How to convert Euler Equations to Lagrangian Form?

[Mr. Cosmos]

I am not entirely sure how to convert the conservation of mass and momentum equations into the Lagrangian form using the mass coordinate h. The one dimensional Euler equations given by,
$$\frac{\partial \rho}{\partial t} + u\frac{\partial \rho}{\partial x} + \rho\frac{\partial u}{\partial x} = 0$$
$$\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + \frac{1}{\rho}\frac{\partial p}{\partial x} = 0$$
need to be converted to,
$$\frac{\partial v}{\partial t} - \frac{\partial u}{\partial h} = 0$$
$$\frac{\partial u}{\partial t} + \frac{\partial p}{\partial h} = 0$$
where the Lagrangian mass coordinate has the relation,
$$\frac{\partial h}{\partial x} = \rho$$
and
$$v = \frac{1}{\rho}$$
Thanks.

 

[Mr. Cosmos]

So I played around with the equations and with the aid of my fluid mechanics book I figured it out. One must realize that the Lagrangian time derivative is related to the Eulerian time derivative by,
$$\left(\frac{\partial f}{\partial t}\right)_L = \left(\frac{\partial f}{\partial t}\right)_E + \vec{V} \cdot \left(\nabla f\right)$$ where f is a flow property like density, pressure, velocity, etc.
Therefore, in one dimension the Euler equations immediately reduce to the Lagrangian equivalent of,
$$\frac{\partial \rho}{\partial t} + \frac{1}{v} \frac{\partial u}{\partial x} = 0 \\ \frac{\partial u}{\partial t} + v \frac{\partial p}{\partial x} = 0$$ Now realizing that,
$$\frac{\partial \rho}{\partial t} = -\frac{1}{v^2} \frac{\partial v}{\partial t}$$ and from the partial derivative chain rule,
$$\frac{\partial u}{\partial x} = \frac{\partial h}{\partial x} \frac{\partial u}{\partial h} = \frac{1}{v} \frac{\partial u}{\partial h} \\ \frac{\partial p}{\partial x} = \frac{\partial h}{\partial x} \frac{\partial p}{\partial h} = \frac{1}{v} \frac{\partial u}{\partial h}$$ Substituting into the the Lagrangian form yields,
$$-\frac{1}{v^2} \frac{\partial v}{\partial t} + \frac{1}{v^2} \frac{\partial u}{\partial h} = 0 \\ \frac{\partial u}{\partial t} + v \frac{1}{v} \frac{\partial p}{\partial h} = 0$$ or equivalently,
$$\frac{\partial v}{\partial t} - \frac{\partial u}{\partial h} = 0 \\ \frac{\partial u}{\partial t} + \frac{\partial p}{\partial h} = 0$$ where h is the mass coordinate.