Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 105 / 3 / 2 / f / Solution

Put $h=|u-v|$, $q=Q(u,v)$ and $h_0=|u_0-v_0|$. The mean value theorem gives $|q|\le Mh$ on the bounded state range. Fix a final time $t$ and let $L(\tau )=a-M(t-\tau )$ and $R(\tau )=b+M(t-\tau )$. Choose a smooth increasing function $\rho$ equal to zero on $(-\infty ,0]$ and one on $[1,\infty )$. For small $\delta >0$, defineIts two nonnegative derivative contributions imply $(W_{\delta }{)}_{\tau }+M|(W_{\delta }{)}_x|\le 0$. Therefore $h(W_{\delta }{)}_{\tau }+q(W_{\delta }{)}_x\le 0$.

Use the nonnegative test $\phi (\tau ,x)={\eta }_{ϵ}(\tau )W_{\delta }(\tau ,x)$, where ${\eta }_{ϵ}(0)=1$, it is one up to just before $s$, decreases smoothly to zero near $s$, and remains zero afterward. The Kato inequality for scalar conservation laws givesbecause the other interior contribution is nonpositive. At Lebesgue times $s$, the temporal approximate identity tends to $\int h(s)W_{\delta }(s)$; take a countable sequence $\delta \downarrow 0$ and use dominated convergence on the bounded cone. This yieldsfor almost every $s\in [0,t]$. This is local L1 contraction for scalar conservation laws. It uses only local integrability, so the bounded data need not have finite global $L^1$ norm. If $M=0$, the flux difference is zero and the same proof uses a stationary interval.