 Solution

The material surface is $F(x,y,t)=y-\eta (x,t)=0$. Its material derivative vanishes, giving the exact kinematic conditionFor irrotational flow $u=\nabla \varphi$, the Euler equations and $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$ integrate spatially to ${\varphi }_t+|\nabla \varphi {|}^2/2+gy+p/\rho =C(t)$. Adding a suitable function of time to $\varphi$ changes no velocity and makes the right side $p_a/\rho$. ThusTake $p_a$ as the air-side pressure and use the given capillary pressure jump on the fluid side. At first order in wave amplitude, the product ${\varphi }_x{\eta }_x$ and kinetic-energy term are quadratic, the curvature denominator is one to the required order, and replacing $y=\eta$ by $y=0$ changes first-order quantities only at second order. ConsequentlyFor a harmonic wave $\eta =A\cos (kx-\omega t)$, the harmonic potential decaying with depth isIt satisfies the kinematic condition. Substituting into the dynamic condition givesThe phase speed is $c=\omega /k$ and the group velocity of a narrow wave packet is $c_g=d\omega /dk$. For the positive-frequency branch and $\sigma =T/\rho$,For positive surface tension this ratio exceeds one exactly when $\sigma k^2>g$, proving