I need some help to solve this transport equation $$ \frac{\partial u}{\partial t} + \frac{\partial}{\partial x} (u\ \text{sign}(x)) =-1$$ with initial condition $$ u_0(x) = \begin{cases} 1 &\text{if } x\in [-2,-1]\cup [1,2] \\ 0 &\text{otherwise}\end{cases} $$
The solution is computable and, for our data, all I have to do is compute this expression $$ u(t,x)= u_0(X(0;t,x))J(0;t,x) +\int_0^t -1\ ds = u_0(X(0;t,x))J(0;t,x) -t$$ where $X(\xi;t,x)$ is the characteristic curve with initial condition $X(t)=x$, that is, $$ (1)\ \ \begin{cases} \frac{dX(\xi;t,x)}{d\xi} = \text{sign}(X(\xi)) \\ X(t)=x \end{cases} $$ and $J(\xi;t,x)$ is, in this case, $J(\xi;t,x) = \det (\frac{dX(\xi;t,x)}{dx})$. I hace problems to solve (1) (of course in the weak sense), and therefore, also to calculate $J(\xi;t,x)$. Any help will be appreciated!
This won't be a very rigorous answer, but let's see intuitively what happens. Using the definition of the sign function, we have \begin{aligned} u_t + u_x &= -1 &&\text{if}\quad x>0 \\ u_t \phantom{+u_x} &= -1 &&\text{if}\quad x=0 \\ u_t - u_x &= -1 &&\text{if}\quad x<0 \end{aligned} Thus, using the method of characteristics in a piece-wise manner, we find \begin{aligned} u(x,t) &= u_0(x-t) - t &&\text{if}\quad x>0 \\ u(0,t) &= -t &&\text{if}\quad x=0 \\ u(x,t) &= u_0(x+t) -t &&\text{if}\quad x<0 \end{aligned} Finally, the provided initial data $u_0$ leads to $$ u(x,t) = \left\lbrace \begin{aligned} &1-t &&\text{if } x\in [-2-t, -1-t]\cup [1+t, 2+t] \\ &{-t} &&\text{otherwise} \end{aligned}\right. $$