11Laplace's and Poisson's equations
IA Vector Calculus
11.1 Uniqueness theorems
Theorem.
Consider
∇
2
ϕ
=
−ρ
for some
ρ
(
r
) on a bounded volume
V
with
S = ∂V being a closed surface, with an outward normal n.
Suppose ϕ satisfies either
(i) Dirichlet condition, ϕ(r) = f(r) on S
(ii) Neumann condition
∂ϕ(r)
∂n
= n · ∇ϕ = g(r) on S.
where f, g are given. Then
(i) ϕ(r) is unique
(ii) ϕ(r) is unique up to a constant.
This theorem is practically important - if you find a solution by any magical
means, you know it is the only solution (up to a constant).
Since the proof of the cases of the two different boundary conditions are very
similar, they will be proved together. When the proof is broken down into (i)
and (ii), it refers to the specific cases of each boundary condition.
Proof.
Let
ϕ
1
(
r
) and
ϕ
2
(
r
) satisfy Poisson’s equation, each obeying the boundary
conditions (N) or (D). Then Ψ(
r
) =
ϕ
2
(
r
)
− ϕ
1
(
r
) satisfies
∇
2
Ψ = 0 on
V
by
linearity, and
(i) Ψ = 0 on S; or
(ii)
∂Ψ
∂n
= 0 on S.
Combining these two together, we know that Ψ
∂Ψ
∂n
= 0 on the surface. So using
the divergence theorem,
Z
V
∇ · (Ψ∇Ψ) dV =
Z
S
(Ψ∇Ψ) · dS = 0.
But
∇ · (Ψ∇Ψ) = (∇Ψ) · (∇Ψ) + Ψ ∇
2
Ψ
|{z}
=0
= |(∇Ψ)|
2
.
So
Z
V
|∇Ψ|
2
dV = 0.
Since
|∇
Ψ
|
2
≥
0, the integral can only vanish if
|∇
Ψ
|
= 0. So
∇
Ψ = 0. So Ψ =
c
,
a constant on V . So
(i) Ψ = 0 on S ⇒ c = 0. So ϕ
1
= ϕ
2
on V .
(ii) ϕ
2
(r) = ϕ
1
(r) + C, as claimed.
We’ve proven uniqueness. How about existence? It turns out it isn’t difficult
to craft a boundary condition in which there are no solutions.
For example, if we have
∇
2
ϕ
=
−ρ
on
V
with the condition
∂ϕ
∂n
=
g
, then by
the divergence theorem,
Z
V
∇
2
ϕ dV =
Z
∂S
∂ϕ
∂n
dS.
Using Poisson’s equation and the boundary conditions, we have
Z
V
ρ dV +
Z
∂V
g dS = 0
So if ρ and g don’t satisfy this equation, then we can’t have any solutions.
The theorem can be similarly proved and stated for regions in
R
2
, R
3
, ···
, by
using the definitions of grad, div and the divergence theorem. The result also
extends to unbounded domains. To prove it, we can take a sphere of radius
R
and impose the boundary conditions
|
Ψ(
r
)
|
=
O
(1
/R
) or
|
∂Ψ
∂n
(
r
)
|
=
O
(1
/R
2
) as
R → ∞. Then we just take the relevant limits to complete the proof.
Similar results also apply to related equations and different kinds of boundary
conditions, eg
D
or
N
on different parts of the boundary. But we have to analyse
these case by case and see if the proof still applies.
The proof uses a special case of the result
Proposition (Green’s first identity).
Z
S
(u∇v) · dS =
Z
V
(∇u) · (∇v) dV +
Z
V
u∇
2
v dV,
By swapping u and v around and subtracting the equations, we have
Proposition (Green’s second identity).
Z
S
(u∇v −v∇u) · dS =
Z
V
(u∇
2
v −v∇
2
u) dV.
These are sometimes useful, but can be easily deduced from the divergence
theorem when needed.