# Showing existence of positive weak solution of a PDE by CoV

Given the following PDE $$\begin{cases} -\Delta u+\alpha=u^q &x\in\Omega\\ u=0 &x\in\partial\Omega \end{cases}$$ where $\Omega\subset\mathbb R^3$ is open bounded with smooth boundary, $1<q<5$, and $\alpha>0$ is a constant.

I am trying to show this PDE has at least a positive weak solution. Here is what I tried so far:

Define the energy functional $$E[u]=\frac{1}{2}\int_\Omega |\nabla u|^2dx +\alpha \int_\Omega |u|\,dx$$ and the admissible set $$\mathcal A:=\{u\in H_0^1(\Omega),\,\|u\|_{L^{q+1}(\Omega)}=1 \}$$ Hence by directly method we have a minimizer $u$ over admissible set $\mathcal A$. Also, notice that $E[u]=E[|u|]$ and hence we could assume the minimizer $u$ is non-negative. Therefore, by Lagrange Multiplier we have, for some $k\in\mathbb R$, $$\int_\Omega\nabla u\nabla v\,dx+\alpha\int_\Omega \frac{u}{u}v\,dx=k\int_\Omega u^qv\,dx$$ for all $v\in H_0^1(\Omega)$.

Now the last step is to show that $u\neq 0$ a.e. and hence the term $$\alpha\int_\Omega \frac{u}{u}v\,dx\tag 1$$ is well-judged. However, I got stuck here... Any help is really welcome!

PS: I have another way to obtain a positive (classical) solution of this PDE but it takes too many pages but I don't want my paper to contain too many tactical details about this part...

• Why is this the last step? You would get that $-\Delta u+\alpha=ku^q$ weakly, which doesn't seem to be exactly what you want. – JLA Jan 25 '15 at 23:51
• @JLA the problem is that if $u=0$ on a positive set, then term $(1)$ does not make sense right? – JumpJump Jan 25 '15 at 23:53
• @JLA weak solution is enough, since we could use boot-strap to get classical solution – JumpJump Jan 25 '15 at 23:53
• Sure, but I thought you wanted a solution of $-\Delta u+\alpha=u^q\,,$ not $-\Delta u+\alpha=ku^q$ for some $k\,.$ – JLA Jan 25 '15 at 23:54
• @JLA Oh I see. Yes, but that $k$ does not matter for me now. All I need is a weak solution. – JumpJump Jan 25 '15 at 23:55