Let the problem, where $\Omega$ is an open set of $\mathbb{R}^3$ and $h_1$ and $h_2$ are regular given functions
\begin{equation}\nonumber%\label{eq:Pe}\tag{$P_{\varepsilon}$}
\left\{
\begin{array}[c]{ll}
\Delta^2\chi = 0, & \text{in } \Omega, \\
\chi= h_1, & \text{on }\partial\Omega\\
\Delta\chi = h_2, & \text{on }\partial\Omega
\end{array}
\right.
\end{equation}
My point is, this problem have solution?
What can I say about $$\|\chi\|_{\infty} \;\; \text{in }\;\; \Omega?$$
Thanks for any help!
Best Answer
The idea is to proceed by steps, considering equations of second order and then making a "substitution".
Let $\theta$ be the solution of $$ \left\{ \begin{array}{lr} - \Delta \theta = 0 & \quad \text{ in } \Omega \\ \theta = h_2 & \quad \text{ on } \partial \Omega \end{array} \right. $$
It exists by Perron's method, see section 2.7 in Gilbarg-Trudinger. In particular, Theorem 2.14. It is smooth, since it is harmonic, and attains the minimum and maximum on the boundary.
Now consider the problem $$ \Delta \chi = \theta \text{ in } \Omega, \qquad \chi = h_1 \text{ on } \partial \Omega. $$ This problem has a $C^\infty$ solution, by Theorem 8.9 and Corollary 8.11 in Gilbarg-Trudinger, which can be given explicitly by Green's representation formula (Evans, pages 34-35). As for the bound, I leave it to you to analise the Green's formula. I have the impression that such a bound indeed appears. In fact, as the solution is $C^{\infty}(\Omega)$ and bounded on $\partial \Omega$, I think that this immediately implies that $u \in L^\infty(\Omega)$.