UMD PDE Qualifying Exams/Aug2005PDE

Problem 1

a) Let$c>0$ be a constant. State what is meant by a weak solution of the PDE $u_t+c{u}_x=0$ on ${ℝ}^2$. b) Show that if $f(\cdot )$ is a continuous function of one real variable, then $f(x-ct)$ is a weak solution of the PDE $u_t+c{u}_x=0$ on ${ℝ}^2$.

Problem 4

Let $U\subset {ℝ}^n$ be a bounded open set with smooth boundary $\partial U$. Let $x\mapsto P(x):\overline{U}\to {ℝ}^{n\times n}$ be a smooth family of symmetric $n\times n$ real matrices that are uniformly positive definite. Let $c\ge 0$ and $f(x)$ be smooth functions on $\overline{U}$. Define the functional $I[u]=\underset{U}{\int }(\frac{1}{2}⟨\mathrm{\nabla }u,P(x)\mathrm{\nabla }u⟩+\frac{1}{2}c(x)u^2-f(x)u)dx$ where $⟨\cdot ,\cdot ⟩$ is the scalar product on ${ℝ}^n$. Suppose that $u\in C^2(U)\cap C^0(\overline{U})$ is a minimizer of this functional subject to the Dirichlet condition $u=g$ on $\partial U$, with $g$ continuous. a) Show that $u$ satisfies the variational equation $\underset{U}{\int }⟨\mathrm{\nabla }u,P(x)\mathrm{\nabla }v⟩+c(x)uv=\underset{U}{\int }fv$ for any $v\in V=\{v\in \mathrm{Lip}(U):{v|}_{\partial U}=0\}$. b) What is the PDE satisfied by $u$? c) Suppose $f,g\ge 0$. Show that $v=\min (u,0)$ is an admissible test function, and use this to conclude that $u\ge 0$. (Hint: You may use the fact $\mathrm{\nabla }v={\chi }_{\{u<0\}}\mathrm{\nabla }u$ a.e.) d) Show that there can be only one minimizer of $I[u]$ or, equivalently, only one solution $u\in C^2(U)\cap C^0(\overline{U})$ of the corresponding PDE.

Solution

4a

For $v\in V$, define $\varphi (t)=I[u+tv]$. Then since $u$ minimizes the functional, ${\varphi }^{\prime }(0)=0$. We can calculate ${\varphi }^{\prime }$ (by exploiting the symmetry of $P$):

$\begin{array}{cc}{\varphi }^{\prime }(t)=& \frac{d}{dt}\underset{U}{\int }\frac{1}{2}⟨\mathrm{\nabla }(u+tv),P(x)\mathrm{\nabla }(u+tv)⟩+\frac{1}{2}c(x)(u+tv{)}^2-f(x)(u+tv)dx\\ =& \frac{d}{dt}\underset{U}{\int }\frac{1}{2}⟨\mathrm{\nabla }u,P\mathrm{\nabla }u⟩+t⟨\mathrm{\nabla }u,P\mathrm{\nabla }v⟩+\frac{1}{2}{t}^2⟨\mathrm{\nabla }v,P\mathrm{\nabla }v⟩+\frac{1}{2}c(x)(u+2tuv+t^2{v}^2)-f(x)(u+tv)dx\\ =& \underset{U}{\int }⟨\mathrm{\nabla }u,P(x)\mathrm{\nabla }v⟩+t⟨\mathrm{\nabla }v,P(x)\mathrm{\nabla }v⟩+c(x)(uv+t{v}^2)-f(x)vdx\end{array}$

And so ${\varphi }^{\prime }(0)=\underset{U}{\int }⟨\mathrm{\nabla }u,P(x)\mathrm{\nabla }v⟩+c(x)uv-f(x)vdx=0$ which proves the result.

4b

We have

$\begin{array}{cc}\underset{U}{\int }fv=& \underset{U}{\int }⟨\mathrm{\nabla }u,P(x)\mathrm{\nabla }v⟩+c(x)uv\\ =& \underset{U}{\int }P(x)\mathrm{\nabla }u\cdot \mathrm{\nabla }v+c(x)uv\\ =& -\underset{U}{\int }\mathrm{div}(P(x)\mathrm{\nabla }u)v+\underset{\partial U}{\int }P(x)\mathrm{\nabla }uv+\underset{U}{\int }c(x)uv\end{array}$

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

$\{\begin{array}{cc}-\mathrm{div}(P(x)\mathrm{\nabla }u)+c(x)u=f& \text{ in }U\\ u=g& \text{ on }\partial U.\end{array}$

4c

First we need to show that $v=\min (u,0)\in V$. Firstly, on $\partial U$, $v=\min (g,0)=0$ since we've assumed $g\ge 0$. Secondly, $u$, hence $v$, must be Lipschitz continuous since $u\in C^2(U)$ and $U$ is a bounded domain in ${ℝ}^n$ (i.e. $\mathrm{\nabla }u\in C^1(U)$) and so $|\mathrm{\nabla }u|$ must achieve a (finite) maximum in $U$, hence the derivative is bounded, hence $u$ is Lipschitz. Therefore, $v\in V$.

This gives $\underset{U}{\int }fv=\underset{U}{\int }⟨\mathrm{\nabla }u,P(x)\mathrm{\nabla }v⟩+c(x)uv=\underset{\{u<0\}}{\int }⟨\mathrm{\nabla }u,P(x)\mathrm{\nabla }u⟩+c(x)u^2$.

But notice that since $P(x)$ is uniformly positive definite, then $\underset{\{u<0\}}{\int }⟨\mathrm{\nabla }u,P(x)\mathrm{\nabla }u⟩=\underset{\{u<0\}}{\int }(\mathrm{\nabla }u{)}^{T}P(x)(\mathrm{\nabla }u)\ge 0.$

Therefore, we have

$0\le \underset{\{u<0\}}{\int }(\mathrm{\nabla }u{)}^{T}P(x)(\mathrm{\nabla }u)+c{u}^2=\underset{\{u<0\}}{\int }fu<0$

a contradiction, unless $m(\{u<0\})=0$, i.e. $u\ge 0$ a.e.

4d

Suppose $u_1,u_2$ are two distinct such solution. Let $w=u_1-u_2$. Then $w$ is Lipschitz (since $u_1,u_2$ must both be) and $w=0$ on $\partial U$. Therefore, the variational equation gives

$\underset{U}{\int }⟨\mathrm{\nabla }w,P(x)\mathrm{\nabla }w⟩+c(x)w^2=0$. Since $P(x)$ is positive definite, this gives

$\underset{U}{\int }c{w}^2\le 0$, a contradiction unless $w=0$ a.e.