Eigenvalues of the Laplacian

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July 7th 2011, 11:05 AM
Jose27

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.
July 8th 2011, 02:23 AM
girdav

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.
July 8th 2011, 09:25 AM
Jose27

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.