= Solution
With $W=0$ and $V>0$, $Q=Q_0$ and $\phi=e^{-kx}$, where $k=V/Q_0$. No carrier <momentum> is withdrawn; hence the integrated <momentum flux> is exactly constant:
$$
\boxed{\frac{Q_0^2}{h}+\frac12g'_0e^{-kx}h^2=K,\qquad K=\frac{Q_0^2}{h_0}+\frac12g'_0h_0^2.}
$$
On an initially small-<Froude number> branch, the <pressure> term dominates. At leading order
$$
\boxed{h(x)\sim h_0e^{kx/2},\qquad u(x)\sim u_0e^{-kx/2}.}
$$
A slightly more accurate large-distance amplitude is $\sqrt{2K/g'_0}$, so $h\sim\sqrt{2K/g'_0}\,e^{kx/2}$. The <Froude number> satisfies $\operatorname{Fr}^2\sim\operatorname{Fr}_0^2e^{-kx/2}$ 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 $\delta=\operatorname{Fr}_0^{-2}\ll1$ and $y=h/h_0$:
$$
\frac1y+\frac\delta2e^{-kx}y^2=1+\frac\delta2,
$$
so
$$
\boxed{\frac h{h_0}=1+\frac\delta2(e^{-kx}-1)+O(\delta^2).}
$$
The depth decreases slightly and tends exactly to $h_\infty=Q_0^2/K=h_0/(1+\delta/2)$. Its <Froude number> grows like $\operatorname{Fr}_0e^{kx/2}$. Both approximations agree with the exact depth equation $h'=kh/[2(1-\operatorname{Fr}^2)]$: the <subcritical flow> deepens, while the <supercritical flow> thins.
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