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Thread: PDE problem

  1. #1
    Super Member Showcase_22's Avatar
    Sep 2006
    The raggedy edge.

    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 $\displaystyle \partial \Omega$


    $\displaystyle -\Delta u_2=f$ 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?
    Last edited by Showcase_22; May 19th 2010 at 01:51 PM.
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