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Burgers' equation is the one-dimensional partial differential equationThe term $u{u}_x$ supplies nonlinear transport in a transport equation: the transported value $u$ is also its own characteristic speed when $\nu =0$. Larger values catch up with smaller values ahead of them, steepening a decreasing profile. The diffusion term $\nu u_{xx}$ opposes this steepening. Positive $\nu$ gives the viscous Burgers equation; setting $\nu =0$ gives the Inviscid Burgers equation.

This small equation makes three different behaviors accessible in one model: Burgers characteristic breaking from continuous decreasing data, an entropy shock for a downward step, and a rarefaction wave for an upward step. The Burgers Riemann problem with positive flux and the Viscous Burgers travelling shock below give exact examples. These are mathematical calculations with chosen data; the figures contain no experimental measurements.

Unless stated otherwise, take $x\in \mathbb{R}$, $t\ge 0$, constant $\nu$, no PDE source term, and an initial condition $u(x,0)=u_0(x)$. There are no finite endpoints requiring boundary conditions. Where a profile has different limiting states at $x=\pm \infty$, those states are specified explicitly. On a finite interval, boundary conditions become part of the problem: positive diffusion normally requires a condition at each endpoint, whereas the Inviscid Burgers equation requires appropriate inflow transport boundary conditions. With dimensional variables, $u$ has units of velocity and $\nu$ of kinematic viscosity; all numerical examples below use dimensionless quantities.

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]

A channel with width $b(z)=\alpha z^2$ has wetted area $A=\alpha h^3/3$. Its inviscid shallow water equations have characteristic speeds $u\pm c$ and the displayed Riemann invariants. Releasing a reservoir of depth $H$ at rest into a dry channel gives a left-family rarefaction wave: $c=(6c_0-x/t)/7$, $u=6(c_0+x/t)/7$, $h/H=[(6-x/(c_{0}t))/7{]}^2$ between $x=-c_{0}t$ and $x=6c_{0}t$, where $c_0=\sqrt{gH/3}$. This width law differs from a geometrically parabolic trough, whose width is proportional to $\sqrt{z}$.

A semi-infinite resting layer of depth $H$ and constant reduced gravity $b$ released into a deep ambient develops a rarefaction wave satisfying $u+2\sqrt{bh}=2\sqrt{bH}$. The head closure $u_f=F\sqrt{b{h}_f}$ gives $h_f=4H/(F+2{)}^2$ and $u_f=2F\sqrt{bH}/(F+2)$. A constant-depth shelf lies between the rarefaction tail speed $u_f-\sqrt{b{h}_f}$ and the front speed $u_f$; its width is positive for every finite positive $F$.

An impermeable wall at $x=-\ell$ first receives the upstream-moving head at $t_1$. The leading downstream characteristic curve of the reflected adjustment obeys $\dot{x}=u+c=(12c_0+5x/t)/7$ in the unmodified rarefaction wave, giving the displayed curve. It stays behind the inviscid dry tip $x_f=6c_{0}t$ at every finite time. The finite reservoir drains and its upstream depth decreases, while only the region behind $x_r$ has yet received the wall signal. Bottom drag or other tip physics can change the front law.

When a rarefaction wave attaches to a saturation shock, its terminal characteristic speed must match the Rankine-Hugoniot condition speed. This gives a tangent from the initial state $(s_0,F(s_0))$ to the fractional-flow curve at $s_s$. It selects the upstream shock saturation; the actual dimensional shock speed still includes total flux divided by porosity.

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

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.

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]

A one-dimensional particle suspension settling without particle diffusion or overturning obeys the scalar conservation law ${\varphi }_t+{\partial }_{z}j(\varphi )=0$. The upward flux is negative for downward settling. Characteristic curves, compressive shocks and rarefaction waves describe concentration transport rather than individual particle paths.

The viscous Burgers step solution with negative flux is symmetric around $\theta =-Uz/2$ after subtracting $U/2$. Its Gaussian-tail ratio equals one precisely at this midpoint, giving $f=U/2$ for either sign of $U$. For $U>0$ this point becomes the entropy-shock center in the vanishing viscosity approximation; for $U<0$ it is the center of a rarefaction wave. The same trajectory therefore does not by itself identify a shock.

For water initially at rest at depth $H$ on one side of a dam and a dry bed on the other, a channel with $b(z)=C\sqrt{z}$ has $c_0=\sqrt{2gH/3}$. The centered rarefaction wave has $u+3c=3c_0$ and $x/t=u-c$. Thus $c=(3c_0-x/t)/4$ and $u=3(c_0+x/t)/4$ within $-c_{0}t<x<3c_{0}t$. The reservoir remains undisturbed to the left and the bed dry to the right. This is an outer shallow-water approximation before finite-reservoir boundary signals return.

The volume alone does not specify the early release geometry. For a concrete sketch, use a rectangular lock initially occupying $0<x<{\ell }_0$ with depth $h_0=V/{\ell }_0$, at rest, and remove its gate at $t=0$. Let $c_0=\sqrt{g^{\prime }{h}_0}$ and impose a gravity-current front condition $u_f=F\sqrt{g^{\prime }{h}_f}$, with fixed $F$. The Benjamin deep-ambient front condition gives $F=\sqrt{2}$ for the usual deep-ambient idealization.

A rarefaction wave propagates back from the release. Before information returns from the closed wall, its invariant is $u+2c=2c_0$. With $s=(x-{\ell }_0)/t$, the fan has $u=2(c_0+s)/3$, $c=(2c_0-s)/3$. The front-adjacent shelf hasThe fan extends from $s=-c_0$ to $s=u_f-c_f$; a uniform shelf connects its tail to the finite-depth head at $x={\ell }_0+u_{f}t$. A small nonhydrostatic head displaces ambient fluid; the hydrostatic interior model does not resolve its overturning structure.

The fan head first reaches the rear wall at $t_r={\ell }_0/c_0$. The wall condition is $u=0$, and the returned positive characteristic curve begins to alter the interior. In the still-undisturbed fan it solves $dx/dt=(4c_0+(x-{\ell }_0)/t)/3$, henceIt reaches the fan's uniform shelf at $t_{\ast }=t_r[(F+2)/2{]}^{3/2}$, then propagates with speed $u_f+c_f$ to the head at $t_{\mathrm{hit}}=2t_{\ast }$. These times refer to this particular initial lock and front closure. After that first arrival, the finite supply influences the front: the constant-speed slumping stage gives way to deceleration and eventually the finite-volume inertial gravity current with $L\propto t^{2/3}$, depth proportional to $t^{-2/3}$ and speed proportional to $t^{-1/3}$. The initial fan and reflected signal are shown below; later characteristic trajectories change as the current adjusts.

Integrate the initial Cole-Hopf transformation and choose a continuous positive normalization:Split the heat kernel convolution at zero, and complete the square in the positive half-line contribution. DefineThe resulting heat equation solution isDifferentiating the moving endpoints gives ${\partial }_{\theta }{I}_{-}=-e^{-{\theta }^2/(4ϵZ)}$ and $E{\partial }_{\theta }{I}_{+}=e^{-{\theta }^2/(4ϵZ)}$. These terms cancel. Hence ${\psi }_{\theta }=(4\pi ϵZ{)}^{-1/2}UE{I}_{+}/(2ϵ)$, and the viscous Burgers step solution with negative flux isThe lower limit of $I_{-}$ is $\theta$, as printed in the original PDF. The converted TeX's $-\theta$ does not give the correct convolution or endpoint cancellation.

For the vanishing-viscosity limit, the conservative flux is $F(q)=-q^2/2$ and the characteristic speed is $F^{\prime }(q)=-q$. If $U>0$, the characteristics from the right move left and collide with those on the left. The Rankine-Hugoniot condition gives shock speed $[F]/[q]=-U/2$. ThusThe viscous transition has width $O(ϵ/U)$, and its centre value is exactly $U/2$ by the midpoint symmetry of a viscous Burgers step.

If $U<0$, the right characteristics move right and separate from the left ones. The limit is a rarefaction wave,This follows either from the Burgers Riemann problem with negative flux or the large-argument Gaussian-tail asymptotics of the exact formula. The two signs therefore give a shock and a spreading rarefaction respectively, not two shocks.

Let $s_{\ast }$ be the unique sonic point, $f^{\prime }(s_{\ast })=0$. For a differentiable convex function, $f^{\prime }$ is nondecreasing, negative below $s_{\ast }$ and positive above it. Split the flux into the parts propagating right and left:Thus $f(u)=f(s_{\ast })+f^{+}(u)+f^{-}(u)$; explicitly, $f^{+}(u)=f(u)-f(s_{\ast })$ for $u>s_{\ast }$ and zero below, while $f^{-}(u)=f(u)-f(s_{\ast })$ for $u<s_{\ast }$ and zero above. The Engquist-Osher flux takes the right-going contribution from the left state and the left-going contribution from the right state:It is consistent, $F(u,u)=f(u)$. For a transonic rarefaction wave with $a\le s_{\ast }\le b$ it gives $f(s_{\ast })$; reversed transonic states give $f(a)+f(b)-f(s_{\ast })$. This is sonic-point splitting of a convex Engquist-Osher flux.

On cells of width $h$ let $U_j^n$ approximate the cell average and $\lambda =\Delta t/h$. The explicit Engquist-Osher method is the finite volume methodThe shared interface flux makes it conservative: sums telescope for periodic data, and the same applies on the infinite grid when boundary contributions vanish. To prove stability, choose an interval $[m,M]$ containing the initial states and impose the CFL conditionThe partial derivatives of the update areTherefore the update is a monotone conservative scheme. Because $H(q,q,q)=q$, componentwise comparison with the constant states $m,M$ proves $m\le U_j^{n+1}\le M$. Induction gives a mesh-independent maximum bound for every time level, and ensures the same CFL speed bound remains valid.

There is also a stronger L1 contraction of a monotone conservative scheme. Let $T$ denote one whole-grid update, and let $W=\min (U,V)$ and $Z=\max (U,V)$ componentwise. Monotonicity gives $TW\le TU,TV\le TZ$. Hence $|TU-TV|\le TZ-TW$ componentwise, and conservation yieldsThis applies to periodic grids or summable perturbations with vanishing end fluxes. Apply the same result to $U$ and its one-cell translate; translation commutes with $T$, soThus the method is a total variation diminishing scheme. Under the stated CFL condition it preserves the initial range, contracts differences in discrete $L^1$, and cannot increase total variation. These bounds give a precise nonlinear stability result, including across the sonic point.

Convexity alone does not remove the time-step restriction omitted in the question. For the Inviscid Burgers equation $f(u)=u^2/2$, linearize about a positive constant $U$. The update becomes $v_j^{n+1}=(1-\lambda U)v_j^n+\lambda U{v}_{j-1}^n$. Its alternating Fourier mode has multiplier $1-2\lambda U$, whose modulus exceeds one if $\lambda U>1$. Hence unrestricted explicit time steps are unstable even for a convex flux with one sonic point. This is CFL necessity for explicit Engquist-Osher stability.

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.