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Let $\Omega$ be a bounded open subset of $\mathbb{R^3}$, and $f$ be in $L^2(\Omega)$. Does there exist a weak solution in $W^{1,2}_0(\Omega)$ to the following equation:

\begin{cases} \Delta u+\dfrac{1}{1+u^{2}}=f & \mathrm{in}\;\Omega\\ u=0 & \mathrm{on\;}\partial\Omega \end{cases}

Help me with some hints to start.

Thanks in advanced.

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up vote 4 down vote accepted

In fact the idea of guacho works:

Let $\varphi(v)\in W^{1,2}_0(\Omega)$ be a weak solution of $$ \Delta u+\frac{1}{1+v^2}=f. $$ Such a weak solution exists as this means for $\varphi(v)$ that $$ -\int_\Omega \nabla\varphi(v)\cdot\nabla w\,dx= \int_{\Omega}\left(f-\frac{1}{1+v^2}\right)w\,dx \quad\text{for all $w\in W^{1,2}_0(\Omega)$}. $$ And since $\ell(w)=-\int_{\Omega}\left(f-\frac{1}{1+v^2}\right)w\,dx$ is a bounded linear functional on $W^{1,2}_0(\Omega)$, then there exists a $\varphi(v)\in W^{1,2}_0(\Omega)$, such that $$ \ell(w)=\langle w,\varphi(v)\rangle_{W^{1,2}_0(\Omega)}=\int_\Omega \nabla\varphi(v)\cdot\nabla w\,dx. $$ Don't forget that $W^{1,2}_0(\Omega)$ is a Hilbert space with inner product $\langle w,w'\rangle_{W^{1,2}_0(\Omega)}=\int_\Omega \nabla w\cdot\nabla w'\,dx$.

Also, $\|\varphi(v)\|_{W^{1,2}_0(\Omega)}=\|\ell\|\le \|f\|_{L^2}+\|1\|_{L^2}=M.$ Hence the nonlinear functional $\varphi$ maps $L^2(\Omega)$ into $$ B=\{w\in {W^{1,2}_0(\Omega)}: \|w\|_{{W^{1,2}_0(\Omega)}}\le M\}. $$ Now $B\subset \{u\in L^2(\Omega) : \|u\|_{L^2}\le N\}$, for some $N>0$, due to Poincaré inequality. In particular, $\varphi$ maps $B$ into $B$, and $\varphi[B]\subset B$. But $B$ is compact subset of $L^2(\Omega)$, due to Rellich compactness theorem. Hence, Schauder fixed point theorem guarantees a fixed point $u$ for $\varphi$. Clearly $u\in L^2(\Omega)$, but $u=\varphi(u)\in B\subset W^{1,2}_0(\Omega)$.

Ὅπερ ἔδει δεῖξαι (=quod erat demostrantum).

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Thank you very much – chuyenvien94 Jan 22 '14 at 1:04
I deleted because you said that it was wrong... and I'm too lazy to double check :-) – guacho Jan 22 '14 at 1:20

Let's try this: If you consider $$ \Delta u + \frac{1}{1+v^2}=f $$ for $v\in W^{1,2}_0$ you can apply the usual Lax-Milgram theorem for the weak formulation. Then notice that the bound for $u$ doesn't depend in $v$. Thus there exists a weak limit in $W^{1,2}_0$. Now use the compactness. Everything should be right with this approach, isn't it?

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