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How to convert Euler Equations to Lagrangian Form?

  1. Sep 7, 2015 #1
    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,
    [tex] \frac{\partial \rho}{\partial t} + u\frac{\partial \rho}{\partial x} + \rho\frac{\partial u}{\partial x} = 0 [/tex]
    [tex] \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + \frac{1}{\rho}\frac{\partial p}{\partial x} = 0 [/tex]
    need to be converted to,
    [tex] \frac{\partial v}{\partial t} - \frac{\partial u}{\partial h} = 0 [/tex]
    [tex] \frac{\partial u}{\partial t} + \frac{\partial p}{\partial h} = 0 [/tex]
    where the Lagrangian mass coordinate has the relation,
    [tex] \frac{\partial h}{\partial x} = \rho [/tex]
    and
    [tex] v = \frac{1}{\rho} [/tex]
    Thanks.
     
    Last edited: Sep 7, 2015
  2. jcsd
  3. Sep 8, 2015 #2
    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,
    [tex] \left(\frac{\partial f}{\partial t}\right)_L = \left(\frac{\partial f}{\partial t}\right)_E + \vec{V} \cdot \left(\nabla f\right) [/tex] 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,
    [tex] \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 [/tex] Now realizing that,
    [tex] \frac{\partial \rho}{\partial t} = -\frac{1}{v^2} \frac{\partial v}{\partial t} [/tex] and from the partial derivative chain rule,
    [tex] \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}[/tex] Substituting into the the Lagrangian form yields,
    [tex] -\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 [/tex] or equivalently,
    [tex] \frac{\partial v}{\partial t} - \frac{\partial u}{\partial h} = 0 \\
    \frac{\partial u}{\partial t} + \frac{\partial p}{\partial h} = 0 [/tex] where h is the mass coordinate.
     
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