In a textbook of functional analysis I found this equation derived from Green's first identity
$\int _{ \Omega }^{ }{ u{ \nabla }^{ 2 }ud\tau } =\int _{ \partial \Omega }^{ }{ u\frac { \partial u }{ \partial n } ds } -\int _{ \Omega }^{ }{ \left| \nabla u \right| ^{2}d\tau }$
Then it goes on saying that if the boundary conditions on u are such that the integral over the boundary vanishes then the operator $ -\nabla^{2}$ is positive definite. Why ?
What I can see is that $\int _{ \Omega }^{ }{ u{ \nabla }^{ 2 }u+{ \left| \nabla u \right| }^{ 2 }d\tau } =0$ and what I'd need to declare that the operator is positive definite is : $\left< -{ \nabla }^{ 2 }u,u \right> >0\Leftrightarrow \int _{ \Omega }^{ }{ -{ \nabla }^{ 2 }u\bar { u } d\tau } >0$
So far I don't see how to prove that the operator is positive definite... Thanks for any kind of help.