# UMD PDE Qualifying Exams/Aug2005PDE

## Problem 1

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

## Problem 4

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

### Solution

#### 4a

For ${\displaystyle v\in V}$, define ${\displaystyle \phi (t)=I[u+tv]}$. Then since ${\displaystyle u}$ minimizes the functional, ${\displaystyle \phi '(0)=0}$. We can calculate ${\displaystyle \phi '}$ (by exploiting the symmetry of ${\displaystyle P}$):

{\displaystyle {\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 ${\displaystyle \phi '(0)=\int _{U}\langle \nabla u,P(x)\nabla v\rangle +c(x)uv-f(x)v\,dx=0}$ which proves the result.

#### 4b

We have

{\displaystyle {\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 ${\displaystyle v\in V}$ and we've obtained a weak form of the PDE. Thus, ${\displaystyle u}$ is a solution to the following PDE:

${\displaystyle \left\{{\begin{array}{rl}-\operatorname {div} (P(x)\nabla u)+c(x)u=f&{\text{ in }}U\\u=g&{\text{ on }}\partial U.\end{array}}\right.}$

#### 4c

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

This gives ${\displaystyle \int _{U}fv=\int _{U}\langle \nabla u,P(x)\nabla v\rangle +c(x)uv=\int _{\{u<0\}}\langle \nabla u,P(x)\nabla u\rangle +c(x)u^{2}}$.

But notice that since ${\displaystyle P(x)}$ is uniformly positive definite, then ${\displaystyle \int _{\{u<0\}}\langle \nabla u,P(x)\nabla u\rangle =\int _{\{u<0\}}(\nabla u)^{T}P(x)(\nabla u)\geq 0.}$

Therefore, we have

${\displaystyle 0\leq \int _{\{u<0\}}(\nabla u)^{T}P(x)(\nabla u)+cu^{2}=\int _{\{u<0\}}fu<0}$

a contradition, unless ${\displaystyle m(\{u<0\})=0}$, i.e. ${\displaystyle u\geq 0}$ a.e.

#### 4d

Suppose ${\displaystyle u_{1},u_{2}}$ are two distinct such solution. Let ${\displaystyle w=u_{1}-u_{2}}$. Then ${\displaystyle w}$ is Lipschitz (since ${\displaystyle u_{1},u_{2}}$ must both be) and ${\displaystyle w=0}$ on ${\displaystyle \partial U}$. Therefore, the variational equation gives

${\displaystyle \int _{U}\langle \nabla w,P(x)\nabla w\rangle +c(x)w^{2}=0}$. Since ${\displaystyle P(x)}$ is positive definite, this gives

${\displaystyle \int _{U}cw^{2}\leq 0}$, a contradiction unless ${\displaystyle w=0}$ a.e.