= 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 $\nabla\phi$, this gives the exact <kinematic boundary condition for a free-surface graph>
$$
\boxed{\eta_t+\phi_x\eta_x=\phi_y\quad\text{on }y=\eta.}
$$
At 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 interface>
$$
\boxed{\phi_t+\frac12(\phi_x^2+\phi_y^2)+g\eta=F(t)\quad\text{on }y=\eta.}
$$
There is no surface-tension term in this model.
Write $\phi=Ux+\varphi$ and linearize in the small perturbation $\varphi$ 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. Thus
$$
\boxed{(\partial_t+U\partial_x)\eta=\varphi_y,\qquad
(\partial_t+U\partial_x)\varphi+g\eta=0\quad(y=0).}
$$
<Incompressibility> requires <Laplace's equation> $\varphi_{xx}+\varphi_{yy}=0$. For $k>0$, the complex perturbation $b e^{ky}e^{i(kx-\omega t)}$ satisfies this equation and decays with depth. Put $\eta=\operatorname{Re}\{a e^{i(kx-\omega t)}\}$ and $\Omega=\omega-kU$. The linear <boundary conditions> become
$$
-i\Omega a=kb,\qquad -i\Omega b+ga=0.
$$
They are consistent precisely when
$$
\boxed{\Omega^2=gk,\qquad
\eta=\operatorname{Re}\left\{\frac{i(\omega-kU)b}{g}e^{i(kx-\omega t)}\right\}.}
$$
Consequently the two laboratory <phase velocities> of a <deep-water gravity wave> are
$$
\boxed{\frac\omega k=U\pm\sqrt{\frac gk}.}
$$
If propagation of a wave packet is intended, its <group velocities> are instead $d\omega/dk=U\pm\tfrac12\sqrt{g/k}$.
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