Initial- and boundary-value problem for non-linear conservation law I'm trying to understand the actual solution which is computed here (Sec. 4.7 of Ref. (1)),
trying to resolve 
$$
    \begin{cases}
    \displaystyle \frac{\partial u}{\partial \tau} + \frac{\partial q}{\partial x} = 0, \\
    u(0,x) = \frac{c_0}{c_{max}} = \begin{cases}
    0, \quad \text{if} \quad x \le 0\\
    u_0, \quad \text{if} \quad  x \in ]0,1[ \\
    1, \quad \text{if} \quad x \ge 1
    \end{cases}\\
u(\tau, 0) = 0\\
u(\tau ,h) = 1, \forall\tau > 0
\end{cases}
$$
where $q(u) = u(1-u)(1-\beta u)$ with $\frac{1}{2} < \beta < 1$.
First, he's trying to connect $0$ to $u_0$ with a shock wave: 
$$x(\tau) = \sigma(0,u_0)(\tau) = \dfrac{q(u_0)}{u_0}\tau$$
so the solution becomes $ u(t,x)  = 0$ if $x <\sigma(0,u_0)\tau$, $u(t,x)  = u_0$ if $1 > x >\sigma(0,u_0)\tau$, and $u(t,x)  = 1$ if $x \ge 1$.
Then he's trying to connect $u_0$ to $1$ by introducing an intermediate state which is $u_1$. How does it change the exact solution? He's only showing the final stationary state at the end. 
the characteristics:
\begin{center}
\begin{tikzpicture}
\draw[-stealth] (-4,0)--(4,0) node[below]{$x$}; 
\draw[-stealth] (-1,0)--(-1,4) node[left]{$t$}; 
\node[below] at (0,0) {$1$};
\node[below] at (1,0) {$2$};
\clip (-4,0) rectangle (4,4);
\foreach \xi in {-8,-7.75,...,0} \draw[green] (\xi,0) -- (\xi+4,4);
\foreach \xi in {0,0.25,...,1} \draw[violet] (\xi,0) -- (\xi+4*1.575,4);
\foreach \xi in {1,1.15,...,5} \draw[blue] (\xi,0) -- (\xi-4*0.1,4);
\end{tikzpicture}
\end{center}


(1) S Salsa: Partial Differential Equations in Action: From Modelling to Theory, 3rd Ed. Springer, 2016. doi:10.1007/978-3-319-31238-5
 A: As described in the book, there is a rarefaction starting at $x=0$, and a semi-shock starting at $x=1$. The state $u_1$ introduced in the book is where the semi-shock wave transitions from a discontinuity to a smooth rarefaction wave. See this post for a presentation of these waves.
A: This is not a full answer. This is only the general solution of the PDE (Without boundary and initial condition). Hopping that it helps.
$$\frac{\partial u}{\partial \tau} + \frac{\partial q}{\partial x} = 0
\quad\text{with}\quad q(u) = u(1-u)(1-\beta u)$$
I suppose that $\beta=$constant.
$\frac{\partial q}{\partial x}=\big(3\beta u^2-2(\beta+1)u+1\big)\frac{\partial u}{\partial x}$
The PDE explicitly written is :
$$\boxed{\frac{\partial u}{\partial \tau} + \big(3\beta u^2-2(\beta+1)u+1\big)\frac{\partial u}{\partial x} = 0}\tag 1$$
The Charpit-Lagrange system of characteristic ODEs is :
$$\frac{d\tau}{1}=\frac{dx}{3\beta u^2-2(\beta+1)u+1}=\frac{du}{0}$$
Finite $\frac{du}{0}$ implies $du=0$. Thus a first characteristic equation is 
$$u=c_1$$
A second characteristic equation comes from solving $\frac{d\tau}{1}=\frac{dx}{3\beta c_1^2-2(\beta+1)c_1+1}$
$$\big(\beta c_1^2-2(\beta+1)c_1+1\big)\tau-x=c_2$$
The general solution of the PDE expressed on the form of implicit equation $c_2=\Phi(c_1)$ or $c_1=\Psi(c_2)$ is :
$$\boxed{u=\Psi\Big(\big(\beta u^2-2(\beta+1)u+1\big)\tau-x\Big)}\tag 2$$
with arbitrary function $\Psi$ (or $\Phi$). The function $\Psi$ (or $\Phi$) has to be determined so that the specified boundary and initial conditions be satisfied.
