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I'm consideirng the example of $-\Delta u + V(x) u = 0$ in $\Omega$ with $u = 0$ on $\partial \Omega$. I'm trying to see if it's true that if $-\lambda_1 < V(x) < -\lambda_2 < 0$ on $\overline{\Omega}$ that we do not have existence of non-trivial solutions to this equation. Here $\lambda_1$ and $\lambda_2$ are two distinct eigenvalues of $-\Delta u$, both of which are of course positive.

In the special case that $V$ is constant, this is certainly true since otherwise $V$ would itself be an eigenvalue. I have tried to see if I can use a sort of comparison principle to sohw that $u \equiv 0$ if $-\Delta u + V(x)u = 0$ in the case that $V$ is not constant but I can't seem to establish this.

For simplicity of course assume that $V$ is as smooth and as bounded as you'd like.

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I assume $\Omega$ is a bounded open subset of $\mathbb{R}^n$ and, say, $V\in L^\infty(\Omega)$. What you say is correct, but of course you need to assume that the eigenvalues are also consecutive (otherwise e.g. $V$ itself could be another eigenvalue in between [edit: as you actually said]). The comparison principle you are seeking comes from the Courant–Fischer–Weyl variational characterization of the eigenvalues of $-\Delta + V,$ by which the $k$-th eigenvalue $\lambda_k(-\Delta + V\\ )$ in the increasing order is expressed as a certain min-max of the Rayleigh quotient, which is monotone wrto $V:$ $$\frac{\int_\Omega \left(|\nabla u|^2+V(x)u^2 \right)dx }{\int_\Omega u^2 dx }$$ (for a precise statement see e.g. Courant-Hilbert, or Reed-Simon, or Gilbarg-Trudinger, &c).

So if we denote $\lambda_k:=\lambda_k(-\Delta)$ and assume $-\lambda_{k+1} < V(x) < -\lambda_{k}\\ ,$ it follows by the monotonicity $$\lambda_k(-\Delta+V)<\lambda_k(-\Delta - \lambda_k)=0 =\lambda_{k+1}(-\Delta - \lambda_{k+1})< \lambda_{k+1}(-\Delta +V),$$ so that $0$ is not an eigenvalue of $-\Delta +V.$

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Is there an easy way to see why we should have monotonicity with respect to $V$? –  Dorian Sep 6 '10 at 21:19
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At least the weak inequality is immediate: for a fixed $u\in H^1_0$ the Rayleigh quotient is increasing wrto $V$. Since the above formula express $\lambda_k$ as a certain min-max over the same sets, it follows that $\lambda_k(-\Delta+V)\leq \lambda_k(-\Delta+W)$ whenever $V\leq W$. For the strict inequality it's jus a bit more delicate, but the idea is the same. –  Pietro Majer Sep 6 '10 at 21:42
    
Note also that if you make the stronger assumption $V(x)+\epsilon\leq -\lambda_k$ with $\epsilon>0$, since $\lambda_k(-\Delta+V+\epsilon)=\lambda_k(-\Delta+V)+\epsilon$, the above weak form of monotonicity is sufficient to conclude $\lambda_k(-\Delta+V)\leq-\epsilon$; analogously for the other inequality. –  Pietro Majer Sep 6 '10 at 22:45
    
Pietro, this is so cool! I haven't seen a comparison principle used in good 15 years, and the min-max principle for eigenvalues (except $\lambda_1$) for comparable time. –  Victor Protsak Sep 7 '10 at 5:04
    
This is really fantastic Pietro! The whole minmax formulation of the eigenvalues was something that I had just skipped in my studies as it didn't seem important but wow what a nice application! I'm at least put at ease that what I was thinking was indeed correct. –  Dorian Sep 7 '10 at 14:09

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