Shock-wave solution for PDE $u_t+(u-1)u_x=2$

Question

I want to solve the following PDE initial value problem

$u_t+(u-1)u_x=2$

and

$u (x,0)=\begin{cases} 1 & \text{for } x <0,\\ 1-x & \text{for } 0<x <1\\ 0 & \text{for } 1 <x \end{cases}$

However, I find that I have intersecting characteristics between $x=t^2$ and $x=t^2-t+1$.

How would I apply the shockwave method in this case since the PDE is given? Is it possible to solve this PDE as is?

Answer 1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=2$ , letting $u(0)=u_0$ , we have $u=u_0+2s=u_0+2t$

$\dfrac{dx}{ds}=u-1=u_0+2s-1$ , letting $x(0)=f(u_0)$ , we have $x=s^2+(u_0-1)s+f(u_0)=t^2+(u-2t-1)t+f(u-2t)=(u-1)t-t^2+f(u-2t)$ , i.e. $u=2t+F(x+t^2-(u-1)t)$

$u(x,0)=\begin{cases}1&\text{when}~x<0\\1-x&\text{when}~0<x<1\\0&\text{when}~1<x\end{cases}$ :

$\therefore u=\begin{cases}1&\text{when}~x+t^2-(u-1)t<0\\2t+1-x-t^2+(u-1)t&\text{when}~0<x+t^2-(u-1)t<1\\0&\text{when}~1<x+t^2-(u-1)t\end{cases}$

Hence $u=\begin{cases}1&\text{when}~x<-t^2\\\dfrac{x+t^2-t-1}{t-1}&\text{when}~0<x+t^2-\dfrac{(x+t^2-t-1)t}{t-1}+t<1\\0&\text{when}~x>1-t^2-t\end{cases}$

Answer 2

@user393349 : Sorry, something has gone wrong during re-typing my answer . I first closed it and later I re-open it to continue the typing. The main part was done when abruptly all disappeared. Probably it was closed by someone else.

All my latex is lost. It took me a so long time that I will not write it again.

The only think remaining is a graph where the results are summarized. This shows where and when a blow-up point appears. I don't know if this can help you (I hope so).

If this figure is of no interest for you, I suppose that the answer will be definitively deleted. Good continuation.