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Use the linearized boundary condition: evaluate at $z=0$ and discard products of perturbations. The kinematic boundary condition and Young–Laplace equation becomeIncompressible flow and irrotational flow give the Laplace equation for each potential. For a normal mode, the wall conditions selectWith $\eta =B{e}^{i(kx-\omega t)}$, the kinematic conditions give $A_1=i\omega B/[k\sinh (k{L}_1)]$ and $A_2=-i\omega B/[k\sinh (k{L}_2)]$. Hence ${\hat{\varphi }}_1(0)=i\omega B\coth (k{L}_1)/k$ and ${\hat{\varphi }}_2(0)=-i\omega B\coth (k{L}_2)/k$. Substitution into the dynamic condition gives the finite-depth Rayleigh-Taylor dispersion relation:Its denominator is positive. Assuming $g>0$ and $\gamma >0$, negative ${\omega }^2$ therefore produces a growing branch $\omega =i\sigma$ with $\sigma >0$, as in Rayleigh-Taylor instability.

Put $A=g({\rho }_1-{\rho }_2)>0$. In the deep-layer limit, the finite-depth Rayleigh-Taylor dispersion relation gives ${\omega }^2=(\gamma k^3-Ak)/({\rho }_1+{\rho }_2)$. Its sign changes atThe Rayleigh-Taylor instability band is $0<k<k_c$; $k=k_c$ is neutral and $k>k_c$ is oscillatory.

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.

For an infinitely deep lower layer and $k_c{L}_1\ll 1$, write $A=g({\rho }_1-{\rho }_2)$. Uniformly over the unstable band, the finite-depth Rayleigh-Taylor dispersion relation gives ${\sigma }^2\sim (L_1/{\rho }_1)(A{k}^2-\gamma k^4)$. Its maximum occurs at $k_m\sim k_c/\sqrt{2}$, with the displayed growth rate. In two deep layers the maximum instead occurs at $k_c/\sqrt{3}$ and obeys ${\sigma }_m^2=2A{k}_c/[3\sqrt{3}({\rho }_1+{\rho }_2)]$. Thus confinement strongly reduces growth while leaving the cutoff unchanged. These thin-layer expressions are leading asymptotics, not exact finite-depth identities.