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

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.