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