11Laplace's and Poisson's equations
IA Vector Calculus
11.2 Laplace’s equation and harmonic functions
Definition
(Harmonic function)
.
A harmonic function is a solution to Laplace’s
equation ∇
2
ϕ = 0.
These have some very special properties.
11.2.1 The mean value property
Proposition
(Mean value property)
.
Suppose
ϕ
(
r
) is harmonic on region
V
containing a solid sphere defined by
|r−a| ≤ R
, with boundary
S
R
=
|r−a|
=
R
,
for some R. Define
¯ϕ(R) =
1
4πR
2
Z
S
R
ϕ(r) dS.
Then ϕ(a) = ¯ϕ(R).
In words, this says that the value at the center of a sphere is the average of
the values on the surface on the sphere.
Proof.
Note that
¯ϕ
(
R
)
→ ϕ
(
a
) as
R →
0. We take spherical coordinates (
u, θ, χ
)
centered on r = a. The scalar element (when u = R) on S
R
is
dS = R
2
sin θ dθ dχ.
So
dS
R
2
is independent of R. Write
¯ϕ(R) =
1
4π
Z
ϕ
dS
R
2
.
Differentiate this with respect to
R
, noting that d
S/R
2
is independent of
R
.
Then we obtain
d
dR
¯ϕ(R) =
1
4πR
2
Z
∂ϕ
∂u
u=R
dS
But
∂ϕ
∂u
= e
u
· ∇ϕ = n · ∇ϕ =
∂ϕ
∂n
on S
R
. So
d
dR
¯ϕ(R) =
1
4πR
2
Z
S
R
∇ϕ · dS =
1
4πR
2
Z
V
R
∇
2
ϕ dV = 0
by divergence theorem. So
¯ϕ
(
R
) does not depend on
R
, and the result follows.
11.2.2 The maximum (or minimum) principle
In this section, we will talk about maxima of functions. It should be clear that
the results also hold for minima.
Definition
(Local maximum)
.
We say that
ϕ
(
r
) has a local maximum at
a
if
for some ε > 0, ϕ(r) < ϕ(a) when 0 < |r − a| < ε.
Proposition
(Maximum principle)
.
If a function
ϕ
is harmonic on a region
V
,
then ϕ cannot have a maximum at an interior point of a of V .
Proof.
Suppose that
ϕ
had a local maximum at
a
in the interior. Then there is
an ε such that for any r such that 0 < |r − a| < ε, we have ϕ(r) < ϕ(a).
Note that if there is an
ε
that works, then any smaller
ε
will work. Pick an
ε
sufficiently small such that the region
|r − a| < ε
lies within
V
(possible since
a
lies in the interior of V ).
Then for any r such that |r − a| = ε, we have ϕ(r) < ϕ(a).
¯ϕ(ε) =
1
4πR
2
Z
S
R
ϕ(r) dS < ϕ(a),
which contradicts the mean value property.
We can understand this by performing a local analysis of stationary points
by differentiation. Suppose at
r
=
a
, we have
∇ϕ
= 0. Let the eigenvalues of the
Hessian matrix
H
ij
=
∂
2
∂x
i
∂x
j
be
λ
i
. But since
ϕ
is harmonic, we have
∇
2
ϕ
= 0,
i.e.
∂
2
ϕ
∂x
i
∂x
i
=
H
ii
= 0. But
H
ii
is the trace of the Hessian matrix, which is the
sum of eigenvalues. So
P
λ
i
= 0.
Recall that a maximum or minimum occurs when all eigenvalues have the
same sign. This clearly cannot happen if the sum is 0. Therefore we can only
have saddle points.
(note we ignored the case where all
λ
i
= 0, where this analysis is inconclusive)