The constant term of the real cubic <dispersion relation> is
$$
c=\nu k^2\left(3\beta\Omega^2-2\pi G\Sigma_0k+v_s^2k^2\right).
$$
A negative $c$ guarantees a positive real root because the cubic tends to $+\infty$ as $\lambda\to+\infty$.

If $\beta<0$, then $c<0$ at sufficiently small positive $k$, producing a <viscous instability of an accretion disk>. This is the local form of the global thin-disk diffusion equation
$$
\frac{\partial\Sigma}{\partial t}
=\frac3r\frac{\partial}{\partial r}
\left[r^{1/2}\frac{\partial}{\partial r}
(\nu\Sigma r^{1/2})\right].
$$
Since $d(\nu\Sigma)/d\Sigma<0$, the effective diffusion reverses sign and amplifies surface-density variations.

If $\beta>0$, instability is still possible when the quadratic expression in parentheses is negative. Its minimum occurs at $k=\pi G\Sigma_0/v_s^2$, so the condition is
$$
3\beta\Omega^2-\frac{(\pi G\Sigma_0)^2}{v_s^2}<0,
$$
or
$$
\boxed{Q<\frac1{\sqrt{3\beta}}}.
$$
This is a <secular gravitational instability of an astrophysical disk>: viscosity allows self-gravity to overcome rotational support even in part of the range that is stable by the inviscid $Q<1$ criterion.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 321 / 1 / e / Solution

Secular gravitational instability uses dissipation to weaken the epicyclic support that stabilizes a disk dynamically. It can therefore grow more slowly than an orbital time in parameter ranges where the inviscid disk is stable.


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 Incoming links: Secular gravitational instability of an astrophysical disk