Quasi-linear PDE with Cauchy conditions: Am attempting to solve the partial differential equation given by
$$u_t + uu_x = 0$$
for some wave function $u(x, t)$, subject to the conditions
$$u(x, 0) = \phi(x) = 1-x^2$$
I started by using the method of characteristics, and hence obtaining the system
$$dt = \frac{dx}{u} = \frac{du}{0}$$
And I can hence solve a system of ordinary differential equations given by
$$\frac{du}{dt} = 0$$
and $$\frac{dx}{dt} = u$$
If the first ode tells us that $u$ is constant in $t$, then we can integrate the second ode such that
$$x = ut + c$$
which, considering a characteristic curve emerging from the $x$-axis at $(\xi, 0)$ we can then get
$$x = ut + \xi$$
So now I figure I need to use those Cauchy conditions to solve the equation. If it's true that 
$$u(x, t) = \phi(\xi)$$ then I get that
$$x = (1-\xi^2)t + \xi$$
I am now slightly unclear on the best method to proceed. Factoring the final equation I have written for $\xi$? If I do that, however, I will be still left with a $\xi$ in the remaining expression, by virtue of the presence of the $\xi^2$ term.
Any thoughts are much appreciated.
 A: $$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0$$
$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial t}dt$
Change of variable : $x=f(u,t)$
$dx=\frac{\partial f}{\partial u}du+\frac{\partial f}{\partial t}dt=\frac{\partial f}{\partial u}\left(\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial t}dt \right)+\frac{\partial f}{\partial t}dt$
$\begin{cases}
     1=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}\\
     0=\frac{\partial f}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial f}{\partial t} 
   \end{cases} \rightarrow 
\begin{cases}
     \frac{\partial u}{\partial x}=\frac{1}{\frac{\partial f}{\partial u}}\\
     \frac{\partial u}{\partial t}=-\frac{\frac{\partial f}{\partial t}}{\frac{\partial f}{\partial u}} 
   \end{cases} \rightarrow 
\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=-\frac{\frac{\partial f}{\partial t}}{\frac{\partial f}{\partial u}}+u \frac{1}{\frac{\partial f}{\partial u}}=0 
$
$\frac{\partial f}{\partial t}=u \rightarrow f=ut+F(u) $ any derivable function $F$
$f(u,t)=x=ut+F(u)\rightarrow F(u)=x-ut $
or, with any derivable function $G$ :
 $$u=G(x-ut)$$
Condition $u(x,0)=\phi(x) \rightarrow  G(x)=\phi(x)=1-x^2$
$G(x-ut)=1-(x-ut)^2$
$$u=1-(x-ut)^2$$
$t^2u^2+(1-2tx)u-1+x^2=0$
$$u=\frac{1}{2t^2}\left(2tx-1\pm\sqrt{1-4tx+4t^2} \right)$$
This is consistent with the method of characteristics :
with your notations : $$u(x,t)=\phi(\xi)=\phi(x-ut)=1-(x-ut)^2$$
