# Non-linear transport equation with shock and rarefaction.

This is an old PDE qual problem and I seem to be lacking proper background to solve.

Solve $u_t+uu_x = 0$, where the initial data is given by $$f(x) = \begin{cases}2, &0<x<1\\ 0, & \text{otherwise} \end{cases}.$$ Moreover, show that the solution has both rarefaction and shock wave, and decide whether the rarefaction wave catches the shock. Characterize the location and velocity of the waves.

What I am able to do is to use the method of characteristics to start as: $$u(t,x) = \begin{cases}2,&ut<x<ut+1\\0,&x\leq ut \,\,\,\vee \,\,x\geq ut+1 \end{cases}.$$

After this, how does one find the shock and rarefaction waves? I am following Introduction to Partial Differential equations by Peter Olver and the text says rarefaction and shock occurs when $f'(x)>0$ and $f'(x)<0$, respectively. But this does not help at all in this problem.

The present PDE is the inviscid Burgers' equation. A sketch of the characteristic curves in the $x$-$t$ plane is
Along these curves, $u$ is constant and equal to its value at $t=0$, deduced from the initial data $f(x)$ (similar to a rectangular function).
• One can observe that the characteristic curves separate at $x=0$: the edges are the lines $x=0$ on the left, and $x=2t$ on the right. According to the Lax entropy condition, a rarefaction wave occurs. Such a wave is a self-similar continuous solution, i.e. $u(x,t) = v(x/t)$. In the case of Burgers' equation, one shows that $v(x/t) = x/t$ (cf. e.g. this post).
• At $x=1$, characteristics cross: a shock wave occurs. Starting from the intersection of characteristics, the method of characteristics is not valid anymore. The speed of shock $s$ is given by the Rankine-Hugoniot condition $s = \frac{1}{2}(2 + 0) = 1$. The shock wave is located on the line $\frac{x-1}{t} = s$, i.e. $x = 1+t$.
Therefore, as long as both waves do not interact, the solution is u(x,t) = \left\lbrace \begin{aligned} &0 & &\text{if}\quad x\leq 0\\ &x/t & &\text{if}\quad 0\leq x \leq 2t\\ &2 & &\text{if}\quad 2t\leq x<1+t\\ &0 & &\text{if}\quad 1+t<x \, . \end{aligned} \right. Both waves will interact at some time $t^*$ such that $2t^* = 1+t^*$, i.e. $t^* = 1$. The rarefaction wave catches the shock.
• The region for which $u(x,t) = 2$ is given by $2t<x<1+t$, but why is that? Which theorem/result imply that? Right now it looks like it come out of nowhere. The shock speed from Rankine-Hugoniot condition seems to be playing a role, too. Can you elaborate little more on those? – dezdichado Jan 9 '18 at 23:53
• @dezdichado The location of the edges of the rarefaction follows from the expression of characteristic curves. In particular, the right edge has equation $x=2t$. The curve $x=1+t$ where the shock is located is deduced from R-H. Between those two curves, one has $u=2$, since $u$ is constant along characteristics. – EditPiAf Jan 10 '18 at 0:55