Fourth order PDE and the maximum principle.

functional-analysismaximum-principlepartial differential equations

In the problem, with $\Omega \subset \mathbb{R}^3$ a smooth bounded domain, and the given functions $h_1, h_2 \in C(\partial\Omega)$
$$
\nonumber%\label{eq:Pe}\tag{$P_{\varepsilon}$}
\begin{cases}
-\Delta \chi + \Delta^2 \chi = 0, & \text{in } \Omega, \\
\chi = h_1, \; \Delta \chi = h_2, & \text{on } \partial\Omega\\
\end{cases}
$$
which admits a unique and regular solution $\chi \in H^2(\Omega)$, by the maximum principle we have
$$\|\chi\|_\infty = \|h_1\|_\infty$$

but how can I argue $$\|\Delta\chi\|_\infty = \|h_2\|_\infty\text{?}$$

Thanks for any help!

Best Answer

Apply maximum principle on $f:=\Delta \chi$: $f$ satisfies

$$ \Delta f = f, \ \ \ f|_{\partial \Omega} = h_2.$$

If the maximum of $f$ is attained inside $\Omega$, then at the maximum

$$0 \ge \Delta f =f\Rightarrow f\le 0.$$

Similarly, if the minimum is in the interior, then $f \ge 0$.

Thus there are only several cases:

  • $ f\le 0$, $f$ not identically zero,
  • $f\ge 0$, $f$ not identically zero,
  • $f$ is identically zero.
  • Both the maximum and minimum are not attained in the interior.

In all cases we have $$\|f\|_\infty = \|h_2\|_\infty.$$

Related Question