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Thread: PDE problem

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

    PDE problem

    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


    -\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?
    Last edited by Showcase_22; May 19th 2010 at 02:51 PM.
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