# Minimizing a norm to get a solution of a pde

Let $\Omega$ be a regular bounded open subset of $\mathbb{R}^3$. The problem is to solve the following pde:

$$\left\{\begin{array}{c c}-\Delta u = u^3 & (\Omega)\\u = 0 &(\partial\Omega)\end{array}\right.$$

Here are the questions (this is not homework but last year's exam):

1. Prove the existence of a solution for (ie. a $v$ such that it achieves the minimum): $$\inf \;\left\{ \int_\Omega |\nabla v|^2 : v\in H^1_0(\Omega), \int v^4 =1 \right\}$$
2. Prove that if $v$ solves $(1)$ then there is a $\lambda > 0$ such that $-\Delta v = \lambda v^3$.
3. Conclude.

I was able to prove 1 by considering a sequence $(v_n)$ such that $\|\nabla v\|^2$ converges towards the $\inf$, obtain a weak limit by Banach-Alaoglu, Rellich-Kondrakov to get a limit in $L^2$ and Fatou's lemma to conclude. And if I were able to prove 2, I would have an answer for 3 by simply scaling $v$. But I can't figure out how to prove 2. It sounds an awful lot like Stampacchia's theorem, but unfortunately the unit sphere of $L^4$ isn't convex, and even if it were I don't see how the $\lambda$ would appear or how it would relate to $v^3$; besides Stampacchia's theorem applies only for bilinear forms. How should I go about proving this?

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@Davide: Could you elaborate? There seem to be many Green's formulas (and I don't know any of them). –  nik May 19 '12 at 9:16
@Davide: Ah, I see, thanks. But how would I prove that the equality holds? –  nik May 19 '12 at 9:46
I think I found the answer, which was much simpler than what I expected. What should I do? –  nik May 19 '12 at 11:55
Question 2 is simply an application of the theory of Lagrange multipliers: en.wikipedia.org/wiki/Lagrange_multipliers_on_Banach_spaces –  Siminore May 19 '12 at 12:31

Define $J(u) = \int |\nabla u|^2$. Then $\forall u \in X, J(u) \geq J(v)$. Now for all $\phi \in H^1_0(\Omega)$ and $\epsilon > 0$, we have that:

$$f(\epsilon,\phi)={v + \epsilon\phi \over \left(\int (v+\epsilon\phi)^4\right)^{1/4}} \in X$$

Therefore $J(f(\epsilon,\phi)) \geq J(v)$. By expanding $J(v+\epsilon\phi)$ and $\left( \int (v+\epsilon\phi)^4\right)^{1/4}$ and applying dominated convergence, we get that:

$$\epsilon^{-1}(J(f(\epsilon,\phi))-J(v)) \xrightarrow[\epsilon \rightarrow 0]{} \int(\nabla v \cdot \nabla\phi - \lambda v^3 \phi) \geq 0$$

Where $\lambda = J(v)$. By replacing $\phi$ by $-\phi$ we get the reverse inequality, and by integration by parts we get that:

$$\forall \phi \in H^1_0(\Omega), \int(-\Delta v- \lambda v^3)\phi=0$$

And therefore $v$ is a weak solution of the pde.

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