4Partial differential equations
IB Methods
4.2 Laplace’s equation in the unit disk in R
2
Let Ω = {(x, y) ∈ R
2
: x
2
+ y
2
≤ 1}. Then Laplace’s equation becomes
0 = ∇
2
φ =
∂
2
φ
∂x
2
+
∂
2
φ
∂y
2
=
∂
2
φ
∂z∂¯z
,
where we define z = x + iy, ¯z = x −iy. Then the general solution is
φ(z, ¯z) = ψ(z) + χ(¯z)
for some ψ, χ.
Suppose we want a solution obeying the Dirichlet boundary condition
φ
(
z, ¯z
)

∂Ω
=
f
(
θ
), where the boundary
∂
Ω is the unit circle. Since the do
main is the unit circle S
1
, we can Fourierexpand it! We can write
f(θ) =
X
n∈Z
ˆ
f
n
e
inθ
=
ˆ
f
0
+
∞
X
n=1
ˆ
f
n
e
inθ
+
∞
X
n=1
ˆ
f
−n
e
−inθ
.
However, we know that
z
=
re
iθ
and
¯z
=
re
iθ
. So on the boundary, we know
that z = e
iθ
and ¯z = e
−iθ
. So we can write
f(θ) =
ˆ
f
0
+
∞
X
n=1
(
ˆ
f
n
z
n
+
ˆ
f
−n
¯z
n
)
This is defined for z = 1. Now we can define φ on the unit disk by
φ(z, ¯z) =
ˆ
f
0
+
∞
X
n=1
ˆ
f
n
z
n
+
∞
X
n=1
ˆ
f
−n
¯z
n
.
It is clear that this obeys the boundary conditions by construction. Also,
φ
(
z, ¯z
)
certainly converges when
z <
1 if the series for
f
(
θ
) converged on
∂
Ω. Also,
φ
(
z, ¯z
) is of the form
ψ
(
z
) +
χ
(
¯z
). So it is a solution to Laplace’s equation on
the unit disk. Since the solution is unique, we’re done!