Proving an inequality using the divergence theorem 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 $t &gt; 0$.
  
  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.
You could try integration by parts on $w \frac{\partial^2 w}{\partial^2 t}$, as this would then lead to a square term.