Example in Detail:
Solve Laplace’s equation on an annulus, where the radius is a ≤ r ≤ b, and the angle is the
full set, −π < θ < π, and the boundary conditions are:
ur (a, θ) = 0
u(b, θ) = g(θ)
along with the usual periodic boundary conditions. Laplace’s equation in polar coordinates
is:
∂u
1 ∂ 2u
1 ∂
r
+ 2 2 =0
r ∂r
∂r
r ∂θ
SOLUTION: First, we note how the boundary conditions will effect the solution to the ODEs
we get after separation of variables. If u = RT , then
u(r, −π) = u(r, π)
and
uθ (r, −π) = uθ (r, π)
are periodic conditions on T : T (π) = T (−π) and T 0 (−π) = T 0 (π). The other ones:
ur (a, θ) = 0
⇒
R0 (a) = 0
And u(b, θ) = g(θ) basically just implies that R(b) should not be zero.
Now we apply the separation of variables, and we’ll solve for T first:
⇒
u = RT
Multiply both sides by
r2
RT
1
1 ∂
(rR0 T ) + 2 RT 00 = 0
r ∂r
r
to get:
r(rR0 )0 T 00
+
=0
R
T
(Recall that the prime notation for R will denote the derivative in terms of r, and the prime
notation for T denotes derivative in terms of θ). Expand a bit and separate variables:
T 00
r(rR00 + R0 )
=−
=λ
R
T
so that
r2 R00 + rR0 − λR = 0
T 00 + λT = 0
Let’s go ahead and think about the case where λ = 0. It might be easier to go back to a
previous form,
r(rR0 )0 = 0
T 00 = 0
For the DE in R,
(rR0 )0 = 0
⇒
rR0 = C1
⇒
R0 =
1
C1
r
⇒
R = C1 ln(r) + C2
For T , we get T (θ) = C3 θ + C4 . Now, considering the boundary conditions,
⇒
T (−π) = T (π)
C3 (−π) + C4 = C3 (π) + C4
⇒
C3 = 0
That also settles the derivatives, since T 0 (θ) = 0.
Considering R(r), we have R0 (a) = 0, which implies that
R0 (a) =
C1
=0
a
⇒
C1 = 0
SUMMARY: R0 T0 is a constant.
We’ll skip the case where λ < 0, but you should be able to provide the details. For the
last case where λ > 0, we have:
√
√
T 00 + λT = 0 ⇒ T (θ) = A cos( λθ) + B sin( λθ)
With T (−π) = T (π), we get the following, which is simplified using the fact that sine is odd
and cosine is even:
√
√
√
√
√
A cos( λ(−π)) + B sin( λ(−π)) = A cos( λπ) + B sin( λπ) ⇒ 2B sin( λπ) = 0
Similarly, using T 0 (−π) = T 0 (π), we get
√
√
−2A λ sin( λπ) = 0
Therefore, we solve both equations if
√
λπ = nπ, for n = 1, 2, 3, · · ·
⇒
λ = n2 for n = 1, 2, 3, · · ·
Therefore, Tn = An cos(nθ) + Bn sin(nθ). Solving for R, we get
r2 R00 + rR0 − n2 R = 0
⇒
r = ±n
Therefore, R(r) = Crn + Dr−n . With R0 (a) = 0, we have:
n−1
nCa
−nDa
−n−1
⇒
=0
n−1
nCa
nD
− n+1 = 0
a
nCa2n − nD
=0
an+1
⇒
⇒
Ca2n = D
We can simplify R a bit:
Rn = C(rn + a2n r−n )
Now we can write the full solution:
u(r, θ) = A0 +
∞
X
2n −n
n
An (r + a r
) cos(nθ) +
n=1
∞
X
Bn (rn + a2n r−n ) sin(nθ)
n=1
with
g(θ) = u(b, θ) = A0 +
∞
X
n
2n −n
An (b + a b
n=1
) cos(nθ) +
∞
X
n=1
2
Bn (bn + a2n b−n ) sin(nθ)
And
and
1
A0 =
2π
Z
π
g(θ) dθ
−π
π
1
An =
n
n(b + a2n b−n
Z
1
Bn =
n
n(b + a2n b−n
Z
3
g(θ) cos(nθ) dθ
−π
π
g(θ) sin(nθ) dθ
−π