Quasilinear PDE $u_t + (u^2)_x = 0$ cauchy problem The problem I am trying to solve is:
\begin{equation}\label{eq:3.1}
\begin{cases}
 \partial_t u + \partial_x(u^2)=0 & x\in \mathbb{R}, t \in (0,\infty]\\
    u(x,0)=
    \begin{cases}
  0 & x\leq 0\\
  x & 0<x\leq 1\\
  1 & x>1
    \end{cases}           
\end{cases} 
\end{equation}

What I have done is:
We will try to reduce the problem to ODEs on a curve $x(t)$ on the $(t,x)$ plane. The equation can be compared with the canonical form,
\begin{equation}
a\frac{\partial u}{\partial x} +b\frac{\partial }{\partial t} = c,
\end{equation}
where $a = 2u$, $b= 1$ and $c=0$. From the Lagrange-Charpit equations, we have,
\begin{align}\label{eq:3.2}
&\frac{dx}{a}=\frac{dt}{b}=\frac{du}{c} & \text{ substituting we have,}\nonumber\\
\implies &\frac{dx}{2u}=\frac{dt}{1}=\frac{du}{0}&
\end{align}
using second and the third ratio from the equation  we have,
\begin{align}\label{eq:3.3}
&\frac{du}{dt}=0 & \text{integrating we have,} \nonumber\\ 
\implies&u=B,& 
\end{align}
where $B$ is a arbitrary constant. Using the initial conditions,
\begin{equation}\label{eq:3.4}
u(x,0)=
\begin{cases}
  0 & x\leq 0\\
  x & 0<x\leq 1\\
  1 & x>1
\end{cases}           
\end{equation}
where the characteristic curve $x(t)$, passes through $(c,0)$. By substitution we have,
\begin{equation}
B=
\begin{cases}
  0 & x\leq 0\\
  c & 0<x\leq 1\\
  1 & x>1.
\end{cases}           
\end{equation}
Therefore solution can be written as
\begin{equation}\label{eq:3.5}
u=
\begin{cases}
  0 & x\leq 0\\
  c & 0<x\leq 1\\
  1 & x>1.
\end{cases}        
\end{equation}
using first and the second ratios from the equation we have,
\begin{align}\label{eq:3.6}
&\frac{dx}{dt}=2u & \text{substituting we have,} \nonumber\\ 
\implies&\frac{dx}{dt}=
\begin{cases}
  0 & x\leq 0\\
  2c & 0<x\leq 1\\
  2 & x>1.
\end{cases}        
&\text{integrating we have,}\nonumber\\
\implies&x=
\begin{cases}
  B & x\leq 0\\
  2ct+B & 0<x\leq 1\\
  2t+B & x>1.
\end{cases}        
&\nonumber\\
\end{align}
where $B$ is a arbitrary constant. Using the initial conditions , and that the characteristic curve $x(t)$ passes through $(c,0)$ we have,
\begin{equation}
x=
\begin{cases}
  c & x\leq 0\\
  2ct+c & 0<x\leq 1\\
  2t+c & x>1.
\end{cases}        
\end{equation}
Therefore $u$ becomes,
\begin{equation}
u(x,t)=
\begin{cases}
  0 & x\leq 0\\
  \frac{x}{2t+1} & 0<x\leq 1\\
  1 & x>1.
\end{cases}   
\end{equation}
I think I am missing something. The solution should have $t$ dependence in the intervals. Thanks.
 A: This PDE is very similar to Burgers equation, and the solution $u(x,t)$ deduced from the method of characteristics reads $u = f(x-2u t)$ in implicit form, where $f = u(\cdot, t=0)$. Following the steps in the linked post (see also the comments section), we find
$$
u(x,t) = \left\lbrace
\begin{aligned}
&0 & & x\leq 0\\
&\tfrac{x}{1+2t} & & 0< x\leq 1+2t\\
&1 & & x> 1+2t
\end{aligned}\right.
$$
A: The main part that you did seems correct. Except the limits at the end.
$$\begin{equation}
\frac{\partial u}{\partial x} +2u\frac{\partial u}{\partial t} = 0
\end{equation}$$
Your Charpit-lagrange characteristic ODEs are correct :
$$\frac{dx}{2u}=\frac{dt}{1}=\frac{du}{0}$$
A  first characteristic equation comes from $du=0$ :
$$u=c_1$$
A second characteristic equation comes from $\frac{dx}{2c_1}=\frac{dt}{1}$ :
$$x-2c_1t=c_2$$
The general solution of the PDE expressed on implicite form $c_1=F(c_2)$ is :
$$u=F(x-2ut)$$
where $F$ is an arbitrary function to be determined according to the initial condition.
\begin{equation}
u(x,0)=F(x)=
\begin{cases}
  0 & x\leq 0\\
  x & 0<x\leq 1\\
  1 & x>1
\end{cases}           
\end{equation}
So, the function $F$ is determined any variable $\chi$ :
\begin{equation}
F(\chi)=
\begin{cases}
  0 & \chi\leq 0\\
  \chi & 0<\chi\leq 1\\
  1 & \chi>1
\end{cases}           
\end{equation}
We put this function $F(\chi)$ into the above general solution where $\chi=x-2ut$
\begin{equation}
u=F(x-2ut)=
\begin{cases}
  0 & x-2ut\leq 0\\
  x-2ut & 0<x-2ut\leq 1\\
  1 & x-2ut>1
\end{cases}           
\end{equation}
Case $u=0$ and $x-2ut\leq 0\quad\to\quad x\leq 0$ .
Case $u=x-2ut$ and $0<x-2ut\leq 1\quad\to\quad u=\frac{x}{1+2t}$ and $0<x\leq 1+2t$
Case $u=1$ and $x-2ut>1 \quad\to\quad x>1+2t$
The solution is :
\begin{equation}
u(x,t)=
\begin{cases}
  0 & x\leq 0\\
  \frac{x}{1+2t} & 0<x\leq 1+2t\\
  1 & x>1+2t
\end{cases}           
\end{equation}
