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.


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).

  • $\begingroup$ I deleted because you said that it was wrong... and I'm too lazy to double check :-) $\endgroup$ – 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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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