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Thread: Admissible Solutions and Rankine-Huoniot Shock Condition

  1. #1
    Senior Member
    Mar 2009

    Admissible Solutions and Rankine-Huoniot Shock Condition

    I thought these problems were pretty routine, and then I came across one that I am not able to do.

    Find the admissible solution to the equation $\displaystyle u_t + u^2u_x=0$ for $\displaystyle t> 0$ subject to the following initial conditions
    $\displaystyle u=\begin{cases} 1 & x<0\\ 0 & 0<x<1\\ -1 & x>1\end{cases}$

    My attempt: We see that $\displaystyle f(u) = \frac{1}{3}u^3$ and $\displaystyle \frac{dt}{dy} = 1\quad \frac{dx}{dy} = u^2$ so
    $\displaystyle \frac{dt}{dx} = \begin{cases}1 & x<0\\ \infty & 0<x<1\\ -1 & x>1\end{cases}$.
    Thus, initially we have two shock waves that should then combine to create a single shock wave. Using the Rankine-Hugoniot condition we have
    $\displaystyle s = \frac{[f(u)]}{[u]} = \begin{cases} \frac{\frac{1}{3}\cdot 1^3 - \frac{1}{3}\cdot 0^3}{1 - 0} \\
    \frac{\frac{1}{3}\cdot (-1)^3 - \frac{1}{3}\cdot 0^3}{-1 - 0} \end{cases} = \frac{1}{3}$.
    This doesn't make sense to me, because I expected the second shock to be in the opposite direction. I must be doing something wrong.

    EDIT: $\displaystyle f'(u^{\ell}) < f'(u^r)$ so there is actually a rarefaction wave at $\displaystyle x=1$. I am just having a hard time picturing this in my head now.
    EDIT2: LOL, now I feel stupid!
    $\displaystyle \frac{dt}{dx} = \begin{cases}1 & x<0\\ \infty & 0<x<1\\ 1 & x>1\end{cases}$.
    Last edited by lvleph; Jul 16th 2010 at 08:12 AM.
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