## Lax Milgram application

Let and . Consider the bilinear form

1. Check that is a continuous symmetric bilinear form and that implies .

2. Prove that is coercive.

3. Deduce that for every there exists a unique satisfying

What is the corresponding minimization problem?

4. Show that the solution of belongs to (and in particular ). Determine the equation and the boundary conditions satisfied by .

5. Assume that , and let be the solution of . Prove that belongs to for every . Show that if and only if .

6. Determine explicitly the solution of when is a constant.

7. Set , where is the solution of and . Check that is a self-adjoint compact operator from into itself.

8. Study the eigenvalues of .

*H. Brezis, Functional Analysis*

**Solution:** 1. It is easy to see that is symmetric and bilinear. To prove the continuity part apply Holder’s inequality to see that

If such that then we first have , which implies that is constant and secondly which implies that .

2. Suppose there exists a sequence such that and . Using the compacity of the unit ball in the weak convergence it is not restrictive to assume that there exists with in and in .

Since we have we get that , and therefore is a constant.

Since in implies that it follows that . Note that we have

which is a contradiction to our assumption. Therefore is coercive.

3. To get the desired result we apply Lax Milgram Theorem. Since is symmetric, the solution is also a minimizer for the problem

4. Let be the solution of . Then for every we have

where . This implies that and therefore . The equation satisfied by is

and the necessary boundary conditions can be obtained by multiplying the equation with and integrating on . The boundary conditions are

5. The differential equation satisfied by can be written as

First note that since we know that for all . Therefore . Moreover, since is continuous, using the differential equation above we can see that for every . Therefore for .

We have the following equivalences:

7. Let and denote . Then we have

which proves that is self-adjoint.

Suppose that and . Denote by and note that by the coercivity of we have

This means that is bounded in and therefore it contains a subsequence which converges weakly to which implies that in . Therefore contains a subsequence converging strongly in . This proves that is compact.