Shock solution using conservation law

[lhy56839]

Homework Statement

I need to find the shock solution for the initial value problem

$$u_t-u^2 u_x =0$$

with

$$u(x,0)=g(x)=\begin{cases}-\frac{1}{2}\quad x\leq 0 \\ 1\quad 0<x<1 \\ \frac{1}{2} \quad x\geq 0\end{cases}$$

Homework Equations

The Attempt at a Solution

Using the conservation law of the form $$u_t + \Phi_x = f(x,t)$$

we have

[tex] \Phi^{'}(u)=\frac{[\Phi(u)]}{} [/tex]

The flux $$\Phi(u)$$ for this problem is given by
$$\Phi(u)=-\frac{1}{3}u^3$$
since
$$\Phi_x=\Phi^{'}(u)u_x$$

The discontinuity occurs at x=0 and x=1. For the discontinuity at x=0 we have

[tex]-[u(0)]^2 = \frac{[\Phi(u)]}{} = \frac{\frac{1}{3}((-\frac{1}{2})^3-1^3)}{-\frac{1}{2}-1}=\frac{3}{4}[/tex]

and for x=1,

[tex]-[u(1)]^2 = \frac{[\Phi(u)]}{} = \frac{\frac{1}{3}(1^3-(-\frac{1}{2})^3)}{1-\frac{1}{2}}=\frac{3}{4}[/tex]I am not sure whether I am on the right track or not, and not sure how I obtain the shock solution from this. Any advice or help on this would be appreciated.

 

[mighty2000]

Thank you for your post. It seems like you are on the right track in solving this initial value problem. To obtain the shock solution, we need to consider the Rankine-Hugoniot condition, which relates the jump in the solution across the shock to the jump in the flux. This condition is given by:

_{shock} = \frac{[\Phi(u)]_{shock}}{_{shock}}

In this case, we have two shocks at x=0 and x=1, so we need to consider the Rankine-Hugoniot conditions at both of these points. This will give us two equations, which we can then solve for the shock solutions _{shock}.

For the shock at x=0, we have:

_{shock} = \frac{[\Phi(u)]_{shock}}{_{shock}} = \frac{\frac{1}{3}((-\frac{1}{2})^3-1^3)}{-\frac{1}{2}-1} = \frac{3}{4}

For the shock at x=1, we have:

_{shock} = \frac{[\Phi(u)]_{shock}}{_{shock}} = \frac{\frac{1}{3}(1^3-(-\frac{1}{2})^3)}{1-\frac{1}{2}} = \frac{3}{4}

Solving these two equations, we get the shock solutions _{shock} = \frac{3}{4} for both shocks.

I hope this helps. Let me know if you have any further questions or need clarification.