Section 4.5 Perron’s method
We can now state and prove the main theorem of this chapter.
Theorem 4.17. Existence and uniqueness for the Dirichlet problem.
Let \(\Omega\) be bounded, connected, and satisfy the exterior ball property, and let \(g \in C^0(\partial \Omega)\text{.}\) Then the Dirichlet problem
\begin{equation}
\left\{
\begin{alignedat}{2}
\Delta u \amp= 0 \amp\quad\amp \ina \Omega \\
u \amp= g \amp\quad\amp \ona \partial \Omega
\end{alignedat}
\right.\tag{4.16}
\end{equation}
has a unique solution \(u \in C^2(\Omega) \cap C^0(\overline \Omega)\text{.}\)
Proof.
Uniqueness follows from Corollary 3.5, and so we only need to show existence.
To construct a solution, we use the so-called Perron method. Let \(S_g\) denote the set of subharmonic functions \(v \in C^0(\overline \Omega)\) satisfying \(v \le g\) on \(\partial\Omega\text{.}\) We easily check that the constant function \(\min_{\partial \Omega} g\) lies in \(S_g\text{,}\) and so the set is nonempty. Moreover, by Proposition 4.14 we have \(v \le
\max_{\partial \Omega} g\) for all \(v \in S_g\text{.}\) Thus the formula
\begin{gather}
u(x) = \sup_{v \in S_g} v(x)\tag{4.17}
\end{gather}
defines a function \(u \maps \Omega \to \R\) satisfying the upper and lower bounds
\begin{equation*}
\min_{\partial \Omega} g \le u \le \max_{\partial\Omega} g \text{.}
\end{equation*}
Step 1: \(u\) is harmonic.
We are about to use Theorem 4.6 to repeatedly take convergent subsequences of harmonic functions. To keep the notation from getting out of hand, we will follow the convention introduced in Section 2.8 and simply replace a sequence with the desired subsequence when necessary. We also note that, thanks to Lemma 4.4, the \(C^4\) requirement in that theorem can be reduced to \(C^2\text{.}\) We will also repeatedly use both parts of Lemma 4.16.
Fix \(x \in \Omega\text{,}\) and let \(v_n \in S_g\) be a sequence such that \(v_n(x)
\to u(x)\text{.}\) By replacing \(v_n\) with \(\max(v_n,\min_{\partial\Omega} g)\text{,}\) we may assume that \(v_n \ge \min_{\partial\Omega} g\text{.}\) Now let \(\overline{B_r(x)} \subseteq
\Omega\) and let \(V_n\) be the harmonic lifting of \(v_n\) on \(B_r(x)\text{.}\) Then \(V_n \in S_g\) and \(V_n \ge v_n\text{,}\) and so we have the chain of inequalities
\begin{equation}
\min_{\partial\Omega} g \le v_n \le V_n \le u \le \max_{\partial\Omega} g\text{.}\tag{4.18}
\end{equation}
Since \(v_n(x) \to u(x)\text{,}\) this implies that \(V_n(x) \to u_n(x)\) at the centre of our ball, and also that the functions \(V_n\) are uniformly bounded. Taking a subsequence using Theorem 4.6, we can therefore assume that \(V_n\) converges uniformly on compact subsets of \(B_r(x)\) to some harmonic function \(V \in C^2(B_r(x))\text{.}\) Taking limits in (4.18) we find that \(V \le u\) in \(B_r(x)\) and \(u(x)=V(x)\text{.}\) We aim to show that in fact \(V=u\) on \(B_r(x)\text{.}\)
Assume for the sake of contradiction that \(V(y) \lt u(y)\) for some \(y
\in B_r(x)\text{.}\) Then there exists some \(w \in S_g\) such that \(V(y) \lt
w(y)\text{.}\) Let \(w_n = \max (w, V_n) \in S_g\text{,}\) and let \(W_n \in S_g\) be the harmonic lifting of \(w_n\) with respect to \(B_r(x)\text{.}\) Then we have on the one hand that
\begin{equation}
W_n(y) \ge w_n(y) \ge w(y) \gt V(y)\text{,}\tag{4.19}
\end{equation}
and on the other hand that
\begin{equation}
v_n \le V_n \le w_n \le W_n \le u\tag{4.20}
\end{equation}
in \(\Omega\text{.}\) Again taking subsequences using Theorem 4.6, we can assume that \(W_n\) converges uniformly on compact subsets of \(B_r(x)\) to a harmonic function \(W \in C^2(B_r(x))\text{.}\) Taking limits in (4.20) and using \(v_n(x) \to u(x)\) we find \(V \le
W\) in \(B_r(x)\) and \(V(x)=W(x)=u(x)\text{.}\) Since \(V\) and \(W\) are harmonic in \(B_r(x)\text{,}\) the strong maximum principle implies \(V \equiv W\) in \(B_r(x)\text{.}\) But (4.19) implies that \(W(y) \ge w(y) \gt V(y)\text{,}\) and so this is a contradiction. As the centre \(x \in \Omega\) of our ball was arbitrary, this completes the proof that \(u \in C^2(\Omega)\) is harmonic.
Step 2: \(u\) satisfies the boundary condition.
We can trivially satisfy the boundary condition by simply defining \(u
\maps \overline \Omega \to \R\) by
\begin{align*}
u(x) =
\begin{cases}
\sup_{v \in S_g} v(x) \amp x \in \Omega, \\
g(x) \amp x \in \partial\Omega;
\end{cases}
\end{align*}
the question is whether the resulting function \(u\) lies in \(C^0(\overline\Omega)\text{.}\) By the previous step and the assumptions on \(g\text{,}\) we certainly know \(u \in C^0(\Omega) \cap C^0(\partial\Omega)\text{.}\) It therefore suffices to show that for any \(\xi \in \partial\Omega\) we have
\begin{gather*}
u(x) \to g(\xi) \text{ as }x \to \xi,\, x \in \Omega\text{.}
\end{gather*}
So fix \(\xi \in
\partial\Omega\) and \(\varepsilon \gt 0\text{.}\) We will find comparison functions \(u_\pm \in
C^\infty(\overline\Omega)\) such that
\begin{gather}
u_- \le u \le u_+ \text{ in } \Omega,\tag{4.21}\\
u_\pm(\xi) = g(\xi) \pm \varepsilon\text{.}\tag{4.22}
\end{gather}
The claim will then follow by a simple analysis argument.
By assumption, we can find a exterior ball \(B = B_r(y)\) at \(\xi\) satisfying \(B \cap \Omega = \varnothing\) and \(\overline B \cap \overline \Omega = \{ \xi
\}.\) The existence of this exterior ball allows us to define a barrier function \(w\) as follows. In Exercise 2.3.2 we showed that the radially symmetric function
\begin{align*}
\Phi(x) =
\begin{cases}
\abs x^{2-N} \amp N \ge 3\\
-\log \abs x \amp N = 2
\end{cases}
\end{align*}
is harmonic away from the origin. Note also that \(\Phi\) is decreasing as a function of \(\abs x\text{.}\) Set
\begin{equation*}
w(x)= \Phi(r) - \Phi(x-y),
\end{equation*}
and notice that, since \(y\) lies outside \(\overline \Omega\text{,}\) \(w\) is harmonic on all of \(\overline \Omega\text{.}\) Moreover, \(w \gt 0\) on \(\overline \Omega \setminus \{\xi\}\) and \(w(\xi) = 0\text{.}\)
Since \(g \in C^0(\partial\Omega)\text{,}\) we can find \(\delta \gt 0\) so that \(\abs{g(x) -
g(\xi)} \lt \varepsilon\) for all \(x \in \partial \Omega \cap B_\delta(\xi)\text{.}\) Since \(w \gt 0\) on the compact set \(\partial\Omega \without B_\delta(\xi)\text{,}\) we can also find \(K \gt 0\) so that \(w \ge 2M/K\) on \(\partial \Omega \without B_\delta(\xi)\text{,}\) where here \(M=\max_{\partial\Omega} \abs g\text{.}\)
Consider the functions \(u_\pm = g(\xi) \pm \varepsilon \pm Kw\text{.}\) Clearly they are harmonic, lie in \(C^2(\Omega) \cap C^0(\overline \Omega)\text{,}\) and satisfy \(u_\pm(\xi) =
g(\xi) \pm \varepsilon\text{.}\) It therefore remains to show that they satisfy \(u_- \le u
\le u_+\) in \(\Omega\text{.}\) With the comparison principle in mind, let us first show that \(u_- \le g \le u_+\) on \(\partial \Omega\text{.}\) For \(x\in \partial\Omega \cap B_\delta(\xi)\) we have
\begin{align*}
u_-(x)
\amp= g(\xi)-\varepsilon-Kw(x) \\
\amp\le g(x)-K w(x) \\
\amp\le g(x),
\end{align*}
while for \(x \in \partial\Omega \without B_\delta(\xi)\) we have
\begin{align*}
u_-(x)
\amp= g(\xi)-\varepsilon-Kw(x)\\
\amp\le g(\xi)- \varepsilon -2M \\
\amp\le - M-\varepsilon \\
\amp\le g(x).
\end{align*}
In particular, \(u_- \in S_g\) and so \(u_- \le u\) in \(\Omega\text{.}\) The inequalities for \(u_+\) are proven similarly. By Proposition 4.14, we have \(v \le u_+\) for all \(v \in S_g\text{,}\) and hence \(u \le u_+\text{.}\) Thus (4.21) holds and the proof is complete.
