Controlling the Second Eigenvalue of a Schrödinger Operator

Consider a bounded domain $\Omega$ (with smooth boundary) in some Riemannian $n$-manifold $M^n$.

Let $L$ be the operator $$L=\Delta+V$$ where $\Delta$ is the Laplace-beltrami operator on $M$ (so is formally negative) and $V$ is some smooth potential (we are mostly interested in $V\geq 0$).

A well known result of Fischer-Colbrie and Schoen says that if $L$ has a positive solution in $\Omega$ then $\lambda_1(\Omega,L)$ is non-negative. That is the first Dirichlet eigenvalue of $L$ is non-negative (i.e. with our sign convention $L \phi=-\lambda_1 \phi$ for some $\lambda_1\geq 0, \phi\neq 0$, $\phi=0$ on $\partial \Omega$).

Suppose instead that we we know only that $L$ has a solution $u$ in $\Omega$ so that both $\Omega^+=\lbrace u>0 \rbrace\cap \Omega \subset \Omega$ and $\Omega^-=\lbrace u<0 \rbrace\cap \Omega \subset \Omega$ are connected subsets. My question is: given this set up is anything known about $\lambda_2(\Omega,L)$ the second Dirichlet eigenvalue?

We note given this set up, if $n=2$ then the nodal line $\lbrace u=0 \rbrace\cap \Omega \subset \Omega$ must have a single component and be smooth. Hence, an argument with nodal lines should imply that if $u=0$ on $\partial \Omega$ then $\lambda_2(\Omega,L)\geq 0$. Also, Fischer-Colbrie and Schoen's result implies that for $n=1$, the one dimensional problem, $\lambda_2(\Omega,L)\geq 0$.

• What do you mean ? If $L$ has a positive solution (I understand $Lw=0$ with $w>0$), then $\lambda_1(\Omega,L)=0$. If it is what you mean then Fischer-Colbrie and Schoen theorem is nothing but the Krein-Rutman theorem. – Denis Serre Jan 15 '11 at 9:16
• @Denis Note I am most definitely NOT assuming $w=0$ on $\partial \Omega$ so one does not have that it is a Dirichlet eigenfunction. The not at all a priori obvious fact is that this doesn't hurt i.e. the spectrum can only go up. – Rbega Jan 15 '11 at 17:31