Eigenvalues of the Laplacian

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

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

$\displaystyle \ \ \ \ u=0 \ \ \ \ \ on \ \ \partial U$

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

Note: Justify all your arguments.

Re: Eigenvalues of the Laplacian

Let $\displaystyle u\neq 0$ a solution of $\displaystyle \left\{\begin{array}{rlc}-\Delta u &=\lambda u&\mbox{ in }U\\ u&=0&\mbox{ on }\partial U\end{array}\right.$. We can find $\displaystyle x_1\in U$ such that $\displaystyle u(x_1)>0$ (if it's $\displaystyle u(x_1)<0$, we consider $\displaystyle -u$). Since $\displaystyle \overline{U}$ is compact and $\displaystyle u\in C(\overline{U})$, we have $\displaystyle \sup_{x\in\overline U}u(x)\in\mathbb R$. Let $\displaystyle x_0\in U$ such that $\displaystyle u(x_0) = \max_{x\in \overline U}u(x)$ ($\displaystyle x$ can't be in $\displaystyle \partial U$ because $\displaystyle u=0$ on $\displaystyle \partial U$). Since $\displaystyle x_0$ is a critical point, we have $\displaystyle \nabla u(x_0)=0$ and by the Taylor's theorem we have $\displaystyle u(x_0+h) =u(x_0)+\frac 12 {}^thH_u(x_0)h +\lVert h\rVert ^2\varepsilon(h),$ where $\displaystyle \lim_{h\to 0}\varepsilon(h)=0$ and $\displaystyle H_u(x_0)$ is the Hessian matrix ($\displaystyle H_u(x_0)_{ij} = \dfrac{\partial^2 u}{\partial x_i\partial x_j}(x_0)$). Since $\displaystyle u(x_0+h)-u(x_0)\leq 0$, we have $\displaystyle {}^thH_u(x_0)h +2\lVert h\rVert^2\varepsilon(h) \leq 0$. We deduce that $\displaystyle {}^txH_u(x_0)x\leq 0$ for all $\displaystyle x\in\mathbb R^n$ and each eigenvalue of $\displaystyle H_u(x_0)$ is $\displaystyle \leq 0$. Hence

$\displaystyle 0\geq \mathrm{Tr}(H_u(x_0)) =\Delta u(x_0) =-\lambda u(x_0)$ and we get that $\displaystyle \lambda \geq 0$. We have to show that we can have $\displaystyle \lambda =0$, but it's a "well-known result" about harmonic functions.

Re: Eigenvalues of the Laplacian

Here's my approach:

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

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

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

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

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

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

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

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

which immediately gives $\displaystyle \lambda >0$.

It should be noted that in case we assume $\displaystyle u\in C^1 (\overline{U})$ then Green's identities give the result immediately.