 but the limits lim t→t (t − t0 )p(t) and lim t→t (t − t0 )2 q(t) exist, then we are tacitly assuming 0 0 that the functions being considered are deﬁned to equal these limits at t = t0 . In many important cases, the functions p(t) and q(t) are rational functions (that

8.4

Solutions Near a Regular Singular Point and the Method of Frobenius

543

is, ratios of polynomials). As previously noted, a rational function is analytic at every point where the denominator polynomial is nonzero. In this case, to show that a singular point t0 is a regular singular point, it sufﬁces to show that the limits lim t→t (t − t0 )p(t) and lim t→t (t − t0 )2 q(t) both exist. 0

0

E X A M P L E

1

For each differential equation, identify the singular points and classify them as regular or irregular singular points. In these equations, ν and μ are constants. t 2 y + ty + (t 2 − ν 2 )y = 0

(a) Bessel’s equation4 :

(b) Legendre’s equation5 : (1 − t 2 )y − 2ty + [ν(ν + 1) − μ2 (1 − t 2 )−1 ] y = 0 π (c) (t − 1)y + tan t y + t5/3 y = 0, −2 < t < 2 2 Solution: (a) For Bessel’s equation, p(t) =

1 t

q(t) =

t2 − ν2

. t2 Therefore, t = 0 is a singular point. All other points are ordinary points. Because p(t) and q(t) are rational functions, we need only determine whether lim t→0 tp(t) and lim t→0 t 2 q(t) exist in order to establish analyticity. Since and

lim tp(t) = 1 t→0

and

lim t 2 q(t) = −ν 2 , t→0

it follows that t = 0 is a regular singular point. (b) For Legendre’s equation, p(t) =

−2t (1 − t 2 )

and

q(t) =

ν(ν + 1) (1 − t 2 )

−

μ2 (1 − t 2 )2

.

Therefore, the points t = ±1 are singular points. All other points are ordinary points. We ﬁrst consider the singular point t = 1. Since p(t) and q(t) are rational functions, we check the two limits lim (t − 1)p(t) = lim t→1

t→1

2t =1 1+t

and lim (t − 1)2 q(t) = lim t→1

t→1

ν(ν + 1)(1 − t 2 ) − μ2 (1 + t)2

=

−μ2 . 4 (continued)

4

Friedrich Wilhelm Bessel (1784–1846) was a German scientist noted for important contributions to the ﬁelds of astronomy, celestial mechanics, and mathematics. His mathematical analysis of what is now known as the Bessel function arose during his studies of planetary motion. Bessel’s achievements seem even more remarkable when one realizes that his formal education ended at age 14. 5 Adrien-Marie Legendre (1752–1833) was a French scientist whose research interests included projectile dynamics, celestial mechanics, number theory, and analysis. What are today called Legendre polynomials appeared in a 1784 paper on celestial mechanics. Legendre authored inﬂuential textbooks on Euclidean geometry and number theory.

544

CHAPTER 8

Series Solutions of Linear Differential Equations (continued)

Since both limits exist, t = 1 is a regular singular point. The analysis of the other singular point, t = −1, is very similar, and we ﬁnd that t = −1 is also a regular singular point. (c) We have π π sin t tan t t5/3 2 = 2 and q(t) = . p(t) = π t−1 t−1 (t − 1) cos t 2 Note that p(t), viewed as p(t) = sin(π t/2)/[(t − 1) cos(π t/2)], is a quotient of functions analytic in −2 < t < 2. Therefore, p(t) is analytic at all points in the interval −2 < t < 2 except t = 1 (where the denominator is zero). The function q(t) fails to be analytic at t = 0 (where t5/3 is not analytic) and t = 1 (where the denominator is zero). Consider ﬁrst the singular point t = 0; t 2 q(t) = (t − 1)−1 t11/3 is not analytic at t = 0, and therefore t = 0 is an irregular singular point. Next, consider the singular point t = 1. In this case, p(t) and q(t) are both quotients of functions analytic at t = 1. Therefore, t = 1 is a regular singular point if both of the limits exist: lim (t − 1)p(t) and t→1

lim (t − 1)2 q(t). t→1

The ﬁrst limit, π t, 2 does not exist. Therefore, t = 1 is also an irregular singular point. ❖ lim (t − 1)p(t) = lim tan t→1

t→1

The Method of Frobenius Just as the Euler equation served as a model for deﬁning regular singular points, the general solution of the Euler equation will serve to introduce the method of Frobenius.6 This method prescribes the type of solution to look for near a regular singular point. For simplicity, consider solution behavior near t = 0. Suppose we know that the differential equation y + p(t)y + q(t)y = 0

(1a) 2

has a regular singular point at t = 0. Then tp(t) and t q(t) are analytic at t = 0. Let the Maclaurin series for tp(t) and t 2 q(t) be tp(t) = α0 + α1 t + α2 t 2 + · · · =

∞

αn t n

n=0

and t 2 q(t) = β0 + β1 t + β2 t 2 + · · · =

∞

βn t n ,

n=0

6

Ferdinand Georg Frobenius (1849–1917), a German mathematician, served on the faculties at the University of Zurich and the University of Berlin. He is remembered for his contributions to group theory and differential equations.

8.4

Solutions Near a Regular Singular Point and the Method of Frobenius

545

where both series converge in some interval −R < t < R. Multiply differential equation (1a) by t 2 to obtain t 2 y + t[tp(t)] y + [t 2 q(t)] y = 0.

(1b)

2

Inserting the Maclaurin series for tp(t) and t q(t), we arrive at t 2 y + t[α0 + α1 t + α2 t 2 + · · ·] y + [β0 + β1 t + β2 t 2 + · · ·] y = 0.

(2)

The method of Frobenius can be motivated by a heuristic argument. Very close to the singular point t = 0, we expect that α0 + α1 t + α2 t 2 + · · · ≈ α0

and

β0 + β1 t + β2 t 2 + · · · ≈ β0 .

(3)

If we use approximations (3) in equation (1b), we recover the Euler equation. Thus, we reason: If the two differential equations are nearly the same near t = 0, shouldn’t the solutions themselves have similar behavior near t = 0? Recall that solutions of the Euler equation were obtained by looking for solutions of the form y(t) = |t|λ , t = 0. For deﬁniteness, we consider t > 0. The method of Frobenius consists in looking for solutions in which the factor t λ is multiplied by an inﬁnite series. In other words, near the regular singular point t = 0, we look for solutions of y + p(t)y + q(t)y = 0 that have the form λ

2

y(t) = t [a0 + a1 t + a2 t + · · ·] =

∞

an t λ+n .

(4)

n=0

In representation (4), λ is a constant (possibly complex-valued) that is to be determined, along with the constants a0 , a1 , . . . . Also, since λ has not been speciﬁed, we can assume without any loss of generality that a0 = 0. [That is, there must be a “ﬁrst nonzero term” in series (4), and we simply take that term to be a0 t λ .]

Implementing the Method of Frobenius Substituting (4) into differential equation (1b) creates the following three terms: t 2 y = t 2

∞

an (λ + n)(λ + n − 1)t λ+n−2 =

n=0

t[tp(t)] y =

∞

αm t

m

m=0

[t 2 q(t)] y =

an (λ + n)(λ + n − 1)t λ+n (5a)

n=0

∞

∞ m=0

∞

an (λ + n)t λ+n

(5b)

n=0

βm t m

∞

an t λ+n .

(5c)

n=0

The three terms in (5) must be added, and the sum equated to zero. The series products are computed using the Cauchy product deﬁned in Section 8.1, and, as before, coefﬁcients of like powers of t are grouped together. We obtain t 2 y + t[tp(t)] y + [t 2 q(t)] y = [λ(λ − 1) + α0 λ + β0 ]a0 t λ + [λ(λ + 1)a1 + α0 a1 (λ + 1) + α1 a0 λ + β0 a1 + β1 a0 ]t λ+1 + [(λ + 2)(λ + 1)a2 + α0 a2 (λ + 2) + α1 a1 (λ + 1) + α2 a0 λ + β0 a2 + β1 a1 + β2 a0 ]t λ+2 +· · · ,

t > 0. (6)

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Series Solutions of Linear Differential Equations

When we equate the right-hand side of (6) to zero and invoke the uniqueness property of power series representations, it follows that the coefﬁcient of each power of t must vanish. Setting the ﬁrst coefﬁcient equal to zero leads to [λ2 + (α0 − 1)λ + β0 ]a0 = 0.

(7)

Since a0 = 0, it follows that representation (4) is a solution of the differential equation only if the exponent λ is a root of the equation F(λ) = 0, where F(λ) = λ2 + (α0 − 1)λ + β0 .

(8)

The equation F(λ) = 0 is called the characteristic equation or the indicial equation. (The latter term is used more frequently in this context, and we will use it as well.) Note that this equation (8) is consistent with the heuristic argument used to motivate representation (4). The equation F(λ) = 0 is precisely the indicial equation for the Euler equation that results from using the two approximations tp(t) ≈ α0 and t 2 q(t) ≈ β0 . Once we choose a root λ of the indicial equation, we set the coefﬁcients of the higher powers of t equal to zero in (6). This gives us a recurrence relation for ﬁnding the coefﬁcients a1 , a2 , a3 , . . . in terms of a0 . For instance, setting the coefﬁcient of t λ+1 equal to zero, we have [(λ + 1)2 + (α0 − 1)(λ + 1) + β0 ]a1 + [α1 λ + β1 ]a0 = 0, or a1 = −

[α1 λ + β1 ]a0 . F(λ + 1)

(9a)

Similarly, knowing a1 , we determine a2 from [(λ + 2)2 + (α0 − 1)(λ + 2) + β0 ]a2 + [α1 (λ + 1) + β1 ]a1 + [α2 λ + β2 ]a0 = 0, obtaining a2 = −

[α1 (λ + 1) + β1 ]a1 + [α2 λ + β2 ]a0 . F(λ + 2)

(9b)

Note the difference between the procedure for constructing the general solution near an ordinary point and the method of Frobenius for constructing the general solution near a regular singular point. For example, if t = 0 is an ordinary point, then we look for a solution of the form y(t) =

∞

an t n .

n=0

The recurrence relation obtained determines the coefﬁcients an in terms of two of these coefﬁcients, typically a0 and a1 . Since a0 and a1 are arbitrary, both members of a fundamental set of solutions are obtained concurrently. If t = 0 is a regular singular point, however, the method of Frobenius leads to the indicial equation, F(λ) = 0. Let the roots of this equation be denoted by λ1 and λ2 . The recurrence relation found using (6) is used twice, ﬁrst with λ = λ1 and then with λ = λ2 . In this way, we seek two linearly independent solutions, y1 (t) =

∞ n=0

λ1 + n a(1) n t

and

y2 (t) =

∞ n=0

λ2 + n a(2) , n t

8.4

Solutions Near a Regular Singular Point and the Method of Frobenius

547

(1) (1) (2) (2) (2) where the coefﬁcients a(1) 1 , a2 , a3 , . . . and a1 , a2 , a3 , . . . are generated, re(1) (2) spectively, in terms of arbitrary constants a0 and a0 . There are cases where the method of Frobenius, as outlined in the foregoing text, does not produce the general solution. An obvious case occurs when the two roots of the indicial equation are equal. In this repeated-root case, the foregoing procedure yields only one member of the fundamental set. A second (less obvious) case occurs when the indicial equation possesses two (real) roots that differ by an integer. These special cases will be discussed (and the entire method summarized) in the next section. We conclude with an example in which the procedure previously outlined can be used to obtain the general solution.

E X A M P L E

2

Use the method of Frobenius to obtain the general solution of 2t 2 y − ty + (1 + t 2 )y = 0

for t > 0.

Solution: The point t = 0 is a singular point. Since p(t) and q(t) are rational functions and since lim t→0 tp(t) = − 12 and lim t→0 t 2 q(t) = 12 , it follows that t = 0 is a regular singular point. Using the method of Frobenius, we look for solutions of the form y(t) =

∞

an t λ+n ,

t > 0.

n=0

Substituting this series into the differential equation leads to ∞ n=0

an (λ + n)(λ + n − 1)t λ+n −

∞

∞

1 1 an (λ + n)t λ+n + a t λ+n 2 n=0 2 n=0 n

∞

+

1 a t λ+n+2 = 0. 2 n=0 n

λ+n+2 λ+n Rewriting the last series as ∞ = ∞ and combining terms n=0 an t n=2 an−2 t where possible, we obtain ∞ ∞ 1 1 1 (10) (λ + n)(λ + n − 1) − (λ + n) + an−2 t λ+n = 0, an t λ+n + 2 2 2 n=0 n=2 or 1 1 3 3 2 λ 2 a t + (λ + 1) − (λ + 1) + a t λ+1 λ − λ+ 2 2 0 2 2 1 ∞ 1 3 1 2 + (λ + n) − (λ + n) + an + an−2 t λ+n = 0. 2 2 2 n=2 The indicial equation, F(λ) = λ2 − 32 λ + 12 = 0, has roots λ1 = 12 and λ2 = 1. We now examine the recurrence relations associated with each of these roots. (continued)

548

CHAPTER 8

Series Solutions of Linear Differential Equations (continued)

Case 1 Let λ = 12 in (10). We set the coefﬁcient of t λ+n = t(1/2)+n equal to zero for each value of n and use a(1) n to denote the coefﬁcients associated with this 3/2 value of λ. The coefﬁcient a(1) must 0 is arbitrary. The coefﬁcient multiplying t be zero, and thus 2 3 (1) 1 + 1 − 32 12 + 1 + 12 a(1) 1 = F 2 a1 = 0. 2 Since F have

3 2

1 2

= 0, it follows that a(1) 1 = 0. As for the remaining coefﬁcients, we 2 +n −

3 2

1 2

1 (1) + n + 12 a(1) n + 2 an−2 = 0,

and so an(1) = −

a(1) n−2 , 2F 12 + n

n = 2, 3, 4, . . . ,

n = 2, 3, 4, . . . .

(11)

Since F( 12 + n) = 0 for all n ≥ 2, recurrence relation (11) is well deﬁned. The coefﬁcients an(1) for n even can ultimately be expressed in terms of a(1) 0 . The (1) (1) coefﬁcients an for n odd are all zero because a1 = 0. Thus, we ﬁnd a solution y1 (t): 1/2 1 5/2 1 9/2 − t + t − · · · , t > 0. y1 (t) = a(1) t 0 6 168 Case 2 Let λ = 1 in (10). We repeat the sequence of computations just completed with this new value of λ and with the coefﬁcients now denoted by an(2) . As in case 1, it follows that a(2) 1 = 0 since F(2) = 0. In general, 1 (2) (1 + n)2 − 32 (1 + n) + 12 a(2) n = 2, 3, 4, . . . , n + 2 an−2 = 0, and thus a(2) n =−

a(2) n−2 , 2F(1 + n)

n = 2, 3, 4, . . . .

The solution obtained is

y2 (t) = a(2) t− 0

1 3 t 10

+

1 5 t 360

(12)

− ··· .

(13)

It is clear that the two solutions obtained are linearly independent on t > 0 since one solution is not a constant multiple of the other. Therefore, the two solutions form a fundamental set, and the general solution is 1 9/2 y(t) = a(1) t − ··· t1/2 − 16 t5/2 + 168 0 + a(2) 0

t−

1 3 t 10

+

1 5 t 360

− ··· ,

(14)

t > 0.

Note that the differential equation possesses the symmetries discussed in Theorem 8.2; that is, p(t) = −2/t is an odd function, and q(t) = (1 + t 2 )/t 2 is an even function. Therefore, to ﬁnd a solution for t < 0, we need only replace t by −t in (14). As a ﬁnal observation, note that recurrence relations (11) and (12), together with the ratio test, can be used to show that the two series in general solution (14) converge absolutely in 0 < t < ∞. ❖

8.4

Solutions Near a Regular Singular Point and the Method of Frobenius

549

EXERCISES Exercises 1–10: In each exercise, ﬁnd the singular points (if any) and classify them as regular or irregular. 1. ty + (cos t)y + y = 0

2. t 2 y + (sin t)y + y = 0

3. (t 2 − 1)y + (t − 1)y + y = 0

4. (t 2 − 1)2 y + (t + 1)y + y = 0

2

5. t y + (1 − cos t)y + y = 0

6. |t|y + y + y = 0

7. (1 − et )y + y + y = 0

8. (4 − t 2 )y + (t + 2)y + (4 − t 2 )−1 y = 0

9. (1 − t 2 )1/3 y + y + ty = 0

10. y + y + t1/3 y = 0

Exercises 11–13: In each exercise, determine the polynomial P(t) of smallest degree that causes the given differential equation to have the stated properties. 11. y +

sin 2t y + y=0 P(t)

There is a regular singular point at t = 0 and irregular singular points at t = ±1. All other points are ordinary points.

12. y +

1 1 y + y=0 t P(t)

There is a regular singular point at t = 0. All other points are ordinary points.

13. y +

1 1 y + 3y = 0 tP(t) t

There are irregular singular points at t = 0 and t = ±1. All other points are ordinary points.

Exercises 14–15: In each exercise, the exponent n in the given differential equation is a nonnegative integer. Determine the possible values of n (if any) for which (a) t = 0 is a regular singular point. (b) t = 0 is an irregular singular point. 14. y +

1 1 y + y=0 tn 1 + t2

15. y +

1 1 y + ny = 0 sin t t

Exercises 16–23: In each exercise, (a) Verify that t = 0 is a regular singular point. (b) Find the indicial equation. (c) Find the recurrence relation. (d) Find the ﬁrst three nonzero terms of the series solution, for t > 0, corresponding to the larger root of the indicial equation. If there are fewer than three nonzero terms, give the corresponding exact solution. 16. 2t 2 y − ty + (t + 1)y = 0

17. 4t 2 y + 4ty + (t − 1)y = 0

18. 16t 2 y + t 2 y + 3y = 0

19. t 2 y + ty + (t − 9)y = 0

20. ty + (t + 2)y − y = 0

21. t 2 y + 3ty + (2t + 1)y = 0

22. t 2 y + t(t − 1)y − 3y = 0

23. ty + (t − 2)y + y = 0

Exercises 24–27: In each exercise, (a) Verify that t = 0 is a regular singular point. (b) Find the indicial equation.

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Series Solutions of Linear Differential Equations (c) Find the ﬁrst three terms of the series solution, for t > 0, corresponding to the larger root of the indicial equation.

8.5

24. t 2 y − (2 sin t)y + (2 + t)y = 0

25. ty − 4y + et y = 0

26. (sin t)y − y + y = 0

27. (1 − et )y + 12 y + y = 0

The Method of Frobenius Continued: Special Cases and a Summary There are two important special cases where the method of Frobenius (as described in the previous section) may not yield the general solution of y + p(t)y + q(t)y = 0 near a regular singular point. These special cases arise when the indicial equation has two real roots that are equal or that differ by an integer. We will use Bessel’s equation as a vehicle to illustrate these cases. We then conclude this section by summarizing, for all cases, the structure of the general solution near a regular singular point.

The General Solution of Bessel’s Equation Consider Bessel’s equation, t 2 y + ty + (t 2 − ν 2 )y = 0,

t > 0,

where ν is a real nonnegative constant. As we saw in Section 8.4, t = 0 is a regular singular point. In addition, note that the coefﬁcient functions t2 − ν2 1 and q(t) = t t2 are odd and even functions, respectively. Therefore, by Theorem 8.2, we can obtain solutions for t < 0 by ﬁnding solutions for t > 0 and replacing t with −t. Weapply the method of Frobenius to Bessel’s equation. Substituting λ+n leads to y(t) = ∞ n=0 an t p(t) =

[λ2 − ν 2 ]a0 t λ + [(λ + 1)2 − ν 2 ]a1 t λ+1 +

∞

[{(λ + n)2 − ν 2 }an + an−2 ]t λ+n = 0,

t > 0.

(1)

n=2

Without loss of generality, we assume a0 = 0. The indicial equation, F(λ) = λ2 − ν 2 = 0,

(2)

has roots λ1 = ν and λ2 = −ν. From (1), we also obtain the equations F(λ + 1)a1 = 0

(3a)

F(λ + n)an + an−2 = 0.

(3b)

and Equation (3b) leads to the recurrence relation −an−2 , n = 2, 3, 4, . . . . an = F(λ + n)

(3c)

The special cases that arise can be illustrated by selecting particular values for the constant ν.

8.5

The Method of Frobenius Continued: Special Cases and a Summary

551

Equal Roots (λ1 = λ2 ) Consider Bessel’s equation with ν = 0. In this case, λ1 = λ2 = 0. Since the root is repeated, the method of Frobenius gives us only one member of the fundamental set of solutions; second linearly independent solution has a structure the λ+n a t . different from ∞ n=0 n Since ν = 0, the indicial polynomial reduces to F(λ) = λ2 . Therefore, F(1) is nonzero, and we see from (3a) that a1 = 0. From (3c), we have −an−2 an = , n = 2, 3, 4, . . . . (4) n2 Since a1 = 0, it follows from (4) that all odd-indexed coefﬁcients are zero. The even-indexed coefﬁcients can be expressed as multiples of a0 : a2 =

−1 2

2

a0 ,

a4 =

−1 4

2

a2 =

(−1)2 4

2 (2!)

2

a0 ,

a6 =

−1 6

2

a4 =

(−1)3 26 (3!)2

a0 ,

and, in general, a2n =

(−1)n 22n (n!)2

a0 ,

n = 0, 1, 2, . . . .

The solution we thus obtain is

∞ t2 t4 t6 (−1)n t2n + − + · · · = a0 1 + y1 (t) = a0 1 − = a0 J0 (t), 4 64 2304 22n (n!)2 n=1

(5a)

where J0 (t) = 1 +

∞ (−1)n t2n n=1

22n (n!)2

.

(5b)

The function J0 (t) is called the Bessel function of the ﬁrst kind of order zero. Having one solution, we could, in principle, use the method of reduction of order (see Section 3.4) to construct a second solution. We shall, however, simply state a form of the second solution that is commonly used in applications— the Bessel function of the second kind of order zero (also called Weber’s function). The Bessel function of the second kind of order zero is given by 2 n t ∞ 2 2 t 4 Y0 (t) = (−1)n+1 kn , t > 0. (6) γ + ln J0 (t) + π 2 π n=1 (n!)2 In (6), k1 = 1 and, in general, kn = 1 +

1 1 1 1 + + + ··· + , 2 3 4 n

n = 2, 3, . . . .

The constant γ in equation (6) is known as the Euler-Mascheroni constant7 and is deﬁned by the limit ⎡ ⎤ n 1 − ln n⎦ ≈ 0.5772 . . . . γ = lim ⎣ n→∞ j j=1 7

Lorenzo Mascheroni (1750–1800) was an ordained priest, poet, and teacher of mathematics and physics. He became professor of algebra and geometry at the University of Pavia in 1786 and later became its rector. In 1790, Mascheroni correctly calculated the ﬁrst 19 decimal places of Euler’s constant. This accomplishment has caused his name to be linked with the constant.

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Note that Y0 (t) can ultimately be expressed as a nconstant multiple of J0 (t)ln t = y1 (t)ln t added to a series of the form ∞ n=0 bn t . Since J0 (0) = 1, Y0 (t) has a logarithmic singularity at t = 0. If we recall the heuristic argument used to motivate the method of Frobenius, the presence of the function ln t in (6) and the corresponding logarithmic singularity at t = 0 is not surprising since the general solution of the Euler equation having repeated roots λ1 = λ2 = 0 is y(t) = c1 + c2 ln t, t > 0. Figure 8.3 shows graphs of J0 (t) and Y0 (t). y

y

1

1 0.5

0.5 5

10

15

20

25

30

t

–0.5 5

10

15

20

25

t

30

–1 –1.5

–0.5

(a)

(b) FIGURE 8.3

(a) The graph of J0 (t), the Bessel function of the ﬁrst kind of order zero. (b) The graph of Y0 (t), the Bessel function of the second kind of order zero. Note that J0 (t) is deﬁned for all t, whereas Y0 (t) has a logarithmic singularity at the regular singular point t = 0.

Roots Differing by Unity (λ1 − λ2 = 1) As an illustration of the case where roots differ by unity, let ν = equation. The indicial equation,

1 2

in Bessel’s

F(λ) = λ2 − 14 , has roots λ1 = 12 and λ2 = − 12 , and therefore λ1 − λ2 = 1. Consider ﬁrst the larger root, λ1 = 12 , which corresponds to a solution of the form y1 (t) =

∞

(1/2)+n a(1) , n t

t > 0.

n=0

3 We can assume a(1) 0 = 0. Since F(λ1 + 1) = F 2 = 0, it follows from (3a) that a(1) 1 = 0. From (3c), the recurrence relation is a(1) n =

−a(1) n−2 2 n + 12 −

1 4

=

−a(1) n−2 , n(n + 1)

n = 2, 3, 4, . . . .

(7)

Equation (7) allows us to determine all the even-indexed coefﬁcients as multiples of a(1) 0 and implies that all the odd-indexed coefﬁcients are zero. Solving

8.5

The Method of Frobenius Continued: Special Cases and a Summary

recurrence relation (7) leads to the solution

5/2 9/2 13/2 t t t + − + ··· t1/2 − y1 (t) = a(1) 0 3! 5! 7!

t3 t5 t7 (1) −1/2 = a0 t + − + ··· t− 3! 5! 7! −1/2 = a(1) sin t, 0 t

553

(8)

t > 0.

Consider now the smaller root, λ2 = − 12 , where we look for a solution of the form y2 (t) =

∞

−(1/2)+n a(2) , n t

t > 0.

n=0

As before, F(λ) = λ2 − 14 . This time, however, 2 F(λ2 + n) = n − 12 − 14 = n(n − 1), and we see that F(λ2 + 1) = F(λ1 ) = F 12 = 0. Therefore [see equation (3a)], a(2) 1 need not be zero. From (3c), the recurrence relation a(2) n =

−a(2) n−2 1 2 n− 2 −

1 4

=

−a(2) n−2 , n(n − 1)

n = 2, 3, 4, . . .

(9)

allows us to determine all even-indexed coefﬁcients as multiples of a(2) 0 and all odd-indexed coefﬁcients as multiples of a(2) . We ﬁnd 1 a(2) 2k =

(−1)k (2) a (2k)! 0

(2) and a2k+1 =

(−1)k (2) a , (2k + 1)! 1

k = 1, 2, 3, . . . .

The resulting solution y2 (t) has the form

t4 t6 t2 (2) −1/2 + − + ··· 1− y2 (t) = a0 t 2! 4! 6!

t5 t7 t3 (2) −1/2 + a1 t + − + ··· t− 3! 5! 7! −1/2 −1/2 = a(2) cos t + a(2) sin t, 0 t 1 t

(10)

(11)

t > 0.

We therefore obtain a second linearly independent solution, t−1/2 cos t, added to a multiple of the solution previously obtained, t−1/2 sin t. In this example, the method of Frobenius has produced both members of a fundamental set of solutions. The Bessel functions of order one-half are deﬁned to be 2 2 sin t and J−1/2 (t) = cos t, t > 0. J1/2 (t) = πt πt The general solution of Bessel’s equation in this case is usually expressed as y(t) = c1 J1/2 (t) + c2 J−1/2 (t).

(12)

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Note that J1/2 (t) is well behaved near t = 0 since sin t lim+ √ = 0. t t→0 By contrast, lim t→0+ J−1/2 (t) = +∞. Figure 8.4 shows the behavior of these two functions. y

y

1

1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2 5

10

15

20

25

t

30

5

–0.2

–0.2

–0.4

–0.4

10

(a)

15

20

25

30

t

(b) FIGURE 8.4

(a) The graph of the Bessel function J1/2 (t). (b) The graph of the Bessel function J−1/2 (t).

Roots Differing by an Integer Greater than 1 (λ1 − λ2 = N ≥ 2) Let ν = M ≥ 1 in Bessel’s equation. The two roots of the indicial equation, λ2 − M2 = 0, are λ1 = M and λ2 = −M. The roots differ by an integer greater than 1, since λ1 − λ2 = 2M ≥ 2. In this case, the method of Frobenius provides us with a solution corresponding to the larger root, λ1 = M. It will fail, however, when we try to apply the method to the smaller root, λ2 = −M. Consider ﬁrst the larger root, λ1 = M. We look for a solution of the form y1 (t) =

∞

M+n a(1) , n t

t > 0.

n=0

We assume a(1) 0 = 0. Since F(λ1 + 1) = F(M + 1) = 0, we know from equation = 0. Using equation (3c), we obtain the recurrence relation (3a) that a(1) 1 a(1) n =

−a(1) n−2 [(n + M)2 − M2 ]

=

−a(1) n−2 , [n(n + 2M)]

n = 2, 3, 4, . . . .

(13)

Equation (13) allows us to determine all the even-indexed coefﬁcients as multiples of a(1) 0 and tells us that the odd-indexed coefﬁcients are zero. Therefore,

8.5

The Method of Frobenius Continued: Special Cases and a Summary

we obtain the solution M y1 (t) = a(1) 0 t

−

1−

t2

22 (M + 1)

+

555

t4 24 2!(M + 2)(M + 1)

t6 26 3!(M + 3)(M + 2)(M + 1)

+ ··· +

(−t 2 )k M! 22k k!(M + k)!

(14)

+ ··· .

Choosing a0(1) = (1/2M )M! in (14) leads to the Bessel function of the ﬁrst kind of order M: 2 k −t M ∞ t 4 . JM (t) = (15) 2 k!(M + k)! k=0

Note that JM (t) is analytic at t = 0 and the series (15) has an inﬁnite radius of convergence. The function JM (t) vanishes at t = 0 when M ≥ 1; JM (t) is an even function when M is an even integer and an odd function when M is an odd integer. Suppose we now consider the smaller root, λ2 = −M, and look for a solution of the form y2 (t) =

∞

−M+n a(2) . n t

n=0

Assume, without loss of generality, that a(2) 0 = 0. Since M ≥ 1, F(λ2 + 1) = F(−M + 1) = 0 and a1(2) = 0. The difﬁculty arises when we try to use equa(2) (2) (2) tion (3c) to evaluate a(2) 2 , a4 , a6 , . . . in terms of a0 . Using λ = −M and setting n = 2k, we have for the recurrence relation (3c) a(2) 2k =

−a(2) 2k−2 , 4k(k − M)

k = 1, 2, 3, . . . .

(16)

The trouble occurs when we try to evaluate (16) for k = M. The breakdown of recurrence relation (16) at k = M tells us that the assumed form of the solution is incorrect and that the second linearly independent solution does not have the structure assumed by the method of Frobenius. The second linearly independent solution of Bessel’s equation customarily used is called the Bessel function of the second kind of order M and is denoted by YM (t). It is deﬁned as −M t k M−1 (M − k − 1)! t 2 2 t 2 YM (t) = γ + ln J (t) − π 2 M π k! 4 k=0

2 n M t t − ∞ 2 4 , − (−1)n kM+n + kn π n!(M + n)! n=0

(17)

t > 0,

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Series Solutions of Linear Differential Equations

where the constants kn are deﬁned as k0 = 0, kn = 1 + 2−1 + 3−1 + · · · + n−1 , n ≥ 1 and where γ is the Euler-Mascheroni constant. Figure 8.5 displays graphs of J1 (t) and Y1 (t). y

y

0.6

0.4

0.4

0.2 5

0.2

10

15

20

25

30

t

–0.2 5

10

15

20

25

30

–0.2

t –0.4 –0.6

–0.4

–0.8

(a)

(b) FIGURE 8.5

(a) The graph of J1 (t). (b) The graph of Y1 (t).

REMARK: For reasonably large values of t (say t > 5), all of the Bessel functions plotted in Figures 8.3–8.5 seem to behave like damped sinusoids—that is, sine or cosine waves having amplitudes that decrease with increasing t (see Exercises 25 and 26).

Summary of the General Solution Near a Regular Singular Point Our study of Bessel’s equation for different values of ν provided some examples of the behavior of solutions near a regular singular point. Although these examples are representative, they are neither exhaustive nor completely general since Bessel’s equation possesses a speciﬁc structure that is not necessarily present in the general case. The following summary, which we present without proof, describes the general behavior of solutions of y + p(t)y + q(t)y = 0. Consider the differential equation y + p(t)y + q(t)y = 0, where we assume t = 0 is a regular singular point. Let tp(t) and t 2 q(t) be real-valued analytic functions with Maclaurin series ∞ ∞ αn t n and t 2 q(t) = βn t n tp(t) = n=0

n=0

that converge in −R < t < R. The corresponding indicial equation is λ2 + (α0 − 1)λ + β0 = (λ − λ1 )(λ − λ2 ) = 0. The roots λ1 and λ2 either are real or form a complex conjugate pair. In the event that λ1 and λ2 are real, we assume λ1 ≥ λ2 .

8.5

The Method of Frobenius Continued: Special Cases and a Summary

557

Then, in either of the intervals (−R, 0) or (0, R), (a) There exists a solution having the form y1 (t) = |t|λ1

∞

an t n ,

a0 = 0,

(18)

n=0

where the series converges at least in |t| < R. (b) The form of the second linearly independent solution, y2 (t), depends on the difference λ1 − λ2 . (i) If λ1 − λ2 is not an integer, then y2 (t) = |t|λ2

∞

bn t n ,

b0 = 0,

(19)

n=0

where the series converges at least in |t| < R. (ii) If λ1 = λ2 , then y2 (t) = y1 (t)ln|t| + |t|λ2

∞

cn t n ,

(20)

n=0

where the series converges at least in |t| < R. (iii) If λ1 − λ2 equals a positive integer, then y2 (t) = Cy1 (t)ln|t| + |t|λ2

∞

dn t n ,

d0 = 0,

(21)

n=0

where C is a constant, possibly zero (if C = 0, there is no logarithmic term). Moreover, the series converges at least in |t| < R.

REMARKS: 1. We have assumed, for convenience, that t = 0 is a regular singular point. In general, if t = t0 is a regular singular point, then the results in the summary are valid when t − t0 replaces t in the formulas. 2. When the roots λ1 and λ2 form a complex conjugate pair—for instance, λ1 = γ + iδ, λ2 = γ − iδ—the difference λ1 − λ2 = 2δi is purely imaginary and thus not equal to zero or a positive integer. In that case, the two complex-valued solutions y1 (t) = |t|γ +iδ

∞ n=0

an t n

and

y2 (t) = |t|γ −iδ

∞

bn t n ,

n=0

obtained using the recurrence relations, can be used to create an equivalent real-valued fundamental set. Using Euler’s formula, we have |t|γ +iδ = |t|γ [cos(δ ln|t|) + i sin(δ ln|t|)]. As mentioned in Section 8.1, linear differential equations with variable coefﬁcients play an important role in applications. Because of their importance,

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the functions that emerge as solutions of these equations, usually collectively referred to as special functions, have been studied exhaustively. There are books devoted to a particular special function,8 as well as general handbooks for special functions.9 Modern software packages have many of these functions available as built-in functions—as accessible as the familiar exponential, logarithmic, and trigonometric functions. In that sense, these so-called special functions are fortunately becoming less and less special. Linear differential equations with variable coefﬁcients, such as Bessel’s equation and Legendre’s equation, arise when the technique called separation of variables is applied to the partial differential equations that model certain physical problems. The Projects at the end of this chapter study steady-state heat conduction between concentric cylinders and provide a brief introduction to this circle of ideas.

EXERCISES Exercises 1–12: In each exercise, (a) Verify that the given differential equation has a regular singular point at t = 0. (b) Determine the indicial equation and its two roots. (These roots are often called the exponents at the singularity.) (c) Determine the recurrence relation for the series coefﬁcients. (d) Consider the interval t > 0. If the two exponents obtained in (c) are unequal and do not differ by an integer, determine the ﬁrst two nonzero terms in the series for each of the two linearly independent solutions. If the exponents are equal or differ by an integer, obtain the ﬁrst two nonzero terms in the series for the solution having the larger exponent. (e) When the given differential equation is put in the form y + p(t)y + q(t)y = 0, note that tp(t) and t 2 q(t) are polynomials. Do the series, whose initial terms were found in part (d), converge for all t, 0 < t < ∞? Explain. 1. 2ty − (1 + t)y + 2y = 0

2. 2ty + 5y + 3ty = 0

3. 3t 2 y − ty + (1 + t)y = 0

4. 6t 2 y + ty + (1 − t)y = 0

5. t 2 y − 5ty + (9 + t 2 )y = 0

6. 4t 2 y + 8ty + (1 + 2t)y = 0

7. t 2 y − 2ty + (2 + t)y = 0

8. ty + 4y − 2ty = 0

2

9. t y + ty − (1 + t )y = 0

10. t 2 y + 5ty + (4 − t 2 )y = 0

11. t 2 y + ty − (16 + t)y = 0

12. 8t 2 y + 6ty − (1 − t)y = 0

2

Exercises 13–16: In each exercise, (a) Determine all singular points of the given differential equation and classify them as regular or irregular singular points.

8

See, for example, George N. Watson, A Treatise on the Theory of Bessel Functions, 2nd ed. (Cambridge: Cambridge University Press, 1966). 9 Milton Abramowitz and Irene A. Stegun, editors, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables (New York: Dover Publications, 1970) and Wilhelm Magnus, Fritz Oberhettinger, and Raj Pal Soni, Formulas and Theorems for the Special Functions of Mathematical Physics (Berlin and New York: Springer-Verlag, 1966).

8.5

The Method of Frobenius Continued: Special Cases and a Summary

559

(b) At each regular singular point, determine the indicial equation and the exponents at the singularity. 13. (t3 + t)y − (1 + t)y + y = 0 2

2

15. (t − 4) y − y + y = 0

14. t 2 y + (sin 3t)y + (cos t)y = 0 16. t 2 (1 − t)1/3 y + ty − y = 0

17. The Legendre differential equation (1 − t 2 )y − 2ty + α(α + 1)y = 0 has regular singular points at t = ±1; all other points are ordinary points. (a) Determine the indicial equation and the exponent at the singularity t = 1. (b) Assume that α = 0, 1. Find the ﬁrst three nonzero terms of the series solution in powers of t − 1 for t − 1 > 0. [Hint: Rewrite the coefﬁcient functions in powers of t − 1. For example, 1 − t 2 = −(t − 1)(t + 1) = −(t − 1)((t − 1) + 2).] (c) What is an exact solution of the differential equation when α = 1? 18. The Chebyshev differential equation (1 − t 2 )y − ty + α 2 y = 0 has regular singular points at t = ±1; all other points are ordinary points. (a) Determine the indicial equation and the exponent at the singularity t = 1. (b) Assume α is nonzero and not an integer multiple of 12 . Find two linearly independent solutions for t − 1 > 0. (Use the hint in Exercise 17.) (c) On what interval of the form 0 < t − 1 < R do the solutions found in part (b) converge? (d) What is an exact solution of the differential equation when α = 12 ? 19. The Laguerre10 differential equation ty + (1 − t)y + αy = 0 has a regular singular point at t = 0. (a) Determine the indicial equation and show that the roots are λ1 = λ2 = 0. (b) Find the recurrence relation. Show that if α = N, where N is a nonnegative integer, then the series solution reduces to a polynomial. Obtain the polynomial solution when N = 5. The polynomial solutions of this differential equation, when properly normalized, are called Laguerre polynomials. (c) Is the polynomial obtained in part (b) for α = N = 5 an even function, an odd function, or neither? Would you expect even and odd solutions of the differential equation based on its structure and the conclusions of Theorem 8.2? Explain.

Exercises 20–23: In each exercise, use the stated information to determine the unspeciﬁed coefﬁcients in the given differential equation. 20. t 2 y + t(α + 2t)y + (β + t 2 )y = 0.

t = 0 is a regular singular point. The roots of the indicial equation at t = 0 are λ1 = 1 and λ2 = 2.

21. t 2 y + αty + (β + t − t3 )y = 0.

t = 0 is a regular singular point. The roots of the indicial equation at t = 0 are λ1 = 1 + 2i and λ2 = 1 − 2i.

22. t 2 y + αty + (2 + βt)y = 0.

t = 0 is a regular singular point. One root of the indicial equation at t = 0 is λ = 2. The recurrence relation for the series solution corresponding to this root is (n2 + n)an − 4an−1 = 0, n = 1, 2, . . . .

10

Edmond Laguerre (1834–1886) attended the Ecole Polytechnique in Paris and returned there after ten years of service as a French artillery ofﬁcer. He worked in the areas of analysis and geometry and is best remembered for his study of the polynomials that bear his name.

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Series Solutions of Linear Differential Equations 23. ty + (1 + αt)y + βty = 0.

The recurrence relation for a series solution is n2 an − (n − 1)an−1 + 3an−2 = 0, n = 2, 3, . . . .

24. Modiﬁed Bessel Equation The differential equation t 2 y + ty − (t 2 + ν 2 )y = 0 is known as the modiﬁed Bessel equation. Its solutions, usually denoted by Iν (t) and Kν (t), are called modiﬁed Bessel functions. This equation arises in solving certain partial differential equations involving cylindrical coordinates. (a) Do you anticipate that the modiﬁed Bessel equation will possess solutions that are even and odd functions of t? Explain. (b) The point t = 0 is a regular singular point of the modiﬁed Bessel equation; all other points are ordinary points. Determine the indicial equation for the singularity at t = 0 and ﬁnd the exponents at the singularity. (c) Obtain the recurrence relation for the modiﬁed Bessel equation. How do the exponents and recurrence relation for this equation compare with their counterparts for Bessel’s equation? 25. Consider Bessel’s equation, t 2 y + ty + (t 2 − ν 2 )y = 0 for t > 0. (a) Deﬁne a new dependent variable u(t) by the relation y(t) = t−1/2 u(t). Show that u(t) satisﬁes the differential equation

2 1 ν −4 u + 1− u = 0. t2 (b) Solve the differential equation in part (a) when ν 2 = 14 . What is the corresponding solution of Bessel’s equation in this case? (c) Suppose that t is large enough to justify neglecting the term (ν 2 − 14 )/t 2 in the differential equation obtained in part (a). Show that neglecting (ν 2 − 14 )/t 2 leads to the approximation y(t) ≈ t1/2 R cos(t − δ) when t is large. 26. This exercise asks you to√use computational software to show that Bessel functions behave like R cos(t − δ)/ t for appropriate choices of constants R and δ and for t large enough. We restrict our attention to J0 (t). (a) Locate the abscissa of the ﬁrst maximum of J0 (t) in t > 0; call this point tm . Since J0 (t) = −J1 (t), this point can be found by applying a root-ﬁnding routine to J1 (t). √ (b) Evaluate the constants R and δ by setting tm − δ = 2π and R = tm J0 (tm ). √ (c) Plot the two functions J0 (t) and R cos(t − δ)/ t on the same graph for tm ≤ t ≤ 50. How do the two graphs compare? 27. For the special case ν = 12 , consider the modiﬁed Bessel equation for t > 0, t 2 y + ty − (t 2 + 14 )y = 0. (a) Deﬁne a new dependent variable u(t) by the relation y(t) = t−1/2 u(t). Show that u(t) satisﬁes the differential equation u − u = 0. (b) Show that the differential equation has a fundamental set of solutions sinh t √ , t

cosh t √ , t

t > 0.

Projects

561

PROJECTS Project 1: The Simple Centrifuge Revisited We revisit a problem that was studied numerically in Project 2 of Chapter 3. At time t = 0, a horizontally mounted tube of length l begins to rotate with a constant (positive) angular acceleration α. At that time, a particle of mass m is located a radial distance r0 from the pivot and moving with a radial velocity r0 . The radial distance of the particle from the pivot, r(t), is a solution of the following initial value problem: r − α 2 t2 r = 0,

r(0) = r0 ,

r (0) = r0 .

(1)

We now develop a series solution of this problem. The point t = 0 is an ordinary point. Assume a solution of the form ∞ r(t) = an t n . n=0

1. Develop the recurrence relation for the coefﬁcients {an }∞ n=0 . What is the radius of convergence of the series obtained? 2. The series solution will have the form r(t) = a0 r1 (t) + a1 r2 (t), with a0 = r0 and a1 = r0 . Determine the ﬁrst four nonvanishing terms for each of the series r1 (t) and r2 (t). 3. Do the functions deﬁned by series r1 (t) and r2 (t) possess even or odd symmetry? Could Theorem 8.2 have been used to predict the existence of even and odd solutions of differential equation (1)? Explain. 4. Assume the following numerical values: α = π rad/s2 ,

l = 2 m,

r0 = 10 cm,

r0 = 20 cm/s.

Use computer software and the polynomial approximation developed in part 2 to estimate the time at which the particle will exit the tube and the exit velocity. Since we are using a truncated series for these computations, the exit time and exit velocity computed will be approximate values. Is it possible to use this computed information to get a rough estimate of the exit time error? Explain. 5. The results of parts 1 and 2 show that the solution r(t) is actually a function of α 1/2 t. Show that this could have been anticipated by studying the structure of the differential equation itself. Speciﬁcally, make the change of independent variable τ = α p t (or t = α −p τ ). Show that with this change of variable, the choice p = 12 transforms the differential equation into the equation d2 r − τ 2 r = 0. dτ 2 What does the initial condition r (0) = r0 transform into?

Project 2: Steady-State Heat Flow between Concentric Cylinders This exercise gives you a brief glimpse into an application involving variable coefﬁcient linear differential equations. Consider the two concentric cylinders shown in Figure 8.6. The inner cylinder has radius a > 0, while the outer cylinder has radius b > a. Assume that these cylinders represent the inner and outer surfaces of a pipe and that the pipe itself is designed to function as part of a simple cooling system. Heat, or thermal energy, is drawn from the region exterior to the pipe by the presence of coolant ﬂowing within the pipe. We assume that the coolant is “well-stirred” and that the inner surface of the pipe is maintained at the coolant temperature. Suppose we know the temperature of the region outside the

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pipe as well as the temperature of the coolant. For ﬁxed values of the radii a and b, we would like to know the rate at which heat is drawn from the exterior region. For a ﬁxed coolant temperature, we also want to know how the heat transfer varies as a function of the two pipe radii. y

b a z x

FIGURE 8.6

The concentric cylinders represent the inner and outer surfaces of a pipe. Heat is drawn from the exterior region by the presence of a coolant ﬂowing through the pipe.

In general, a temperature reading depends on where and when the reading is taken. Thus, if T represents temperature (measured, perhaps, in degrees Centigrade), then T = T(x, y, z, t). We shall assume, however, that our cooling system is operating in a steady-state mode; that is, system operation has “settled down” to the point where the temperature everywhere remains constant in time. We further assume that temperature does not change in the axial or z-direction. With these assumptions, T = T(x, y). Because of the cylindrical geometry, we introduce polar coordinates. We set x = r cos θ and y = r sin θ , as in Figure 8.7. We view temperature as a function of the polar variables, T = T(r, θ). The domain of interest is the annular region between the cylinders, described in polar coordinates by a ≤ r ≤ b, 0 ≤ θ < 2π. We assume that the temperature at the outer radius of the pipe is known; that is, T(b, θ) = Tb (θ),

0 ≤ θ < 2π,

where Tb is a known function of the angle θ. We also assume that the temperature at the inner pipe radius is a known constant; that is, T(a, θ) = Ta ,

0 ≤ θ < 2π.

Since the coolant is to draw heat from the exterior region, we assume that Tb (θ) > Ta , 0 ≤ θ < 2π. The problem is to determine the rate at which the coolant draws heat through the pipe from the exterior region. We ﬁrst determine the temperature T(r, θ) within the annular cross-section of the pipe and then use this information to compute the required heat ﬂow. Within the annular region of the pipe, the steady-state temperature must be a solution of a partial differential equation known as Laplace’s equation. In polar coordinates, Laplace’s equation is ∂ 2T 1 ∂ 2T 1 ∂T + 2 2 = 0. + 2 r ∂r ∂r r ∂θ

(2)

To determine the steady-state temperature within the annular pipe region, we need to

Projects

563

y

r a

x

b

FIGURE 8.7

A cross-section of the pipe shown in Figure 8.6. We introduce polar coordinates into the problem because of the cylindrical geometry. solve partial differential equation (2) subject to the boundary conditions T(a, θ) = Ta ,

T(b, θ) = Tb (θ),

0 ≤ θ < 2π.

(3)

It should be emphasized that the problems we ultimately solve in this exercise are not initial value problems since constraints are imposed at two different values of the independent variable. Once we ﬁnd the temperature distribution within the annular region, T(r, θ),

a ≤ r ≤ b,

0 ≤ θ < 2π,

we will compute the heat transfer rate into the coolant at the inner radius a. Heat ﬂows “downhill,” from hotter to cooler regions. Moreover, the rate of heat ﬂow is proportional to the temperature gradient—the steeper the gradient, the greater the rate of heat transfer. At a point on the inner pipe radius, the rate of heat transfer per unit surface area is given by κ

∂T(a, θ) , ∂r

(4)

where κ is a positive constant (known as the thermal conductivity) that depends on the nature of the pipe material. It follows that the rate of heat transfer per unit axial length of the pipe can be found by integrating (4) around the inner pipe radius, obtaining 2π 2π ∂T(a, θ) ∂T(a, θ) a dθ = κa dθ. (5) κ ∂r ∂r 0 0 1. The Case of Constant Exterior Temperature Assume that the temperature at the outer pipe radius is constant; that is, Tb (θ) = Tb ,

0 ≤ θ < 2π.

Since neither boundary condition varies with angle, we expect the temperature in the annular pipe region to likewise be independent of θ. Assume a solution of the form T = T(r), a ≤ r ≤ b. (a) Substitute T = T(r) into Laplace’s equation (2), obtaining d2 T(r) 1 dT(r) = 0, + r dr dr 2

a < r < b.

(6)

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Obtain the general solution of this Euler equation, and impose the boundary constraints T(a) = Ta ,

T(b) = Tb .

(7)

The constraints (7) will determine the two arbitrary constants in the general solution. (b) Compute dT/dr, and evaluate the integral (5). (c) Let Ta = 40◦ F, Tb = 120◦ F, a = 1 in., and b = 1.5 in. For these parameter values, plot T(r) versus r for a ≤ r ≤ b. Do the maximum and minimum temperatures occur where you expect them to occur? 2. The Case of Varying Exterior Temperature Assume now that the temperature distribution at the outer pipe radius is not constant. As a speciﬁc case, assume that T(b, θ) = (1 + α sin θ) Tb ,

0 ≤ θ < 2π,

◦

where Tb = 120 F and where 0 ≤ α < 1. (a) Do you think the heat transfer rate will differ from that obtained in part (a) of the constant exterior temperature case? (b) Assume a solution of Laplace’s equation of the form T(r, θ) = T0 (r) + T1 (r) sin θ

(8)

within the annular pipe region. The unknown functions T0 (r) and T1 (r) must be determined. Substitute (8) into (2), obtaining

1 d2 T0 (r) 1 dT0 (r) d2 T1 (r) 1 dT1 (r) + − + + T (r) sin θ = 0, 1 r dr r dr dr 2 dr 2 r2 a < r < b,

0 ≤ θ < 2π.

Assume for the moment that the radial variable has an arbitrary but ﬁxed value. The set of functions {1, sin θ} is linearly independent in 0 ≤ θ < 2π , and so this equation implies, for the particular ﬁxed value of r, that d2 T0 (r) 1 dT0 (r) =0 + r dr dr 2

(9a)

and d2 T1 (r) 1 dT1 (r) 1 − 2 T1 (r) = 0. + (9b) r dr dr 2 r Since r is assumed to be arbitrary, these equations must hold for a < r < b. Find the general solution for each of the two Euler equations (9a) and (9b). (c) Apply the boundary constraints (3). In particular, we have T(a, θ ) = T0 (a) + T1 (a) sin θ = Ta T(b, θ ) = T0 (b) + T1 (b) sin θ = Tb + Tb α sin θ,

0 ≤ θ < 2π.

Use the same linear independence argument employed in (b) to obtain boundary conditions for the functions T0 and T1 . Impose these boundary conditions on the general solutions of the Euler equations obtained in (b), and determine T(r, θ). (d) Determine the heat transfer rate at the inner pipe radius. Is the heat transfer rate the same as that obtained for the constant exterior temperature case?

C H A P T E R

9

Second Order Partial Diﬀerential Equations and Fourier Series CHAPTER OVERVIEW 9.1

Introduction

9.2

Heat Flow in a Thin Bar; Separation of Variables

9.3

Series Solutions

9.4

Calculating the Solution

9.5

Fourier Series

9.6

The Wave Equation

9.7

Laplace’s Equation

9.8

Higher-Dimensional Problems; Nonhomogeneous Equations

9.1

Introduction In this chapter, we begin a discussion of partial differential equations. Such equations involve a dependent variable (often denoted by u) that is a function of two or more independent variables. Frequently, the independent variables are time t and one or more of the spatial variables x, y, z. For example, u(x, y, z, t) might represent the temperature of a three-dimensional solid at spatial point (x, y, z) and time t. A partial differential equation is an equation involving a dependent variable and its partial derivatives. Partial differential equations often arise when we model phenomena that change in both space and time. Although the problems we consider in this chapter may seem complicated, the techniques we use 565

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to solve them are familiar. The method of separation of variables, discussed in this chapter, and the method of characteristics, discussed in Chapter 10, ultimately reduce to problems involving ordinary differential equations. The present chapter focuses on second order partial differential equations and the technique of separation of variables. Chapter 10 is concerned with ﬁrst order partial differential equations. To provide maximum ﬂexibility, we have written Chapters 9 and 10 so that they can be read in any order. In Section 9.2, we introduce partial differential equations with a discussion of the heat equation. Section 9.2 also features a qualitative discussion of what to expect of solutions of the heat equation and a brief introduction to the main solution technique in this chapter, separation of variables. In Section 9.3, we examine the eigenvalue problems that arise from separation of variables and determine the building-block solutions that will ultimately form the basis for an inﬁnite series representation of solutions of the heat equation. Sections 9.4 and 9.5 complete the discussion of the heat equation, describing the theory of Fourier series, the Fourier convergence theorem, and the Gibbs phenomenon. Having used the heat equation to introduce the major ideas of Fourier series and separation of variables, we then consider other boundary value problems: the wave equation in Section 9.6 and Laplace’s equation in Section 9.7. Finally, in Section 9.8 and its Exercises, we look at higher-dimensional problems and problems involving nonhomogeneous partial differential equations.

Second Order Partial Differential Equations The order of a partial differential equation is the order of the highest partial derivative appearing in the equation. For example, ∂u(x, t) ∂u(x, t) − = t sin x ∂x ∂t is a ﬁrst order partial differential equation, while ∂u(x, t) ∂u2 (x, t) − =0 ∂t ∂x2 is a second order partial differential equation. Consider the special case where the dependent variable u is a function of two independent variables, a spatial variable x and a temporal variable t. In this case, the general second order linear partial differential equation has the form a11 (x, t)

∂ 2 u(x, t) ∂x2

+ a12 (x, t)

∂ 2 u(x, t) ∂ 2 u(x, t) + a22 (x, t) ∂x∂t ∂t2

∂u(x, t) ∂u(x, t) + b2 (x, t) + c(x, t)u(x, t) = f (x, t). + b1 (x, t) ∂x ∂t

(1a)

In equation (1a), a11 (x, t), a12 (x, t), a22 (x, t), b1 (x, t), b2 (x, t), c(x, t),and f (x, t) are known functions of the independent variables x and t. In order to simplify notation, it is common to use subscripts to denote partial derivatives. Thus, we may write equation (1a) as a11 (x, t)uxx (x, t) + a12 (x, t)uxt (x, t) + a22 (x, t)utt (x, t) + b1 (x, t)ux (x, t) + b2 (x, t)ut (x, t) + c(x, t)u(x, t) = f (x, t).

(1b)

9.1

Introduction

567

Note the special structure of the partial differential equation (1b) that makes it a linear partial differential equation. The coefﬁcient functions as well as the right-hand side, f (x, t), are functions only of the independent variables. Moreover, the dependent variable u and its partial derivatives occur only to the ﬁrst power. If f (x, t) = 0 in (1b), the equation is called a homogeneous equation. If f (x, t) = 0 in (1b), the equation is nonhomogeneous. In many of the applications we consider, the coefﬁcient functions are constant.

Superposition Linear homogeneous partial differential equations satisfy the same principle of superposition as linear homogeneous ordinary differential equations. In particular, whenever we form a linear combination of solutions, the resulting function is again a solution. For example, suppose u1 (x, t), u2 (x, t), . . . , uN (x, t) are N solutions of the linear homogeneous equation a11 (x, t)uxx (x, t) + a12 (x, t)uxt (x, t) + a22 (x, t)utt (x, t) + b1 (x, t)ux (x, t) + b2 (x, t)ut (x, t) + c(x, t)u(x, t) = 0.

(2)

If k1 , k2 , . . . , kN are N arbitrary constants, then the function u(x, t) = k1 u1 (x, t) + k2 u2 (x, t) + · · · + kN uN (x, t) is likewise a solution of the linear homogeneous equation (2).

Examples of Linear Partial Differential Equations The class of second order linear partial differential equations contains some of the most important equations of mathematical physics. Three such equations discussed in this chapter are the heat equation, the wave equation, and Laplace’s equation. The Heat Equation

κuxx (x, t) − ut (x, t) = 0

or ut (x, t) = κuxx (x, t)

(3)

In (3), the dependent variable u(x, t) represents the temperature at position x and time t in a structure (such as a thin, laterally insulated bar) in which heat is constrained to ﬂow in only one dimension (along the x-axis). The positive constant κ, called the diffusivity, depends on the thermal properties of the material. The appendix of Section 9.2 outlines a derivation of the heat equation. 1

The Wave Equation

uxx (x, t) −

Laplace’s Equation

uxx (x, y) + uyy (x, y) = 0

2

utt (x, t) = 0

or

utt (x, t) = c2 uxx (x, t)

(4)

c The wave equation arises in modeling phenomena, such as the small-amplitude vibrations of a taut string. In contrast to the heat equation, the wave equation involves a second order partial derivative with respect to time, utt (x, t). In this model, the dependent variable u(x, t) represents the displacement of the string at position x and time t. As we shall see, the positive constant c is the speed of the wave. (5)

In equation (5), the dependent variable u(x, y) is not a function of time; instead, it is a function of two spatial variables. In applications, solutions of

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Laplace’s equation often represent potentials, such as electrostatic or gravitational potentials. Laplace’s equation also arises in modeling steady-state (that is, time-independent) heat ﬂow in two or more dimensions. (Recall Project 2 in Chapter 8, which studied steady-state heat ﬂow in a cylindrical pipe.) The heat equation, the wave equation, and Laplace’s equation each have straightforward generalizations to higher spatial dimensions. For example, let u(x, y, z, t) represent the temperature within a solid object at location (x, y, z) and time t. The three-dimensional heat equation modeling temperature within the object is ut (x, y, z, t) = κ[uxx (x, y, z, t) + uyy (x, y, z, t) + uzz (x, y, z, t)]. Similarly, three-dimensional versions of the wave equation and Laplace’s equation are, respectively, utt (x, y, z, t) = c2 [uxx (x, y, z, t) + uyy (x, y, z, t) + uzz (x, y, z, t)] and uxx (x, y, z) + uyy (x, y, z) + uzz (x, y, z) = 0. [When a steady-state temperature exists—that is, when ut (x, y, z, t) = 0—the heat equation reduces to Laplace’s equation.]

Separation of Variables We begin by describing separation of variables, a technique for solving partial differential equations. The discussion of Fourier series in Section 9.5 addresses some of the mathematical questions arising from using separation of variables. Our discussion of the relevant mathematical theory is not complete, however. We do not present a detailed discussion of two important issues that are typically addressed in more advanced studies; in particular, we do not address the question of existence and uniqueness of solutions. The existenceuniqueness theory for the range of problems we will study is too extensive to permit a meaningful and succinct summary. Instead, we tacitly assume the problem under consideration has a unique solution and simply concentrate on the task of computing it. In addition, we sidestep most of the mathematical questions that arise from forming an inﬁnite series of functions. For example, the solutions we construct in this chapter typically have the form u(x, t) =

∞

an un (x, t).

(6)

n=0

Separation of variables will tell us how to form the functions un (x, t), and the theory of Fourier series will tell us how to determine the values of the coefﬁcients an . Some questions then arise: What kind of function is actually deﬁned by the inﬁnite series (6)? Does the series converge for all points (x, t) in the region of interest? If so, is the function deﬁned by the inﬁnite series a solution of the partial differential equation? When we apply the method of separation of variables, each of the functions un (x, t) in equation (6) will be a solution of the given homogeneous partial differential equation. With respect to series (6), however, recall that the superpo-

9.1

Introduction

569

sition result cited earlier applies to a ﬁnite sum of solutions. Demonstrating that the inﬁnite sum of solutions (6) is also a solution requires that we justify interchanging partial differentiation and inﬁnite summation operations. The validity of such interchanges is typically addressed in multivariable advanced calculus. In this chapter, we will simply assume that interchanging operations is permissible and that the series solution we construct is, in fact, the solution of the problem of interest.

EXERCISES Exercises 1–10: Determine the values of the constant α, if any, for which the speciﬁed function is a solution of the given partial differential equation. 1. u(x, t) = 4t − αx2 , ut − uxx = 0

2. u(x, t) = e−αt sin 2x, ut − uxx = 0

3. u(x, t) = e−2αt cos αx, ut − uxx = 0

4. u(x, t) = sin(x + αt), utt − 4uxx = 0

5. u(x, t) = 2 cos(x + αt), utt − uxx + 2u = 0 6. u(x, y) = ex sin αy, uxx + uyy = 0 7. u(x, y, z) = ex+αy sin z, uxx + uyy + uzz = 0 8. u(x, y, t) = eαt sin(x) cos(2y), uxx + uyy − ut = 0 9. u(x, t) = sin(αx) cos(2t), uxx − utt − 4αu = 0 10. u(x, y) = α + e−x cos y, uxx + uyy − 2u = 4

Exercises 11–15: In each exercise, (a) Show by direct substitution that the linear combination of functions is a solution of the given homogeneous linear partial differential equation. (b) Determine values of the constants so that the linear combination satisﬁes the given supplementary condition. 11. u(x, t) = c1 e−t sin x + c2 e−4t sin 2x, uxx − ut = 0; u(x, 0) = 3 sin 2x − sin x 12. u(x, t) = c1 + c2 e−t cos x + c3 e−4t cos 2x, uxx − ut = 0; u(x, 0) = 2 − cos 2x 13. u(x, t) = c1 sin x sin 2t + c2 sin x cos 2t, 4uxx − utt = 0; u(x, 0) = −2 sin x, ut (x, 0) = 6 sin x 14. u(x, t) = c1 e−3t sin x + c2 e−6t sin 2x, uxx − ut − 2u = 0; u(x, 0) = sin x − 4 sin 2x 15. u(x, t) = c1 + c2 (x − t) + c3 (x + t), uxx − utt = 0; u(x, 0) = 1 + 2x, ut (x, 0) = 0 16. Let u1 (x, t) and u2 (x, t) be solutions of the linear homogeneous partial differential equation (2). Show that the linear combination u(x, t) = c1 u1 (x, t) + c2 u2 (x, t) is also a solution of equation (2). 17. Let u1 (x, t) be a solution of the linear homogeneous partial differential equation (2), and let u2 (x, t) be a solution of the linear nonhomogeneous partial differential equation (1b). Show, for any constant c1 , that u(x, t) = c1 u1 (x, t) + u2 (x, t) is also a solution of the nonhomogeneous equation.

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Exercises 18–21: In each exercise, the function u is known to be a solution of the given nonhomogeneous partial differential equation. Determine the function f . 18. u(x, t) = 2 sin(πx) cos(2π t) − x2 t, 4uxx − utt = f (x, t) 19. u(x, y) = xy + 2x2 y3 + 2 sinh x sin y, uxx + uyy = f (x, y) 20. u(x, y, t) = ex+2y + e−5t sin x sin 2y, uxx + uyy − ut = f (x, y, t) 21. u(x, y) = 3 + x − y + 2xy − x2 y3 , uxx + uyy = f (x, y)

9.2

Heat Flow in a Thin Bar; Separation of Variables In this section, we consider two different problems involving heat ﬂow. Besides being of intrinsic interest, these problems serve to introduce a useful solution technique known as separation of variables. Once it is clear how this technique is used to solve the heat equation, we will apply variations of the same basic idea to other partial differential equations, such as the wave equation and Laplace’s equation. This section and Section 9.3 concentrate on the one-dimensional heat equation, ut (x, t) = κuxx (x, t).

(1)

The remaining sections of this chapter focus on other partial differential equations, using separation of variables to solve them.

Problem Formulation Consider a bar having constant cross-sectional area A and length l, as shown in Figure 9.1. We assume that the bar is (a) thin (its length l is much larger than its cross-sectional area A), (b) homogeneous (it has uniform heat-conducting properties characterized by its thermal diffusivity constant κ), and (c) laterally insulated (so that no heat can ﬂow through its sides). Bar

Insulation

l x=0

x=l

x Cross-sectional area A

FIGURE 9.1

A thin bar having cross-sectional area A and length l.

For such a thin bar, it is reasonable to assume, at a given point x on the longitudinal axis, that temperature is constant throughout the corresponding crosssectional area A. Given this assumption, the temperature in the bar depends only on a single spatial variable x and on time t. Since no heat can enter or leave through the lateral surface, heat is constrained to ﬂow along the bar’s axis. Let u(x, t) represent the temperature of the bar at axial location x and time t. We assume that at an initial time, say t = 0, the temperature distribution along the

9.2

Heat Flow in a Thin Bar; Separation of Variables

571

bar is known. That is, u(x, 0), 0 ≤ x ≤ l, is a known function of x. Our goal is to determine how the initial temperature distribution evolves in time. The appendix to this section gives a derivation of the one-dimensional heat equation. In addition to partial differential equation (1), two supplementary conditions are needed to completely specify the problem. The ﬁrst condition is the initial temperature distribution in the bar, u(x, 0), 0 ≤ x ≤ l. The second is a description of how temperature behaves at the bar ends. That is, what happens at x = 0 and x = l? We consider two different endpoint constraints. In Problem 1, we assume that the temperature at both ends is maintained at the value zero for all t ≥ 0. In Problem 2, we assume that the ends of the bar are insulated so that no heat can enter or leave. The mathematical statements of these two problems follow. Solve the heat equation

Problem 1 (Zero Temperature Ends) ut (x, t) = κuxx (x, t),

0 < x < l,

0