# PDE problem

#### Showcase_22

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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