Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 5 / 4 / b

Insert the travelling wave into the viscous scalar conservation law. With $s=x-\sigma t$ the equation becomesand one integration givesFor a nonconstant profile, $Q(v(s))$ never vanishes. Indeed, the autonomous ordinary differential equation $v^{\prime }=Q(v)/\epsilon$ has unique local solutions since $Q$ is $C^1$; reaching an equilibrium would force the whole solution to be constant. Separation and $c=v(0)$ therefore giveA different choice of reference point gives $s-s_0$ on the left, expressing the translation freedom of the travelling wave.

The printed formula needs a nonconstant-profile qualification. Constant profiles also solve the PDE, but their denominator vanishes at their constant value, so the separated integral is not defined. They must be included separately as equilibrium solutions of the integrated ordinary differential equation.

The integrated travelling wave equation is $v^{\prime }=Q(v)/\epsilon$. A finite limiting value $a$ at either end must satisfy $Q(a)=0$. Otherwise continuity of $Q$ makes $v^{\prime }$ eventually have a fixed sign and an absolute value bounded below, which is incompatible with convergence to $a$. ThereforeSubtracting yields the Rankine-Hugoniot conditionFor distinct end states this givesDistinctness is needed for the printed quotient. If $u_l=u_r$, the identity is $0=0$. A nonconstant global profile is strictly monotone by the scalar ordinary differential equation, so cannot have equal finite end states. The equal-state profiles here are constant and their representation allows any $\sigma$.

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.