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The Engquist-Osher method is a conservative finite volume method built by separating positive and negative characteristic speeds. Take $f\in C^1$, a uniform cell width $d$, a time step $k$, and cell averages $U_j^n$. Write $\lambda =k/d$ and define the numerical fluxThe Engquist-Osher flux uses the left state for right-going transport and the right state for left-going transport. Its conservative update isIt is consistent because $F(u,u)=f(u)$. The integrals have their ordinary oriented meaning even for negative states. For the Inviscid Burgers equation, $f(u)=u^2/2$ gives the useful exampleIn a region where all characteristic speeds have one sign, the flux reduces to the corresponding upwind flux. In smooth regions, this basic piecewise-constant, forward-time version is first order in space and time; its main virtue is robust nonlinear stability across shocks.

Here is a precise stability proof. Assume the data lie in a bounded interval $J=[m,M]$ and chooseWork on a periodic grid, or on the infinite grid with summable differences; on a finite interval the inflow boundary data and boundary fluxes must be treated monotonically as well. Define the three-point update mapWriting $f_{+}^{\prime }=\max (f^{\prime },0)$ and $f_{-}^{\prime }=\min (f^{\prime },0)$, its three partial derivatives areThus it is a monotone conservative scheme. Since $H(s,s,s)=s$, comparison with the constant states provesThe interval is invariant, so the same Courant–Friedrichs–Lewy condition remains valid at every step. This is an $L^{\infty }$ bound relative to constant states, not a claim that arbitrary pairs of solutions are contractive in $L^{\infty }$.

For a stronger perturbation estimate, let $\mathcal{T}$ denote the global update and use componentwise maxima. Monotonicity gives $\mathcal{T}(U\vee V)\ge \mathcal{T}U,\mathcal{T}V$. ThereforeSum over the grid. The shared interface flux cancels telescopically, so conservation givesInterchanging $U,V$ and adding yields the L1 contraction of a monotone conservative scheme:This is a mesh-independent nonlinear stability estimate. On an infinite grid the telescoping argument is justified by summability of the differences and the bounded characteristic-speed range, or by truncation followed by a limit. For unequal prescribed boundary data, additional boundary flux terms enter the estimate.

The method is also a total variation diminishing scheme. Let $\mathcal{S}$ shift a grid sequence by one cell. Translation invariance gives $\mathcal{T}\mathcal{S}=\mathcal{S}\mathcal{T}$. Applying the preceding $L^1$ contraction to $U$ and $\mathcal{S}U$ showsHence initially finite discrete total variation cannot grow.

Finally, monotonicity supplies a discrete entropy inequality, explaining why the stability is compatible with the physically admissible entropy solution. For a constant $c\in J$, defineComparison of $H$ with the clipped triples above and below $c$ givesIndeed $H(U\vee c)\ge \max (H(U),c)$ and $H(U\wedge c)\le \min (H(U),c)$, and subtracting these two update formulas yields exactly the right side. Also $Q(u,u;c)=\mathrm{sgn}(u-c)(f(u)-f(c))$, the Kruzhkov entropy flux. Under the displayed CFL condition the method therefore has an invariant state range, $L^1$ contraction, decreasing discrete total variation and entropy dissipation. These estimates remain meaningful at discontinuities, where linearized Fourier analysis alone cannot establish the corresponding nonlinear stability.

For the negative-flux Inviscid Burgers equation, the method of characteristics givesThe characteristic curve starting at ${\theta }_0$ therefore has $\theta ={\theta }_0-f_0({\theta }_0)z$, and henceFor smooth data this describes a single-valued classical solution as long as the characteristic flow map is invertible. Its Jacobian is $1-z{f}_0^{\prime }({\theta }_0)$; after characteristic crossing, one must instead select a weak solution with the appropriate entropy solution condition.

The conservation form is $f_z+G(f{)}_{\theta }=0$, with conservation law flux $G(f)=-f^2/2$. Integrating this scalar conservation law across a moving discontinuity, or differentiating its Heaviside representation as a distribution, yields the Rankine-Hugoniot conditionFor distinct one-sided limits it simplifies toThis jump-speed relation is exact for the Burgers conservation law; an additional entropy condition is needed to distinguish a physical compressive shock wave from an expansion discontinuity.

The step gives the Burgers Riemann problem with negative flux. If $U>0$, characteristic curves from the left have speed zero, while those from the right have speed $-U$: they converge. The entropy shock wave has speed $-U/2$ and thusCharacteristics enter this shock from both sides, since $0>-U/2>-U$.

If $U<0$, the right-hand speed $-U$ is positive and the two families separate. A smooth steep approximation to the initial step spreads into a rarefaction wave. In the fan, the self-similar characteristic curve relation is $-f=\theta /z$, givingThe endpoint values match continuously. The discontinuity moving at $-U/2$ would satisfy the jump condition even for $U<0$, but its characteristic curves leave the discontinuity and it fails entropy admissibility. For $U=0$ the solution is identically zero.

Set $q=\log \psi$ and use the sign-appropriate Cole-Hopf transformation $f=2\epsilon q_{\theta }$. Direct differentiation givesThus the heat equation ${\psi }_z=\epsilon {\psi }_{\theta \theta }$ implies the viscous Burgers equation with the required negative nonlinear sign. Conversely, vanishing of the Burgers residual makes the final quotient depend only on $z$, and a multiplicative time-dependent factor in $\psi$ removes it. The algebraic substitution works for any nonzero $\epsilon$ where $\psi \ne 0$.

For a forward dissipative initial-value solution, take $\epsilon >0$ and $z>0$. This is essential for the supplied heat kernel: at $\epsilon <0$, $z>0$, its real Gaussian integral diverges even for $\psi (0,\varphi )=1$. More generally that kernel requires $\epsilon z>0$. The physical viscosity convention selects the positive case; the mere condition $\epsilon \ne 0$ does not justify a forward Gaussian function formula.

Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\epsilon )$ gives a positive continuous initial factor, normalized asIts heat kernel convolution converges for both signs of $U$, since a Gaussian function dominates the one-sided exponential. DefineThe left half-line contributes $J(\theta )$; completing the square on the right half-line gives $e^{S}J(-\theta -Uz)$. ThusWhen differentiating, the two moving-endpoint Gaussian function terms cancel, becauseOnly ${\partial }_{\theta }{e}^S=(U/(2\epsilon ))e^S$ remains in the logarithmic derivative. The viscous Burgers step solution with negative flux is thereforeThe denominator is strictly positive, so $f$ lies between $0$ and $U$, irrespective of the sign of $U$. Its initial limits away from $\theta =0$ are the required step.

For the spatial limits, the Gaussian function tails must be compared with $e^S$; inspecting that exponential alone gives the wrong inference when $U<0$. As $b\to +\infty$, the complementary error function asymptotic givesAt $\theta \to +\infty$, $J(\theta )e^{-S}$ tends to zero even for $U<0$, because its quadratic decay dominates any exponential linear in $\theta$. Hence $\alpha e^{-S}\to 0$. At $\theta \to -\infty$, use the identity above to obtainwhile $J(\theta )$ approaches its full Gaussian integral. Thusfor both signs of $U$.

Since $J$ is strictly decreasing, $\alpha =1$ holds exactly when $\theta =-\theta -Uz$. At this point $S=0$, so the midpoint symmetry of a viscous Burgers step givesMore generally the exact symmetry is $f(z,-Uz-\theta )=U-f(z,\theta )$. For $U>0$, the midpoint follows the inviscid shock trajectory. For $U<0$, it is instead the center of an expanding fan, not a shock.

The vanishing viscosity approximation makes the comparison precise. For fixed $z>0$ and $U>0$, near $\theta =-Uz/2$ both tail integrals tend to the full Gaussian integral, so $\alpha \to 1$ andThe layer has width $O(\epsilon /U)$ and tends to the entropy shock, with value $U/2$ at its center. Outside it the limits are $0$ and $U$.

For $U<0$, inside $0<\theta <-Uz$ both Gaussian function tails have large positive lower limits. Their asymptotics giveOutside this interval the limits are $0$ on the left and $U$ on the right. Thus positive viscosity selects the rarefaction wave, including the same midpoint value as the inviscid fan. The two sign cases agree with the preceding entropy solution construction, while a negative diffusivity would not furnish this dissipative selection.

For the sharp energy estimate, $C_0=0$ stays uniform while the available bound $C_1=M^2/\epsilon$ diverges as $\epsilon \downarrow 0$ when $M>0$. If the coarser estimate is used for the first part, both displayed bounds $M^2/\epsilon$ diverge, but the first divergence is only an artifact of discarding an exact cancellation.

The method of characteristics explains why uniform control of the gradient cannot generally persist for the inviscid scalar conservation law. Before characteristic crossing, with initial data $u_0$,If $F^{\prime \prime }(u_0(\xi ))u_0^{\prime }(\xi )<0$ somewhere, the denominator reaches zero in finite positive time. The solution steepens and the smooth description breaks down; an entropy solution can subsequently contain shocks. Positive viscosity replaces such a discontinuity by a thin smooth layer, which can have large gradient even while its $L^2$ norm remains controlled. The divergent bound does not assert that every flux and every initial datum form a shock; a linear flux, for example, has no such steepening.

Write the negative-flux Inviscid Burgers equation as $f_Z+{\partial }_{\theta }F(f)=0$, with $F(f)=-f^2/2$. Along a characteristic curve,so a characteristic starting at ${\theta }_0$ has $\theta ={\theta }_0-f_0({\theta }_0)Z$ andThis formula is a classical solution only while the method of characteristics is one-to-one; after characteristic crossing one must select an entropy solution.

For a discontinuity with left and right values $f_{-}$ and $f_{+}$, the Rankine-Hugoniot condition follows by integrating conservation across a moving small interval:For a nonzero jump, factor the difference of squares to obtainThe negative flux makes increasing jumps compressive. Decreasing jumps spread into rarefaction waves; they cannot be retained as nonphysical expansion shock waves.

In the central ramp of each period, $f_0({\theta }_0)={\theta }_0-2m$, so its method of characteristics is $\theta -2m=(1-Z)({\theta }_0-2m)$. For $0<Z<1$, the retained ramp therefore occupies $|\theta -2m|<1-Z$. At the odd boundary $r=2m+1$, the initial limiting values are $+1$ on the left and $-1$ on the right. Their characteristic speeds are $-1$ and $+1$, giving the fan $f=-(\theta -r)/Z$ on $|\theta -r|<Z$. Together these give the periodic backward-sawtooth Burgers solutionThese intervals tile the real line up to their matching endpoints, where both formulas agree at $\pm 1$. The increasing ramp steepens, but the wave's maximum magnitude remains one before breaking. A steep continuous regularization of the original downward jump produces exactly the limiting fan, as suggested by the characteristic construction.

At $Z=1$, each increasing ramp collapses at $\theta =2m$. The fans on either side meet there with values $-1$ and $+1$, producing a compressive stationary shock wave. For all $Z\ge 1$ the fan between successive shock waves remains centered at the odd point, givingAt a shock wave the limiting values are $f_{-}=-1/Z$ and $f_{+}=1/Z$. Their average is zero, so the Rankine-Hugoniot condition keeps the shock wave fixed. Their characteristic speeds satisfy $1/Z>0>-1/Z$, confirming compression into the shock wave. The arbitrary pointwise value at a shock wave does not affect the weak solution. The post-breaking amplitude decays as $1/Z$, despite the lack of explicit viscosity, because shock wave dissipate the wave.

For the requested sketches, $Z=1/3$ has ramp slope $3/2$ on $|\theta -2m|<2/3$ and fan slope $-3$ around odd points on width $2/3$. It is still continuous. At $Z=3$, the profile decreases linearly with slope $-1/3$ between even points and jumps from $-1/3$ to $+1/3$ at each even point:

Saturation equation. Introduce intrinsic permeability of a porous medium $K$ and porosity $\varphi$. The relative porous permeabilities depend on the wetting fluid saturation. Define the phase mobilities and fractional flowUse the conventional capillary pressure $p_c=p_{nw}-p_w$, generally decreasing as wetting saturation increases. Neglect gravity in this horizontal model. The two Darcy fluxes are $q_w=-K{\lambda }_w{p}_{w,x}$ and $q_n=-K{\lambda }_n(p_{w,x}+p_c^{\prime }(s)s_x)$. Eliminate the common pressure gradient using $q_w+q_n=Q$ to obtainThe wetting-phase mass conservation law is consequentlyFor $p_c^{\prime }\le 0$, $D_c\ge 0$ and the capillary pressure provides nonlinear diffusion. The initially uniform saturation is $s_0$. Pure wetting-fluid injection prescribes $q_w(0,t)=Q$ and $q_n(0,t)=0$; in the zero-capillarity, zero-residual-nonwetting idealization this corresponds to inlet saturation one. With residual nonwetting fluid, replace one by the appropriate maximum accessible saturation.

Shock formation and its speed. Neglect capillarity first. Smooth saturation values travel along characteristic curves at speed $(Q/\varphi )F^{\prime }(s)$. When the trailing values have larger characteristic speeds than the values ahead, the curves intersect, and the Buckley-Leverett equation requires a shock selected as an entropy solution. This is typical for the increasing-convex part of a physical fractional-flow curve; shocks are not inevitable for every possible $F$.

Integrating mass conservation across a shock from upstream $s_s$ to downstream $s_0$ gives the Rankine-Hugoniot conditionIf an upstream rarefaction wave joins the shock smoothly in similarity coordinates, its terminal characteristic speed equals $V_s$. Therefore the saturation at that join obeys the fractional-flow tangent constructionThis is the printed equality. It is a tangent-selection condition, not the dimensional speed by itself; the speed includes $Q/\varphi$. An arbitrary shock between prescribed constant states satisfies the jump condition but need not satisfy this additional tangency relation.

Positive capillary diffusion spreads a jump into a transition layer. For a travelling profile $s(x-V_{s}t)$, integration givesThe endpoint states still satisfy the same Rankine-Hugoniot condition. Thus small capillarity primarily gives the front a finite thickness rather than changing the selected limiting speed; vanishing phase mobility at an endpoint can make the regularization degenerate.

The specified reciprocal fractional flow. On its stated branch, $F^{\prime }=1/s^2$ and $F^{\prime \prime }=-2/s^3<0$. If $F$ extends continuously to the initial saturation and the inlet is at one, the decreasing saturation from behind to ahead makes characteristic speeds increase forwards: there is a rarefaction wave, not a compressive shock. Write $a=Q/\varphi$ and $\xi =x/t$. The self-similar solution isThe fan connects continuously to the initial state and broadens linearly in time. This interpretation requires $F(s_0)=2-1/s_0$. A nonnegative physical fractional flow on the entire interval $[s_0,1]$ also requires $s_0\ge 1/2$.

The PDF specifies the formula only for $s>s_0$ and earlier calls $s_0$ residual. If that means an immobile wetting phase with $F(s_0)=0$, the endpoint is an extra constraint and the continuous-branch answer applies without conflict only at $s_0=1/2$. For $0<s_0<1/2$, the printed branch would give negative fractional flow for $s_0<s<1/2$, so it cannot be the complete physical constitutive law. With an additional admissible extension, for example $F=0$ up to $1/2$ and the given positive branch above it, a compound fan–shock is possible. The tangent condition then givesThe fan $s=\sqrt{a/\xi }$ ends at $\xi =V_s$, followed by a jump to $s_0$. This is a conditional physical extension, not data supplied by the question. For $s_0>1/2$ an immobile residual endpoint is likewise incompatible with a continuous version of the stated branch. These distinctions identify what is, and is not, determined by the printed constitutive assumption.

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.

Let $h=\log \psi$ and $f=2\alpha h_{\theta }$. Direct substitution into the viscous Burgers equation givesIf ${\psi }_Z=\alpha {\psi }_{\theta \theta }$, then $h_Z=\alpha (h_{\theta \theta }+h_{\theta }^2)$, proving the Cole-Hopf transformation. A function of $Z$ inside the parentheses can be removed by rescaling $\psi$ by a time-dependent factor, which leaves $f$ unchanged.

For the forward heat equation and the displayed Gaussian kernel we require $\alpha >0$ and $Z>0$. The algebraic transformation works for nonzero $\alpha$, but the printed convolution is not a forward solution for negative $\alpha$: its Gaussian then grows and the step-data integral diverges.

Integrating the initial logarithmic derivative fixes a convenient positive initial heat datum,Split the heat kernel convolution at zero and complete the square in the positive half. With $h_{\ast }=U(2\theta +UZ)/(4\alpha )$, putThen $\psi =(A_0+e^{h_{\ast }}{C}_0)/\sqrt{4\pi \alpha Z}$. On differentiation, the moving-limit terms cancel because $e^{h_{\ast }}{e}^{-(\theta +UZ{)}^2/(4\alpha Z)}=e^{-{\theta }^2/(4\alpha Z)}$. HenceThis is the viscous Burgers step solution with negative flux. Both integrals can be written as $\sqrt{\pi \alpha Z}$ times a complementary error function.

For fixed $Z>0$, the Gaussian-tail asymptotics give $f\to U$ as $\theta \to +\infty$ and $f\to 0$ as $\theta \to -\infty$. For example $J{e}^{-h_{\ast }}$ tends to zero on the right with a Gaussian factor $e^{-(\theta +UZ{)}^2/(4\alpha Z)}$, while on the left it diverges with a Gaussian factor $e^{{\theta }^2/(4\alpha Z)}$. This holds for either sign of $U$; for $U=0$ the solution is already zero.

The Gaussian tail is strictly decreasing in its lower limit, so $J=1$ occurs exactly at $\theta =-UZ/2$. There $h_{\ast }=0$, and $f=U/2$. For $U>0$, the two lower limits are both far into the negative tail in the mature shock region, so $J\simeq 1$ throughout its thin transition. The solution is then approximately the traveling viscous frontcentred at the inviscid shock position, with thickness of order $\alpha /U$.

In the vanishing-viscosity limit, for $U>0$ the solution tends to the compressive entropy solution of part (a), away from its shock. For $U<0$, it tends instead to the rarefaction wave. To see the latter explicitly inside $0<\theta <-UZ$, both tails have positive lower limits, and their leading asymptotics give $J{e}^{-h_{\ast }}\to (-\theta -UZ)/\theta$. Thus $f\to -\theta /Z$ in the fan, with the constant states outside. Although $J=1$ still marks the fan midpoint, it is not approximately one throughout the expanding fan; replacing it by one there would create an inadmissible compressive-front approximation.

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.

When the velocity depends on $u$, different initial values travel at different speeds. The resulting characteristic crossing can destroy a classical solution in finite time, even though the initial function is smooth and bounded. For the Inviscid Burgers equation, take $u_0(a)=-\tanh a$. Before crossing,andAt $a=0$ this tends to $-\infty$ as $t\uparrow 1$, while $|u|\le 1$. Thus $\boxed{\text{bounded states can undergo finite-time gradient blow-up and shock formation}.}$ The bounded-derivative Burgers flux agrees with this equation on the whole state range, if a global bounded-speed flux is desired. After crossing, arbitrary weak solutions need not be unique; an entropy solution condition is needed to select the admissible continuation.

If $u_0=v_0$, the right-hand side of the local L1 contraction for scalar conservation laws estimate is zero. Applying it on a countable collection of rational intervals and final times proves $u=v$ almost everywhere in space-time. Thus

To deduce nonnegativity one needs a one-sided comparison, rather than merely putting $v=0$ in the absolute-value estimate. The difference $w=u-v$ satisfies the weak equation with flux $f(u)-f(v)$. Subtract its weak integral identity from the Kato inequality for scalar conservation laws and divide by two. The result is the same inequality forWhen $w\ge 0$ both expressions vanish; when $w<0$, $|q_{-}|\le M{w}_{-}$. The shrinking-interval proof therefore bounds the negative part by its initial negative part. This is order preservation for scalar entropy solutions.

The constant $v=0$ is an entropy solution for every $f$, even if $f(0)\ne 0$, because the constant flux has zero spatial derivative. If $u_0\ge 0$, its initial negative part is zero, and comparison givesThis explicitly establishes the additional sign conclusion without incorrectly identifying uniqueness alone with order preservation.

With both ice accumulation and ice ablation removed, the equation is $h_t+x^{-1}(xD{h}^3{)}_x=0$ and the conserved volume is $V_0=2\pi \cos \alpha M$, where $M={\int }_0^{x_N}xhdx$.

If the thickness and extent scales are $H(t)$ and $X(t)$, mass conservation gives $H{X}^2\sim M$, while the flow equation gives $H/t\sim D{H}^3/X$. Therefore $X\propto t^{1/5}$ and $H\propto t^{-2/5}$. Set $h={\tau }^{-2/5}f(\xi )$, $\xi =x{\tau }^{-1/5}$, with a possible virtual time origin $\tau =t+t_0$. The similarity solution satisfiesIntegration and regular zero total flux at the apex give $D\xi f^3={\xi }^{2}f/5$, hence the positive profile iswith dry bed beyond the front. Volume normalization givesso the volume-conserving conical ice-current similarity solution hasThe terminus has finite thickness and is a shock wave in the gravity-only kinematic wave equation. The Rankine-Hugoniot condition gives ${\dot{x}}_N=q_N/h_N=D{h}_N^2=x_N/(5\tau )$, exactly agreeing with the similarity extent. The characteristic speed behind the front is $3D{h}_N^2$, larger than its speed, while the dry-bed characteristic speed is zero, so the front is compressive and gives an entropy solution.

This solution describes the long-time spreading, rather than exactly matching the earlier steady profile at the instant snowfall stops. The initial transient can be described by the characteristic transformation for conical ice drainage: with starting point $s$, put $K(s)=s^{1/3}{h}_s(s)$; thenwhere characteristics remain smooth. Subsequent crossings are resolved by the same conservation and entropy conditions. Restoring the neglected local pressure gradient would smooth the idealized front.

Write the Inviscid Burgers equation in conservation form asA bounded function $u$ is a weak solution with initial datum $u_0$ whenfor every compactly supported $C^1$ test function $\phi$.

Across a straight discontinuity $x=\sigma t$, integration by parts on its two sides shows that the boundary terms cancel exactly when the Rankine-Hugoniot condition holds:when $u_{-}\ne u_{+}$. For every $c>0$, defineThe three jumps have left and right states $(0,-c)$, $(-c,c)$, and $(c,0)$, so their Rankine-Hugoniot speeds are respectively $-c/2$, $0$, and $c/2$, exactly the speeds of the displayed lines. Hence each $u_c$ satisfies the weak equation away from the origin and across every jump. Moreover, its nonzero support at time $t$ has length $ct$, so $u_c(t,\cdot )\to 0$ in $L_{\mathrm{loc}}^1$ as $t\downarrow 0$; its initial datum is therefore zero in the weak identity.

The zero function and all the distinct functions $u_c$ are bounded weak solutions with the same zero initial datum. Thus weak solutions are not unique. The central jump from $-c$ to $c$ is an expansion shock, which an entropy condition would exclude.

For the negative-flux equation $f_Z-f{f}_{\theta }=0$, a period-two increasing initial ramp develops rarefaction fans at odd points and steepens at even points. Before $Z=1$, its central ramp is $f=(\theta -2m)/(1-Z)$ on $|\theta -2m|<1-Z$, and its odd-point fans are $f=-(\theta -(2m+1))/Z$. After $Z=1$, stationary compressive shocks occupy the even points and $f=(2m+1-\theta )/Z$ between consecutive shocks. The amplitude decays as $1/Z$. The Rankine-Hugoniot condition and incoming characteristic speeds certify the post-breaking entropy solution.

A Riemann problem for a one-dimensional system of conservation laws ${\partial }_{t}U+{\partial }_{x}f(U)=0$ has two constant initial states, $U_L$ for $x<0$ and $U_R$ for $x>0$. The data and equation are invariant under $(x,t)\mapsto (\lambda x,\lambda t)$. When the physically admissible entropy solution is unique, this invariance makes it a self-similar solution depending on $x/t$. For ideal-gas compressible flow, the characteristic speeds are $u-c,u,u+c$. The two acoustic families carry shock waves or rarefaction waves, while the middle family carries a contact discontinuity. Sufficiently strong separating velocities can instead create a vacuum region.