Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is the functional $F:H^{1}_{0}(\Omega) \longrightarrow \mathbb{R}$ given by \begin{equation} F(u) = \int_{\Omega} \langle (A_1(x)\chi_{\{u>0\}}+A_2(x)\chi_{\{u\le0\}}) \nabla u, \nabla u \rangle \, dx \end{equation} where $A_i,i=1,2$ is a matrix satisfying \begin{equation} \lambda |\xi|^2 \le \langle A_i(x) \xi,\xi \rangle \le \Lambda |\xi|^2, i=1,2. \end{equation} with $\lambda>0$ weak lower semicontinuous? I will appreciate any hint. Thank you.

share|improve this question
    
At this rate of setting bounties you are going to spend all your reputation pretty fast. More to the point: what is the dependence of $A_i$ on $x$: smooth, continuous, or just measurable? –  user31373 Aug 21 '12 at 1:08
    
Yf you get with just measurable, great. If not, you can assume $A_i$ Holder or what you need. In fact, this is a good question. What regularity on $A_i$ is possible to obtain a satisfactory answer? –  user29999 Aug 21 '12 at 1:19
    
Let's start with the simplest case: is $\int |\nabla u^+|^2$ weakly lower semicontinuous? –  user31373 Aug 21 '12 at 3:48
    
I think the answer is no in general. Let us say $\Omega$ is 2-dimensional, $A_1=1$, and $A_2=2$. Take $u$ to be nonpositive (or $u=0$ for concreteness), and $u_k=u+\phi_k$, where $\phi_k$ is an appropriately chosen positive function of height $1$, supported in a disk of radius $\varepsilon=1/k$. One can check that $u_k$ converges to $u$ weakly in $H^1$, but $F(u_k)$ stays strictly below $F(u)$. –  timur Aug 23 '12 at 4:00
    
On a more positive note, there maybe some hope for weak lower semicontinuity along minimizing sequences. The fact that $A_i$ are matrix functions complicates matters. Would it be worth studying what happens when $A_i$ are just constants? –  timur Aug 23 '12 at 4:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.