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[SOLVED] shock wave

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Feb 1, 2012
1,673
$\frac{dt}{ds} =1\Rightarrow t = s + c$ 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
Feb 1, 2012
1,673
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
Feb 1, 2012
1,673
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
Feb 1, 2012
1,673
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?


What is this asking?


Determine the location in space-time where the shocks intersect.
Again use the concept of the jump condition to determine the shock speed of the single shock front that results from the merging of the two initial shocks.
 
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