Let $\Omega$ be an open domain in $\mathbb {R^n}$ and $f \in C^{\infty}(\Omega)$ then how can we prove there is a weak solution $u \in ‎‎ W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) \cap L^{\infty}(\Omega) $ for the problem \begin{cases} ‎\Delta^2u=‎‎f & in‎\hspace{.2cm}‎ \Omega \\ u>0 & in ‎\hspace{.2cm}‎ \Omega \\ u=\Delta u =0 & on‎\hspace{.2cm}‎ \partial \Omega ‎ \end{cases}‎

my attempt:

In this case $u$ must satisfy

$$\int_{\Omega} (-\Delta u )(-\Delta \phi ) \mathrm{d}x = \int_{\Omega} f \phi \mathrm{d}x $$ for all $$\phi \in W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$$

I know that $W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$ is a Hilbert space with inner product $$\langle u,v\rangle=\int_{\Omega} \Delta u \Delta v dx $$ and I know $$‎\Delta ‎^2: W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) \to W_0^{-2,2}(\Omega) ‎$$ is coercive.

So by the Lax-Milgram Theorem I can prove existence of $u \in W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) $ .

But my question is why also we have that $$u \in L^{\infty}(\Omega)$$ $$ u>0 $$

  • $\begingroup$ Please correct me if I'm wrong but if we set $\Delta u=v$, then $v$ solves $\Delta v=f$ and $v=0$ on $\partial\Omega$. Since $f\in C_{0}^{\infty}\left( \Omega\right) $, we have $v\in C_{0}^{\infty}\left( \Omega\right) $. Now, $u$ solves $\Delta u=v$ and $u=0$ on $\partial\Omega.$ Since $v\in C_{0}^{\infty}\left( \Omega\right) $, we have $u\in C_{0}^{\infty}\left( \Omega\right) $. $\endgroup$ – user72012 Jul 11 '15 at 12:49

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