UMD PDE Qualifying Exams/Jan2013PDE

Problem 1Edit

Find the explicit solution,  , of

  subject to  .

SolutionEdit

Note: For notational purposes, let's put the time variable last. i.e.   so that   is the first variable,   is the second variable.

We then write our PDE as  .

We write the characteristic ODEs  

This gives  

Notice that this gives   and   which means that   and   must have the following form:

 

where the coefficients are chosen so that  .

Also since,  , then  .

Now, given any  , we need to find   such that  . Clearly, we need  . This means that we just need to solve the following system for  

 

Solving the second equation for   gives  . Substitute this into the first equation and we can solve for  . We should get (after simplifying)  .

Therefore,  .


Problem 2Edit

Let   be a   function. Define

 .

a). Show that  .

b). Let   solve   for some continuous  . Assume that   for every  , and that   for  . Prove that if   at some  , then   for every  .

SolutionEdit

aEdit

We perform a change of variables   which gives:

 .

So then differentiating and the use of Green's Formula gives:

 

bEdit

Notation: I use   to denote the average integral value symbol (dashed integral). The usual symbol used in Evans would not typeset on this wikibook.


Since  ,  . Therefore,  , that is,   is a supersolution to Laplace's equation.

Suppose  . Then by Part a,  . So   is a decreasing function in  .

Now,

 

This estimate must hold for all  . This necessarily implies   since nonconstant supersolutions tend to   as  .

Problem 3Edit

Let   solve the nonlinear eigenvalue problem

 

Here   is a 1-periodic function in all variables (that is,   is the  -dimensional torus) with   and  .

a. Prove that  .

b. Prove that there exists no sequence of eigen-solutions   such that   and  . Hint: Prove b by contradiction.

SolutionEdit

aEdit

Multiply both sides of the PDE by   and integrate.

 .

Integrate by parts to obtain:

 .

The boundary term vanishes by the periodicity of   in all variables.

Thus   implies that  .

bEdit

Assuming   and our result from part a, we get

 

This gives

  where the last inequality is due to Jensen's Inequality.

So if  , this contradicts the above inequality, i.e. we would have  .