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{\bf Question}

A perfect fluid has an equation of state given by $p =
\rho^\gamma$, where $p$ is the pressure, $\rho$ is the density and
$\gamma$ is a positive constant.  Show that in one dimension its
density and speed $v(x,t)$ satisfy: $$\frac{\partial
\rho}{\partial t} + v\frac{\partial \rho}{\partial x} = -\rho
\frac{\partial v }{\partial x}$$ $$\frac{\partial v}{\partial t} +
v \frac{\partial v}{\partial x} = -\gamma \rho^{\gamma -
2}\frac{\partial p}{\partial x}.$$

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{\bf Answer}

Using conservation of mass gives: $\ds\frac{\partial
\rho}{\partial t} + \frac{\partial }{\partial x}(v \rho)  = 0,$

which gives $\ds\frac{\partial \rho}{\partial t} + v
\frac{\partial \rho}{\partial x} + \rho \frac{\partial v}{\partial
x}  =  0 $

Using conservation of momentum gives: $\ds \frac{\partial
v}{\partial t} + v \frac{\partial v}{\partial x}  =
-\frac{1}{\rho} \frac{\partial p}{\partial x}.$

Now $\ds p = \rho^\gamma,$ so $\ds\frac{\partial v}{\partial t} +
v \frac{\partial v}{\partial x} =  -\frac{1}{\rho} \frac{\partial
(\rho^\gamma) }{\partial x}  =  -\gamma \rho^{\gamma - 2}
\frac{\partial \rho}{\partial x}.$





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