= Solution
The <free surface> is a material surface, so its <kinematic boundary condition> is
$$
\eta_t+\phi_x\eta_x=\phi_y\quad\text{at }y=\eta.
$$
With constant atmospheric pressure $p_a$ and no <surface tension>, the <dynamic boundary condition for an inviscid interface> follows from <Unsteady Bernoulli equation>:
$$
\phi_t+\tfrac12|\nabla\phi|^2+g\eta=F(t)-p_a/\rho\quad\text{at }y=\eta.
$$
Write $\phi=Ux+\varphi$, where $\varphi$ and $\eta$ are small. Choose the time-dependent potential gauge so the undisturbed right-hand side is $U^2/2$. Retaining first-order terms, evaluated at $y=0$, gives
$$
\boxed{\eta_t+U\eta_x=\varphi_y,\qquad \varphi_t+U\varphi_x+g\eta=0.}
$$
The <incompressibility> condition also gives <Laplace's equation> for $\varphi$.
Set $\Omega=\omega-kU$. Substitution of the specified <normal mode> into the two linearized <boundary conditions> yields $-i\Omega a=ikb$ and $\Omega b+ga=0$. Eliminating $b$ gives the <dispersion relation>
$$
\boxed{(\omega-kU)^2=gk.}
$$
The decay factor $e^{ky}$ in a fluid extending to $y=-\infty$ uses $k>0$. With signed <wavenumber> the decaying factor is $e^{|k|y}$ and the right-hand side is $g|k|$. The frequency shift is the <advection> of a <deep-water gravity wave> by the uniform current.
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