UMD PDE Qualifying Exams/Aug2005PDE

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Problem 1[edit]

a) Letc>0 be a constant. State what is meant by a weak solution of the PDE u_t+cu_c=0 on \mathbb{R}^2.

b) Show that if f(\cdot) is a continuous function of one real variable, then f(x-ct) is a weak solution of the PDE u_t +cu_x=0 on \mathbb R^2.

Solution[edit]

Problem 4[edit]

Let U\subset \mathbb R^n be a bounded open set with smooth boundary \partial U. Let x\mapsto P(x) : \bar{U}\to \mathbb{R}^{n\times n} be a smooth family of symmetric n\times n real matrices that are uniformly positive definite. Let c\geq 0 and f(x) be smooth functions on \bar U. Define the functional

I[u]=\int_U \left( \frac{1}{2} \langle \nabla u,P(x)\nabla u\rangle +\frac{1}{2} c(x)u^2-f(x)u\right)\,dx where \langle\cdot,\cdot\rangle is the scalar product on \mathbb R^n. Suppose that u\in C^2(U)\cap C^0(\bar U) is a minimizer of this functional subject to the Dirichlet condition u=g on \partial U, with g continuous.

a) Show that u satisfies the variational equation

\int_U\langle\nabla u,P(x)\nabla v\rangle + c(x) uv = \int_U fv for any v\in V=\{v \in \operatorname{Lip}(U): \left. v\right|_{\partial U}=0\}.

b) What is the PDE satisfied by u?

c) Suppose f,g\geq 0. Show that v=\min(u,0) is an admissible test function, and use this to conclude that u\geq 0. (Hint: You may use the fact \nabla v = \chi_{\{u<0\}}\nabla u a.e.)

d) Show that there can be only one minimizer of I[u] or, equivalently, only one solution u\in C^2(U)\cap C^0(\bar U) of the corresponding PDE.


Solution[edit]

4a[edit]

For v\in V, define \phi(t) = I[u+tv]. Then since u minimizes the functional, \phi'(0)=0. We can calculate \phi' (by exploiting the symmetry of P):

\begin{align}
\phi'(t) =& \frac{d}{dt} \int_U \frac{1}{2}\langle \nabla (u+tv),P(x)\nabla(u+tv)\rangle +\frac{1}{2} c(x)(u+tv)^2 - f(x) (u+tv)\,dx\\
=&\frac{d}{dt} \int_U \frac{1}{2}\langle \nabla u,P\nabla u\rangle + t \langle \nabla u, P\nabla v\rangle + \frac{1}{2}t^2\langle \nabla v,P\nabla v\rangle + \frac{1}{2} c(x) (u+2tuv+t^2v^2)-f(x)(u+tv)\,dx\\
=& \int_U \langle \nabla u, P(x)\nabla v\rangle +  t \langle \nabla v, P(x)\nabla v\rangle +c(x) (uv+tv^2)-f(x)v\,dx
\end{align}

And so \varphi'(0)=\int_U \langle \nabla u,P(x)\nabla v\rangle + c(x)uv-f(x)v\,dx=0 which proves the result.

4b[edit]

We have

\begin{align}
\int_U fv =& \int_U\langle\nabla u,P(x)\nabla v\rangle + c(x) uv \\
=&  \int_U P(x)\nabla u \cdot \nabla v + c(x) uv\\
=& -\int_U \operatorname{div}(P(x)\nabla u)v + \int_{\partial U} P(x)\nabla u v + \int_Uc(x) uv
\end{align}

The boundary terms vanish since v\in V and we've obtained a weak form of the PDE. Thus, u is a solution to the following PDE:

\left\{ \begin{array}{rl}
-\operatorname{div}(P(x)\nabla u) +c(x)u = f & \text{ in }U\\
u=g & \text{ on }\partial U.
\end{array}\right.

4c[edit]

First we need to show that v=\min(u,0)\in V. Firstly, on \partial U, v=\min(g,0)=0 since we've assumed g\geq 0. Secondly, u, hence v, must be Lipschitz continuous since u\in C^2(U) and U is a bounded domain in \mathbb R^n (i.e. \nabla u\in C^1(U)) and so |\nabla u| must achieve a (finite) maximum in U, hence the derivative is bounded, hence u is Lipschitz. Therefore, v\in V.

This gives \int_U fv=\int_U \langle \nabla u, P(x)\nabla v\rangle + c(x)uv = \int_{\{u<0\}}\langle \nabla u, P(x)\nabla u\rangle + c(x)u^2.

But notice that since P(x) is uniformly positive definite, then \int_{\{u<0\}}\langle \nabla u, P(x)\nabla u\rangle = \int_{\{u<0\}} (\nabla u)^T P(x)(\nabla u) \geq 0.

Therefore, we have

0\leq \int_{\{u<0\}} (\nabla u)^T P(x)(\nabla u) + cu^2 = \int_{\{u<0\}} fu <0

a contradition, unless m(\{u<0\})=0, i.e. u\geq 0 a.e.


4d[edit]

Suppose u_1,u_2 are two distinct such solution. Let w=u_1-u_2. Then w is Lipschitz (since u_1,u_2 must both be) and w=0 on \partial U. Therefore, the variational equation gives

\int_U \langle \nabla w,P(x)\nabla w\rangle +c(x)w^2 = 0. Since P(x) is positive definite, this gives

\int_U c w^2 \leq0 , a contradiction unless w=0 a.e.