= Solution
For <potential flow>, each <velocity potential> satisfies <Laplace's equation> in its fluid. The exact <kinematic boundary condition> on $y=\eta$ is $\eta_t+\phi_{j,x}\eta_x=\phi_{j,y}$. Keeping only first-order terms gives
$$
\phi_{1,y}=\phi_{2,y}=\eta_t\qquad(y=0).
$$
The linearized <Bernoulli equation> is $p_j=p_0-\rho_j(\phi_{j,t}+g\eta)$ at the interface, so <pressure> <continuity> gives
$$
\boxed{\rho_1\phi_{1,t}-\rho_2\phi_{2,t}+g(\rho_1-\rho_2)\eta=0\qquad(y=0).}
$$
For a decaying disturbance away from the interface, put $\psi=kx-\omega t$ and take
$$
\phi_1=\frac{A\omega}{k}e^{ky}\sin\psi\quad(y<0),\qquad
\phi_2=-\frac{A\omega}{k}e^{-ky}\sin\psi\quad(y>0).
$$
They satisfy <Laplace's equation> and the <kinematic boundary conditions> for $\eta=A\cos\psi$. Substituting in the dynamic condition gives the <dispersion relation>
$$
\boxed{\omega^2=gk\frac{\rho_1-\rho_2}{\rho_1+\rho_2},\qquad
c^2=\frac gk\frac{\rho_1-\rho_2}{\rho_1+\rho_2}.}
$$
If the heavier fluid is on top, $\omega^2<0$. Instead of oscillation, a Fourier disturbance has an exponentially growing component with growth rate $\sqrt{gk(\rho_2-\rho_1)/(\rho_1+\rho_2)}$. This is the <Rayleigh-Taylor instability> in the absence of <surface tension>.
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