Chapt_14

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1414 Boundary-Value Problems inOther Coordinate Systems
EXERCISES 14.1
Problems in Polar Coordinates
1. We have
A0 =
1
2π
∫ π
0
u0 dθ =
u0
2
An =
1
π
∫ π
0
u0 cosnθ dθ = 0
Bn =
1
π
∫ π
0
u0 sinnθ dθ =
u0
nπ
[1− (−1)n]
and so
u(r, θ) =
u0
2
+
u0
π
∞∑
n=1
1− (−1)n
n
rn sinnθ.
2. We have
A0 =
1
2π
∫ π
0
θ dθ +
1
2π
∫ 2π
π
(π − θ) dθ = 0
An =
1
π
∫ π
0
θ cosnθ dθ +
1
π
∫ 2π
π
(π − θ) cosnθ dθ = 2
n2π
[(−1)n − 1]
Bn =
1
π
∫ π
0
θ sinnθ dθ +
1
π
∫ 2π
π
(π − θ) sinnθ dθ = 1
n
[1− (−1)n]
and so
u(r, θ) =
∞∑
n=1
rn
[
(−1)n − 1
n2π
cosnθ +
1− (−1)n
n
sinnθ
]
.
3. We have
A0 =
1
2π
∫ 2π
0
(2πθ − θ2) dθ = 2π
2
3
An =
1
π
∫ 2π
0
(2πθ − θ2) cosnθ dθ = − 4
n2
Bn =
1
π
∫ 2π
0
(2πθ − θ2) sinnθ dθ = 0
and so
u(r, θ) =
2π2
3
− 4
∞∑
n=1
rn
n2
cosnθ.
755
14.1 Problems in Polar Coordinates
4. We have
A0 =
1
2π
∫ 2π
0
θ dθ = π
An =
1
π
∫ 2π
0
θ cosnθ dθ = 0
Bn =
1
π
∫ 2π
0
θ sinnθ dθ = − 2
n
and so
u(r, θ) = π − 2
∞∑
n=1
rn
n
sinnθ.
5. As in Example 1 in the text we have R(r) = c3rn + c4r−n. In order that the solution be bounded as r →∞ we
must define c3 = 0. Hence
u(r, θ) = A0 +
∞∑
n=1
r−n(An cosnθ + Bn sinnθ)
A0 =
1
2π
∫ 2π
0
f(θ) dθwhere
An =
cn
π
∫ 2π
0
f(θ) cosnθ dθ
Bn =
cn
π
∫ 2π
0
f(θ) sinnθ dθ.
6. Using the same reasoning as in Example 1 in the text we obtain
u(r, θ) = A0 +
∞∑
n=1
rn(An cosnθ + Bn sinnθ).
The boundary condition at r = c implies
f(θ) =
∞∑
n=1
ncn−1(An cosnθ + Bn sinnθ).
Since this condition does not determine A0, it is an arbitrary constant. However, to be a full Fourier series on
[0, 2π] we must require that f(θ) satisfy the condition A0 = a0/2 = 0 or
∫ 2π
0
f(θ) dθ = 0. If this integral were
not 0, then the series for f(θ) would contain a nonzero constant, which it obviously does not. With this as a
necessary compatibility condition we can then make the identifications
ncn−1An = an and ncn−1Bn = bn
or
An =
1
ncn−1π
∫ 2π
0
f(θ) cosnθ dθ and Bn =
1
ncn−1π
∫ 2π
0
f(θ) sinnθ dθ.
7. Proceeding as in Example 1 in the text and again using the periodicity of u(r, θ), we have
Θ(θ) = c1 cosαθ + c2 sinαθ
where α = n for n = 0, 1, 2, . . . . Then
R(r) = c3rn + c4r−n.
756
14.1 Problems in Polar Coordinates
[We do not have c4 = 0 in this case since 0 < a ≤ r.] Since u(b, θ) = 0 we have
u(r, θ) = A0 ln
r
b
+
∞∑
n=1
[(
b
r
)n
−
(r
b
)n]
[An cosnθ + Bn sinnθ] .
From
u(a, θ) = f(θ) = A0 ln
a
b
+
∞∑
n=1
[(
b
a
)n
−
(a
b
)n]
[An cosnθ + Bn sinnθ]
we find
A0 ln
a
b
=
1
2π
∫ 2π
0
f(θ) dθ,
[(
b
a
)n
−
(a
b
)n]
An =
1
π
∫ 2π
0
f(θ) cosnθ dθ,
and [(
b
a
)n
−
(a
b
)n]
Bn =
1
π
∫ 2π
0
f(θ) sinnθ dθ.
8. Substituting u(r, θ) = v(r, θ) + ψ(r) into the partial differential equation we obtain
∂2v
∂r2
+ ψ′′(r) +
1
r
[
∂v
∂r
+ ψ′(r)
]
+
1
r2
∂2v
∂θ2
= 0.
This equation will be homogeneous provided
ψ′′(r) +
1
r
ψ′(r) = 0 or r2ψ′′(r) + rψ′(r) = 0.
The general solution of this Cauchy-Euler differential equation is
ψ(r) = c1 + c2 ln r.
From
u0 = u(a, θ) = v(a, θ) + ψ(a) and u1 = u(b, θ) = v(b, θ) + ψ(b)
we see that in order for the boundary values v(a, θ) and v(b, θ) to be 0 we need ψ(a) = u0 and ψ(b) = u1. From
this we have
ψ(a) = c1 + c2 ln a = u0
ψ(b) = c1 + c2 ln b = u1.
Solving for c1 and c2 we obtain
c1 =
u1 ln a− u0 ln b
ln(a/b)
and c2 =
u0 − u1
ln(a/b)
.
Then
ψ(r) =
u1 ln a− u0 ln b
ln(a/b)
+
u0 − u1
ln(a/b)
ln r =
u0 ln(r/b)− u1 ln(r/a)
ln(a/b)
.
From Problem 7 with f(θ) = 0 we see that the solution of
∂2v
∂r2
+
1
r
∂v
∂r
+
1
r2
∂2v
∂θ2
= 0, 0 < θ < 2π, a < r < b,
v(a, θ) = 0, v(b, θ) = 0, 0 < θ < 2π
is v(r, θ) = 0. Thus the steady-state temperature of the ring is
u(r, θ) = v(r, θ) + ψ(r) =
u0 ln(r/b)− u1 ln(r/a)
ln(a/b)
.
757
14.1 Problems in Polar Coordinates
9. This is similar to the solution to Problem 7 above. When n = 0, Θ(θ) = c5θ + c6 and R(r) = c7 + c8 ln r.
Periodicity implies c5 = 0 and the insulation condition at r = a implies c8 = 0. Thus, we take u0 = A0 = c6c7.
Then, for n = 1, 2, 3, . . . , Θ(θ) = c1 cosnθ + c2 sinnθ and R(r) = c3rn + c4r−n. From R′(a) = 0 we get
c3na
n−1 − c4na−n−1 = 0, which implies c4 = c3a2n. Then
R(r) = c3(rn + a2nr−n) = c3
r2n + a2n
rn
and
u(r, θ) = A0 +
∞∑
n=1
r2n + a2n
rn
(An cosnθ + Bn sinnθ).
Taking r = b we have
f(θ) = A0 +
∞∑
n=1
b2n + a2n
bn
(An cosnθ + Bn sinnθ),
which implies
A0 =
a0
2
=
1
2π
∫ 2π
0
f(θ) dθ
and
b2n + a2n
bn
An =
1
π
∫ 2π
0
f(θ) cosnθ dθ and
b2n + a2n
bn
Bn =
1
π
∫ 2π
0
f(θ) sin 2nθ dθ.
Hence
An =
bn
π(a2n + b2n)
∫ 2π
0
f(θ) cosnθ dθ and Bn =
bn
π(a2n + b2n)
∫ 2π
0
f(θ) sinnθ dθ.
10. We solve
∂2u
∂r2
+
1
r
∂u
∂r
+
1
r2
∂2u
∂θ2
= 0, 0 < θ <
π
2
, 0 < r < c,
u(c, θ) = f(θ), 0 < θ <
π
2
,
u(r, 0) = 0, u(r, π/2) = 0, 0 < r < c.
Proceeding as in Example 1 in the text we obtain the separated differential equations
r2R′′ + rR′ − λR = 0
Θ′′ + λΘ = 0.
Taking λ = α2 the solutions are
Θ(θ) = c1 cosαθ + c2 sinαθ
R(r) = c3rα + c4r−α.
Since we want R(r) to be bounded as r → 0 we require c4 = 0. Applying the boundary conditions Θ(0) = 0
and Θ(π/2) = 0 we find that c1 = 0 and α = 2n for n = 1, 2, 3, . . . . Therefore
u(r, θ) =
∞∑
n=1
Anr
2n sin 2nθ.
From
u(c, θ) = f(θ) =
∞∑
n=1
Anc
n sin 2nθ
we find
An =
4
πc2n
∫ π/2
0
f(θ) sin 2nθ dθ.
758
14.1 Problems in Polar Coordinates
11. Referring to the solution of Problem 10 above we have
Θ(θ) = c1 cosαθ + c2 sinαθ
R(r) = c3rα.
Applying the boundary conditions Θ′(0) = 0 and Θ′(π/2) = 0 we find that c2 = 0 and α = 2n for
n = 0, 1, 2, . . . . Therefore
u(r, θ) = A0 +
∞∑
n=1
Anr
2n cos 2nθ.
From
u(c, θ) =
{
1, 0 < θ < π/4
0, π/4 < θ < π/2
= A0 +
∞∑
n=1
Anc
2n cos 2nθ
we find
A0 =
1
π/2
∫ π/4
0
dθ =
1
2
and
c2nAn =
2
π/2
∫ π/4
0
cos 2nθ dθ =
2
nπ
sin
nπ
2
.
Thus
u(r, θ) =
1
2
+
2
π
∞∑
n=1
1
n
sin
nπ
2
(r
c
)2n
cos 2nθ.
12. We solve
∂2u
∂r2
+
1
r
∂u
∂r
+
1
r2
∂2u
∂θ2
= 0, 0 < θ < π/4, r > 0
u(r, 0) = 0, r > 0
u(r, π/4) = 30, r > 0.
Proceeding as in Example 1 in the text we find the separated ordinary differential equations to be
r2R′′ + rR′ − λR = 0
Θ′′ + λΘ = 0.
With λ = α2 > 0 the corresponding general solutions are
R(r) = c1rα + c2r−α
Θ(θ) = c3 cosαθ + c4 sinαθ.
The condition Θ(0) = 0 implies c3 = 0 so that Θ = c4 sinαθ. Now, in order that the temperature be bounded
as r → ∞ we define c1 = 0. Similarly, in order that the temperature be bounded as r → 0 we are forced to
define c2 = 0. Thus R(r) = 0 and so no nontrivial solution exists for λ > 0. For λ = 0 the separated differential
equations are
r2R′′ + rR′ = 0 and Θ′′ = 0.
Solutions of these latter equations are
R(r) = c1 + c2 ln r and Θ(θ) = c3θ + c4.
Θ(0) = 0 still implies c4 = 0, whereas boundedness as r → 0 demands c2 = 0. Thus, a product solution is
u = c1c3θ = Aθ.
759
14.1 Problems in Polar Coordinates
From u(r, π/4) = 0 we obtain A = 120/π. Thus, a solution to the problem is
u(r, θ) =
120
π
θ.
13. We solve
∂2u
∂r2
+
1
r
∂u
∂r
+
1
r2
∂2u
∂θ2
= 0, 0 < θ < π , a < r < b,
u(a, θ) = θ(π − θ), u(b, θ) = 0, 0 < θ < π ,
u(r, 0) = 0, u(r, π) = 0,a < r < b.
Proceeding as in Example 1 in the text we obtain the separated differential equations
r2R′′ + rR′ − λR = 0
Θ′′ + λΘ = 0.
Taking λ = α2 the solutions are
Θ(θ) = c1 cosαθ + c2 sinαθ
R(r) = c3rα + c4r−α.
Applying the boundary conditions Θ(0) = 0 and Θ(π) = 0 we find that c1 = 0 and α = n for n = 1, 2, 3, . . . .
The boundary condition R(b) = 0 gives
c3b
n + c4b−n = 0 and c4 = −c3b2n.
Then
R(r) = c3
(
rn − b
2n
rn
)
= c3
(
r2n − b2n
rn
)
and
u(r, θ) =
∞∑
n=1
An
(
r2n − b2n
rn
)
sinnθ.
From
u(a, θ) = θ(π − θ) =
∞∑
n=1
An
(
a2n − b2n
an
)
sinnθ
we find
An
(
a2n − b2n
an
)
=
2
π
∫ π
0
(θπ − θ2) sinnθ dθ = 4
n3π
[1− (−1)n].
Thus
u(r, θ) =
4
π
∞∑
n=1
1− (−1)n
n3
r2n − b2n
a2n − b2n
(a
r
)n
sinnθ.
14. Letting u(r, θ) = v(r, θ) + ψ(θ) we obtain ψ′′(θ) = 0 and so ψ(θ) = c1θ + c2. From ψ(0) = 0 and ψ(π) = u0
we find, in turn, c2 = 0 and c1 = u0/π. Therefore ψ(θ) =
u0
π
θ. Now u(1, θ) = v(1, θ) + ψ(θ) so that
v(1, θ) = u0 − u0
π
θ. From
v(r, θ) =
∞∑
n=1
Anr
n sinnθ and v(1, θ) =
∞∑
n=1
An sinnθ
we obtain
An =
2
π
∫ π
0
(
u0 − u0
π
θ
)
sinnθ dθ =
2u0
πn
.
760
14.1 Problems in Polar Coordinates
Thus
u(r, θ) =
u0
π
θ +
2u0
π
∞∑
n=1
rn
n
sinnθ.
15. We solve
∂2u
∂r2
+
1
r
∂u
∂r
+
1
r2
∂2u
∂θ2
= 0, 0 < θ < π , 0 < r < 2,
u(2, θ) =
{
u0, 0 < θ < π/2
0, π/2 < θ < π
∂u
∂θ
∣∣∣∣
θ=0
= 0,
∂u
∂θ
∣∣∣∣
θ=π
= 0, 0 < r < 2.
Proceeding as in Example 1 in the text we obtain the separated differential equations
r2R′′ + rR′ − λR = 0
Θ′′ + λΘ = 0.
Taking λ = α2 the solutions are
Θ(θ) = c1 cosαθ + c2 sinαθ
R(r) = c3rα + c4r−α.
Applying the boundary conditions Θ′(0) = 0 and Θ′(π) = 0 we find that c2 = 0 and α = n for n = 0, 1, 2, . . . .
Since we want R(r) to be bounded as r → 0 we require c4 = 0. Thus
u(r, θ) = A0 +
∞∑
n=1
Anr
n cosnθ.
From
u(2, θ) =
{
u0, 0 < θ < π/2
0, π/2 < θ < π
= A0 +
∞∑
n=1
An2n cosnθ
we find
A0 =
1
2
2
π
∫ π/2
0
u0 dθ =
u0
2
and
2nAn =
2u0
π
∫ π/2
0
cosnθ dθ =
2u0
π
sinnπ/2
n
.
Therefore
u(r, θ) =
u0
2
+
2u0
π
∞∑
n=1
1
n
(
sin
nπ
2
) (r
2
)n
cosnθ.
16. (a) From Problem 1 in this section, with u0 = 100,
u(r, θ) = 50 +
100
π
∞∑
n=1
1− (−1)n
n
rn sinnθ.
761
1 2 3 4 5 6
q
20
40
60
80
100
u
r=0.9
r=0.7
r=0.5
r=0.3
r=0.1
14.1 Problems in Polar Coordinates
(b)
(c) We could use S5 from part (b) of this problem to compute the approximations, but in a CAS it is just as
easy to compute the sum with a much larger number of terms, thereby getting greater accuracy. In this
case we use partial sums including the term with r99 to find
u(0.9, 1.3) ≈ 96.5268 u(0.9, 2π − 1.3) ≈ 3.4731
u(0.7, 2) ≈ 87.871 u(0.7, 2π − 2) ≈ 12.129
u(0.5, 3.5) ≈ 36.0744 u(0.5, 2π − 3.5) ≈ 63.9256
u(0.3, 4) ≈ 35.2674 u(0.3, 2π − 4) ≈ 64.7326
u(0.1, 5.5) ≈ 45.4934 u(0.1, 2π − 5.5) ≈ 54.5066
(d) At the center of the plate u(0, 0) = 50. From the graphs in part (b) we observe that the solution curves
are symmetric about the point (π, 50). In part (c) we observe that the horizontal pairs add up to 100, and
hence average 50. This is consistent with the observation about part (b), so it is appropriate to say the
average temperature in the plate is 50◦.
17. Let u1 be the solution of the boundary-value problem
∂2u1
∂r2
+
1
r
∂u1
∂r
+
1
r2
∂2u1
∂θ2
= 0, 0 < θ < 2π, a < r < b
u1(a, θ) = f(θ), 0 < θ < 2π
u1(b, θ) = 0, 0 < θ < 2π,
and let u2 be the solution of the boundary-value problem
∂2u2
∂r2
+
1
r
∂u2
∂r
+
1
r2
∂2u2
∂θ2
= 0, 0 < θ < 2π, a < r < b
u2(a, θ) = 0, 0 < θ < 2π
u2(b, θ) = g(θ), 0 < θ < 2π.
762
14.2 Problems in Cylindrical Coordinates
Each of these problems can be solved using the methods shown in Problem 7 of this section. Now if u(r, θ) =
u1(r, θ) + u2(r, θ), then
u(a, θ) = u1(a, θ) + u2(a, θ) = f(θ)
u(b, θ) = u1(b, θ) + u2(b, θ) = g(θ)
and u(r, θ) will be the steady-state temperature of the circular ring with boundary conditions u(a, θ) = f(θ)
and u(b, θ) = g(θ).
EXERCISES 14.2
Problems in Cylindrical Coordinates
1. Referring to the solution of Example 1 in the text we have
R(r) = c1J0(αnr) and T (t) = c3 cos aαnt + c4 sin aαnt
where the αn are the positive roots of J0(αc) = 0. Now, the initial condition u(r, 0) = R(r)T (0) = 0 implies
T (0) = 0 and so c3 = 0. Thus
u(r, t) =
∞∑
n=1
An sin aαntJ0(αnr) and
∂u
∂t
=
∞∑
n=1
aαnAn cos aαntJ0(αnr).
From
∂u
∂t
∣∣∣∣
t=0
= 1 =
∞∑
n=1
aαnAnJ0(αnr)
we find
aαnAn =
2
c2J21 (αnc)
∫ c
0
rJ0(αnr) dr x = αnr, dx = αn dr
=
2
c2J21 (αnc)
∫ αnc
0
1
α2n
xJ0(x) dx
=
2
c2J21 (αnc)
∫ αnc
0
1
α2n
d
dx
[xJ1(x)] dx see (4) of Section 12.6 in text
=
2
c2α2nJ
2
1 (αnc)
xJ1(x)
∣∣∣∣αnc
0
=
2
cαnJ1(αnc)
.
Then
An =
2
acα2nJ1(αnc)
and
u(r, t) =
2
ac
∞∑
n=1
J0(αnr)
α2nJ1(αnc)
sin aαnt.
2. From Example 1 in the text we have Bn = 0 and
An =
2
J21 (αn)
∫ 1
0
r(1− r2)J0(αnr) dr.
763
14.2 Problems in Cylindrical Coordinates
From Problem 10, Exercises 12.6 we obtained An =
4J2(αn)
α2nJ
2
1 (αn)
. Thus
u(r, t) = 4
∞∑
n=1
J2(αn)
J21 (αn)
cos aαntJ0(αnr).
3. Referring to Example 2 in the text we have
R(r) = c1J0(αr) + c2Y0(αr)
Z(z) = c3 coshαz + c4 sinhαz
where c2 = 0 and J0(2α) = 0 defines the positive eigenvalues λn = α2n. From Z(4) = 0 we obtain
c3 cosh 4αn + c4 sinh 4αn = 0 or c4 = −c3 cosh 4αnsinh 4αn .
Then
Z(z) = c3
[
coshαnz − cosh 4αnsinh 4αn sinhαnz
]
= c3
sinh 4αn coshαnz − cosh 4αn sinhαnz
sinh 4αn
= c3
sinhαn(4− z)
sinh 4αn
and
u(r, z) =
∞∑
n=1
An
sinhαn(4− z)
sinh 4αn
J0(αnr).
From
u(r, 0) = u0 =
∞∑
n=1
AnJ0(αnr)
we obtain
An =
2u0
4J21 (2αn)
∫ 2
0
rJ0(αnr) dr =
u0
αnJ1(2αn)
.
Thus the temperature in the cylinder is
u(r, z) = u0
∞∑
n=1
sinhαn(4− z)J0(αnr)
αn sinh 4αnJ1(2αn)
.
4. (a) The boundary condition ur(2, z) = 0 implies R′(2) = 0 or J ′0(2α) = 0. Thus α = 0 is also an eigenvalue
and the separated equations are in this case rR′′+R′ = 0 and z′′ = 0. The solutions of these equations are
then
R(r) = c1 + c2 ln r, Z(z) = c3z + c4.
Now Z(0) = 0 yields c4 = 0 and the implicit condition that the temperature is bounded as r → 0 demands
that we define c2 = 0. Thus we have
u(r, z) = A1z +
∞∑
n=2
An sinhαnzJ0(αnr). (1)
At z = 4 we obtain
f(r) = 4A1 +
∞∑
n=2
An sinh 4αnJ0(αnr).
764
14.2 Problems in Cylindrical Coordinates
Thus from (17) and (18) of Section 12.6 in the text we can write with b = 2,
A1 =
1
8
∫ 2
0
rf(r) dr (2)
An =
1
2 sinh 4αnJ20 (2αn)
∫ 2
0
rf(r)J0(αnr) dr. (3)
A solution of the problem consists of the series (1) with coefficients A1 and An defined in (2) and (3),
respectively.
(b) When f(r) = u0 we get A1 = u0/4 and
An =
u0J1(2αn)
αn sinh 4αnJ20 (2αn)
= 0
since J ′0(2α) = 0 is equivalent to J1(2α) = 0. A solution of the problem is then u(r, z) =
u0
4
z.
5. Letting u(r, t) = R(r)T (t) and separating variables we obtain
R′′ + 1rR
′
R
=
T ′
kT
= −λ and R′′ + 1
r
R′ + λR = 0, T ′ + λkT = 0.
From the last equation we find T (t) = e−λkt. If λ < 0, T (t) increases without bound as t→∞. Thus we assume
λ = α2 > 0. Now
R′′ +
1
r
R′ + α2R = 0
is a parametric Bessel equation with solution
R(r) = c1J0(αr) + c2Y0(αr).
Since Y0 is unbounded as r → 0 we take c2 = 0. Then R(r) = c1J0(αr) and the boundary condition
u(c, t) = R(c)T (t) = 0 implies J0(αc) =0. This latter equation defines the positive eigenvalues λn = α2n. Thus
u(r, t) =
∞∑
n=1
AnJ0(αnr)e−α
2
nkt.
From
u(r, 0) = f(r) =
∞∑
n=1
AnJ0(αnr)
we find
An =
2
c2J21 (αnc)
∫ c
0
rJ0(αnr)f(r) dr, n = 1, 2, 3, . . . .
6. If the edge r = c is insulated we have the boundary condition ur(c, t) = 0. Referring to the solution of
Problem 5 above we have
R′(c) = αc1J ′0(αc) = 0
which defines an eigenvalue λ = α2 = 0 and positive eigenvalues λn = α2n. Thus
u(r, t) = A0 +
∞∑
n=1
AnJ0(αnr)e−α
2
nkt.
From
u(r, 0) = f(r) = A0 +
∞∑
n=1
AnJ0(αnr)
765
14.2 Problems in Cylindrical Coordinates
we find
A0 =
2
c2
∫ c
0
rf(r) dr
An =
2
c2J20 (αnc)
∫ c
0
rJ0(αnr)f(r) dr.
7. Referring to Problem 5 above we have T (t) = e−λkt and R(r) = c1J0(αr). The boundary condition
hu(1, t) + ur(1, t) = 0 implies hJ0(α) + αJ ′0(α) = 0 which defines positive eigenvalues λn = α
2
n. Now
u(r, t) =
∞∑
n=1
AnJ0(αnr)e−α
2
nkt
where
An =
2α2n
(α2n + h2)J20 (αn)
∫ 1
0
rJ0(αnr)f(r) dr.
8. We solve
∂2u
∂r2
+
1
r
∂u
∂r
+
∂2u
∂z2
= 0, 0 < r < 1, z > 0
∂u
∂r
∣∣∣∣
r=1
= −hu(1, z), z > 0
u(r, 0) = u0, 0 < r < 1.
assuming u = RZ we get
R′′ + 1rR
′
R
= −Z
′′
Z
= −λ
and so
rR′′ + R′ + λ2rR = 0 and Z ′′ − λZ = 0.
Letting λ = α2 we then have
R(r) = c1J0(αr) + c2Y0(αr) and Z(z) = c3e−αz + c4eαz.
We use the exponential form of the solution of Z ′′−α2Z = 0 since the domain of the variable z is a semi-infinite
interval. As usual we define c2 = 0 since the temperature is surely bounded as r → 0. Hence R(r) = c1J0(αr).
Now the boundary-condition ur(1, z) + hu(1, z) = 0 is equivalent to
αJ ′0(α) + hJ0(α) = 0. (4)
The eigenvalues αn are the positive roots of (4) above. Finally, we must now define c4 = 0 since the temperature
is also expected to be bounded as z →∞. A product solution of the partial differential equation that satisfies
the first boundary condition is given by
un(r, z) = Ane−αnzJ0(αnr).
Therefore
u(r, z) =
∞∑
n=1
Ane
−αnzJ0(αnr)
is another formal solution. At z = 0 we have u0 = AnJ0(αnr). In view of (4) above we use equations (17) and
(18) of Section 12.6 in the text with the identification b = 1:
An =
2α2n
(α2n + h2) J20 (αn)
∫ 1
0
rJ0(αnr)u0 dr
766
14.2 Problems in Cylindrical Coordinates
=
2α2nu0
(α2n + h2)J20 (αn)α2n
tJ1(t)
∣∣∣∣αn
0
=
2αnu0J1(αn)
(α2n + h2)J20 (αn)
. (5)
Since J ′0 = −J1 [see equation (6) of Section 11.5 in the text] it follows from (4) above that αnJ1(αn) = hJ0(αn).
Thus (5) above simplifies to
An =
2u0h
(α2n + h2)J0(αn)
.
A solution to the boundary-value problem is then
u(r, z) = 2u0h
∞∑
n=1
e−αnz
(α2n + h2) J0(αn)
J0(αnr).
9. Substituting u(r, t) = v(r, t) + ψ(r) into the partial differential equation gives
∂2v
∂r2
+
1
r
∂v
∂r
+ ψ′′ +
1
r
ψ′ =
∂v
∂t
.
This equation will be homogeneous provided ψ′′ + 1rψ
′ = 0 or ψ(r) = c1 ln r + c2. Since ln r is unbounded as
r → 0 we take c1 = 0. Then ψ(r) = c2 and using u(2, t) = v(2, t) + ψ(2) = 100 we set c2 = ψ(2) = 100.
Therefore ψ(r) = 100. Referring to Problem 5 above, the solution of the boundary-value problem
∂2v
∂r2
+
1
r
∂v
∂r
=
∂v
∂t
, 0 < r < 2, t > 0,
v(2, t) = 0, t > 0,
v(r, 0) = u(r, 0)− ψ(r)
is
v(r, t) =
∞∑
n=1
AnJ0(αnr)e−α
2
nt
where
An =
2
22J21 (2αn)
∫ 2
0
rJ0(αnr)[u(r, 0)− ψ(r)] dr
=
1
2J21 (2αn)
[∫ 1
0
rJ0(αnr)[200− 100] dr +
∫ 2
1
rJ0(αnr)[100− 100] dr
]
=
50
J21 (2αn)
∫ 1
0
rJ0(αnr) dr x = αnr, dx = αn dr
=
50
J21 (2αn)
∫ αn
0
1
α2n
xJ0(x) dx
=
50
α2nJ
2
1 (2αn)
∫ αn
0
d
dx
[xJ1(x)] dx see (5) of Section 12.6 in text
=
50
α2nJ
2
1 (2αn)
(
xJ1(x)
) ∣∣∣∣αn
0
=
50J1(αn)
αnJ21 (2αn)
.
Thus
u(r, t) = v(r, t) + ψ(r) = 100 + 50
∞∑
n=1
J1(αn)J0(αnr)
αnJ21 (2αn)
e−α
2
nt.
10. Letting u(r, t) = u(r, t)+ψ(r) we obtain rψ′′+ψ′ = −βr. The general solution of this nonhomogeneous Cauchy-
Euler equation is found with the aid of variation of parameters: ψ = c1 + c2 ln r − βr2/4. In order that this
767
14.2 Problems in Cylindrical Coordinates
solution be bounded as r → 0 we define c2 = 0. Using ψ(1) = 0 then gives c1 = β/4 and so ψ(r) = β(1− r2)/4.
Using v = RT we find that a solution of
∂2v
∂r2
+
1
r
∂v
∂r
=
∂v
∂t
, 0 < r < 1, t > 0
v(1, t) = 0, t > 0
v(r, 0) = −ψ(r), 0 < r < 1
is
v(r, t) =
∞∑
n=1
Ane
−α2ntJ0(αnr)
where
An = −β4
2
J21 (αn)
∫ 1
0
r(1− r2)J0(αnr) dr
and the αn are defined by J0(α) = 0. From the result of Problem 10, Exercises 12.6 (see also Problem 2 of this
exercise set) we get
An = − βJ2(αn)
α2nJ
2
1 (αn)
.
Thus from u = v + ψ(r) it follows that
u(r, t) =
β
4
(1− r2)− β
∞∑
n=1
J2(αn)
α2nJ
2
1 (αn)
e−α
2
ntJ0(αnr).
11. (a) Writing the partial differential equation in the form
g
(
x
∂2u
∂x2
+
∂u
∂x
)
=
∂2u
∂t2
and separating variables we obtain
xX ′′ + X ′
X
=
T ′′
gT
= −λ.
Letting λ = α2 we obtain
xX ′′ + X ′ + α2X = 0 and T ′′ + gα2T = 0.
Letting x = τ2/4 in the first equation we obtain dx/dτ = τ/2 or dτ/dx = 2τ . Then
dX
dx
=
dX
dτ
dτ
dx
=
2
τ
dX
dτ
and
d2X
dx2
=
d
dx
(
2
τ
dX
dτ
)
=
2
τ
d
dx
(
dX
dτ
)
+
dX
dτ
d
dx
(
2
τ
)
=
2
τ
d
dτ
(
dX
dτ
)
dτ
dx
+
dX
dτ
d
dτ
(
2
τ
)
dτ
dx
=
4
τ2
d2X
dτ2
− 4
τ3
dX
dτ
.
Thus
xX ′′ + X ′ + α2X =
τ2
4
(
4
τ2
d2X
dτ2
− 4
τ3
dX
dτ
)
+
2
τ
dX
dτ
+ α2X =
d2X
dτ2
+
1
τ
dX
dτ
+ α2X = 0.
This is a parametric Bessel equation with solution
X(τ) = c1J0(ατ) + c2Y0(ατ).
768
14.2 Problems in Cylindrical Coordinates
(b) To insure a finite solution at x = 0 (and thus τ = 0) we set c2 = 0. The condition u(L, t) = X(L)T (t) = 0
implies X
∣∣
x=L
= X
∣∣
τ=2
√
L
= c1J0(2α
√
L ) = 0, which defines positive eigenvalues λn = α2n. The solution
of T ′′ + gα2T = 0 is
T (t) = c3 cos(αn
√
g t) + c4 sin(αn
√
g t).
The boundary condition ut(x, 0) = X(x)T ′(0) = 0 implies c4 = 0. Thus
u(τ, t) =
∞∑
n=1
An cos(αn
√
g t)J0(αnτ).
From
u(τ, 0) = f(τ2/4) =
∞∑
n=1
AnJ0(αnτ)
we find
An =
2
(2
√
L )2J21 (2αn
√
L )
∫ 2√L
0
τJ0(αnτ)f(τ2/4) dτ v = τ/2, dv = dτ/2
=
1
2LJ21 (2αn
√
L )
∫ √L
0
2vJ0(2αnv)f(v2)2 dv
=
2
LJ21 (2αn
√
L )
∫ √L
0
vJ0(2αnv)f(v2) dv.
The solution of the boundary-value problem is
u(x, t) =
∞∑
n=1
An cos(αn
√
g t)J0(2αn
√
x ).
12. (a) First we see that
R′′Θ +
1
r
R′Θ +
1
r2
RΘ′′
RΘ
=
T ′′
a2T
= −λ.
This gives T ′′ + a2λT = 0 and from
R′′ +
1
r
R′ + λR
−R/r2 =
Θ′′
Θ
= −ν
we get Θ′′ + νΘ = 0 and r2R′′ + rR′ + (λr2 − ν)R = 0.
(b) With λ = α2 and ν = β2 the general solutions of the differential equations in part (a) are
T = c1 cos aαt + c2 sin aαt
Θ = c3 cosβθ + c4 cosβθ
R = c5Jβ(αr) + c6Yβ(αr).
(c) Implicitly we expect u(r, θ, t) = u(r, θ + 2π, t) and so Θ must be 2π-periodic. Therefore β = n, n = 0, 1,
2, . . . . The corresponding eigenfunctions are 1, cos θ, cos 2θ, . . . , sin θ, sin 2θ, . . . . Arguing that u(r, θ, t)
is bounded as r → 0 we then define c6 = 0 and so R = c3Jn(αr). But R(c) = 0 gives Jn(αc) = 0; this
equation defines the eigenvalues λn = α2n. For each n, αni = xni/c, i = 1, 2, 3, . . . . The corresponding
eigenfunctions are Jn(λnir) = 0.
769
2 4 6 8 10
r
-1
-0.5
0.5
1 u
t=0
t=4
t=10
t=12
t=20
14.2 Problems in Cylindrical Coordinates
(d) u(r, θ,t) =
n∑
i=1
(A0i cos aα0it + B0i sin aα0it)J0(α0ir)
+
∞∑
n=1
∞∑
i=1
[
(Ani cos aαnit + Bni sin aαnit) cosnθ
+ (Cni cos aαnit + Dni sin aαnit) sinnθ
]
Jn(αnir)
13. (a) With c = 10 in Example 1 in the text the eigenvalues are λn = α2n = x
2
n/100 where xn is a positive root
of J0(x) = 0. From a CAS we find that x1 = 2.4048, x2 = 5.5201, and x3 = 8.6537, so that the first three
eigenvalues are λ1 = 0.0578, λ2 = 0.3047, and λ3 = 0.7489. The corresponding coefficients are
A1 =
2
100J21 (x1)
∫ 10
0
rJ0(x1r/10)(1− r/10) dr = 0.7845,
A2 =
2
100J21 (x2)
∫ 10
0
rJ0(x2r/10)(1− r/10) dr = 0.0687,
and
A3 =
2
100J21 (x3)
∫ 10
0
rJ0(x3r/10)(1− r/10) dr = 0.0531.
Since g(r) = 0, Bn = 0, n = 1, 2, 3, . . . , and the third partial sum of the series solution is
S3(r, t) =
∞∑
n=1
An cos(xnt/10)J0(xnr/10)
= 0.7845 cos(0.2405t)J0(0.2405r) + 0.0687 cos(0.5520t)J0(0.5520r)
+ 0.0531 cos(0.8654t)J0(0.8654r).
(b)
14. Because of the nonhomogeneous boundary condition u(c, t) = 200 we use the substitution u(r, t) = v(r, t)+ψ(r).
This gives
k
(
∂2v
∂r2
+
1
r
∂v
∂r
+ ψ′′ +
1
r
ψ′
)
=
∂v
∂t
.
This equation will be homogeneous provided ψ′′+(1/r)ψ′ = 0 or ψ(r) = c1 ln r + c2. Since ln r is unbounded as
r → 0 we take c1 = 0. Then ψ(r) = c2 and using u(c, t) = v(c, t) + c2 = 200 we set c2 = 200, giving v(c, t) = 0.
Referring to Problem 5 in this section, the solution of the boundary-value problem
k
(
∂2v
∂r2
+
1
r
∂v
∂r
)
=
∂v
∂t
, 0 < r < c, t > 0
770
5000 10000
t
-100
100
200
u
r�0
r�5
14.2 Problems in Cylindrical Coordinates
v(c, t) = 0, t > 0
v(r, 0) = −200, 0 < r < c
is
v(r, t) =
∞∑
n=1
AnJ0(αnr)e−α
2
nkt,
where the separation constant is −λ = −α2. The eigenvalues are λn = α2 = x2n/c2 where xn is a positive root
of J0(x) = 0 and the coefficients An are
An =
2
c2J21 (αnc)
∫ c
0
rJ0(αnr)(−200)dr = − 400
c2J21 (αnc)
∫ c
0
rJ0(αnr)dr.
Taking c = 10 and k = 0.1 we have
u(r, t) = v(r, t) + 200 = 200 +
∞∑
n=1
AnJ0(xnr/10)e−0.01x
2
nt/100
where
An = − 4
J21 (xn)
∫ 10
0
rJ0(xnr/10) dr.
Using a CAS we find that the first five values of xn are
x1 = 2.4048, x2 = 5.5201, x3 = 8.6537, x4 = 11.7915,
and x5 = 14.9309, with corresponding eigenvalues
λ1 = 0.0578, λ2 = 0.3047, λ3 = 0.7489, λ4 = 1.3904,
and λ5 = 2.2293. The first five values of An are
A1 = −320.4, A2 = 213.0, A3 = −170.3, A4 = 145.9,
and A5 = −129.7. Using a root finding application in
a CAS we find that u(5, t) = 100 when t ≈ 1331 and
u(0, t) = 100 when t ≈ 2005. Since u = 200 is an asymptote for the graphs of u(0, t) and u(5, t) we solve
u(5, t) = 199.9 and u(0, t) = 199.9. This gives t ≈ 13,265 and t ≈ 13,958, respectively.
15. (a) The boundary-value problem is
a2
(
∂2u
∂r2
+
1
r
∂u
∂r
)
=
∂2u
∂t2
, 0 < r < 1, t > 0
u(1, t) = 0, t > 0
u(r, 0) = 0,
∂u
∂t
∣∣∣
t=0
=
{−v0, 0 ≤ r < b
0, b ≤ r < 1 , 0 < r < 1,
and the solution is
u(r, t) =
∞∑
n=1
(An cos aαnt + Bn sin aαnt)J0(αnr),
where the eigenvalues λn = α2n are defined by J0(α) = 0 and An = 0 since f(r) = 0. The coefficients Bn
771
-1 -0.5 0.5 1 r
-0.2
-0.1
0.1
0.2
u
-1 -0.5 0.5 1 r
-0.2
-0.1
0.1
0.2
u
-1 -0.5 0.5 1 r
-0.2
-0.1
0.1
0.2
u
-1 -0.5 0.5 1 r
-0.2
-0.1
0.1
0.2
u
-1 -0.5 0.5 1 r
-0.2
-0.1
0.1
0.2
u
-1 -0.5 0.5 1 r
-0.2
-0.1
0.1
0.2
u
14.2 Problems in Cylindrical Coordinates
are given by
Bn =
2
aαnJ21 (αn)
∫ b
0
rJ0(αnr)g(r) dr = − 2v0
aαnJ21 (αn)
∫ b
0
rJ0(αnr)dr
let x = αnr
= − 2v0
aαnJ21 (αn)
∫ αnb
0
x
αn
J0(x)
1
αn
dx = − 2v0
aα3nJ
2
1 (αn)
∫ αnb
0
xJ0(x) dx
= − 2v0
aα3nJ
2
1 (αn)
(
xJ1(x)
) ∣∣∣αnb
0
= − 2v0
aα3nJ1(αn)
(
αnbJ1(αnb)
)
= −2v0b
aα2n
J1(αnb)
J21 (αn)
.
Thus,
u(r, t) =
−2v0b
a
∞∑
n=1
1
α2n
J1(αnb)
J21 (αn)
sin(aαnt)J0(αnr).
(b) The standing wave un(r, t) is given by un(r, t) = Bn sin(aαnt)J0(αnr), which has frequency fn = aαn/2π,
where αn is the nth positive zero of J0(x). The fundamental frequency is f1 = aα1/2π. The next two
frequencies are
f2 =
aα2
2π
=
α2
α1
(
aα1
2π
)
=
5.520
2.405
f1 = 2.295f1
and
f3 =
aα3
2π
=
α3
α1
(
aα1
2π
)
=
8.654
2.405
f1 = 3.598f1.
(c) With a = 1, b = 14 , and v0 = 1, the solution becomes
u(r, t) = −1
2
∞∑
n=1
1
α2n
J1(αn/4)
J21 (αn)
sin(αnt)J0(αnr).
The graphs of S5(r, t) for t = 1, 2, 3, 4, 5, 6 are shown below.
772
14.3 Problems in Spherical Coordinates
(d) Three frames from the movie are shown.
EXERCISES 14.3
Problems in Spherical Coordinates
1. To compute
An =
2n + 1
2cn
∫ π
0
f(θ)Pn(cos θ) sin θ dθ
we substitute x = cos θ and dx = − sin θ dθ. Then
An =
2n + 1
2cn
∫ −1
1
F (x)Pn(x)(−dx) = 2n + 12cn
∫ 1
−1
F (x)Pn(x) dx
where
F (x) =
{
0, −1 < x < 0
50, 0 < x < 1
= 50
{
0, −1 < x < 0
1, 0 < x < 1
.
The coefficients An are computed in Example 3 of Section 12.6. Thus
u(r, θ) =
∞∑
n=0
Anr
nPn(cos θ)
= 50
[
1
2
P0(cos θ) +
3
4
(
r
c
)
P1(cos θ)− 716
(r
c
)3
P3(cos θ) +
11
32
(r
c
)5
P5(cos θ) + · · ·
]
.
2. In the solution of the Cauchy-Euler equation,
R(r) = c1rn + c2r−(n+1),
we define c1 = 0 since we expect the potential u to be bounded as r →∞. Hence
un(r, θ) = Anr−(n+1)Pn(cos θ)
u(r, θ) =
∞∑
n=0
Anr
−(n+1)Pn(cos θ).
When r = c we have
f(θ) =
∞∑
n=0
Anc
−(n+1)Pn(cos θ)
773
14.3 Problems in Spherical Coordinates
so that
An = cn+1
(2n + 1)
2
∫ π
0
f(θ)Pn(cos θ) sin θ dθ.
The solution of the problem is then
u(r, θ) =
∞∑
n=0
(
2n + 1
2
∫ π
0
f(θ)Pn(cos θ) sin θdθ
) ( c
r
)n+1
Pn(cos θ).
3. The coefficients are given by
An =
2n + 1
2cn
∫ π
0
cos θPn(cos θ) sin θ dθ =
2n + 1
2cn
∫ π
0
P1(cos θ)Pn(cos θ) sin θ dθ
x = cos θ, dx = − sin θ dθ
=
2n + 1
2cn
∫ 1
−1
P1(x)Pn(x) dx.
Since Pn(x) and Pm(x) are orthogonal for m �= n, An = 0 for n �= 1 and
A1 =
2(1) + 1
2c1
∫ 1
−1
P1(x)P1(x) dx =
3
2c
∫ 1
−1
x2dx =
1
c
.
Thus
u(r, θ) =
r
c
P1(cos θ) =
r
c
cos θ.
4. The coefficients are given by
An =
2n + 1
2cn
∫ π
0
(1− cos 2θ)Pn(cos θ) sin θ dθ.
These were computed in Problem 18 of Section 12.6. Thus
u(r, θ) =
4
3
P0(cos θ)− 43
(r
c
)2
P2(cos θ).
5. Referring to Example 1 in the text we have
Θ = Pn(cos θ) and R = c1rn + c2r−(n+1).
Since u(b, θ) = R(b)Θ(θ) = 0,
c1b
n + c2b−(n+1) = 0 or c1 = −c2b−2n−1,
and
R(r) = −c2b−2n−1rn + c2r−(n+1) = c2
(
b2n+1 − r2n+1
b2n+1rn+1
)
.
Then
u(r, θ) =
∞∑
n=0
An
b2n+1 − r2n+1
b2n+1rn+1
Pn(cos θ)
where
b2n+1 − a2n+1
b2n+1an+1
An =
2n + 1
2
∫ π
0
f(θ)Pn(cos θ) sin θ dθ.
6. Referring to Example 1 in the text we have
R(r) = c1rn and Θ(θ) = Pn(cos θ).
774
14.3 Problems in Spherical Coordinates
Now Θ(π/2) = 0 implies that n is odd, so
u(r, θ) =
∞∑
n=0
A2n+1r
2n+1P2n+1(cos θ).
From
u(c, θ) = f(θ) =
∞∑
n=0
A2n+1c
2n+1P2n+1(cos θ)
we see that
A2n+1c
2n+1 = (4n + 3)
∫ π/2
0
f(θ) sin θ P2n+1(cos θ) dθ.
Thus
u(r, θ) =
∞∑
n=0
A2n+1r
2n+1P2n+1(cos θ)
where
A2n+1 =
4n + 3
c2n+1
∫ π/2
0
f(θ) sin θ P2n+1(cos θ) dθ.
7. Referring to Example 1 in the text we have
r2R′′ + 2rR′ − λR = 0
sin θ Θ′′ + cos θ Θ′ + λ sin θ Θ = 0.
Substituting x = cos θ, 0 ≤ θ ≤ π/2, the latter equation becomes
(1− x2) d
2Θ
dx2
− 2x dΘ
dx
+ λΘ = 0, 0 ≤ x ≤ 1.
Taking the solutions of this equation to be theLegendre polynomials Pn(x) corresponding to
λ = n(n + 1) for n = 1, 2, 3, . . . , we have Θ = Pn(cos θ). Since
∂u
∂θ
∣∣∣∣
θ=π/2
= Θ′(π/2)R(r) = 0
we have
Θ′(π/2) = −(sinπ/2)P ′n(cosπ/2) = −P ′n(0) = 0.
As noted in the hint, P ′n(0) = 0 only if n is even. Thus Θ = Pn(cos θ), n = 0, 2, 4, . . . . As in Example 1,
R(r) = c1rn. Hence
u(r, θ) =
∞∑
n=0
A2nr
2nP2n(cos θ).
At r = c,
f(θ) =
∞∑
n=0
A2nc
2nP2n(cos θ).
Using Problem 19 in Section 12.6, we obtain
c2nA2n = (4n + 1)
∫ 0
π/2
f(θ)P2n(cos θ)(− sin θ) dθ
and
A2n =
4n + 1
c2n
∫ π/2
0
f(θ) sin θ P2n(cos θ) dθ.
775
14.3 Problems in Spherical Coordinates
8. Referring to Example 1 in the text we have
R(r) = c1rn + c2r−(n−1) and Θ(θ) = Pn(cos θ).
Since we expect u(r, θ) to be bounded as r →∞, we define c1 = 0. Also Θ(π/2) = 0 implies that n is odd, so
u(r, θ) =
∞∑
n=0
A2n+1r
−2(n+1)P2n+1(cos θ).
From
u(c, θ) = f(θ) =
∞∑
n=0
A2n+1c
−2(n+1)P2n+1(cos θ)
we see that
A2n+1c
−2(n+1) = (4n + 3)
∫ π/2
0
f(θ) sin θ P2n+1(cos θ) dθ.
Thus
u(r, θ) =
∞∑
n=0
A2n+1r
−2(n+1)P2n+1(cos θ)
where
A2n+1 = (4n + 3)c2(n+1)
∫ π/2
0
f(θ) sin θ P2n+1(cos θ) dθ.
9. Checking the hint, we find
1
r
∂2
∂r2
(ru) =
1
r
∂
∂r
[
r
∂u
∂r
+ u
]
=
1
r
[
r
∂2u
∂r2
+
∂u
∂r
+
∂u
∂r
]
=
∂2u
∂r2
+
2
r
∂u
∂r
.
The partial differential equation then becomes
∂2
∂r2
(ru) = r
∂u
∂t
.
Now, letting ru(r, t) = v(r, t) + ψ(r), since the boundary condition is nonhomogeneous, we obtain
∂2
∂r2
[v(r, t) + ψ(r)] = r
∂
∂t
[
1
r
v(r, t) + ψ(r)
]
or
∂2v
∂r2
+ ψ′′(r) =
∂v
∂t
.
This differential equation will be homogeneous if ψ′′(r) = 0 or ψ(r) = c1r + c2. Now
u(r, t) =
1
r
v(r, t) +
1
r
ψ(r) and
1
r
ψ(r) = c1 +
c2
r
.
Since we want u(r, t) to be bounded as r approaches 0, we require c2 = 0. Then ψ(r) = c1r. When r = 1
u(1, t) = v(1, t) + ψ(1) = v(1, t) + c1 = 100,
and we will have the homogeneous boundary condition v(1, t) = 0 when c1 = 100. Consequently, ψ(r) = 100r.
The initial condition
u(r, 0) =
1
r
v(r, 0) +
1
r
ψ(r) =
1
r
v(r, 0) + 100 = 0
implies v(r, 0) = −100r. We are thus led to solve the new boundary-value problem
∂2v
∂r2
=
∂v
∂t
, 0 < r < 1, t > 0,
v(1, t) = 0, lim
r→0
1
r
v(r, t) <∞,
v(r, 0) = −100r.
776
14.3 Problems in Spherical Coordinates
Letting v(r, t) = R(r)T (t) and using the separation constant −λ we obtain
R′′ + λR = 0 and T ′ + λT = 0.
Using λ = α2 > 0 we then have
R(r) = c3 cosαr + c4 sinαr and T (t) = c5e−α
2t.
The boundary conditions are equivalent to R(1) = 0 and limr→0 R(r)/r <∞. Since
lim
r→0
cosαr
r
does not exist we must have c3 = 0. Then R(r) = c4 sinαr, and R(1) = 0 implies α = nπ for n = 1, 2, 3, . . . .
Thus
vn(r, t) = Ane−n
2π2t sinnπr
for n = 1, 2, 3, . . . . Using the condition limr→0 R(r)/r < ∞ it is easily shown that there are no eigenvalues
for λ = 0, nor does setting the common constant to −λ = α2 when separating variables lead to any solutions.
Now, by the superposition principle,
v(r, t) =
∞∑
n=1
Ane
−n2π2t sinnπr.
The initial condition v(r, 0) = −100r implies
−100r =
∞∑
n=1
An sinnπr.
This is a Fourier sine series and so
An = 2
∫ 1
0
(−100r sinnπr) dr = −200
[
− r
nπ
cosnπr
∣∣∣∣1
0
+
∫ 1
0
1
nπ
cosnπr dr
]
= −200
[
−cosnπ
nπ
+
1
n2π2
sinnπr
∣∣∣∣1
0
]
= −200
[
− (−1)
n
nπ
]
=
(−1)n200
nπ
.
A solution of the problem is thus
u(r, t) =
1
r
v(r, t) +
1
r
ψ(r) =
1
r
∞∑
n=1
(−1)n 20
nπ
e−n
2π2t sinnπr +
1
r
(100r)
=
200
πr
∞∑
n=1
(−1)n
n
e−n
2π2t sinnπr + 100.
10. Referring to Problem 9 we have
∂2v
∂r2
+ ψ′′(r) =
∂v
∂t
where ψ(r) = c1r. Since
u(r, t) =
1
r
v(r, t) +
1
r
ψ(r) =
1
r
v(r, t) + c1
we have
∂u
∂r
=
1
r
vr(r, t)− 1
r2
v(r, t).
When r = 1,
∂u
∂r
∣∣∣∣
r=1
= vr(1, t)− v(1, t)
777
14.3 Problems in Spherical Coordinates
and
∂u
∂r
∣∣∣∣
r=1
+ hu(1, t) = vr(1, t)− v(1, t) + h[v(1, t) + ψ(1)] = vr(1, t) + (h− 1)v(1, t) + hc1.
Thus the boundary condition
∂u
∂r
∣∣∣∣
r=1
+ hu(1, t) = hu1
will be homogeneous when hc1 = hu1 or c1 = u1. Consequently ψ(r) = u1r. The initial condition
u(r, 0) =
1
r
v(r, 0) +
1
r
ψ(r) =
1
r
v(r, 0) + u1 = u0
implies v(r, 0) = (u0 − u1)r. We are thus led to solve the new boundary-value problem
∂2v
∂r2
=
∂v
∂t
, 0 < r < 1, t > 0,
vr(1, t) + (h− 1)v(1, t) = 0, t > 0,
lim
r→0
1
r
v(r, t) <∞,
v(r, 0) = (u0 − u1)r.
Separating variables as in Problem 9 leads to
R(r) = c3 cosαr + c4 sinαr and T (t) = c5e−α
2t.
The boundary conditions are equivalent to
R′(1) + (h− 1)R(1) = 0 and lim
r→0
1
r
R(r) <∞.
As in Problem 6 we use the second condition to determine that c3 = 0 and R(r) = c4 sinαr. Then
R′(1) + (h− 1)R(1) = c4α cosα + c4(h− 1) sinα = 0
and the αn are the consecutive nonnegative roots of tanα = α/(1− h). Now
v(r, t) =
∞∑
n=1
Ane
−α2nt sinαnr.
From
v(r, 0) = (u0 − u1)r =
∞∑
n=1
An sinαnr
we obtain
An =
∫ 1
0
(u0 − u1)r sinαnr dr∫ 1
0
sin2 αnr dr
.
We compute the integrals∫ 1
0
r sinαnr dr =
(
1
α2n
sinαnr − 1
αn
cosαnr
) ∣∣∣∣1
0
=
1
α2n
sinαn − 1
αn
cosαn
and ∫ 1
0
sin2 αnr dr =
(
1
2
r − 1
4αn
sin 2αnr
) ∣∣∣∣1
0
=
1
2
− 1
4αn
sin 2αn.
778
14.3 Problems in Spherical Coordinates
Using αn cosαn = −(h− 1) sinαn we then have
An = (u0 − u1)
1
α2n
sinαn − 1αn cosαn
1
2 − 14αn sin 2αn
= (u0 − u1) 4 sinαn − 4αn cosαn2α2n − αn2 sinαn cosαn
= 2(u0 − u1) sinαn + (h− 1) sinαn
α2n + (h− 1) sinαn sinαn
= 2(u0 − u1)h sinαn
α2n + (h− 1) sin2 αn
.
Therefore
u(r, t) =
1
r
v(r, t) +
1
r
ψ(r) = u1 + 2(u0 − u1)h
∞∑
n=1
sinαn sinαnr
r[α2n + (h− 1) sin2 αn]
e−α
2
nt.
11. We write the differential equation in the form
a2
1
r
∂2
∂r2
(ru) =
∂2u
∂t2
or a2
∂2
∂r2
(ru) = r
∂2u
∂t2
,
and then let v(r, t) = ru(r, t). The new boundary-value problem is
a2
∂2v
∂r2
=
∂2v
∂t2
, 0 < r < c, t > 0
v(c, t) = 0, t > 0
v(r, 0) = rf(r),
∂v
∂t
∣∣∣∣
t=0
= rg(r).
Letting v(r, t) = R(r)T (t) and using the separation constant −λ = −α2 we obtain
R′′ + α2R = 0
T ′′ + a2α2T = 0
and
R(r) = c1 cosαr + c2 sinαr
T (t) = c3 cos aαt + c4 sin aαt.
Since u(r, t) = v(r, t)/r, in order to insure boundedness at r = 0 we define c1 = 0. Then R(r) = c2 sinαr and
the condition R(c) = 0 implies α = nπ/c. Thus
v(r, t) =
∞∑
n=1
(
An cos
nπa
c
t + Bn sin
nπa
c
t
)
sin
nπ
c
r.
From
v(r, 0) = rf(r) =
∞∑
n=1
An sin
nπ
c
r
we see that
An =
2
c
∫ c
0
rf(r) sin
nπ
c
r dr.
From
∂v
∂t
∣∣∣∣
t=0
= rg(r) =
∞∑
n=1
(
Bn
nπa
c
)
sin
nπ
c
r
we see that
Bn =
c
nπa
· 2
c
∫ c
0
rg(r) sin
nπ
c
r dr =
2
nπa
∫ c
0
rg(r) sin
nπ
c
r dr.
The solution is
u(r, t) =
1
r
∞∑
n=1
(
An cos
nπa
c
t + Bn sin
nπa
c
t
)
sin
nπ
c
r,
779
14.3 Problems in Spherical Coordinates
where An and Bn are given above.
12. Proceeding as in Example 1 we obtain
Θ(θ) = Pn(cos θ) and R(r) = c1rn + c2r−(n+1)
so that
u(r, θ) =
∞∑
n=0
(Anrn + Bnr−(n+1))Pn(cos θ).
To satisfy limr→∞ u(r, θ) = −Er cos θ we must have An = 0 for n = 2, 3, 4, . . . . Then
lim
r→∞u(r, θ) = −Er cos θ = A0 · 1 + A1r cos θ,
so A0 = 0 and A1 = −E. Thus
u(r, θ) = −Er cos θ +
∞∑
n=0
Bnr
−(n+1)Pn(cos θ).
Now
u(c, θ) = 0 = −Ec cos θ +
∞∑
n=0
Bnc
−(n+1)Pn(cos θ)so
∞∑
n=0
Bnc
−(n+1)Pn(cos θ) = Ec cos θ
and
Bnc
−(n+1) =
2n + 1
2
∫ π
0
Ec cos θ Pn(cos θ) sin θ dθ.
Now cos θ = P1(cos θ) so, for n �= 1, ∫ π
0
cos θ Pn(cos θ) sin θ dθ = 0
by orthogonality. Thus Bn = 0 for n �= 1 and
B1 =
3
2
Ec3
∫ π
0
cos2 θ sin θ dθ = Ec3.
Therefore,
u(r, θ) = −Er cos θ + Ec3r−2 cos θ.
780
CHAPTER 14 REVIEW EXERCISES
CHAPTER 14 REVIEW EXERCISES
1. We have
A0 =
1
2π
∫ π
0
u0 dθ +
1
2π
∫ 2π
π
(−u0) dθ = 0
An =
1
cnπ
∫ π
0
u0 cosnθ dθ +
1
cnπ
∫ 2π
π
(−u0) cosnθ dθ = 0
Bn =
1
cnπ
∫ π
0
u0 sinnθ dθ +
1
cnπ
∫ 2π
π
(−u0) sinnθ dθ = 2u0
cnnπ
[1− (−1)n]
and so
u(r, θ) =
2u0
π
∞∑
n=1
1− (−1)n
n
(r
c
)n
sinnθ.
2. We have
A0 =
1
2π
∫ π/2
0
dθ +
1
2π
∫ 2π
3π/2
dθ =
1
2
An =
1
cnπ
∫ π/2
0
cosnθ dθ +
1
cnπ
∫ 2π
3π/2
cosnθ dθ =
1
cnnπ
[
sin
nπ
2
− sin 3nπ
2
]
Bn =
1
cnπ
∫ π/2
0
sinnθ dθ +
1
cnπ
∫ 2π
3π/2
sinnθ dθ =
1
cnnπ
[
cos
3nπ
2
− cos nπ
2
]
and so
u(r, θ) =
1
2
+
1
π
∞∑
n=1
(r
c
)n [ sin nπ2 − sin 3nπ2
n
cosnθ +
cos 3nπ2 − cos nπ2
n
sinnθ
]
.
3. The conditions Θ(0) = 0 and Θ(π) = 0 applied to Θ = c1 cosαθ + c2 sinαθ give c1 = 0 and α = n,
n = 1, 2, 3, . . . , respectively. Thus we have the Fourier sine-series coefficients
An =
2
π
∫ π
0
u0(πθ − θ2) sinnθ dθ = 4u0
n3π
[1− (−1)n].
Thus
u(r, θ) =
4u0
π
∞∑
n=1
1− (−1)n
n3
rn sinnθ.
4. In this case
An =
2
π
∫ π
0
sin θ sinnθ dθ =
1
π
∫ π
0
[cos(1− n)θ − cos(1 + n)θ] dθ = 0, n �= 1.
For n = 1,
A1 =
2
π
∫ π
0
sin2 θ dθ =
1
π
∫ π
0
(1− cos 2θ) dθ = 1.
Thus
u(r, θ) =
∞∑
n=1
Anr
n sinnθ
781
CHAPTER 14 REVIEW EXERCISES
reduces to
u(r, θ) = r sin θ.
5. The insulation conditions are uθ(r, 0) = 0 and uθ(r, π) = 0. The eigenvalue λ = 0 corresponds to a constant
eigenfunction. The other eigenvalues are λ = n2, n = 1, 2, 3, . . . , with corresponding eigenfunctions Θ(θ) =
c1 cosnθ. Also, assuming a bounded solution at r = 0, R(r) = c3rn and so
u(r, θ) = A0 +
∞∑
n=1
Anr
n cosnθ.
At r = c
f(θ) = A0 +
∞∑
n=1
Anc
n cosnθ,
from which we see that
A0 =
a0
2
=
1
π
∫ π
0
f(θ) dθ and An =
2
π
∫ π
0
f(θ) cosnθ dθ.
6. Two of the boundary conditions are u(r, 0) = 0 and uθ(r, π) = 0 which imply Θ(0) = 0 and Θ′(π) = 0. For
λ = [(2n − 1)/2]2, n = 1, 2, 3, . . . , we have Θ(θ) = c2 sin[(2n − 1)/2]θ. Assuming a bounded solution at r = 0
we have R(r) = c3rn, so
u(r, θ) =
∞∑
n=1
Anr
n sin
(
2n− 1
2
θ
)
and f(θ) =
∞∑
n=1
Anc
n sin
(
2n− 1
2
θ
)
.
This is not a Fourier series but it is an orthogonal series expansion of f , so
An = c−n

∫ π
0
f(θ) sin
(
2n− 1
2
θ
)
dθ∫ π
0
sin2
(
2n− 1
2
θ
)
dθ
 .
7. We solve
∂2u
∂r2
+
1
r
∂u
∂r
+
1
r2
∂2u
∂θ2
= 0, 0 < θ <
π
4
,
1
2
< r < 1,
u(r, 0) = 0, u(r, π/4) = 0,
1
2
< r < 1,
u(1/2, θ) = u0, ur(1, θ) = 0, 0 < θ <
π
4
.
Proceeding as in Example 1 in Section 14.1 using the separation constant λ = α2 we obtain
r2R′′ + rR′ − λR = 0
Θ′′ + λΘ = 0
with solutions
Θ(θ) = c1 cosαθ + c2 sinαθ
R(r) = c3rα + c4r−α.
Applying the boundary conditions Θ(0) = 0 and Θ(π/4) = 0 gives c1 = 0 and α = 4n for
n = 1, 2, 3, . . . . From Rr(1) = 0 we obtain c3 = c4. Therefore
u(r, θ) =
∞∑
n=1
An
(
r4n + r−4n
)
sin 4nθ.
782
CHAPTER 14 REVIEW EXERCISES
From
u(1/2, θ) = u0 =
∞∑
n=1
An
(
1
24n
+
1
2−4n
)
sin 4nθ
we find
An
(
1
24n
+
1
2−4n
)
=
2
π/4
∫ π/4
0
u0 sin 4nθ dθ =
2u0
nπ
[1− (−1)n]
or
An =
2u0
nπ(24n + 2−4n)
[1− (−1)n].
Thus the steady-state temperature in the plate is
u(r, θ) =
2u0
π
∞∑
n=1
[r4n + r−4n][1− (−1)n]
n[24n + 2−4n]
sin 4nθ.
8. The boundary-value problem in polar coordinates is
∂2u
∂r2
+
1
r
∂u
∂r
+
1
r2
∂2u
∂θ2
= 0, 0 < θ < β, a < r < b,
u(a, θ) = 0, u(b, θ) = f(θ), 0 < θ < β,
u(r, 0) = u0, u(r, β) = u1, a < r < b.
Since the boundary conditions are nonhomogeneous at θ = 0 and θ = β we let u(r, θ) = v(r, θ) + ψ(θ).
Substituting into the partial differential equation we obtain
vrr +
1
r
vr +
1
r2
(vθθ + ψ′′) = 0.
This equation will be homogeneous if we require ψ′′ = 0. In this case ψ(θ) = c1 + c2θ. The boundary conditions
ψ(0) = u0 and ψ(β) = u1 then give c1 = u0 and c2 = (u1 − u0)/β. Hence
ψ(θ) = u0 +
u1 − u0
β
θ.
The homogeneous boundary-value problem for v(r, θ) is then
∂2v
∂r2
+
1
r
∂v
∂r
+
1
r2
∂2v
∂θ2
= 0, 0 < θ < β, a < r < b,
v(a, θ) = −u0 − u1 − u0
β
θ, 0 < θ < β,
v(b, θ) = f(θ)− u0 − u1 − u0
β
θ, 0 < θ < β,
v(r, 0) = 0, v(r, β) = 0, a < r < b.
Letting v(r, θ) = R(r)Θ(θ) and separating variables gives
Θ(θ) = c3 cosαθ + c4 sinαθ and R(r) = c5rα + c6r−α.
The boundary conditions Θ(0) = 0 and Θ(β) = 0 give c3 = 0 and α = nπ/β. Thus, Θ(θ) = c4 sinnπθ/β and
v(r, θ) =
∞∑
n=1
(Anrnπ/β + Bnr−nπ/β) sin
nπ
β
θ.
At r = a we have
−u0 − u1 − u0
β
θ =
∞∑
n=1
(Ananπ/β + Bna−nπ/β) sin
nπ
β
θ,
783
CHAPTER 14 REVIEW EXERCISES
so
Ana
nπ/β + Bna−nπ/β = − 2
β
∫ β
0
(
u0 +
u1 − u0
β
θ
)
sin
nπ
β
θ dθ.
Similarly, at r = b we have
Anb
nπ/β + Bnb−nπ/β =
2
β
∫ β
0
(
f(θ)− u0 − u1 − u0
β
θ
)
sin
nπ
β
θ dθ.
Solving the above two simultaneous equations for An and Bn we get
An =
− 2
βbnπ/β
∫ β
0
(
u0 +
u1 − u0
β
θ
)
sin
nπ
β
θ dθ − 2
βanπ/β
∫ β
0
(
f(θ)− u0 − u1 − u0
β
θ
)
sin
nπ
β
θ dθ(a
b
)nπ/β
−
(
b
a
)nπ/β
and
Bn =
−2a
nπ/β
β
∫ β
0
(
f(θ)− u0 − u1 − u0
β
θ
)
sin
nπ
β
θ dθ +
2bnπ/β
β
∫ β
0
(
u0 +
u1 − u0
β
θ
)
sin
nπ
β
θ dθ(a
b
)nπ/β
−
(
b
a
)nπ/β .
9. The boundary-value problem in polar coordinates is
∂2u
∂r2
+
1
r
∂u
∂r
+
1
r2
∂2u
∂θ2
= 0, 0 < θ < 2π, 1 < r < 2,
u(1, θ) = sin2 θ,
∂u
∂r
∣∣∣∣
r=2
= 0, 0 < θ < 2π.
Letting u(r, θ) = R(r)Θ(θ) and separating variables we obtain
r2R′′ + rR′ − λR = 0 and Θ′′ + λΘ = 0.
For λ = 0 we have Θ′′ = 0 and r2R′′ + rR′ = 0. This gives Θ = c1 + c2θ and R = c3 + c4 ln r. The
periodicity assumption (as mentioned in Example 1 of Section 14.1 in the text) implies c2 = 0, while the
boundary condition R′(2) = 0 implies c4 = 0. Thus, for λ = 0, u = c1c3 = A0. Now, for λ = α2, the
differential equations become Θ′′ + α2Θ = 0 and r2R′′ + rR′ − α2R = 0. The corresponding solutions are
Θ = c5 cosαθ + c6 sinαθ and R = c7rα + c8r−α. In this case the periodicity assumption implies α = n,
n = 1, 2, 3, . . . , while the boundary condition R′(2) = 0 implies R = c7(rn + 4nr−n). The product of the
solutions is un = (rn + 4nr−n)(An cosnθ + Bn sinnθ) and the superposition principle implies
u(r, θ) = A0 +
∞∑
n=1
(rn + 4nr−n)(An cosnθ + Bn sinnθ).
Using the boundary condition at r = 1 we have
sin2 θ =
1
2
(1− cos 2θ) = A0 +
∞∑
n=1
(1 + 4n)(An cosnθ + Bn sinnθ).
From this we conclude that Bn = 0 for all integers n, A0 = 12 , A1 = 0, A2 = − 134 , and Am = 0 for
m = 3, 4, 5, . . . . Therefore
u(r, θ) = A0 + A2 cos 2θ =
1
2
− 1
34
(r2 + 16r−2) cos 2θ =
1
2
−
(
1
34
r2 +
8
17
r−2
)
cos 2θ.
784
CHAPTER 14 REVIEW EXERCISES
10. We solve
∂2u
∂r2
+
1
r
∂u
∂r
+
1
r2
∂2u
∂θ2
= 0, r > 1, 0 < θ < π,
u(r, 0) = 0, u(r, π) = 0, r > 1,
u(1, θ) = f(θ), 0 < θ < π.
Separating variables we obtain
Θ(θ) = c1 cosαθ + c2 sinαθ
R(r)= c3rα + c4r−α.
Applying the boundary conditions Θ(0) = 0, and Θ(π) = 0 gives c1 = 0 and α = n for n = 1, 2, 3, . . . .
Assuming f(θ) to be bounded, we expect the solution u(r, θ) to also be bounded as r →∞. This requires that
c3 = 0. Therefore
u(r, θ) =
∞∑
n=1
Anr
−n sinnθ.
From
u(1, θ) = f(θ) =
∞∑
n=1
An sinnθ
we obtain
An =
2
π
∫ π
0
f(θ) sinnθ dθ.
11. Letting u(r, t) = R(r)T (t) and separating variables we obtain
R′′ + 1rR
′ − hR
R
=
T ′
T
= λ
so
R′′ +
1
r
R′ − (λ + h)R = 0 and T ′ − λT = 0.
From the second equation we find T (t) = c1eλt. If λ > 0, T (t) increases without bound as t → ∞. Thus we
assume λ = −α2 < 0. Since h > 0 we can take µ = −α2 − h. Then
R′′ +
1
r
R′ + α2R = 0
is a parametric Bessel equation with solution
R(r) = c1J0(αr) + c2Y0(αr).
Since Y0 is unbounded as r → 0 we take c2 = 0. Then R(r) = c1J0(αr) and the boundary condition u(1, t) =
R(1)T (t) = 0 implies J0(α) = 0. This latter equation defines the positive eigenvalues λn. Thus
u(r, t) =
∞∑
n=1
AnJ0(αnr)e(−α
2
n−h)t.
From
u(r, 0) = 1 =
∞∑
n=1
AnJ0(αnr)
we find
An =
2
J21 (αn)
∫ 1
0
rJ0(αnr) dr x = αnr, dx = αn dr
=
2
J21 (αn)
∫ αn
0
1
α2n
xJ0(x) dx.
785
CHAPTER 14 REVIEW EXERCISES
From recurrence relation (5) in Section 12.6 of the text we have
xJ0(x) =
d
dx
[xJ1(x)].
Then
An =
2
α2nJ
2
1 (αn)
∫ αn
0
d
dx
[xJ1(x)] dx =
2
α2nJ
2
1 (αn)
(
xJ1(x)
) ∣∣∣∣αn
0
=
2α1J1(αn)
α2nJ
2
1 (αn)
=
2
αnJ1(αn)
and
u(r, t) = 2e−ht
∞∑
n=1
J0(αnr)
αnJ1(αn)
e−α
2
nt
12. Letting λ = α2 > 0 and proceeding in the usual manner we find
u(r, t) =
∞∑
n=1
An cos aαntJ0(αnr)
where the eigenvalues λn = α2n are determined by J0(α) = 0. Then the initial condition gives
u0J0(xkr) =
∞∑
n=1
AnJ0(αnr)
and so
An =
2
J21 (αn)
∫ 1
0
r (u0J0(xkr))J0(αnr) dr.
But J0(α) = 0 implies that the eigenvalues are the positive zeros of J0, that is, αn = xn for n = 1, 2, 3, . . . .
Therefore
An =
2u0
J21 (αn)
∫ 1
0
rJ0(αkr)J0(αnr) dr = 0, n �= k
by orthogonality. For n = k,
Ak =
2u0
J21 (αk)
∫ 1
0
rJ20 (αk) dr = u0
by (7) of Section 12.6. Thus the solution u(r, t) reduces to one term when n = k, and
u(r, t) = u0 cos aαktJ0(αkr) = u0 cos axktJ0(xkr).
13. Letting the separation constant be λ = α2 and referring to Example 2 in Section 14.2 in the text we have
R(r) = c1J0(αr) + c2Y0(αr)
Z(z) = c3 coshαz + c4 sinhαz
where c2 = 0 and the positive eigenvalues λn are determined by J0(2α) = 0. From Z ′(0) = 0 we obtain c4 = 0.
Then
u(r, z) =
∞∑
n=1
An coshαnzJ0(αnr).
From
u(r, 4) = 50 =
∞∑
n=1
An cosh 4αnJ0(αnr)
we obtain (as in Example 1 of Section 14.1)
An cosh 4αn =
2(50)
4J21 (2αn)
∫ 2
0
rJ0(αnr) dr =
50
αnJ1(2αn)
.
786
CHAPTER 14 REVIEW EXERCISES
Thus the temperature in the cylinder is
u(r, z) = 50
∞∑
n=1
coshαnzJ0(αnr)
αn cosh 4αnJ1(2αn)
.
14. Using u = RZ and −λ as a separation constant and then letting λ = α2 > 0 leads to
r2R′′ + rR′ + α2r2R = 0, R′(1) = 0, and Z ′′ − α2Z = 0.
Thus
R(r) = c1J0(αr) + c2Y0(αr)
Z(z) = c3 coshαz + c4 sinhαz
for α > 0. Arguing that u(r, z) is bounded as r → 0 we define c2 = 0. Since the eigenvalues are defined by
J ′0(α) = 0 we know that λ = α = 0 is an eigenvalue. The solutions are then
R(r) = c1 + c2 ln r and Z(z) = c3z + c4
where boundedness again dictates that c2 = 0. Thus,
u(r, z) = A0z + B0 +
∞∑
n=1
(An sinhαnz + Bn coshαnz)J0(αnr).
Finally, the specified conditions z = 0 and z = 1 give, in turn,
B0 = 2
∫ 1
0
rf(r) dr
Bn =
2
J20 (αn)
∫ 1
0
rf(r)J0(αnr) dr
A0 = −B0 + 2
∫ 1
0
rg(r) dr
An =
1
sinhαn
[
−Bn coshαn + 2
J20 (αn)
∫ 1
0
rg(r)J0(αnr) dr
]
.
15. Referring to Example 1 in Section 14.3 of the text we have
u(r, θ) =
∞∑
n=0
Anr
nPn(cos θ).
For x = cos θ
u(1, θ) =
{
100 0 < θ < π/2
−100 π/2 < θ < π = 100
{−1, −1 < x < 0
1, 0 < x < 1
= g(x).
From Problem 22 in Exercise 12.6 we have
u(r, θ) = 100
[
3
2
rP1(cos θ)− 78r
3P3(cos θ) +
11
16
r5P5(cos θ) + · · ·
]
.
16. Since
1
r
∂2
∂r2
(ru) =
1
r
∂
∂r
[
r
∂u
∂r
+ u
]
=
1
r
[
r
∂2u
∂r2
+
∂u
∂r
+
∂u
∂r
]
=
∂2u
∂r2
+
2
r
∂u
r
the differential equation becomes
1
r
∂2
∂r2
(ru) =
∂2u
∂t2
or
∂2
∂r2
(ru) = r
∂2u
∂t2
.
787
CHAPTER 14 REVIEW EXERCISES
Letting v(r, t) = ru(r, t) we obtain the boundary-value problem
∂2v
∂r2
=
∂2v
∂t2
, 0 < r < 1, t > 0
∂v
∂r
∣∣∣∣
r=1
− v(1, t) = 0, t > 0
v(r, 0) = rf(r),
∂v
∂t
∣∣∣∣
t=0
= rg(r), 0 < r < 1.
If we separate variables using v(r, t) = R(r)T (t) and separation constant −λ then we obtain
R′′
R
=
T ′′
T
= −λ
so that
R′′ + λR = 0
T ′′ + λT = 0.
Letting λ = α2 > 0 and solving the differential equations we get
R(r) = c1 cosαr + c2 sinαr
T (t) = c3 cosαt + c4 sinαt.
Since u(r, t) = v(r, t)/r, in order to insure boundedness at r = 0 we define c1 = 0. Then R(r) = c2 sinαr. Now
the boundary condition R′(1)−R(1) = 0 implies α cosα− sinα = 0. Thus, the eigenvalues λn are determined
by the positive solutions of tanα = α. We now have
vn(r, t) = (An cosαnt + Bn sinαnt) sinαnr.
For the eigenvalue λ = 0,
R(r) = c1r + c2 and T (t) = c3t + c4,
and boundedness at r = 0 implies c2 = 0. We then take
v0(r, t) = A0tr + B0r
so that
v(r, t) = A0tr + B0r +
∞∑
n=1
(an cosαnt + Bn sinαnt) sinαnr.
Now
v(r, 0) = rf(r) = B0r +
∞∑
n=1
An sinαnr.
Since {r, sinαnr} is an orthogonal set on [0, 1],∫ 1
0
r sinαnr dr = 0 and
∫ 1
0
sinαnr sinαnr dr = 0
for m �= n. Therefore ∫ 1
0
r2f(r) dr = B0
∫ 1
0
r2 dr =
1
3
B0
and
B0 = 3
∫ 1
0
r2f(r) dr.
788
CHAPTER 14 REVIEW EXERCISES
Also ∫ 1
0
rf(r) sinαnr dr = An
∫ 1
0
sin2 αnr dr
and
An =
∫ 1
0
rf(r) sinαnr dr∫ 1
0
sin2 αnr dr
.
Now ∫ 1
0
sin2 αnr dr =
1
2
∫ 1
0
(1− cos 2αnr) dr = 12
[
1− sin 2αn
2αn
]
=
1
2
[1− cos2 αn].
Since tanαn = αn,
1 + α2n = 1 + tan
2 αn = sec2 αn =
1
cos2 αn
and
cos2 αn =
1
1 + α2n
.
Then ∫ 1
0
sin2 αnr dr =
1
2
[
1− 1
1 + α2n
]
=
α2n
2(1 + α2n)
and
An =
2(1 + α2n)
α2n
∫ 1
0
rf(r) sinαnr dr.
Similarly, setting
∂v
∂t
∣∣∣∣
t=0
= rg(r) = A0r +
∞∑
n=1
Bnαn sinαnr
we obtain
A0 = 3
∫ 1
0
r2g(r) dr
and
Bn =
2(1 + α2n)
α3n
∫ 1
0
rg(r) sinαnr dr.
Therefore, since v(r, t) = ru(r, t) we have
u(r, t) = A0t + B0 +
∞∑
n=1
(An cosαnt + Bn sinαnt)
sinαnr
r
,
where the αn are solutions of tanα = α and
A0 = 3
∫ 1
0
r2g(r) dr
B0 = 3
∫ 1
0
r2f(r) dr
An =
2(1 + α2n)
α2n
∫ 1
0
rf(r) sinαnr dr
Bn =
2(1 + α2n)
α3n
∫ 1
0
rg(r) sinαnr dr
for n = 1, 2, 3, . . . .
789
CHAPTER 14 REVIEW EXERCISES
17. We note that the differential equation can be expressed in the form
d
dx
[xu′] = −α2xu.
Thus
un
d
dx
[xu′m] = −α2mxumun
and
um
d
dx
[xu′n] = −α2nxunum.
Subtracting we obtain
un
d
dx
[xu′m]− um
d
dx
[xu′n] = (α
2
n − α2m)xumun
and ∫ b
a
un
d
dx
[xu′m] dx−
∫ b
a
um
d
dx
[xu′n] = (α
2
n − α2m)
∫ b
a
xumun dx.
Using integration by parts this becomes
unxu
′
m
∣∣∣b
a
−
∫ b
a
xu′mu
′
n dx− umxu′n
∣∣∣b
a
+
∫ b
a
xu′nu
′
m dx
= b[un(b)u′m(b)− um(b)u′n(b)]− a[un(a)u′m(a)− um(a)u′n(a)]
= (α2n − α2m)
∫ b
a
xumun dx.
Since
u(x) = Y0(αa)J0(αx)− J0(αa)Y0(αx)
we have
un(b) = Y0(αna)J0(αnb)− J0(αna)Y0(αnb) = 0
by the definition of the αn. Similarlyum(b) = 0. Also
un(a) = Y0(αa)J0(αna)− J0(αna)Y0(αna) = 0
and um(a) = 0. Therefore∫ b
a
xumun dx =
1
α2n − α2m
(b[un(b)u′m(b)− um(b)u′n(b)]− a[un(a)u′m(a)− um(a)u′n(a)]) = 0
and the un(x) are orthogonal with respect to the weight function x.
18. Letting u(r, t) = R(r)T (t) and the separation constant be −λ = −α2 we obtain
rR′′ + R′ + α2rR = 0
T ′ + α2T = 0,
with solutions
R(r) = c1J0(αr) + c2Y0(αr)
T (t) = c3e−α
2t.
Now the boundary conditions imply
R(a) = 0 = c1J0(αa) + c2Y0(αa)
R(b) = 0 = c1J0(αb) + c2Y0(αb)
790
CHAPTER 14 REVIEW EXERCISES
so that
c2 = −c1J0(αa)
Y0(αa)
and
c1J0(αb)− c1J0(αa)
Y0(αa)
Y0(αb) = 0
or
Y0(αa)J0(αb)− J0(αa)Y0(αb) = 0.
This equation defines αn for n = 1, 2, 3, . . . . Now
R(r) = c1J0(αr)− c1 J0(αa)
Y0(αa)
Y0(αr) =
c1
Y0(αa)
[
Y0(αa)J0(αr)− J0(αa)Y0(αr)
]
and
un(r, t) = An
[
Y0(αna)J0(αnr)− J0(αna)Y0(αnr)
]
e−α
2
nt = Anun(r)e−α
2
nt.
Thus
u(r, t) =
∞∑
n=1
Anun(r)e−α
2
nt.
From the initial condition
u(r, 0) = f(r) =
∞∑
n=1
Anun(r)
we obtain
An =
∫ b
a
rf(r)un(r) dr∫ b
a
ru2n(r) dr
.
19. We use the superposition principle for Laplace’s equation discussed in Section 13.5 and shown schematically in
Figure 13.15 in the text. That is,
Solution u = Solution u1 of Problem 1 + Solution u2 of Problem 2,
where in Problem 1 the boundary condition on the top and bottom of the cylinder is u = 0, while on the lateral
surface r = c it is u = h(z), and in Problem 2 the boundary condition on the top of the cylinder z = L is
u = f(r), on the bottom z = 0 it is u = g(r), and on the lateral surface r = c it is u = 0.
20. Solution for u1(r, z)
Using λ as a separation constant we have
R′′ + 1rR
′
R
= −Z
′′
Z
= λ,
so
rR′′ + R′ − λrR = 0 and Z ′′ + λZ = 0.
The differential equation in Z, together with the boundary conditions Z(0) = 0 and Z(L) = 0 is a Sturm-
Liouville problem. Letting λ = α2 > 0 we note that the above differential equation in R is a modified parametric
Bessel equation which is discussed in Section 5.3 in the text. Also, we have Z(z) = c1 cosαz + c2 sinαz. The
boundary conditions imply c1 = 0 and sinαL = 0. Thus, αn = nπ/L, n = 1, 2, 3, . . . , so λn = n2π2/L2 and
R(r) = c3I0
(nπ
L
r
)
+ c4K0
(nπ
L
r
)
.
791
CHAPTER 14 REVIEW EXERCISES
Now boundedness at r = 0 implies c4 = 0, so R(r) = c3I0(nπr/L) and
u1(r, z) =
∞∑
n=1
AnI0
(nπ
L
r
)
sin
(nπ
L
z
)
.
At r = c for 0 < z < L we have
h(z) = u1(c, z) =
∞∑
n=1
AnI0
(nπ
L
c
)
sin
(nπ
L
z
)
which gives
An =
2
LI0(nπc/L)
∫ L
0
h(z) sin
(nπ
L
z
)
dz.
Solution for u2(r, z)
In this case we use −λ as a separation constant which leads to
R′′ + 1rR
′
R
= −Z
′′
Z
= −λ,
so
rR′′ + R′ + λrR = 0 and Z ′′ − λZ = 0.
The differential equation in R is a parametric Bessel equation. Using λ = α2 we find R(r) = c1J0(αr)+c2Y0(αr).
Boundedness at r = 0 implies c2 = 0 so R(r) = c3J0(αr). The boundary condition R(c) = 0 then gives the
defining equation for the eigenvalues: J0(αc) = 0. Let λn = α2n where αnc = xn are the roots. The solution of
the differential equation in Z is Z(z) = c4 coshαnz + c5 sinhαnz, so
u2(r, z) =
∞∑
n=1
(Bn coshαnz + Cn sinhαnz)J0(αnr).
At z = 0, for 0 < r < c, we have
f(r) = u2(r, 0) =
∞∑
n=1
BnJ0(αnr),
so
Bn =
2
c2J21 (αnc)
∫ c
0
rf(r)J0(αnr) dr.
At z = L, for 0 < r < c, we have
g(r) = u2(r, L) =
∞∑
n=1
(Bn coshαnL + Cn sinhαnL)J0(αnr),
so
Bn coshαnL + Cn sinhαnL =
2
c2J21 (αnc)
∫ c
0
rg(r)J0(αnr) dr
and
Cn = −Bn coshαnLsinhαnL +
2
c2(sinhαnL)J21 (αnc)
∫ c
0
rg(r)J0(αnr) dr.
By the superposition principle the solution of the original problem is
u(r, z) = u1(r, z) + u2(r, z).
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