UMD PDE Qualifying Exams/Jan2005PDE

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Problem 1[edit]

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.


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}^2we have

|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}}

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[edit]

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




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.

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.


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{align} 1& x<0\\ -2 & x>0 \end{align}\right.

Problem 3[edit]

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.



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

\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.

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.


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[edit]

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):

u_t - \nabla \cdot (a(x)\nabla u) + b(x) u = q,& x\in\Omega,t>0\\
u_t+\partial u/\partial n +u =0,& x\in \partial\Omega,t>0

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.


Suppose u,v are two distinct solutions. Then w=u-v is a smooth solution to

w_t - \nabla \cdot (a(x)\nabla w) + b(x) u = 0,& x\in\Omega,t>0\\
w_t+\partial w/\partial n +w =0,& x\in \partial\Omega,t>0

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

\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

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[edit]

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.


(\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)\geq0. 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

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.