Burgers' equation with piecewise constant initial conditions

Question

I'm trying to solve Burgers' inviscid equation with these initial conditions:

$$u(0,x)= \begin{cases} a, & \text{if $x<0$} \\ b, & \text{if $x>0$} \end{cases}$$

Here is my work so far: $$u(0,x)= \begin{cases} a, & \text{if $x<0$} \\ b, & \text{if $x>0$} \end{cases}$$ $$x(t)= u(0, x(0))t+x(0)$$ $$x(t)=\begin{cases} at+x(0), & \text{if $x<0$} \\ bt+x(0), & \text{if $x>0$} \end{cases}$$

$$t= \begin{cases} \frac1a (x-x(0)), & \text{if $x(0)<0$} \\ \frac1b(x-x(0)), & \text{if $x(0)>0$} \end{cases}$$

If all this is correct, I am a bit confused as to how to put them all together, this was my attempt:

$$u(t,x) = \begin{cases} a, & \text{if $x<at$} \\ \frac xt, & \text{if $at<x<bt$} \\ b, & \text{if $x>bt$} \end{cases}$$

Answer 1

Your solution is correct, assuming $a\le b$. The situation becomes transparent once one sketches the characteristic lines:

• The lines $x = x_0+at$, $x_0<0$, sweep out the region $x<at$; so in this region $u=a$.
• The lines $x = x_0+bt$, $x_0>0$, sweep out the region $x>bt$; so in this region $u=b$.
• In the remaining part of the plane the characteristic lines must go to $(0,0)$, since there is no other place they can reach without crossing other characteristics. Hence they are of the form $x = ct$, and $u=c$ on such a characteristic. In other words, $u=x/t$. This region is the rarefaction wave.

However, if $a>b$, the above does not apply. In this case one must be aware that a shock wave forms at the discontinuity, and propagates with the velocity $(a+b)/2$. Hence,

• $u=a$ when $x<(a+b)t/2$
• $u=b$ when $x>(a+b)t/2$

with the solution being discontinuous at $x=(a+b)t/2$.