1. ## PDE problem

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?