Energy estimate for a friedrich system with relaxation and periodic boundary conditions

Question

I'm considering the following system of pdes $$ \partial_t u + \sum\limits_{i=1}^3 \mathcal{A}_i\partial_{i} u = -\mathcal{R} u. $$ Where the matrices $\mathcal{A}_i$ are symmetrics, $u$ is valued in $\mathbb{R}^p$, on $\Omega = \left[0,1\right]^3$ with periodic boundary conditions. The matrix $\mathcal{R}$ is diagonal with positive coefficient. And i would like to show that the energy norm of $u$ is decreasing $$\partial_t \| u \| _{L^2(\Omega)^p} \leq 0.$$ Here is what i've done so far : I took the scalar product with $u$ in $\mathbb{R}^p$ and i integrated over $\Omega$, $$ \frac{d}{dt} \frac{1}{2} \int_{\Omega} \vert u(t,x) \vert ^2 dx +\sum\limits_{i=1}^3\int_{\Omega} \left(\mathcal{A}_i \partial_i u(t,x),u(t,x) \right) dx = - \int_{\Omega} \left(\mathcal{R}u(t,x),u(t,x) \right) dx $$ And i know that i have to use the periodic boundary condition to show that $\int_{\Omega} \left(\mathcal{A}_i \partial_i u(t,x),u(t,x) \right) dx = 0$ but i'm stucked, i only managed to show that $$ \int_{\Omega} \left(\mathcal{A}_i \partial_i u(t,x),u(t,x) \right) dx = - \int_{\Omega} \left(\mathcal{A}_iu(t,x),\partial_i u(t,x) \right) dx $$ which isn't that helpful. Any help would be greatly appreciated.

Answer 1

What you've derived is helpful. Since $\mathcal{A}_i$ is symmetric you have that $$ \int_\Omega (\mathcal{A}_i u, \partial_i u) = \int_\Omega ( u, \mathcal{A}_i\partial_i u) = \int_\Omega ( \mathcal{A}_i\partial_i u, u) $$ and so you've derived an equality of the form $X = -X$, which implies that $X=0$. Now you can conclude.