# First eigenvalue of laplacian

I know the laplacian $\Delta$ has only positive eigenvalues, but why there is a first one? Assume $\Delta$ is acting on an appropriate set of real valued functions on the bounded domain $\Omega \subset \mathbb{R}^n$, $n \geq 2$, which vanish on $\partial \Omega$.

• The inverse of $\Delta$ is a compact operator, whence the discrete spectrum of $\Delta$ going to infinity May 24, 2016 at 19:19
• @user8268 What are the domain and codomain under consideration? May 24, 2016 at 20:41

Consider

$$\left\{\begin{array}{c} −\Delta u = f & \text{on }\Omega \\ u(x)=0, \partial\Omega \end{array}\right.,$$

where $\Omega \subset \mathbb{R}^{N}$ is open, bounded and $f \in L^{2}(\Omega)$

See that $u \in H^{1}_{0}(\Omega)$ is a weak solution for the problem above if

$(u,v)_{H^{1}_{0}(\Omega)} = \displaystyle\int_{\Omega} \nabla u \nabla v = \displaystyle\int_{\Omega} fv$ for all $v \in H^{1}_{0}(\Omega)$.

Defining $g(v) = \displaystyle\int_{\Omega} fv$, use the Riesz Representation Theorem to conclude that problem has a unique solution. Moreover, is well defined the "solution operador":

$S: L^{2}(\Omega) \rightarrow H^{1}_{0}(\Omega)$, where for each $f_{0} \in L^{2}(\Omega)$, we have $S(f_{0})=u_{0}$ where

$$\left\{\begin{array}{c} −\Delta u_{0} = f_{0} & \text{on }\Omega \\ u_{0}(x)=0, \partial\Omega \end{array}\right.,$$

One can show that $S$ is linear, compact and self-adjoint. Then by Spectral Theory, exists a sequence $(\mu_{n})$ of postives eigenvalues of S, such that:

$\displaystyle\lim_{n \rightarrow \infty} \mu_{n} = 0$ and $\mu_{1} < \mu_{2} < ... < \mu_{n} < ...$

we say that $\lambda$ is a eigenvalue of Laplacian when

$\displaystyle\int_{\Omega} \nabla \phi \nabla v = \displaystyle\int_{\Omega} (\lambda \phi)v$ for all $v \in H^{1}_{0}(\Omega)$.

Note that if $\mu$ is eigenvalue of $S$ then $\lambda = \dfrac{1}{\mu}$ is a eigenvalue of Laplacian. Then, exits a sequence $(\lambda_{n})$ of eigenvalues of Laplacian such that

$\lambda_{1} > \lambda_{2} > ... >\lambda_{n} > ...$