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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.

For zero viscosity the Inviscid Burgers equation is $q_X-q{q}_{\theta }=0$. Along a characteristic curve, $dq/dX=0$ and $d\theta /dX=-q$. Labelling a characteristic by its initial position $\eta$ givesWhere $f$ is differentiable, $q_{\theta }=f^{\prime }(\eta )/(1-X{f}^{\prime }(\eta ))$. The first characteristic crossing occurs at $X=1/\max f^{\prime }$ if the maximum positive slope exists. A multivalued continuation is not a physical wave: rewrite the equation as the conservation law $q_X+{\partial }_{\theta }(-q^2/2)=0$ and replace the overturning part by an entropy-admissible shock wave. The Rankine-Hugoniot conditions give its trajectoryThe states on either side are supplied by the characteristics meeting the shock. Conservation supplies the equal-area construction, and the entropy condition requires characteristics to enter the shock. Since the flux is concave, a jump with $q_L<q_R$ is compressive.

For the rising step, $q_L=0$ and $q_R=U>0$, the left characteristics have slope zero and the right ones slope $-U$. They overlap immediately. The Burgers Riemann problem with negative flux is solved byIt is a shock, not a rarefaction wave; the minus sign in the nonlinear flux is decisive.

For positive viscosity, substitute $q=2ϵ{\partial }_{\theta }\log \psi$ into the viscous Burgers equation. Writing $w=\log \psi$ givesAn $X$-dependent multiplicative factor in $\psi$ removes the possible integration function. The Cole-Hopf transformation therefore gives ${\psi }_X=ϵ{\psi }_{\theta \theta }$. Normalize its initial value continuously at zero:Split the heat kernel integral at zero. Completing the square in the positive-half integral yieldsWhen differentiating, the Gaussian terms from the two complementary error functions cancel, becauseThus the exact viscous Burgers step solution with negative flux isFor $X>0$ its denominator is positive, its tails tend to $0$ and $U$, and it approaches the initial step as $X\downarrow 0$. At $\theta =-UX/2$ it equals $U/2$. Its vanishing-viscosity limit gives the entropy shock above. At large $U\sqrt{X/ϵ}$ near the moving shock center it approaches the travelling profile $U/[1+e^{-U(\theta +UX/2)/(2ϵ)}]$, with thickness of order $ϵ/U$.

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.

For positive diffusivity, the Cole-Hopf transformation with $f=2\alpha {\partial }_{\theta }\log \psi$ converts the negative-flux equation into the heat equation. Step data give a ratio of two Gaussian-tail integrals $J$. At $\theta =-UZ/2$, $J=1$ and $f=U/2$. Its vanishing-viscosity limit selects the shock or rarefaction wave in the Burgers Riemann problem with negative flux, depending on the sign of $U$.