Home > Partial Differential Equations, shape optimization, Sobolev Spaces > First Dirichlet eigenvalue is simple for connected domains

## First Dirichlet eigenvalue is simple for connected domains

Suppose ${\Omega \subset \Bbb{R}^N}$ is a connected open set and consider the first two eigenvalues of the Laplace operator with Dirichlet boundary conditions ${\lambda_1(\Omega),\lambda_2(\Omega)}$. Then ${\lambda_1(\Omega)<\lambda_2(\Omega)}$.

Proof: First step is to prove that we can choose ${u_1}$, the first eigenfunction, to be positive. Using the definition and the weak formulation we get

$\displaystyle \lambda_1(\Omega)=\frac{\int_\Omega |\nabla u_1|^2 }{\int_{\Omega} u_1^2 }=\inf_{ u \in H_0^1(\Omega) \setminus\{0\}} \frac{\int_\Omega |\nabla u|^2}{\int_\Omega u^2 }.$

On the other hand if we denote ${u^+}$ and ${u^-}$ the positive and negative parts of ${u}$ we have

$\displaystyle \min \left\{\frac{\int_\Omega |\nabla u_1^+|^2 }{\int_{\Omega} (u_1^+)^2 },\frac{\int_\Omega |\nabla u_1^+|^2 }{\int_{\Omega} (u_1^-)^2 } \right\} \leq \frac{\int_\Omega |\nabla u_1|^2 }{\int_{\Omega} u_1^2 }.$

This means that we can choose the positive or negative part of ${u_1}$ and still get an eigenfunction (because at least one of them is non-zero). Because the first eigenfunction is analytic and non identically zero, it must be positive in the interior of ${\Omega}$ (here we have used the fact that ${\Omega}$ is connected).

Suppose now, that ${\lambda_1(\Omega)=\lambda_2(\Omega)}$, and choose ${u_2}$ an eigenfunction for ${\lambda_2(\Omega)}$ which is orthogonal to ${u_1}$. Using the same trick as above, and because ${\lambda_2(\Omega)}$ is characterized here by the same minimization problem as ${\lambda_1(\Omega)}$, we can apply the same reasoning as above, and see that we can choose ${u_2}$ non-negative.

Therefore we have found two orthogonal functions in ${L^2(\Omega)}$ one which is strictly positive and the other one non-negative. This cannot happen unless one of them is zero, and this contradicts the fact that ${u_1,u_2}$ are eigenfunctions.

As a consequence we deduce that ${\lambda_1(\Omega) \neq \lambda_2(\Omega)}$ if ${\Omega}$ is connected.

I’m still a little confused. The positive/negative part of $u_1$ might not be in $H_0^1$?
Yes, the positive and negative parts of a function $u$ which is $H_0^1$ are again $H_0^1$. In fact, a more general theorem holds: If $G: \Bbb{R} \to \Bbb{R}$ is a Lipschitz function with $G(0)=0$ then $G(u)$ is $H_0^1$ if $u$ is $H_0^1$. For a reference and proofs, look in A. Henrot, M. Pierre, Variation et Optimisation de Formes, Proposition 3.1.11.