= Solution
The stationary contribution travels with <group velocity>
$$
V=\omega'(k_0)=\frac{3S}{2}\sqrt{k_0},
$$
so $k_0=4V^2/(9S^2)$. The condition
$$
\frac{\epsilon V^2}{S^2}\ll1
$$
is therefore equivalent, up to a constant factor, to $k_0\epsilon\ll1$: the dominant <wavelength> is much longer than the impacted interval. Its Fourier transform can then be replaced by
$$
-\frac{2W\sin(k\epsilon)}k\sim-2W\epsilon,
$$
which is the total initial vertical velocity integrated over space. The far field cannot resolve $W$ and $\epsilon$ separately and depends only on their product $W\epsilon$.
Back to article page
