Consider the PDE: $$u_t=ku_{xx}-\mu u,~0<x<L,~t>0,$$ where $k,~\mu>0$ and $u(x,0)=\phi(x),~\phi(0)=\phi(L)=0,~\phi(x)>0$ in $(0,L)$. Let $$\displaystyle E(t)=\int\limits_{0}^{L}u^2(x,t)dx.$$ If either $u(0,t)=u(L,t)=0$, or $u_x(0,t)=u_x(L,t)=0$, then prove that $E(t) \rightarrow 0,~t\rightarrow +\infty.$
Attempt I have noticed (by simple calculations) that, in either case, $E$ must be strictly decreasing, i.e. $E'(t)<0,~t>0$, but I don't know if and how this leads be to the solution.
Thank you!
Multiply the PDE by $u$ then integrate by parts in the spatial variable to derive the following first-order linear ODE in $E(t)$: \begin{equation*} E'(t) + 2\mu E(t) = f(t), \end{equation*} where $f(t)\leq0$ for all $t$ (the explicit form of $f$ is not needed for your question). Multiply the above differential equation by $e^{2\mu t}$ and use the estimate on $f$ to obtain the differential inequality \begin{equation*} \frac{d}{dt}\left( e^{2\mu t}E(t)\right)\leq 0. \end{equation*} The Fundamental Theorem of Calculus now gives $E(t)\leq e^{-2\mu t}E(0)$.