# Is this inequality true?

Let $\Omega$ be a bounded domain with smooth boundary. Let $$S=\{u\in C^2(\overline \Omega): \frac{\partial u}{\partial n}=0 \text{ on } \partial\Omega \}.$$ Fix $\Phi\in S$ with $\Phi(x)>0$ for all $x\in\overline\Omega$. Can will find $\delta>0$ such that $$f(u)=_{df} \int_\Omega u\frac{\Delta\Phi}{\Phi} + \Phi \frac{\Delta u}{u}\ge 0$$ for all $u\in B_\delta(\Phi)$, where $B_\delta(\Phi)=\{u\in S: \|u-\Phi\|_{C^1(\overline\Omega)}<\delta \}$?

I only know that $f(\Phi)=0$ and $f(\Phi+\epsilon)\ge 0$ if $\epsilon$ is small. Can any one prove or find a counterexample or suggest me some possible methods to approach it? Thanks....

This certainly depends upon $\phi$. A necessary condition is that the differential of $f$ at $\phi$ vanishes (it does) and the Hessian be non-negative. After two integration by parts, I obtain $$Hf_{\phi}(v)=\int_\Omega\left(2\frac{|\nabla v|^2}{\phi}+v^2(3\frac{\Delta\phi}\phi-2\frac{|\nabla\phi|^2}{\phi^3})\right)dx.$$ So the question can be more or less rephrased as: is it true that this integral be always non-negative. If it is positive definite, then the answer to you question is Yes.

• I thought that just because the Hessian is positive definite this isn't enough to imply a minimum. I could be sorely mistaken and will gladly delete this comment if you tell me otherwise. – k3thomps Oct 25 '14 at 20:03
• Thank you for your anser and discussion. It is very helpful for me. It seems that I need a strong condition on $\phi$... – user60904 Oct 25 '14 at 20:46