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The Inviscid Burgers equation here has the negative quadratic flux $F(f)=-f^2/2$. By the method of characteristics, $df/dZ=0$ and $d\theta /dZ=-f$, so a characteristic beginning at ${\theta }_0$ satisfiesThis is a classical solution until characteristic curves intersect. Across a shock wave, the Rankine-Hugoniot condition givesIt uses the one-sided limiting values and agrees with the weak-shock expression.

For $U>0$, the right-state characteristic curves travel left with speed $-U$ while the left-state characteristic curves have speed zero. They converge into a compressive shock wave with speed $-U/2$. The entropy solution isThe characteristic speeds satisfy $0>-U/2>-U$, so both sides enter the shock.

For $U<0$, right-state characteristic curves travel right with speed $-U>0$, leaving a gap. A steep continuous transition fills that gap with a rarefaction wave. The entropy solution isThe middle value is fixed by $\theta /Z=F^{\prime }(f)=-f$. A discontinuity joining the two states would satisfy the jump equation but violate entropy admissibility. If $U=0$, the solution is identically zero.

Write $z=(x,v)$ and let $\Phi (t,s,z)$ denote the Hamiltonian flow from time $s$ to time $t$. The characteristic curves satisfy Hamilton's equations:The Hamiltonian Liouville equation then reduces along each curve toThe signs and derivative variables here are those in the PDF.

The global characteristic flow for a Hamiltonian with bounded Hessian follows, for example, from $H\in C^2(\mathbb{R}\times {\mathbb{R}}^{2d})$ and, for every finite $T$,Thus the Hamiltonian vector field $b=({\nabla }_{v}H,-{\nabla }_{x}H)$ is globally Lipschitz continuous in $z$ on each finite time interval and satisfies $|b(t,z)|\le A_T+L_T|z|$. The Picard-Lindelöf theorem gives local existence and uniqueness, while the Gronwall inequality gives, for example,This excludes finite-time escape. There is a unique Hamiltonian flow for all finite forward and backward times, and $\Phi (s,t)$ is the inverse of $\Phi (t,s)$. These sufficient conditions are deliberately stronger than necessary.

The chain rule along a smooth curve isDifferentiating the circular parametrization gives the tangent vectorThus the chain rule turns the angular derivative of $u$ into $-y{u}_x+x{u}_y$. The given partial differential equation makes this equal to $u$, so along every portion of the circle inside the upper half-plane, $U(t)=u(\mathbf{x}(t))$ satisfies the ordinary differential equation $U^{\prime }=U$. Its solution is $U(t)=U(t_0)e^{t-t_0}$.

For the two specified points take $a=\sqrt{2}$. The initial point corresponds to $t_0=\pi /4$ and the terminal point to $t_1=3\pi /4$. The entire intervening arc lies in $y>0$, so no continuation outside the domain is used. The requested value isThese circular characteristic curves also show locally that the general solution has the polar-coordinate form $u(r,\theta )=C(r)e^{\theta }$, with $0<\theta <\pi$; there is no requirement of angular periodicity on the upper half-plane.

For the method of characteristics, parameterize by $x$; the characteristic curves satisfyThus $x^2+y=C$ and $u-\sin x$ is constant on each characteristic curve. The general form in the quadrant is $u=\sin x+F(x^2+y)$. The boundary condition gives $F(s^2)=\cos s-\sin s$ for every $s\ge 0$.

The characteristic curve through $(x,y)$ meets the boundary at $(s,0)$ with $s=\sqrt{x^2+y}\ge x$. Integrating backwards from that boundary value yieldsIndeed ${\partial }_x-2x{\partial }_y$ annihilates $x^2+y$ and sends $\sin x$ to $\cos x$, while on $y=0$ the expression reduces to $\cos x$. Every characteristic in the quadrant meets the prescribed boundary, so its value is determined. The displayed solution is smooth away from the origin. At the origin the boundary is characteristic and the solution is continuous but has a singular inward $y$ derivative; the differential equation is interpreted in the interior, with the boundary data attained continuously.

Take the primitive state vector $\mathbf{U}=(\rho ,p,u_x,u_y,u_z,B_x,B_y,B_z{)}^T$, with $\rho >0$. In the absence of gravity, the ideal magnetohydrodynamic equations are first order: expanding the material derivatives, pressure gradient and Lorentz force makes every term linear in a spatial derivative of $\mathbf{U}$, with coefficients depending on $\mathbf{U}$. Dividing momentum balance by $\rho$ therefore gives a quasilinear partial differential equation system ${\partial }_t\mathbf{U}+A_i(\mathbf{U}){\partial }_i\mathbf{U}=0$ with eight-by-eight matrices. The solenoidal magnetic-field constraint is an additional constraint on initial data; the ideal magnetohydrodynamic induction equation preserves it because the divergence of a curl vanishes.

For a one-dimensional simple wave in magnetohydrodynamics, write $\mathbf{U}=\mathbf{U}(q(x,t))$ with a nonzero state-space tangent. Substitution gives ${\mathbf{U}}_q{q}_t+A_x{\mathbf{U}}_q{q}_x=0$. A nonconstant profile requires the tangent to be a right eigenvector of $A_x$:The chain rule then gives the wave-speed equation:This is the Inviscid Burgers equation, including the special case of constant $v$. Its characteristic curves are $x=a+v_0(a)t$, with $v=v_0(a)$. As long as the mapping remains invertible,Thus a region with $v_0^{\prime }<0$ steepens: faster characteristics catch slower ones, and the first gradient catastrophe occurs at $t_{\ast }=-1/{\min }_a{v}_0^{\prime }(a)$ when the minimum is negative. A genuinely nonlinear compressive simple wave therefore forms a shock wave. Rarefactive profiles can spread instead; steepening is not inevitable for every initial profile.

To continue past this time, use weak solutions of the conservative mass, momentum, fluid total-energy equation and ideal magnetohydrodynamic induction equation. The Rankine-Hugoniot conditions fix the jumps and shock velocity, while physical entropy production selects admissible magnetohydrodynamic shocks. Diffusive shock wave layers are replaced by moving discontinuities, so their microscopic structure need not be explicitly resolved. In particular, the smooth adiabatic pressure equation must be replaced by total-energy conservation when imposing the ideal magnetohydrodynamic shock conditions.

For the Alfvén waves, impose the one-dimensional solenoidal magnetic-field constraint, so $B_x$ is constant and $\delta B_x=0$. Put ${\mathbf{B}}_{\perp }=(B_y,B_z)$ and $c=v-u_x$. The transverse components of the linearized ideal magnetohydrodynamic equations giveThe longitudinal momentum equation additionally gives $\delta p+{\mathbf{B}}_{\perp }\cdot \delta {\mathbf{B}}_{\perp }/{\mu }_0=0$. The Alfvén eigenvectors have $\delta \rho =\delta p=\delta u_x=0$, hence their transverse polarization is perpendicular to ${\mathbf{B}}_{\perp }$. For $s=\pm 1$, their speed and explicit right Alfvén characteristic eigenvectors areFor ${\mathbf{B}}_{\perp }\ne 0$ one may choose $(h_y,h_z)=(B_z,-B_y)$, up to a nonzero scalar factor. If ${\mathbf{B}}_{\perp }=0$, either transverse polarization is allowed and this characteristic speed is degenerate. These expressions describe the two propagating Alfvén branches for $B_x\ne 0$; if $B_x=0$, they coalesce with advected, nonpropagating transverse disturbances.

Integrating along an Alfvén simple wave leaves $\rho ,p,u_x,B_x$ unchanged and gives $d{\mathbf{u}}_{\perp }=-sd{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and ${\mathbf{B}}_{\perp }\cdot d{\mathbf{B}}_{\perp }=0$. Consequently the finite-amplitude nonlinear Alfvén wave relations areThe constant-magnitude condition is essential: otherwise a varying magnetic pressure would drive longitudinal compression. An arbitrary smooth phase profile gives an explicit family, $f_3=B_{\perp }\cos \Theta (x-v_{s}t)$ and $f_4=B_{\perp }\sin \Theta (x-v_{s}t)$, with $f_1,f_2$ as above. Direct substitution in the transverse momentum and ideal magnetohydrodynamic induction equation gives $(u_x-v_s){\mathbf{u}}_{\perp }^{\prime }=B_x{\mathbf{B}}_{\perp }^{\prime }/({\mu }_0\rho )$ and $(u_x-v_s){\mathbf{B}}_{\perp }^{\prime }=B_x{\mathbf{u}}_{\perp }^{\prime }$, confirming the solution without relying on the eigenvector argument. Longitudinal momentum holds because $p+B_{\perp }^2/(2{\mu }_0)$ is constant. The speed $v_s$ is independent of amplitude along this family: these linearly degenerate characteristic fields translate without steepening.

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.

Put $a={\varphi }_1$, $b={\varphi }_2=a/2$, $z_m=4H/5$, $L=H-z_m$ and $D=(1-b)z_m$. The quadratic hindered-settling flux is convex. At the initial internal interface its lower particle volume fraction is smaller than its upper particle volume fraction, so characteristic curves diverge: the entropy solution has a settling rarefaction fan, not a compression shock. Within this fan,The two bounding straight shocks initially haveThe upper front's nominal meeting with the fan's upper edge occurs atThe lower front meets the fan's lower edge atThe lower meeting is always physical in the stated range. The upper meeting need not happen before the suspension disappears; this distinction matters for sufficiently small $a$.

For the curved settling fronts within a rarefaction fan, the Rankine-Hugoniot condition gives ${\dot{z}}_u=u_s(\varphi -1)$ at the clear-fluid front and ${\dot{z}}_d=u_s\varphi$ at the deposit front. Substitution of the fan particle volume fraction and matching to the straight sections givesThe first curve applies after $t_{c1}$ if that meeting occurs, and the second after $t_{c2}$ until it leaves the fan or deposition ends. If both fronts meet inside the fan, their intersection isThe last surviving characteristic of a settling fan joins $(z_m,0)$ to $(h_c,t_c)$. It has constant particle volume fractionIt is swallowed simultaneously by the clear-fluid and deposit shocks at final deposition. This requires $b\le {\varphi }_{\ast }\le a$, which in the given geometry is equivalent toEquality places the last characteristic curve on the upper fan edge. In the generic fan-intersection regime $a>a_{\ast }$, both nominal meetings and the two curved fronts are realized.

For $0<a<a_{\ast }$ there is early extinction of a settling rarefaction fan. The deposit shock reaches the upper fan edge before the clear-fluid front does, atIt then travels through the remaining uniform particle volume fraction $a$ with speed $u_{s}a$. The upper front retains $V_1$ until the final intersection. Since the final bed height is still $h_c=3aH/5$, the actual stopping time isIn this regime $(h_{c1},t_{c1})$ is only an extrapolated intersection; for $a<1/6$ its extrapolated height is even negative. The line joining the initial internal interface to the final point would have a particle volume fraction greater than $a$, so it is not a characteristic curve emitted by that fan. The single fan-characteristic curve stopping-time construction does not apply to the entire printed interval $0<a<1/2$; the missing subregime must be included.

Both panels label the actual final deposit height and time; the second shows the deposit shock leaving the fan while the upper front remains straight. Finally, the density inversion in a settling suspension puts more particles, hence denser fluid, above lighter fluid. Rayleigh-Taylor instability and overturning can accompany this profile in a real container. They are excluded by the one-dimensional quiescent kinematic sedimentation model used for these calculations.

For real initial data, separation of variables givesThis includes $z_0=0$, whose solution is identically zero. Every other real $z_0$ has a pole at $t=1/z_0$, so only $z_0=0$ gives a finite solution on the whole real line. If complex-valued initial data are allowed, a nonreal $z_0$ has no real-time pole; the whole-real-line conclusion here uses the usual real-valued differential-equation convention.

Along a characteristic curve $y=u+ax$, the chain rule gives $df(x,u+ax)/dx=f_x+a{f}_y$. For the homogeneous transport equation this is zero, so $f(x,y)=F(y-ax)$. For the nonlinear equation, the same method of characteristics reduces the problem to $df/dx=f^2$, with initial value $g(u)$. ThereforeFor differentiable real $g$, this is the classical solution wherever its denominator is nonzero, on the characteristic intervals containing the initial line. If $g(u)\ne 0$ for some $u$, that characteristic blows up at $x=1/g(u)$, $y=u+a/g(u)$. Thus the only real-valued solution bounded on all of ${\mathbb{R}}^2$ has $g\equiv 0$, giving $f\equiv 0$.