UMD PDE Qualifying Exams/Jan2005PDE

Problem 1

Let $u(x,y)$ be a harmonic function on ${ℝ}^2$ and suppose that $\int \underset{{ℝ}^2}{\int }|\mathrm{\nabla }u{|}^2(x,y)dxdy<\mathrm{\infty }.$ Show that $u$ is a constant function.

Solution

Let $C=\int \underset{{ℝ}^2}{\int }|\mathrm{\nabla }u{|}^2(x,y)dxdy<\mathrm{\infty }.$

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

Then by the mean value property of subharmonic functions, for any $x_0\in {ℝ}^2$we have

$\begin{array}{ccc}|v(x_0)|& \le & \frac{1}{\pi r^2}\underset{B(x_0,r)}{\int }|v|dxdy\\ & \le & \frac{1}{\pi r^2}{\left(\underset{B(x_0,r)}{\int }1dxdy\right)}^{1/2}{(\underset{B(x_0,r)}{\int }|v{|}^{2}dxdy)}^{1/2}\\ & \le & \frac{C}{\sqrt{\pi r^2}}\end{array}$

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

This estimate hold for all $r>0$. Therefore if we send $r\to \mathrm{\infty }$ we see that for all $x_0\in {ℝ}^2,$$|v(x_0)|=|\mathrm{\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 $u_t+f(u{)}_x=0,-\mathrm{\infty }<x<\mathrm{\infty },t>0.$ a) Derive the Rankine-Hugoniot conditions at a discontinuity of the solution. b)Find a piecewise smooth solution to the IVP $u_t+(u^2+u{)}_x=0,-\mathrm{\infty }<x<\mathrm{\infty },t>0$ $u(x,0)=\{\begin{array}{cc}1,& x<0\\ -2& x>0.\end{array}$

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 $ℝ\times (0,\mathrm{\infty })$. Then by an integration by parts:

$\begin{array}{cc}0=& \underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(u_t+f(u{)}_x)vdxdt\\ =& -\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}u{v}_{t}dxdt+{\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}vudx|}_{t=0}^{t=\mathrm{\infty }}-\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(u)v_{x}dxdt+{\underset{0}{\overset{\mathrm{\infty }}{\int }}f(u)vdt|}_{x=-\mathrm{\infty }}^{x=\mathrm{\infty }}\\ =& -\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}u{v}_{t}dxdt-\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}vu(x,0)dx-\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(u)v_{x}dxdt.\end{array}$

Let $V_l$ denote the open region in $ℝ\times (0,\mathrm{\infty })$ 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 $0=\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(u_t+f(u{)}_x)vdxdt=\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}u{v}_t+f(u)v_{x}dxdt.$

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,

$u(x,t)=\{\begin{array}{cc}1& x<0\\ -2& x>0\end{array}$

Problem 3

Consider the evolution equation with initial data $u_{tt}-u_{xx}+u_t=(u_{xt}^3{)}_x,-\mathrm{\infty }<x<\mathrm{\infty },t>0$ $u(x,0)=f(x),u_t(x,0)=g(x),u(0,t)=u(1,t)=0.$ 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 $E(t)=\frac{1}{2}\underset{0}{\overset{1}{\int }}{u}_t^2+u_x^{2}dx$. Then $\dot{E}(t)=\underset{0}{\overset{1}{\int }}{u}_t{u}_{tt}+u_x{u}_{xt}dx$. Integrate by parts to get $\dot{E}(t)=\underset{0}{\overset{1}{\int }}{u}_t{u}_{tt}-u_{xx}{u}_{t}dx+{u_t{u}_x|}_0^1$. 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

$\begin{array}{ccc}\dot{E}(t)& =& \underset{0}{\overset{1}{\int }}{u}_t((u_{xt}^3{)}_x-u_t)dx\\ & =& \underset{0}{\overset{1}{\int }}{u}_t(u_{xt}^3{)}_x-u_t^2)dx\\ & =& \underset{0}{\overset{1}{\int }}-u_{xt}^4-u_t^{2}dx+{u_{xt}^3{u}_t|}_0^1.\end{array}$

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

$w_{tt}-w_{xx}+w_t=(w_{xt}^3{)}_x,-\mathrm{\infty }<x<\mathrm{\infty },t>0$

$w(x,0)=0,w_t(x,0)=0,w(0,t)=w(1,t)=0.$

This tells us that at $t=0$, $w_x=w_t=0$. Therefore, $E(0)=0$. Since $E(t)\le E(0)$ then $E(t)=0$ for all $t$. This implies $w\equiv 0$. That is, $u=v$.

Problem 4

Let $\mathrm{\Omega }\subseteq {ℝ}^n$ be a bounded open set with smooth boundary $\partial \mathrm{\Omega }$. Consider the initial boundary value problem for $u(x,t)$: $\{\begin{array}{cc}{u}_t-\mathrm{\nabla }\cdot (a(x)\mathrm{\nabla }u)+b(x)u=q,& x\in \mathrm{\Omega },t>0\\ u(x,0)=f(x),& x\in \mathrm{\Omega }\\ u_t+\partial u/\partial n+u=0,& x\in \partial \mathrm{\Omega },t>0\end{array}$ where $\partial u/\partial n$ is the exterior normal derivative. Assume that $a,b\in C^1(\overline{\mathrm{\Omega }})$ and that $a(x)\ge 0$ for $x\in \mathrm{\Omega }$. 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

$\{\begin{array}{cc}{w}_t-\mathrm{\nabla }\cdot (a(x)\mathrm{\nabla }w)+b(x)w=0,& x\in \mathrm{\Omega },t>0\\ w(x,0)=0,& x\in \mathrm{\Omega }\\ w_t+\partial w/\partial n+w=0,& x\in \partial \mathrm{\Omega },t>0\end{array}$

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

$\begin{array}{cc}\dot{E}(t)=& \underset{\mathrm{\Omega }}{\int }w{w}_{t}dx+\underset{\partial \mathrm{\Omega }}{\int }a(x)w{w}_{t}dS\\ =& \underset{\mathrm{\Omega }}{\int }w\mathrm{\nabla }\cdot (a(x)\mathrm{\nabla }w)-b(x)w^{2}dx+\underset{\partial \mathrm{\Omega }}{\int }a(x)w{w}_{t}dS\\ =& \underset{\mathrm{\Omega }}{\int }-\mathrm{\nabla }w\cdot (a(x)\mathrm{\nabla }w)-b(x)w^{2}dx+\underset{\partial \mathrm{\Omega }}{\int }wa(x)\frac{\partial w}{\partial n}dS+\underset{\partial \mathrm{\Omega }}{\int }a(x)w{w}_{t}dS\\ =& \underset{\mathrm{\Omega }}{\int }-a(x)\mathrm{\nabla }w\cdot \mathrm{\nabla }w-b(x)w^{2}dx+\underset{\partial \mathrm{\Omega }}{\int }-a(x)w^2-a(x)w{w}_{t}dS+\underset{\partial \mathrm{\Omega }}{\int }a(x)w{w}_{t}dS\\ =& \underset{\mathrm{\Omega }}{\int }-a(x)\mathrm{\nabla }w\cdot \mathrm{\nabla }w-b(x)w^{2}dx+\underset{\partial \mathrm{\Omega }}{\int }-a(x)w^{2}dS\\ \le & \underset{\mathrm{\Omega }}{\int }-b(x)w^{2}dx\\ \le & \Vert b{\Vert }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}\underset{\mathrm{\Omega }}{\int }{w}^{2}dx\end{array}$

Therefore $\dot{E}(t)\le \Vert b{\Vert }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}E(t)$ implies $E(t)\le E(0)e^{\Vert b{\Vert }_{L^{\mathrm{\infty }}(\mathrm{\Omega })}t}=0$ for all $t$. Thus, $E(t)=0$ for all $t$ which implies $w\equiv 0$

Problem 5

Let $\mathrm{\Omega }\subset {ℝ}^n$ be a bounded open set with smooth boundary. Let $K=\{u\in H_0^1(\mathrm{\Omega }):u\ge 0\text{ in }\mathrm{\Omega }\}$. Let $f\in L^2(\mathrm{\Omega })$ and define the functional $I(u)=\frac{1}{2}\underset{\mathrm{\Omega }}{\int }|\mathrm{\nabla }u{|}^{2}dx-\underset{\mathrm{\Omega }}{\int }fudx$. Show that $u$ is a minimizer of $I$ over $K$ if and only if $u$ satisfies the variational inequality $\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }u\cdot \mathrm{\nabla }(u-v)dx\le \underset{\mathrm{\Omega }}{\int }f(u-v)dx$ for all $v\in K$.

Solution

$(\Rightarrow )$ Suppose $u$ minimizes $I$, i.e. $I[u]\le I[w]\mathrm{\forall }w\in K$. Then for any fixed $w\in K$, if we let $g(t)=(1-t)u+tv$ then $I[g(0)]\le I[g(t)]\mathrm{\forall }0\le t\le 1$. Let $i(t)=I[g(t)]$; then we can say that $i^{\prime }(0)\ge 0$. Now we must compute $i^{\prime }(0)$. We have

$i(t)=\frac{1}{2}\underset{\mathrm{\Omega }}{\int }{|(1-t)\mathrm{\nabla }u+t\mathrm{\nabla }v|}^2-f((1-t)u+tv)dx$

$i^{\prime }(t)=\underset{\mathrm{\Omega }}{\int }-(1-t)|\mathrm{\nabla }u{|}^2+(1-2t)\mathrm{\nabla }u\cdot \mathrm{\nabla }v+t|\mathrm{\nabla }v{|}^2-(1-t)fu-tfvdx$

$i^{\prime }(0)=\underset{\mathrm{\Omega }}{\int }-|\mathrm{\nabla }u{|}^2+\mathrm{\nabla }u\cdot \mathrm{\nabla }v-fudx$

Since we know $i^{\prime }(0)\ge 0$ then

$\underset{\mathrm{\Omega }}{\int }f(u-v)dx\ge \underset{\mathrm{\Omega }}{\int }|\mathrm{\nabla }u{|}^2-\mathrm{\nabla }u\cdot \mathrm{\nabla }v=\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }u\cdot \mathrm{\nabla }(u-v)dx$ as desired.

$(\Leftarrow )$ Conversely suppose

$\underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }u\cdot \mathrm{\nabla }(u-v)dx=\underset{\mathrm{\Omega }}{\int }|\mathrm{\nabla }{|}^2-\mathrm{\nabla }u\cdot \mathrm{\nabla }vdx\le \underset{\mathrm{\Omega }}{\int }f(u-v)dx.$

Then $\begin{array}{ccc}\underset{\mathrm{\Omega }}{\int }|\mathrm{\nabla }u{|}^2-fudx& \le & \underset{\mathrm{\Omega }}{\int }\mathrm{\nabla }u\cdot \mathrm{\nabla }v-fvdx\\ & \le & \underset{\mathrm{\Omega }}{\int }|\mathrm{\nabla }u||\mathrm{\nabla }v|-fvdx\\ & \le & \underset{\mathrm{\Omega }}{\int }1/2|\mathrm{\nabla }u{|}^2+1/2|\mathrm{\nabla }v{|}^2-fvdx\end{array}$

Therefore, $1/2\underset{\mathrm{\Omega }}{\int }|\mathrm{\nabla }u{|}^2-fudx\le 1/2\underset{\mathrm{\Omega }}{\int }|\mathrm{\nabla }v{|}^2-fvdx$ for all $v\in K$. That is, $I[u]\le I[v]$ for all $v\in K$, as desired.