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 Neumann boundary condition $(A(x,t)\nabla u(x,t))\nu=0$for $x\in\partial\Omega$. Show that \begin{equation} \lVert u(t)-\frac{1}{|\Omega|}\int _{\Omega} u_0(x) dx\rVert _{L^2(\Omega)}\leq Ce^{- \mu t} \end{equation} for some $C,\mu >0$.
I tried the following.
First I subtracted the mean value of u over $\Omega$ from both sides of the equation and defined \begin{equation} v=u-\frac{1}{|\Omega|}\int _{\Omega} u_0(x) dx. \end{equation} Then $v$ satisfies \begin{equation} \partial _t v-\nabla (A(x,t)\nabla v)=-\frac{1}{|\Omega|}\int _{\Omega} \partial _t u_0(x) dx\ in\ \Omega \times (0,\infty) \end{equation} \begin{equation} v(x,0)=0\ in\ \Omega \end{equation} \begin{equation} (A(x,t)\nabla v)n=0,\ on\ \partial\Omega \end{equation} I know from the first part of the exercise that \begin{equation} \lVert u(t) \rVert _{L^2(\Omega)}\leq Ce^{- \lambda t} \end{equation} for some $C,\lambda >0$. Therefore, it holds that \begin{equation} \lVert v(t) \rVert _{L^2(\Omega)}\leq Ce^{- \lambda t} \end{equation} for some $C,\lambda >0$. My next step was to combine the two results \begin{equation} C_p \lVert \nabla v(t)\rVert \geq \lVert v(t)\rVert \geq 0. \end{equation} This implies that $\nabla v=0$. Thus, $v$ is a constant function. As the mean value of $v$ is zero, I know that $v\equiv 0$. Therefore, \begin{equation} \lVert \lVert u(t)-\frac{1}{|\Omega|}\int _{\Omega} u_0(x) dx\rVert\rVert _{L^2(\Omega)}=\lVert v(t)\rVert _{L^2(\Omega)}=0. \end{equation} This shows that \begin{equation} \lVert \lVert u(t)-\frac{1}{|\Omega|}\int _{\Omega} u_0(x) dx\rVert\rVert _{L^2(\Omega)}\leq Ce^{\mu t}. \end{equation} Is my solution correct or are there any mistakes?
Let $v = u - \frac{1}{\vert\Omega\vert} \int_\Omega u_0(x) \,dx$. Then $v$ is a solution to \begin{align*} \partial_t v - \nabla\cdot(A\nabla v) &= 0 \quad \text{in} \ \Omega\times (0,\infty), \\ v(x,0) &= u_0(x) - \frac{1}{\vert\Omega\vert} \int_\Omega u_0(x) \,dx \quad \text{in} \ \Omega, \\ (A\nabla v )\cdot\nu &= 0 \quad\text{on}\ \partial\Omega\times (0,\infty). \end{align*} As $u$ is mass preserving, i.e. $$ \frac{d}{dt} \int_\Omega u(t,x) \,dx = 0$$ for all $t > 0$ (Prove it!), we have in particular $$ \int_\Omega u(t,x) \, dx = \int_\Omega u_0(x) \, dx$$ for all $t > 0$. This implies that $$ \int_\Omega v(x,t) \,dx = \int_\Omega u(x,t) \,dx - \int_\Omega u_0(x) \, dx = 0$$ for all $t>0$. If we then define the energy $E:[0,\infty)\rightarrow\mathbb{R}$, $E(t) = \frac{1}{2} \int_\Omega v(t,x)^2 \,dx$, we obtain \begin{align} \frac{d}{dt} E(t) &= \int_\Omega v\partial_t v \,dx = - \int_\Omega A\nabla v \cdot\nabla v \, dx \\ &\leq - \frac{1}{\xi} \int_\Omega \vert\nabla v \vert^2 \, dx \\ &\leq -\frac{c}{\xi} \int_\Omega v^2 \, dx \\ &= -\frac{2c}{\xi} E(t). \end{align} Here we used the Poincare inequality, which is applicable since $\int_\Omega v \, dx = 0$. We conclude using Gronwall's inequality.