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.

Lemma: For the domain  , if   then  .

Proof of Lemma: Consider the domain   and the function  . Since   is the fundamental solution to Laplaces equation, then clearly   is harmonic and one can easily verify that   on  .

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.