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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

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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.

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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

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