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

[BloonAinte]

TL;DR Summary

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 [itex](x,t)[/itex] plane, or more properly the half-plane [itex]t > 0[/itex]. If you're having difficulty expressing them in the form [itex]t(x)[/itex], then use [itex]x(t)[/itex], which you already have.

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

 

[BloonAinte]

Thank you so much!