= Solution
Let $S>0$ denote volume loss per unit downstream length and take $Q_0>0$. Mass balance gives $Q_x=-S$, so
$$
\boxed{Q(x)=Q_0-Sx.}
$$
Assume fluid is removed carrying its local axial <velocity>, with no additional axial force on the retained flow. Under this inviscid seepage approximation, the horizontal momentum equation and its Bernoulli head are unchanged. On the flat bed, <hydraulic control with prescribed seepage> therefore uses
$$
\boxed{E(h;x)=h+\frac{(Q_0-Sx)^2}{2gb_0^2[1+(x/L)^2]^2h^2}=\mathcal B.}
$$
The calculation applies on a reach with positive remaining discharge. Constant loss cannot be extended indefinitely past $x=Q_0/S$ while preserving this positive-flow model.
At a regular control, $F=1$ and the same differentiated energy condition requires $q_x=0$, where $q=(Q_0-Sx)/b(x)$. Differentiating gives
$$
q_x=\frac{Sx^2-2Q_0x-SL^2}{b_0L^2[1+(x/L)^2]^2}.
$$
Write $D=\sqrt{Q_0^2+S^2L^2}$. The two algebraic roots are $(Q_0\pm D)/S$. At the plus root $Q=-D<0$, so it is outside the stipulated positive-flow branch. The admissible control candidate is
$$
\boxed{x_c=\frac{Q_0-D}{S}=-\frac{SL^2}{Q_0+D}<0.}
$$
It lies upstream of the geometric throat. At this point $Q(x_c)=D$ and
$$
b(x_c)=\frac{2b_0D}{Q_0+D},\qquad q_c=\frac{Q_0+D}{2b_0}.
$$
It is a maximum of positive $q$: differentiating the numerator of $q_x$ at $x_c$ gives $2Sx_c-2Q_0=-2D<0$. Consequently
$$
\boxed{h_c=\left[\frac{(Q_0+D)^2}{4gb_0^2}\right]^{1/3}.}
$$
A control exists only if the reach includes this point and boundary conditions supply the critical conserved head $\mathcal B=3h_c/2$ and a suitable transcritical connection. Merely prescribing $Q_0$ and $S$ does not force that head. At $S=0$ the limiting location is zero and the usual throat depth is recovered.
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