I like to proof the exponential stability of a damped wave equation with the following form.
Let $G \subset \mathbb{R}^n$ be bounded with smooth margin. For $\gamma > 0$ we are given the damped wave equation:
$u_{tt} – \triangle u + \gamma u_t = 0$
$u(0,\cdot) = u_0(\cdot)$
$u_t(0,\cdot) = u_1(\cdot)$
$u|_{\partial G} = 0$
We are given the energy of this equation as $E(t) = \Vert u_t(t,\cdot) \Vert_2^2 + \Vert \nabla u(t,\cdot) \Vert_2^2$. I want to proof that $\exists C,c > 0$ such that $E(t) \leq C e^{-ct} E(0)$.
My approach is to use a lyapunov functional of the form $L(t) = E(t) + f(t)$ such that $\exists \alpha , \beta > 0: \alpha E(t) \leq L(t) \leq \beta E(t)$ and $\dfrac{d}{dt} L(t) \leq -dE(t)$ for $d>0$. With that we could esaly get the desired solution.
I obtained the derivative of $E(t)$ by multiplying the pde with $u_t$ and then integraiting it which gives me $\dfrac{d}{dt} E(t) = -2\gamma \int \limits _{G} u_t^2(t,x)dx$. But I can't think of a fitting definition for $L(t)$. Any suggestion or hint would be appreciated.
Best Answer
From the inequality you obtained it follows in particular that $$ \frac{dE}{dt} \leq -2\gamma E(t). $$ The desired estimate then follows from Gronwall's inequality. More concretely, note that if you move all the terms to the left and then multiply by $e^{2\gamma t}$, then the inequality is equivalent to $$ \frac{d}{dt}(e^{2\gamma t}E(t)) \leq 0. $$ Integrating, it follows that for $t>0$, $$ e^{2\gamma t}E(t) = E(0) + \int_0^t \frac{d}{ds}(e^{2\gamma s}E(s))~ds \leq E(0). $$ Therefore $$ E(t) \leq e^{-2\gamma t}E(0). $$ If you want to do it your way in terms of Lyapunov functionals, you could just take $L = E$.