UMD PDE Qualifying Exams/Jan2013PDE
Find the explicit solution, , of
subject to .
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
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) .
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 .
We perform a change of variables which gives:
So then differentiating and the use of Green's Formula gives:
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 .
This estimate must hold for all . This necessarily implies since nonconstant supersolutions tend to as .
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.
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 .
Assuming and our result from part a, we get
where the last inequality is due to Jensen's Inequality.
So if , this contradicts the above inequality, i.e. we would have .