= Solution
Seek the periodic response of a <surface gravity wave forced by a traveling bottom>, with the same <wavenumber> $k$ and <angular frequency> $kc$ as the bottom. The linear equations have constant coefficients in $x,t$, so they preserve each such harmonic. There is no dissipative phase lag in this model; away from <resonance>, the surface response is in phase or in opposition with the bottom. Thus write $\eta=b\sin\theta$, $\theta=k(x-ct)$, with a signed coefficient $b$. Its physical <amplitude> is $|b|$. Arbitrary additional free <surface gravity waves> depend on initial data; the paper's expression selects the forced traveling component.
For $k>0$, $c\ne0$, use $\phi=F(y)\cos\theta$. The <Laplace equation> gives $F''-k^2F=0$. At $y=0$, the two <linearized free-surface boundary conditions> imply
$$
kcF(0)+gb=0,\qquad F'(0)=-kcb.
$$
These determine both constants in the vertical solution:
$$
\boxed{\phi(x,y,t)=-\frac b{kc}\left[g\cosh(ky)+kc^2\sinh(ky)\right]\cos(k(x-ct)).}
$$
Its vertical derivative at the bottom is
$$
\phi_y(x,-d,t)=\left[\frac{gb}{c}\sinh(kd)-kcb\cosh(kd)\right]\cos\theta.
$$
Matching it to $-akc\cos\theta$ and multiplying by $c$ gives
$$
b\left[kc^2\cosh(kd)-g\sinh(kd)\right]=akc^2,
$$
and therefore
$$
\boxed{b=\frac{akc^2}{kc^2\cosh(kd)-g\sinh(kd)}.}
$$
An equivalent expression for the <velocity potential>, which makes the bottom condition immediate, is
$$
\phi=\left[D\cosh(k(y+d))-ac\sinh(k(y+d))\right]\cos\theta,\qquad
D=\frac c{\sinh(kd)}\left[a\cosh(kd)-b\right].
$$
Let $\omega_f=kc$ be the forcing <angular frequency>, and let $\omega_0^2=gk\tanh(kd)$ be the natural <dispersion relation> from <Airy wave theory>. Then
$$
\frac ba=\frac{\omega_f^2\operatorname{sech}(kd)}{\omega_f^2-\omega_0^2}.
$$
The denominator vanishes at $c^2=(g/k)\tanh(kd)$, when the forcing travels at the free-wave <phase velocity>. For nonzero forcing there is \b[no bounded periodic forced solution at <resonance>]; the undamped linear response develops secular growth. Below that speed $b<0$, giving a phase reversal, while above it $b>0$. At $c=0$ the bottom is stationary: the rest solution has $\phi=0$ and $\eta=0$, consistent with the zero limit of $b$, though the formula containing $1/c$ must be interpreted through that limit.
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