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.

enter image description here

Approach : Find energy functional : $\int_\Omega \nabla u \nabla v \int_\Omega |u| u.v -\int_\Omega fu=0 \forall u \in H_0^1(\Omega) \cap L^3(\Omega)$

$\implies E(u)=\int_\Omega\frac{1}{2} |\nabla u|^2+\frac{1}{3} |u|^3 -fu dx$

If $u$ solves $\min_{u \in A} E(u)$ with $A={}u\in H_0^1( \Omega)\cap L^3(\Omega)$ then for any $v\in A$ we have

$0=\frac{d}{d\epsilon}E(u+\epsilon v)=\int\frac {d}{d\epsilon}|\nabla(u+\epsilon v)|^2 +\frac{1}{3} |u+\epsilon v|^3 -f(u+\epsilon v)dx$ $$=\int(\nabla u. \nabla v +|u|v -fv) dx$$

Remark: There seems to be a problem while differentiating with respect to $\epsilon$ the term $|u+\epsilon v|^3$ , how do i resolve it ?

Next : Can i say that $u$ is a weak solution now ? If i can then i proceed further this way , $A$ is not a null set , then $\exists (u_k)_{k\in \mathbb N} \subset A$

$$\lim_{k\to\infty}u_k=\inf_{u\in A} E(u)$$

Assume $u_k$ is not bounded in $H_0^1 $ $$E(u_k)\ge \int_\Omega \frac {|\nabla u_k|^2}{2}-fu_k dx \ge \frac{1}{2} ||\nabla u||_{L^2}-C ||f||_{L^2} ||\nabla u_k||_{L^2(\Omega)}$$

I am stuck now , How do i proceed further ? Thanks

share|improve this question
add comment

1 Answer

Concerning the differentiation of $\epsilon\mapsto |u+\epsilon v|^3$. The function $g(x)=|x|^3$ is in $C^2(\mathbb R)$ with $g\,'(x)=3x|x|$ and $g\,''(x)=6|x|$. Notably, $g$ is convex, which makes the entire functional $E$ convex; we need this for its lower semicontinuity with respect to the weak convergence in $A$.

But to begin with, we want to show that $E$ is bounded from below on $A$; otherwise the entire approach is doomed. The only question is what to do with $\int-fu$: suppressing the urge to use Holder's inequality, I would write $-fu\ge -\frac12f^2-\frac12 u^2$. Here the integral of $f^2$ is a finite constant and $u^2\le \frac12|u|^3+8$ pointwise. It follows that $E$ is bounded from below: $E(u)\ge \int (\frac12|\nabla u|^2+\frac{1}{12}|u|^3-\frac12f^2-8)$ or something of the kind. Note that I bounded $u^2$ by a small multiple of $|u|^3$ in order to have some of the cubic term left over. This will be useful in a moment.

Pick a minimizing sequence $u_k\in A$ as you've done already. The lower bound for $E(u)$ in the previous paragraph tells us that $\int|\nabla u_k|^2$ and $\int|u_k|^3$ are uniformly bounded. Thus, the sequences is bounded in $H_0^1$ and in $L^3$. Pick a weakly convergent subsequence in $H_0^1$; extract from it a weakly convergent subsequence in $L^3$. The lower semicontinuity of $E$ implies $E(u)=\min_A E$ where $u$ is the weak limit.

And yes, having $\int (\nabla u\cdot\nabla v +\frac13 u|u|v-fv)=0$ for all test functions $v$ (you only need to consider $C^\infty$-smooth $v$ with compact support) means that $u$ is a weak solution; this is what weak solution means. The boundary condition $u=0$ is automatically fulfilled since we work in $H_0^1$ all the time.

share|improve this answer
    
Sir , what is meant by lower semicontinuous functional ? Is there a relation with convexity of the functional ? –  Theorem Jul 25 '12 at 22:28
    
@Theorem It means a functional that is lower semicontinuous. And yes, there is a relation, –  user31373 Jul 26 '12 at 0:08
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.