Proving Euler's identity (fluid mechanics) by differentiating the determinant $J=\partial(x,y,z)/\partial(X,Y,Z)$

Question

I am looking at some fluid mechanics lecture notes. In this context, $X,Y,Z$ are Lagrangian variables (at time $t = 0$) and $x,y,z$ are Eulerian variables (at time $t > 0$). Euler's identity says that $$\frac{\mathrm{D}J}{\mathrm{D}t}=J\mathbf{\nabla}\cdot \mathbf{u}$$ where $\mathbf{u}=\mathbf{u}(x(t),y(t),z(t),t)$ is the velocity of the fluid, $J=\frac{\partial (x,y,z)}{\partial(X,Y,Z)}$ is the Jacobian determinant and $\frac{\mathrm{D}}{\mathrm{D}t}=\frac{\partial }{\partial t}+\mathbf{u \cdot \nabla}$ is the convective differential operator.

The notes say this:

Whilst this seems to look quite straightforward, I am not sure what property has been used here or why it is permissible for us to be able to differentiate each row of the determinant separately and add up the resulting determinants. Could somebody please offer an input or key result that is being used?

Answer 1

Consider the case $n=2$. Observe \begin{align} \frac{d}{dt} \begin{vmatrix} a_{11}(t) & a_{12}(t)\\ a_{21}(t) & a_{22}(t) \end{vmatrix} =&\ \frac{d}{dt} \sum_{\sigma \in S_2}(-1)^{\operatorname{sgn}\sigma} a_{1\sigma(1)}(t) a_{2\sigma(2)}(t)\\ =&\ \sum_{\sigma \in S_2}(-1)^{\operatorname{sgn}\sigma} [a_{1\sigma(1)}'(t) a_{2\sigma(2)}(t)+a_{1\sigma(1)}(t) a_{2\sigma(2)}'(t)]\\ =&\ \sum_{\sigma \in S_2}(-1)^{\operatorname{sgn}\sigma} a_{1\sigma(1)}'(t) a_{2\sigma(2)}(t)+\sum_{\sigma \in S_2}(-1)^{\operatorname{sgn}\sigma} a_{1\sigma(1)}(t) a_{2\sigma(2)}'(t)\\ =&\ \begin{vmatrix} a_{11}'(t) & a_{12}'(t)\\ a_{21}(t) & a_{22}(t) \end{vmatrix} + \begin{vmatrix} a_{11}(t) & a_{12}(t)\\ a_{21}'(t) & a_{22}'(t) \end{vmatrix}. \end{align} Note I have used the definition: If $A=(a_{ij})$ is an $n\times n$ matrix then \begin{align} \det A = \sum_{\sigma \in S_n} (-1)^{\operatorname{sgn}\sigma} a_{1 \sigma(1)}\cdots a_{n\sigma(n)} \end{align} where $S_n$ is the permutation group of $n$ objects.

Answer 2

The determinant of a matrix $J$ is the volume of the parallelepiped made up of its columns (or, equivalently, of its rows).

If $R_i$ are the rows of J, the determinant can then be written as $det(J) = \sum_{ijk}\epsilon_{ijk}R_1^iR_2^jR_3^k$ where $\epsilon_{ijk}$ is the Levi-Civita symbol (which is either 1, -1 or 0 depending on the value of the indexes).

Once you write it like that, you use the linearity of the derivative to bring it inside the sum and then use Leibniz's rule to derive term by term.

If you think this through, you will see that this gives exactly what you have written.

Answer 3

An additional resource which I found was useful for this problem are those slides:

http://www3.dicca.unige.it/rrepetto/linked-files/fluid-dynamics-lecture-notes.pdf

The appendix (Slide 154 - 156) it explains the topic as well.

The approach there is as follows:

$$J=\operatorname{det}\left(\begin{array}{ccc}\frac{\partial x_{1}}{\partial \xi_{1}} & \frac{\partial x_{1}}{\partial \xi_{2}} & \frac{\partial x_{1}}{\partial \xi_{3}} \\ \frac{\partial x_{2}}{\partial \xi_{1}} & \frac{\partial x_{2}}{\partial \xi_{2}} & \frac{\partial x_{2}}{\partial \xi_{3}} \\ \frac{\partial x_{3}}{\partial \xi_{1}} & \frac{\partial x_{3}}{\partial \xi_{2}} & \frac{\partial x_{3}}{\partial \xi_{3}}\end{array}\right)$$

Definition of the material derivative:

$$\frac{D}{D t}\left(\frac{\partial x_{i}}{\partial \xi_{j}}\right)=\frac{\partial}{\partial \xi_{j}} \frac{D x_{j}}{D t}=\frac{\partial u_{i}}{\partial \xi_{j}}$$

Apply the chain rule:

$$\frac{\partial u_{i}}{\partial \xi_{j}}=\frac{\partial u_{i}}{\partial x_{1}} \frac{\partial x_{1}}{\partial \xi_{j}}+\frac{\partial u_{i}}{\partial x_{2}} \frac{\partial x_{2}}{\partial \xi_{j}}+\frac{\partial u_{i}}{\partial x_{3}} \frac{\partial x_{3}}{\partial \xi_{j}}=\frac{\partial u_{i}}{\partial x_{k}} \frac{\partial x_{k}}{\partial \xi_{j}}$$

As an example for the first row, insert the previous results:

$$ \operatorname{det}\left(\begin{array}{ccc} \frac{\partial u_{1}}{\partial x_{k}} \frac{\partial x_{k}}{\partial \xi_{1}} & \frac{\partial u_{1}}{\partial x_{k}} \frac{\partial x_{k}}{\partial \xi_{2}} & \frac{\partial u_{1}}{\partial x_{k}} \frac{\partial x_{k}}{\partial \xi_{3}} \\ \frac{\partial x_{2}}{\partial \xi_{1}} & \frac{\partial x_{2}}{\partial \xi_{2}} & \frac{\partial x_{2}}{\partial \xi_{3}} \\ \frac{\partial x_{3}}{\partial \xi_{1}} & \frac{\partial x_{3}}{\partial \xi_{2}} & \frac{\partial x_{3}}{\partial \xi_{3}} \end{array}\right) $$

Simplify the result:

$$ \operatorname{det} = \left(\begin{array}{ccc}\frac{\partial u_{1}}{\partial x_{1}} \frac{\partial x_{1}}{\partial \xi_{1}} & \frac{\partial u_{1}}{\partial x_{1}} \frac{\partial x_{1}}{\partial \xi_{2}} & \frac{\partial u_{1}}{\partial x_{1}} \frac{\partial x_{1}}{\partial \xi_{3}} \\ \frac{\partial x_{2}}{\partial \xi_{1}} & \frac{\partial x_{2}}{\partial \xi_{2}} & \frac{\partial x_{2}}{\partial \xi_{3}} \\ \frac{\partial x_{3}}{\partial \xi_{1}} & \frac{\partial x_{3}}{\partial \xi_{2}} & \frac{\partial x_{3}}{\partial \xi_{3}}\end{array}\right)=\frac{\partial u_{1}}{\partial x_{1}}J$$