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Math Help - 2-Dimensional Boundary Value Problem

  1. #1
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    2-Dimensional Boundary Value Problem

    Let D be the unit disk in \mathbb{R}^2 and let \partial D denote the boundary. Show that the only solution of the boundary value problem:

    \Delta u - u = 0 in D
    u = 0 on \partial D

    is u=0. The hint given is to apply the 2-D divergence theorem on the vector field u\nabla u, but I don't see how that even enters in.
    Last edited by Ackbeet; May 3rd 2011 at 08:31 AM. Reason: Used [tex][/tex] tags.
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  2. #2
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    Quote Originally Posted by mathematicalbagpiper View Post
    Let D be the unit disk in \mathbb{R}^2 and let \partial D denote the boundary. Show that the only solution of the boundary value problem:

    \Delta u - u = 0 in D
    u = 0 on \partial D

    is u=0. The hint given is to apply the 2-D divergence theorem on the vector field u\nabla u, but I don't see how that even enters in.
    You are going to want to calculate an integral two different ways

    First note that

    \Delta u=u \implies u\Delta u=u^2

    The right hand side of the above equation is non negative so we get that

    \iint_{D}u\Delta udA= \iint_{D}u^2dA \ge 0

    Now consider the flux integral out of the disk

    \oint_{\partial D} (u \nabla u)\cdot d\mathbf{s}=0

    This is zero because u=0 on the boundary.

    Now if we use the divergence theorem we get

    0=\oint_{\partial D} (u \nabla u)\cdot d\mathbf{s}=\iint_{D}\nabla \cdot (u \nabla u)dA=\iint_{D}|\nabla u|^2dA+\iint_{D}u\Delta u dA

    This gives

    -\iint_{D}u\Delta u dA = \iint_{D}|\nabla u|^2dA \ge 0 \iff \iint_{D}u\Delta u dA \le 0

    Combining these two equations gives

    0 \le  \iint_{D}u\Delta u dA \le 0 \implies \iint_{D}u\Delta u dA=0

    So the integrand must be equal to zero so

    u\Delta u=0 \iff u=0 \text{ or } \Delta u=0

    If it is the first cases we are done if it is the plug that into the pde and conclue.
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    Behold, the power of SARDINES!
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    I am now doubting the above the very last line does not hold!

    0 \le  \iint_{D}u\Delta u dA \le 0 \implies \iint_{D}u\Delta u dA=0

    So the integrand must be equal to zero
    This is only true if the integrand does not change signs and I have not shown that!

    There are many counter examples. Sorry
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  4. #4
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    Ahh okay here is the fix. It is true that

    0 \le  \iint_{D}u\Delta u dA \le 0 \implies \iint_{D}u\Delta u dA=0

    Now if we sub this back into the above equation we get

    0=\iint_{D}|\nabla u|^2dA+\iint_{D}u\Delta u dA  \iff 0=\iint_{D}|\nabla u|^2dA

    Now this integrand is non negative so it must be zero this gives

    |\nabla u|^2=0 \iff \nabla u=0 \implies u=c

    is a constant but u on the boundary is zero so  u \equiv 0
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