= Solution
For $k>0$, try
$$
\eta=A\cos(kx-\omega t),\qquad
\phi=B\cosh(k(y+h))\sin(kx-\omega t).
$$
The potential satisfies Laplace's equation, and its derivative with respect to $y$ vanishes at $y=-h$. The surface kinematic condition gives
$$
Bk\sinh(kh)=A\omega.
$$
The dynamic condition gives $-B\omega\cosh(kh)+gA=0$. Eliminating the nonzero amplitudes yields
$$
\boxed{\omega^2=gk\tanh(kh).}
$$
Thus choosing either frequency sign satisfying this relation and $B=A\omega/(k\sinh(kh))$ produces the required harmonic wave.
Back to article page
