# UMD PDE Qualifying Exams/Jan2013PDE

## Contents

## Problem 1[edit]

Find the explicit solution, , of subject to . |

### Solution[edit]

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 2[edit]

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 . |

### Solution[edit]

#### a[edit]

We perform a change of variables which gives:

.

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

#### b[edit]

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 3[edit]

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. |

### Solution[edit]

#### a[edit]

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 .

#### b[edit]

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 .