= Solution
In the inviscid incompressible irrotational model, write the total velocity as $(U+\phi_x,\phi_y)$. The perturbation potential obeys <Laplace's equation> in $-h<y<\eta(x,t)$. The complete boundary conditions, neglecting surface tension and taking the air pressure constant, are
$$
\phi_y=0\quad\text{at }y=-h,
$$
$$
\eta_t+(U+\phi_x)\eta_x=\phi_y,\qquad
\phi_t+U\phi_x+\frac12(\phi_x^2+\phi_y^2)+g\eta=0
\quad\text{at }y=\eta(x,t).
$$
The first surface condition expresses that the <free surface> is material; the second is the <Bernoulli equation> with atmospheric pressure and the constant background kinetic <energy> absorbed into the potential's time gauge.
At first order, evaluation is on $y=0$ and products of perturbations vanish. Hence
$$
\boxed{\eta_t+U\eta_x=\phi_y,\qquad\phi_t+U\phi_x+g\eta=0\quad(y=0),\qquad\phi_y=0\quad(y=-h).}
$$
For a mode $e^{i(\omega t-kx)}$, the bottom condition selects $\phi=C\cosh[k(y+h)]e^{i(\omega t-kx)}$. Let $\Omega=\omega-kU$. The linear surface conditions are
$$
i\Omega a=Ck\sinh(kh),\qquad i\Omega C\cosh(kh)+ga=0.
$$
Eliminating $C$ gives the <dispersion relation>
$$
\boxed{(\omega-kU)^2=gk\tanh(kh).}
$$
It is the intrinsic gravity-wave frequency squared, with the uniform current supplying the Doppler shift. For either sign of real $k$, $k\tanh(kh)$ is nonnegative.
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