Burgers equations with piecewise initial data Given the equation
$$u_t + uu_x = 0, x \in \mathbb{R}, t > 0$$
with the initial conditions:
$$u(x,0)=u_0(x)=
\begin{cases}
1 & \text{if} \ x < -1 \\ 
x^2 & \text{if} \ -1 < x < 0 \\
0 & \text{if} \ x > 0
\end{cases}$$
Using the method of characteristics I obtain the following solutions in a parametrized form:
$$(x,t,u) = (tu_0(s)+s,t,u_0(s))$$
and hence
$$u(x,t) = 
\begin{cases}
1 & \text{if} \ x - t < 1 \\
\frac{x}{t} & \ \text{if} \ 0 < \frac{x}{t} < 1 \\
0 & \text{if} \ x > 0 
\end{cases}$$ 
Thus the shocks occur "everywhere" and I do not know how to use Rankine-Hogoniot condition to handle the shocks. Even if I just formally use Rankine condition, does it lose many informations of $u(x,t)$ in the region $0 < x/t < 1$? 
 A: Starting with the method of characteristics, we know $u = u_0(x-ut)$. Hence, for small positive $t$,


*

*$u=1$ if $x-t<-1$;

*$u=\frac{1+2tx-\sqrt{1+4tx}}{2t^2}$ if $-1<x-ut<0$;

*$u=0$ if $x>0$
Note that a shock occurs from the intersection of characteristics at the breaking time $$t_b=\frac{-1}{\inf u'_0}=\frac12 .$$
A: The shock wave starts at the minimal $t$ where for a given point $(x,t)$ there are two solutions to $x=x_0+tu_0(x_0)$, which happens on the line $x=-1+t$ when it is equal for infinitesimal $h$ to $-1+h+t(-1+h)^2=-1+t+h(1-2t+h)$, thus at $t=\frac12$, $x=-\frac12$.
The shock wave follows a curve $x=v(t)$, where the characteristic lines with $v(t)=x_{01}+t$, $x_{01}<-1$ and $v(t)=x_{02}+tx_{02}^2$, $x_{02}\in(-1,0)$ meet, that is, 


*

*$x_{01}(t)=v(t)-t$ with slope $1$ and 

*$1+4v(t)t=(1+2x_{02}t)^2$, $x_{02}(t)=-\frac1{2t}(1-\sqrt{1+4tv(t)})=\frac{2v(t)}{1+\sqrt{1+4tv(t)}}$ with speed $x_{02}^2=\frac{v(t)-x_{02}}{t}$. 


The Rankine-Hogoniot equation for the speed of the shock now gives 
\begin{align}
v'(t)&=\frac12(1+x_{02}(t)^2)\\
&=(1+2tx_{02})x_{02}'(t)+x_{02}(t)^2
\end{align}
Solve this (numerically) for the inverse function $t_s(x)$ with 
$$
t_s'(x)=1/x_{02}'(t_s(x))=\frac{2(1+2t_s(x)x)}{1-x^2}
$$ 
with initial conditions $t_s(-1)=0.5$ over the interval $[-1,0]$. This now is a linear DE with integrating factor $(1-x^2)^{2}$. Carrying out the integration ends with
\begin{align}
(1-x^2)^2t_s(x)&=\int 2(1-x^2)\,dx=2x-\frac23x^3+C\\
0&=-2+\frac23+C\implies C=\frac43\\
t_s(x) &=\frac23\frac{3(x+1)-(x+1)(x^2-x+1)}{(1-x^2)^2}
\\
&= \frac{2(2-x)}{3 (1-x)^2}
\end{align}
At time $T=t_s(0)=\frac43$ the middle phase has collapsed and the outer phases meet in a shock wave with speed $0.5$. 
Constructing the solution picture from the characteristic curves ending at the shock curves gives the plot

