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Thread: 2-Dimensional Boundary Value Problem

  1. #1
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    2-Dimensional Boundary Value Problem

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

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

    is $\displaystyle u=0$. The hint given is to apply the 2-D divergence theorem on the vector field $\displaystyle u\nabla u$, but I don't see how that even enters in.
    Last edited by Ackbeet; May 3rd 2011 at 07:31 AM. Reason: Used [tex][/tex] tags.
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  2. #2
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    Quote Originally Posted by mathematicalbagpiper View Post
    Let $\displaystyle D$ be the unit disk in $\displaystyle \mathbb{R}^2$ and let $\displaystyle \partial D$ denote the boundary. Show that the only solution of the boundary value problem:

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

    is $\displaystyle u=0$. The hint given is to apply the 2-D divergence theorem on the vector field $\displaystyle u\nabla u$, 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

    $\displaystyle \Delta u=u \implies u\Delta u=u^2$

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

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

    Now consider the flux integral out of the disk

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

    This is zero because u=0 on the boundary.

    Now if we use the divergence theorem we get

    $\displaystyle 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

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

    Combining these two equations gives

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

    So the integrand must be equal to zero so

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

    If it is the first cases we are done if it is the plug that into the pde and conclue.
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  3. #3
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    I am now doubting the above the very last line does not hold!

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

    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

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

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

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

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

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

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