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I am currently working through Alexandre Chorin's Mathematical Introduction to Fluid Mechanics. In the first chapter, he treats the change in Kinetic energy of a fluid region $W\subset D$ subject to the fluid flow map $\varphi_t:\mathbf{x}\mapsto\varphi(\mathbf{x},t)$ in the following manner ($\frac{D}{Dt}$ denoting the material derivative):

\begin{aligned} \frac { d } { d t } E _ { \text { kinetic } } & = \frac { d } { d t } \left[ \frac { 1 } { 2 } \int _ { W _ { t } } \rho \| \mathbf { u } \| ^ { 2 } d V \right] \\ & = \frac { 1 } { 2 } \int _ { W _ { t } } \rho \frac { D \| \mathbf { u } \| ^ { 2 } } { D t } d V \\ & = \int _ { W _ { t } } \rho \left( \mathbf { u } \cdot \left( \frac { \partial \mathbf { u } } { \partial t } + ( \mathbf { u } \cdot \nabla ) \mathbf { u } \right) \right) d V \end{aligned}

As best as I can determine, an implicit assumption seems to be made that the fluid has a density constant with time in this derivation. Specifically, it appears that the assumption $\frac{\partial}{\partial t}(\rho\| \mathbf { u } \| ^ { 2 }) = \rho\frac{D}{Dt}(\| \mathbf { u } \| ^ { 2 })$ is being made rather than obeying the typical product rule. I am not quite sure why. I would greatly appreciate it if someone could help shed some light on what is going on here!

If more context is needed, I can provide it upon request, or you may reference the presentation which is given on page 12 of Chorin's book.

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  • $\begingroup$ Has he mentioned that the derivation is for an incompressible fluid? $\endgroup$ – Deep Feb 14 at 5:11
  • $\begingroup$ The book is here , p.24 as the PDF is numbered. $\endgroup$ – Keith McClary Feb 14 at 5:16
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    $\begingroup$ No, there is no such assumption, the density can evolve in time. Just the continuity equation of mass has been considered. You are forgetting that also $W_t$ depends on time so that you cannot simply swap the derivative and the integral symbols. $\endgroup$ – Valter Moretti Feb 14 at 7:10
  • $\begingroup$ Yes, I was overlooking that! Thank you for bringing that to my attention, it hadn’t even crossed my mind. $\endgroup$ – Kyle Poe Feb 14 at 8:45
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Indeed, there is no assumption.

You can use the Reynolds transport theorem https://en.wikipedia.org/wiki/Reynolds_transport_theorem

With the Green theorem , it gives : $\frac{d}{dt}\int\limits_{\Omega (t)}{f\rho d\tau }=\int\limits_{\Omega (t)}{\left( \frac{\partial f\rho }{\partial t}+\nabla \cdot (f\rho \overrightarrow{v}) \right)d\tau }$

Then :

$\overrightarrow{\nabla }\cdot (f\rho \overrightarrow{v})=\rho \overrightarrow{v}\cdot \overrightarrow{\nabla }\cdot (f)+f\overrightarrow{\nabla }\cdot (\rho \overrightarrow{v})$

$\frac{\partial f\rho }{\partial t}+\nabla \cdot (f\rho \overrightarrow{v})=\rho \left( \frac{\partial f}{\partial t}+\overrightarrow{v}\cdot \overrightarrow{\nabla }\cdot (f) \right)+f\underbrace{\left( \frac{\partial \rho }{\partial t}+\overrightarrow{\nabla }\cdot (\rho \overrightarrow{v}) \right)}_{0}$

$\frac{d}{dt}\int\limits_{\Omega (t)}{f\rho d\tau }=\int\limits_{\Omega (t)}{\left( \frac{\partial f}{\partial t}+\vec{v}\cdot \vec{\nabla }\cdot (f) \right)\rho d\tau }=\int\limits_{\Omega (t)}{\left( \frac{\partial f}{\partial t}+(\vec{v}\cdot \vec{\nabla })\cdot (f) \right)dm}=\int\limits_{\Omega (t)}{\left( \frac{Df}{Dt} \right)dm}$

It is rather intuitive. You apply Newton law to a small fluid particle and you add all.

Sorry for my english !

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  • $\begingroup$ I see now. Today was my first time seeing the transport theorem, and it evidently hasn’t sunk in yet... Unfortunately I didn’t realize that was where we were to invoke the transport theorem... but upon looking at the next step it is obviously just simple differentiation! Thank you! $\endgroup$ – Kyle Poe Feb 14 at 8:42

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