Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 70 / 3 / b / Solution

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.