= Solution
Without <gas drag>, the <dispersion relation> factorizes as $s(s^2+\omega^2)=0$. There is a neutral branch and two density-wave branches. If $\omega^2>0$, the waves have $s=\pm i\omega$; if $\omega^2<0$, one root is $\boxed{s=\sqrt{-\omega^2}>0}$ and the dust layer is gravitationally unstable.
For $c>0$, complete the square in the wavenumber magnitude:
$$
\omega^2=c^2\left(|k|-\frac{\pi G\sigma_0}{c^2}\right)^2
+\Omega^2-\frac{(\pi G\sigma_0)^2}{c^2}
=c^2\left(|k|-\frac{\pi G\sigma_0}{c^2}\right)^2+\Omega^2(1-Q^{-2}).
$$
The minimum occurs at $|k|=\pi G\sigma_0/c^2$. Thus the <Toomre stability criterion> is
$$
\boxed{Q=\frac{c\Omega}{\pi G\sigma_0}<1\quad\Longleftrightarrow\quad
\text{some axisymmetric wavelength is dynamically unstable}.}
$$
At $Q=1$ the minimizing mode is marginal. For $Q>1$ there is no growing axisymmetric wave in this razor-thin, pressure-supported, drag-free model. The unstable band for $Q<1$ is
$$
\frac{\pi G\sigma_0}{c^2}\left(1-\sqrt{1-Q^2}\right)<|k|
<\frac{\pi G\sigma_0}{c^2}\left(1+\sqrt{1-Q^2}\right).
$$
In a finite layer, this band must contain an admissible mode; the continuum statement assumes an adequately large domain.
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