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The example 1.7 in "An introduction to Theoretical Fluid Mechanics" by Stephen Childress, he calculated the material derivative of the Jacobian determinant $Det \ \textbf{J}$ of the Lagrangian map

\begin{align*} X:\Omega_0 \times [0,+\infty) \to \Omega_t \subset \mathbb{R}^N, \ \ \ (\alpha,t) \mapsto X(\alpha,t)=\textbf{x}. \end{align*}

I think that the definition of $\textbf{J}$ is: \begin{align*} \textbf{J}=\begin{pmatrix} \frac{\partial X_1}{\partial \alpha_1} & \frac{\partial X_1}{\partial \alpha_2} & \cdots & \frac{\partial X_1}{\partial \alpha_N} \\ \frac{\partial X_2}{\partial \alpha_1} & \frac{\partial X_2}{\partial \alpha_2} & \cdots & \frac{\partial X_2}{\partial \alpha_N}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial X_N}{\partial \alpha_1} & \frac{\partial X_N}{\partial \alpha_2} & \cdots & \frac{\partial X_N}{\partial \alpha_N} \end{pmatrix} \end{align*}.

Then, he concluded that \begin{align*} \frac{D}{Dt}(Det \ \textbf{J})= div(\text{u})(Det \ \textbf{J}), \end{align*} where \begin{align*} \frac{D}{Dt}=\frac{\partial}{\partial t}+ \textbf{u}\cdot \nabla, \end{align*} is the material derivative.

On the other hand, in the book "A Mathematical Introduction to Fluid Mechanics" Third Edition by Chorin and Marsden, in the lemma of page 8, they asserts that (in the Childress notation) \begin{align*} \frac{\partial}{\partial t} Det \ \textbf{J}=(Det \ \textbf{J})div(\textbf{u}). \end{align*}

So, like \begin{align*} \frac{D}{Dt} Det \ \textbf{J}=\frac{\partial}{\partial t} (Det \ \textbf{J})+\textbf{u}\cdot \nabla(Det \ \textbf{J}), \end{align*} for the above, we can conclude that \begin{align*} \frac{\partial}{\partial t} (Det \ \textbf{J})+\textbf{u}\cdot \nabla(Det \ \textbf{J})=div(\textbf{u})(Det \ \textbf{J}), \end{align*} then \begin{align*} div(\textbf{u})(Det \ \textbf{J})+\textbf{u}\cdot \nabla(Det \ \textbf{J})=div(\textbf{u})(Det \ \textbf{J}), \end{align*} therefore \begin{align*} \textbf{u}\cdot\nabla (Det \ \textbf{J})=0. \end{align*}

It is right? Or am I misunderstanding the concept of material derivative?

Hapa
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    The correct result is $\frac{\partial}{\partial t} Det \ \textbf{J}=(Det \ \textbf{J})div(\textbf{u})$. You ae on a wild goose chase equating the correct result to the incorrect result to conclude that $\textbf{u}\cdot\nabla (Det \ \textbf{J})=0.$. – RRL Oct 06 '22 at 20:35
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    I derive the result in Chorin and Marsden in this answer. – RRL Oct 06 '22 at 20:38
  • Thank you so much, it is just that in the Childress' book appear that result, it is to say, $\frac{D}{Dt} Det \ \textbf{J}=div(\textbf{u})(Det \ \textbf{J})$, https://www.math.nyu.edu/~childres/fluidsbook.pdf page 9 – Hapa Oct 06 '22 at 21:10
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    To use $\frac{D}{Dt}$ rather than $\frac{\partial}{\partial t}$ as Childress is doing is very unconventional. I would normally say immediately it is a typographical error but, Childress brings up the material derivative just before. The problem is Childress doesn't show a careful proof and there is no need for this form when later proving the Reynolds transport theorem -- which is why this lemma is needed in the first place. Is it possible that $\textbf{u}\cdot\nabla ( |\textbf{J}|) = 0$? I would need to go back and check. See if you can follow my proof first. – RRL Oct 06 '22 at 22:00
  • In fact, I don't know if that result is true, I can follow your proof for $\frac{\partial}{\partial t}|\textbf{J}|=div(\textbf{u})|\textbf{J}|$, may be you are right and it is typo error. Thank you. – Hapa Oct 06 '22 at 22:22

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