Suppose $u_t=u_{xx}$ on $(0,1)\times(0,\infty)$ and $\int_0^1u(x,0)dx=0$ with Neumann boundary condition $u_x(0,t)=u_x(1,t)=0$. Show that $\int_0^1u^2(x,t)dx\to0$ as $t\to\infty$.
The $t$-derivative of $\int_0^1u(x,t)dx=0$ is 0, so it is constantly 0. Also, $\int_0^1 u^2(x,t)dx$ is decreasing, but why should it go to 0?