Consider
$\begin{cases} u_y+uu_x=0\\ u(x,0)=h(x) \tag{1} \end{cases}$
$$h(x)=\begin{cases} 1 &, x\leq 0 \\ 1-\frac{x}{\alpha} & ,0 \leq x \leq \alpha\\ 0 &, x\geq \alpha \end{cases}$$
whereas $x$ is the spatial coordinated, and $y$ is the time.
The critical time is $y_c=\alpha$
For the classical solution i.e. $y\in [0,\alpha)$, we use the M.o.C on each of the three cases and get
$$ u(x,y)=\begin{cases} 1 &, x\leq y\\ \frac{\alpha-x}{\alpha-y} &, y\leq x \leq \alpha \\ 0 &, x\geq \alpha \end{cases}\tag{2} $$
Now the weak part comes into play. So we basically look for a shock wave using Rankine-Hugoniot.
$\gamma'(y)=\frac{f(u^+)-f(u^{-})}{u^+-u^-}=\frac{1}{2}\tag{3}$
So we integrate this and get
$\gamma(y)=\frac{1}{2}y+C \tag{4}$
By the condition that the shock starts at the point $(\alpha, \alpha)$ we get $C=\frac{\alpha}{2}$. So we can write down the weak solution: (for $y\in [\alpha,\infty)$
$$ u(x,y)=\begin{cases} 1 &, x < \gamma(y) \\ 0 &, x > \gamma(y) \end{cases}\tag{5} $$
out overall solution is:
$$ u(x,y)=\begin{cases} 1 &, x\leq y, y\in[0,\alpha)\\ \frac{\alpha-x}{\alpha-y} &, y\leq x \leq \alpha , y\in[0,\alpha)\\ 0 &, x\geq \alpha , y\in[0,\alpha)\\ 1 &, x < (y+\alpha)/2, y\in [\alpha, \infty) \\ 0 &, x>(y+\alpha)/2, y \in [\alpha, \infty) \end{cases}\tag{2} $$
I don't 100% understand how we know that the shock wave starts at $(\alpha, \alpha)$ or rather: To get $C$ in $\gamma$ I need a point I can plug into $\gamma$ - how do I usually get this point (in this case $(\alpha, \alpha)$)?