= Solution
Take the current to be dilute, vertically well mixed and <Boussinesq approximation>, with constant channel width $w$, a horizontal absorbing bed, negligible ambient <fluid entrainment>, negligible bed friction, and no resuspension. The <shallow-water approximation> neglects vertical inertia and gives <hydrostatic pressure>. Particle loss alters <reduced gravity> but does not alter the leading-order carrier-fluid volume or inertia; otherwise additional mass and momentum terms would be necessary. The current occupies depth $h(x,t)$, has depth-uniform speed $u(x,t)$, and has <particle volume fraction> $c(x,t)$, with $g'=g_oc$ and constant $g_o>0$.
<Volume conservation> of a short channel segment gives $h_t+(uh)_x=0$. Its suspended particle amount per unit bed area is $hc$, and the downward <particle deposition flux> is $v_sc$. Thus $(hc)_t+(uhc)_x=-v_sc$, which becomes $(hg')_t+(uhg')_x=-v_sg'$ on multiplying by $g_o$.
At vertical coordinate $y$ above the bed, excess <hydrostatic pressure> divided by reference <mass density> is $g'(h-y)$. Integrating across the depth gives the depth-integrated <force> flux $g'h^2/2$. The horizontal momentum per unit bed area is $hu$ and the advected kinematic <momentum flux> is $hu^2$. The resulting <variable-buoyancy shallow water equations> are
$$
\boxed{h_t+(uh)_x=0,\qquad (hg')_t+(uhg')_x=-v_sg',\qquad (hu)_t+(hu^2+\tfrac12g'h^2)_x=0.}
$$
The derivative of the pressure term acts on both $h$ and $g'$; treating <reduced gravity> as horizontally constant while particles settle would lose part of the force.
For a steady interior, let $q=Q/w=uh>0$ denote <volume flux per unit width>. Then the particle balance is a <first-order linear differential equation>,
$$
\boxed{\frac{dg'}{dx}=-\frac{v_s}{q}g'=-\frac{v_sw}{Q}g',\qquad g'(x)=g'_0\exp\left(-\frac{v_sw}{Q}x\right).}
$$
Here $g'_0=g'(0)>0$, $v_s\geq0$ is constant, and deposited particles are not returned to the current. In particular the decay length is $q/v_s=Q/(wv_s)$ for $v_s>0$, and the cancellation of depth follows from $uh=q$: dilution and transit time compensate in the deposition balance.
The steady momentum equation also gives
$$
\boxed{K=u^2h+\frac12g'h^2=\frac{q^2}{h}+\frac12g'h^2=\text{constant}.}
$$
The inlet depth selects $K=q^2/h_0+g'_0h_0^2/2$. Depth and speed at every downstream position are determined by the continuous positive <root of a polynomial> in
$$
\boxed{\frac12g'_0e^{-v_sx/q}h^3-Kh+q^2=0,\qquad u=\frac qh.}
$$
This implicit expression also makes clear why the inlet condition is needed: there are generally two possible positive depths.
The <depth branches of a steady depositing gravity current> are distinguished by the <Froude number> $\operatorname{Fr}^2=q^2/(g'h^3)$. Differentiating the momentum invariant, and using the <exponential decay> of <reduced gravity>, gives
$$
\left(g'h-\frac{q^2}{h^2}\right)h_x=\frac{v_sg'h^2}{2q},\qquad \boxed{h_x=\frac{v_sh}{2q(1-\operatorname{Fr}^2)}.}
$$
For $v_s>0$, <supercritical flow> has $\operatorname{Fr}>1$ and decreasing depth, with $h\to q^2/K$ and $u\to K/q$ as $x\to\infty$. <Subcritical flow> has $\operatorname{Fr}<1$ and increasing depth, with $h\sim\sqrt{2K/g'(x)}\propto e^{v_sx/(2q)}$. This indefinitely deepening branch eventually violates the assumed shallow or confined-channel geometry. Hence the depth does not universally decrease: that conclusion needs a supercritical inlet.
For fixed $g'>0$, $q^2/h+g'h^2/2$ has its unique minimum at $h_c=(q^2/g')^{1/3}$, with $K_c=3q^{4/3}(g')^{1/3}/2$. The two positive roots when $K>K_c$ lie below and above $h_c$. Since $K_c$ decreases downstream while $K$ is constant, an initially strict branch remains separate from the other branch and cannot cross a critical point. A critical inlet with $v_s>0$ has no finite depth derivative: at $\operatorname{Fr}=1$ the first differentiated equation would equate zero to a positive value. It requires a singular entrance adjustment or physics beyond this steady model. If $v_s=0$, both <reduced gravity> and the continuous depth branch are constant. These steady solutions describe the supplied current interior, not a moving finite-release front.
For two particle populations, write $g'=g'_1+g'_2$ and $g'_i=g_{0i}c_i$, allowing different constant particle-density conversion factors. The bidisperse <particle-laden gravity current> equations are
$$
\boxed{h_t+(uh)_x=0,\qquad (hg'_i)_t+(uhg'_i)_x=-v_i g'_i\ (i=1,2),\qquad (hu)_t+\bigl(hu^2+\tfrac12h^2(g'_1+g'_2)\bigr)_x=0.}
$$
For the steady case,
$$
g'_i(x)=g'_{i0}e^{-v_ix/q},\qquad c_i(x)=c_{i0}e^{-v_ix/q},\qquad K=\frac{q^2}{h}+\frac12h^2\bigl(g'_{10}e^{-v_1x/q}+g'_{20}e^{-v_2x/q}\bigr).
$$
Thus the depth is obtained from the same cubic <polynomial> invariant with the sum of the two exponentially decreasing <reduced gravity> contributions.
The local deposit consists of the two <particle deposition fluxes> $v_1c_1$ and $v_2c_2$. For a positive total deposition rate, the <bidisperse gravity-current deposition> fraction by particle volume is
$$
\boxed{f_1(x)=\frac{v_1c_{10}e^{-v_1x/q}}{v_1c_{10}e^{-v_1x/q}+v_2c_{20}e^{-v_2x/q}}.}
$$
If the particles have a common material density, this is also their local mass fraction. If they have common $g_o$, it can be written in the source's buoyancy notation as
$$
\boxed{f_1(x)=\left[1+\frac{v_2g'_{20}}{v_1g'_{10}}\exp\left(\frac{(v_1-v_2)wx}{Q}\right)\right]^{-1}.}
$$
For unequal $g_{0i}$ replace each $g'_{i0}$ by $g'_{i0}/g_{0i}$. The exponent sign matters: with both populations present and $v_1>v_2>0$, the faster-settling population has decreasing local deposit fraction downstream, since $f_1'=(v_2-v_1)f_1(1-f_1)/q<0$. Equal positive settling speeds give a constant fraction. A nonsettling population contributes no deposit; if the entire local deposition rate is zero the deposit fraction is undefined.
If “fraction of particles” means number fraction rather than particle-volume or mass fraction, let $a_i$ be a single particle's volume in population $i$. The correct weights are then $v_ic_i/a_i$, so
$$
\boxed{f_1^{\mathrm{number}}(x)=\frac{v_1c_{10}e^{-v_1x/q}/a_1}{v_1c_{10}e^{-v_1x/q}/a_1+v_2c_{20}e^{-v_2x/q}/a_2}.}
$$
Settling speeds alone do not determine $a_i$ without particle properties and a drag law. Finally, if the intended deposit is cumulative over the bed from $0$ to $x$, rather than local at $x$, integrating <particle deposition flux> gives the species volumes per unit channel width $D_i(x)=q c_{i0}(1-e^{-v_ix/q})$. The cumulative particle-volume fraction is
$$
\boxed{F_1(x)=\frac{c_{10}(1-e^{-v_1x/q})}{c_{10}(1-e^{-v_1x/q})+c_{20}(1-e^{-v_2x/q})}.}
$$
For both $v_i>0$, its limit at infinity is the incoming particle-volume fraction, whereas its limit as $x\downarrow0$ is the local deposition fraction at the inlet. Local sorting and the cumulative inventory are different observables.
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