Tài liệu Boundary Value Problems, Sixth Edition: and Partial Differential Equations pptx

Preface ix CHAPTER 0 Ordinary Differential Equations 1 0.1 Homogeneous Linear Equations 1 0.2 Nonhomogeneous Linear Equations 14 0.3 Boundary Value Problems 26 0.4 Singular Boundary Valu

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B O U N D A R Y VALUE PROBLEMS

FIFTH EDITION

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B O U N D A R Y VALUE PROBLEMS AND PARTIAL DIFFERENTIAL EQUATIONS

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Preface ix

CHAPTER 0 Ordinary Differential Equations 1

0.1 Homogeneous Linear Equations 1

0.2 Nonhomogeneous Linear Equations 14

0.3 Boundary Value Problems 26

0.4 Singular Boundary Value Problems 38

0.5 Green’s Functions 43

Chapter Review 51 Miscellaneous Exercises 51

CHAPTER 1 Fourier Series and Integrals 59

1.1 Periodic Functions and Fourier Series 59

1.2 Arbitrary Period and Half-Range Expansions 64

1.3 Convergence of Fourier Series 73

1.4 Uniform Convergence 79

1.5 Operations on Fourier Series 85

1.6 Mean Error and Convergence in Mean 90

1.7 Proof of Convergence 95

1.8 Numerical Determination of Fourier Coefﬁcients 100

1.9 Fourier Integral 106

1.10 Complex Methods 113

1.11 Applications of Fourier Series and Integrals 117

1.12 Comments and References 124

v

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vi Contents

CHAPTER 2 The Heat Equation 135

2.1 Derivation and Boundary Conditions 135

2.2 Steady-State Temperatures 143

2.3 Example: Fixed End Temperatures 149

2.4 Example: Insulated Bar 157

2.5 Example: Different Boundary Conditions 163 2.6 Example: Convection 170

2.7 Sturm–Liouville Problems 175

2.8 Expansion in Series of Eigenfunctions 181

2.9 Generalities on the Heat Conduction Problem 184 2.10 Semi-Inﬁnite Rod 188

2.11 Inﬁnite Rod 193

2.12 The Error Function 199

CHAPTER 3 The Wave Equation 215

3.1 The Vibrating String 215

3.2 Solution of the Vibrating String Problem 218 3.3 d’Alembert’s Solution 227

3.4 One-Dimensional Wave Equation: Generalities 233 3.5 Estimation of Eigenvalues 236

3.6 Wave Equation in Unbounded Regions 239

CHAPTER 4 The Potential Equation 255

4.1 Potential Equation 255

4.2 Potential in a Rectangle 259

4.3 Further Examples for a Rectangle 264

4.4 Potential in Unbounded Regions 270

4.5 Potential in a Disk 275

4.6 Classiﬁcation and Limitations 280

CHAPTER 5 Higher Dimensions and Other Coordinates 295

5.1 Two-Dimensional Wave Equation: Derivation 295 5.2 Three-Dimensional Heat Equation 298

5.3 Two-Dimensional Heat Equation: Solution 303

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Contents vii 5.4 Problems in Polar Coordinates 308

5.5 Bessel’s Equation 311

5.6 Temperature in a Cylinder 316

5.7 Vibrations of a Circular Membrane 321

5.8 Some Applications of Bessel Functions 329

5.9 Spherical Coordinates; Legendre Polynomials 335

5.10 Some Applications of Legendre Polynomials 345

CHAPTER 6 Laplace Transform 363

6.1 Deﬁnition and Elementary Properties 363

6.2 Partial Fractions and Convolutions 369

6.3 Partial Differential Equations 376

6.4 More Difﬁcult Examples 383

Miscellaneous Exercises 389

CHAPTER 7 Numerical Methods 397

7.1 Boundary Value Problems 397

Appendix: Mathematical References 435

Answers to Odd-Numbered Exercises 441

Index 495

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This text is designed for a one-semester or two-quarter course in partial ferential equations given to third- and fourth-year students of engineering andscience It can also be used as the basis for an introductory course for graduatestudents Mathematical prerequisites have been kept to a minimum — calculusand differential equations Vector calculus is used for only one derivation, andnecessary linear algebra is limited to determinants of order two A reader needsenough background in physics to follow the derivations of the heat and waveequations

dif-The principal objective of the book is solving boundary value problemsinvolving partial differential equations Separation of variables receives thegreatest attention because it is widely used in applications and because it pro-vides a uniform method for solving important cases of the heat, wave, andpotential equations One technique is not enough, of course D’Alembert’s so-lution of the wave equation is developed in parallel with the series solution,and the distributed-source solution is constructed for the heat equation Inaddition, there are chapters on Laplace transform techniques and on numeri-cal methods

The second objective is to tie together the mathematics developed and thestudent’s physical intuition This is accomplished by deriving the mathemati-cal model in a number of cases, by using physical reasoning in the mathemat-ical development, by interpreting mathematical results in physical terms, and

by studying the heat, wave, and potential equations separately

In the service of both objectives, there are many fully worked examples andnow about 900 exercises, including miscellaneous exercises at the end of eachchapter The level of difficulty ranges from drill and verification of details

to development of new material Answers to odd-numbered exercises are in

ix

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x Preface

the back of the book An Instructor’s Manual is available both online and inprint (ISBN: 0-12-369435-3), with the answers to the even-numbered prob-lems A Student Solutions Manual is available both online and in print (ISBN:0-12-088586-7), that contains detailed solutions of odd-numbered problems.There are many ways of choosing and arranging topics from the book toprovide an interesting and meaningful course The following sections formthe core, requiring at least 14 hours of lecture: Sections 1.1–1.3, 2.1–2.5, 3.1–3.3, 4.1–4.3, and 4.5 These cover the basics of Fourier series and the solutions

of heat, wave, and potential equations in finite regions My choice for the nextmost important block of material is the Fourier integral and the solution ofproblems on unbounded regions: Sections 1.9, 2.10–2.12, 3.6, and 4.4 Theserequire at least six more lectures

The tastes of the instructor and the needs of the audience will govern thechoice of further material A rather theoretical flavor results from including:Sections 1.4–1.7 on convergence of Fourier series; Sections 2.7–2.9 on Sturm–Liouville problems, and the sequel, Section 3.4; and the more difficult parts ofChapter 5, Sections 5.5–5.10 on Bessel functions and Legendre polynomials

On the other hand, inclusion of numerical methods in Sections 1.8 and 3.5and Chapter 7 gives a very applied flavor

Chapter 0 reviews solution techniques and theory of ordinary differentialequations and boundary value problems Equilibrium forms of the heat andwave equations are derived also This material belongs in an elementary differ-ential equations course and is strictly optional However, many students haveeither forgotten it or never seen it

For this fifth edition, I have revised in response to students’ changing needsand abilities Many sections have been rewritten to improve clarity, provideextra detail, and make solution processes more explicit In the optional Chap-ter 0, free and forced vibrations are major examples for solution of differentialequations with constant coefficients In Chapter 1, I have returned to derivingthe Fourier integral as a “limit” of Fourier series New exercises are includedfor applications of Fourier series and integrals Solving potential problems on arectangle seems to cause more difficulty than expected A new section 4.3 givesmore guidance and examples as well as some information about the Poissonequation New exercises have been added and old ones revised throughout

In particular I have included exercises based on engineering research tions These provide genuine problems with real data

publica-A new feature of this edition is a CD with auxiliary materials: animations

of convergence of Fourier series; animations of solutions of the heat and waveequations as well as ordinary initial value problems; color graphics of solu-tions of potential problems; additional exercises in a workbook style; reviewquestions for each chapter; text material on using a spreadsheet for numericalmethods All files are readable with just a browser and Adobe Reader, availablewithout cost

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Preface xi

I wish to acknowledge the skillful work of Cindy Smith, who was the LaTeXcompositor and corrected many of my mistakes, the help of Academic Presseditors and consultants, and the guidance of reviewers for this edition:Darryl Yong, Harvey Mudd College

Ken Luther, Valparaiso University

Alexander Kirillov, SUNY at Stony Brook

James V Herod, Georgia Tech University

Hilary Davies, University of Alaska Anchorage

Catherine Crawford, Elmhurst College

Ahmed Mohammed, Ball State University

I also wish to acknowledge the guidance of reviewers for the previous tion:

edi-Linda Allen, Texas Tech University

Ilya Bakelman, Texas A&M University

Herman Gollwitzer, Drexel University

James Herod, Georgia Institute of Technology

Robert Hunt, Humboldt State University

Mohammad Khavanin, University of North Dakota

Jeff Morgan, Texas A&M University

Jim Mueller, California Polytechnic State University

Ron Perline, Drexel University

William Royalty, University of Idaho

Lawrence Schovanec, Texas Tech University

Al Shenk, University of California at San Diego

Michael Smiley, Iowa State University

Monty Strauss, Texas Tech University

Kathie Yerion, Gonzaga University

David L Powers

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Ordinary Differential

0

0.1 Homogeneous Linear Equations

The subject of most of this book is partial differential equations: their physicalmeaning, problems in which they appear, and their solutions Our principalsolution technique will involve separating a partial differential equation intoordinary differential equations Therefore, we begin by reviewing some factsabout ordinary differential equations and their solutions

We are interested mainly in linear differential equations of first and secondorders, as shown here:

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2 Chapter 0 Ordinary Differential Equations

This equation can be solved by isolating u on one side and then integrating:

It is easy to check directly that the last expression is a solution of the differential

equation for any value of c That is, c is an arbitrary constant and can be used

to satisfy an initial condition if one has been specified

The most common case of this differential equation has k(t) = k constant.

The differential equation and its general solution are

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Chapter 0 Ordinary Differential Equations 3 Principle of Superposition. If u1(t) and u2(t) are solutions of the same linear homogeneous equation (6), then so is any linear combination of them: u(t)=

c1u1(t) + c2u2(t).

This theorem, which is very easy to prove, merits the name of principle

be-cause it applies, with only superficial changes, to many other kinds of linear,homogeneous equations Later, we will be using the same principle on partialdifferential equations To be able to satisfy an unrestricted initial condition, weneed two linearly independent solutions of a second-order equation Two so-

lutions are linearly independent on an interval if the only linear combination of

them (with constant coefficients) that is identically 0 is the combination with 0for its coefficients There is an alternative test: Two solutions of the same linearhomogeneous equation (6) are independent on an interval if and only if their

is nonzero on that interval

If we have two independent solutions u1(t), u2(t) of a linear second-order

homogeneous equation, then the linear combination u(t) = c1u1(t) + c2u2(t)

is a general solution of the equation: Given any initial conditions, c1and c2can

be chosen so that u(t) satisfies them.

1 Constant coefﬁcients

The most important type of second-order linear differential equation that can

be solved in closed form is the one with constant coefficients,

d2u

dt2 + k du

dt + pu = 0 (k, p are constants). (8)

There is always at least one solution of the form u(t) = e mtfor an appropriate

constant m To find m, substitute the proposed solution into the differential

equation, obtaining

m2e mt + kme mt + pe mt = 0,

or

(since e mt is never 0) This is called the characteristic equation of the

differ-ential equation (8) There are three cases for the roots of the characteristicequation (9), which determine the nature of the general solution of Eq (8).These are summarized in Table 1

This method of assuming an exponential form for the solution works forlinear homogeneous equations of any order with constant coefficients In all

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Roots of Characteristic General Solution of Differential

Real, distinct: m1= m2 u(t) = c1e m1t + c2e m2t

Real, double: m1= m2 u(t) = c1e m1t + c2te m1t

Conjugate complex: u(t) = c1e αt cos(βt) + c2e αt sin(βt)

We include two important examples First, consider the differential equation

d2u

dt2 + λ2u = 0, (12)

where λ is constant The characteristic equation is m2+ λ2= 0, with roots

m = ±iλ The third case of Table 1 applies if λ = 0; the general solution of the

The characteristic equation now is m2− λ2= 0, with roots m = ±λ If λ = 0,

the first case of Table 1 applies, and the general solution is

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Chapter 0 Ordinary Differential Equations 5

Figure 1 Mass–spring–damper system

test shows them to be independent Therefore, we may equally well write

u(t) = c

1cosh(λt) + c

2sinh(λt)

as the general solution of Eq (14), where c1 and c2 are arbitrary constants

Example: Mass–Spring–Damper System.

The displacement of a mass in a mass–spring–damper system (Fig 1) is scribed by the initial value problem

The equation is derived from Newton’s second law Coefficients b and ω2

are proportional to characteristic constants of the damper and the spring, spectively The characteristic equation of the differential equation is

The nature of the solution, and therefore the motion of the mass, is determined

by the relation between b/2 and ω.

b = 0: undamped. The roots are±iω and the general solution of the

differ-ential equation is

u(t) = c1cos(ωt) + c2sin(ωt).

The mass oscillates forever

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0 < b/2 < ω: underdamped. The roots are complex conjugates α ±iβ with

α = −b/2, β = ω2− (b/2)2 The general solution of the differential tion is

equa-u(t) = e −bt/2

c1cos(βt) + c2sin(βt)

.

The mass oscillates, but approaches equilibrium as t increases.

b/2 = ω: critically damped. The roots are both equal to b/2 The general

solution of the differential equation is

u(t) = e −bt/2 (c1+ c2t).

The mass approaches equilibrium as t increases and may pass through librium (u(t) may change sign) at most once.

equi-b/2 > ω: overdamped. Both roots of the characteristic equation are real,

say, m1and m2 The general solution of the differential equation is

u(t) = c1e m1t + c2e m2t

.

The mass approaches equilibrium as t increases, and u(t) may change sign at

for this equation is quite similar to the preceding: Assume that a solution has

the form u(t) = t m , and then find m Substituting u in this form into Eq (17)

leads to

t2m(m − 1)t m−2+ ktmt m−1+ pt m = 0, or

m(m − 1) + km + p = 0 (k, p are constants). (18)This is the characteristic equation for Eq (17), and the nature of its roots de-termines the solution, as summarized in Table 2

One important example of the Cauchy–Euler equation is

t2d

2u

dt2 + t du

dt − λ2u = 0, (19)

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Roots of Characteristic General Solution of Differential

Real, distinct roots: m1= m2 u(t) = c1t m1+ c2t m2

Real, double root: m1= m2 u(t) = c1t m1+ c2( ln t)t m1

Conjugate complex roots: u(t) = c1t α cos(β ln t) + c2t α sin(β ln t)

m1= α + iβ, m2= α − iβ

Table 2 Solutions of t 2 d2u

dt2 + kt du

dt + pu = 0

where λ > 0 The characteristic equation is m(m −1)+m−λ2= m2−λ2= 0

The roots are m = ±λ, so the first case of Table 2 applies, and

u(t) = c1t λ + c2t −λ (20)

is the general solution of Eq (19)

For the general linear equation

d2u

dt2 + k(t) du

dt + p(t)u = 0, any point where k(t) or p(t) fails to be continuous is a singular point of the

differential equation At such a point, solutions may break down in various

ways However, if t0is a singular point where both of the functions

(t − t0)k(t) and (t − t0)2p(t) (21)

have Taylor series expansions, then t0 is called a regular singular point The

Cauchy–Euler equation is an example of an important differential equation

having a regular singular point (at t0= 0) The behavior of its solution nearthat point provides a model for more general equations

t=1

z , u(t)=1

z v(z).

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Here are the details The second derivative of u has to be replaced by its pression in terms of v, using the chain rule Start by finding

C Second Independent Solution

Although it is not generally possible to solve a second-order linear neous equation with variable coefficients, we can always find a second inde-

homoge-pendent solution if one solution is known This method is called reduction of

Assume that u2(t) = v(t)u1(t) is a solution We wish to find v(t) so that u2

is indeed a solution However, v(t) must not be constant, as that would not supply an independent solution A straightforward substitution of u2= vu1

into the differential equation leads to

vu1+ 2vu + vu+ k(t)(vu1+ vu) + p(t)vu1= 0.

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Now collect terms in the derivatives of v The preceding equation becomes

which is a first-order linear equation for v Thus, a nonconstant v can be

found, at least in terms of some integrals

Linear homogeneous equations of order higher than 2 — especially order 4 —

occur frequently in elasticity and fluid mechanics A general, nth-order

homo-geneous linear equation may be written

u (n) + k1(t)u (n −1) + · · · + k n−1(t)u ( 1) + k n (t)u = 0, (26)

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Root Multiplicity Contribution

Table 3 Contributions to general solution

in which the coefficients k1(t), k2(t), etc., are given functions of t The

tech-niques of solution are analogous to those for second-order equations In ticular, they depend on the Principle of Superposition, which remains validfor this equation That principle allows us to say that the general solution

par-of Eq (26) has the form par-of a linear combination par-of n independent solutions

u1(t), u2(t), , u n (t) with arbitrary constant coefficients,

differen-m n + k1m n−1+ · · · + k n−1m + k n = 0, (28)

called the characteristic equation of the differential equation (27).

Each distinct root of the characteristic equation contributes as many

inde-pendent solutions as its multiplicity, which might be as high as n Recall also

that the polynomial equation (28) may have complex roots, which will occur

in conjugate pairs if — as we assume — the coefficients k1, k2, etc., are real.When this happens, we prefer to have real solutions, in the form of an expo-nential times sine or cosine, instead of complex exponentials The contribution

of each root or pair of conjugate roots of Eq (28) is summarized in Table 3

Since the sum of the multiplicities of the roots of Eq (28) is n, the sum of the contributions produces a solution with n terms, which can be shown to be the

general solution

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Chapter 0 Ordinary Differential Equations 11 Example.

Find the general solution of this fourth-order equation

u ( 4) + 3u ( 2) − 4u = 0.

The characteristic equation is m4+ 3m2− 4 = 0, which is easy to solve because

it is a biquadratic We find that m2= −4 or 1, and thus the roots are m = ±2i,

±1, all with multiplicity 1 From Table 3 we find that a cos(2t) + b sin(2t) responds to the complex conjugate pair, m = ±2i, while e t and e −tcorrespond

cor-to m = 1 and m = −1 Thus we build up the general solution,

u(t) = a cos(2t) + b sin(2t) + c1e t + c2e −t

u(t) = (c1+ c2t)e t + (c3+ c4t)e −t

With sinh(t) = (e t − e −t )/ 2 and cosh(t) = (e t + e −t )/2, the terms of the ceding combination can be rearranged to give the general solution in a differ-ent form,

pre-u(t) = (C1+ C2t) cosh(t) + (C3+ C4t) sinh(t).

Some important equations and their solutions.

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12. Compare and contrast the form of the solutions of these three differential

equations and their behavior as t→ ∞

a. d2u

dt2 + u = 0; b. d2u

dt2 = 0; c.d2u

dt2 − u = 0.

In Exercises 13–15, use the “exponential guess” method to find the general

solution of the differential equations (λ is constant).

13. d4u

dx4 − λ4u= 0

14. d4u

dx4 + λ4u= 0

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mechanism of rolling instability and chatter” [Y.-J Lin et al., J of

Manu-facturing Science and Engineering, 125 (2003): 778–786], the authors find

that the distance between rollers is well approximated by h + y, where h

is the nominal output thickness and y is the solution of the differential equation y+ 2αy+ σ2y= 0 The elasticity of the sheet and the rollersprovides the restoring force, and the plastic deformation of the sheet ef-fectively provides damping

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For high-speed operation, the system is underdamped Solve the initialvalue problem consisting of the differential equation and the initial con-

ditions y(0) = −0.001h, y( 0)= 0

24. (Continuation) For an input speed of 25.4 m/s, it is observed that σ ∼=

600 Hz or 1200π radians/s and α = 0.103σ Using these values, obtain

a graph of the solution of the preceding exercise, over the range 0 < t < 0.02 s How far does the sheet move in 0.02 s?

25. (Continuation) The damping constant α referred to in the previous ercises appears to depend on v, the speed of the sheet into the rollers, according to the relation α/σ = A/v, where A is a constant From the information given previously, the value of A is about 2.62 Assuming this is correct, find the speed v at which damping is critical.

ex-0.2 Nonhomogeneous Linear Equations

In this section, we will review methods for solving nonhomogeneous linearequations of first and second orders,

Of course, we assume that the inhomogeneity f (t) is not identically 0 The

simplest nonhomogeneous equation is

We have used an indefinite integral and have written c as a reminder that there

is an arbitrary additive constant in the general solution of Eq (1) A moreprecise way to write the solution is

u(t)=

t

t0

f (z) dz + c. (3)Here we have replaced the indefinite integral by a definite integral with vari-able upper limit The lower limit of integration is usually an initial time Note

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0.2 Nonhomogeneous Linear Equations 15

that the name of the integration variable is changed from t to something else (here, z) to avoid confusing the limit with the dummy variable of integration.

The simple second-order equation

d2u

can be solved by two successive integrations

The two theorems that follow summarize some properties of linear tions that are useful in constructing solutions

equa-Theorem 1. The general solution of a nonhomogeneous linear equation has the form u(t) = u p (t) + u c (t), where u p (t) is any particular solution of the nonhomogeneous equation and u c (t) is the general solution of the corresponding homo-

Theorem 2. If u p1 (t) and u p2 (t) are particular solutions of a differential equation with inhomogeneities f1(t) and f2(t), respectively, then k1u p1 (t) + k2u p2 is a particular solution of the differential equation with inhomogeneity k1f1(t) + k2f2(t) (k1, k2are constants).

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Inhomogeneity, f (t) Form of Trial Solution, u p (t)

(a0t n + a1t n−1+ · · · + a n )e αt (A0t n + A1t n−1+ · · · + A n )e αt

(a0t n + · · · + a n )e αt cos(βt)

+ (b0t n + · · · + b n )e αt sin(βt)

(A0t n + · · · + A n )e αt cos(βt) + (B0t n + · · · + B n )e αt sin(βt)

Table 4 Undetermined coefficients

general solution of the given equation is

u(t)= 1 −1

2e

−t + c1cos(t) + c2sin(t).

If two initial conditions are given, then c1and c2are available to satisfy them

Of course, an initial condition applies to the entire solution of the given

Now we turn our attention to methods for finding particular solutions ofnonhomogeneous linear differential equations

A Undetermined Coefﬁcients

This method involves guessing the form of a trial solution and then finding theappropriate coefficients Naturally, it is limited to the cases in which we canguess successfully: when the equation has constant coefficients and the inho-mogeneity is simple in form Table 4 offers a summary of admissible inhomo-geneities and the corresponding forms for particular solution The parameters

n, α, β and the coefficients a0, , a n , b0, , b n are found by inspecting the

given inhomogeneity The table compresses several cases For instance, f (t) in line 1 is a polynomial if α = 0 or an exponential if n = 0 and α = 0 In line 2,

both sine and cosine must be included in the trial solution even if one is absent

from f (t); but α = 0 is allowed, and so is n = 0.

Example.

Find a particular solution of

d2u

dt2 + 5u = te −t .

We use line 1 of Table 4 Evidently, n = 1 and α = −1 The appropriate form

for the trial solution is

u p (t) = (A0t + A1)e −t

When we substitute this form into the differential equation, we obtain

(A0t + A1− 2A0)e −t + 5(A0t + A1)e −t = te −t .

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Now, equating coefficients of like terms gives these two equations for the ficients:

6t+ 118

A trial solution from Table 4 will not work if it contains any term that is a lution of the homogeneous differential equation In that case, the trial solutionhas to be revised by the following rule

so-Revision Rule. Multiply by the lowest positive integral power of t such that no term in the trial solution satisfies the corresponding homogeneous equation.

The trial solution contains a term (A1e −t )that is a solution of the

homoge-neous equation Multiplying the trial solution by t eliminates the problem.

Thus, the trial solution is

has to be revised The solution of the corresponding homogeneous equation

is u c (t) = c1e −t + c2te −t The trial solution from the table has to be multiplied

by t2to eliminate solutions of the homogeneous equation

Example: Forced Vibrations.

The displacement u(t) of a mass in a mass–spring–damper system, starting

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Figure 2 Mass–spring–damper system with an external force

from rest, with an external sinusoidal force (see Fig 2) is described by thisinitial value problem:

See the Section 1 example on the mass–spring–damper system The

coeffi-cient f0is proportional to the magnitude of the force There are three tant cases

impor-b = 0, µ = ω: undamped, no resonance. The form of the trial solution is

u p (t) = A cos(µt) + B sin(µt).

Substitution and simple algebra lead to the particular solution

u0(t)= f0

ω2− µ2cos(µt) (that is, B= 0) The general solution of the differential equation is

u(t)= f0

ω2− µ2cos(µt) + c1cos(ωt) + c2sin(ωt).

Applying the initial conditions determines c1 and c2 Finally, the solution ofthe initial value problem is

u(t)= f0

ω2− µ2

cos(µt) − cos(ωt).

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b = 0, µ = ω: resonance. Now, since ω = µ, the trial solution must be

2µ t sin(µt) + c1cos(µt) + c2sin(µt).

(Remember that b = 0 and ω = µ.) The initial conditions give c1= c2= 0, sothe solution of the initial value problem is

u(t)= f0

2µ t sin(µt).

The presence of the multiplier t means that the amplitude of the oscillation is

increasing This is the phenomenon of resonance

b > 0: damped motion. The ideas are straightforward applications of the

techniques developed earlier The trial solution is a combination of cos(µt) and sin(µt) Somewhat less simple algebra gives

u p (t)= f0

(ω2− µ2) cos(µt) + µb sin(µt),

where = (ω2−µ2)2+µ2b2 The general solution of the differential equation

may take different forms, depending on the relation between b and ω (See

Section 1.) Assuming the underdamped case holds, we have

u(t)= f0

(ω2− µ2) cos(µt) + µb sin(µt)+ e −bt/2

c1cos(γ t) + c2sin(γ t)

for the general solution of the differential equation Here, γ = ω2− (b/2)2

is real because we assumed underdamping

Applying the initial conditions gives, after some nasty algebra,

ω2+ µ2

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Notice that, as t increases, the terms that come from the complementary

solution approach 0, while the terms that come from the particular solutionpersist These cases are illustrated with animation on the CD

B Variation of Parameters

Generally, if a linear homogeneous differential equation can be solved, the responding nonhomogeneous equation can also be solved, at least in terms ofintegrals

The latter is a nonhomogeneous equation of simplest type, which can be

solved for v(t) in one integration.

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or, after canceling 5ve 5tfrom both sides and simplifying, we find

e −5t

From here, we obtain u p (t) = v(t) · e 5t= −1

5t+ 1 25

dv1

dt u1+dv2

dt u2= 0, (12)then we find that

v1u1+ v

2u2+ v1(u1+ k(t)u

1+ p(t)u1) + v2(u2+ k(t)u

2+ p(t)u2) = f (t) This simplifies further: The multipliers of v1and v2are both 0, because u1and

u satisfy the homogeneous Eq (10)

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Thus, we are left with a pair of simultaneous equations,

the Wronskian of u1 and u2 Since these were to be independent solutions of

Eq (10), their Wronskian is nonzero, and we may solve for v1(t) and v2(t) and

hence for v1and v2

These equations are to be integrated to find v1and v2, and then u p (t).

Finally, we note that v1(t) and v2(t) can be found from Eqs (12) and (15)

u1(t)f (t) W(t) dt. (21)

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Now, Eq (11) may be used to form a particular solution of the neous equation (9)

nonhomoge-We may also obtain v1and v2by using definite integrals with variable upperlimit:

v1(t)= −

t

t0

u2(z)f (z) W(z) dz, v2(t)=

t

t0

u1(z)f (z) W(z) dz. (22)

The lower limit is usually the initial value of t, but may be any convenient

value The particular solution can now be written as

u p (t) = −u1(t)

t

t0

u2(z)f (z) W(z) dz + u2(t)

t

t0

u1(z)f (z) W(z) dz.

Furthermore, the factors u1(t) and u2(t) can be inside the integrals (which are not with respect to t), and these can be combined to give a tidy formula, as

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r du dr

in two cases: (a) µ = ω, (b) µ = ω.

In Exercises 13–19, use variation of parameters to find a particular solution ofthe differential equation Be sure that the differential equation is in the correctform

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23. In “Model for temperature estimation of electric couplings suffering heavy

lightning currents” [A.D Polykriti et al., IEE Proceedings — Generation,

Transmission and Distribution, 151 (2004): 90–94], the authors model the

temperature rise above ambient in a coupling with this initial value lem:

prob-ρc dT

dt = i2(t)R(1 + αT), T(0) = 0.

Parameters: ρ is density, c is specific heat, i(t) is the current due to a ening strike, R is the resistance of the coupling at ambient temperature, and the factor (1 + αT) shows how resistance increases with temperature.

light-Simplify the differential equation algebraically to get

dT

dt = Ki2(t)(β + T), T(0) = 0, and identify β and K in terms of the other parameters.

24. (Continuation) The authors model the lightning current with the

func-tion i(t) = Imax(e −λt − e −µt )/n, where n is a factor to make Imax the tual maximum Obtain graphs of this function and the simpler function

ac-i(t) = Imaxe −λt , using these values: Imax = 100 kA, n = 0.93, λ = 2.1,

µ = 150 The unit for time is milliseconds Graph for t from 0 to 2 ms,

which is the range of interest

25. (Continuation) Solve the initial value problem using the simpler function

for current (Don’t forget to square.) Graph the result for t from 0 to 2 ms, using β = 0.26 and K = 13.

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0.3 Boundary Value Problems

A boundary value problem in one dimension is an ordinary differential

equa-tion together with condiequa-tions involving values of the soluequa-tion and/or its atives at two or more points The number of conditions imposed is equal tothe order of the differential equation Usually, boundary value problems of anyphysical relevance have these characteristics: (1) The conditions are imposed

deriv-at two different points; (2) the solution is of interest only between those twopoints; and (3) the independent variable is a space variable, which we shall

represent as x In addition, we are primarily concerned with cases where the

differential equation is linear and of second order However, problems in ticity often involve fourth-order equations

elas-In contrast to initial value problems, even the most innocent lookingboundary value problem may have exactly one solution, no solution, or aninfinite number of solutions Exercise 1 illustrates these cases

When the differential equation in a boundary value problem has a knowngeneral solution, we use the two boundary conditions to supply two equationsthat are to be satisfied by the two constants in the general solution If the dif-ferential equation is linear, these are two linear equations and can be easilysolved, if there is a solution

In the rest of this section we examine some physical examples that are rally associated with boundary value problems

natu-Example: Hanging Cable.

First we consider the problem of finding the shape of a cable that is fastened

at each end and carries a distributed load The cables of a suspension bridge

provide an important example Let u(x) denote the position of the centerline

of the cable, measured upward from the x-axis, which we assume to be zontal (See Fig 3.) Our objective is to find the function u(x).

hori-The shape of the cable is determined by the forces acting on it In our sis, we consider the forces that hold a small segment of the cable in place (SeeFig 4.) The key assumption is that the cable is perfectly flexible This meansthat force inside the cable is always a tension and that its direction at everypoint is the direction tangent to the centerline

analy-Figure 3 The hanging cable

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0.3 Boundary Value Problems 27

We suppose that the cable is not moving Then by Newton’s second law,the sum of the horizontal components of the forces on the segment is 0, and

likewise for the vertical components If T(x) and T(x+ x) are the magnitudes

of the tensions at the ends on the segment, we have these two equations:

T(x + x) cosφ(x + x)− T(x) cosφ(x)

= 0 ( Horizontal), (1)

T(x + x) sinφ(x + x)− T(x) sinφ(x)

− f (x) x = 0 (Vertical) (2)

In the second equation, f (x) is the intensity of the distributed load, measured

in force per unit of horizontal length, so f (x) x is the load borne by the small

segment

From Eq (1) we see that the horizontal component of the tension is the same

at both ends of the segment In fact, the horizontal component of tension has

the same value — call it T — at every point, including the endpoints where

the cable is attached to solid supports By simple algebra we can now find thetension in the cable at the ends of our segment,

Before going further we should note (Fig 4) that φ(x) measures the angle

between the tangent to the centerline of the cable and the horizontal As the

position of the centerline is given by u(x), tan(φ(x)) is just the slope of the cable at x From elementary calculus we know

Tiêu đề	Boundary Value Problems Fifth Edition
Tác giả	David L. Powers
Trường học	Clarkson University
Chuyên ngành	Boundary Value Problems and Partial Differential Equations
Thể loại	Sách giáo trình
Năm xuất bản	2006
Thành phố	Burlington

Định dạng
Số trang	515
Dung lượng	2,45 MB