= Solution
Irrotationality gives $\mathbf u_i=\nabla\phi_i$ and $\nabla^2\phi_i=0$. At $z=\zeta$, the exact kinematic conditions are
$$
\zeta_t+\phi_{ix}\zeta_x=\phi_{iz},
$$
and pressure continuity gives $p_1=p_2$. <Bernoulli equation> in each fluid is
$$
p_i=-\rho_i\left(\phi_{it}+\tfrac12|\nabla\phi_i|^2+gz\right)+C_i(t).
$$
Linearize at $z=0$ and write $\zeta=Ze^{i(kx-\omega t)}$. Decay at infinity gives
$$
\phi_1=A_1e^{-kz}e^{i(kx-\omega t)},
\qquad
\phi_2=A_2e^{kz}e^{i(kx-\omega t)}.
$$
The kinematic conditions give $-i\omega Z=-kA_1=kA_2$, while pressure continuity gives
$$
\rho_1(-i\omega A_1+gZ)=
ho_2(-i\omega A_2+gZ).
$$
Elimination yields
$$
\boxed{
\omega^2=gk\frac{\rho_2-\rho_1}{\rho_1+\rho_2},
\qquad F(r)=\frac{1-r}{1+r}}.
$$
Stable waves require the lower fluid to be denser, $\rho_2>\rho_1$; otherwise this is the <Rayleigh-Taylor instability>.
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