I have the following integral over the domain $\Omega$ with boundary $\Gamma$
and $ u,\phi \in \Omega$:
$\int_{\Omega} \nabla^2u \nabla^2\phi\, \mathrm{d}\Omega$.
The problem is to get rid of the Laplacian in front of $\phi$, presumably by using integration by parts and the divergence theorem.
Could someone give a hint as to how this is done? I know how to do this using integration by parts and the divergence theorem for: $\int_{\Omega} \nabla u \cdot \nabla \phi\, \mathrm{d}\Omega$.
Best Answer
It is easy to check that $$\nabla\cdot(\nabla^2u\nabla\phi)=\nabla(\nabla^2u)\cdot\nabla\phi+\nabla^2u\nabla^2\phi,$$ and $$\nabla\cdot(\nabla(\nabla^2u)\phi)=\nabla^2(\nabla^2u)\phi+\nabla(\nabla^2u)\cdot\nabla\phi.$$ Thus, we obtain, $$\nabla^2u\nabla^2\phi=\nabla\cdot(\nabla^2u\nabla\phi)-\nabla\cdot(\nabla(\nabla^2u)\phi)+\nabla^2(\nabla^2u)\phi.$$ Integrating on $\Omega$ and using the the divergence theorem, $$\int_{\Omega}\nabla^2u\nabla^2\phi=\int_{\Gamma}\nabla^2u\frac{\partial\phi}{\partial n}-\int_{\Gamma}\frac{\partial\nabla^2u}{\partial n}\phi+\int_{\Omega}\nabla^2(\nabla^2u)\phi.$$