Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 52 / 4 / c / Solution

With $W=0$ and $V>0$, $Q=Q_0$ and $\varphi =e^{-kx}$, where $k=V/Q_0$. No carrier momentum is withdrawn; hence the integrated momentum flux is exactly constant:On an initially small-Froude number branch, the pressure term dominates. At leading orderA slightly more accurate large-distance amplitude is $\sqrt{2K/g_0^{\prime }}$, so $h\sim \sqrt{2K/g_0^{\prime }}{e}^{kx/2}$. The Froude number satisfies ${\mathrm{Fr}}^2\sim {\mathrm{Fr}}_0^2{e}^{-kx/2}$ and remains small. The growing depth can eventually violate confinement or the shallow water assumption; this does not make the asymptotic momentum calculation inconsistent within its intended regime.

On an initially large-Froude number branch, horizontal momentum flux dominates instead. The leading depth is constant, $\boxed{h\sim h_0}$, while the decaying reduced gravity makes the Froude number still larger. To resolve the small change, put $\delta ={\mathrm{Fr}}_0^{-2}\ll 1$ and $y=h/h_0$:soThe depth decreases slightly and tends exactly to $h_{\infty }=Q_0^2/K=h_0/(1+\delta /2)$. Its Froude number grows like ${\mathrm{Fr}}_0{e}^{kx/2}$. Both approximations agree with the exact depth equation $h^{\prime }=kh/[2(1-{\mathrm{Fr}}^2)]$: the subcritical flow deepens, while the supercritical flow thins.