- #1

BloonAinte

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- 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.

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.

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