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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.

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 strictly hyperbolic system of conservation laws $U_t+F(U{)}_x=0$ with ordered characteristic speeds ${\lambda }_1<\cdots <{\lambda }_n$, a Lax family-$j$ shock has ${\lambda }_j(U_R)<s<{\lambda }_j(U_L)$. The remaining families must stay on the same side of $s$: ${\lambda }_{j-1}(U_L)<s$ and $s<{\lambda }_{j+1}(U_R)$ where those neighbors exist. These inequalities imply that the lower families have speed below $s$ on both sides and the higher families speed above $s$ on both sides. There are $n+1$ incoming characteristic directions: two from the compressed family and one from each other family. The Rankine-Hugoniot conditions supply the jump balance, while these inequalities supply admissibility. For a convex scalar flux this reduces to the entropy shock condition.

The Rankine-Hugoniot condition for the flux $f(u)=u^2/2$ gives, for distinct states,The characteristic speed is $f^{\prime }(u)=u$. A causal entropy shock has characteristics entering from both sides, so $u_{-}>\dot{s}>u_{+}$, equivalentlyAn upward jump instead resolves into a rarefaction wave; the jump condition alone does not select the physical weak solution.

The upward jump at $-1/2$ creates a rarefaction wave with $u=(x+1/2)/t$, bounded by the characteristic lines $x=-1/2$ and $x=t-1/2$. The downward jump at zero joins the states $1,0$, so part (a) gives the shock position $s(t)=t/2$. The intervening plateau exists exactly while $t-1/2<t/2$, or $t<1$. The fan satisfies the Inviscid Burgers equation classically and matches the constant branches continuously; the shock satisfies the Rankine-Hugoniot condition and is causal because $1>1/2>0$. Its initial trace is the prescribed pulse.

At $t=1$, the fan reaches the shock at $x=1/2$. Subsequently the states immediately to its left and right are $(s(t)+1/2)/t$ and zero, soSolving gives $s(t)+1/2=\sqrt{t}$. Thus the continuation isValues assigned precisely at the shock do not affect the weak solution. Its speed $1/(2\sqrt{t})$ lies strictly between the left state $1/\sqrt{t}$ and zero, so it remains a causal entropy shock. The area under the triangular profile is $\frac{1}{2}(\sqrt{t})(1/\sqrt{t})=1/2$, checking conservation of the initial mass. This is the rectangular pulse solution of the inviscid Burgers equation.

The pressure-free equation has the scalar conservation law formA smooth travelling ansatz $u=U(x-Vt)$ obeys $(U-V)U^{\prime }=0$, so it cannot continuously connect two distinct constant states. The inviscid shock wave must be interpreted as a discontinuous weak solution. Its Rankine-Hugoniot condition isFor $u_1\ne u_2$,The result is independent of which state is called upstream. For a physical compressive entropy shock of the Inviscid Burgers equation, the left and right states satisfy $u_L>u_R$ and therefore $u_L>V>u_R$: characteristic curves run into the shock wave from both sides. The ordering $u_L<u_R$ instead produces a rarefaction wave.

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.

Fix $u_l>u_r$ and the Rankine-Hugoniot condition speed from the preceding part. The new hypothesis is a uniformly convex scalar flux, $F^{\prime \prime }\ge \kappa >0$; it replaces the globally bounded-$F^{\prime }$ hypothesis of part (a). Choose $b=\sigma u_r-F(u_r)$. Thensince a strictly convex function lies below the chord between its two endpoint values. Also $Q(u_l)=Q(u_r)=0$ andThe zeros at the endpoints are simple, so the separated integral diverges logarithmically there. Thus the travelling wave is a decreasing connection defined for all $s$, unique up to translation.

To specify a limit, fix a number $c\in (u_r,u_l)$ independently of $\epsilon$ and normalize $v_{\epsilon }(0)=c$. If $V^{\prime }=Q(V)$ and $V(0)=c$, uniqueness gives $v_{\epsilon }(s)=V(s/\epsilon )$. ConsequentlyAt $x=\sigma t$ the normalized profile equals $c$ for every $\epsilon$. This single-line value is immaterial to the weak solution. Convergence holds pointwise off the line and in $L_{\mathrm{loc}}^1$ by bounded convergence; it cannot be uniform across a nonzero jump. The transition has thickness of order $\epsilon$.

This is the vanishing viscosity approximation to a compressive entropy shock. The Rankine-Hugoniot condition makes the step a weak solution of the inviscid scalar conservation law, while $F^{\prime }(u_r)<\sigma <F^{\prime }(u_l)$ means characteristic curves enter the shock from both sides. For every smooth convex function $\eta$ used as an entropy, with entropy flux for a scalar conservation law $q^{\prime }={\eta }^{\prime }{F}^{\prime }$, the viscous equation givesAgainst compactly supported tests the right side tends to zero, since $u_{\epsilon }$ stays in $[u_r,u_l]$. Passing to the $L_{\mathrm{loc}}^1$ limit yields the entropy inequality, explaining the direction selected by positive viscosity.

A translation must be fixed to obtain this particular limit. An $\epsilon$-dependent translate $V((s-a_{\epsilon })/\epsilon )$ can converge to a shock at a different location, to a constant if its center escapes, or fail to converge if the centers oscillate. Thus existence of profiles alone does not specify a single vanishing-viscosity limit without a phase normalization.

Take period-one data equal to one on $[k-1/2,k]$ and zero on $(k,k+1/2)$. Before $t=1$, the separate rectangular pulse solutions of the inviscid Burgers equation do not overlap in their interiors. At $t=1$ each shock meets the two neighboring fans. Afterward, on the interval between shocks $s_{k-1}$ and $s_k$, the solution remains $(x-k+1/2)/t$. The Rankine-Hugoniot condition gives $s_k^{\prime }=(s_k-k)/t$ and $s_k(1)=k+1/2$, hence $s_k=k+t/2$.

The title formula uses the fractional part and chooses right-hand shock values. Its left and right states are $1/2+1/(2t)$ and $1/2-1/(2t)$, making every shock a causal entropy shock of speed $1/2$. Therefore ${\sup }_x|u(x,t)-1/2|\le 1/(2t)$: the periodic entropy solution converges uniformly to its conserved mean even though it has a jump at every finite time. Periodizing a nonlinear solution is valid here only before the supports interact; after interaction the matching shock calculation is essential.

Start the Inviscid Burgers equation with height $a>0$ on $[x_L,x_L+L]$, $L>0$, and zero elsewhere. The upward left jump gives a rarefaction wave $u=(x-x_L)/t$, while the downward right jump gives an entropy shock at $x_L+L+at/2$. They meet at $t_{\ast }=2L/a$. Afterward the entire nonzero profile is the triangular fan on $x_L<x<s(t)$, with $s(t)=x_L+\sqrt{2aLt}$ and zero outside. To prove the continuation, the Rankine-Hugoniot condition gives $s^{\prime }=(s-x_L)/(2t)$, and matching at $t_{\ast }$ fixes the integration constant. The triangle has conserved mass $aL$, and its left shock state $(s-x_L)/t$ is twice the shock speed, certifying causality.

This approximation studies the limit of a viscous scalar conservation law as its positive diffusion coefficient tends to zero. Under hypotheses giving compactness and appropriate convergence, the viscous entropy dissipation selects an entropy solution of the inviscid scalar conservation law. Fixed-phase decreasing travelling waves for a strictly convex flux converge to an entropy shock. Translating each profile differently can change or destroy the limit, so phase or initial data must specify the family.

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.