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Math Help - Eigenvalues of the Laplacian

  1. #1
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    Eigenvalues of the Laplacian

    Let U\subset \mathbb{R}^n be a smooth bounded region (open, connected with smooth boundary). Assume u\in C^2(U) \cap C(\overline{U}) solves

    -\Delta u = \lambda u \ \ \ \ in \ \ U

    \ \ \ \ u=0 \ \ \ \ \ on \ \ \partial U

    with \lambda \in \mathbb{R}. Prove that if u is not identically 0 then \lambda >0.

    Note: Justify all your arguments.
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  2. #2
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    Re: Eigenvalues of the Laplacian

    Let u\neq 0 a solution of \left\{\begin{array}{rlc}-\Delta u &=\lambda u&\mbox{ in }U\\ u&=0&\mbox{ on }\partial U\end{array}\right.. We can find x_1\in U such that u(x_1)>0 (if it's u(x_1)<0, we consider -u). Since \overline{U} is compact and u\in C(\overline{U}), we have \sup_{x\in\overline U}u(x)\in\mathbb R. Let x_0\in U such that u(x_0) = \max_{x\in \overline U}u(x) ( x can't be in \partial U because u=0 on \partial U). Since x_0 is a critical point, we have \nabla u(x_0)=0 and by the Taylor's theorem we have u(x_0+h) =u(x_0)+\frac 12 {}^thH_u(x_0)h +\lVert h\rVert ^2\varepsilon(h), where \lim_{h\to 0}\varepsilon(h)=0 and H_u(x_0) is the Hessian matrix ( H_u(x_0)_{ij} = \dfrac{\partial^2 u}{\partial x_i\partial x_j}(x_0)). Since u(x_0+h)-u(x_0)\leq 0, we have {}^thH_u(x_0)h +2\lVert h\rVert^2\varepsilon(h) \leq 0. We deduce that {}^txH_u(x_0)x\leq 0 for all x\in\mathbb R^n and each eigenvalue of H_u(x_0) is \leq 0. Hence
    0\geq \mathrm{Tr}(H_u(x_0)) =\Delta u(x_0) =-\lambda u(x_0) and we get that \lambda \geq 0. We have to show that we can have \lambda =0, but it's a "well-known result" about harmonic functions.
    Last edited by girdav; July 8th 2011 at 12:13 PM.
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  3. #3
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    Re: Eigenvalues of the Laplacian

    Here's my approach:

    Since u\in C(\overline{U}) we have \Delta u \in C(\overline{U}) and so we can extend both to \mathbb{R}^n as 0 outside U. Take u_{\varepsilon} be the standard regularization (convolution with an appropiate test function). For small enough \varepsilon we con find R>0 such that supp (u_{\varepsilon}) \subset B_R(0). Now, integrating by parts in this ball we get

    \int_{B_R(0)} |\nabla u_{\varepsilon}|^2 = \int _{B_R(0)} u_{\varepsilon} \Delta u_{\varepsilon} \leq \frac{1}{2}\left(  \int _{B_R(0)} u_{\varepsilon}^2 + \int_{B_R(0)} \Delta u_{\varepsilon} ^2 \right)

    where for the last inequality we use Cauchy-Schwarz and Young's inequality. Take V\subset \subset U then u_{\varepsilon}, \ \nabla u_{\varepsilon}, \ \Delta u_{\varepsilon} all converge uniformly to their respective limits in V (and the first and last in B_R(0)) so we have

    \int_V |\nabla u|^2 \leq \frac{1}{2} \left( \int_U u^2 + \int_U \Delta u^2 \right)

    Take now a sequence (V_k) with V_k \subset int(V_{k+1}) and \cup_k V_k=U and take w_k= |\nabla u|^2 |_{V_k} then w_k\leq w_{k+1} \rightarrow |\nabla u|^2 and their integrals are bounded by the last inequality so we get that \nabla u \in L^2(U, \mathbb{R}^n). Now, for any f\in C_c^\infty (U) we have

    \lambda \int_Uuf= -\int_Uf\Delta u = \int_U \langle \nabla f, \nabla u \rangle

    By approximation ( u\in H_0^1(U)) we can change f for u to get

    \lambda \int_U u^2 = \int_U |\nabla u|^2

    which immediately gives \lambda >0.

    It should be noted that in case we assume u\in C^1 (\overline{U}) then Green's identities give the result immediately.
    Last edited by Jose27; July 8th 2011 at 09:40 AM.
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