= Solution
Differentiate the specific energy at fixed discharge:
$$
\frac{dE}{dx}=(1-F^2)h'-F^2h\frac{b'}b.
$$
In the lossless flat channel $E$ is constant. A smooth <hydraulic control> has $F=1$, so regularity requires $b'=0$. For the prescribed width this occurs at the minimum width $x=0$. The required minimum specific energy is largest at the narrowest section; a controlled flow supplies exactly that head.
Far upstream $b\to\infty$, so the velocity of a finite-discharge flow with depth tending to $H$ tends to zero. Thus the <Bernoulli equation> gives $E=H$. At the control $u_c^2=g'h_c$ and $E=3h_c/2$, hence
$$
\boxed{x_c=0,\qquad h_c=\frac{2H}{3},\qquad
u_c=\sqrt{\frac{2g'H}{3}},\qquad Q=b(0)\sqrt{g'}\left(\frac{2H}{3}\right)^{3/2}.}
$$
The printed width normalization has $b(0)=1$. Downstream on the accelerating controlled branch the flow becomes shallower and supercritical.
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