shock waves characteristics I'm trying to solve $u_t + u^2u_x = 0$ with $u(x, 0) = 2 + x$.
I'm thinking to proceed by characteristics where we have above that $\frac{dx}{dt} = 1$ and $dy/dt = u^2$, but not sure if this will help. This is from shock waves idea. 
Here's what I have
$u_t + u^2 u_x = 0$
$q +u^2 p =0$
$u^2 p +q =0$
$\frac{dx}{u^2} = \frac{dy}{1} $
and $\frac{u^{-1}}{-1} + C = y $
then $u(x,t)^{-1} + C = y$
is this correct?
 A: The quasilinear first order PDE
$$
a\big(x,y,u(x,y)\big) u_x(x,y) + b\big(x,y,u(x,y)\big)u_y(x,y) = c\big(x,y,u(x,y)\big)
$$
where $a,\,b,\,c \in C^1$ with data $\mathcal{C}(\xi) = \big(x(\xi), y(\xi), u(\xi)\big) \in C^1$ and with
$$
\begin{vmatrix} \frac{dx}{d\xi} & a \\ \frac{dy}{d\xi} & b\end{vmatrix} \neq 0
$$
has a unique solution near $\mathcal{C}$ given by
\begin{align}
 \frac{d x}{d \eta} &= a  & x\big|_{\eta = 0}&= x(\xi)\\
 \frac{d y}{d \eta} &= b  & y\big|_{\eta = 0}&= y(\xi)\\
 \frac{d u}{d \eta} &= c  & u\big|_{\eta = 0}&= u(\xi)\\
\end{align}
For proof and geometrical interpretation, see F. John's Partial Differential Equations (§1.4)
In your case, $\mathcal{C}(\xi) = \big(\xi,0,\xi+2\big)$. Near $\eta \sim 0$ 
$$
\begin{vmatrix} \frac{dx}{d\xi} & a \\ \frac{dy}{d\xi} & b\end{vmatrix} = \begin{vmatrix} 1 & u^2 \\ 0 & 1\end{vmatrix} = 1
$$
and the solution is unique.
The system of ODE's is
\begin{align}
 \frac{d x}{d \eta} &= u^2  & x\big|_{\eta = 0}&= \xi\\
 \frac{d t}{d \eta} &= 1    & t\big|_{\eta = 0}&= 0\\
 \frac{d u}{d \eta} &= 0    & u\big|_{\eta = 0}&= \xi + 2\\
\end{align}
with solution
$$
t = \eta, \quad u = \xi + 2, \quad x = (\xi + 2)^2 \eta + \xi.
$$
The characteristics are $t = \frac{x - \xi}{(\xi + 2)^2}$ hence $\xi = -2$ is a special point. As $\xi \rightarrow \infty$, $t \rightarrow 0$. As $\xi \rightarrow -\infty$, $t \rightarrow 0$. As $\xi \rightarrow -2$, $t \rightarrow \infty$.

This of course, means that there is no solution when the characteristics meet.
A simple explanation for this is that the transformation
$$
(x,t) \rightarrow (\xi,\eta)
$$
is invertible iff
$$
\begin{vmatrix} \partial_\xi x & \partial_\eta x \\ \partial_\xi t & \partial_\eta t \end{vmatrix} = 1 + 4\eta + 2\xi \eta \neq 0
$$
meaning there is no solution when $\xi = -\frac{1 + 4 \eta}{2\eta}$ or, inverting the transformation, when
$$
t = - \frac{1}{4(x+2)}
$$
$\hskip.75in$
Lastly, inverting for $\xi$
$$
\xi = \frac{-(1 + 4t) \pm \sqrt{1 + 4t(2 + x)}}{2 t}
$$
and
$$
u(x,t) = \frac{-1 \pm \sqrt{1 + 4t(2 + x)}}{2 t}.
$$
In order to determine the correct sign, we must look at the initial condition. For the minus sign $\lim_{t \rightarrow 0} u(x,t) = -\infty$, while the plus sign gives the correct answer. Hence
$$
u(x,t) = \frac{-1 + \sqrt{1 + 4t(2 + x)}}{2 t}.
$$
