= Solution
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 $\tau=V_0t$ 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> is
$$
\boxed{s(z,t)=\begin{cases}
0.2,&z<-1-0.6\tau,\\
\displaystyle\frac12+\frac{0.3z}{1+0.6\tau},&|z|\leq1+0.6\tau,\\
0.8,&z>1+0.6\tau.
\end{cases}}
$$
The 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,
$$
s(z,t)=\begin{cases}
0.8,&z<-1+0.6\tau,\\
\displaystyle\frac12-\frac{0.3z}{1-0.6\tau},&|z|\leq1-0.6\tau,\\
0.2,&z>1-0.6\tau.
\end{cases}
$$
All the ramp <characteristic curves> meet at $z=0$ at
$$
\boxed{t_*=\frac{5}{3V_0}.}
$$
Thereafter a stationary <sedimentation shock> separates $s_L=0.8$ below from $s_R=0.2$ above. Indeed the <Rankine-Hugoniot condition> gives
$$
\dot z_s=\frac{f(0.2)-f(0.8)}{0.2-0.8}=0,
$$
and 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-52-settling-ramps.png]
{title=Spreading and shock formation for quadratic hindered-settling flux}
The dependence on $\alpha$ is controlled by the <hindered-settling flux inflection>:
$$
f''(s)=V_0\alpha(1-s)^{\alpha-2}\bigl[2-(\alpha+1)s\bigr].
$$
For any smooth initial profile, the <characteristic flow map> has <derivative>
$$
\frac{\partial z}{\partial\xi}=1+t f''(s_0(\xi))s'_0(\xi).
$$
This formula gives the first <characteristic crossing>: minimize the positive values of $-1/[f''(s_0)s'_0]$. At $\alpha=0$ neither ramp changes shape; both move down at $V_0$. For $0<\alpha\leq3/2$, the flux is convex throughout $0.2\leq s\leq0.8$: the increasing ramp spreads and the decreasing ramp steepens, although its entire ramp generally no longer focuses simultaneously. For $\alpha\geq9$, 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 $s_*$ satisfies the tangent-chord relation $f'(s_*)=[f(s_b)-f(s_*)]/(s_b-s_*)$, 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.
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