Can anyone explain to me how I can conclude $\frac{d}{dt}E(f)=-\int _M (\Delta f)\dot{f} d\mu$ by using integration by parts and $\langle f_1 ,f_2 \rangle_\mu:=\int_M f_1 f_2 d\mu$?
Where $M$ is a compact Manifold, $E(f)=\frac{1}{2}\int _M |\nabla f|^2d\mu$ and $\dot{f}=\frac{d}{dt}f$.
Thanks.
Well, you must have either $M $ being closed (i.e. $\partial M = \phi $) or some Neumann conditions like $ \partial f/\partial \nu $ vanishes on boundary to have the following relation from Green's formula $$ \int_M \dot{f}\Delta f \ d\mu + \int_M \langle \nabla f,\nabla\dot{f} \rangle d\mu = \int_{\partial M}\dot{f}\frac{\partial f}{\partial \nu}dS = 0 $$ This immidiately gives $$ \frac{d}{dt}E(f) = \frac{1}{2} \int_M \frac{d}{dt}\Bigl(|\nabla f|^2\Bigr)d\mu = \int_M \langle \nabla f, \nabla\dot{f}\rangle d\mu = -\int_M \dot{f}\Delta f\ d\mu $$