 Solution

Take $z$ upward and $u_s>0$ downward. The quadratic hindered-settling flux is $j(\varphi )=-u_s\varphi (1-\varphi )$. Particle mass conservation gives the kinematic sedimentation equationon a smooth characteristic curve. Integrate the scalar conservation law across a moving discontinuity: $V[\varphi ]=[j]$. The Rankine-Hugoniot condition gives the settling shock speedwhere $L$ is the lower side and $R$ the upper side. Since $j^{\prime \prime }=2u_s>0$, a compressive shock has ${\varphi }_L>{\varphi }_R$ and its neighboring characteristic curves point into it.

Put $a={\varphi }_1$, so ${\varphi }_2=2a$, with $0<a<1/4$. The two-layer sedimentation with a compression shock hasThese respectively separate upper suspension from clear fluid, lower suspension from upper suspension, and deposit from lower suspension. Their initial positions are $H,z_m,0$. The characteristic curves in the two suspended states have slopes $u_s(2a-1)$ and $u_s(4a-1)$; clear-fluid and deposit characteristic curves have slopes $-u_s$ and $u_s$.

For $z_m=H/2$, $V_2$ and $V_3$ meet first. Their intersection givesThe potential meeting of the upper two fronts would take $(H-z_m)/(2u_{s}a)$, which is later in the stated range. The new front separates particle volume fraction one from particle volume fraction $a$, soIt follows $z=h_{c1}+u_{s}a(t-t_{c1})$ until it meets $z=H-u_s(1-a)t$. The final mass-conserving intersection isFor $z_m=H/2$, these are $H(1-3a/2)/[u_s(1-a)]$ and $3aH/2$.

The PDF prints a plus sign before $a{z}_m/H$ in the stopping-time numerator. That sign contradicts both the top-front trajectory and particle mass conservation; the correct sign is minus, as derived here. Once all particles are deposited, the remaining $1$-to-$0$ front is stationary because both limiting fluxes vanish.

The first panel shows the two-front merger and the final intersection, with straight characteristic curves in each suspended layer. The second panel shows the tuned three-front intersection.