 Solution

For fixed ${\omega }^2>0$ and weak gas drag, expand a wave root as $s=s_0+ϵs_1+O({ϵ}^2)$ with $s_0=\pm i\omega$. At order $ϵ$, the dispersion relation yieldsSince $s_0^2=-{\omega }^2$, this reduces to $-2{\omega }^2{s}_1-\Omega ({\omega }^2+{\Omega }^2)=0$. The weak-drag dust density waves therefore haveThe first-order real part is negative: the two density waves are weakly damped, with their frequencies unchanged to first order. Gas drag removes their perturbation energy relative to the fixed gas flow. The expansion is an asymptotic result at fixed positive ${\omega }^2$; it is not uniform close to ${\omega }^2=0$, where the nominal damping can become comparable to the oscillation frequency and a different scaling is needed.