 Solution

Irrotationality gives ${\mathbf{u}}_i=\nabla {\varphi }_i$ and ${\nabla }^2{\varphi }_i=0$. At $z=\zeta$, the exact kinematic conditions areand pressure continuity gives $p_1=p_2$. Bernoulli equation in each fluid is

Linearize at $z=0$ and write $\zeta =Z{e}^{i(kx-\omega t)}$. Decay at infinity givesThe kinematic conditions give $-i\omega Z=-k{A}_1=k{A}_2$, while pressure continuity givesElimination yieldsStable waves require the lower fluid to be denser, ${\rho }_2>{\rho }_1$; otherwise this is the Rayleigh-Taylor instability.