Past exam of the applied partial differential equations course of the University of Oxford / 2023 / 2 / c

The initial condition is a period-one square wave: height one on each interval $[k-1/2,k]$ and zero on each gap $(k,k+1/2)$, for integer $k$. Each period has mean $1/2$.

For $0<t<1$, translated copies of the preceding pulse have disjoint interiors of support. In the cell $k-1/2\le x<k+1/2$, write $r=x-k$. The solution isAt $t=0$ use the given initial function. Each fan starts at $k-1/2$, and each shock has position $x=k+t/2$. At $t=1$, the plateau disappears: its left fan catches its right shock at $k+1/2$, which is also the starting edge of the next fan. The shocks must henceforth connect two nonconstant fan branches rather than a fan to a zero interval.

Let $s_k(t)$ be the shock separating the fans from $k-1/2$ and $k+1/2$. Their traces giveThus $\boxed{s_k(t)=k+t/2}$. Between consecutive shocks, $u=(x-k+1/2)/t$. Writing $\{r\}=r-\lfloor r\rfloor$ for the fractional part, the period-one solution can be expressed, with right-hand shock traces, asTo put it in the requested fixed-support sum, takeThe half-open cells partition the line, so exactly one summand contributes at each point. This is a representation of the periodic solution, not a superposition principle for arbitrary solutions of a nonlinear equation.

The left and right shock states are $1/2+1/(2t)$ and $1/2-1/(2t)$; both average to the shock speed $1/2$, and the strict inequalities certify causality. The periodic square-wave relaxation for the inviscid Burgers equation has the uniform estimateThe limit is the conserved mean $1/2$.