 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 $u_j=\nabla {\varphi }_j$, and incompressibility gives $\Delta {\varphi }_j=0$ in each fluid. The unsteady Bernoulli equation iswhere the arbitrary time functions can be absorbed into the potentials. On $z=\zeta (x,t)$ the material-interface and normal-stress conditions areDisturbance velocities decay as $z\to \pm \infty$. The equilibrium pressures are $p_{\ast }-{\rho }_{j}gz$.

Linearize at the resting flat interface. The equations become $\Delta {\varphi }_j=0$, ${\zeta }_t={\varphi }_{j,z}$ on $z=0$, andFor $k>0$, write $\zeta =\eta e^{i(kx-\omega t)}$. Decay selects ${\varphi }_1=A_1{e}^{-kz}{e}^{i(kx-\omega t)}$ above and ${\varphi }_2=A_2{e}^{kz}{e}^{i(kx-\omega t)}$ below. The kinematic conditions give $A_1=i\omega \eta /k$ and $A_2=-i\omega \eta /k$. Substituting into the dynamic condition givesFor 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.