# PDE problem

• May 19th 2010, 01:40 PM
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PDE problem
Quote:

Let $\Omega \subset \mathbb{R}^n$ be a bounded set with smooth boundary. Assume that $u_1$ and $u_2$ are $C^2(\overline{\Omega})$ and that they satisfy:

$-\Delta u_1=f$ in $\Omega$
$u_1(x)=g_1(x)$ on $\partial \Omega$

and

$-\Delta u_2=f$ in $\Omega$
$u_2(x)=g_2(x)$ on $\partial \Omega$

where $f, \ g_1$ and $g_2$ are continuous functions.

Assume that $g_1(x) \leq g_2(x)$ for $x$ on $\partial \Omega$. Prove that

$u_1(x) \leq u_2(x)$ for $x \in \Omega$

State any theorems used in the proof.

By the maximum principle I know that $u_1(x)$ attains it's maximum on the boundary. Hence I know that the maximum of $u_1(x)$ and $u_2(x)$ are $g_1(x)$ and $g_2(x)$ respectively.

Since $g_1(x) \leq g_2(x)$ I know that $\max u_1(x) \leq \max u_2(x)$ for $x \in \Omega$.

I need to justify that $\max u_1(x) \leq \max u_2(x) \ \Rightarrow \ u_1(x) \leq u_2(x)$ where $x \in \Omega$.

Since the laplacian is negative I know that $u_1(x) \leq \max u_1(x)$ and $u_2(x) \leq \max u_2(x)$.

This is where I get my problem. $u_1(x) \leq \max u_1(x)$, $u_2(x) \leq \max u_2(x)$ and $\max u_1(x) \leq \max u_2(x)$ do not necessarily imply that $u_1(x) \leq u_2(x)$ for $x \in \Omega$.

Can anyone help?