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

For $L_2\to \infty$, the denominator is ${\rho }_1\coth (k{L}_1)+{\rho }_2>0$. The numerator is unchanged, so the cutoff is exactly independent of depth:This remains exact before taking the thin-layer limit in the finite-depth Rayleigh-Taylor dispersion relation.

Write $A=g({\rho }_1-{\rho }_2)$ and $\epsilon =k_c{L}_1\ll 1$. Throughout the unstable band, $k{L}_1\le \epsilon$ and $\coth (k{L}_1)=1/(k{L}_1)+O(k{L}_1)$. Since ${\rho }_2<{\rho }_1$,The leading polynomial is maximized at $k^2=A/(2\gamma )$, where it equals $L_1{A}^2/(4{\rho }_1\gamma )$. Thus the thin-layer Rayleigh-Taylor growth asymptotics giveThis is a leading asymptotic expression; the exact maximum for a small but nonzero layer depth has corrections.

Maximize the leading thin-layer squared growth rate, $(L_1/{\rho }_1)(A{k}^2-\gamma k^4)$. Its positive stationary point obeys $2Ak-4\gamma k^3=0$ and is the unique interior maximum. ThereforeThe maximizing point stays within the thin-layer regime because $k_m{L}_1\sim k_c{L}_1/\sqrt{2}\ll 1$.

Use superscripts $T$ and $D$ for the thin-upper-layer and two-deep-layer cases. The cutoff is identical, whileThus confinement suppresses the Rayleigh-Taylor instability growth strongly, but moves the most amplified disturbance to a somewhat larger wavenumber. The cutoff depends on the gravity-capillary balance, whereas growth depends also on the inertia of the moving layers.