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.

  • $\begingroup$ 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? $\endgroup$ – dezdichado Jan 9 '18 at 23:53
  • 2
    $\begingroup$ @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. $\endgroup$ – EditPiAf Jan 10 '18 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.