PDE problem

Sep 2006
782
100
The raggedy edge.
Let \(\displaystyle \Omega \subset \mathbb{R}^n\) be a bounded set with smooth boundary. Assume that \(\displaystyle u_1\) and \(\displaystyle u_2\) are \(\displaystyle C^2(\overline{\Omega})\) and that they satisfy:

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

and

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

where \(\displaystyle f, \ g_1\) and \(\displaystyle g_2\) are continuous functions.

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

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

State any theorems used in the proof.
By the maximum principle I know that \(\displaystyle u_1(x)\) attains it's maximum on the boundary. Hence I know that the maximum of \(\displaystyle u_1(x)\) and \(\displaystyle u_2(x)\) are \(\displaystyle g_1(x)\) and \(\displaystyle g_2(x)\) respectively.

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

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

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

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

Can anyone help?
 
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