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This one is probably simple, but I just can't prove the result.

Suppose that $\mathop {\max }\limits_{x \in \overline \Omega } u\left( x \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega } {u^ + }\left( x \right)$ and $\mathop {\min }\limits_{x \in \overline \Omega } u\left( x \right) \geqslant \mathop {\min }\limits_{x \in \partial \Omega } {u^ - }\left( x \right)$, where $\overline \Omega $ is closure of $\Omega $ and ${\partial \Omega }$ is boundary of $\Omega $, ${u^ + } = \max \left\{ {u,0} \right\}$ and ${u^ - } = \min \left\{ {u,0} \right\}$ (notice that $\left| u \right| = {u^ + } - {u^ - }$).

How to show that $\mathop {\max }\limits_{x \in \overline \Omega } \left| {u\left( x \right)} \right| = \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right|$?

So far, I've got

$\mathop {\max }\limits_{x \in \overline \Omega } \left| {u\left( x \right)} \right| = \mathop {\max }\limits_{x \in \overline \Omega } \left( {{u^ + }\left( x \right) - {u^ - }\left( x \right)} \right) \leqslant \mathop {\max }\limits_{x \in \overline \Omega } {u^ + }\left( x \right) + \mathop {\max }\limits_{x \in \overline \Omega } \left( { - {u^ - }\left( x \right)} \right) \leqslant \mathop {\max }\limits_{x \in \overline \Omega } {u^ + }\left( x \right) - \mathop {\min }\limits_{x \in \overline \Omega } {u^ - }\left( x \right)$

and

$\mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right| = \mathop {\max }\limits_{x \in \partial \Omega } \left( {{u^ + }\left( x \right) - {u^ - }\left( x \right)} \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega } {u^ + }\left( x \right) - \mathop {\min }\limits_{x \in \partial \Omega } {u^ - }\left( x \right)$

Edit: $u \in {C^2}\left( \Omega \right) \cap C\left( {\overline \Omega } \right)$, although I don't see how that helps. $Lu=0$, which gives us $\mathop {\max }\limits_{x \in \overline \Omega } u\left( x \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega } {u^ + }\left( x \right)$ and $\mathop {\min }\limits_{x \in \overline \Omega } u\left( x \right) \geqslant \mathop {\min }\limits_{x \in \partial \Omega } {u^ - }\left( x \right)$.

(Renardy, Rogers, An introduction to partial differential equations, p 103)

Edit 2: Come on, this should be super easy, the author didn't even comment on how the equality follows from those two.

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Where does $u$ live? Assume $u \in C^2(\Omega)\cap C(\bar{\Omega})$? Also, are we assuming anything for $Lu$? –  Euler....IS_ALIVE Feb 2 '13 at 1:55

1 Answer 1

up vote 2 down vote accepted

$\left. \begin{gathered} \mathop {\max }\limits_{x \in \overline \Omega } u\left( x \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega } {u^ + }\left( x \right)\mathop \leqslant \limits^{{u^ + } \leqslant \left| u \right|} \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right| \\ \mathop {\max }\limits_{x \in \overline \Omega } \left( { - u\left( x \right)} \right) \leqslant \mathop {\max }\limits_{x \in \partial \Omega } - {u^ - }\left( x \right)\mathop \leqslant \limits^{ - {u^ - } \leqslant \left| u \right|} \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right| \\ \end{gathered} \right\} \Rightarrow \mathop {\max }\limits_{x \in \overline \Omega } \left| {u\left( x \right)} \right| \leqslant \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right|$.

On the other hand, $\partial \Omega \subseteq \overline \Omega \Rightarrow \mathop {\max }\limits_{x \in \partial \Omega } \left| {u\left( x \right)} \right| \leqslant \mathop {\max }\limits_{x \in \overline \Omega } \left| {u\left( x \right)} \right|$

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