UMD PDE Qualifying Exams/Aug2005PDE

Problem 1Edit

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  .


Problem 4Edit

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.



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


And so   which proves the result.


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:



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.


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.