UMD PDE Qualifying Exams/Jan2010PDE

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.

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

$$\begin{align} 0= \int_U w \Delta w &= -\int_U |Dw|^2 +\int_\gamma w \frac{\partial w}{\partial \nu}\,dS + \lim_{r\to\infty} \int_{\partial B(0,r)} w\frac{\partial w}{\partial \nu}\,dS\\ &=-\int_U |Dw|^2 +0 + \lim_{r\to\infty} \int_{\partial B(0,r)} w\frac{\partial w}{\partial \nu}\,dS \end{align}$$

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.