Let's assume system $$ \tag 1 \frac{\partial \mu}{\partial t} = \gamma (\mathbf B \cdot \mathbf E), $$ $$ \tag 2 [\nabla \times \mathbf E] = -\frac{\partial \mathbf B}{\partial t}, $$ $$ \tag 3 [\nabla \times \mathbf B ] = \sigma \mathbf E + \frac{\alpha}{\pi}\mu \mathbf B, $$ with initial conditions $$ \mathbf B(0) = \mathbf B_{0}, \quad \mu (0) \neq \frac{\alpha}{\pi} k|\mathbf B_{0}|. $$ System $(1)-(3)$ has exact solution for ansatz $\mathbf B = b(t)(sin(kz), cos(kz), 0)$: $$ b(t) = \frac{B_{0}\sqrt{\delta +B_{0}^{2}\gamma}e^{t(\delta + B_{0}^{2}\gamma)}}{\sqrt{B_{0}^{2}\gamma e^{2t(\delta + B_{0}^{2}\gamma)} + \delta}}. $$ So at infinity
$b(t) \to const , \quad \mu (t) \to const, \quad \mathbf E (t) \to 0$.
Let's then assume that there are perturbations $\delta \mu , \quad \delta \mathbf E, \quad \delta \mathbf B $ with trivial initial conditions $\delta \mu (0)=\delta \mathbf E(0) = \delta \mathbf B (0) = 0$. If we perturb $(1)-(3)$ by substitutions $\mu \to \mu_{5} + \delta \mu_{5}$, we will get system which has trivial solution $\delta \mu = \delta \mathbf B = \delta \mathbf E = 0$.
I need to check if this solution is stable. I've searched for some books which contains discussion about methods of verification the stability of trivial solutions of nonlinear system of differential equations in partial derivatives like $(3)$, but I haven't found them. Books which I've found contain cases which don't coincide with my case.
Can you help me with searching? Or, maybe, you know the methods for checking stability of perturbed trivial solution of $(3)$?
Edit.
Since $\mathbf B , \mathbf E$ are EM field tensor components, in Coulomb gauge I can write $\mathbf E = -\frac{\partial \mathbf A}{\partial t}, \mathbf B = [\nabla \times \mathbf A]$, $\nabla \cdot \mathbf A = 0$, so perturbed system $(1)-(3)$ may be written in a form $$ \tag 4 \frac{\partial \mu}{\partial t} = -\gamma \left( \frac{\partial \delta \mathbf A}{\partial t} \cdot \mathbf B + \frac{1}{k}\frac{\partial \mathbf B}{\partial t} \cdot [\nabla \times \delta \mathbf A]\right), $$ $$ \tag 5 \sigma \frac{\partial \delta \mathbf A}{\partial t}=\Delta \mathbf A + \frac{\alpha}{\pi }\left( [\nabla \times \delta \mathbf A] \mu + \delta \mu \mathbf B\right), $$ $$ \tag 6 \delta \mathbf A (0) = \delta \mu (0) = 0. $$
I wrote up some notes on how one can start doing such an analysis. It is far from complete, but hopefully it can be useful.
We start by perturbing the system of equations to find
$$\frac{d\delta\mu}{dt} = \gamma[B \cdot \delta E + \delta B \cdot E]$$ $$\nabla\times\delta E = -\frac{d\delta B}{dt}$$ $$\nabla\times\delta B = \sigma\delta E + \frac{\alpha}{\pi}[\mu \delta B + \delta\mu B]$$
Taking the divergence of the two equations gives us: $$\nabla\cdot\delta B = 0$$ $$\nabla\cdot\delta E = -\frac{\alpha}{\pi\sigma}[B \cdot\nabla \delta\mu]$$ where we have used that $\mu = \mu(t)$ and $\nabla\cdot B = 0$. Taking the curl gives us
$$\nabla(\nabla\cdot\delta E)-\nabla^2\delta E = -\frac{d}{dt}\nabla\times\delta B$$ $$-\nabla^2\delta B = -\sigma\frac{d\delta B}{dt} + \frac{\alpha}{\pi}[\mu \nabla\times\delta B + \nabla\delta\mu\times B + \delta\mu\nabla\times B]$$
It follows that (the general solution has an additional term $\delta \tilde{E}$ satisfying $\nabla\cdot \delta \tilde{E} =0$ which we for simplicity here just assume is zero)
$$\delta E = -\frac{\alpha}{\pi\sigma}\delta\mu B$$
and
$$\nabla\times\delta B = \frac{\alpha}{\pi}\mu \delta B$$
Taking the curl of the equation above gives us
$$\nabla^2\delta B = -\left(\frac{\alpha}{\pi}\mu\right)^2 \delta B$$
This equation has the solution (for simplicity we only include a single $\cos$ term in the solution)
$$\delta B = A\cos\left(\frac{\alpha\mu}{\pi}n \cdot r\right)$$
for some constant vector $A$, unit vector $n$ with $n\cdot A = 0$ and where $r=(x,y,z)$. This is an oscillating (non-growing) solution and thus the $B$-field is therefore stable againgst perturbations (but the perturbations does not vanish with time). It now remains to solve for $\delta E$ and $\delta \mu$ and see if they are also stable.