UMD PDE Qualifying Exams/Jan2010PDE

Problem 1

Let $U=\{x\in {ℝ}^n:|x|>1\}$. Suppose $u\in C^2(U)\cap C(\overline{U})$ is a bounded solution of the following Dirichlet problem: $\mathrm{\Delta }u=0$ in $U$, and $u=f$ on $\mathrm{\Gamma }=\{x\in {ℝ}^n:|x|=1\}$ where $f(x)\in C(\mathrm{\Gamma })$. (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 $u\pm ϵ\log |x|$. (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 $U=\{x\in {ℝ}^n:|x|>1\}$, if $\mathrm{\Delta }u=0$ then $\underset{\overline{U}}{\max }u=\underset{\mathrm{\Gamma }}{\max }u$. Proof of Lemma: Consider the domain $U_r=\{1<|x|<r\}$ and the function $u_{ϵ}=u-ϵ\log |x|$. Since $\log |x|$ is the fundamental solution to Laplaces equation, then clearly $u_{ϵ}$ is harmonic and one can easily verify that $u_{ϵ}=f$ on $\mathrm{\Gamma }$. Then since the domain $U_r$ is bounded, we can use the ordinary maximum principle and say that $\underset{\overline{U_r}}{\max }{u}_{ϵ}=\underset{\mathrm{\Gamma }}{\max }{u}_{ϵ}\vee \underset{\partial B(0,r)}{\max }{u}_{ϵ}$. We know that $\underset{\mathrm{\Gamma }}{\max }{u}_{ϵ}=\Vert f{\Vert }_{L^{\mathrm{\infty }}(\mathrm{\Gamma })}$. Now if $r$ is sufficiently large then we can guarantee that $\underset{\partial B(0,r)}{\max }{u}_{ϵ}<\Vert f{\Vert }_{L^{\mathrm{\infty }}(\mathrm{\Gamma })}$. Sending $r\to \mathrm{\infty }$ gives us $\underset{\overline{U}}{\sup }{u}_{ϵ}\le \underset{\mathrm{\Gamma }}{\max }{u}_{ϵ}$. Now send $ϵ\to 0$ and we get $\underset{\overline{U}}{\max }u\le \underset{\mathrm{\Gamma }}{\max }u$ which proves the lemma.

If we replace $u-ϵ\log |x|$ with $u+ϵ\log |x|$ 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 $\mathrm{\Delta }w=0$ in $U$ and $w=0$ on $\mathrm{\Gamma }$. Then by our maximum principle lemma, $\underset{\overline{U}}{\sup }w\le \underset{\mathrm{\Gamma }}{\max }w=0$. Similarly by the minimum principle, $\underset{\overline{U}}{\inf }u\ge \underset{\mathrm{\Gamma }}{\min }w=0$. 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 $\mathrm{\Gamma }$. 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 $\mathrm{\Delta }w=0$ in $U$ and $w=0$ on $\mathrm{\Gamma }$. Integration by parts shows that

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

So if we imposed $\underset{|x|\to \mathrm{\infty }}{\lim }w(x)=0$ then we would have $\underset{U}{\int }|Dw{|}^2=0$ 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.