Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 75 / 3 / b / Solution

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.