Characteristic curves for ##u_t + (1-2u)u_x = -1/4, u(x,0) = f(x)##

[BloonAinte]

TL;DR

Characteristic curves for $u_t + (1-2u)u_x = -1/4, u(x,0) = f(x)$ where $f(x) = \begin{cases} \frac{1}{4} & x > 0 \\ \frac{3}{4} & x < 0 \end{cases}$

I woud like to find the characteristic curves for $u_t + (1-2u)u_x = -1/4, u(x,0) = f(x)$ where $f(x) = \begin{cases} \frac{1}{4} & x > 0 \\ \frac{3}{4} & x < 0 \end{cases}$.

By using the method of chacteristics, I obtain the Charpit-Lagrange system of ODEs: $dt/ds = 1$, $dx/ds = 1 - 2u$, $du/ds = -1/4$. I then solve to get $t = s$, $u = - 1/4t + \xi$, and $x = t + t^2/4 - 2tf(\xi) + \xi$. I then rearrange and use the quadratic formula to get $$t = -2a \pm \sqrt{4a^2 - 4(\xi - x)}$$ where $a = 1-2f(\xi) = \begin{cases} \frac{1}{2} & \xi > 0 \\ -\frac{1}{2} & \xi < 0 \end{cases}$. I think that this is correct so far. However, I am unsure on how to select the square roots appropriately. I would be grateful for any help! Thank you for your time.

 

[pasmith]

I'm not sure why you need to do that. The characteristics are curves in the $(x,t)$ plane, or more properly the half-plane $t > 0$. If you're having difficulty expressing them in the form $t(x)$, then use $x(t)$, which you already have.

You can see that the characteristics are $x = \frac14t^2 - \frac12t + \xi$ for $\xi < 0$ and $x = \frac14t^2 + \frac12t + \xi$ for $\xi > 0$. This gives you three regions of the half-plane:
(1) For $x < \frac14t^2 - \frac12t$, $\xi < 0$ so $u(x,t) = (3 - t)/4$.
(2) For $x > \frac14t^2 + \frac12t$, $\xi > 0$ so $u(x,t) = (1 - t)/4$.
(3) For $\frac14t^2 - \frac12t < x < \frac14t^2 + \frac12t$ we are between two characteristics which both have $\xi = 0$; in this region we have an expansion fan.

 

[BloonAinte]

Thank you so much!