Section 4.2 The Dirichlet problem in a ball
We are now ready to solve the Dirichlet problem (4.1) in the special case where the domain \(\Omega\) is a ball. There are several ways to do this, but we will use an ingenious trick involving polynomials, the Weierstrass approximation theorem, and the estimates from the previous section.
Lemma 4.7.
Let \(B \subset \R^N\) be a ball and let \(g \maps \R^N \to \R\) be a polynomial. Then the Dirichlet problem
\begin{equation}
\left\{
\begin{alignedat}{2}
\Delta u \amp= 0 \amp\quad\amp \ina B \\
u \amp= g \amp\quad\amp \ona \partial B
\end{alignedat}
\right.\tag{4.6}
\end{equation}
has a unique solution \(u\text{,}\) which is also a polynomial.
Proof.
Uniqueness follows from Corollary 3.5. By shifting and scaling coordinates (as in Exercise 2.3.1 and Exercise 3.1.2), we can assume that \(B=B_1(0)\) is the unit ball centred at the origin. Set
\begin{gather}
u = (1-\abs x^2)p + g \tag{4.7}
\end{gather}
where \(p\) is a polynomial to be determined. This automatically satisfies the boundary condition \(u=g\) on \(\partial B = \partial B_1(0)\text{,}\) and so it remains to choose \(p\) so that \(\Delta u = 0\text{,}\) i.e.
\begin{gather}
\Delta[(1-\abs x^2)p] = -\Delta g. \tag{4.8}
\end{gather}
Since \(g\) is a polynomial, \(\Delta g\) is also a polynomial. Let \(d\) be the degree of \(\Delta g\text{,}\) and let \(P_d\) be the space of polynomials with degree \(\le d\text{.}\) Consider the linear mapping \(T\) defined by the left hand side of (4.8),
\begin{equation*}
T \maps p \mapsto \Delta [(1-\abs x^2) p],
\qquad
T \maps P_d \to P_d\text{.}
\end{equation*}
Since multiplication by \(1-\abs x^2\) increases the degree of a polynomial by 2, and applying \(\Delta\) decreases it by at least 2, we easily check that \(T\) is well defined.
We claim that \(T\) is invertible. Since \(P_d\) is a finite dimensional vector space, by basic linear algebra this is equivalent to the kernel of \(T\) being trivial. So suppose that \(Tp = 0\) for some \(p \in P_d\text{,}\) and let \(v=(1-\abs x^2) p\text{.}\) Then by assumption \(\Delta v =
0\text{,}\) and moreover we have \(v=0\) on \(\partial B\text{.}\) Thus by Corollary 3.5 we have \(v \equiv 0\) in \(B\text{,}\) which in turn forces \(p \equiv 0\) as desired.
Theorem 4.8.
Let \(B \subset \R^N\) be a ball and let \(g \in C^0(\partial B)\text{.}\) Then the Dirichlet problem
\begin{equation}
\left\{
\begin{alignedat}{2}
\Delta u \amp= 0 \amp\quad\amp \ina B \\
u \amp= g \amp\quad\amp \ona \partial B
\end{alignedat}
\right.\tag{4.10}
\end{equation}
has a unique solution \(u \in C^2(B) \cap C^0(\overline B)\text{.}\)
Proof.
As usual, uniqueness follows from Corollary 3.5. Since \(\partial B\) is compact, by Theorem 2.22 there exists a sequence \(g_n\) of polynomials converging uniformly to \(g\) on \(\partial B\text{.}\) By Lemma 4.7, there also exist a sequence of harmonic polynomials \(u_n\) with \(u_n=g_n\) on \(\partial B\text{.}\) Applying the maximum principle to the harmonic functions \(\pm(u_n - u_m)\text{,}\) we find that
\begin{equation*}
\sup_B \abs{u_n - u_m}
\le \sup_{\partial B} \abs{g_n-g_m} \to 0
\end{equation*}
as \(n,m \to \infty\text{,}\) and hence by the completeness of \(C^0(\overline B)\) that \(u_n\) converges uniformly to some limit \(u\) in \(C^0(\overline B)\text{.}\) By our choice of \(g_n\) we also have \(u=g\) on \(\partial B\text{.}\) Finally, Lemma 4.5 implies that \(u\) is \(C^2(B)\) and harmonic.
Exercises Exercises
1. (PS6) Explicitly solving a Dirichlet problem.
When \(N=2\) and \(d=1\text{,}\) the space \(P_1\) of polynomials in the proof of Lemma 4.7 has the basis \(\{1,x_1,x_2\}\text{.}\) Express the linear mapping \(T\) in that proof in terms of this basis, and use this to solve (4.6) with data \(g=x_1^2 x_2\text{.}\)
Hint.
Use (4.9).
Solution.
We calculate \(T1 = -4\) and \(Tx_i=-8x_i\text{.}\) Thus with respect to the basis \(\{1,x_1,x_2\}\text{,}\) the operator \(T\) is represented by the diagonal matrix
\begin{equation*}
T =
\begin{pmatrix}
-4 \amp 0 \amp 0 \\
0 \amp -8 \amp 0 \\
0 \amp 0 \amp -8
\end{pmatrix}\text{,}
\end{equation*}
which is simple to invert. The data \(g = x_1^2 x_2\) satisfies \(\Delta g = 2x_2\text{.}\) Thus, the solution \(u\) of the associated Dirichlet problem (4.6) is given by (4.9) as
\begin{align*}
u \amp = (1-\abs x^2) T^{-1}(-\Delta g) + g \\
\amp = (1-\abs x^2) T^{-1}(-2x_2) + x_1^2 x_2\\
\amp = \tfrac 14(1-\abs x^2) x_2 + x_1^2 x_2\\
\amp = \tfrac 14(1+3x_1^2-x_2^2)x_2\text{.}
\end{align*}
Once this formula has been obtained, we can of course easily verify that it solves (4.6).
2. (PS6) Harmonic functions are \(C^\infty\).
(a)
Let \(\ell \in \N\text{.}\) Explain how, by using the bounds in Exercise 4.1.3 for \(0 \le k \le \ell\text{,}\) we can prove a generalisation of Lemma 4.5 where \(C^2,C^4\) are replaced by \(C^\ell,C^{\ell+2}\text{.}\) No need to write out a complete argument, just explain what parts of the original proof need to be modified.
Solution.
Consider the setting of Lemma 4.5, but assume now that the functions \(f_n\) are \(C^{\ell+2}\text{.}\) Using the bounds in Exercise 4.1.3 for \(0 \le k \le \ell\text{,}\) we find that
\begin{equation*}
\n{f_n-f_m}_{C^\ell(B_{r/2}(x))}
\le C \sup_{B_r(x)} \abs{f_n-f_m} \to 0
\end{equation*}
as \(n,m \to \infty\text{,}\) where here the constant \(C\) depends on \(N\text{,}\) \(r\text{,}\) and \(\ell\text{.}\) Appealing to Theorem 2.20, we deduce that \(f_n \to f\) in \(C^\ell(\overline{B_{r/2}(x)})\text{.}\) Since \(x\) was arbitrary, this in particular implies that \(f \in C^\ell(\Omega)\text{.}\)
(b)
Deduce that the function \(u\) in Theorem 4.8 lies in \(C^\ell(B)\) for any \(\ell\text{,}\) i.e. \(u \in C^\infty(B)\text{.}\)
Solution.
Using our new and upgraded version of Lemma 4.5 in the proof of Theorem 4.8, we see that the sequence \(u_n\) in fact converges uniformly to some limit \(\tilde u \in
C^\ell(\Omega)\text{,}\) which by the uniqueness of limits must be the same as \(u\text{.}\) Note that the functions \(u_n\) are polynomials, and so lie in \(C^\infty(\Omega)\text{.}\)
(c)
Finally, let \(u \in C^2(\Omega)\) be an arbitrary harmonic function. Then for any ball \(B\) with \(\overline B \subset \Omega\text{,}\) we trivially have that \(u\) solves (4.10) with \(g = u|_{\partial B}
\in C^0(\partial B)\text{.}\) Conclude that \(u \in C^\infty(B)\text{.}\)
Solution.
If \(u \in C^0(\Omega)\text{,}\) then clearly \(g = u|_{\partial B} \in C^0(\partial B)\) as \(\partial
B \subset \Omega\text{.}\) Thus, for any \(\ell\text{,}\) our improved version of Theorem 4.8 there exists a harmonic function \(\tilde u = C^\ell(B) \cap
C^0(\overline B)\) which equals \(g\) on the boundary. By uniqueness we must have \(\tilde u = u|_{\overline B}\text{.}\) Since \(B\) and \(\ell\) were arbitrary, we must have \(u \in C^\infty(\Omega)\text{.}\)
3. Smooth but unbounded harmonic functions.
In dimension \(N \ge 2\text{,}\) give an example of a ball \(B\) and a harmonic function \(u \in C^\infty(B)\) which does not lie in \(C^0(\overline B)\text{.}\)
Hint.
Solution.
Let \(\Phi\) be the function from Exercise 2.3.2, and note that while \(\Phi\) is \(C^\infty\) and harmonic away from \(x = 0\text{,}\) it is unbounded in the limit as \(x \to 0\text{.}\) So let \(B=B_1(0)\) and \(u(x)=\Phi(x-e_1)\text{.}\) Then \(u \in C^\infty(B)\) and harmonic in \(B\text{,}\) but does not lie in \(C^0(\overline
B)\text{.}\) Indeed, as \(x \in B\) tends to \(e_1 \in \partial B\text{,}\) we have \(\abs{u(x)} \to
+\infty\text{.}\)
