I am not quite sure how to deal with discrete IVP
Find self-similar solution \begin{equation} u_t=u u_x\qquad -\infty <x <\infty,\ t>0 \end{equation}
satisfying initial conditions
\begin{equation} u|_{t=0}=\left \{\begin{aligned} -1& &x\le 0,\\ 1& &x> 0 \end{aligned}\right. \end{equation}
Here is my attempt
Characteristic equation: \begin{align} \frac{dt}{1} &=\ \frac{dx}{-u} = \frac{du}{0}\\ \frac{du}{dt}&=\ 0 \implies u=f(C)\\ \frac{dx}{dt} &=\ -u = -f(C)\\ x &=\ C - tf(C)\\ \end{align}
Impose boundary condition: $t=0$ and $x = C$, \begin{align} u &=\ f(C) = u|_{t=0}=\left\{\begin{aligned} -1& &x\le 0,\\ 1& &x> 0 \end{aligned}\right. \end{align} \begin{align} x = C+t \left\{\begin{aligned} -1& &x\le 0,\\ 1& &x> 0 \end{aligned}\right. \end{align} Where the characteristic lines are never across each other
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$\dfrac{dx}{ds}=-u=-u_0$ , letting $x(0)=f(u_0)$ , we have $x=f(u_0)-u_0s=f(u)-ut$ , i.e. $u=F(x+ut)$
$u(x,0)=\begin{cases}-1&\text{when}~x\leq0\\1&\text{when}~x>0\end{cases}$ :
$\therefore u=\begin{cases}-1&\text{when}~x+ut\leq0\\1&\text{when}~x+ut>0\end{cases}=\begin{cases}-1&\text{when}~x-t\leq0\\1&\text{when}~x+t>0\end{cases}$
Hence $u(x,t)=\begin{cases}-1&\text{when}~x\leq t\\1&\text{when}~x>-t\\c&\text{otherwise}\end{cases}$