If $v(x,y)$ solves Laplace's equation $v_{xx} + v_{yy} = 0$ on a bounded domain $S$, and $u(x,y,t)$ solves $u_t = u_{xx} + u_{yy}$ on $S$, with $u=v$ on $\partial S$ for all $t$, one can show that $\phi(t):= \int_S (u-v)^2$ is decreasing in $t$. Is it true that $\phi(t) \to 0$ as $t\to \infty$? Will appreciate any hints, thanks.
(It probably isn't needed, but for the first part, $\phi'(t) = \int_S 2(u-v)\nabla^2 u = \int_S 2(u-v)\nabla^2 (u-v)$ $= \int_S \nabla^2 (u-v)^2 - 2\|\nabla(u-v)\|^2 \leq \oint_{\partial S} \nabla(u-v) \cdot dz = 0$.)
The computations become a little cleaner if you introduce $w=u-v$, which also solves the heat equation and has zero boundary conditions. You have shown that $$\frac{d}{dt}\int_S w^2 \le -\int_S |\nabla w|^2 \tag{1}$$ What is needed is to relate the right hand side to $\int_S w^2$. This is what the Poincaré inequality does: $$\int_S w^2 \le C \int_S |\nabla w|^2 \tag{2}$$ Combine (2) with (1), and you'll get a differential inequality from which the desired conclusion follows quickly.