Powers D.L. Boundary Value Problems and Partial Differential Equations
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448 Answers to Odd-Numbered Exercises
Section 1.3
1. a. sectionally smooth; b, c, d, e are not;
b: vertical tangent at 0; c: vertical asymptote at ±π/2; d, e: vertical asymp-
tote at π/2.
3. To f (x) everywhere.
5. b. Graph consists of straight-line segments. c. x = 1, sum = 1/2; x = 2,
sum =0; x = 9.6, sum =−0.6; x =−3.8, sum = 0.2. Use periodicity.
7. B = 0, A =−π
2
/12, C = 1/4.
9. a.
√
1 −x
2
;b.a
0
=π/4; c. No; d. nothing.
Section 1.4
1. (c), (d), (f), (g) have uniformly convergent Fourier series.
3. All of the cosine series converge uniformly. The sine series converges uni-
formly only in case (b).
5. (a), (b), (d) converge uniformly; (c) does not.
Section 1.5
1.
∞
n=1
1
n
2
=
π
2
6
.
3. f
(x) = 1, 0 < x <π. The sine series cannot be differentiated, because the
odd periodic extension of f is not continuous. But the cosine series can be
differentiated.
5. For the sine series: f (0+) = 0andf (a−) = 0. For the cosine series no
additional condition is necessary.
7. No. The function ln |2cos(
x
2
)| is not even sectionally continuous.
9. Since f is odd, periodic, and sectionally smooth, (c) follows, and also
b
n
→ 0asn →∞.Then
∞
n=1
|n
k
b
n
e
−n
2
t
| converges for all integers k
(t > 0) by the comparison test and ratio test:
n
k
b
n
e
−n
2
t
≤Mn
k
e
−n
2
t
for some M
and
M(n + 1)
k
e
−(n+1)
2
t
Mn
k
e
−n
2
t
=
n +1
n
k
e
−(2n+1)t
→0
as n →∞. Then by Theorem 7, (a) is valid. Property (b) follows by direc-
tion substitution.
Chapter 1 449
Section 1.6
1.
1
π
π
−π
ln
2cos
x
2
2
dx =
∞
n=1
1
n
2
=
π
2
6
.
3. a. Coefficients tend to zero.
b. Coefficients tend to zero, although
1
−1
|x|
−1
dx is infinite.
5. The integral must be infinite, because
∞
n=1
a
2
n
+b
2
n
=∞.
Section 1.7
1. Theequalitytobeprovedis
2sin
1
2
y
1
2
+
N
n=1
cos(ny)
=sin
N +
1
2
y
.
The left-hand side is transformed as follows:
2sin
1
2
y
1
2
+
N
n=1
cos(ny)
=sin
1
2
y
+
N
n=1
2sin
1
2
y
cos(ny)
=sin
1
2
y
+
N
n=1
sin
n +
1
2
y
−sin
n −
1
2
y
=sin
1
2
y
+
N
n=1
sin
n +
1
2
y
−
N−1
n=0
sin
n +
1
2
y
=sin
N +
1
2
y
because all other terms cancel.
3. φ(0+) = 1, φ(0−) =−1. See Fig. 1.
5. a. f
(x) =
3
4
x
−1/4
for 0 < x <π (and f
is an odd function). Thus, f has a
vertical tangent at x = 0, although it is continuous there.
b. φ(y) =
|y|
3/4
2sin(
1
2
y)
cos
1
2
y
, −π<y <π
450 Answers to Odd-Numbered Exercises
Figure 1 Graph for Exercise 3, Section 1.7.
is a product of continuous functions and is therefore continuous, except
perhaps where the denominator is 0. At y = 0, cos(
1
2
y)
∼
=
1, 2sin(
1
2
y)
∼
=
y,
so φ(y)
∼
=
|y|
3/4
/y =±|y|
−1/4
near y =0.
c. Now,
π
−π
φ
2
(y)dy is finite, so the Fourier coefficients of φ approach
zero.
Section 1.8
1. ˆa
6
=−0.00701, a
6
=−0.00569.
3. ˆa
0
=1.367,
ˆa
1
=−0.844,
ˆ
b
1
=−0.043,
ˆa
2
=0.208,
ˆ
b
2
=−0.115,
ˆa
3
=0.050,
ˆ
b
3
=−0.050,
ˆa
4
=0.042,
ˆ
b
4
=0.00,
ˆa
5
=−0.0064,
ˆ
b
5
=0.043,
ˆa
6
=0.0167.
Section 1.9
1. Each function has the representations (for x > 0)
f (x ) =
∞
0
A(λ) cos(λx )dλ =
∞
0
B(λ) sin(λx)dλ.
a. A(λ) = 2/π(1 +λ
2
), B(λ) = 2λ/π(1 +λ
2
);
b. A(λ) = 2sin(λ)/πλ, B(λ) = 2(1 −cos(λ))/πλ;
c. A(λ) = 2(1 −cos(λπ))/λ
2
π, B(λ) = 2(πλ −sin(λπ ))/πλ
2
.
3. a.
1
1 +x
2
=
∞
0
e
−λ
cos(λx)dλ;
Chapter 1 451
b.
sin(x)
x
=
∞
0
A(λ) cos(λx )dλ,whereA(λ) =
1, 0 < x < 1,
0, 1 < x.
5. a. A(λ) ≡ 0, B(λ) =
2sin(λπ )
π(1 −λ
2
)
;
b. A(λ) =
1 +cos(λπ )
π(1 −λ
2
)
, B(λ) =
sin(λπ)
π(1 −λ
2
)
;
c. A(λ) =
2(1 +cos(λπ ))
π(1 −λ
2
)
, B(λ) ≡ 0.
7. Change variable from x to λ with x = λz.
Section 1.10
1. e
αx
=2
sinh(απ)
π
1
2α
+
∞
n=1
(−1)
n
α
2
+n
2
α cos(nx) −n sin(nx)
.
3. f (x) =
∞
−∞
C(λ)e
iλx
dλ.
a. C(λ) =
1
2π(1 +iλ)
;b.C(λ) =
1 +e
−iλπ
2π(1 −λ
2
)
.
5. a. 1 +
∞
n=1
r
n
cos(nx) = Re
∞
0
re
ix
n
=Re
1
1 −re
ix
;
b.
∞
n=1
sin(nx)
n!
=Im
∞
n=1
e
inx
n!
=Im exp
e
ix
.
7. a. f (x) =
2sin(x)
x
;b.f (x) =
2
1 +x
2
.
Section 1.11
1. u(t) = A
0
+
∞
n=1
A
n
cos(nt/2) + B
n
sin(nt/2),
A
0
=
1
2.08
, A
n
=
0.4/π
(1.04 −n
2
)
2
+(0.4n)
2
,
B
n
=−
1
nπ
1.04 −n
2
(1.04 −n
2
)
2
+(0.4n)
2
.
3. u(x) =
∞
n=1
B
n
sin(nπx/L), B
n
=
8K sin(nπ/2)
((nπ/L)
2
+γ
2
)n
2
π
2
,
K =w/EI, γ
2
=T/EI.
452 Answers to Odd-Numbered Exercises
Chapter 1 Miscellaneous Exercises
1. f (x) =
∞
n=1
b
n
sin(nx),
b
n
=
0, n even,
4sin(nα)
παn
2
, n odd.
3. Yes. As α → 0, sin(nα)/nα → 1.
5. f (x) =
∞
n=1
b
n
sin(nπx/a),
b
n
=
2h
π
2
sin(nπα)
n
2
1
α
+
1
1 −α
.
7. a. b
n
=0, a
n
=0, a
0
=1;
b.
∞
n=1
b
n
sin(nπx/a), b
n
=
2(1 −cos(nπ))
nπ
;
c. and d. same as a;
e. same as b;
f. a
0
+
∞
n=1
a
n
cos(nπx/a) +b
n
sin(nπx/a),
a
0
=
1
2
, a
n
=0, b
n
=
1 −cos(nπ)
nπ
.
9. f (x) = a
0
+
∞
n=1
a
n
cos(nπx/a) +b
n
sin(nπx/a),
a
0
=
1
2
a, a
n
=−
2a(1 −cos(nπ))
n
2
π
2
, b
n
=−
2a cos(nπ)
nπ
,
x =−a, −a/2, 0, a, 2a,
sum = a, 0, 0, a, 0.
11. f (x) = a
0
+
∞
n=1
a
n
cos(nx),
a
0
=
3
4
, a
n
=
sin(nπ/2)
nπ
,
x = 0,π/2,π,3π/2, 2π,
sum = 1,
3
4
,
1
2
,
3
4
, 1.
Chapter 1 453
13. f (x) =
∞
n=1
b
n
sin(nπx), b
n
=2
1 +cos(nπ)
/nπ.
15. f (x) =
∞
n=1
b
n
sin(nx),
b
2
=
1
2
,otherb
n
=
4sin(nπ/2)
π(4 −n
2
)
.
17.
N
1
cos(nx) = Re
N
1
e
inx
=Re
e
ix
−e
iNx
1 −e
ix
=Re
e
ix/2
−e
i(2N−1)x/2
e
−ix/2
−e
ix/2
.
The denominator is now −2i sin(x/2).
19. f (x) =
∞
n=1
b
n
sin(nx), b
n
=
2a sin(na +π)
n
2
a
2
−π
2
.
21. f (x) =
∞
0
sin(λa)
λπ
cos(λx) +
1 −cos(λa)
λπ
sin(λx)
dλ.
23. f (x) =
∞
0
2sin(λπ )
π(1 −λ
2
)
sin(λx)d λ (x > 0).
29. Use
∞
0
sin(λt)
λ
dλ =
π
2
.
31. These answers are not unique.
a.
∞
n=1
b
n
sin(nπx), b
n
=2/nπ ;
b. a
0
+
∞
n=1
a
n
cos(nπx), a
0
=
1
2
, a
n
=2
1 −cos(nπ)
/n
2
π
2
;
c.
∞
0
B(λ) sin(λx)dλ, B(λ) = 2
λ −sin(λ)
πλ
2
;
d.
∞
0
A(λ) cos(λx )dλ, A(λ) = 2
1 −cos(λ)
πλ
2
.
The integrals of parts c. and d. converge to 0 for x > 1.
33. Use s = 6inEq.(7)ofSection8.
ˆa
0
=0.78424, ˆa
4
=−0.00924,
ˆa
1
=0.22846, ˆa
5
=0.00744,
ˆa
2
=−0.02153, ˆa
6
=−0.00347,
ˆa
3
=0.01410.
454 Answers to Odd-Numbered Exercises
35. a
0
=
a
6
, a
n
=
2a
n
2
π
2
cos
2nπ
3
−cos
nπ
3
.
37. a
0
=
5
8
, a
n
=
2
n
2
π
2
3cos
nπ
2
−2 − cos(nπ)
.
39. a
0
=
1
2
, a
n
=
2
n
2
π
2
1 −cos(nπ)
.
41. a
0
=
a
2
6
, a
n
=
−2a
2
n
2
π
2
1 +cos(nπ)
.
43. a
0
=
1
2
, a
n
=
−1
nπ
2sin
nπ
2
.
45. b
n
=
1 +cos(nπ/2) − 2cos(nπ)
nπ
.
47. b
n
=a
2sin(nπ/2)
n
2
π
2
−
cos(nπ)
nπ
.
49. b
n
=
2
nπ
cos
nπ
4
−cos
3nπ
4
.
51. b
n
=2nπ
(1 −e
ka
cos(nπ))
(a
2
k
2
+n
2
π
2
)
.
53. A(λ) =
2
π(1 +λ
2
)
.
55. A(λ) =
2sin(λb)
πλ
.
57. A(λ) =
2(1 −cos(λ))
πλ
2
.
59. B(λ) =
2λ
π(1 +λ
2
)
.
61. B(λ) =
2(1 −cos(λb))
λπ
.
63. B(λ) =
2(λ −sin(λ))
λ
2
π
.
65. The term a
n
cos(nx) + b
n
sin(nx) appears in S
n
, S
n+1
,...,S
N
,andthus
N +1 −n times in σ
N
.
67. Use Eq. (13) of Section 7 and the identity in Exercise 66.
69. a. Use x =0; b. x = 1/2; c. x = 0.
Chapter 2 455
Chapter 2
Section 2.1
1. One possibility: u(x, t) is the temperature in a rod of length a whose lat-
eral surface is insulated. The temperature at the left end is held constant
at T
0
. The right end is exposed to a medium at temperature T
1
.Initially
the temperature is f (x).
3. A xg=hCx(U −u(x, t)),whereh is a constant of proportionality and
C is the circumference. Eq. (4) becomes
∂
2
u
∂x
2
+
hC
κA
(U −u) =
1
k
∂u
∂t
.
5. If
∂x
∂u
(0, t) is positive, then heat is flowing to the left, so u(0, t) is greater
than T(t).
7. The second factor is approximately constant if T is much larger than u or
if T and u are approximately equal.
Section 2.2
1. v
−γ
2
(ν −U) = 0, 0 < x < a,
v(0) = T
0
, v(a) = T
1
,
v(x) = U +A cosh(γ x) +B sinh(γ x),
A =T
0
−U, B =
(T
1
−U) −(T
0
−U) cosh(γ a)
sinh(γ a)
.
One interpretation: u is the temperature in a rod, with convective heat
transfer from the cylindrical surface to a medium at temperature U.
3. v(x) = T. Heat is being generated at a rate proportional to u − T.Ifγ =
π/a, the steady-state problem does not have a unique solution.
5. v(x) = A ln(κ
0
+βx) + B, A = (T
1
−T
0
)/ ln(1 +aβ/κ
0
),
B =T
0
−A ln(κ
0
).
7. v(x) = T
0
+r(2a −x)x/2.
9. Du
−Su
=0, 0 < x < a; u(0) =U, u(a) = 0,
u(x) = U(e
Sx/D
−e
Sa/D
)/(1 −e
Sa/D
).
456 Answers to Odd-Numbered Exercises
Section 2.3
1. w(x, t) =−
2
π
(T
0
+T
1
) sin
πx
a
exp
−
π
2
kt
a
2
−
2
π
T
0
−T
1
2
sin
2πx
a
exp
−
4π
2
kt
a
2
−···
3. The partial differential equation is
∂
2
U
∂ξ
2
=
∂U
∂τ
, 0 <ξ<1, 0 <τ.
5. w(x, t) =
∞
n=1
b
n
sin
nπx
a
exp
−n
2
π
2
kt/a
2
, b
n
=T
0
2(1 −cos(nπ))
πn
.
7. w(x, t) as in the answer to Exercise 5, with b
n
=
2βa
π
·
1
n
.
9. a. v(x) = C
1
;
b.
∂w
∂t
=D
∂
2
w
∂x
2
,0< x < a,0< t,
w(0, t) = 0, w(a, t) = 0, 0 < t,
w(x, 0) = C
0
−C
1
;
c. C(x, t) = C
1
+
∞
n=1
b
n
sin
nπx
a
exp
−n
2
π
2
kt/a
2
,
b
n
=(C
0
−C
1
)
2(1 −cos(nπ))
πn
;
d. t =
−a
2
Dπ
2
ln
π
40
;
e. t =6444 s = 107.4min.
Section 2.4
1. a
0
=T
1
/2, a
n
=2T
1
(cos(nπ)−1)/(nπ)
2
.
3. u(x, t) asgiveninEq.(9),withλ
n
= nπ/a, a
0
= T
0
/2, and a
n
=
4T
0
(2cos(nπ/2) −1 −cos(nπ))/n
2
π
2
.
5. a. The general solution of the steady-state equation is v(x) = c
1
+ c
2
x .
The boundary conditions are c
2
= S
0
, c
2
= S
1
; thus there is a solution
if S
0
= S
1
. If heat flux is different at the ends, the temperature cannot
approach a steady state. If S
0
=S
1
,thenv(x) = c
1
+S
0
x , c
1
undefined.
Chapter 2 457
c. A = (S
1
−S
0
)/a, B = S
0
.IfS
0
=S
1
,then
∂u
∂t
=kA for all t.
7. φ
+λ
2
φ = 0, 0 < x < a,
φ(0) = 0, φ(a) = 0.
Solution: φ
n
=sin(λ
n
x), λ
n
=nπ/a(n = 1, 2,...).
9. The series
∞
n=1
|A
n
(t
1
)| converges.
11. No. u(0, t) is constant if u
t
(0, t) = 0.
Section 2.5
1. v(x, t) = T
0
.
3. u(x, t) = T
0
+
∞
n=1
b
n
sin(λ
n
x) exp
−λ
2
n
kt
, λ
n
=(2n −1)π/2a,
b
n
=
8T(−1)
n+1
π
2
(2n −1)
2
−
4T
0
π(2n −1)
.
5. The steady-state solution is v(x) = T
0
− Tx(x − 2a)/2a
2
. The transient
satisfies Eqs. (5)–(8) with
g(x) = T
0
−v(x) =
Tx(x −2a)
2a
2
.
7. u(x, t) = T
0
+
∞
n=1
c
n
cos(λ
n
x) exp
−λ
2
n
kt
,
λ
n
=(2n −1)π/2a, c
n
=
4(T
1
−T
0
)(−1)
n+1
π(2n −1)
.
9. u(x, t) = T
1
cos(πx/2a) exp
−
π
2a
2
kt
.
11. The graph of G in the interval 0 < x < 2a is made by reflecting the graph
of g in the line x =a (likeanevenextension).
13. a. u(x, t) = T
0
+
∞
n=1
b
n
sin(λ
n
x) exp
−λ
2
n
kt
, λ
n
=(2n −1)π/2a,
b
n
=
1
a
2a
0
g(x) sin
nπx
2a
dx .