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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]

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 $f(a)+f(b)-f(s_{\ast })$.

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.

A single rarefaction wave enforces two matching equalities: equal specific entropy and one common Riemann invariant. The inequality $u_R>u_L$ does not enforce either. If the entropies differ, one smooth isentropic fan cannot connect them; even if they agree, arbitrary end velocities and pressures need not satisfy either acoustic-family invariant.

The general positive-gas Riemann problem has a left acoustic wave, a middle contact discontinuity, and a right acoustic wave. Across the contact the two intermediate states have the same velocity $u_{\ast }$ and pressure $p_{\ast }$, but can have different mass densities and entropies. Each acoustic wave is a rarefaction wave or a shock wave, depending on the required intermediate pressure relative to its adjacent initial state. Shocks satisfy the conservative jump conditions and increase entropy. A sufficiently large pressure contrast can produce a shock on one side even when the initial velocities obey $u_R>u_L$.

For sufficiently rapid separation, the two rarefactions instead exhaust the gas and create vacuum between them. Their vacuum-edge velocities are $u_L+2c_L/(\gamma -1)$ and $u_R-2c_R/(\gamma -1)$, so a finite vacuum gap occurs whenEquality is the limiting case in which the vacuum edges meet. Thus the additional features can be two acoustic waves, a contact carrying an entropy jump, shocks, or a vacuum region, rather than one fan joining the two original uniform states.

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.

Consider the advection equation $u_t+a{u}_x=0$ with constant speed $a>0$, no sources, and periodic boundary conditions on $[0,1]$. Use $N$ equal cells of width $h=1/N$, time step $k>0$, and indices modulo $N$. The exact solution is the periodic translation $u(x,t)=u_0(x-at)$. At a face, the Riemann problem transports the left state to the right, so the two-state upwind numerical flux is $F(L,R)=aL$, with $F(U,U)=aU$.^[ref]

The finite volume method with this numerical flux and an explicit Euler method step givesHere $U_j^n$ represents a cell average; the recurrence also occurs in the upwind finite difference scheme. Impose the specific CFL condition $0<\nu \le 1$. Each updated cell average is then a convex combination of two old ones. This proves preservation of their initial minimum and maximum, including nonnegativity for nonnegative data, and contraction of the maximum norm of the difference of two solutions on this periodic grid. These statements concern this constant-speed scalar update with this time step.

For an original synthetic example, take $N=8$, $a=1$, $h=1/8$, $k=1/16$, hence Courant number $\nu =1/2$. Choose a function constant on each successive cell, with the following values, so these are exact initial cell averages. One step givesFor example, cell zero receives from cell seven across the periodic face: $U_0^1=(0+1)/2=1/2$. All eight face transfers cancel in the total by shared face flux cancellation. More generally this periodic recurrence conserves the total algebraically for any $\nu$, even though the CFL condition is needed for the convex combination argument. With the same initial data and $\nu =3/2$, one step gives $U_2^1=-1$ and $U_4^1=5/2$, despite preserving the total.

After four steps, $t=1/4$, the exact translated profile has moved two cells to the right. Repeated averaging instead gives the explicit binomial coefficient formulawith all indices modulo eight. The diagram compares these computed cell averages with the exact translated cell averages. Their integrals agree, but the numerical jumps spread across cells. The finite volume method is useful for jump discontinuities because it retains the integral balance even at a jump; conservation alone does not guarantee a sharp profile. At $\nu =1$ this particular update is an exact one-cell shift, whereas for $0<\nu <1$ it mixes neighboring states. No accuracy conclusion at a jump follows just from the conserved total.