Ordinary Differential Equations

Transcription

1 Ordinary Differential Equations

2

3 Ordinary Differential Equations: Methods and Applications W. T. Ang and Y. S. Park Universal Publishers Boca Raton

4 Ordinary Differential Equations: Methods and Applications Copyright 2008 W. T. Ang and Y. S. Park All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the publisher Universal Publishers Boca Raton, Florida USA 2008 ISBN-10: /ISBN-13: (paper) ISBN-10: /ISBN-13: (ebook) Library of Congress Cataloging-in-Publication Data Ang, W. T., Ordinary differential equations : methods and applications / W.T. Ang and Y.S. Park. p. cm. Includes bibliographical references and index. ISBN (pbk. : alk. paper) 1. Differential equations. I. Park, Y. S., II. Title. QA372.A '.352--dc

5 To our parents Everywhere, we learn from those whom we love Johann Wolfgang von Goethe Preface This introductory course in ordinary differential equations is intended for junior undergraduate students in applied mathematics, science and engineering. It focuses on methods of solutions and applications rather than theoretical analyses. Applications drawn mainly from dynamics, population biology and electric circuit theory are used to show how ordinary differential equations appear in the formulation of problems in science and engineering. The calculus required to comprehend this course is rather elementary, involving differentiation, integration and power series representation of only real functions of one variable. A basic knowledge of complex numbers and their arithmetic is also assumed, so that elementary complex functions which can be used for working out easily the general solutions of certain ordinary differential equations can be introduced. The pre-requisites just mentioned aside, the course is mainly self-contained. The course comprises six chapters. Chapter 1 gives the basic concepts of ordinary differential equations, explaining what an ordinary differential equation is and what is involved in solving such an equation. It also illustrates how ordinary differential equations can be derived from physical laws or basic principles for two specific examplesof problems. In Chapter 2, methods of solution are given for some first order ordinary differential equations. The equations studied include those which can be written in separable form, those which are linear, and the nonlinear Bernoulli differential equation. Mathematical models which describe population growth are given as examples of applications involving first order ordinary differential equations. v

6 In Chapter 3, the mathematical theory for constructing general solutions of second order linear ordinary differential equations is studied. It is applied to obtain general solutions of second order linear ordinary differential equations with constant coefficients and the Euler-Cauchy equations. Also discussed is the extension of the theory to higher order linear ordinary differential equations. Chapter 4 shows how linear ordinary differential equations with constant coefficients arise in the formulation of problems involving electric circuits and spring-mass systems. Specific examples of problems are solved. Chapter 5 introduces the power series method and the Frobenius method for deriving series solutions of rather general homogeneous second order linear ordinary differential equations. The methods studied can be applied to solve some well known ordinary differential equations in mathematical physics, such as the Legendre s equation and the Bessel s equation, giving rise to particular special functions, but those equations and the associated special functions are not examined in this course. Chapter 6 describes some simple numerical methods for solving first and second order ordinary differential equations. For a particular example of applications, the second order nonlinear ordinary differential equation which governs the motion of a swinging pendulum is solved numerically. Exercises are set not only to test the understanding of students but sometimes also to impart additional insights into the materials studied. Suggested solutions to all the exercises are given at the end of the chapters. To promote the use of this course for self-study, the solutions provided are by and large complete with details. W. T. Ang and Y. S. Park, Singapore, 2008, 2014 vi

7 Contents 1 Basic concepts WhatisanODE? SolvinganODE Generalsolution Particularsolution Exactclosedformsolution ExerciseI WhystudyODEs? ODEforabodyinmotion ODEforapursuitproblem ExerciseII SolutionstoExerciseI SolutionstoExerciseII First order ODEs Preamble FirstorderODEsinseparableform Linear1storderODEs Homogeneous linear 1st order ODEs Nonhomogeneous linear 1st order ODEs Bernoulli differentialequation Populationdynamics Malthus theory of unlimited growth Verhulsttheoryoflimitedgrowth ExerciseIII SolutionstoExerciseIII vii

8 3 Second order linear ODEs Preamble General solution of homogeneous 2nd order linearode Linearlyindependentfunctions Constructionofgeneralsolution Homogeneous 2nd order linear ODEs with constant coefficients Euler-Cauchyequations ExerciseIV SolvingnonhomogeneousODEs Finding a particular solution by guesswork Methodofvariationofparameters ExtensiontohigherorderlinearODEs General N-thorderlinearODEs General solution of a homogeneous ODE General solution of a nonhomogeneous linearode ExerciseV SolutionstoExerciseIV SolutionstoExerciseV Circuits and springs Preamble Electriccircuits Basicelectricalcomponents Voltage across an electric component ODEsinelectriccircuits ExerciseVI Spring-masssystems Asimplespring-masssystem A more complicated spring-mass system ExerciseVII SolutionstoExerciseVI SolutionstoExerciseVII viii

9 5 Series solutions Preamble Reviewofpowerseries PowerseriesmethodforODEs ExerciseVIII Frobeniusmethod ExerciseIX SolutionstoExerciseVIII SolutionstoExerciseIX Numerical methods Preamble Euler smethodfor1storderodes SecondorderODEs Oscillation of a pendulum NonlinearODE ODE for very small oscillation Numerical solution for larger oscillation Numericalprudence ExerciseX SolutionstoExerciseX ix

10 Chapter 1 Basic concepts 1.1 What is an ODE? An equation which contains the derivative(s) of a yet to be determined function y(x) (a function of one variable) is called an ordinary differential equation (ODE) in y(x). Below are some examples of ODEs in y(x): (1) dy 2x =0 dx (2) d 2 y dx 2 +3dy 2y(x) =x5 dx (3) x 3 d3 y dx 3 +6x2 d2 y 3xy(x) =sin(2x) dx2 (4) 2x 2 (y(x)) 10 d4 y dx 4 +3x d2 y dx 2 = xy(x) An ODE in y(x) issaidtobeofordern if d N y/dx N is the highest order derivative of y(x) present in the ODE. In the examples above, (1) is an ODE of order 1 (or 1st order ODE); (2) is of order 2; (3) is of order 3; and (4) is of order 4. It may be sometimes convenient to use the notation y 0 (x) = dy dx,y00 (x) = d2 y dx 2,y000 (x) = d3 y dx 3,,y(N) (x) = dn y dx N. 10

11 Thus, we may write the 4th order ODE in (4) above as or even more simply as 2x 2 (y(x)) 10 y 0000 (x)+3xy 00 (x) =xy(x), 2x 2 y 10 y xy 00 = xy if it is already understood that y is a function of x. Notethat y 10 above refers to y raised to the power 10, nottobemistaken as the 10th order derivative y (10) = d 10 y/dx 10. More generally, we may regard an ODE in y(x) asanequation of the general form F (x, y, y 0,y 00,y 000,,y (N 1),y (N) )=0. Here F denotes a mathematical expression involving x, y, y 0, y 00,y 000,,y (N 1) and y (N). 1.2 Solving an ODE Given an ODE in y(x), we are interested in finding functions y(x) that satisfy the equation, that is, we are interested in solving the ODE. For the purpose of illustration, let us now consider solving the ODE which may be rewritten as y 00 6x =0 d 2 y dx 2 =6x. We may solve the ODE above by directly integrating it twice with respect to x, that is, Z dy dx = 6xdx =3x 2 + C Z y = (3x 2 + C)dx = x 3 + Cx + D. 11

12 Here C and D are constants which may have any values (arbitrary constants). Thus, we find that y(x) =x 3 + Cx + D (where C and D are arbitrary constants) satisfies the ODE y 00 6x =0. Not all ODEs may be easily solved by direct integration like in the example above. In subsequent chapters, we will look at various methods for solving ODEs General solution From the example above, it appears that solving an ODE is really undoing the derivatives in the ODE. Roughly speaking, if an ODE is of order N, we are required to integrate it N times in order to solve it. Consequently, N arbitrary constants appear in the solution obtained. For our purpose, we regard any function y(x) with N arbitrary constants in it as a general solution of an N-th order ODE, if the function satisfies the ODE. Thus, we may regard y(x) =x 3 + Cx + D as a general solution of the ODE y 00 6x = Particular solution If some or all of the arbitrary constants in a general solution of an ODE assume specific values, we obtain a particular solution of the ODE. Examples of particular solutions of the ODE y 00 6x =0 are y(x) =x 3 + x + D, y(x) =x 3 + Cx 2andy(x) =x 3 3x Exact closed form solution We regard a solution of an ODE to be in exact closed form if the solution can be directly expressed in terms of elementary functions 1. 1 For our purpose, a function is regarded as elementary if it can be calculated directly using the function keys of an ordinary scientific hand calculator. Thus, cos(x), sin(x), exp(x),g(x) =x 3 1andf(x) =cos 3 (x 2 + 1) exp( x 2 ) are elementary. 12

13 Thus, the general solution y(x) =x 3 + Cx+ D of the ODE y 00 6x = 0 is in exact closed form. If the values of C and D are given, we may readily evaluate y(x) =x 3 + Cx+ D for any value of x by using an ordinary hand calculator. We may not be able to find exact closed form solutions of some ODEs. For example, take the ODE dy dx = sin(x) x whose solution is theoretically given by Z sin(x) y = x dx. It is, however, not possible to express the integral (on the right hand side) in terms of elementary functions. Thus, the ODE does not have an exact closed form solution. 1.3 Exercise I 1. Check by direct substitution whether y(x) =e x +2e x is a solution of each of the following ODEs in y(x) ornot. (Substitute y(x) =e x +2e x intothelefthandsideofa given ODE and simplify to see if it is possible to obtain the right hand side.) (a) y 00 y =0 (b) y y 00 2y 0 =2e x 4e x (c) 2y 00 +2y 0 3y = e 2x +5e x (d) y y 00 y 0 3y =0 2. Find out whether each of the following functions is a solution of the ODE p 0000 (x) 5p 00 (x) 36p(x) =0ornot. (a) (b) (c) (d) p(x) = 2 sinh(3x) p(x) =cosh(2x) p(x) =cos(3x) sin(3x) p(x) =2sin(2x)+3cos(2x) 13

14 3. If y(x) =α sin(2x)+β cos(2x) is a solution of the ODE y 00 +4y 0 +3y =3sin(2x), find the constants α and β. (Hint. Substitute y(x) = α sin(2x) + β cos(2x) into the ODE and choose the constants α and β in such a way that the equation is satisfied for all x.) 4. If y (x) =α + βx + γx 2 is a solution of the ODE x 3 y x 2 y 00 y = x 2 +2x +1, find the constants α, β and γ. 5. Each of the following ODEs in y(x), if rewritten in an appropriate form, can be solved by direct integration. Find general solutions of these ODEs. (a) xy 0 = x 6 +5x (b) y 0 e x =5 (c) y 00 +6x 2 2=0 (d) xy 0 + y =4x 3 +2x (Hint. d dy (xy) =x dx dx + y) 6. Verify that s 5 y(x) = 2cos(x)+4sin(x)+10e 2x is a solution of the ODE y 0 + y y 3 sin(x) =0. (Hint. Differentiate the solution with respect to x to obtain y 0 = (sin(x) 2cos(x) 10e 2x )y 3 /5. You may do this by squaring both sides of the solution first.) 1.4 Why study ODEs? Some problems in science and engineering may be formulated in terms of ODEs. Below are two examples of such problems. More examples will be given in later chapters ODE for a body in motion Consider the motion of a body which is dropped at some height above the ground. 14

15 We assume that the body, which is pulled by gravity towards the earth, moves along a straight line which is perpendicular to the surface of the earth (that is, it moves along a vertical path). In addition to gravity, the body is also acted upon by a force due to air resistance. A sketch of the situation is given in Figure 1.1. Let s(t) be the vertical downward displacement (in meter) of the body from a fixed point P (which may lie on the path of motion) 2. We are interested in finding s(t), that is, the position of the body at time t. Figure 1.1 All motion obeys Newton s law which may be stated as follows. If the mass of a moving body is constant, then the total force (in newton) acting on the body is equal to the product of the mass (in kilogram) and the 2 We define s(t) to be the downward displacement of the object from P at time t. This means that s(t) > 0iftheobjectisbelowP at time t, and s(t) < 0ifitisaboveP. It is, of course, possible to choose P in such a way that s(t) > 0atalltimet during which the object is falling. 15

16 acceleration (in meter per second per second) of the body. Force and acceleration are vectors. In simple terms, a vector is a quantity which has a magnitude and a direction. Thus, Newton s law as stated above implies that the force and the acceleration are in the same direction. For the situation depicted in Figure 1.1, Newton s law is simply: (total downward force acting on body) = (mass of body) (downward acceleration of body). As mentioned above, we assume that there are only two types of forces acting on the body, namely gravity and air resistance. The gravitational force has a magnitude given by mg, where m is the mass of the body and g is the acceleration due to gravity (that is, g ' 9.81 meter per second per second near planet earth). The gravitational force acts downward. The magnitude of the force due to air resistance is taken to be given by k s 0 (t). Here k is a positive coefficient which depends on, among other things, the shape of the body. The derivative s 0 (t) is the downward velocity 3 of the body. In this case, s 0 (t) is always greater than zero at any time t during the motion as the body is always moving towards the ground. The force due to air resistance acts upward against the motion of the body. It is obvious that (total downward force on body) = mg k s 0 (t) = mg ks 0 (t) (since s 0 (t) > 0). The downward acceleration of the body is given by s 00 (t). Newton s law now becomes mg ks 0 (t) =ms 00 (t). 3 Downward velocity implies that s 0 (t) > 0 if the object is moving downward, and s 0 (t) < 0ifitismovingupward. 16

17 The above equation is an ODE in s(t) whichmayberewritten as s 00 (t)+ k m s0 (t) g =0. If we can solve the ODE for s(t), we can predict the position of the body at any time t during which it is moving towards the earth. For the case in which k is a non-zero constant, the ODE may be solved using methods of solution in later chapters (see Problem 3 in Exercise V on page 84). For now, let us consider the special case in which the effect of air resistance on the motion is so small that it may be ignored, that is, we take k/m in the ODE to be zero. For such a case, the ODE which describes Newton s law of motion reduces to s 00 (t) g =0 or s 00 (t) =g. This simpler ODE may be solved by directly integrating it twice with respect to t. Bearing in mind that g is a constant, we obtain Z ds dt = gdt = gt + C Z s(t) = (gt + C)dt = 1 2 gt2 + Ct + D. Here C and D are arbitrary constants. To calculate the position of the body, we need to know the values of the constants C and D. The values of these constants may be determined if the displacement and velocity of the body are known at a certain time. For example, if we are told that s 0 (t) = 3 (meter per second) when t = 0 (second) then 3=g 0+C C =3. Furthermore, if s(t) = 2(meter) at t = 0, we obtain 2= 1 2 g D D =2. 17

18 Thus, if there is no air resistance, and if the displacement and velocity at t = 0 are 2 and 3 (in the appropriate units) respectively, then the displacement of the body is given by s(t) = 1 2 gt2 +3t +2 at any time t during which the body is falling towards the ground ODE for a pursuit problem With reference to a Cartesian coordinate system denoted by Oxy, consider the following pursuit problem. At time t =0, a navy ship is at (a, 0), where a>0, and it spots a small boat at (0, 0). The boat moves with a constant speed u along the positive y axis. The navy ship decides to chase after the boat by moving at a constant speed v andinadirectionthatis always directed at the boat. We are interested in finding the path of the navy ship. Figure 1.2 A geometrical sketch of the problem is given in Figure 1.2. Let the path be given by the curve y = p(x), where p(x) isan 18

19 unknown function yet to be determined. The position of the boat at time t is B (0,ut)(alongthey axis). At time t, the navy ship is located at N (x, p(x)). Now the line NB (dotted line in Figure 1.2) is tangential to the path y = p(x) atn because the navy ship is always directed at the boat. Thus,the gradient of NB is equal to the gradient of the tangent to the curve y = p(x) atthepointn (x, p(x)), that is, p 0 p(x) ut (x) = x xp 0 (x) =p(x) ut. Since the ship is moving, x and y are functions of t. Let us differentiate the above equation once with respect to t. Using the chain rule for differentiation, we obtain which simplifies to xp 00 (x) dx dt + p0 (x) dx dt = p0 (x) dx dt u xp 00 (x) dx dt = u. Note that dx/dt and dy/dt are the x and y components of the velocity of the navy ship. Since its speed is v, we may write r ( dx dt )2 +( dy dt )2 = v. From y = p(x) and the chain rule for differentiation, we find that Thus, dy dt = p0 (x) dx dt. r ( dx dt )2 (1 + (p 0 (x)) 2 )=v dx dt = ± v p 1+(p 0 (x)). 2 19

20 From Figure 1.2, it is clear that dx/dt < 0asthex-coordinate of the ship decreases during the pursuit of the boat. Thus, we take dx dt = v p 1+(p 0 (x)). 2 Substituting the above expression for dx/dt into the equation xp 00 (x)dx/dt = u, we obtain the 2nd order ODE in p(x) given by xp 00 (x) u v p 1+(p 0 (x)) 2 =0 If we can solve this ODE for p(x), we obtain a formula for the path taken by the navy ship when it chases after the small boat. The two arbitrary constants in the general solution of the ODE may be determined from the conditions p(a) =0andp 0 (a) =0. (Can you see how the conditions come about?) The ODE may be converted into a 1st order ODE and solved as explained in Chapter 2 (on page 30). 1.5 Exercise II 1. An extremely thin wire lies along the x-axis from x =0 to x = ` (` is the length of the wire). The temperature in the wire varies from one point to the next along the wire, that us, it is a function of x. If it is denoted by T (x), then that under certain conditions T (x) satisfies the simple 2nd order ODE d 2 T dx 2 =0. If temperature is given by T 0 and T` at x =0andx = ` respectively, find the temperature throughout the whole wire. 2. The tangent to a curve at the point (x, y) has gradient given by x 3 +2x +1. Iftheequationofthecurveisgiven by y = p(x), writedownanodeinp(x). Find the curve given that (1, 2) is a point on it. 20

21 3. The tangent to a curve at the point (x, y) has gradient given by (x y)/x. Iftheequationofthecurveisgiven by y = p(x), writedownanodeinp(x). Find the curve given that (1, 1) is a point on it. (Hint. Usetheformula d dp (xp(x)) = x + p(x) to help you to solve the ODE.) dx dx 4. A body is moving to the right along a horizontal line. Its speed at time t 0isgivenby8t 3 +3t If w(t) is the distance of the body from a fixed point P on the line, writedownanodeinw(t) for the motion of the body. Where is the body at time t = 2 given that it is a unit distance away from P at time t =0? 5. A body of 1 kilogramme is moving to the right along a horizontal line. It is acted upon by a rightward force given by F (t) =1+e 2t (in newton) at time t (in second). If s(t) istherightward displacement (in meter) of the body from a fixed point P on the line, use Newton s law of motion to write down an ODE in s(t). Find the displacementofthebodyattimet =1giventhatitisat rest at the point P at time t =0. 6. The uniform temperature of a body at time t is given by T (t). The rate of change of the body temperature per unit time is given by c(t (t) T ambient ), where T ambient is the constant temperature of the surrounding atmosphere and c is a constant. Express the statement in the last sentence by writing down an ODE in T (t). What can you say about the constant c? (Hint. Tofind out something about c, ask whether the body should be gaining or losing heat to its surrounding when T (t) >T ambient. What about when T (t) <T ambient?) 21

22 1.6 Solutions to Exercise I 1. From y = e x +2e x, we have y 0 = e x 2e x, y 00 = e x +2e x = y and y 000 = e x 2e x = y 0. Thus: (a) y 00 y =0(fromy 00 = y). Yes. (y = e x +2e x is a solution of the ODE.) (b) y y 00 2y 0 =2e x +8e x 6=2e x 4e x. No. (y = e x +2e x is not a solution of the ODE.) (c) 2y 00 +2y 0 3y = e x 6e x (d) y y 00 y 0 3y 6= e 2x +5e x. No. = y 0 +3y y 0 3y =0. Yes. 2. (a) p = 2 sinh(3x), p 0 =6cosh(3x), p 00 =18sinh(3x) = 9p, p 000 =9p 0 and p 0000 =9p 00 =81p. Thus, p p 00 36p =81p 45p 36p =0. Yes, p = 2 sinh(3x) is a solution of the given ODE. (b) p =cosh(2x), p 0 = 2 sinh(2x), p 00 = 4 cosh(2x) =4p, p 000 =4p 0 and p 0000 =16p. Thus, p p 00 36p =16p 20p 36p 6= 0. No, p =2cosh(2x) is not a solution of the given ODE. (c) p(x) =cos(3x) sin(3x) is not a solution. (d) p(x) =2sin(2x) +3cos(2x) isasolution. 3. Substitute y = α sin(2x)+β cos(2x) intoode,weobtain ( α 8β)sin(2x)+( β +8α)cos(2x) =3sin(2x). Thus, for the ODE to be satisfied for all x, we require α +8β = 3 and8α β =0. This gives α = 3/65 and β = 24/65. 22

23 4. Substitute y = α + βx + γx 2 into ODE to obtain α βx + γx 2 = x 2 +2x +1. This gives α = 1, β = 2 andγ =1. 5. (a) Rewrite ODE as dy dx = x11/2 +5x 1/2 and integrate directly once to obtain y = 2 13 x13/ x3/2 + C. (b) Rewrite ODE as and integrate to obtain (c) Rewrite ODE as y 0 =5e x y = 5e x + C. d 2 y dx 2 = 6x2 +2 and integrate twice to obtain y = 1 2 x4 + x 2 + Cx + D. (d) Rewrite ODE as Integrateoncetoobtain d dx (xy) =4x3 +2x. xy = x 4 + x 2 + C y = x 3 + x + C x. 23

24 6. It may be very tedious to proceed by directly differentiating s 5 y(x) = 2cos(x)+4sin(x)+10e 2x. Instead, rewrite the above as y 2 = 5 (2 cos(x)+4sin(x)+10e 2x ) y 2 (2 cos(x)+4sin(x)+10e 2x )=5 and differentiate both sides with respect to x to obtain 2yy 0 ( 5 y 2 )+y2 ( 2sin(x)+4cos(x)+20e 2x )=0 10y 0 = y 3 ( 2sin(x)+4cos(x)+20e 2x ) 10y 0 +10y 10y 3 sin(x) = y[ 5(2 sin(x) 4cos(x) 20e2x ) 2cos(x)+4sin(x)+10e 2x sin(x) + 2cos(x)+4sin(x)+10e 2x ] = 0 y 0 + y y 3 sin(x) =0(asrequired). 1.7 Solutions to Exercise II 1. The conditions are T (0) = T 0 and T (`) =T`. Solving the ODE, we find that d 2 T dt =0 = C T = Cx + D. dx2 dx Applying the conditions, we obtain T (0) = T 0 = D T (`) = T` = C` + T 0 C = T` T 0. ` Thus, the required temperature is T (x) = T` T 0 x + T 0. ` 24

25 2. The ODE giving the tangent of the curve is Integrating, we obtain dp dx = x3 +2x +1. p(x) = 1 4 x4 + x 2 + x + C. From the point (1, 2), we know p(1) = 2. Thus, The required curve is 3. The ODE is C =2 C = 1 4. y = 1 4 x4 + x 2 + x 1 4. dp dx = x p x Integrating, we obtain xp 0 + p = x d (xp) =x. dx xp = 1 2 x2 + C p = 1 2 x + C x. From the point (1, 1), we have p(1) = 1, that is, C = 1/2. Thus, the required curve is 4. The ODE is y = 1 2 x 1 2x (for x>0). dw dt =8t3 +3t 2 +1 which can be integrated to obtain w =2t 4 + t 3 + t + C. From w(0) = 1, we find that C =1. Thus, w(2) = 42+1 =