6
$\begingroup$

I am trying to rigorously go through some fluid mechanics proofs and theorems. I am currently going through a proof related to the transport theorem and I am having trouble with a step.

The steps in question are the following, first the variables are transformed according to:

$$ \frac{d}{dt}\int_{W_t} \rho \mathbf{u} dV = \frac{d}{dt}\int_W (\rho \mathbf{u})(\phi (\mathbf{x},t),t)J(\mathbf{x},t)dV $$

where $\phi(\mathbf{x},t)$ is the trajectory function of the particle found at $\mathbf{x}$ at time $t=0$. $W$ is the volume of the fluid under consideration at $t=0$ and $W_t$ is that volume at time $t$. $J$ is the Jacobian determinant of the transformation, followed by differenciating under the integral sign (RHS)

$$ \frac{\partial}{\partial t}(\rho \mathbf{u}) (\phi(\mathbf{x},t),t) = \left( \frac{D}{Dt} \rho \mathbf{u} \right) (\phi (\mathbf{x},t),t) $$

In the context of the first equation, why is the second equation true? When I try to prove it to myself I get:

$$ \frac{D}{Dt}(\rho \mathbf{u}) = \frac{\partial}{\partial t}(\rho \mathbf{u}) + \mathbf{u}\cdot \nabla (\rho \mathbf{u}) $$

and I don't know how to eliminate the second term. I think it has something to do with the trajectory function $\phi$ but I am not familiar enough with this idea to know how to proceed.

$\endgroup$
  • $\begingroup$ What happened to the derivative of $J$ (which depends explicitly on time)? $\endgroup$ – Doru Constantin Dec 30 '13 at 19:49
2
$\begingroup$

Short answer:

I think the notation is the main problem here. In your second equation, the LHS $\rho\mathbf{u}$ is a function of $\mathbf{x}_0$ and $t$, while your RHS $\rho\mathbf{u}$ is a function of $\mathbf{x}$ and $t$. The subtle difference is that $\mathbf{x}_0$ should be treated as a particle label, not an actual position. As you suspected, the bridge between the two formulations of $\rho\mathbf{u}$ is related by $\phi$: $$ (\rho\mathbf{u})(\mathbf{x},t) = (\rho\mathbf{u})\left[\phi(\mathbf{x},t),t\right] $$ In your second equation, the time derivative on the LHS is with respect to a fixed collection of particles, whereas on the RHS it should be with respect to a fixed locations in space. In fact, the material derivative $\frac{D}{Dt}$ applies to functions of $\mathbf{x}$ and $t$. Using it on a function of particle label and time does not make sense. So the second equation comes from taking the time derivative of the expression above.

$$ \begin{align} \frac{d}{dt}(\rho\mathbf{u})\left[\phi(\mathbf{x},t),t\right] & = \frac{d}{dt}(\rho\mathbf{u})(\mathbf{x},t) \\ & = \frac{D}{Dt}(\rho\mathbf{u})(\mathbf{x},t) \end{align} $$

Long Answer/Derivation:

To keep the derivation more readable, we will define $\mathbf{f}$ as the transport property we are interested in. In your case, $\mathbf{f} = \rho\mathbf{u}$. The change of variables that you want to perform transforms $\mathbf{f}$ between two different frames of reference.

The Eulerian frame of reference looks at changes in $\mathbf{f}$ from the point of view of an outside observer watching the entire flow field. The property of the fluid depends on its location and time: $$ \mathbf{f} = \mathbf{f}(\mathbf{x}, t) $$ The Eulerian frame is convenient for experiments and operations involving spatial gradients. However, applying the laws of classical mechanics in this frame of reference is not as straight forward, since the particles that occupy a location $\mathbf{x}$ at different times are usually not the same. In order to use the conservation laws, we switch to a more suitable reference frame.

The Lagrangian reference frame will monitor the change of $\mathbf{f}$ from the point of view of a fluid particle. In such a reference frame, each particle has an associated $\mathbf{f}(t)$. Let's name each particle in our continuum with a label $\mathbf{\xi}$, so that $$ \mathbf{f} = \mathbf{f}(\mathbf{\xi}, t) $$ gives the property of particle $\mathbf{\xi}$ at time $t$. To distinguish properties in the two reference frames, we will accent the Lagrangian properties with a tilde. $$ \tilde{\mathbf{f}} = \tilde{\mathbf{f}}(\mathbf{\xi}, t) $$

Now we need a map between the $\tilde{\mathbf{f}}$ and $\mathbf{f}$. Let's define $\phi$ as the mapping that takes the particle's location $\mathbf{x}$ at some time $t$ and returns the label of the particle $\mathbf{\xi}$ $$ \mathbf{\xi} = \phi(\mathbf{x},t) $$ If we used the location of a particle at $t=0$ as its label, then we get the trajectory function you used $$ \mathbf{x}_0 = \phi(\mathbf{x}, t) $$ It is important to note that $\mathbf{x}_0$ serves as a particle label, so while $$ \tilde{\mathbf{f}}(\mathbf{x}_0,0) = \mathbf{f}(\mathbf{x}_0,0) $$ , in general $$ \tilde{\mathbf{f}}(\mathbf{x},t) \ne \mathbf{f}(\mathbf{x},t) $$

Instead, the two reference frames are related by $$ \mathbf{f}(\mathbf{x},t) = \tilde{\mathbf{f}}\left[\phi(\mathbf{x},t),t\right] $$ and the Jacobian should be a function of particle label and time: $$ J = J(\mathrm{\xi},t) $$

Using this notation, the first integral in your question becomes $$ \begin{align} \frac{d}{dt}\int_{W(t)} \mathbf{f}(\mathbf{x},t) dV & = \frac{d}{dt}\int_W \tilde{\mathbf{f}}\left[\phi(\mathbf{x},t),t\right]J(\mathbf{\xi},t)dV \\ & = \int_W \left[ \frac{d\tilde{\mathbf{f}}}{dt}J + \tilde{\mathbf{f}}\frac{dJ}{dt} \right] dV \end{align} $$

The second equation in your question is not applied until the integral above is transformed back from the $(\mathbf{\xi},t)$ frame to the $(\mathbf{x},t)$ frame. Noting that the time derivative of this Jacobian can be written as (this is a whole other derivation): $$ \frac{dJ}{dt} = (\nabla \cdot \mathbf{u})J(\mathbf{x},t) $$

we have:

$$ \begin{align} \int_W \left[ \frac{d\tilde{\mathbf{f}}}{dt}J + \tilde{\mathbf{f}}\frac{dJ}{dt} \right] dV & = \int_{W(t)} \left[\frac{d}{dt}\mathbf{f}(\mathbf{x},t) + \mathbf{f}(\mathbf{x},t)(\nabla \cdot \mathbf{u})\right] dV \end{align} $$

The first term in the integral can be expanded using the chain rule $$ \frac{d}{dt}\mathbf{f}(\mathbf{x},t) = \frac{\partial \mathbf{f}}{\partial t} + \frac{\partial \mathbf{x}}{\partial t} \cdot \frac{\partial \mathbf{f}}{\partial \mathbf{x}} $$

which is just the definition of the material derivative $\frac{D}{Dt}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.