= Solution
Let $W\geq0$ be the downward fluid suction speed and $V\geq0$ the downward particle <settling velocity> relative to that fluid. Set $Q=uh$ and $g'=g'_0\phi$. The carrier <volume conservation> and particle-volume balances per unit channel width are
$$
\boxed{Q'=-W,\qquad (Q\phi)'=-(W+V)\phi.}
$$
The particle sink includes suction as well as relative settling: the particles follow the well-mixed withdrawn fluid, in addition to settling through it. Subtracting $\phi Q'$ yields $Q\phi'=-V\phi$. Thus for $W>0$, as long as $Q>0$,
$$
\boxed{Q=Q_0-Wx,\qquad\phi=\left(1-\frac{Wx}{Q_0}\right)^{V/W}.}
$$
For $W=0$ the continuous limit is $Q=Q_0$ and $\phi=e^{-Vx/Q_0}$. In particular suction alone does not change the well-mixed particle fraction: $V=0$ implies $\phi=1$.
Withdrawn fluid carries its local horizontal <momentum>, so the <momentum flux> and <hydrostatic pressure> force obey
$$
\boxed{\left(hu^2+\frac12g'_0\phi h^2\right)'=-Wu.}
$$
Expand this and use $Q'=-W$. The suction <momentum> terms cancel in the primitive acceleration equation, leaving
$$
uu'+g'_0\phi h'+\frac12g'_0h\phi'=0.
$$
These are the governing equations for a <particle-laden current with suction>. Substituting $u=Q/h$ and the <concentration> equation gives the convenient depth equation
$$
\boxed{(1-\operatorname{Fr}^2)h'=\frac{Wu}{g'_0\phi}+\frac{Vh}{2Q},\qquad \operatorname{Fr}^2=\frac{u^2}{g'_0\phi h}.}
$$
The steady carrier flux is exhausted at the formal length $\boxed{x_*=Q_0/W}$ when $W>0$. Positive $Q$ cannot continue beyond it. Near that limit a chosen depth branch, mixing or shallow-layer assumptions may fail, so this is the mass-budget extent, not a guarantee that every arbitrary inlet state has a regular shallow continuation to it. When $W=0$, settling makes the driving <reduced gravity> exponentially small but does not remove carrier fluid; this ideal steady model has no finite mass-budget runout.
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