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For the negative-flux viscous Burgers equation, the Cole-Hopf transformation uses $f=2\alpha {\partial }_{\theta }\log \psi$. Initial N-wave data give ${\psi }_0=\exp [U(L^2-{\theta }^2)/(4\alpha )]$ inside $[-L,L]$ and one outside. Completing the square in the heat kernel convolution gives, with $a=1+UZ$,The Gaussian interval masses and their exponentially weighted tails must be treated together when taking the small-diffusion limit.

Let $h=\log \psi$, with $\psi >0$, and set $f=2\alpha h_{\theta }$. Direct differentiation givesThe heat equation ${\psi }_Z=\alpha {\psi }_{\theta \theta }$ implies $h_Z=\alpha (h_{\theta \theta }+h_{\theta }^2)$, so the bracket vanishes. Conversely its being independent of $\theta$ can be absorbed into a $Z$-dependent multiplicative normalization of $\psi$, leaving $f$ unchanged. This proves the Cole-Hopf transformation with the positive sign appropriate to the negative-flux Burgers convention. Although the algebra works for nonzero $\alpha$, the given forward Gaussian function diffusion kernel and a physical vanishing-viscosity limit require $\alpha >0$.

Take $U,L>0$ for the stated Burgers N-wave. Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\alpha )$ and normalizing the exterior value to one givesThis function is continuous at $\pm L$; its logarithmic derivative has the specified jumps. Convolution with the heat kernel is positive and solves the heat equation for $Z>0$. Splitting the integral into the exterior baseline and the interior correction yieldsPut $a=1+UZ$. Completing the square givesChanging the interior integration variable to $\eta =a\phi$ changes its limits to $\pm La$ and the Gaussian function width to $Za$. The Jacobian and normalization leave the factor $a^{-1/2}$. Thus the Cole-Hopf solution for a Burgers N-wave isHere $I_{\alpha }$ is the normalized Gaussian function mass of its indicated interval. Its explicit error function representation isFor fixed $w>0$, the Gaussian function concentrates at $\phi =\theta$ as $\alpha \downarrow 0$. Consequentlyaway from the endpoints; at $\theta =\pm L$ the limit is $1/2$. The transition layer has width $O(\sqrt{\alpha w})$. This is an approximate identity argument, not a uniform step approximation across the endpoints.

For fixed $Z>0$ and $|\theta |\ll L$, both interval masses tend to one, and the exponentially large positive $\hat{\psi }$ dominates the exterior correction. Its logarithmic derivative therefore givesFor $|\theta |\gg La$, both interval masses are exponentially small and the weighted interior integral is also negligible, while the exterior contribution tends to one. Therefore $\boxed{f(Z,\theta )\simeq 0}$ in the specified far exterior.

An exponentially weighted Gaussian function tail should not be discarded solely because its unweighted interval mass tends to zero. In fact, comparing the order-one exterior term with $\hat{\psi }$ gives the sharper inviscid Burgers N-wave fronts $|\theta |=L\sqrt{1+UZ}$, not $La$. Away from these fronts, the vanishing-viscosity limit isInside $|\theta |<La$, this follows from the sign of $L^2-{\theta }^2/a$ in the exponential. Outside $La$, the constrained Gaussian function maximum lies at an interval endpoint and has negative exponent. The right shock wave's speed is $UL/(2\sqrt{a})$, equal both to the derivative of $L\sqrt{a}$ and to minus half the sum of its two limiting states; the left shock wave is its reflection. This checks consistency with the Rankine-Hugoniot condition and shows that the requested near-center and far-exterior approximations are compatible with the full entropy limit.

Let $h=\log \psi$ and $f=2\alpha h_{\theta }$. Direct substitution into the viscous Burgers equation givesIf ${\psi }_Z=\alpha {\psi }_{\theta \theta }$, then $h_Z=\alpha (h_{\theta \theta }+h_{\theta }^2)$, proving the Cole-Hopf transformation. A function of $Z$ inside the parentheses can be removed by rescaling $\psi$ by a time-dependent factor, which leaves $f$ unchanged.

For the forward heat equation and the displayed Gaussian kernel we require $\alpha >0$ and $Z>0$. The algebraic transformation works for nonzero $\alpha$, but the printed convolution is not a forward solution for negative $\alpha$: its Gaussian then grows and the step-data integral diverges.

Integrating the initial logarithmic derivative fixes a convenient positive initial heat datum,Split the heat kernel convolution at zero and complete the square in the positive half. With $h_{\ast }=U(2\theta +UZ)/(4\alpha )$, putThen $\psi =(A_0+e^{h_{\ast }}{C}_0)/\sqrt{4\pi \alpha Z}$. On differentiation, the moving-limit terms cancel because $e^{h_{\ast }}{e}^{-(\theta +UZ{)}^2/(4\alpha Z)}=e^{-{\theta }^2/(4\alpha Z)}$. HenceThis is the viscous Burgers step solution with negative flux. Both integrals can be written as $\sqrt{\pi \alpha Z}$ times a complementary error function.

For fixed $Z>0$, the Gaussian-tail asymptotics give $f\to U$ as $\theta \to +\infty$ and $f\to 0$ as $\theta \to -\infty$. For example $J{e}^{-h_{\ast }}$ tends to zero on the right with a Gaussian factor $e^{-(\theta +UZ{)}^2/(4\alpha Z)}$, while on the left it diverges with a Gaussian factor $e^{{\theta }^2/(4\alpha Z)}$. This holds for either sign of $U$; for $U=0$ the solution is already zero.

The Gaussian tail is strictly decreasing in its lower limit, so $J=1$ occurs exactly at $\theta =-UZ/2$. There $h_{\ast }=0$, and $f=U/2$. For $U>0$, the two lower limits are both far into the negative tail in the mature shock region, so $J\simeq 1$ throughout its thin transition. The solution is then approximately the traveling viscous frontcentred at the inviscid shock position, with thickness of order $\alpha /U$.

In the vanishing-viscosity limit, for $U>0$ the solution tends to the compressive entropy solution of part (a), away from its shock. For $U<0$, it tends instead to the rarefaction wave. To see the latter explicitly inside $0<\theta <-UZ$, both tails have positive lower limits, and their leading asymptotics give $J{e}^{-h_{\ast }}\to (-\theta -UZ)/\theta$. Thus $f\to -\theta /Z$ in the fan, with the constant states outside. Although $J=1$ still marks the fan midpoint, it is not approximately one throughout the expanding fan; replacing it by one there would create an inadmissible compressive-front approximation.

For the heat kerneldirect partial differentiation givesThus $K$ solves the heat equation. For bounded continuous $f$, differentiation under the integral sign givesWith $Y=(x-y)/\sqrt{4t}$,The Gaussian integral makes the weight have total mass one, and the dominated convergence theorem gives $u(x,t)\to f(x)$. This is the Gaussian approximate identity.

For the viscous Burgers equation with unit viscosity, the Cole-Hopf transformation $w=-2u_x/u$ reduces the equation to $u_t=u_{xx}$. To obtain initial value $g$, choose(up to an irrelevant positive constant), and set $u=K_t\ast f$. ThereforeEquivalently,and the approximate-identity limit gives $w(x,t)\to g(x)$.

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 $U$.