1. The Perron method for the Dirichlet
problem: proving existence of a solution on
arbitrarily-shaped regions using sequences
of subharmonic functions
Michael Maltese
May 5, 2014

2. The Dirichlet problem
Theorem
If Ω has a barrier at every boundary point, then for any
g ∈ C(∂Ω), there exists a unique u ∈ C(Ω) ∩ C∞(Ω) such that
∆u = 0 in Ω
u = g on ∂Ω.
Main trick: We can find sequences of subharmonic functions which
converge to harmonic u, and then show that this in fact solves the
problem.

3. Uniqueness
Given two solutions u1 and u2, let u∗ = u1 − u2. Then u∗ = 0 on
∂Ω, so by the maximum and minimum principle of harmonic
functions, u∗ = 0 everywhere on Ω.

4. Existence
We’ll use:
the solution for the ball
some properties of subharmonic functions
the Arzelà-Ascoli Theorem: any uniformly bounded set of
(harmonic) functions has a convergent subsequence

5. Claim
Let
m = inf g
M = sup g
σg = {subharmonic v ∈ C(Ω) : v ≤ g on ∂Ω}.
Then σg contains the constant function m and is non-empty, and
u(x) = sup
v∈σg
v(x)
is well-defined and the solution to our problem.

6. We’ll prove this in multiple steps. First, we’ll prove the following:
Claim (Harmonic function on a countable sequence in a ball)
For any Br (a) ⊂ Ω and {xi } ⊂ Br (a), there is a harmonic function
h on Br (a) such that u(xi ) = h(xi ) for all i ∈ N.

7. We’ll prove this in multiple steps. First, we’ll prove the following:
Claim (Harmonic function on a countable sequence in a ball)
For any Br (a) ⊂ Ω and {xi } ⊂ Br (a), there is a harmonic function
h on Br (a) such that u(xi ) = h(xi ) for all i ∈ N.
Then we’ll use that to show:
Claim (u is a solution)
u ∈ C(Ω) ∩ C∞(Ω), ∆u = 0 on Ω, and u = g on ∂Ω.

8. Harmonic function on a countable sequence in a ball
Lemma (Step 1: Maximum of two functions is in σg )
If v1, v2 ∈ σg , then w = max v1, v2 ∈ σg .
Recall that to be in σg , w must be continuous on the closure of Ω,
less than or equal to g on the boundary, and subharmonic.
We can show that w is subharmonic by local sub-meanvalue
inequalities. For any Br (a) ⊂ Ω, we have
w(a) = max v1(a), v2(a) ≤ max M
Br (a)
(v1), M
Br (a)
(v2) ≤ M
Br (a)
(w),
so w is subharmonic.

9. Harmonic function on a countable sequence in a ball
Lemma (Step 2: Solution on a ball)
For v ∈ σg and Br (a) ⊂ Ω, there exists a va,r ∈ σg such that
v ≤ va,r
∆va,r = 0 on Br (a)
va,r = v on Ω Br (a)
By solution of the Dirichlet problem for a ball, and by maximum
principle for harmonic and subharmonic functions. In σg because
harmonic functions are subharmonic functions.

10. Harmonic function on a countable sequence in a ball
Lemma (Step 3: Harmonic function on a countable sequence
in a ball)
For any Br (a) ⊂ Ω and {xi } ⊂ Br (a), there is a harmonic function
h on Br (a) such that u(xi ) = h(xi ) for all i ∈ N.
Let vj,i ∈ σg such that vj,i (xi ) increases and converges to u(xi ) as
j → ∞.
Let Vj = max{inf g, vj,1, vj,2, . . . , vj,j}. By the previous lemma, we
know there exists a Uj ≥ Vj which is harmonic on Br (a) and is in
σg .
By Arzelà-Ascoli, because Uj is bounded by inf g and sup g, there
is a subsequence Uj which converges to a function h, and h is
harmonic.
Then
h(xi ) = lim
j→∞
Uj (xi ) ≥ lim
j→∞
Vj (xi ) = u(xi ),
but h ∈ σg so h(xi ) = u(xi ).

11. We just proved:
Claim (Harmonic function on a countable sequence in a ball)
For any Br (a) ⊂ Ω and {xi } ⊂ Br (a), there is a harmonic function
h on Br (a) such that u(xi ) = h(xi ) for all i ∈ N.

12. We just proved:
Claim (Harmonic function on a countable sequence in a ball)
For any Br (a) ⊂ Ω and {xi } ⊂ Br (a), there is a harmonic function
h on Br (a) such that u(xi ) = h(xi ) for all i ∈ N.
Now we’ll show:
Claim (u is a solution)
u ∈ C(Ω) ∩ C∞(Ω), ∆u = 0 on Ω, and u = g on ∂Ω.

13. u is a solution
Lemma (Step 4: u ∈ C(Ω))
For any point a ∈ Ω, any ball in Ω and any convergent sequence
zi → a in the ball. Apply the previous lemma with
{xi } = {a, z1, z2, . . .} to get
u(a) = h(a) = lim
i→∞
h(xi ) = lim
i→∞
u(xi ),
which is a definition of continuity.

14. u is a solution
Lemma (Step 5: u is harmonic on Ω)
For any ball whose closure is in Ω, apply the lemma two steps back
with {xi } a countable dense subset of the ball (we can do this
because we’re in a bounded region in real space). Then we have a
harmonic function h which coincides with u on {xi }, and since u is
continuous {xi } is dense, h = u on the ball.

15. u is a solution
Definition (Barrier)
For b ∈ ∂Ω, a function Qb ∈ C(Ω) is a barrier at b if Qb
subharmonic on Ω, Qb(b) = 0, and Qb < 0 on ∂Ω {b}.

16. u is a solution
Lemma (Step 6: Boundary values)
For any b ∈ ∂Ω, lim infx→b u(x) ≥ g(b).
For any ε, we can create a function
v(x) = g(b) − ε + KQb(x) for x ∈ Ω
which is continuous and subharmonic. We can pick a δ > 0 so that
g(x) > g(b) − ε ⇒ v(x) ≤ g(x) on ∂Ω ∩ Bδ(b), and we can pick a
K > 0 so that v(x) ≤ g(x) on ∂Ω Bδ(b).
Then v ∈ σg ⇒ v ≤ u, and
g(b) − ε = v(b) ≤ lim inf
x→b
u(x).

17. u is a solution
Lemma (Step 7: Boundary values)
For any b ∈ ∂Ω, lim supx→b u(x) ≤ g(b).
Let ˜u(x) = sup−w∈σ−g −w(x). Then by the previous lemma,
lim infx→b ˜u(x) ≥ −g(b). For any v ∈ σg , −w ∈ σ−g , v − w ≤ 0
on ∂Ω, so v − w ≤ 0 on Omega by the maximum principle. We
can take supremums to get u ≤ ˜u, so
lim sup
x→b
u(x) ≤ lim sup
x→b
˜u(x) = − lim inf
x→b
˜ux ≤ g(b).
.

18. u is a solution
Lemma (Step 8)
There is no step 8.
Thanks for listening! Questions?