I am not sure whether I should ask for help here or math stackexchange. I got trouble with an inequality involving the Schrodinger operator on manifolds. Any suggestion is appreciated!

Let $(M,g)$ be a closed Riemannian manifold, and two functions $\phi, \psi\in C^{\infty}(M)$ satisfying $$\phi+\psi>0\ \mathrm{on}\ M.$$

Is it possible to find a positive function $u>0$ such that \begin{equation} (-\Delta+\phi)u\geq0\ \mathrm{on}\ M, \end{equation} or \begin{equation} (-\Delta+\psi)u\geq0\ \mathrm{on}\ M? \end{equation}

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    $\begingroup$ what about $u=1$? $\endgroup$ – username Jun 17 '14 at 18:49
  • $\begingroup$ $u=1$ requires either $\phi>0$ or $\psi>0$. $\endgroup$ – littlelittlelittle Jun 18 '14 at 1:18

Here is an alternative formulation of a question related to yours. Consider the case when $M=\Omega\subset\mathbb{R}^{n}$, a (smooth) open bounded domain. Suppose that $u>0$ and $$ -\Delta u+\psi u=f \mbox{ in }\Omega. $$

Integrate against $\frac{w^{2}}{u}$ and rearranging you obtain for all $w\in H_{0}^{1}(\Omega),$ $$ \int\nabla w\cdot\nabla w-\int_{\Omega}\left|\frac{u\left|\nabla w\right|-w\left|\nabla u\right|}{u}\right|^{2}+\int_{\Omega}\psi w^{2}=\int f\frac{w^{2}}{u} $$ therefore for all $w\in H_{0}^{1}(\Omega)$ $$ \int\nabla w\cdot\nabla w+\int_{\Omega}\psi w^{2}\geq\int f\frac{w^{2}}{u} $$ therefore, if $\lambda_{\psi}$ is the first dirichlet eigenvalue of $-\Delta+\psi$, you obtain $$ \lambda_{\psi}\int w^{2}\geq\int f\frac{w^{2}}{u}. $$ So if there exists a positive solution such that $-\Delta u+\psi u>0$, then $\lambda_{\psi}>0$. The converse is obsiously true, as we can choose the first eigenvector to satisfy $u_{\psi}>0$ in $\Omega$, and therefore $-\Delta u+\psi u=\lambda_{\psi}u>0$ in $\Omega$. So your question could be written in this case:

Is it true that $$ \mbox{if }\quad \psi+\phi>0, \mbox{ then } \lambda_{\psi}>0 \mbox{ or } \lambda_{\phi}>0\quad? $$

It is not obvious, because $\lambda_{f}$ is concave in $f$: the fact that $\lambda_{\frac{1}{2}\psi+\frac{1}{2}\phi}>0$ does not help..

  • $\begingroup$ Why is $\phi u_0+\psi v_0\ge \min(u_0,v_0)(\phi+\psi)$. I see why this is true if $\phi$ and $\psi$ are positive, but what if one of them is negative? $\endgroup$ – Michael Renardy Jun 18 '14 at 10:06
  • $\begingroup$ Is it obvious that positive eigenfunctions exist? $\endgroup$ – timur Jun 18 '14 at 12:22
  • $\begingroup$ @timur yes, it is Krein-Rutman's Theorem. $\endgroup$ – username Jun 18 '14 at 12:23
  • $\begingroup$ @MichaelRenardy Corrected thanks, I modified my answer to simply state my thought. $\endgroup$ – username Jun 18 '14 at 12:24
  • $\begingroup$ @littlelittlelittle you are welcome, but I did not answer your question! $\endgroup$ – username Jun 23 '14 at 14:30

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