UMD PDE Qualifying Exams/Jan2010PDE

Problem 1

Let $U=\{x\in \mathbb {R} ^{n}:|x|>1\}$. Suppose $u\in C^{2}(U)\cap C({\bar {U}})$ is a bounded solution of the following Dirichlet problem: $\Delta u=0$ in $U$, and $u=f$ on $\Gamma =\{x\in \mathbb {R} ^{n}:|x|=1\}$ where $f(x)\in C(\Gamma )$. (a) Consider $n=2$. Show that there exists at most one solution of the above problem. Hint: First, you might want to consider an appropriate maximum principle in $U$ by using $u\pm \epsilon \log |x|$. (b) Now consider $n=3$. Show that it is possible to have more than one bounded solutions of the above problem. What additional condition should you impose so that the solution $u$ be unique in this case?

Solution

Solution 1a

We can't use the ordinary maximum principle for harmonic functions since our domain $U$ is unbounded. We hope to use what we would expect the maximum principle would be for this domain, but that requires proof.

Lemma: For the domain $U=\{x\in \mathbb {R} ^{n}:|x|>1\}$, if $\Delta u=0$ then $\max _{\bar {U}}u=\max _{\Gamma }u$. Proof of Lemma: Consider the domain $U_{r}=\{1<|x|<r\}$ and the function $u_{\epsilon }=u-\epsilon \log |x|$. Since $\log |x|$ is the fundamental solution to Laplaces equation, then clearly $u_{\epsilon }$ is harmonic and one can easily verify that $u_{\epsilon }=f$ on $\Gamma$. Then since the domain $U_{r}$ is bounded, we can use the ordinary maximum principle and say that $\max _{\bar {U_{r}}}u_{\epsilon }=\max _{\Gamma }u_{\epsilon }\vee \max _{\partial B(0,r)}u_{\epsilon }$. We know that $\max _{\Gamma }u_{\epsilon }=\|f\|_{L^{\infty }(\Gamma )}$. Now if $r$ is sufficiently large then we can guarantee that $\max _{\partial B(0,r)}u_{\epsilon }<\|f\|_{L^{\infty }(\Gamma )}$. Sending $r\to \infty$ gives us $\sup _{\bar {U}}u_{\epsilon }\leq \max _{\Gamma }u_{\epsilon }$. Now send $\epsilon \to 0$ and we get $\max _{\bar {U}}u\leq \max _{\Gamma }u$ which proves the lemma.

If we replace $u-\epsilon \log |x|$ with $u+\epsilon \log |x|$ then by the same methods we can prove a minimum principle for our domain $U$.

Now suppose $u_{1},u_{2}$ are two bounded solutions to the Dirichlet problem. Let $w=u_{1}-u_{2}$. Then $w$ solves $\Delta w=0$ in $U$ and $w=0$ on $\Gamma$. Then by our maximum principle lemma, $\sup _{\bar {U}}w\leq \max _{\Gamma }w=0$. Similarly by the minimum principle, $\inf _{\bar {U}}u\geq \min _{\Gamma }w=0$. This implies $w\equiv 0$, i.e. $u_{1}\equiv u_{2}$, which proves that bounded solutions to this Dirichlet problem are unique.

Solution 1b

Now when $n=3$ we can have more than one bounded solution to the Dirichlet problem on $U$. Suppose $u$ is one such bounded $C^{2}$ solution. Recall that $1/|x|$ is the fundamental solution to Laplace's equation in dimension 3. Therefore $v=(u-1)+1/|x|$ is also harmonic and one can verify that $v=f$ on $\Gamma$. It is also easy to verify that $v$ is also bounded on $U$. Therefore both $u,v$ are distinct, bounded solutions. Therefore solution is not unique.

How can we get a unique solution? Recall that $w$ solves $\Delta w=0$ in $U$ and $w=0$ on $\Gamma$. Integration by parts shows that

${\begin{aligned}0=\int _{U}w\Delta w&=-\int _{U}|Dw|^{2}+\int _{\gamma }w{\frac {\partial w}{\partial \nu }}\,dS+\lim _{r\to \infty }\int _{\partial B(0,r)}w{\frac {\partial w}{\partial \nu }}\,dS\\&=-\int _{U}|Dw|^{2}+0+\lim _{r\to \infty }\int _{\partial B(0,r)}w{\frac {\partial w}{\partial \nu }}\,dS\end{aligned}}$

So if we imposed $\lim _{|x|\to \infty }w(x)=0$ then we would have $\int _{U}|Dw|^{2}=0$ which implies that $w$ is constant. But the boundary conditions tell us that $w\equiv 0$. In other words, then the solution would be unique.