4

I'm trying to follow the steps starting at the last section of page 2 here. I'm asked to show that if $|J(x,y,z,t)|$ is the (spatial) Jacobian determinant of the flow map $\boldsymbol{\varphi}(x,y,z,t)$, then $$\frac{\partial |J|}{\partial t}=|J| \operatorname{div}(\mathbf{v}) $$ where $\mathbf{v}=\partial \boldsymbol{\varphi}/\partial t $ is the velocity field of the flow.

I couldn't prove it for the 3D case, so I tried the 1D analogue, which boils down to the equation

$$\frac{\partial^2 \varphi}{\partial t \partial x}= \frac{\partial \varphi}{\partial x} \frac{\partial^2 \varphi}{\partial x \partial t}.$$ I can't see why this simpler equation must hold either. I'd like some help with both of this cases please. Thank you.

user1337
  • 24,381

2 Answers2

6

Changing notation from $(x,y,z)$ to $(x_1,x_2,x_3)$, the flow map $\boldsymbol{\varphi}$ maps the coordinates $(x_1,x_2,x_3)$ of a fluid particle at time $t= 0$ onto the coordinates $(\varphi_1, \varphi_2,\varphi_3)$ at time $t$.

The Jacobian determinant $|J|$ can be expressed as

$$|J| = \sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial\varphi_1}{\partial x_{j_1}}\frac{\partial\varphi_2}{\partial x_{j_2}}\frac{\partial\varphi_3}{\partial x_{j_3}},$$

where the sum is taken over all permutations $(j_1,j_2,j_3)$ of $(1,2,3)$ and $s(j_1,j_2,j_3)$ is the sign of the permutation.

Since

$$\frac{\partial}{\partial t}\frac{\partial \varphi_k}{\partial x_{j_k}} = \frac{\partial}{\partial x_{j_k}}\frac{\partial \varphi_k}{\partial t} = \frac{\partial v_k}{\partial x_{j_k}}$$

we have

$$\tag{*}\begin{align}\frac{\partial}{\partial t} |J| &= \sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial v_1}{\partial x_{j_1}}\frac{\partial\varphi_2}{\partial x_{j_2}}\frac{\partial\varphi_3}{\partial x_{j_3}} \\ &+ \sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial \varphi_1}{\partial x_{j_1}}\frac{\partial v_2}{\partial x_{j_2}}\frac{\partial\varphi_3}{\partial x_{j_3}} \\ &+ \sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial \varphi_1}{\partial x_{j_1}}\frac{\partial\varphi_2}{\partial x_{j_2}}\frac{\partial v_3}{\partial x_{j_3}} \end{align}.$$

By the chain rule,

$$\frac{\partial v_1}{\partial x_{j_1}} = \frac{\partial v_1}{\partial \varphi_1}\frac{\partial \varphi_1}{\partial x_{j_1}} + \frac{\partial v_1}{\partial \varphi_2}\frac{\partial \varphi_2}{\partial x_{j_1}} + \frac{\partial v_3}{\partial \varphi_1}\frac{\partial \varphi_3}{\partial x_{j_1}}. $$

Applying this to (*) the first sum on the RHS becomes

$$\begin{align}\sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial v_1}{\partial x_{j_1}}\frac{\partial\varphi_2}{\partial x_{j_2}}\frac{\partial\varphi_3}{\partial x_{j_3}} &= \frac{\partial v_1}{\partial \varphi_1}\sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial \varphi_1}{\partial x_{j_1}}\frac{\partial\varphi_2}{\partial x_{j_2}}\frac{\partial\varphi_3}{\partial x_{j_3}} \\ &+ \frac{\partial v_2}{\partial \varphi_1}\sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial \varphi_2}{\partial x_{j_1}}\frac{\partial\varphi_2}{\partial x_{j_2}}\frac{\partial\varphi_3}{\partial x_{j_3}} \\ &+ \frac{\partial v_3}{\partial \varphi_1}\sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial \varphi_3}{\partial x_{j_1}}\frac{\partial\varphi_2}{\partial x_{j_2}}\frac{\partial\varphi_3}{\partial x_{j_3}} \\ \end{align} $$

The second and third sums on the RHS of (**) are determinants with identical rows and must vanish.

Hence,

$$\sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial v_1}{\partial x_{j_1}}\frac{\partial\varphi_2}{\partial x_{j_2}}\frac{\partial\varphi_3}{\partial x_{j_3}} = \frac{\partial v_1}{\partial \varphi_1}\sum_{(j_1,j_2,j_3)}s(j_1,j_2,j_3) \frac{\partial \varphi_1}{\partial x_{j_1}}\frac{\partial\varphi_2}{\partial x_{j_2}}\frac{\partial\varphi_3}{\partial x_{j_3}} = \frac{\partial v_1}{\partial \varphi_1} |J|$$

Applying the chain rule, similarly, to the second and third sums on the RHS of (*) and adding we obtain

$$\frac{\partial}{\partial t} |J|= \left(\frac{\partial v_1}{\partial \varphi_1} + \frac{\partial v_2}{\partial \varphi_2} + \frac{\partial v_3}{\partial \varphi_3} \right)|J| = \operatorname{div}(\mathbf{v}) |J|$$

RRL
  • 90,707
0

This answer is adapted from About the derivative of the Jaobian in fluid dynamics with new variables naming.

A straightforward derivation is to use the differential of a determinant that can be written as \begin{eqnarray} d|\mathbf{J}| &=& |\mathbf{J}| \mathrm{tr} \left( \mathbf{J}^{-1} d\mathbf{J} \right) = |\mathbf{J}| \mathrm{tr} \left( \mathbf{J}^{-1} \dot{\mathbf{J}} \right) dt \tag{1} \end{eqnarray} @time $t=0$ the fluid particle is at position $\mathbf{x}$ and at a later time, it is at position $\mathbf{\varphi}(\mathbf{x},t)$ (which is a 3-D vector). The velocity of the particle (in the moving frame) is denoted $\mathbf{v}$ with tht $i$-th component given by $v_i = \frac{\partial \varphi_i}{\partial t}$.

The key observation is $$ (\dot{\mathbf{J}})_{ij} = \frac{\partial}{\partial t} \left( \frac{\partial \varphi_i}{\partial x_j} \right) = \frac{\partial}{\partial x_j} \left( \frac{\partial \varphi_i}{\partial t} \right) = \frac{\partial v_i}{\partial x_j} = \sum_k \frac{\partial v_i}{\partial \varphi_k} \frac{\partial \varphi_k}{\partial x_j} = \sum_k \frac{\partial v_i}{\partial \varphi_k} J_{kj} $$ In matrix form, this writes $$ \dot{\mathbf{J}} = \begin{pmatrix} \frac{\partial v_1}{\partial \varphi_1} & \frac{\partial v_1}{\partial \varphi_2} & \frac{\partial v_1}{\partial \varphi_3} \\ \frac{\partial v_2}{\partial \varphi_1} & \frac{\partial v_2}{\partial \varphi_2} & \frac{\partial v_2}{\partial \varphi_3} \\ \frac{\partial v_3}{\partial \varphi_1} & \frac{\partial v_3}{\partial \varphi_2} & \frac{\partial v_3}{\partial \varphi_3} \end{pmatrix} \mathbf{J} =\mathbf{A}\mathbf{J} $$

The relation (1) writes \begin{eqnarray} d|\mathbf{J}| &=& |\mathbf{J}| \mathrm{tr} \left( \mathbf{J}^{-1} \mathbf{A}\mathbf{J} \right) dt = |\mathbf{J}| \mathrm{tr} \left( \mathbf{A} \right) dt = |\mathbf{J}| \operatorname{div}(\mathbf{v}) dt \end{eqnarray}

and thus $$ \frac{\partial |\mathbf{J}|}{\partial t} = |\mathbf{J}| \operatorname{div}(\mathbf{v}) $$

Steph
  • 3,665