# 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 $u_{t}+cu_{x}=0$ on $\mathbb {R} ^{2}$.
* 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 $u_{t}+cu_{x}=0$ on $\mathbb {R} ^{2}$.

### Solution
## Problem 4

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

### Solution
#### 4a

For $v\in V$, define $\phi (t)=I[u+tv]$. Then since $u$ minimizes the functional, $\phi '(0)=0$. We can calculate $\phi '$ (by exploiting the symmetry of $P$):
${\begin{aligned}\phi '(t)=&{\frac {d}{dt}}\int _{U}{\frac {1}{2}}\langle \nabla (u+tv),P(x)\nabla (u+tv)\rangle +{\frac {1}{2}}c(x)(u+tv)^{2}-f(x)(u+tv)\,dx\\=&{\frac {d}{dt}}\int _{U}{\frac {1}{2}}\langle \nabla u,P\nabla u\rangle +t\langle \nabla u,P\nabla v\rangle +{\frac {1}{2}}t^{2}\langle \nabla v,P\nabla v\rangle +{\frac {1}{2}}c(x)(u+2tuv+t^{2}v^{2})-f(x)(u+tv)\,dx\\=&\int _{U}\langle \nabla u,P(x)\nabla v\rangle +t\langle \nabla v,P(x)\nabla v\rangle +c(x)(uv+tv^{2})-f(x)v\,dx\end{aligned}}$
And so $\phi '(0)=\int _{U}\langle \nabla u,P(x)\nabla v\rangle +c(x)uv-f(x)v\,dx=0$ which proves the result.

#### 4b

We have
${\begin{aligned}\int _{U}fv=&\int _{U}\langle \nabla u,P(x)\nabla v\rangle +c(x)uv\\=&\int _{U}P(x)\nabla u\cdot \nabla v+c(x)uv\\=&-\int _{U}\operatorname {div} (P(x)\nabla u)v+\int _{\partial U}P(x)\nabla uv+\int _{U}c(x)uv\end{aligned}}$
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:
$\left\{{\begin{array}{rl}-\operatorname {div} (P(x)\nabla u)+c(x)u=f&{\text{ in }}U\\u=g&{\text{ on }}\partial U.\end{array}}\right.$

#### 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 $g\geq 0$. Secondly, $u$, hence $v$, must be Lipschitz continuous since $u\in C^{2}(U)$ and $U$ is a bounded domain in $\mathbb {R} ^{n}$ (i.e. $\nabla u\in C^{1}(U)$) and so $|\nabla u|$ must achieve a (finite) maximum in $U$, hence the derivative is bounded, hence $u$ is Lipschitz. Therefore, $v\in V$.
This gives $\int _{U}fv=\int _{U}\langle \nabla u,P(x)\nabla v\rangle +c(x)uv=\int _{\{u<0\}}\langle \nabla u,P(x)\nabla u\rangle +c(x)u^{2}$.
But notice that since $P(x)$ is uniformly positive definite, then $\int _{\{u<0\}}\langle \nabla u,P(x)\nabla u\rangle =\int _{\{u<0\}}(\nabla u)^{T}P(x)(\nabla u)\geq 0.$
Therefore, we have
$0\leq \int _{\{u<0\}}(\nabla u)^{T}P(x)(\nabla u)+cu^{2}=\int _{\{u<0\}}fu<0$
a contradiction, unless $m(\{u<0\})=0$, i.e. $u\geq 0$ 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
$\int _{U}\langle \nabla w,P(x)\nabla w\rangle +c(x)w^{2}=0$. Since $P(x)$ is positive definite, this gives
$\int _{U}cw^{2}\leq 0$, a contradiction unless $w=0$ a.e.