UMD PDE Qualifying Exams/Aug2005PDE

From Wikibooks, open books for an open world
Jump to: navigation, search

Problem 1[edit]

a) Let be a constant. State what is meant by a weak solution of the PDE on .

b) Show that if is a continuous function of one real variable, then is a weak solution of the PDE on .

Solution[edit]

Problem 4[edit]

Let be a bounded open set with smooth boundary . Let be a smooth family of symmetric real matrices that are uniformly positive definite. Let and be smooth functions on . Define the functional

where is the scalar product on . Suppose that is a minimizer of this functional subject to the Dirichlet condition on , with continuous.

a) Show that satisfies the variational equation

for any .

b) What is the PDE satisfied by ?

c) Suppose . Show that is an admissible test function, and use this to conclude that . (Hint: You may use the fact a.e.)

d) Show that there can be only one minimizer of or, equivalently, only one solution of the corresponding PDE.


Solution[edit]

4a[edit]

For , define . Then since minimizes the functional, . We can calculate (by exploiting the symmetry of ):

And so which proves the result.

4b[edit]

We have

The boundary terms vanish since and we've obtained a weak form of the PDE. Thus, is a solution to the following PDE:

4c[edit]

First we need to show that . Firstly, on , since we've assumed . Secondly, , hence , must be Lipschitz continuous since and is a bounded domain in (i.e. ) and so must achieve a (finite) maximum in , hence the derivative is bounded, hence is Lipschitz. Therefore, .

This gives .

But notice that since is uniformly positive definite, then

Therefore, we have

a contradition, unless , i.e. a.e.


4d[edit]

Suppose are two distinct such solution. Let . Then is Lipschitz (since must both be) and on . Therefore, the variational equation gives

. Since is positive definite, this gives

, a contradiction unless a.e.