Beginning partial differential equations, 3rd edition

vi CONTENTS4.5.1 Uniqueness and Existence 614.5.2 Neumann Problem for a Rectangle 624.5.3 Neumann Problem for a Disk 63 4.7 An Existence Theorem for the Dirichlet Problem 65 5 Fourier In

Solutions Manual for Beginning Partial

Differential Equations

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Library of Congress Cataloging in Publication Data is available.

ISBN 978-1-118-63009-9

Printed in the United States of America

10 9 8 7 6 5 4 3 2 1

1.4 A Proof of the Convergence Theorem 14

2 Solutions of the Heat Equation 15

2.1 Solutions on an Interval [0, L] 15

3 Solutions of the Wave Equation 25

3.1 Solutions on Bounded Intervals 25

3.2.2 The Cauchy Problem on a Half Line 363.2.3 Characteristic Triangles and Quadrilaterals 413.2.4 A Cauchy Problem with a Forcing Term 413.2.5 String with Moving Ends 423.3 The Wave Equation in Higher Dimensions 463.3.1 Vibrations in a Membrane with Fixed Frame 463.3.2 The Poisson Integral Solution 473.3.3 Hadamard’s Method of Descent 47

4 Dirichlet and Neumann Problems 49

4.1 Laplace’s Equation and Harmonic Functions 494.2 The Dirichlet Problem for a Rectangle 504.3 The Dirichlet Problem for a Disk 524.4 Properties of Harmonic Functions 57

4.4.2 Representation Theorems 584.4.3 The Mean Value Theorem and the Maximum Principle 60

v

vi CONTENTS

4.5.1 Uniqueness and Existence 614.5.2 Neumann Problem for a Rectangle 624.5.3 Neumann Problem for a Disk 63

4.7 An Existence Theorem for the Dirichlet Problem 65

5 Fourier Integral Methods of Solution 67

5.1 The Fourier Integral of a Function 675.2 The Heat Equation on the Real Line 705.3 The Debate Over the Age of the Earth 73

5.5 The Cauchy Problem for the Wave Equation 745.6 Laplace’s Equation on Unbounded Domains 76

6 Solutions Using Eigenfunction Expansions 79

6.1 A Theory of Eigenfunction Expansions 79

6.3 Applications of Bessel Functions 876.3.1 Temperature Distribution in a Solid Cylinder 876.3.2 Vibrations of a Circular Drum 876.4 Legendre Polynomials and Applications 90

7 Integral Transform Methods of Solution 97

vii

Chapter 1

First Ideas

2 Verifying that the function is a solution of the heat equation is a

straight-forward exercise in differentiation One way to show that u(x, t) is bounded is to observe that if t > 0 and x = 2 √

It is routine to verify that u tt = c2u xx

7 One way to show that the transformation is one to one is to evaluate theJacobian 

Solutions Manual to Accompany Beginning Partial Differential Equations,

Third Edition Peter V O’Neil.

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 2014 John Wiley & Sons, Inc Published 2014 by John Wiley & Sons, Inc.

1

2 CHAPTER 1 FIRST IDEAS

Finally, solve ξ = a + at, η = x + bt for x and t to obtain the inverse

This, coupled with the fact that H(x, t, u, u x , u t) transforms to some

func-tion K(ξ, η, V, V ξ , V η), yields the conclusion

9 From the solution of problem 8, the transformed equation is hyperbolic if

C = 0 because in that case we can choose a and b to make the coefficients

of V ξξ and V ηη vanish This is done by choosing a and b to be the distinct

1.1 TWO PARTIAL DIFFERENTIAL EQUATIONS 3

We do not need u tt , because C = 0 in this case Now we obtain

Au xx + Bu xt + Cu tt=− B2

A V ξη ,

yielding a hyperbolic canonical form

V ξη + K(ξ, η, V, V ξ , V η) = 0

of the given partial differential equation

10 In this case suppose B2− 4AC = 0 Now let

with two terms on the next to last line vanishing because B2− 4AC = 0.

This gives the canonical form

V ξξ + K(ξ, η, V, V ξ , V η) = 0

for the original partial differential equation when B2− 4AC = 0.

11 Suppose now that B2− 4AC < 0 Let the roots of Ca2+ Ba + A = 0 be

4 CHAPTER 1 FIRST IDEAS

Now we need some information about p and q Because of the way p + iq

V ξξ + V ηη + K(ξ, η, V, V ξ , V η) = 0for this case

12 The diffusion equation is parabolic and the wave equation is hyperbolic

14 B2− 4AC = 33 > 0, so the equation is hyperbolic With

16 With A = 1, B = 0, and C = 0, B2− 4AC = −36 < 9, so the equation is

elliptic Solve 9a2+ 1 = 0 to get a = ±i/3 Thus use the transformation

converging to 1− |x| for −2 ≤ x ≤ 2 Figure 1.1 compares a graph of f(x)

with the fifth partial sum of the series

1.2 FOURIER SERIES 5

Figure 1.1: f(x) and the 5th partial sum of the Fourier series in Problem 4.

6 The Fourier series is

Figure 1.2 compares a graph of the function with the fifth partial sum ofthe series

8 The Fourier series converges to

10 The series converges to

6 CHAPTER 1 FIRST IDEAS

Figure 1.2: f(x) and the 5th partial sum of the Fourier series in Problem 6.

12 The series converges to

1.2 FOURIER SERIES 7Then L

Upon division by L, this yields Parseval’s equation.

16 The cosine series is

converging to 1 for 0 ≤ x < 1, to −1 for 1 < x ≤ 2, and to 0 at x = 1.

Figure 1.3 compares the function to the 100th partial sum of this cosineexpansion

The sine series is

1 and 1 < x < 2 Figure 1.4 is the 100th partial sum of this sine series.

18 The cosine expansion is

This converges to f(x) on [0, 1] Figure 1.5 compares the function with the

10th partial sum of this cosine series

Figure 1.3: f(x) and the 100th partial sum of the cosine series in Problem 16.

8 CHAPTER 1 FIRST IDEAS

Figure 1.4: f(x) and the 100th partial sum of the sine expansion in Problem 16.

Figure 1.5: f(x) and the 10th partial sum of the cosine series in Problem 18.

The sine expansion is

compares the function with the 50th partial sum of this sine expansion

1.2 FOURIER SERIES 9

Figure 1.6: f(x) and the 50th partial sum of the sine expansion in Problem 18.

Figure 1.7: f(x) and the 10th partial sum of the cosine series in Problem 20.

20 The cosine expansion is

10th partial sum of this series

10 CHAPTER 1 FIRST IDEAS

Figure 1.8: f(x) and the 50th partial sum of the sine expansion in Problem 20.

The sine expansion is

This series converges to 0 at x = 0 and at x = 1, and to e −x for 0 < x < 1.

Figure 1.8 shows the 50th partial sum

22 The cosine expansion is

the 10th partial sum of this cosine series

The sine series is

converging to f(x) on [0, 2] The function and the 10th partial sum of this

sine series are shown in Figure 1.10

23 Expand f(x) = sin(x) in a cosine series on [0, π]:

1.2 FOURIER SERIES 11

Figure 1.9: f(x) and the 10th partial sum of the cosine series in Problem 22.

Figure 1.10: f(x) and the 50th partial sum of the sine expansion in Problem 22.

Since 1 + (−1) n = 0 if n is odd, we need only to retain the even positive integers in the sum Replace n with 2n to write

Now choose x = π/2.

12 CHAPTER 1 FIRST IDEAS

2 Eigenvalues and eigenfunctions are

Figure 1.11: f(x) and the 10th partial sum.

Figure 1.12: f(x) and the 25th partial sum.

1.3 TWO EIGENVALUE PROBLEMS 13

Figure 1.13: f(x) and the 50th partial sum.

Figure 1.14: f(x) and the 100th partial sum.

4 Eigenvalues and eigenfunctions are

λ n = α2n , X n (x) = sin(α n x),

where α n is the nth positive root (in increasing order) of the equation tan(αL) = −2α.

14 CHAPTER 1 FIRST IDEAS

The Fourier series of f(x) on [ −1, 1] is

Chapter 2

Solutions of the Heat

Equation

2.1 Solutions on an Interval [0, L]

2 By inspection the solution is u(x, t) = T.

4 By equation 2.2 the solution is

u(x, t) = sin(πx)e −kπ2t

6 Let u(x, t) = U(x, t) + ψ(x) For U to satisfy the standard heat equation choose ψ(x) so that ψ  (x) = 0 For homogeneous boundary conditions on the problem for U(x, t), we also want ψ(0) = 3 and ψ(5) = √

Solutions Manual to Accompany Beginning Partial Differential Equations,

Third Edition Peter V O’Neil.

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 2014 John Wiley & Sons, Inc Published 2014 by John Wiley & Sons, Inc.

15

16 CHAPTER 2 SOLUTIONS OF THE HEAT EQUATION

v t = kv xx , v(0, t) = v(L, t) = 0, v(x, 0) = e −hx/2k f(x).

This problem has the solution

The solution for v is

4

n2(1 + (−1) n ) cos(nx/2)e −kn2t/4

The solution of the original problem is u(x, t) = e −8t v(x, t).

16 By inspection, the solution is u(x, t) = B.

18 The problem is

u t = ku xx , u(0, t) = u x (L, t) = 0,

u(x, 0) = B.

Upon letting u(x, t) = X(x)T(t), we obtain

X  + λX = 0, X(0) = X  (L) = 0

T  + kλX = 0.

18 CHAPTER 2 SOLUTIONS OF THE HEAT EQUATION

The eigenvalues and eigenfunctions are

Using the same informal reasoning used to derive the Fourier coefficients,

multiply the series for u(x, 0) by sin((2m − 1)πx/2l) and integrate term

u(x, t) = w(x, t) + T = e −At U(x, t) + T.

22 Multiply the heat equation by u and integrate to get

Integrate the right side of this equation by parts and rewrite the left side

as the integral of a partial derivative to obtain

b

a

12

20 CHAPTER 2 SOLUTIONS OF THE HEAT EQUATION

For any t ≥ 0, this is the Fourier cosine expansion of F(x, t) on [0, L],

thinking of F(x, t) as a function of x Therefore the coefficients are

A n (t) = 2

L

L

0 F(ξ, t) cos(nπξ/L) dξ.

the nth Fourier cosine coefficient of f(x) on [0, L] Thus T n (t) is determined

as the solution of the problem

where the a n ’s are the Fourier cosine coefficients of f(x) on [0, L].

This results in the solution

22 CHAPTER 2 SOLUTIONS OF THE HEAT EQUATION

Carry out an analysis like that done in this section (substitute for u t (ξ, t)

and integrate by parts, using the boundary conditions) to derive theexpression

2.2 A NONHOMOGENEOUS PROBLEM 23Thus show that

4 The solution reduces to a single term

u(x, t) = sin(x) cos(4t).

Solutions Manual to Accompany Beginning Partial Differential Equations,

Third Edition Peter V O’Neil.

c

 2014 John Wiley & Sons, Inc Published 2014 by John Wiley & Sons, Inc.

25

26 CHAPTER 3 SOLUTIONS OF THE WAVE EQUATION

To obtain solutions e αt, substitute this into the differential equation and

solve for α To retain the dependence on n, denote the solutions for α as

4

Therefore, for n = 1, 2, , T n (t) has the form

T n (t) = a n e −At/2 cos(β n t) + b n sin(β n t).

Now attempt a solution

3.1 SOLUTIONS ON BOUNDED INTERVALS 27

β n b n − A

2a n= 0so

Consideration of cases on λ shows that 0 is not an eigenvalue, and there

is no negative eigenvalue Set λ = k2with k > 0 to obtain solutions for X

−kc sin(kL) + kd cos(kL) + α(c cos(kL) + d sin(kL)) = 0.

From these we obtain

tan(kL) = 2αk

j2− α2.

If we think of the left and right sides of this equation as functions of k, the

straight line graph (right side) intersects the graph of the tangent function

(right side) infinitely many times with k > 0 The first coordinate of each such point is an eigenvalue of this problem If k n is the nth such first

28 CHAPTER 3 SOLUTIONS OF THE WAVE EQUATION

coordinate (counting from left to right), then the eigenvalues are λ n = k2n

Although we cannot solve for k n in an exact algebraic expression, wecan approximate these numbers to any degree of accuracy we need The

problem for T is now

which satisfy the partial differential equation, both boundary conditions,

and the zero initial velocity condition To satisfy u(x, 0) = ϕ(x), attempt

This reminds one of a Fourier series, but here the functions we are

ex-panding ϕ(x) in terms of are

Upon integrating term by term, all terms on the right vanish except

pos-sibly the n = m term, yielding

3.1 SOLUTIONS ON BOUNDED INTERVALS 29

18 Let u(x, t) = v(x, t) + f(x) and substitute into the wave equation to obtain

.

Then u(x, t) = v(x, t) + f(x).

30 CHAPTER 3 SOLUTIONS OF THE WAVE EQUATION

20 Suppose α is a positive number that is not an integer Let u(x, t) = v(x, t)+

f(x) to obtain the solution

3.1 SOLUTIONS ON BOUNDED INTERVALS 31Integrate by parts to get

Rearrangement of this equation yields the conclusion to be proved

27 Let u(x, t) and v(x, t) be solutions and let w(x, t) = u(x, t) − v(x, t) Then

w is a solution of the problem

w x and w t must be zero This means that w(x, t) must be constant But

w(x, 0) = 0, so w(x, t) = 0 and u(x, t) = v(x, t).

32 CHAPTER 3 SOLUTIONS OF THE WAVE EQUATION

28

u(x, t) = e −t/2

cos(√

3.2 THE CAUCHY PROBLEM 33

11 The solution with ϕ(x) = sin(x) is

|u2(x, t) − u1(x, t)

On any interval 0≤ x ≤ T, this difference has magnitude not exceeding

13 Let v(x, t) =t

0w(x, t, T) dT and show that u is a solution of the Cauchy

problem with the given initial conditions Compute

34 CHAPTER 3 SOLUTIONS OF THE WAVE EQUATION

16 With zero initial velocity, the solution is

u(x, t) = 1

2(ϕ(x − t) + ϕ(x + t)),

the sum of a forward and backward wave, respectively Figures 3.1–3.5

show the wave at times t = 0, t = 0.3, t = 0.6, t = 1, and 1.3, respectively.

At t = 1.3 the forward and backward waves have separated.

Figure 3.1: Problem 16, wave at time t = 0.

Figure 3.2: Problem 16, wave at time t = 0.3.

3.2 THE CAUCHY PROBLEM 35

Figure 3.3: Problem 16, wave at time t = 0.6.

Figure 3.4: Problem 16, wave at time t = 1.

18 Figures 3.6–3.9 show the wave at times t = 0, t = 0.3, t = 0.7, and t = 1.3,

respectively

20 Figures 3.10–3.13 show the wave at times t = 0, t = 0.3, t = 0.6, and

t = 1.3.

Solutions Manual to Accompany Beginning Partial Differential Equations,< /small>

Third Edition Peter V O’Neil.

c

... cos(4t).

Solutions Manual to Accompany Beginning Partial Differential Equations,< /small>

Third Edition Peter V O’Neil.

c

... 50th partial sum

22 The cosine expansion is

the 10th partial sum of this cosine series

The sine series is

converging to f(x) on [0, 2] The function and the 10th partial