I am trying to solve the following exercise. Consider the linear parabolic equation \begin{equation} \partial _t u -\nabla (A(x,t)\nabla u)=f(x,t), x \in \Omega,t>0, \end{equation} subject to the initial data $u(x,0)=u_0(x)$ for $x\in\Omega$, where the diffusion matrix A satisfies \begin{equation} \xi ^{-1}|y|^2\leq \sum _{i,j =1} ^n A_{i,j}(x,t)y_i y_j\leq\xi|y|^2, x\in\Omega,t>0,\forall y \in R^n \end{equation} for some $\epsilon>0$. Let $f\equiv 0$ and consider the Dirichlet boundary condition $u(x,t)=0$for $x\in\partial\Omega$. Show that \begin{equation} \lVert u(t)\rVert _{L^2(\Omega)}\leq Ce^{- \lambda t} \end{equation} for some $C,\lambda >0$.
I tried the following.
I used the Dirichlet Boundary Condition \begin{equation} \int _{\Omega} u\partial _t u dx-\int _{\Omega}A(x,t)\nabla u \nabla u dx=0 \end{equation} and integrated the first term by parts \begin{equation} \int _{\Omega} u\partial _t u dx=-\frac{1}{2}\frac{d}{dt}(\int _{\Omega} u^2dx). \end{equation} After that I applied the given inequality for A(x,t) \begin{equation} \int _{\Omega} A(x,t)\nabla u\nabla u dx \geq\xi ^{-1}\int _{\Omega}|\nabla u|^2 dx. \end{equation} By using Youngs inequality I got \begin{equation} \frac{d}{dt}(\int u^2 dx)+\frac{2}{\xi}\int _{\Omega}|\nabla u|^2 dx \leq 0. \end{equation} Then I defined the energy functional \begin{equation} E(t)=\frac{1}{2}\int _{\Omega} u^2 dx. \end{equation} \begin{equation} E(t)\leq E(0)e^{-\frac{t}{\xi}}. \end{equation} Finally I got \begin{equation} \lVert u(t) \rVert _{L^2}=(\int _{\Omega} u^2 dx)^{\frac{1}{2}}\leq (2E(t))^{\frac{1}{2}}\leq Ce^{-\lambda t} \end{equation} Is my solution correct or are there any mistakes?
Testing the PDE with $u$ and then using integration by parts yields $$\int_\Omega u\partial_t u \,dx + \int_\Omega A(x,t)\nabla u\nabla u \, dx = 0.$$ The first integral can be rewritten as $$ \int_\Omega u\partial_t u \, dx = \frac{1}{2} \frac{d}{dt} \int_\Omega u^2 \, dx,$$ while for the second one we use $$\int_\Omega A(x,t)\nabla u\nabla u \, dx \geq \frac{1}{\xi} \int_\Omega \vert \nabla u \vert^2 \, dx.$$ From there we obtain $$ \frac{d}{dt} \int_\Omega u^2 \,dx + \frac{2}{\xi}\int_\Omega\vert\nabla u\vert^2 \, dx \leq 0$$ and you can conclude as above using Gronwall's lemma.
EDIT: There seems to be another mistake when using Gronwall's lemma. With the above inequality we only have $$ \frac{d}{dt} E(t) \leq -\frac{2}{\xi} \int_\Omega \Vert \nabla u \Vert^2 \, dx.$$ Here we need to use Poincare's inequality (I suppose $\Omega$ is bounded) to get for some constant $c>0$ $$ \frac{d}{dt} E(t) \leq -\frac{c}{\xi} E(t).$$ Then we can use Gronwall's lemma to conclude $$ E(t) \leq E(0) e^{-ct/\xi}.$$