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 $\partial \Omega$.
 A: 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
$\lambda_{1} > \lambda_{2} > ... >\lambda_{n} > ...$
