Treat $n^2=N^2/\Omega^2$ as a signed small parameter, with $|n^2|\ll1$, as printed in the original PDF. At $n^2=0$, the <dispersion relation> factors as
$$
(s+\beta)(s^2+\Omega^2)=0.
$$
Thus the two <epicyclic motion> modes have $s_0=\pm i\Omega$, while the <thermal energy mode of a shearing sheet> has
$$
\boxed{s_{E,0}=-\beta=-\xi k^2.}
$$
It decays on the <thermal diffusion time> $\beta^{-1}$. For completeness, its next correction is $s_E=-\beta+\beta N^2/(\beta^2+\Omega^2)+O(\Omega n^4)$ at fixed $\beta/\Omega$.

For either <epicyclic motion> root, the <implicit function theorem> applied to the <dispersion relation> gives
$$
2s_0(s_0+\beta)s_1+\Omega^2s_0=0,\qquad
\boxed{s_1=-\frac{\Omega^2}{2(\beta+s_0)}.}
$$
Hence
$$
s_\pm=\pm i\Omega-\frac{N^2}{2(\beta\pm i\Omega)}+O(\Omega n^4),\qquad
\operatorname{Re}s_\pm=-\frac{N^2\beta}{2(\beta^2+\Omega^2)}+O(\Omega n^4).
$$
Since $\beta>0$, the oscillations undergo <overstability> precisely when
$$
\boxed{N^2<0.}
$$
The <convective overstability growth rate> tends to zero both for very slow diffusion and for very fast diffusion. Differentiating $\beta/(\beta^2+\Omega^2)$ shows that its maximum occurs at $\beta=\Omega$, corresponding to $k^2=\Omega/\xi$. Therefore
$$
\boxed{\operatorname{Re}s_{\rm max}=-\frac{N^2}{4\Omega}+O(\Omega n^4),\qquad\xi k_{\rm max}^2=\Omega.}
$$
This is the growth rate of the perturbation amplitude; a quadratic perturbation energy grows at twice that rate. The <convective overstability> arises when an adverse radial <specific entropy> gradient couples to <epicyclic motion> with a finite <thermal conduction> lag.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 321 / 3 / d / Solution

{title2=$\sigma=-\frac{N^2\beta}{2(\beta^2+\Omega^2)}$}

For $|N^2|\ll\Omega^2$, this leading amplitude-growth rate is largest at <thermal diffusion rate of a Fourier mode> $\beta=\Omega$, giving $\sigma_{\rm max}=-N^2/(4\Omega)$ when $N^2<0$.

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 Incoming links: Convective overstability growth rate