Write $A=g(\rho_1-\rho_2)$ and $\varepsilon=k_cL_1\ll1$. Throughout the unstable band, $kL_1\le\varepsilon$ and $\coth(kL_1)=1/(kL_1)+O(kL_1)$. Since $\rho_2<\rho_1$,
$$
\sigma^2(k)=\frac{Ak-\gamma k^3}{\rho_1\coth(kL_1)+\rho_2}
=\frac{L_1}{\rho_1}(Ak^2-\gamma k^4)[1+O(\varepsilon)].
$$
The leading polynomial is maximized at $k^2=A/(2\gamma)$, where it equals $L_1A^2/(4\rho_1\gamma)$. Thus the <thin-layer Rayleigh-Taylor growth asymptotics> give
$$
\boxed{\sigma_m\sim\frac{g(\rho_1-\rho_2)}2\sqrt{\frac{L_1}{\rho_1\gamma}}.}
$$
This is a leading asymptotic expression; the exact maximum for a small but nonzero layer depth has corrections.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 331 / 1 / c / ii / Solution

{title2=$\sigma_m\sim\tfrac12g(\rho_1-\rho_2)\sqrt{L_1/(\rho_1\gamma)}$}

For an infinitely deep lower layer and $k_cL_1\ll1$, write $A=g(\rho_1-\rho_2)$. Uniformly over the unstable band, the <finite-depth Rayleigh-Taylor dispersion relation> gives $\sigma^2\sim(L_1/\rho_1)(Ak^2-\gamma k^4)$. Its maximum occurs at $k_m\sim k_c/\sqrt2$, with the displayed growth rate. In two deep layers the maximum instead occurs at $k_c/\sqrt3$ and obeys $\sigma_m^2=2Ak_c/[3\sqrt3(\rho_1+\rho_2)]$. Thus confinement strongly reduces growth while leaving the cutoff unchanged. These thin-layer expressions are leading asymptotics, not exact finite-depth identities.

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