# A question on Schrodinger operator

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 $$(-\Delta+\phi)u\geq0\ \mathrm{on}\ M,$$ or $$(-\Delta+\psi)u\geq0\ \mathrm{on}\ M?$$

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

• 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? – Michael Renardy Jun 18 '14 at 10:06
• Is it obvious that positive eigenfunctions exist? – timur Jun 18 '14 at 12:22
• @timur yes, it is Krein-Rutman's Theorem. – username Jun 18 '14 at 12:23
• @MichaelRenardy Corrected thanks, I modified my answer to simply state my thought. – username Jun 18 '14 at 12:24
• @littlelittlelittle you are welcome, but I did not answer your question! – username Jun 23 '14 at 14:30