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For a bounded entropy solution with its initial entropy inequalities, the displayed limit holds on every compact spatial set. Test a constant-level entropy with $(1-t/\delta {)}_{+}\rho (x)$. The flux remainder is $O(\delta )$, giving a bound by $\int \rho |u_0-k|$. A finite smooth partition and constant levels approximate the initial function in local $L^1$; the triangle inequality makes the limsup arbitrarily small. This averaged trace suffices for one-sided time mollifiers supported in $[\delta ,2\delta ]$.

This $C^1$ flux has $|f^{\prime }|\le 1$ globally and agrees with the Inviscid Burgers equation flux on $[-1,1]$. It allows Burgers examples to satisfy a global bounded-speed hypothesis. For Riemann data $-1$ to the left and $1$ to the right, both the stationary expansion discontinuity and the fan $u=x/t$ on $|x|<t$, with exterior values $\pm 1$, are bounded weak solutions. Only the fan is an entropy solution.

Before characteristic crossing, the method of characteristics for the Inviscid Burgers equation transports each initial value without changing it:Here $\xi$ labels the starting position. Suppose $u_0$ is a bounded continuously differentiable function with bounded derivative. As long as $1+t{u}_0^{\prime }(\xi )>0$ for every label, the characteristic flow map is invertible and differentiating it gives the displayed formula for $u_x$.^[ref]

If the negative minimum $m={\min }_{\xi }{u}_0^{\prime }(\xi )$ is attained, the first breaking time is $t_{\ast }=-1/m$. Two labels ${\xi }_1<{\xi }_2$ with $u_0({\xi }_1)>u_0({\xi }_2)$ have formal intersection timeTheir neighboring-label limit explains why the most negative initial slope breaks first. As $t\uparrow t_{\ast }$, the amplitude remains bounded but $u_x$ diverges: this is gradient blow-up. Continuing intersecting lines would assign incompatible values at one position; the post-breaking entropy solution instead uses shock waves. Nondecreasing data have no forward characteristic crossing by this mechanism.

For a concrete calculation in dimensionless quantities on the infinite line, choose $u_0(\xi )=1-\tanh \xi$, with limiting states $2$ and $0$. Then $u_0^{\prime }(\xi )=-(1-{\tanh }^2\xi )$ has minimum $-1$ at $\xi =0$, so $t_{\ast }=1$ and the breaking point is $(x,t)=(1,1)$. Along the central characteristic curve $x=t$, $u=1$ and $u_x=-1/(1-t)$ before breaking. The labels $\xi =\pm 1$ give intersecting formal lines at $t=1/\tanh 1\approx 1.313$, after the classical solution has already broken.

For left state zero and right state $U$, negative-flux Inviscid Burgers equation gives a shock at $\theta =-UZ/2$ when $U>0$, and a rarefaction wave $f=-\theta /Z$ on $0<\theta <-UZ$ when $U<0$. This is the standard positive-flux equation after setting $u=-f$; entropy solution selection reverses which step direction is compressive.

Consider the Riemann problem on $\mathbb{R}$ with initial condition $u_0(x)=u_L$ for $x<0$ and $u_0(x)=u_R$ for $x>0$. No endpoint boundary conditions are imposed. When $u_L>u_R$, its entropy solution is a shock wave with constant speed $s=(u_L+u_R)/2$. In the original example $u_L=2$, $u_R=0$, the Rankine-Hugoniot condition givesThe left characteristic speed is $2$ and the right characteristic speed is $0$, so $2>1>0$: both sides enter the entropy shock. The Viscous Burgers travelling shock gives an exact smooth family tending to this solution.

Reversing the states to $u_L=0$, $u_R=2$ still gives a Rankine-Hugoniot condition speed of $1$, but now characteristic curves move away from the discontinuity. That expansion jump is a weak solution and fails entropy admissibility. For general $u_L<u_R$, the selected entropy solution is the continuous self-similar solutionThis rarefaction wave fills the expanding gap between the two exterior characteristic curves. Inside its fan, $u_t=-x/t^2$ and $u{u}_x=x/t^2$, verifying the Inviscid Burgers equation; its values agree continuously with both exterior states. For $0$ to $2$, the fan occupies $0<x<2t$. If $u_L=u_R$, the solution is constant.^[ref]

Putting $y=x^{5/3}$ and $w=x^{1/3}h$ transforms $h_t+x^{-1}(xD{h}^3{)}_x=0$ into the scalar conservation lawAlong smooth characteristic curves, $w$ is constant and $y$ grows at speed $5D{w}^2$. Starting from $h_0(s)$ givesAfter characteristic crossing, the Rankine-Hugoniot condition and entropy solution select the physical front.

For a scalar conservative monotone update with flux $F$, define $Q(a,b;c)=F(a\vee c,b\vee c)-F(a\wedge c,b\wedge c)$. Monotonicity and preservation of constant states give $|U_j^{n+1}-c|\le |U_j^n-c|-(k/d)(Q_{j+1/2}-Q_{j-1/2})$. At equal states $Q(u,u;c)=\mathrm{sgn}(u-c)(f(u)-f(c))$, the Kruzhkov entropy flux. Such discrete inequalities connect stability of a monotone conservative scheme with the admissibility of its limiting entropy solution.

Apply the entropy solution inequality for $u(t,x)$ with level $v(s,y)$ and for $v(s,y)$ with level $u(t,x)$, integrate both in the other pair of variables, and add them. Symmetry of the Kruzhkov entropy flux produces derivatives ${\partial }_t+{\partial }_s$ and ${\partial }_x+{\partial }_y$. A nonnegative mollifier concentrated near the diagonal then yields the Kato inequality for scalar conservation laws. Boundary-time terms require an initial trace argument, not merely interior translation continuity.

For a scalar equation with strictly convex flux and states $u_l>u_r$, the Rankine-Hugoniot condition determines a discontinuity speed strictly between the right and left characteristic speeds. Characteristic curves enter this compressive discontinuity from both sides. Its entropy flux for a scalar conservation law inequalities make it an entropy solution. It is the limit of a phase-normalized decreasing travelling wave in the vanishing viscosity approximation.

The finite volume method follows the amount of a conserved quantity inside each control volume. Its unknowns are cell averages, rather than values at isolated points: a cell's width times its stored average approximates that cell's total. A numerical flux transfers an amount across a face. Using one shared transfer for both neighboring cells makes the loss from one exactly the gain of the other, as explained in shared face flux cancellation.^[ref]

For a one-dimensional scalar conservation law without sources, let $u_t+{\partial }_{x}f(u)=0$, with conservation law flux $f$. Fix cells $I_j=[x_{j-1/2},x_{j+1/2}]$ of widths $h_j>0$. Define the exact cell average and its numerical approximation byThe integral balance over one cell and one time interval of length $k>0$ isFor a smooth solution this follows by integration of the differential equation; with jump discontinuities, it expresses the conservation law in integral form, using appropriate face traces when they exist. Replacing each time-averaged face conservation law flux by a numerical flux givesThus the finite volume method implements a balance directly, without needing a pointwise derivative at a jump. The tube balance in Riemann Problems and Jupyter Solutions by David I. Ketcheson, Randall J. LeVeque, and Mauricio J. del Razo (Advection, §Conservation of mass) motivates this construction from conservation of mass.

This is useful for shock waves: their speed is constrained by the Rankine-Hugoniot condition, itself an integral balance. A nonconservative approximation can give a visibly plausible jump with the wrong speed. Conservation is one requirement; a nonlinear scalar conservation law also needs the correct entropy solution, and the chosen numerical flux and update need their own stability and accuracy analysis.^[ref] The original periodic upwind finite volume advection example isolates the shared-transfer mechanism in a problem with a known translated solution.

For a scalar conservation law whose flux is a convex function, the Godunov numerical flux is the physical flux at the origin of the entropy solution of the Riemann problem between states $a,b$. It is ${\min }_{u\in [a,b]}f(u)$ for $a\le b$ and ${\max }_{u\in [b,a]}f(u)$ for $a>b$. For a rarefaction wave, the characteristic speed is monotone across its states, so the origin sees either an endpoint or a zero-speed minimizing state. For a shock wave, its speed is $(f(a)-f(b))/(a-b)$; the origin sees the left state if this speed is positive and the right state if negative, selecting the larger endpoint flux. Unlike the Godunov flux, the Engquist-Osher flux at reversed transsonic states may equal $f(a)+f(b)-f(s_{\ast })$.

The Inviscid Burgers equation is the scalar conservation lawFor a classical solution, the chain rule makes this equivalent to $u_t+u{u}_x=0$, a nonlinear transport equation with characteristic speed $u$. The conservative form remains meaningful for a discontinuous weak solution, whereas the product $u{u}_x$ cannot simply be evaluated pointwise at a jump. Its discontinuities obey the Rankine-Hugoniot condition $s=(u_L+u_R)/2$; this condition alone permits inadmissible expansion jumps. For bounded initial conditions on the infinite line, imposing the Kruzhkov entropy inequalities and the initial trace selects a unique entropy solution. For step data, incoming characteristic curves select $u_L>s>u_R$, while $u_L<u_R$ produces a rarefaction wave.^[ref]

For bounded entropy solutions of the same scalar law, $Q(u,v)=\mathrm{sgn}(u-v)(f(u)-f(v))$ satisfies the displayed distributional inequality, with initial value $|u_0-v_0|$ in its test-function form. The doubling of variables for scalar conservation laws, local translation continuity and the averaged initial trace of an entropy solution prove it. Its significance is comparison between two solutions rather than an inequality against a constant state.

A numerical flux estimates the rate of transfer across a cell face from nearby states. In the finite volume method update, $F_{j+1/2}^n$ approximates the conservation law flux averaged over the time step; $k{F}_{j+1/2}^n$ is the transferred amount. For a two-state numerical flux $F(L,R)$ approximating a scalar conservation law with conservation law flux $f$, consistency of a numerical method requires $F(U,U)=f(U)$. Thus equal states recover the physical transfer rate. With constant equal-width cells, this also preserves a constant state because its incoming and outgoing numerical fluxes agree.^[ref]

Consistency and shared face flux cancellation answer different questions: matching $f$ at equal states approximates the intended equation, while sharing one oriented face value preserves the total. The example periodic upwind finite volume advection uses $F(L,R)=aL$ for $a>0$, so the arriving state comes from the left. The Godunov numerical flux uses an entropy solution of a local Riemann problem; the Engquist-Osher flux provides a different construction for a scalar conservation law. A monotone conservative scheme uses a numerical flux nondecreasing in its left state and nonincreasing in its right state, together with a time step that makes the complete update monotone. Neither two-state consistency nor conservation by itself supplies that time-step restriction or a general stability or order of a numerical method guarantee.

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.

The Engquist-Osher method is an upwind finite volume method. For cell values $U_j^n$, mesh width $h$ and time step $k$, put $r=k/h$ and use the conservative updatewhere the numerical flux isConsistency of a numerical method follows from $F(u,u)=f(u)$. Positive characteristic speeds use the left state, and negative speeds use the right state. The flux difference telescopes when summed over cells, giving a discrete conservation law. The method is first order in space and time for smooth solutions; its conservation law form is also appropriate when shock waves develop.

Let $s_{\ast }$ be the unique sonic point of a scalar flux, $f^{\prime }(s_{\ast })=0$. Convexity implies $f^{\prime }<0$ to its left and $f^{\prime }>0$ to its right. Put $f_{\ast }=f(s_{\ast })$ and splitThen $f=f_{\ast }+f^{+}+f^{-}$, $(f^{+}{)}^{\prime }\ge 0$, $(f^{-}{)}^{\prime }\le 0$, andThis is the sonic-point splitting of a convex Engquist-Osher flux. If both states are above $s_{\ast }$, $F=f(a)$; if both are below it, $F=f(b)$. Across a transsonic rarefaction wave, $a\le s_{\ast }\le b$, the flux is the minimum $f_{\ast }$. In the opposite crossing it is $f(a)+f(b)-f_{\ast }$, which in general differs from the exact Godunov numerical flux. This distinction does not affect the monotonicity proof.

Here stability requires the usual Courant–Friedrichs–Lewy condition. Let the initial values lie in $[m,M]$, and chooseA differentiable convex function has a continuous derivative on this interval, so the maximum is finite. Write the three-input update as $H(a,b,c)$. Differentiating givesThus the update $T$ preserves componentwise order. Constants are fixed, since $H(c,c,c)=c$, so comparison with the constants $m,M$ proves $m\le U_j^n\le M$ for every time level. In particular the same CFL condition remains applicable. This is a mesh-independent maximum-principle stability bound.

A stronger stability statement is contraction in discrete $L^1$. Consider two solutions $U,V$ in the same invariant interval, on a periodic mesh or on the infinite lattice with summable differences. Let $Q=U\vee V$ be their componentwise maximum. Monotonicity gives $TQ\ge TU,TV$, henceThe discrete conservation law for the difference gives the equality: on a finite periodic mesh the flux sum cancels exactly, and on the infinite lattice it follows by truncation and the fact that the flux is a Lipschitz function on $[m,M]$. Interchanging $U,V$ and adding provesThis is L1 contraction of a monotone conservative scheme. Since $T$ commutes with the cell shift, applying this contraction to $U$ and its shift gives ${\sum }_j|U_{j+1}^{n+1}-U_j^{n+1}|\le {\sum }_j|U_{j+1}^n-U_j^n|$. The scheme is therefore a total variation diminishing scheme as well.

The same order argument proves a discrete entropy inequality, explaining why this stable discretization selects entropy solutions rather than nonphysical expansion shock waves. For a constant $c\in [m,M]$, setBecause $T(U\vee c)\ge TU,c$ and $T(U\wedge c)\le TU,c$, their componentwise difference is at least $|TU-c|$. Subtracting their updates givesFor constants outside $[m,M]$, the entropy inequality is an equality by the invariant bound and the conservative update. At equal states $Q_c(u,u)=\mathrm{sgn}(u-c)[f(u)-f(c)]$, the Kruzhkov entropy flux. The flux is a Lipschitz function on the invariant interval: writing $L=\max |f^{\prime }|$, the conservative update gives $h{\sum }_j|U_j^{n+1}-U_j^n|\le 2kL{\sum }_j|U_{j+1}^n-U_j^n|$. This time-translation bound, the invariant bound and the total variation estimate give local space-time compactness for bounded-variation initial conditions as the mesh is refined; the conservative update passes to the weak conservation law, and the displayed inequality passes to its Kruzhkov entropy inequality. This identifies the limit as the entropy solution.

Finally the time-step qualification is essential. For the convex Inviscid Burgers equation flux $f(u)=u^2/2$, whose unique sonic point of a scalar flux is zero, linearize about a positive constant $U$. The perturbation update becomes $v_j^{n+1}=(1-rU)v_j^n+rU{v}_{j-1}^n$. At Fourier frequency $\pi$ its multiplier is $1-2rU$, with modulus greater than one if $rU>1$. Wave packets near this phase give linear instability even for summable Cauchy perturbations. Thus the assumptions on $f$ imply the stability results above under the CFL condition, not for arbitrary $k/h$. This is CFL necessity for explicit Engquist-Osher stability.

The initial-value problem is a Riemann problem. Both the one-dimensional Euler equations and the two constant initial states are unchanged under the simultaneous scaling $(x,t)\mapsto (\lambda x,\lambda t)$ for $\lambda >0$. There is no imposed length or time scale. Rescaling a physically admissible entropy solution therefore gives another solution with exactly the same initial data. Uniqueness of the physical Riemann solution makes the two coincide, soThe smooth pieces are consequently those in part (a), while shocks and contacts lie on rays $x=st$. Across shocks one uses the conservative mass, momentum and total-energy equations and the entropy admissibility condition, rather than extending the smooth entropy-advection equation through the discontinuity. Scale invariance by itself, without selecting the unique admissible solution, would not justify the uniqueness step.