How to convert Euler Equations to Lagrangian Form?

Mr. Cosmos
Messages
9
Reaction score
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,
\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.
 
Last edited:
Physics news on Phys.org
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.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top