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Take $z$ increasing upwards and write $s=\varphi /{\varphi }_{\max }$. The particles have downward settling velocity; their upward volume flux is $\varphi V$. Conservation of particle volume in any interval givesThus the kinematic sedimentation equation and its characteristic speed areFor $0<s<1$ this is a scalar hyperbolic partial differential equation: its one characteristic speed is real. The method of characteristics gives $ds/dt=0$ and $dz/dt=a(s)$. A concentration disturbance need not descend at the particle settling velocity; $a(s)$ is the derivative of the transported flux, not $V(s)$. In particular $a(s)$ can be positive even though every particle settles downwards. At $\alpha =0$, $a=-V_0$ and the equation is ordinary translation. At the packing endpoint $s=1$ the differentiability of the chosen constitutive law depends on $\alpha$; none of the initial values used below is at that endpoint.

Use the explicit printed width law $b(z)=\alpha z^2$. Its wetted area and hydrostatic pressure integral areAssuming hydrostatic pressure, a horizontal prismatic bed and cross-sectionally uniform velocity, volume conservation and momentum conservation give the prismatic-channel shallow water equationsEquivalently,The coefficient matrix in variables $(h,u)$ is $\left(\begin{array}{cc}u& h/3\\ g& u\end{array}\right)$. Its eigenvalues and the corresponding gravity-wave speed areFor $h>0$ these eigenvalues are distinct and real, establishing strict hyperbolicity in the hyperbolic partial differential equation sense. Since $dc/dh=c/(2h)$, substitution in the two equations showsFor example $({\partial }_t+(u+c){\partial }_x)(u+6c)=0$. This provides both the slopes and the conserved characteristic quantities. At a dry bed $h=0$, strict hyperbolicity degenerates and the front is interpreted as the limit from positive depth.

The descriptive word “parabolic” alone would normally suggest $b\propto \sqrt{z}$ and different coefficients. The explicit $z^2$ width in the PDF determines the calculation here; it must not be silently replaced by that other geometry.

The chain rule gives ${\partial }_x={\partial }_{\xi }+\beta {\partial }_{\eta }$ and ${\partial }_y=\alpha {\partial }_{\xi }+{\partial }_{\eta }$. The coefficients of $u_{\xi \xi }$ and $u_{\eta \eta }$ are $2+3{\alpha }^2-7\alpha$ and $2{\beta }^2+3-7\beta$. Their integer roots are $\alpha =2$ and $\beta =3$. Thus choose characteristic coordinates $\xi =x+2y$, $\eta =3x+y$, whose Jacobian determinant is $-5\ne 0$. The coefficient of $u_{\xi \eta }$ is $4\beta +6\alpha -7(1+\alpha \beta )=-25$, so the equation becomes $-25u_{\xi \eta }=0$, equivalently $u_{\xi \eta }=0$.

Integrating the transformed hyperbolic partial differential equation gives $u=F(\xi )+G(\eta )$. Along $x=-2y$ the coordinates are $(0,-5y)$, so the zero boundary condition gives $G(\eta )=-F(0)$ for every real $\eta$. Along $x=0$ they are $(2y,y)$, so $F(2y)-F(0)=4y^2$, hence $F(\xi )={\xi }^2+F(0)$. The constants cancel and the unique resulting solution isIt directly satisfies both boundary conditions and $2u_{xx}+3u_{yy}-7u_{xy}=4+24-28=0$.

Expanding the equation givesFor the hyperbolic partial differential equation discriminant, $A=1$, $B=\rho$, $C={\rho }^2-1$, soFor a curve $\rho =\rho (t)$, the characteristic hypersurface equation is $({\rho }^{\prime }{)}^2-2\rho {\rho }^{\prime }+{\rho }^2-1=0$, or ${\rho }^{\prime }=\rho \pm 1$. Therefore the two families of characteristic curves areThe values $C=0$ include the characteristic lines $\rho =-1$ and $\rho =1$. Every curve in the first family approaches $\rho =-1$ as $t\to -\infty$; every curve in the second approaches $\rho =1$. The two curves through $(\rho ,t)=(0,0)$ are $\rho =e^t-1$ and $\rho =1-e^t$, shown as the thick curves below. Equivalently, the characteristic coordinates are $e^{-t}(\rho +1)$ and $e^{-t}(\rho -1)$.

Assume $0<\epsilon <1$ and setThe travel-time coordinate for a one-dimensional variable-speed wave equationsends the initial interval $I_{\epsilon }=(-a,a)$ to $(-S_{\epsilon },S_{\epsilon })$. By the characteristic curves for speed one minus y squared, the two characteristic coordinates are $s-x$ and $s+x$. The finite propagation speed and uniqueness theorem for hyperbolic partial differential equations therefore give the maximal characteristic diamondEquivalently,In the $(x,s)$ plane this is a diamond with vertices $(0,\pm S_{\epsilon })$ and $(\pm S_{\epsilon },0)$; transforming back bends its four sides into the characteristic curves found in part 3(c). Beyond any one of those sides, a point's backward characteristics meet $x=0$ outside $I_{\epsilon }$, where no Cauchy data were prescribed, so uniqueness cannot be extended farther.

The principal symbol of the equation isIn the standard notation $A{u}_{tt}+2B{u}_{tx}+C{u}_{xx}$, we have $A=tx$, $B=0$, and $C=-1$, soA second-order equation is a hyperbolic partial differential equation exactly where this discriminant is positive. Hence the hyperbolic set isthe union of the first and third open quadrants.

For $u_0>u_A$, define the Alfvénic Mach-cone slopeThe equation in part (b) is the hyperbolic partial differential equationwhose characteristic curves are $x\pm mz=\text{constant}$. Let $W(s)$ be one for $|s|<L/2$ and zero otherwise. Selecting the characteristics that trail downstream, toward negative $x$, and imposing part (c) givesThe stationary induction and mass conservation equations then givewith $\delta u_z=\delta u_y=\delta B_y=0$. In the strongly super-Alfvénic limit, $\delta u_x\simeq u_{A}b/B_0$ and $\delta B_z\simeq b{u}_A/u_0$ inside the disturbed region.

The perturbations have support only whereBecause the wing is infinite in $y$, this is the union of two inclined slabs bounded by the characteristic planes $x+m|z|=\pm L/2$. These two trailing slabs are the Alfvén wings generated by the conductor.