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

Let $M=\Vert F^{\prime }{\Vert }_{\infty }$. The sharper energy estimate uses the divergence structure of the viscous scalar conservation law. Multiply by $u$ and integrate over the line. With $G(s)={\int }_0^{s}r{F}^{\prime }(r)dr$,because the decay makes $G(u)$ tend to zero at both ends. Integration by parts in the diffusion term therefore givesOne may take $C_0=0$, uniformly in $\epsilon$. This does not require $F(0)=0$.

If a bound explicitly involving $M$ and $\epsilon$ is desired, retaining the transport term and using the Cauchy-Schwarz inequality and the elementary inequality $ab\le (a^2+b^2)/2$ givesThe Gronwall inequality then gives the valid but weaker choice $C_0=M^2/\epsilon$. The exact cancellation explains why its divergence is unnecessary.

Multiply the viscous scalar conservation law by $-u_{xx}$, rather than estimating $F^{\prime \prime }$ after differentiating. Integration by parts yieldsThus $d\Vert u_x{\Vert }_2^2/dt\le (M^2/\epsilon )\Vert u_x{\Vert }_2^2$. Applying the Gronwall inequality givesOnly the assumed bound on $F^{\prime }$ is used; a global bound on $F^{\prime \prime }$ is not needed for this energy estimate.

For the sharp energy estimate, $C_0=0$ stays uniform while the available bound $C_1=M^2/\epsilon$ diverges as $\epsilon \downarrow 0$ when $M>0$. If the coarser estimate is used for the first part, both displayed bounds $M^2/\epsilon$ diverge, but the first divergence is only an artifact of discarding an exact cancellation.

The method of characteristics explains why uniform control of the gradient cannot generally persist for the inviscid scalar conservation law. Before characteristic crossing, with initial data $u_0$,If $F^{\prime \prime }(u_0(\xi ))u_0^{\prime }(\xi )<0$ somewhere, the denominator reaches zero in finite positive time. The solution steepens and the smooth description breaks down; an entropy solution can subsequently contain shocks. Positive viscosity replaces such a discontinuity by a thin smooth layer, which can have large gradient even while its $L^2$ norm remains controlled. The divergent bound does not assert that every flux and every initial datum form a shock; a linear flux, for example, has no such steepening.

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.