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

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:

$\begin{align} 0 =& \int_0^\infty \int_{-\infty}^\infty (u_t + f(u)_x) v\,dx\,dt \\ =&- \int_0^\infty \int_{-\infty}^\infty u v_t\,dx\,dt +\left.\int_{-\infty}^\infty v u \,dx \right|_{t=0}^{t=\infty} - \int_0^\infty \int_{-\infty}^\infty f(u)v_x\,dx\,dt + \left.\int_0^\infty f(u) v\,dt \right|_{x=-\infty}^{x=\infty}\\ =&- \int_0^\infty \int_{-\infty}^\infty u v_t\,dx\,dt -\int_{-\infty}^\infty v u(x,0) \,dx - \int_0^\infty \int_{-\infty}^\infty f(u)v_x\,dx\,dt. \end{align}$

Let $V_l$ denote the open region in $\mathbb{R}\times(0,\infty)$ to the left of $C$ and simlarly $V_r$ 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 $C$.

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 $\dot{E}(t)=\int_0^1 u_tu_{tt}+u_xu_{xt}\,dx$. Integrate by parts to get $\dot{E}(t)=\int_0^1 u_tu_{tt}-u_{xx}u_{t}\,dx + \left. u_tu_x\right|_0^1$. 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

$\begin{align} \dot{E}(t)&=&\int_0^1 u_t((u_{xt}^3)_x-u_t)\,dx\\ &=&\int_0^1 u_t(u_{xt}^3)_x-u_t^2)\,dx\\ &=&\int_0^1 -u_{xt}^4 -u_t^2\,dx + \left. u_{xt}^3u_t\right|_0^1. \end{align}$

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

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

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

$\begin{align} \dot E(t) =& \int_\Omega w w_t\,dx +\int_{\partial\Omega} a(x) w w_t\,dS\\ =& \int_\Omega w \nabla\cdot(a(x)\nabla w) - b(x) w^2\, dx + \int_{\partial\Omega} a(x) w w_t\,dS\\ =& \int_\Omega -\nabla w \cdot (a(x)\nabla w) - b(x) w^2\, dx + \int_{\partial\Omega} w a(x) \frac{\partial w}{\partial n} \, dS+ \int_{\partial\Omega} a(x) w w_t\,dS\\ =& \int_\Omega -a(x) \nabla w \cdot \nabla w - b(x) w^2\, dx + \int_{\partial\Omega} -a(x) w^2 - a(x) w w_t \, dS+ \int_{\partial\Omega} a(x) w w_t\,dS\\ =& \int_\Omega -a(x) \nabla w \cdot \nabla w - b(x) w^2\, dx + \int_{\partial\Omega} -a(x) w^2 \, dS\\ \leq & \int_\Omega - b(x) w^2\,dx \\ \leq& \| b\|_{L^\infty(\Omega)} \int_\Omega w^2\,dx \end{align}$

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 $w \equiv 0$