Burgers Equation
Consider the initial value problem for Burgers' equation
$$ \begin{align}\begin{cases} u_{t} + u u_{x} = 0 \\ u(x,0) = \phi(x) \end{cases} \end{align} \tag{1}$$
our initial data is given as
$$ \phi(x)= \begin{align}\begin{cases} 1 & x \leq0 \\ 1 -x & 0 < x < 1 \\ 0 & x \geq 1 \end{cases} \end{align} \tag{2}$$
The characteristics are then given by
$$ \frac{dx}{dt} = u \\ \frac{du}{dt} =0 \tag{3} $$
When we solve them we should get
$$ x(t) = ut +x_{0} \tag{4} $$
$$ u(t) = c_{0} $$
then we get
$$ u = c_{0} = \phi(x_{0}) \tag{5} $$
$$ x(t)= \phi(x_{0})t + x_{0} \tag{6} $$
so our solution is given by
$$ u(x,t) = \phi(x_{0}) = \phi(x-ut) \tag{6} $$
$$ u(x,t) = \begin{align}\begin{cases} 1-c_{0}t & x \leq0 \\ 1 -x+c_{0}t & 0 < x < 1 \\ -c_{0}t & x \geq 1 \end{cases} \end{align} \tag{7}$$
So then if we have $u(x,0) =\phi(x-u\cdot 0)$ = $\phi(x)$ we would have
$$ u(x,0) = \begin{align}\begin{cases} 1 – c_{0} \cdot 0 & x \leq0 \\ 1 -x+c_{0}\cdot 0 & 0 < x < 1 \\ -c_{0} \cdot 0 & x \geq 1 \end{cases} \end{align} \tag{8}$$
which is
$$ \phi(x)= \begin{align}\begin{cases} 1 & x \leq0 \\ 1 -x & 0 < x < 1 \\ 0 & x \geq 1 \end{cases} \end{align} \tag{9}$$
Best Answer
The method of characteristics gives the implicit equation $u = \phi(x-ut)$. Therefore, if $x-ut\leq 0$, we have $u=1$. If $x-ut\geq 1$, we have $u=0$. Otherwise, we have $u = \frac{1-x}{1-t}$. To summarize the solution obtained via the method of characteristics, $$ u(x,t) = \left\lbrace \begin{aligned} &1 && x\leq t\\ &\tfrac{1-x}{1-t} && t<x<1\\ &0 && 1\leq x \end{aligned} \right. $$ which is valid for $t<1$. At $t=1$ (breaking time), characteristics intersect and a shock wave occurs. This is illustrated on the following plot of the characteristic curves in the $x$-$t$ plane
See this post for a book reference.