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For a scalar equation with strictly convex flux and states $u_l>u_r$, the Rankine-Hugoniot condition determines a discontinuity speed strictly between the right and left characteristic speeds. Characteristic curves enter this compressive discontinuity from both sides. Its entropy flux for a scalar conservation law inequalities make it an entropy solution. It is the limit of a phase-normalized decreasing travelling wave in the vanishing viscosity approximation.

An entropy solution of a scalar conservation law is a weak solution satisfying additional inequalities that select the admissible continuation through shock waves. For a bounded solution of $u_t+f(u{)}_x=0$ on the real line with $f\in C^1$, require every Kruzhkov entropy inequalityin the sense of distributions, together with the initial condition as a local $L^1$ trace. Equivalently, the interior integral of $|u-k|{\phi }_t+q_k(u){\phi }_x$ is nonnegative for every nonnegative compactly supported test function away from $t=0$.

For a smooth convex function $\eta$, the corresponding entropy flux for a scalar conservation law satisfies $q^{\prime }={\eta }^{\prime }{f}^{\prime }$. A viscous scalar conservation law yields ${\partial }_t\eta (u)+{\partial }_{x}q(u)=\epsilon {\partial }_{xx}\eta (u)-\epsilon {\eta }^{\prime \prime }(u)u_x^2$; the final term is nonpositive. Suitable bounded strongly convergent vanishing viscosity approximations therefore motivate the inequalities. Convexity of the flux itself is unnecessary for the local L1 contraction for scalar conservation laws and the resulting uniqueness; for strictly convex fluxes, the inequalities exclude expansion jumps and select compressive entropy shocks.

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.

Set $q=\log \psi$ and use the sign-appropriate Cole-Hopf transformation $f=2\epsilon q_{\theta }$. Direct differentiation givesThus the heat equation ${\psi }_z=\epsilon {\psi }_{\theta \theta }$ implies the viscous Burgers equation with the required negative nonlinear sign. Conversely, vanishing of the Burgers residual makes the final quotient depend only on $z$, and a multiplicative time-dependent factor in $\psi$ removes it. The algebraic substitution works for any nonzero $\epsilon$ where $\psi \ne 0$.

For a forward dissipative initial-value solution, take $\epsilon >0$ and $z>0$. This is essential for the supplied heat kernel: at $\epsilon <0$, $z>0$, its real Gaussian integral diverges even for $\psi (0,\varphi )=1$. More generally that kernel requires $\epsilon z>0$. The physical viscosity convention selects the positive case; the mere condition $\epsilon \ne 0$ does not justify a forward Gaussian function formula.

Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\epsilon )$ gives a positive continuous initial factor, normalized asIts heat kernel convolution converges for both signs of $U$, since a Gaussian function dominates the one-sided exponential. DefineThe left half-line contributes $J(\theta )$; completing the square on the right half-line gives $e^{S}J(-\theta -Uz)$. ThusWhen differentiating, the two moving-endpoint Gaussian function terms cancel, becauseOnly ${\partial }_{\theta }{e}^S=(U/(2\epsilon ))e^S$ remains in the logarithmic derivative. The viscous Burgers step solution with negative flux is thereforeThe denominator is strictly positive, so $f$ lies between $0$ and $U$, irrespective of the sign of $U$. Its initial limits away from $\theta =0$ are the required step.

For the spatial limits, the Gaussian function tails must be compared with $e^S$; inspecting that exponential alone gives the wrong inference when $U<0$. As $b\to +\infty$, the complementary error function asymptotic givesAt $\theta \to +\infty$, $J(\theta )e^{-S}$ tends to zero even for $U<0$, because its quadratic decay dominates any exponential linear in $\theta$. Hence $\alpha e^{-S}\to 0$. At $\theta \to -\infty$, use the identity above to obtainwhile $J(\theta )$ approaches its full Gaussian integral. Thusfor both signs of $U$.

Since $J$ is strictly decreasing, $\alpha =1$ holds exactly when $\theta =-\theta -Uz$. At this point $S=0$, so the midpoint symmetry of a viscous Burgers step givesMore generally the exact symmetry is $f(z,-Uz-\theta )=U-f(z,\theta )$. For $U>0$, the midpoint follows the inviscid shock trajectory. For $U<0$, it is instead the center of an expanding fan, not a shock.

The vanishing viscosity approximation makes the comparison precise. For fixed $z>0$ and $U>0$, near $\theta =-Uz/2$ both tail integrals tend to the full Gaussian integral, so $\alpha \to 1$ andThe layer has width $O(\epsilon /U)$ and tends to the entropy shock, with value $U/2$ at its center. Outside it the limits are $0$ and $U$.

For $U<0$, inside $0<\theta <-Uz$ both Gaussian function tails have large positive lower limits. Their asymptotics giveOutside this interval the limits are $0$ on the left and $U$ on the right. Thus positive viscosity selects the rarefaction wave, including the same midpoint value as the inviscid fan. The two sign cases agree with the preceding entropy solution construction, while a negative diffusivity would not furnish this dissipative selection.

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.

The viscous Burgers equation combines a nonlinear transport equation with diffusion with constant $\nu >0$. On the infinite line, a bounded initial condition gives a smooth function solving the equation for every positive time; a jump in the initial condition is smoothed rather than retained as a discontinuity. Small $\nu$ can nevertheless produce a steep layer, as the Viscous Burgers travelling shock illustrates. The Cole-Hopf transformation $u=-2\nu {\partial }_x\log \varphi$, with positive $\varphi$, converts this equation to the heat equation ${\varphi }_t=\nu {\varphi }_{xx}$.^[ref]

Positive diffusion also explains the admissibility of the inviscid limit. For the entropy flux for a scalar conservation law associated with $\eta (u)=u^2/2$, direct multiplication by $u$ givesThe last term is nonpositive. With convergence sufficient to pass to the limit, the vanishing viscosity approximation therefore retains an entropy inequality rather than just the weak solution equation. For general bounded data, the full Kruzhkov entropy inequalities specify the entropy solution; the quadratic identity is a useful illustration of dissipation.^[ref]

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.