Solving the inviscid Burgers' equation $u_t + uu_x = 0$ Here, we are given the initial condition:
$$u(0, x) = \begin{cases}
\, 1 & \text{when } x \ge 0 \\
\, -1 & \text{when } x < 0.
\end{cases}$$
I am aware that then solution has the general form $u(t, x) = u(u(x)t + x_0)$.  My question is how to find where the solutions exist? I think it's clear that it will not be the entire $t$-$x$ plane, but I don't know how to think about these recursive situations.
 A: Hint: Here is a weak 1-parameter solution 
$$u(t,x)~=~\theta (x\!-\!t\!+\!t_0)\theta(x) - \theta(-x\!-\!t\!+\!t_0)\theta(-x)+\frac{x}{t\!-\!t_0}\theta(t\!-\!t_0\!-\!|x|) , $$ $$ (t,x)~\in~\mathbb{R}^2\backslash (]-\infty,t_0]\times \{0\} ), \qquad t_0~\in~[0,\infty[,   \tag{1}$$
to the inviscid Burgers' equation 
$$ \frac{\partial u}{\partial t} +\frac{1}{2}\frac{\partial (u^2)}{\partial x}~=~0,\qquad u(t\!=\!0,x)~=~{\rm sgn}(x).\tag{2}$$
The solution (1) is defined and continuous everywhere in the plane $\mathbb{R}^2$ except for a halfline, where it has a kink. Physically, the solution (1) describes the dispersion of a kink at some future time $t=t_0$. If we take the parameter $t_0\to \infty$, we get the constant kink solution
$$ u(t,x)~=~{\rm sgn}(x), \qquad x~\neq~ 0.\tag{3}$$
The solutions (1) & (3) are not the only solutions, cf. above comments by user Mattos & Hans Lundmark. This is due to rarefaction in the future region $t>|x|$ and shock  in the past region $-t>|x|$, cf. e.g. Scott A. Sarra's website.
Sketched check of eq. (1):
$$u^2~=~\theta (|x|\!-\!t\!+\!t_0)+\frac{x^2}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|) .\tag{4}$$
$$  \frac{\partial (u^2)}{\partial x}~=~\left({\rm sgn}(x)-\frac{x|x|}{(t\!-\!t_0)^2} \right)\delta (|x|\!-\!t\!+\!t_0)+\frac{2x}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|)$$ 
$$~=~ \frac{2x}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|).\tag{5}$$
$$  \frac{\partial u}{\partial t}
~=~ -\delta (x\!-\!t\!+\!t_0)\theta(x)+ \delta (x\!+\!t\!-\!t_0)\theta(-x)+\frac{x}{t\!-\!t_0}\delta (|x|\!-\!t\!+\!t_0)-\frac{x}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|)$$ 
$$~=~-\frac{x}{(t\!-\!t_0)^2}\theta(t\!-\!t_0\!-\!|x|).\tag{6}$$
A: This is a Riemann problem which cannot be solved by applying the method of characteristics alone: one should consider weak solutions. For positive times $t>0$, the solution satisfying Lax's entropy condition is a rarefaction wave (simple wave). The latter is a smooth self-similar solution of the form $u(x,t) = v(x/t)$:
$$
u(x,t)=\left\lbrace
\begin{aligned}
&{-1} &&\text{when } x <-t\\
&x/t &&\text{when } {-t}\leq x <t \\
&{1} &&\text{when } x \geq t
\end{aligned}\right.
$$
A: 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 $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$
$u(0,x)=\begin{cases}1&\text{when}~x\geq0\\-1&\text{when}~x<0\end{cases}$ :
$\therefore u=\begin{cases}1&\text{when}~x-ut\geq0\\-1&\text{when}~x-ut<0\end{cases}=\begin{cases}1&\text{when}~x-t\geq0\\-1&\text{when}~x+t<0\end{cases}$
Hence $u(t,x)=\begin{cases}1&\text{when}~x-t\geq0\\-1&\text{when}~x+t<0\\c&\text{otherwise}\end{cases}$
