Math 241
Some solutions for Homework 4
7.9.3a. Separating u(r, θ, z, t) = b(r)q(θ)g(z)h(t) gives
r2 b00 + rb0 + (µr2 − ν)b = 0 b(a) = 0, |b(0)| < ∞
q 00 + νq = 0
q(0) = q( π2 ) = 0
g 00 + (λ − µ)g
g(0) = g(H) = 0
h0 + λkh = 0
Solving for q gives q(θ) = sin(2mθ), with ν = (2m)2 . Then solving for b gives b(r) =
2
√
z
J2m ( µr); using the boundary condition b(a) = 0 gives µmn = (2m)n
, where z(2m)n is the
a
2
pπ
nth positive zero of J2m . Solving for g gives g(z) = sin( H z), with λ − µ = pπ
. Then we
H
z(2m)n 2
pπ 2
−kλ
t
+ H , and h(t) = e mnp .
have λmnp =
a
So the general solution is
∞ X
∞ X
∞
X
m=1 n=1 p=1
pπ
√
Amnp J2m ( µmn r) sin(2mθ) sin( z)e−kλmnp t .
H
The initial conditions give
f (r, θ, z) =
∞ X
∞ X
∞
X
m=1 n=1 p=1
so
RR
Amnp =
pπ
√
Amnp J2m ( µmn r) sin(2mθ) sin( z),
H
√
f (r, θ, z)rJ2m ( µmn r) sin(2mθ) sin( pπ
H z) dr dθ dz
.
RR
√
pπ 2
J2m ( µmn r) sin(2mθ) sin( H z) r dr dθ dz
7.10.2c. Separating u(ρ, θ, φ, t) = w(ρ, θ, φ)h(t) gives
∆w + λw = 0 w(a, θ, φ) = 0
h0 + kλh = 0.
We solved for w in class; we get product solutions
p
cos mθ
−1/2
w=ρ
Jn+ 1 ( λnp ρ)
Pnm (cos φ),
2
sin mθ
where n ≥ m. Here both cos mθ and sin mθ will give us product solutions, and λnp =
z 1 2
(n+ 2 )p
, where z(n+ 1 )p is the pth positive zero of Jn+ 1 .
a
2
Solving for h gives h(t) =
2
e−kλnp .
So
u=
∞ X
∞
∞ X
X
m=0 n=m p=1
ρ−1/2 Jn+ 1 (
2
p
λnp ρ)(Amnp cos mθ + Bmnp sin mθ)Pnm (cos φ)e−kλnp .
If you want, omit the sin mθ term when m = 0, but it doesn’t matter because the initial
conditions are
∞ X
∞ X
∞
X
p
u(ρ, θ, φ, t) =
ρ−1/2 Jn+ 1 ( λnp ρ)(Amnp cos mθ+Bmnp sin mθ)Pnm (cos φ) = F (ρ, φ) cos θ,
2
m=0 n=m p=1
so only the cos θ terms will remain anyway. So Amnp = 0 if m 6= 0, and Bmnp = 0 always. So
our solution becomes
∞ X
∞
X
p
A1np ρ−1/2 Jn+ 1 ( λnp ρ) cos(θ)Pn1 (cos φ)e−kλnp ,
2
n=1 p=1
and
RRR
A1np =
p
F (ρ, φ)ρ−1/2 Jn+ 1 ( λnp ρ) cos(θ)Pn1 (cos φ) cos θρ2 sin φdρ dφ dθ
2
.
2
p
RRR −1/2
1
2
ρ
Jn+ 1 ( λnp ρ) cos(θ)Pn (cos φ) ρ sin φdρ dφ dθ
2
7.10.10b. (I didn’t notice this at the time, but you need to assume that u is bounded when ρ goes to
∞, and the problem doesn’t explicitly say so.)
We want to solve ∆u = 0. Since the boundary conditions are symmetric about the z-axis (i.e.
the axis of the earth), we expect u to be too. So u shouldn’t depend on θ. So we separate
u(ρ, φ) = f (ρ)g(φ), and get
d
2 f 0 − µf = 0
ρ
|f (∞)| < ∞
dρ dg
d
dφ sin φ dφ + µ sin(φ)g = 0 |g(0)| < ∞, |g(π)| < ∞
(Here I’m using µ. Otherwise I would’ve used λ.) Note that there are no homogeneous
boundary conditions explicitly spelled out in the problem statement; we needed to come up
with the boundedness on our own.
Anyway, the solution to the φ equation is Pn0 (cos φ) and µ = n(n + 1), where n ≥ 0. Plugging
this into the ρ equation gives
d
ρ2 f 0 − n(n + 1)f = 0.
dρ
This is a Cauchy-Euler equation whose solution is f (ρ) = c1 ρn + c2 ρ−n−1 . We don’t want f
to blow up when ρ → ∞, so we set c1 = 0. So our product solutions will be Pn0 (cos φ)ρ−n−1 ,
and the general solution is
u(ρ, φ) =
∞
X
An Pn0 (cos φ)ρ−n−1 .
n=0
To find An , we use the BCs:
u(a, φ) =
∞
X
An a−n−1 Pn0 (cos φ) = F (φ).
n=0
So we have
An a
−n−1
Rπ
F (φ)P 0 (cos φ) sin φ dφ
= 0R π 0 n
2
0 Pn (cos φ) sin φ dφ
8.3.1c. For a reference temperature, we use r(x, t) = A(t). Plugging u(x, t) = v(x, t) + A(t) into the
PDE gives
∂
∂2
(v + A(t)) = k 2 (v + A(t)) + Q(x, t),
∂t
∂x
which we rearrange to
vt = kvxx + Q̄(x, t),
where Q̄ = Q − A0 (t). For BCs and ICs, we have
v(0, t) = 0
vx (L, t) = 0
v(x, 0) = f (x) − A(0).
The homogeneous problem is vt = kvxx ; the corresponding eigenvalue problem is φ00 + λφ = 0,
2
(2n−1)π
with φ(0) = φ0 (L) = 0. Then we have φn = sin (2n−1)π
. We expand
x
,
and
λ
=
n
L
L
v in terms of the eigenfunctions:
∞
X
v(x, t) =
an (t)φn (x),
n=1
and while we’re at it, let’s expand Q̄:
Q̄(x, t) =
∞
X
qn (t)φn (x),
n=1
where
RL
qn (t) =
We have
vt =
0
Q(x, t)φn (x) dx
.
RL
2
0 φn (x) dx
∞
X
a0n (t)φn (x)
n=1
and
vxx =
∞
X
an (t)φ00n (x)
n=1
=−
∞
X
λn an (t)φn (x).
n=1
Expanding the equation vt − kvxx = Q̄ gives
∞
X
(a0n (t) + kλn an (t))φn (x) =
n=1
∞
X
qn (t)φn (x).
n=1
0 + kλ a = q ; solving (by integrating factors, for example), gives a (t) = e−kλt (a (0) +
So
n n
n
n
n
n
R t akλτ
e
q
(τ
)
dτ
).
We
solve
for
a
(0)
with
the
initial
conditions
n
n
0
v(x, 0) =
∞
X
n=1
an (0)φn (x) = f (x) − A(0),
so
RL
0
an (0) =
(f (x) − A(0))φn (x) dx
.
RL
2 dx
φ
(x)
n
0
Then u(x, t) = v(x, t) + A(t). So,
u(x, t) =
∞
X
an (t)φn (x) + A(t),
n=1
where an (t) and φn (x) are given above.
8.3.5. We use r(x, t) = 1 − πx . Then vt = vxx + e−2t sin 5x, v(0, t) = v(π, t) = 0, and v(x, 0) = πx − 1.
The corresponding eigenvalue problem is φ00 + λφ = 0, φ(0) = φ(π) = 0. So we have
φn (x) = sin(nx) with λn = n2 .
Expand v into eigenfunctions:
∞
X
v(x, t) =
an (t)φn (x)
n=1
and Q(x, t) = e−2t sin 5x:
−2t
e
sin 5x =
∞
X
qn (t) sin(nx).
n=1
We can see directly that q5 (t) = e−2t , and qn (t) = 0 for n 6= 5. As in the previous problem,
we get
a0n + λn an = qn (t).
For n = 5, we have
a05 (t) + 25a5 (t) = e−2t ;
we can solve by an integrating factor or the method of undetermined coefficients:
a5 (t) = a5 (0)e−25t +
1 −2t
(e
− e−25t ).
23
For n 6= 5, we just get
a0n + n2 an = 0,
2
so an (t) = an (0)e−n t . For initial conditions, we use
v(x, 0) =
∞
X
an (0) sin(nx) =
n=1
so an (0) =
2
π
Rπ
0
sin(nx)(1 − πx ) dx =
v(x, t) =
=
2
nπ .
x
− 1,
π
Then,
∞
X
an (t)φn (x)
n=1
∞
X
2 −n2 t
n=1
nπ
e
sin(nx)
!
+
1 −2t
(e
− e−25t ),
23
and
u(x, t) = v(x, t) + 1 −
=
x
π
∞
X
2 −n2 t
e
sin(nx)
nπ
!
+
n=1
1 −2t
x
(e
− e−25t ) + 1 − ,
23
π
RR
8.6.1d. The total flux in is 0, and the total heat from sourcesRRis
Q(x, y) dx dy. For there to be an
equilibrium, we need no change in the total heat, so
Q(x, y) dx dy = 0.
Let’s expand u in terms of the eigenfunctions of φ00 + λφ = 0, where φ has the same BCs as
u. Separating φ(x, y) = f (x)g(y) gives
f 00 + µf = 0 f 0 (0) = f 0 (L) = 0
g 00 + (λ − µ)g g 0 (0) = g 0 (H) = 0.
mπ 2
nπ 2
, and g(y) = cos( nπ
.
So we have f (x) = cos( mπ
L x) and µ =
L
H y) and λ − µ =
H
mπ
nπ
Here m = 0, 1, 2, . . . and n = 0, 1, 2, . . .. Then φmn = cos( L x) cos( H y), with λmn =
2
mπ 2
+ nπ
.
L
H
Then
∞ X
∞
X
u(x, t) =
Amn cos(
m=0 n=0
nπ
mπ
x) cos( y).
L
H
Setting ∆u = Q(x, y), we have
∞ X
∞
X
(−λmn )Amn cos(
m=0 n=0
If we let
Q(x, y) =
∞ X
∞
X
m=0 n=0
with
mπ
nπ
x) cos( y) = Q(x, y).
L
H
qmn cos(
nπ
mπ
x) cos( y),
L
H
nπ
Q(x, y) cos( mπ
L x) cos( H y) dx dy
,
2
RR
nπ
dx dy
cos( mπ
L x) cos( H y)
RR
qmn =
then −λmn Amn = qmn . Note that λ00 = 0, so we need q00 = 0. But if either m or n is
nonzero, then λmn 6= 0, so we have Amn = − λqmn
. Note that A00 can be anything.
mn