Given the transport equation:
$$u^2u_x + u_y = 0$$
with the initial condition $u(x, 0) = 1$ for $x \leq 0$, $u(x, 0)=0$ for $x \geq 1$ and $u(x, 0) = \sqrt{1-x}$ for $0 < x < 1$.
$$\begin{align} &\frac{dx}{dt} = u^2 \Rightarrow h(s)^2*t + c\\ &\frac{dy}{dt} = 1 \Rightarrow t + c\\ & \frac{du}{dt} = 0 \Rightarrow h(s) \end{align}$$
$s \leq 0$ $$\begin{align} &x(t, s) = s + y(t, s) &\Rightarrow s = x-y\\ &y(t, s) = t &\Rightarrow t = y\\ \end{align}$$
$0 \leq s \leq 1$ $$\begin{align} &x(t, s) = s + (1-s)*t &\Rightarrow s = \frac{x-y}{1-y}\\ &y(t, s) = y &\Rightarrow t = y\\ \end{align}$$
$s \geq 1$ $$\begin{align} &x(t, s) = s &\Rightarrow s = x\\ &y(t, s) = y &\Rightarrow t = y\\ \end{align}$$
How can I find now the critical time where the characteristics intersect? I know the solution is $(x, y) = (1, 1)$ but I couldn't find a place that really shows how to derive it.
Problems involving characteristics are a lot easier to grasp once you draw a picture. IF you divide through by $u^2$, you get the equation in the standard form of an advection equation: $$ u_x+\frac{1}{u^2}u_y=0. $$ From this you can derive the characteristic lines easily since the slope of the lines $y=f(x)$ is given by the coefficient of $u_y$ in the equation. In this case, the characteristics are straight lines so you can solve things graphically.