# UMD PDE Qualifying Exams/Jan2005PDE
## Problem 1

* Let $u(x,y)$ be a harmonic function on $\mathbb {R} ^{2}$ and suppose that
* $\int \int _{\mathbb {R} ^{2}}|\nabla u|^{2}(x,y)\,dx\,dy<\infty .$
* Show that $u$ is a constant function.

### Solution

Let $C=\int \int _{\mathbb {R} ^{2}}|\nabla u|^{2}(x,y)\,dx\,dy<\infty .$
If $u$ is harmonic (i.e. $\Delta u=0$) then so must $v=\nabla u$ (surely, $\Delta \nabla u=0$). Then since the absolute value as an operator is convex, we have that $|v|=|\nabla u|$ is a subharmonic function on $\mathbb {R} ^{2}$.
Then by the mean value property of subharmonic functions, for any $x_{0}\in \mathbb {R} ^{2}$we have
${\begin{array}{lll}|v(x_{0})|&\leq &{\frac {1}{\pi r^{2}}}\int _{B(x_{0},r)}|v|\,dx\,dy\\&\leq &{\frac {1}{\pi r^{2}}}\left(\int _{B(x_{0},r)}1\,dx\,dy\right)^{1/2}\left(\int _{B(x_{0},r)}|v|^{2}\,dx\,dy\right)^{1/2}\\&\leq &{\frac {C}{\sqrt {\pi r^{2}}}}\end{array}}$
where the second inequality is due to Cauchy-Schwarz (Hölder) inequality.
This estimate hold for all $r>0$. Therefore if we send $r\to \infty$ we see that for all $x_{0}\in \mathbb {R} ^{2},$$|v(x_{0})|=|\nabla u(x_{0})|=0$ which gives us that $u$ is constant.

## Problem 2

* Let $u(x,t)$ be a piecewise smooth weak solution of the conservation law
* $u_{t}+f(u)_{x}=0,\quad -\infty <x<\infty ,t>0.$
* a) Derive the Rankine-Hugoniot conditions at a discontinuity of the solution.
* b)Find a piecewise smooth solution to the IVP
* $u_{t}+(u^{2}+u)_{x}=0,\quad -\infty <x<\infty ,t>0$
* $u(x,0)=\left\{{\begin{array}{ll}1,&x<0\\-2&x>0.\end{array}}\right.$

### Solution
#### a)

When we solve the PDE by methods of characteristics, the characteristic curves can cross, causing a shock, or discontinuity. The task at hand, is to find the curve of discontinuity, call it $C$.
Multiply the PDE by $v$, a smooth test function with compact support in $\mathbb {R} \times (0,\infty )$. Then by an integration by parts:
${\begin{aligned}0=&\int _{0}^{\infty }\int _{-\infty }^{\infty }(u_{t}+f(u)_{x})v\,dx\,dt\\=&-\int _{0}^{\infty }\int _{-\infty }^{\infty }uv_{t}\,dx\,dt+\left.\int _{-\infty }^{\infty }vu\,dx\right|_{t=0}^{t=\infty }-\int _{0}^{\infty }\int _{-\infty }^{\infty }f(u)v_{x}\,dx\,dt+\left.\int _{0}^{\infty }f(u)v\,dt\right|_{x=-\infty }^{x=\infty }\\=&-\int _{0}^{\infty }\int _{-\infty }^{\infty }uv_{t}\,dx\,dt-\int _{-\infty }^{\infty }vu(x,0)\,dx-\int _{0}^{\infty }\int _{-\infty }^{\infty }f(u)v_{x}\,dx\,dt.\end{aligned}}$
Let $V_{l}$ denote the open region in $\mathbb {R} \times (0,\infty )$ to the left of $C$ and similarly $V_{r}$ denotes the region to the right of $C$. If the support of $v$ lies entirely in either of these two regions, then all of the above boundary terms vanish and we get $0=\int _{0}^{\infty }\int _{-\infty }^{\infty }(u_{t}+f(u)_{x})v\,dx\,dt=\int _{0}^{\infty }\int _{-\infty }^{\infty }uv_{t}+f(u)v_{x}\,dx\,dt.$
Now suppose the support of $v$ intersects the discontinuity $C$.

#### b)

We can calculate $\sigma ={\frac {F(u_{l})-F(u_{r})}{u_{l}-u_{r}}}={\frac {1^{2}+1-(4-2)}{1+2}}=0$. Therefore, the shock wave extends vertically from the origin. That is,
$u(x,t)=\left\{{\begin{aligned}1&x<0\\-2&x>0\end{aligned}}\right.$

## Problem 3

* Consider the evolution equation with initial data
* $u_{tt}-u_{xx}+u_{t}=(u_{xt}^{3})_{x},\quad -\infty <x<\infty ,t>0$
* $u(x,0)=f(x),u_{t}(x,0)=g(x),u(0,t)=u(1,t)=0.$
* a) What energy quantity is appropriate for this equation? Is it conserved or dissipated?
* b) Show that $C^{3}$ solutions of this problem are unique.

### Solution
#### 3a

Consider the energy $E(t)={\frac {1}{2}}\int _{0}^{1}u_{t}^{2}+u_{x}^{2}\,dx$. Then ${\dot {E}}(t)=\int _{0}^{1}u_{t}u_{tt}+u_{x}u_{xt}\,dx$. Integrate by parts to get ${\dot {E}}(t)=\int _{0}^{1}u_{t}u_{tt}-u_{xx}u_{t}\,dx+\left.u_{t}u_{x}\right|_{0}^{1}$. The boundary terms vanish since $u(0,t)=0$ implies $u_{t}(0,t)=0$ (similarly at $x=1$). Then by the original PDE we get
${\begin{aligned}{\dot {E}}(t)&=&\int _{0}^{1}u_{t}((u_{xt}^{3})_{x}-u_{t})\,dx\\&=&\int _{0}^{1}u_{t}(u_{xt}^{3})_{x}-u_{t}^{2})\,dx\\&=&\int _{0}^{1}-u_{xt}^{4}-u_{t}^{2}\,dx+\left.u_{xt}^{3}u_{t}\right|_{0}^{1}.\end{aligned}}$
where the last equality is another integration by parts. The boundary terms vanish again by the same argument. Therefore, ${\dot {E}}(t)<0$ for all $t$; that is, energy is dissipated.

#### 3b

Suppose $u,v$ are two distinct solutions to the system. Then $w=u-v$ is a solution to
$w_{tt}-w_{xx}+w_{t}=(w_{xt}^{3})_{x},\quad -\infty <x<\infty ,t>0$
$w(x,0)=0,w_{t}(x,0)=0,w(0,t)=w(1,t)=0.$
This tells us that at $t=0$, $w_{x}=w_{t}=0$. Therefore, $E(0)=0$. Since $E(t)\leq E(0)$ then $E(t)=0$ for all $t$. This implies $w\equiv 0$. That is, $u=v$.

## Problem 4

* Let $\Omega \subseteq \mathbb {R} ^{n}$ be a bounded open set with smooth boundary $\partial \Omega$. Consider the initial boundary value problem for $u(x,t)$:
* $\left\{{\begin{array}{ll}u_{t}-\nabla \cdot (a(x)\nabla u)+b(x)u=q,&x\in \Omega ,t>0\\u(x,0)=f(x),&x\in \Omega \\u_{t}+\partial u/\partial n+u=0,&x\in \partial \Omega ,t>0\end{array}}\right.$
* where $\partial u/\partial n$ is the exterior normal derivative. Assume that $a,b\in C^{1}({\bar {\Omega }})$ and that $a(x)\geq 0$ for $x\in \Omega$. Show that smooth solutions of this problem are unique.

### Solution

Suppose $u,v$ are two distinct solutions. Then $w=u-v$ is a smooth solution to
$\left\{{\begin{array}{ll}w_{t}-\nabla \cdot (a(x)\nabla w)+b(x)w=0,&x\in \Omega ,t>0\\w(x,0)=0,&x\in \Omega \\w_{t}+\partial w/\partial n+w=0,&x\in \partial \Omega ,t>0\end{array}}\right.$
Consider the energy $E(t)={\frac {1}{2}}\int _{\Omega }w^{2}\,dx+{\frac {1}{2}}\int _{\partial \Omega }a(x)w^{2}\,dS$. It is easy to verify that $E(0)=0$. Then
${\begin{aligned}{\dot {E}}(t)=&\int _{\Omega }ww_{t}\,dx+\int _{\partial \Omega }a(x)ww_{t}\,dS\\=&\int _{\Omega }w\nabla \cdot (a(x)\nabla w)-b(x)w^{2}\,dx+\int _{\partial \Omega }a(x)ww_{t}\,dS\\=&\int _{\Omega }-\nabla w\cdot (a(x)\nabla w)-b(x)w^{2}\,dx+\int _{\partial \Omega }wa(x){\frac {\partial w}{\partial n}}\,dS+\int _{\partial \Omega }a(x)ww_{t}\,dS\\=&\int _{\Omega }-a(x)\nabla w\cdot \nabla w-b(x)w^{2}\,dx+\int _{\partial \Omega }-a(x)w^{2}-a(x)ww_{t}\,dS+\int _{\partial \Omega }a(x)ww_{t}\,dS\\=&\int _{\Omega }-a(x)\nabla w\cdot \nabla w-b(x)w^{2}\,dx+\int _{\partial \Omega }-a(x)w^{2}\,dS\\\leq &\int _{\Omega }-b(x)w^{2}\,dx\\\leq &\|b\|_{L^{\infty }(\Omega )}\int _{\Omega }w^{2}\,dx\end{aligned}}$
Therefore ${\dot {E}}(t)\leq \|b\|_{L^{\infty }(\Omega )}E(t)$ implies $E(t)\leq E(0)e^{\|b\|_{L^{\infty }(\Omega )}t}=0$ for all $t$. Thus, $E(t)=0$ for all $t$ which implies $w\equiv 0$

## Problem 5

* Let $\Omega \subset \mathbb {R} ^{n}$ be a bounded open set with smooth boundary. Let $K=\{u\in H_{0}^{1}(\Omega ):u\geq 0{\text{ in }}\Omega \}$. Let $f\in L^{2}(\Omega )$ and define the functional
* $I(u)={\frac {1}{2}}\int _{\Omega }|\nabla u|^{2}\,dx-\int _{\Omega }fu\,dx$.
* Show that $u$ is a minimizer of $I$ over $K$ if and only if $u$ satisfies the variational inequality
* $\int _{\Omega }\nabla u\cdot \nabla (u-v)\,dx\leq \int _{\Omega }f(u-v)\,dx$ for all $v\in K$.

### Solution

$(\Rightarrow )$ Suppose $u$ minimizes $I$, i.e. $I[u]\leq I[w]\,\forall w\in K$. Then for any fixed $w\in K$, if we let $g(t)=(1-t)u+tv$ then $I[g(0)]\leq I[g(t)]\,\forall 0\leq t\leq 1$. Let $i(t)=I[g(t)]$; then we can say that $i'(0)\geq 0$. Now we must compute $i'(0)$. We have
$i(t)={\frac {1}{2}}\int _{\Omega }\left|(1-t)\nabla u+t\nabla v\right|^{2}-f((1-t)u+tv)\,dx$
$i'(t)=\int _{\Omega }-(1-t)|\nabla u|^{2}+(1-2t)\nabla u\cdot \nabla v+t|\nabla v|^{2}-(1-t)fu-tfv\,dx$
$i'(0)=\int _{\Omega }-|\nabla u|^{2}+\nabla u\cdot \nabla v-fu\,dx$
Since we know $i'(0)\geq 0$ then
$\int _{\Omega }f(u-v)\,dx\geq \int _{\Omega }|\nabla u|^{2}-\nabla u\cdot \nabla v=\int _{\Omega }\nabla u\cdot \nabla (u-v)\,dx$ as desired.
$(\Leftarrow )$ Conversely suppose
$\int _{\Omega }\nabla u\cdot \nabla (u-v)\,dx=\int _{\Omega }|\nabla |^{2}-\nabla u\cdot \nabla v\,dx\leq \int _{\Omega }f(u-v)\,dx.$
Then ${\begin{array}{lll}\int _{\Omega }|\nabla u|^{2}-fu\,dx&\leq &\int _{\Omega }\nabla u\cdot \nabla v-fv\,dx\\&\leq &\int _{\Omega }|\nabla u||\nabla v|-fv\,dx\\&\leq &\int _{\Omega }1/2|\nabla u|^{2}+1/2|\nabla v|^{2}-fv\,dx\end{array}}$
Therefore, $1/2\int _{\Omega }|\nabla u|^{2}-fu\,dx\leq 1/2\int _{\Omega }|\nabla v|^{2}-fv\,dx$ for all $v\in K$. That is, $I[u]\leq I[v]$ for all $v\in K$, as desired.