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

The Rankine-Hugoniot condition for the flux $f(u)=u^2/2$ gives, for distinct states,The characteristic speed is $f^{\prime }(u)=u$. A causal entropy shock has characteristics entering from both sides, so $u_{-}>\dot{s}>u_{+}$, equivalentlyAn upward jump instead resolves into a rarefaction wave; the jump condition alone does not select the physical weak solution.

The upward jump at $-1/2$ creates a rarefaction wave with $u=(x+1/2)/t$, bounded by the characteristic lines $x=-1/2$ and $x=t-1/2$. The downward jump at zero joins the states $1,0$, so part (a) gives the shock position $s(t)=t/2$. The intervening plateau exists exactly while $t-1/2<t/2$, or $t<1$. The fan satisfies the Inviscid Burgers equation classically and matches the constant branches continuously; the shock satisfies the Rankine-Hugoniot condition and is causal because $1>1/2>0$. Its initial trace is the prescribed pulse.

At $t=1$, the fan reaches the shock at $x=1/2$. Subsequently the states immediately to its left and right are $(s(t)+1/2)/t$ and zero, soSolving gives $s(t)+1/2=\sqrt{t}$. Thus the continuation isValues assigned precisely at the shock do not affect the weak solution. Its speed $1/(2\sqrt{t})$ lies strictly between the left state $1/\sqrt{t}$ and zero, so it remains a causal entropy shock. The area under the triangular profile is $\frac{1}{2}(\sqrt{t})(1/\sqrt{t})=1/2$, checking conservation of the initial mass. This is the rectangular pulse solution of the inviscid Burgers equation.

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 $1/2$.