How to convert Euler Equations to Lagrangian Form?

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The discussion focuses on converting the one-dimensional Euler equations of conservation of mass and momentum into Lagrangian form using the mass coordinate h. The conversion process involves recognizing the relationship between Lagrangian and Eulerian time derivatives, which is crucial for the transformation. Key equations are derived, showing how the Euler equations can be expressed in Lagrangian terms, leading to a simplified form. The transformation highlights the importance of the mass coordinate and its derivatives in the equations. Ultimately, the discussion provides a clear pathway for achieving the conversion from Euler to Lagrangian forms in fluid dynamics.
Mr. Cosmos
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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.
 
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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 \\<br /> \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} \\<br /> \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 \\<br /> \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 \\<br /> \frac{\partial u}{\partial t} + \frac{\partial p}{\partial h} = 0 where h is the mass coordinate.
 

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