Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 50 / 4

Write $L=\log \psi$, with $\psi >0$. Under the Cole-Hopf transformation $q=2ϵL_{\theta }$, direct differentiation givesSince $L_{\theta \theta }+L_{\theta }^2={\psi }_{\theta \theta }/\psi$, Burgers' equation is equivalent toThe ratio is therefore a function $a(Z)$ alone. Multiplying $\psi$ by $\exp [-\int a(Z)dZ]$ leaves $q$ unchanged and removes this scalar freedom. Thus one may choose the normalization for whichThis is the heat equation. The positive sign in $q=2ϵ{\partial }_{\theta }\log \psi$ corresponds to the negative conservative flux $-q^2/2$ used here; the more usual positive-flux convention has the opposite sign.

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.

Set $s=Z+c\theta$ and write $q=Q(s)$. Substitution into the modified Burgers equation with cubic flux givesAfter one integration,A regular constant limiting state has $Q^{\prime }\to 0$. The state $Q=0$ at the negative end sets $C=0$, and the nonzero state $Q=\beta$ at the positive end givesThus the first-order equation is $ϵc^2{Q}^{\prime }=Q(1-Q^2/{\beta }^2)$. Put $W=Q^2$. Its equation is logistic,The heteroclinic travelling wave connecting the prescribed endpoints is consequentlywhere $s_0$ is an arbitrary translation. Taking the sign from $\beta$ is essential; squaring the equation alone loses the negative-front solution. A fixed level propagates with $d\theta /dZ=-1/c=-{\beta }^2/3$, also obtained from the Rankine-Hugoniot condition for flux $-q^3/3$.