How can I solve the problem
\begin{cases} d_t+(1-2d)d_x=0 & \text{if $x \in \mathbb R$ , $t>0$} \\ d(x,0)=d_0(x) & \text{if $x \in \mathbb R$} \end{cases} with the boundary condition
$d_o(x)=\begin{cases} 0,&\text{if $x \leq0$}\\ \frac{x}{2}, & \text{if $0<x\leq 2$}\\1, &\text{if $2<x$} \end{cases}$
I know that I have to make use of the characteristic method, but can someone please explain the method to me, I didn't really get it yet...
Appreciated.
The method of characteristics is explained in many books. It should be too long and it is not usual to give a course on the forum. Moreover, they are several variants to use the method of characteristics and you give no information on what you have learn. An example of approach is presented below.
In order to avoid confusion between your function $d(x,t)$ and the differential symbol d, the notation $d(x,t)$ is replaced by $u(x,t)$. So, the question is rewritten as :
\begin{cases} u_t+(1-2u)u_x=0 & \text{if $x \in \mathbb R$ , $t>0$} \\ u(x,0)=u_0(x) & \text{if $x \in \mathbb R$} \end{cases} with the boundary condition $u_o(x)=\begin{cases} 0,&\text{if $x \leq0$}\\ \frac{x}{2}, & \text{if $0<x\leq 2$}\\ 1, &\text{if $2<x$} \end{cases}$
The set of differential equations for the characteristic functions is : $$\frac{dt}{1}=\frac{dx}{1-2u}=\frac{du}{0}$$
Necessarily the equation of a first characteristic curve comes from $du=0\quad\to\quad u=c_1$
A second equation of characteristic curve comes from $\frac{dt}{1}=\frac{dx}{1-2c_1}\quad\to\quad x-(1-2c_1)t=c_2$
The general solution of the PDE, expressed on the form of implicit equation $\Phi(c_1\:,\:c_2)=0$ is : $$\Phi\left(u\:,\: x-(1-2u)t\right)=0$$ where $\Phi(X,Y)$ is any differentiable function of two variables.
Equivalent forms expressing any relationship between $X$ and $Y$ are $X=F(Y)$ and $Y=G(X)$ where $F$ and $G$ are any differentiable functions (inverse one to the other). So, equivalent forms of the general solution are : $$x-(1-2u)t=F(u)$$ $$u=G\left(x-(1-2u)t\right)$$ The function $F$ or $G$ has to be determined according to the boundary conditions. Then, the last equation can be solved for $u(x,t)$.
$t=0\quad\to\quad u_0=G(x)\quad\to\quad G(x)=\begin{cases} 0,&\text{if $x \leq0$}\\ \frac{x}{2}, & \text{if $0<x\leq 2$}\\ 1, &\text{if $2<x$} \end{cases}$ $t\neq 0\quad\to\quad u=G\left(x-(1-2u)t\right)=\begin{cases} 0,&\text{if $\left(x-(1-2u)t\right) \leq0$}\\ \frac{x-(1-2u)t}{2}, & \text{if $0<\left(x-(1-2u)t\right)\leq 2$}\\ 1, &\text{if $2<\left(x-(1-2u)t\right)$} \end{cases}$
Particular solution of the PDE fitting with the boundary conditions : $$\begin{cases} u=0 &\text{if $(x-t) \leq0$}\\ u=\frac{x-t}{2(1-t)} & \text{if $0<\frac{x-t}{1-t} \leq 2$}\\ u=1 &\text{if $2<(x+t)$} \end{cases}$$