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

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.