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

In summary, the method of characteristics was used to obtain the Charpit-Lagrange system of ODEs which were then solved to get the characteristics as x = \frac14t^2 - \frac12t + \xi for \xi < 0 and x = \frac14t^2 + \frac12t + \xi for \xi > 0. This leads to three regions in the half-plane, each with a different expression for u(x,t).
  • #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.
 
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  • #2
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.
 
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  • #3
Thank you so much!
 

1. What is the purpose of characteristic curves in solving this equation?

The characteristic curves help to visualize the behavior of the solution to the given equation. They are curves that represent the paths along which the solution propagates.

2. How do you determine the direction of the characteristic curves?

The direction of the characteristic curves is determined by the coefficient of the u term in the equation. In this case, since the coefficient is (1-2u), the curves will slope downwards in the direction of decreasing u.

3. Can you explain the significance of the initial condition, u(x,0) = f(x), in this equation?

The initial condition represents the starting point of the solution at time t=0. It is necessary to have this condition in order to uniquely determine the solution to the equation.

4. How does the value of f(x) affect the characteristic curves?

The value of f(x) determines the shape and position of the characteristic curves. It represents the initial state of the solution at time t=0 and therefore influences the behavior of the solution along the characteristic curves.

5. Are there any limitations to using characteristic curves to solve this equation?

While characteristic curves are a useful tool for visualizing the solution to the given equation, they may not always provide a complete solution. In some cases, additional techniques may be needed to find a complete solution.

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