# UMD PDE Qualifying Exams/Jan2005PDE

## Problem 1

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

### Solution

Let ${\displaystyle C=\int \int _{\mathbb {R} ^{2}}|\nabla u|^{2}(x,y)\,dx\,dy<\infty .}$

If ${\displaystyle u}$ is harmonic (i.e. ${\displaystyle \Delta u=0}$) then so must ${\displaystyle v=\nabla u}$ (surely, ${\displaystyle \Delta \nabla u=0}$). Then since the absolute value as an operator is convex, we have that ${\displaystyle |v|=|\nabla u|}$ is a subharmonic function on ${\displaystyle \mathbb {R} ^{2}}$.

Then by the mean value property of subharmonic functions, for any ${\displaystyle x_{0}\in \mathbb {R} ^{2}}$we have

${\displaystyle {\begin{array}{lll}|v(x_{0})|&\leq &{\frac {1}{\pi r^{2}}}\int _{B(x_{0},r)}|v|\,dx\,dy\\&\leq &{\frac {1}{\pi r^{2}}}\left(\int _{B(x_{0},r)}1\,dx\,dy\right)^{1/2}\left(\int _{B(x_{0},r)}|v|^{2}\,dx\,dy\right)^{1/2}\\&\leq &{\frac {C}{\sqrt {\pi r^{2}}}}\end{array}}}$

where the second inequality is due to Cauchy-Schwarz (Hölder) inequality.

This estimate hold for all ${\displaystyle r>0}$. Therefore if we send ${\displaystyle r\to \infty }$ we see that for all ${\displaystyle x_{0}\in \mathbb {R} ^{2},}$${\displaystyle |v(x_{0})|=|\nabla u(x_{0})|=0}$ which gives us that ${\displaystyle u}$ is constant.

## Problem 2

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

### 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 ${\displaystyle C}$. Multiply the PDE by ${\displaystyle v}$, a smooth test function with compact support in ${\displaystyle \mathbb {R} \times (0,\infty )}$. Then by an integration by parts:

{\displaystyle {\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 ${\displaystyle V_{l}}$ denote the open region in ${\displaystyle \mathbb {R} \times (0,\infty )}$ to the left of ${\displaystyle C}$ and similarly ${\displaystyle V_{r}}$ denotes the region to the right of ${\displaystyle C}$. If the support of ${\displaystyle v}$ lies entirely in either of these two regions, then all of the above boundary terms vanish and we get ${\displaystyle 0=\int _{0}^{\infty }\int _{-\infty }^{\infty }(u_{t}+f(u)_{x})v\,dx\,dt=\int _{0}^{\infty }\int _{-\infty }^{\infty }uv_{t}+f(u)v_{x}\,dx\,dt.}$

Now suppose the support of ${\displaystyle v}$ intersects the discontinuity ${\displaystyle C}$.

#### b)

We can calculate ${\displaystyle \sigma ={\frac {F(u_{l})-F(u_{r})}{u_{l}-u_{r}}}={\frac {1^{2}+1-(4-2)}{1+2}}=0}$. Therefore, the shock wave extends vertically from the origin. That is,

{\displaystyle u(x,t)=\left\{{\begin{aligned}1&x<0\\-2&x>0\end{aligned}}\right.}

## Problem 3

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

### Solution

#### 3a

Consider the energy ${\displaystyle E(t)={\frac {1}{2}}\int _{0}^{1}u_{t}^{2}+u_{x}^{2}\,dx}$. Then ${\displaystyle {\dot {E}}(t)=\int _{0}^{1}u_{t}u_{tt}+u_{x}u_{xt}\,dx}$. Integrate by parts to get ${\displaystyle {\dot {E}}(t)=\int _{0}^{1}u_{t}u_{tt}-u_{xx}u_{t}\,dx+\left.u_{t}u_{x}\right|_{0}^{1}}$. The boundary terms vanish since ${\displaystyle u(0,t)=0}$ implies ${\displaystyle u_{t}(0,t)=0}$ (similarly at ${\displaystyle x=1}$). Then by the original PDE we get

{\displaystyle {\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, ${\displaystyle {\dot {E}}(t)<0}$ for all ${\displaystyle t}$; that is, energy is dissipated.

#### 3b

Suppose ${\displaystyle u,v}$ are two distinct solutions to the system. Then ${\displaystyle w=u-v}$ is a solution to

${\displaystyle w_{tt}-w_{xx}+w_{t}=(w_{xt}^{3})_{x},\quad -\infty 0}$

${\displaystyle w(x,0)=0,w_{t}(x,0)=0,w(0,t)=w(1,t)=0.}$

This tells us that at ${\displaystyle t=0}$, ${\displaystyle w_{x}=w_{t}=0}$. Therefore, ${\displaystyle E(0)=0}$. Since ${\displaystyle E(t)\leq E(0)}$ then ${\displaystyle E(t)=0}$ for all ${\displaystyle t}$. This implies ${\displaystyle w\equiv 0}$. That is, ${\displaystyle u=v}$.

## Problem 4

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

### Solution

Suppose ${\displaystyle u,v}$ are two distinct solutions. Then ${\displaystyle w=u-v}$ is a smooth solution to

${\displaystyle \left\{{\begin{array}{ll}w_{t}-\nabla \cdot (a(x)\nabla w)+b(x)u=0,&x\in \Omega ,t>0\\w(x,0)=0,&x\in \Omega \\w_{t}+\partial w/\partial n+w=0,&x\in \partial \Omega ,t>0\end{array}}\right.}$

Consider the energy ${\displaystyle E(t)={\frac {1}{2}}\int _{\Omega }w^{2}\,dx+{\frac {1}{2}}\int _{\partial \Omega }a(x)w^{2}\,dS}$. It is easy to verify that ${\displaystyle E(0)=0}$. Then

{\displaystyle {\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 ${\displaystyle {\dot {E}}(t)\leq \|b\|_{L^{\infty }(\Omega )}E(t)}$ implies ${\displaystyle E(t)\leq E(0)e^{\|b\|_{L^{\infty }(\Omega )}t}=0}$ for all ${\displaystyle t}$. Thus, ${\displaystyle E(t)=0}$ for all ${\displaystyle t}$ which implies ${\displaystyle w\equiv 0}$

## Problem 5

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

### Solution

${\displaystyle (\Rightarrow )}$ Suppose ${\displaystyle u}$ minimizes ${\displaystyle I}$, i.e. ${\displaystyle I[u]\leq I[w]\,\forall w\in K}$. Then for any fixed ${\displaystyle w\in K}$, if we let ${\displaystyle g(t)=(1-t)u+tv}$ then ${\displaystyle I[g(0)]\leq I[g(t)]\,\forall 0\leq t\leq 1}$. Let ${\displaystyle i(t)=I[g(t)]}$; then we can say that ${\displaystyle i'(0)\geq 0}$. Now we must compute ${\displaystyle i'(0)}$. We have

${\displaystyle i(t)={\frac {1}{2}}\int _{\Omega }\left|(1-t)\nabla u+t\nabla v\right|^{2}-f((1-t)u+tv)\,dx}$

${\displaystyle i'(t)=\int _{\Omega }-(1-t)|\nabla u|^{2}+(1-2t)\nabla u\cdot \nabla v+t|\nabla v|^{2}-(1-t)fu-tfv\,dx}$

${\displaystyle i'(0)=\int _{\Omega }-|\nabla u|^{2}+\nabla u\cdot \nabla v-fu\,dx}$

Since we know ${\displaystyle i'(0)\geq 0}$ then

${\displaystyle \int _{\Omega }f(u-v)\,dx\geq \int _{\Omega }|\nabla u|^{2}-\nabla u\cdot \nabla v=\int _{\Omega }\nabla u\cdot \nabla (u-v)\,dx}$ as desired.

${\displaystyle (\Leftarrow )}$ Conversely suppose

${\displaystyle \int _{\Omega }\nabla u\cdot \nabla (u-v)\,dx=\int _{\Omega }|\nabla |^{2}-\nabla u\cdot \nabla v\,dx\leq \int _{\Omega }f(u-v)\,dx.}$

Then ${\displaystyle {\begin{array}{lll}\int _{\Omega }|\nabla u|^{2}-fu\,dx&\leq &\int _{\Omega }\nabla u\cdot \nabla v-fv\,dx\\&\leq &\int _{\Omega }|\nabla u||\nabla v|-fv\,dx\\&\leq &\int _{\Omega }1/2|\nabla u|^{2}+1/2|\nabla v|^{2}-fv\,dx\end{array}}}$

Therefore, ${\displaystyle 1/2\int _{\Omega }|\nabla u|^{2}-fu\,dx\leq 1/2\int _{\Omega }|\nabla v|^{2}-fv\,dx}$ for all ${\displaystyle v\in K}$. That is, ${\displaystyle I[u]\leq I[v]}$ for all ${\displaystyle v\in K}$, as desired.