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.

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  1. September 7, 2013 at 5:50 pm

    I’m still a little confused. The positive/negative part of u_1 might not be in H_0^1?

    • September 7, 2013 at 6:40 pm

      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.

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