# [SOLVED]shock wave

#### dwsmith

##### Well-known member
Consider the quasi-linear 1-D wave equation
$$\frac{\partial\rho}{\partial t} + 2\rho\frac{\partial\rho}{\partial x} = 0$$
with the piecewise constant initial conditions
$$\rho(x,0) = \begin{cases} \rho_1, & x < -x_0\\ \rho_2, & -x_0 < x < x_0\\ \rho_3, & x > x_0 \end{cases}$$
where $\rho_1 > \rho_2 > \rho_3$ and $\rho_i, x_0\in\mathbb{R}$ with $i = 1, 2, 3$.

Argue that two shocks form at $x = \pm x_0$ in this case and sketch the space-time diagram for the density field.

I have no idea on what to do or how to start.

#### dwsmith

##### Well-known member
$\frac{dt}{ds} =1\Rightarrow t = s + c$ but when $t(0) = 0$ $t = s$.

$\frac{dx}{ds} = 2\rho\Rightarrow x = 2\rho t + x_0$ at $t=0$, $x=x_0$. So $x_0 = x-2\rho t$.

$$\rho(x-2\rho t,0) = \begin{cases} \rho_1, & x-2\rho t < -x_0\\ \rho_2, & -x_0 < x-2\rho t < x_0\\ \rho_3, & x-2\rho t > x_0 \end{cases}$$

Is this what I should be doing? If so, what next?

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#### dwsmith

##### Well-known member
Then we obtain $t(r) = r$, $x(r) = 2\rho r + x_0$, and $\rho(r) = c$ when $t = 0$.
Let $s$ be a parameter.
Then $\rho(r;s) = f(s)$, $x(r;s) = 2tf(s) + s$, and $t(r;s) = r$.
In order to determine shock, we need to find the Jacobian.
\begin{alignat*}{3}
\mathcal{J} & = & \det\begin{pmatrix}
x_r & x_s\\
t_r & t_s
\end{pmatrix}\\
& = & \frac{\partial x}{\partial r}\frac{\partial t}{\partial s} - \frac{\partial x}{\partial s}\frac{\partial t}{\partial r}\\
& = & 0 - (2tf'(s) + 1)\\
& = & -2tf'(s) - 1
\end{alignat*}

Correct?

#### dwsmith

##### Well-known member
Since we have that $t = r$, we can make the substitution
$$x = 2t\rho + x_0.$$
Let's put the equation in the form of $y = mx + b$ or in our case $t = mx + x_0$.
So we have
$$t = \frac{x - x_0}{2\rho}.$$
A shock will occur when two characteristic lines intersect or there is a jump discontinuity.
To view that $\pm x_0$ causes shock, for simplicity, let $x_0 = 1$, $\rho_1 = 3$, $\rho_2 = 2$, and $\rho_3 = 1$.
Then the characteristic lines are
\begin{alignat*}{5}
t & = & \frac{x - 1}{6}, & \ \ \text{for} & \ \ x < -1\\
t & = & \frac{x - 1}{4}, & \ \ \text{for} & \ \ -1 < x < 1\\
t & = & \frac{x - 1}{2}, & \ \ \text{for} & \ \ x > 1
\end{alignat*}
At $-x_0 = -1$, we have $t = \frac{-1}{3}$ and $t = \frac{-1}{2}$.
Therefore, we have a jump discontinuity at $-x_0$.
For $x_0 = 1$, $t = 0$, i.e. we have two intersecting characteristic lines.
Since the choice of $x_0$ and $\rho$ were arbitrary, $\pm x_0$ will cause a shock for all choices.

#### dwsmith

##### Well-known member
How can I determine shock velocity?

We know that $q = \rho u$ and that $2\rho = \frac{dq(\rho)}{d\rho}$.
Solving this equation we get that $q = \rho^2$.
The shock velocity is $\frac{dx_s}{dt} = \frac{q(x_{s-},t) - q(x_{s+},t)}{\rho(x_{s-},t) - \rho(x_{s+},t)}$.
The shock velocity for $x = -x_0$ is
$$\frac{dx_s}{dt} = \frac{\rho_1^2 - \rho_2^2}{\rho_1 - \rho_2},$$
and when $x = x_0$, we have
$$\frac{dx_s}{dt} = \frac{\rho_2^2 - \rho_3^2}{\rho_2 - \rho_3}.$$

Correct?