# UMD PDE Qualifying Exams/Jan2010PDE

## Problem 1Edit

Let . Suppose is a bounded solution of the following Dirichlet problem: in , and on where . (a) Consider . Show that there exists at most one solution of the above problem. Hint: First, you might want to consider an appropriate maximum principle in by using . (b) Now consider . 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 be unique in this case? |

### SolutionEdit

#### Solution 1aEdit

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

Then since the domain is bounded, we can use the ordinary maximum principle and say that . We know that . Now if is sufficiently large then we can guarantee that . Sending gives us . Now send and we get which proves the lemma. |

If we replace with then by the same methods we can prove a minimum principle for our domain .

Now suppose are two bounded solutions to the Dirichlet problem. Let . Then solves in and on . Then by our maximum principle lemma, . Similarly by the minimum principle, . This implies , i.e. , which proves that bounded solutions to this Dirichlet problem are unique.

#### Solution 1bEdit

Now when we can have more than one bounded solution to the Dirichlet problem on . Suppose is one such bounded solution. Recall that is the fundamental solution to Laplace's equation in dimension 3. Therefore is also harmonic and one can verify that on . It is also easy to verify that is also bounded on . Therefore both are distinct, bounded solutions. Therefore solution is not unique.

How can we get a unique solution? Recall that solves in and on . Integration by parts shows that

So if we imposed then we would have which implies that is constant. But the boundary conditions tell us that . In other words, *then* the solution would be unique.