Q:

integral of 2 functions satisfying Poisson's equation

Let $u,v:\Omega\subset\mathbb{R}^n\rightarrow\mathbb{R}$ be such that $u=v=0$ on $\partial\Omega$, $-\Delta u=\lambda u$ and $-\Delta v=\mu v$ in $\Omega$ with $\lambda\neq\mu$, and neither $u$ nor $v$ are trivially $0$.  I need to show that $\int_{\Omega}uvdx=0$.  I have already shown that both $\lambda,\mu>0$, but I am not sure if this will help me (it was the first part of the problem).   Can someone please give me a hint as for something to try?   If there were derivatives I could mess around with integration by parts/greens identities, but I haven't had any ideas or any intuition as to why this should be true. 

A:

The Poisson equation gives you the derivative :
$$(\mu-\lambda)\int_\Omega uv = \int_\Omega \Delta u\ v - \int_\Omega u\ \Delta v$$
Then by integration by part, you get
$$(\mu-\lambda)\int_\Omega uv = \int_{\partial \Omega} v \nabla u d\vec{\sigma} - \int_\Omega \nabla u \cdot \nabla v - \int_{\partial \Omega} u \nabla v d\vec{\sigma} + \int_\Omega \nabla u \cdot \nabla v$$
And as $u=v=0$ on $\partial \Omega$,
$$(\mu-\lambda)\int_\Omega uv= 0$$

