# SOLVEDAdmissible Solutions and Rankine-Huoniot Shock Condition

#### lvleph

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
$$\displaystyle s = \frac{[f(u)]}{} = \begin{cases} \frac{\frac{1}{3}\cdot 1^3 - \frac{1}{3}\cdot 0^3}{1 - 0} \\ \frac{\frac{1}{3}\cdot (-1)^3 - \frac{1}{3}\cdot 0^3}{-1 - 0} \end{cases} = \frac{1}{3}$$.
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: