UMD PDE Qualifying Exams/Jan2005PDE

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

$$ \begin{array}{lll} &\leq& \frac{1}{\pi r^2} \left( \int_{B(x_0,r)} 1\,dx\,dy \right)^{1/2} \left( \int_{B(x_0,r)} |v|^2\,dx\,dy\right)^{1/2}\\ &\leq & \frac{C}{\sqrt{\pi r^2}} \end{array} $$
 * v(x_0)|&\leq& \frac{1}{\pi r^2} \int_{B(x_0,r)} |v|\,dx\,dy\\

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.

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:

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

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

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

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

$$w_{tt}-w_{xx}+w_t=(w_{xt}^3)_x,\quad -\infty0$$

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

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, $$u=v$$.

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

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

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 $$w \equiv 0$$

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 $$i'(0)\geq0$$. 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 $$\begin{array}{lll} \int_\Omega |\nabla u|^2 -fu\,dx &\leq& \int_\Omega \nabla u \cdot \nabla v -fv\,dx \\ &\leq & \int_\Omega |\nabla u||\nabla v| -fv\,dx \\ & \leq & \int_\Omega 1/2|\nabla u|^2+1/2|\nabla v|^2 -fv\,dx \end{array}$$

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.