 Solution

The axisymmetric razor-thin astrophysical disk equations areFor the stationary background, mass and azimuthal momentum conservation are automatic, while radial force balance and the Poisson equation giveHere $d{p}_0/dR=0$ under the stated constant-pressure assumption.

Retaining first-order perturbations and using $c_s^2=dp/d\Sigma$ giveswhereis the squared radial epicyclic frequency. In the local $|k|R\gg 1$ limit, a Fourier mode $e^{i(kR-\omega t)}$ obeysEliminating ${\Sigma }^{\prime }$, $u_{\varphi }^{\prime }$, and ${\Phi }^{\prime }$ yields the local dispersion relation

For a Keplerian orbit, $\Omega \propto R^{-3/2}$, soInstability requires ${\omega }^2<0$, equivalently a mode with positive imaginary frequency. Treating the right-hand side as a quadratic in $x=|k|$, its two roots areReal distinct roots, and hence an unstable interval $x_{-}<|k|<x_{+}$, exist exactly when the Toomre stability criterion has $Q<1$.

With $h=c_s/\Omega$,The constant positive term is epicyclic restoration by rotation and stabilizes long wavelengths. The negative term is the razor-thin disk's self-gravity and drives collapse. The positive $k^2$ term is gas-pressure restoration and stabilizes short wavelengths. Gravitational instability can therefore survive only on an intermediate band of scales.