= Solution
A <hydraulic control of a zero-PV rotating channel> is a critical section separating the upstream branch, where long-wave disturbances can propagate upstream, from the downstream branch, where they cannot. In the steady specific-energy description it is a stationary point of $E$ with respect to depth at fixed <volume flux> and local width. The control therefore fixes the discharge for a given upstream <Bernoulli function> and channel geometry; an arbitrary downstream condition cannot generally be imposed independently on the upstream reservoir.
Evaluate all quantities at $x=0$ and keep $Q,b$ fixed while differentiating. With $\Delta=fb$, the preceding branch gives
$$
\frac{\partial u_1}{\partial h_1}=\frac g\Delta\left(1-\frac{h_1}{h_2}\right),\qquad \frac{\partial E}{\partial h_1}=1+\frac{u_1}{\Delta}\left(1-\frac{h_1}{h_2}\right).
$$
At the <hydraulic control of a zero-PV rotating channel>, this derivative vanishes. Using $h_1-h_2=\Delta v/g$ reduces it to $u_1v=gh_2$, where $v=u_1+\Delta/2$. Equivalently,
$$
v\left(v+\frac\Delta2\right)=gh_1,\qquad v_c=\sqrt{gh_1+\frac{\Delta^2}{16}}-\frac\Delta4.
$$
Combining this with $Q=b h_1v-b\Delta v^2/(2g)$ cancels the quadratic term and leaves
$$
\boxed{Q_c=\frac bg\left[\sqrt{gh_1+\frac{b^2f^2}{16}}-\frac{bf}{4}\right]^3.}
$$
At control $u_{1c}=\sqrt{gh_1+\Delta^2/16}-3\Delta/4$ and $h_{2c}=u_{1c}v_c/g$. Both walls must remain wet. Since $v_c>0$, the second depth is positive precisely when $u_{1c}>0$, giving
$$
\boxed{bf<\sqrt{2gh_1},\qquad b<\frac{\sqrt{2gh_1}}{f}.}
$$
Equality is the limiting case $u_{1c}=h_{2c}=0$, with the second wall just dry. For a wider channel, a dry lateral region requires a different solution and the two-wet-wall <hydraulic control of a zero-PV rotating channel> formula cannot be continued. The channel must also remain sufficiently long and slowly varying for the transverse <geostrophic balance> approximation to hold. As a check, $f\to0$ gives $Q_c=b\sqrt g\,h_1^{3/2}$, the familiar rectangular-channel critical discharge.
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