UMD PDE Qualifying Exams/Aug2005PDE

4a
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 $$\phi'(0)=\int_U \langle \nabla u,P(x)\nabla v\rangle + c(x)uv-f(x)v\,dx=0$$ which proves the result.

4b
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
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 contradiction, unless $$m(\{u<0\})=0$$, i.e. $$u\geq 0$$ a.e.

4d
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.