= Solution
At $V=0$, the particle budget gives $\phi=1$, so $g'=g'_0$ is constant. The primitive <momentum conservation> equation integrates to
$$
\frac12u^2+g'_0h=\frac12u_0^2+g'_0h_0,\qquad u_0=Q_0/h_0.
$$
Introduce $q=Q/Q_0=1-Wx/Q_0$, $y=h/h_0$ and $\varepsilon=\operatorname{Fr}_0^2=Q_0^2/(g'_0h_0^3)\ll1$. The exact relation on the <subcritical flow> branch is
$$
y+\frac{\varepsilon q^2}{2y^2}=1+\frac\varepsilon2.
$$
Expanding uniformly for $0\leq q\leq1$ gives
$$
\boxed{\frac{h}{h_0}=1+\frac{\operatorname{Fr}_0^2}{2}(1-q^2)+O(\operatorname{Fr}_0^4),\qquad h\sim h_0.}
$$
The depth increases slightly because horizontal kinetic energy decreases as fluid is removed. In fact the depth equation has $h'\geq0$ on this branch, and
$$
\operatorname{Fr}^2=\varepsilon\frac{q^2}{y^3}\leq\varepsilon\ll1.
$$
Thus the small-<Froude number> approximation becomes better downstream, including as the flux vanishes. At leading order $u=(Q_0-Wx)/h_0$.
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