Solutions Manual – Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th Edition – Richard Haberman – ISBN 9780134995434 (Chapters 1–14 Covered)

Edrftgyihu jiuh




Applied Partial Differential
Equations with Fourier
ST

Series and Boundary Value
UV

Problems, 5th Edition
IA

SOLUTIONS
_A


MANUAL
PP
RO

Richard Haberman
VE

Comprehensive Solutions Manual for Instructors
and Students
||| ||| || ||| || | || ||| | || ||| |
D?

9780134995434

© Richard Haberman. All rights reserved. Reproduction or distribution without permission is
prohibited.




© MEDCONNOISSEUR

, TABLE OF CONTENTS

Solutions Manual – Applied Partial Differential Equations with Fourier Series and Boundary Value
Problems (5th Ed.)
Author: Richard Haberman
ST
ISBN: 9780134995434

PART I: CORE CONCEPTS AND SEPARATION OF VARIABLES

Chapter 1: Heat Equation
UV
Chapter 2: Method of Separation of Variables
Chapter 3: Fourier Series
Chapter 4: Wave Equation: Vibrating Strings and Membranes
Chapter 5: Sturm-Liouville Eigenvalue Problems

PART II: NUMERICAL METHODS AND HIGHER DIMENSIONS
IA
Chapter 6: Finite Difference Numerical Methods for Partial Differential Equations
Chapter 7: Higher Dimensional Partial Differential Equations
Chapter 8: Nonhomogeneous Problems
_A
PART III: TRANSFORM METHODS AND GREEN'S FUNCTIONS

Chapter 9: Green's Functions for Time-Independent Problems
Chapter 10: Infinite Domain Problems: Fourier Transform Solutions
PP
Chapter 11: Green's Functions for Wave and Heat Equations
Chapter 13: Laplace Transform Solution of Partial Differential Equations

PART IV: ADVANCED TOPICS
RO
Chapter 12: The Method of Characteristics for Linear and Quasilinear Wave Equations
Chapter 14: Dispersive Waves: Slow Variations, Stability, Nonlinearity, and Perturbation Methods
VE
D?

, Chapter 1. Heat Equation
Section 1.2
1.2.9 (d) Circular cross section means that P = 2πr, A = πr2 , and thus P/A = 2/r, where r is the radius.
Also γ = 0.
1.2.9 (e) u(x, t) = u(t) implies that
du 2h
cρ =− u.
dt r
ST
The solution of this first-order linear differential equation with constant coefficients, which satisfies the
initial condition u(0) = u0 , is · ¸
2h
u(t) = u0 exp − t .
cρr
UV
Section 1.3
1.3.2 ∂u/∂x is continuous if K0 (x0 −) = K0 (x0 +), that is, if the conductivity is continuous.

Section 1.4
1.4.1 (a) Equilibrium satisfies (1.4.14), d2 u/dx2 = 0, whose general solution is (1.4.17), u = c1 + c2 x. The
IA
boundary condition u(0) = 0 implies c1 = 0 and u(L) = T implies c2 = T /L so that u = T x/L.
1.4.1 (d) Equilibrium satisfies (1.4.14), d2 u/dx2 = 0, whose general solution (1.4.17), u = c1 + c2 x. From
the boundary conditions, u(0) = T yields T = c1 and du/dx(L) = α yields α = c2 . Thus u = T + αx.
1.4.1 (f) In equilibrium, (1.2.9) becomes d2 u/dx2 = −Q/K0 = −x2 , whose general solution (by integrating
_A
twice) is u = −x4 /12 + c1 + c2 x. The boundary condition u(0) = T yields c1 = T , while du/dx(L) = 0
yields c2 = L3 /3. Thus u = −x4 /12 + L3 x/3 + T .
1.4.1 (h) Equilibrium satisfies d2 u/dx2 = 0. One integration yields du/dx = c2 , the second integration
yields the general solution u = c1 + c2 x.
x=0: c2 − (c1 − T ) = 0
PP

x=L: c2 = α and thus c1 = T + α.
Therefore, u = (T + α) + αx = T + α(x + 1).
1.4.7 (a) For equilibrium:
d2 u x2
RO
du
2
= −1 implies u = − + c1 x + c2 and = −x + c1 .
dx 2 dx
From the boundary conditions du du
dx (0) = 1 and dx (L) = β, c1 = 1 and −L + c1 = β which is consistent
2
only if β + L = 1. If β = 1 − L, there is an equilibrium solution (u = − x2 + x + c2 ). If β 6= 1 − L,
there isn’t an equilibrium solution. The difficulty is caused by the heat flow being specified at both
ends and a source specified inside. An equilibrium will exist only if these three are in balance. This
VE
balance can be mathematically verified from conservation of energy:
Z Z L
d L du du
cρu dx = − (0) + (L) + Q0 dx = −1 + β + L.
dt 0 dx dx 0

If β + L = 1, then the total thermal energy is constant and the initial energy = the final energy:
Z L Z Lµ 2 ¶
D?
x
f (x) dx = − + x + c2 dx, which determines c2 .
0 0 2

If β + L 6= 1, then the total thermal energy is always changing in time and an equilibrium is never
reached.

1

, Section 1.5
d
¡ du ¢
1.5.9 (a) In equilibrium, (1.5.14) using (1.5.19) becomes dr r dr = 0. Integrating once yields rdu/dr = c1
and integrating a second time (after dividing by r) yields u = c1 ln r + c2 . An alternate general solution
is u = c1 ln(r/r1 ) + c3 . The boundary condition u(r1 ) = T1 yields c3 = T1 , while u(r2 ) = T2 yields
c1 = (T2 − T1 )/ ln(r2 /r1 ). Thus, u = ln(r21/r1 ) [(T2 − T1 ) ln r/r1 + T1 ln(r2 /r1 )].

1.5.11 For equilibrium, the radial flow at r = a, 2πaβ, must equal the radial flow at r = b, 2πb. Thus β = b/a.
d
¡ 2 du ¢
1.5.13 From exercise 1.5.12, in equilibrium dr r dr = 0. Integrating once yields r2 du/dr = c1 and integrat-
ST
2
ing a second time (after dividing by r ) yields u = −c1 /r + c2 . The boundary conditions ¡ u(4) ¢ = 80
and u(1) = 0 yields 80 = −c1 /4 + c2 and 0 = −c1 + c2 . Thus c1 = c2 = 320/3 or u = 320 3 1 − 1r .
UV
IA
_A
PP
RO
VE
D?

2