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For $\alpha =1$, $f=V_0(s^2-s)$, so the characteristic speed is $a=V_0(2s-1)$ and the conservation law flux is strictly convex. Let $\tau =V_{0}t$ and let $\xi$ be the initial vertical position of a characteristic curve.

For the increasing ramp, the characteristic curves within its sloping part satisfy $z=\xi (1+0.6\tau )$. They spread apart and never cross. The outer plateau characteristic speeds are $-0.6V_0$ and $0.6V_0$, so the complete entropy solution isThe transition widens linearly and its gradient decreases. At large time it approaches a rarefaction wave, with $s\simeq (1+z/\tau )/2$ inside its expanding transition.

For the decreasing ramp, $z=\xi (1-0.6\tau )$ and the characteristic curves converge. Before they meet,All the ramp characteristic curves meet at $z=0$ atThereafter a stationary sedimentation shock separates $s_L=0.8$ below from $s_R=0.2$ above. Indeed the Rankine-Hugoniot condition givesand the incoming characteristic speeds satisfy $a(0.8)=0.6V_0>0>a(0.2)=-0.6V_0$, which makes it an admissible entropy shock. The discontinuity transports no net excess particle flux because both plateau fluxes equal $-0.16V_0$. These are local solutions before a finite container's boundaries affect the transition.

The dependence on $\alpha$ is controlled by the hindered-settling flux inflection:For any smooth initial profile, the characteristic flow map has derivativeThis formula gives the first characteristic crossing: minimize the positive values of $-1/[f^{\prime \prime }(s_0)s_0^{\prime }]$. At $\alpha =0$ neither ramp changes shape; both move down at $V_0$. For $0<\alpha \le 3/2$, the flux is convex throughout $0.2\le s\le 0.8$: the increasing ramp spreads and the decreasing ramp steepens, although its entire ramp generally no longer focuses simultaneously. For $\alpha \ge 9$, the flux is concave throughout that interval and these roles reverse. At the endpoint values the curvature vanishes at one plateau; it does not reverse sign inside the interval.

For $3/2<\alpha <9$, the inflection $s_i=2/(\alpha +1)$ lies between the plateau concentrations. Below $s_i$ the increasing ramp spreads, while above it that ramp compresses; the decreasing ramp has the opposite local behavior. After characteristic crossing, the entropy solution can contain both a shock wave and a rarefaction wave. Quantitatively, an increasing step is selected by the lower convex envelope of $f$ on the interval between its states: curved portions supply rarefaction waves, and straight portions supply shock waves. A decreasing step uses the upper concave envelope. A join between a fan and a shock wave at a state $s_{\ast }$ satisfies the tangent-chord relation $f^{\prime }(s_{\ast })=[f(s_b)-f(s_{\ast })]/(s_b-s_{\ast })$, where $s_b$ is the other shock state. Thus curvature, not simply the magnitude of $\alpha$, decides the wave pattern; the envelope may reduce to one shock if the tangent lies outside the state interval.

Use downward distance $y=H-z$, with $z$ the height above the bottom, and write $W=W_s(a)>0$. This convention will also reproduce the printed negative deposit-front velocity in part (d). The ideal simple-cubic sphere packing has packing fraction $\pi /6$; leaving ${\varphi }_{\max }$ symbolic keeps the jump calculation independent of its value. In suspension, the kinematic sedimentation equation isIntegrate this conservation law across a moving discontinuity to obtain the Rankine-Hugoniot conditionThe brackets denote values on the increasing-$y$ side minus those on the other side. A sedimentation shock is a concentration jump moving at this secant slope of the particle flux. Deposited material is stationary and has zero flux.

For the upper clearing front, the states are $0$ and ${\varphi }_{\max }/8$, giving $U_1=7W/8$ downward. For the lower deposition front, the states are ${\varphi }_{\max }/8$ and ${\varphi }_{\max }$, giving $U_2=-W/8$, upward. ThusIn terms of height above the bottom, the shock paths areThey meet when $H=Wt$. Therefore complete settling occurs atThe final height also follows directly from particle volume fraction conservation, ${\varphi }_{\max }{h}_c=({\varphi }_{\max }/8)H$. The stationary deposit after the meeting carries no particle flux.

The diagram plots time vertically against height, as requested. Its shock slopes have the opposite spatial sign to $U_1,U_2$, because those velocities were defined in the downward coordinate.

Continue with downward coordinate $y=H-z$ and $W=W_s(a)>0$. By Stokes drag law, the small-particle isolated settling velocity is $W/4$. Normalize concentrations by ${\varphi }_{\max }$: write $q_i={\varphi }_i/{\varphi }_{\max }$. In suspension the specified independent hindered settling laws give fluxes $j_1/{\varphi }_{\max }=W{q}_1(1-q_1)$ and $j_2/{\varphi }_{\max }=(W/4)q_2(1-q_2)$. The initial state is $(q_1,q_2)=(1/8,1/8)$.

Two upper sedimentation shocks clear the species separately. The faster front removes large particles, leaving $(0,1/8)$ behind it; the small concentration is unchanged across this front because the species fluxes are independent. The slower front separates that small-only suspension from clear fluid. Their downward velocities areThere is also a growing mixed deposit at the bottom. Let its normalized solid concentrations be $(p_1,p_2)$, with $p_1+p_2=1$, and let its upward speed magnitude be $s=-{\hat{U}}_3>0$. Applying the Rankine-Hugoniot condition to each species givesAdding these relations provesThe deposited flux is zero even though each species occupies only part of the packing: the settling law applies to the suspension, while the deposit is a separate stationary state. These jump relations determine the deposit composition from sedimentation jump conditions.

The large-particle clearing front has height $H-7Wt/8$ and meets the mixed-deposit front at height $35Wt/192$. ThusAt this merger all large particles have entered the deposit. Only the small particles remain suspended, at normalized concentration $1/8$. The new material deposited above the mixed layer is pure small-particle solid, at total concentration ${\varphi }_{\max }$. Its sedimentation shock therefore has downward speedFor $t>t_1$, the deposit top and the remaining clearing front have heightsTheir meeting, or the total solid-volume balance, givesThis is two-stage bidisperse batch sedimentation: the fast species finishes first, followed by deposition of the remaining slow species.

The final solid composition by volume, and also by mass because the particle densities are identical, isThe upper pure-small layer has thickness $9H/116$. As checks, $(29/40)h_1=H/8$ accounts for all large-particle solid volume, and $(11/40)h_1+(h_2-h_1)=H/8$ accounts for all small-particle solid volume, in units of ${\varphi }_{\max }$. If “percentage of particles” means number rather than volume, a large sphere occupies eight times the volume of a small sphere: the mixed bottom layer contains $29/117\simeq 24.8\%$ large particles by number and $88/117\simeq 75.2\%$ small. The upper layer remains entirely small under either convention.

A moving discontinuity of particle volume fraction in kinematic sedimentation. Conservation of particle volume gives the Rankine-Hugoniot condition $U[\varphi ]=[j]$ in a consistently oriented coordinate. A clearing front and a deposit front can both be sedimentation shocks. A stationary deposit uses zero particle flux; its compacted branch need not obey the suspension flux law.

Initially, separate clearing fronts of fast and slow species coexist with a mixed deposition front. After the fast clearing front meets that deposit, the remaining suspension contains only the slow species. Its later sedimentation shock builds a pure-slow layer above the earlier mixed deposit. Final layer thicknesses and composition follow from separate particle-volume conservation.