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Math Help - Really Struggling :(

  1. #1
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    Unhappy Really Struggling :(

    Hi,
    I am really struggling with a subject at my university which is all about partial differential equations. It takes me a long time to understand even the most basic concepts, and I really need some help understanding and completing the following questions.

    Let  \mathcal{R} be a bounded region in \mathbb{R}^3, and suppose  p(x) > 0 on  \mathcal{R}.

    (i) If u is a solution of

    \bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = 0 \ \x \in \partial \mathcal{R}

    show that u \equiv 0 on \mathcal{R}

    (ii) If u is a solution of

    \bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = g(x) \ \x \in \partial \mathcal{R}

    show that u is unique (It can be assumed that part (i) is true).
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  2. #2
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    I haven't made any progress on this one, but this is Poisson's equation, isn't it? And its inhomogenous? I think I need to use Greens identity to solve this. Is that right?
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  3. #3
    Super Member Rebesques's Avatar
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    Quote Originally Posted by OliviaB View Post
    Hi,
    I am really struggling with a subject at my university which is all about partial differential equations. It takes me a long time to understand even the most basic concepts, and I really need some help understanding and completing the following questions.

    Let  \mathcal{R} be a bounded region in \mathbb{R}^3, and suppose  p(x) > 0 on  \mathcal{R}.

    (i) If u is a solution of

    \bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = 0 \ \x \in \partial \mathcal{R}

    show that u \equiv 0 on \mathcal{R}

    (ii) If u is a solution of

    \bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = g(x) \ \x \in \partial \mathcal{R}

    show that u is unique (It can be assumed that part (i) is true).


    Well, it looks like a typical application of the Lax-Milgram theorem to me.

    The weak formulation of this problem on L^2(\mathcal{R}) reads

    \int_{\mathcal{R}}(\bigtriangledown u)^2v=\int_{\mathcal{R}}puv,\ \forall v\in L^2(\mathcal{R})

    or, by integrating the left hand side by parts and using the boundary conditions,

    \int_{\mathcal{R}}[\langle \nabla u,\nabla v\rangle+puv]=0.

    Define the bilinear form on L^2(\mathcal{R})\times L^2(\mathcal{R}),

    \alpha(u,v)=\int_{\mathcal{R}}[\langle \nabla u,\nabla v\rangle+puv].

    To show that \alpha(u,v)=0 is solvable for some u, we can invoke the Lax-Milgram theorem: We need just show that \alpha is bounded and coercive.

    You can show it is bounded, ie there exists a B>0 such that

    |\alpha(u,v)|\leq B||u||||v||, where ||u||^2=\int_{\mathcal{R}}u^2+|\nabla u|^2.

    As for \alpha being coercive, we must show there is C>0 with \alpha(u,u)\geq C||u||^2. We compute

    \alpha(u,u)=\int_{\mathcal{R}}|\nabla u|^2+pu^2\geq\min\{1,\inf_{\mathcal{R}} p\}||u||^2.


    Now the Lax-Milgram theorem applies, and there is a u_0\in L^2(\mathcal{R}) such that
    \alpha(u_0,v)=0 \ \forall v\in L^2(\mathcal{R}).
    And reflexivity implies u_0=0.

    For part ii), consider two solutions u_1,u_2\in L^2(\mathcal{R}) of the problem at hand.
    Then their difference u_1-u_2 is a solution for the problem given in part i), and therefore must be zero.






    Ps. You cannot construct a Green's function here, as zero is an eigenvalue for -\bigtriangledown^2+p on a region.
    Last edited by Rebesques; October 18th 2010 at 09:11 AM.
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