Q:

Method of characteristics with discontinuous source term

Let's say I have the system of equations $$ \frac{\mathrm{d}\rho}{\mathrm{d}t} = \left\{\begin{matrix}
\beta, \hspace{3mm}  x < 0 \\ 
0, \hspace{3mm} x \geq 0 
\end{matrix}\right.,
\\ \frac{\mathrm{d}x}{\mathrm{d}t} = 1-2\rho
$$
With the initial condition $ \rho(x,0) = \left\{\begin{matrix}
1/2, \hspace{3mm}  x < 1 \\ 
1, \hspace{3mm} x \geq 1 
\end{matrix}\right. $
edit: $\beta$ is a positive constant.
How would I go about solving this? A general outline is fine I would not like a detailed working. For context, this is in an attempt to solve the PDE $$\frac{\partial\rho}{\partial t} + (1-2\rho)\frac{\partial\rho}{\partial x} = \left\{\begin{matrix}
\beta, \hspace{3mm}  x < 0 \\ 
0, \hspace{3mm} x \geq 0 
\end{matrix}\right.,$$
using the method of characteristics. 
EDIT: 
As @EditPiAf points out, the characteristics for $x < 0$ are given by $x = x_0 -\beta t^2$, along which $\rho = 1/2 + \beta t$, here the first issue arises. which is what happens after $t = 1/2\beta$? The shock emanating from $x = 1$ will reach the $x < 0$ region at $t =2$, resulting in another shock, defined by $x_s$. But what actually is the form of the characteristics in this region (neighbourhood of $x=0^-, t=2$)? I would think that they are  given by $x = -\beta(t-t_0)^2$ but I can't rigorously justify this. 

A:

This is the traffic flow equation with a car source of strength $β>0$ over $\Bbb R−$, which is somewhat similar to the on-ramp problem. Let's apply the method of characteristics.

characteristic curves $t\mapsto (x(t),t)$ starting at $x_0<0$ satisfy $\dot x=1-2\rho$ with $\dot \rho=\beta$, i.e. the curves are of the form $x = x_0 - \beta t^2$ along which $\rho = \frac12 +\beta t$.
characteristics starting at $0\le x_0<1$ are vertical lines $x = x_0$ along which $\rho = \frac12$.
characteristics starting at $1\le x_0$ are the lines $x = x_0 - t$ along which $\rho = 1$.

Characteristics intersect around $x=1$. A left-going shock wave with Rankine-Hugoniot speed $s = -1/2$ is produced. Characteristics separate around $x=0$. Similarly to the highway entrance problem, the discontinuity of the source term creates an artificial boundary between two road segments with different behavior, that might be expressed by the condition $\rho|_{x=0}=1/2$. 

characteristics starting at $0<t_0$ from the line $x=0$ satisfy $\rho = \frac12 + \beta (t-t_0)$ and $x = -\beta(t-t_0)^2$.

At some point ($t=2$), the shock wave will cross the vertical line $x=0$. Before this happens ($t<2$), the solution reads
$$
\rho(x,t) = \left\lbrace
\begin{aligned}
&\tfrac12 + \beta t & &\text{if} \quad x<-\beta t^2\\
&\tfrac12 + \sqrt{-\beta x} & &\text{if} \quad {-\beta t^2}\le x< 0\\
&\tfrac12 & &\text{if} \quad 0\le x< 1-\tfrac12 t\\
&1 & & \text{if}\quad 1-\tfrac12 t < x 
\end{aligned}
\right.
$$
From $t=2$, the Rankine-Hugoniot shock speed $\dot x_s$ is given by the differential equation
$$
\dot x_s(t) = \frac{\frac14 + \beta x_s(t) - 0}{\frac12 + \sqrt{-\beta x_s(t)} - 1},
\qquad
x_s(2) = 0 \, ,
$$
which solution
$$
x_s(t) = -\frac1{4\beta}\left(1 + W\big({-e}^{-1-\beta(t-2)}\big)\right)^2
$$
is expressed using the product-log function (Lambert's $W$-function).
The solution for times $t>2$ reads
$$
\rho(x,t) = \left\lbrace
\begin{aligned}
&\tfrac12 + \beta t & &\text{if} \quad x<-\beta t^2\\
&\tfrac12 + \sqrt{-\beta x} & &\text{if} \quad {-\beta t^2}\le x< x_s(t)\\
&1 & & \text{if}\quad x_s(t) < x
\end{aligned}
\right.
$$

