I have a question that reads
$$\frac{\partial}{\partial t}w=\nabla^2w$$ for $w(t,\vec x)$ with initial condition $w(0,\vec x) = w(0,x)$ in $V$ , boundary condition $w(t,\vec x) = f(\vec x)$ on $S$ for all $t > 0$.
Show that $$\frac{\partial}{\partial t}\iiint |\nabla w|^2 dV\le0$$
So far I have
$$\frac{\partial}{\partial t}\iiint |\nabla w|^2 = \iiint\frac{\partial}{\partial t}(\nabla.(w\nabla w)-w\nabla^2w)dV$$ $$= \iint\frac{\partial}{\partial t}(w\nabla w).d\vec S-\iiint\frac{\partial}{\partial t}(w\nabla^2 w)dV$$ (divergence theorem) $$=-\iiint\frac{\partial}{\partial t}(w\nabla^2 w) dV$$ ($w$ constant on $S$) $$=-\iiint(\frac{\partial w}{\partial t})^2+(w\frac{\partial^2}{\partial^2 t}w) dV$$ (using the PDE given).
The first term is then clearly negative, but I'm not sure about the second?
You could try integration by parts on $w \frac{\partial^2 w}{\partial^2 t}$, as this would then lead to a square term.