Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 39B / iii

The stationary contribution travels with group velocityso $k_0=4V^2/(9S^2)$. The conditionis therefore equivalent, up to a constant factor, to $k_0ϵ\ll 1$: the dominant wavelength is much longer than the impacted interval. Its Fourier transform can then be replaced bywhich is the total initial vertical velocity integrated over space. The far field cannot resolve $W$ and $ϵ$ separately and depends only on their product $Wϵ$.