= Solution
Use the standard ideal-fluid assumptions underlying the stated relation: both fluids are incompressible, inviscid and unbounded away from the interface, with no surface tension. An <irrotational flow> has $u_j=\nabla\phi_j$, and <incompressibility> gives $\Delta\phi_j=0$ in each fluid. The unsteady <Bernoulli equation> is
$$
\phi_{j,t}+\tfrac12|\nabla\phi_j|^2+\frac{p_j}{\rho_j}+gz=C_j(t),
$$
where the arbitrary time functions can be absorbed into the potentials. On $z=\zeta(x,t)$ the material-interface and normal-stress conditions are
$$
\zeta_t+\phi_{j,x}\zeta_x=\phi_{j,z}\quad(j=1,2),\qquad p_1=p_2.
$$
Disturbance velocities decay as $z\to\pm\infty$. The equilibrium pressures are $p_*-\rho_jgz$.
Linearize at the resting flat interface. The equations become $\Delta\phi_j=0$, $\zeta_t=\phi_{j,z}$ on $z=0$, and
$$
\rho_1(\phi_{1,t}+g\zeta)=\rho_2(\phi_{2,t}+g\zeta).
$$
For $k>0$, write $\zeta=\eta e^{i(kx-\omega t)}$. Decay selects $\phi_1=A_1e^{-kz}e^{i(kx-\omega t)}$ above and $\phi_2=A_2e^{kz}e^{i(kx-\omega t)}$ below. The kinematic conditions give $A_1=i\omega\eta/k$ and $A_2=-i\omega\eta/k$. Substituting into the dynamic condition gives
$$
\rho_1(\omega^2/k+g)\eta=\rho_2(-\omega^2/k+g)\eta,
\qquad \boxed{\omega^2=\frac{\rho_2-\rho_1}{\rho_2+\rho_1}gk.}
$$
For a signed wave number replace $k$ by $|k|$. A heavier lower fluid gives stable <interfacial gravity waves>; a heavier upper fluid gives $\omega^2<0$ and <Rayleigh-Taylor instability>. The density ordering is not silently assumed positive in the derivation.
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