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A conservation law flux $F(U)$ transports the conserved density $U$ across a surface. In one dimension $U_t+{\partial }_{x}F(U)=0$; a moving discontinuity satisfies $s[U]=[F]$. For a stationary shock wave, each conserved flux has the same value on both sides.

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 hindered settling with normalized particle volume fraction $s$, upward coordinate and downward isolated settling velocity magnitude $V_0$, the scalar conservation law has conservation law flux $f(s)=-V_{0}s(1-s{)}^{\alpha }$. For $\alpha >0$, its convexity changes at $s=2/(\alpha +1)$ when this lies in the physical interval. This change distinguishes spreading rarefaction waves from compressing characteristic curves; a flux with an inflection can produce both a shock wave and a rarefaction wave in one transition. At $\alpha =0$ the flux is linear and the profile simply translates.

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.

Let $e=\rho u^2/2+p/(\gamma -1)$ be the total gas energy density, with $\gamma >1$. The one-dimensional continuity and momentum equations areUsing continuity to put the momentum equation into conservative form givesThe internal energy density $e_{\mathrm{int}}=p/(\gamma -1)$ satisfiesby the adiabatic pressure equation. Multiplying the momentum equation by $u$ and using continuity givesAdding these equations combines the pressure work into $-{\partial }_z(pu)$. The conservation law variables and conservation law fluxes are thereforeThe three rows express conservation of mass, momentum and total energy. Their integral conservation laws remain meaningful across a shock wave, where the differential pressure and velocity equations cannot be applied pointwise.

Integrating each conservation law across a vanishingly thin interval around the stationary shock wave leaves equal conservation law fluxes on its two sides. HenceThese are the Rankine-Hugoniot conditions for a perfect gas. Signed velocities may both be negative when the material travels from positive to negative $z$; no sign change is needed in the conservation law fluxes.

Put $r={\rho }_2/{\rho }_1=u_1/u_2$, $P=p_2/p_1$ and $m=M_1^2={\rho }_1{u}_1^2/(\gamma p_1)$. Momentum conservation givesDivide the energy condition by the nonzero mass flux. Equality of kinetic energy plus specific enthalpy givesSubstituting $P$ and multiplying by $2(\gamma -1)r^2/\gamma$ yieldsThe factor $r-1$ is the continuous, no-shock solution. On the nontrivial normal shock wave branch,An admissible compressive gas shock wave has $M_1>1$, so $r>1$ and $P>1$. The algebraic jump equations alone also allow a reversed expansive discontinuity; the entropy production in a perfect-gas shock excludes that branch. At $M_1=1$ the nontrivial branch joins the continuous solution.

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.

A travelling wave of the viscous Burgers equation can connect constant states $u_L>u_R$ through a smooth decreasing layer. Require $u(x,t)=U(z)$, $z=x-st$, $U(-\infty )=u_L$, $U(+\infty )=u_R$, and $U^{\prime }\to 0$ at both ends. Substitution and one integration giveThe speed follows by evaluating the integration constant at both end states. It is precisely the Rankine-Hugoniot condition for conservation law flux $f(u)=u^2/2$. Choosing the center $U(0)=s$ gives the displayed hyperbolic tangent profile. Its characteristic width scales as $\nu /(u_L-u_R)$; a smaller positive $\nu$ makes a thinner smooth layer.

For an original worked example in dimensionless quantities, take $u_L=2$, $u_R=0$ and $s=1$:This is an exact classical solution on $\mathbb{R}$ with its own smooth initial condition $u(x,0)=1-\tanh (x/(2\nu ))$ and limiting states $2,0$. It is a family of moving profiles, not the finite-$\nu$ solution starting from an exact discontinuous step. For that different initial condition, the Cole-Hopf transformation supplies the evolution. At $\nu =1/2$, this front starts from the same $1-\tanh x$ profile as the Burgers characteristic breaking example: positive diffusion maintains a smooth moving layer while the inviscid profile breaks at $t=1$.

To check the partial differential equation explicitly, set $w=(x-t)/(2\nu )$ and $S=1-{\tanh }^{2}w$. ThenThe independent Rankine-Hugoniot condition check gives $s=(0-2)/(0-2)=1$, using $f(2)=2$ and $f(0)=0$. At fixed $x\ne t$, the vanishing viscosity approximation tends to the entropy shock $2$ for $x<t$ and $0$ for $x>t$. At the center the smooth profiles stay at $1$; changing the value on the discontinuity does not change the limiting weak solution.

For $u_L<u_R$, the same integrated ordinary differential equation has $U^{\prime }<0$ between the two states when $\nu >0$, so it cannot supply an increasing connecting front. The inviscid admissible expansion is the rarefaction wave of the Burgers Riemann problem with positive flux.