Consider the transport equation
$$u_y+u^2u_x = 0 \tag{1}$$
with initial condition
$$\begin{cases}1 &, x\leq 0\\ 0 &,x\geq 1\\\sqrt{1-x} &, 0<x<1\end{cases} \tag{2}$$
Find the solution using the method of characteristics. Up to which time is the solution defined in a classical sense?
So we get
$$\begin{align}x_t &= \hat{u}^2 \\ y_t &= 1 \\ \hat{u}_t &= 0\end{align} \tag{3}$$
with
$$\begin{align}x_0(s) &= s \\ y_0(s) &= 0 \\ \hat{u}_0(s) &= u(s,0)\end{align} \tag{4}$$
so we get
$$\begin{align}x(t,s)&=\hat{u}^2\cdot t + s \\ y(t,s)& = t \\ \hat{u}(t,s) &= u(s,0)\end{align} \tag{5}$$
We now invert the map $(s,t) \mapsto (x,y)$:
$\underline{s\leq0}:$
$x=y+s \ \Rightarrow s=x-y \ \ \text{with} \ \ x\leq y \tag{6}$
$\underline{s\geq 1:}$
$ x=s \ \ \text{with} \ \ x\geq 1 \tag{7}$
$\underline{0<s<1:}$
$x=(1-s)\cdot y + s = y-sy+s = y + s(1-y) \quad \Rightarrow \quad s=\frac{x-y}{1-y} \tag{8}$
with
$y<x<1 \ \ \text{if} \ \ y<1 \ \ \text{and} \ \ y>x>1 \ \ \text{if} \ \ y>1 \tag{9}$
The characteristic are intersecting at the point $(x,y)=(1,1)$. Thus, the solution is defined up to $y=1$. We get:
$$u(x,y)=\begin{cases} 1 &, x\leq y \\ 0 &, x\geq 1 \\ \sqrt{\frac{1-x}{1-y}} &, y<x<1 \tag{10}\end{cases} \quad \text{for} \quad y\leq 1$$
Question: How exactly can I see that the lines intersect at $(x,y)=(1,1)$?
Best Answer
You should have written $\hat u(t,s) = u(0,s)$. With this, the characteristics $\binom{x(t,s)}{y(t,s)}$ are $$ \binom{x(t,s)}{y(t,s)} = \begin{cases} \binom{s + t}t & s\le 0 \\ \binom{(1-s)t+s}t & s\in(0,1) \\ \binom st & s\ge 1\end{cases} $$
Now you can study this graph:
With the graph as a guide, It's clear that two lines of the first type $(s,s'\le 0)$ don't intersect; similarly for $s,s'\ge 1$; Two lines for $s,s'\in (0,1)$ do intersect, and they do so at $t=1$:
$$\binom{(1-s)t+s}t = \binom{(1-s')t+s'}t \iff (1-s)t+s = (1-s')t+s' \iff s(t-1) = s'(t-1)$$ so if we have two different lines ($s\neq s'$) then there is a solution at $t=1$. I'll leave the other cases to you, which should be easy to check.