At $\mu=\mu_c$, the real undamped part $F$ vanishes. Put
$$
d=\frac12-\frac{\coth\mu_c}{\mu_c}.
$$
The neutral equation implies $d=1/4-1/\mu_c^2>0$. The radicand is therefore
$$
z=i\epsilon d-\frac{\epsilon^2}{4}.
$$
If $\sqrt z=u+iv$ is chosen with $v>0$, then
$$
v^2=\frac12\left(sqrt{\epsilon^2d^2+\frac{\epsilon^4}{16}}
+\frac{\epsilon^2}{4}\right)>\frac{\epsilon^2}{4}.
$$
Thus $v>\epsilon/2$, and the plus branch has
$$
\operatorname{Im}\widetilde c_+=-\frac\epsilon2+v>0.
$$
For $k\Lambda>0$ this is exponential growth. Lower-boundary damping therefore destabilizes the formerly neutral cutoff mode, an example of <dissipation-induced instability>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 333 / 3 / ii / Solution

Dissipation-induced instability occurs when damping one component of a coupled system moves an eigenvalue into the growing half-plane. Selective damping can alter the phase relation that previously stabilized two interacting waves or oscillators.


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 Incoming links: Dissipation-induced instability