Let $u = u(t,x)$ and $$ \dot{W}(t) \leq -\left(\frac{\sigma \omega \pi^2}{4L^2} + g\right)\int_0^L u^2 dx - \sigma(1-\omega) \int_0^L u_x^2 dx - aμ \int_0^L u^2_t dx $$ $$ -σ^{-1}c^2Q\int_0^L (u_t - σu_{xx}+gu)^2dx + σ^{-1}c^2Q(1 + c^{-2}σ(g-μ))\int_0^L u_t(u_t -σu_{xx} + gu)dx $$ $$ + (1 + σ^{-1}c^2Qg) \int_0^L u(u_t -σu_{xx} + gu)dx $$
where $\sigma,\mu,a, L, c, Q$ are positive, $g \in \mathbb{R}$, $\omega \in (0,1)$ and
$$ W(t) = \frac12\int_0^Lu^2dx + \frac{Q}{2}\int_0^L(u_t - \sigma u_{xx} + gu)^2dx+\frac{a}{2}\int_0^Lu_t^2 + \frac{ac^2}{2}\int_0^Lu_x^2dx $$
It has to be shown that there exists $\varepsilon >0$ such that $\dot{W}(t) \leq -\varepsilon W(t)$.
Attempt:
$$ M= \begin{bmatrix} \frac{\sigma\omega\pi^2}{4L^2} + g & 0 & -\frac{\sigma + c^2Qg}{2\sigma} \\ 0 & a\mu & -\frac{Q}{2}\left(\frac{c^2}\sigma + g - \mu \right) \\ -\frac{\sigma + c^2Qg}{2\sigma} & -\frac{Q}{2}\left(\frac{c^2}\sigma + g - \mu \right)& \frac{c^2Q}{\sigma} \end{bmatrix} \ $$
is the matrix of the quadratic form with respect to $u, u_t, (u_t-\sigma u_{xx}+gu)$ and it can be shown that there exist values of $Q,a >0$ and $g \in \mathbb{R}$ that make $M$ positive definite. Thus for such values, $M$ is diagonalizable.
How can we use this to yield the desired inequality?
Any help would be very appreciated.