A bound for the PDE $\frac{\partial}{\partial t}u=\Delta u+ au$

Question

Consider the following ODE: $$\frac{d}{dt}u(t)=au(t)$$ with $t\geq 0$ and initial value $u(0)=u_0$, then we have $$u(t)=e^{at}u_0.$$ Now if we consider the following PDE $$\frac{\partial}{\partial t}u(t,x)=\Delta u(t,x)+ au(t,x)$$ on a smooth bounded open subset $\Omega\subset\mathbb{R}^N$ and $u=0$ on the boundary $\partial \Omega$. Do we get an estimation of the form $$|u(t)|_{L^2(\Omega)}\leq Me^{\omega t}.$$ If yes how to prove such estimation, or where can I find such results?

Answer 1

Multiplying both sides by $u$ and integrating in $\Omega$, one has $$ \frac{1}{2}\frac{d}{dt}|u(t)|_2^2+\|u(t)\|_2^2=a|u(t)|_2^2. $$ Due to the Poincare Inequality, one attains $$ \frac{1}{2}\frac{d}{dt}|u(t)|_2^2+\lambda_1|u(t)|_2^2\le a|u(t)|_2^2 $$ or $$ \frac{1}{2}\frac{d}{dt}|u(t)|_2^2+(\lambda_1-a)|u(t)|_2^2\le0. $$ Hence $$ |u(t)|_2^2\le |u(0)|_2^2e^{-2(\lambda_1-a)t} $$ or $$ |u(t)|_2\le |u(0)|_2e^{-(\lambda_1-a)t}. $$