 Solution

The free surface is a material surface. For $F(x,y,t)=y-\eta (x,t)$, the material derivative vanishes on $F=0$. Since the velocity is $\nabla \varphi$, this gives the exact kinematic boundary condition for a free-surface graphAt the surface the pressure equals the constant atmospheric pressure. Absorbing that constant into the time-dependent potential gauge in Unsteady Bernoulli equation yields the exact dynamic boundary condition for an inviscid interfaceThere is no surface-tension term in this model.

Write $\varphi =Ux+\phi$ and linearize in the small perturbation $\phi$ and the surface displacement. The base value is $F=U^2/2$, and evaluation at the displaced surface can be replaced by evaluation at zero to first order. ThusIncompressibility requires Laplace's equation ${\phi }_{xx}+{\phi }_{yy}=0$. For $k>0$, the complex perturbation $b{e}^{ky}{e}^{i(kx-\omega t)}$ satisfies this equation and decays with depth. Put $\eta =\mathrm{Re}\{a{e}^{i(kx-\omega t)}\}$ and $\Omega =\omega -kU$. The linear boundary conditions becomeThey are consistent precisely whenConsequently the two laboratory phase velocities of a deep-water gravity wave areIf propagation of a wave packet is intended, its group velocities are instead $d\omega /dk=U\pm \frac{1}{2}\sqrt{g/k}$.