About the derivative of the Jacobian in fluid dynamics

Question

I was studying a book on the mathematics of fluid dynamics in which there was a lemma on how to find the derivative of the Jacobian. The explanation is as follows (Sorry if it's too long):

There is a region $D$ in Euclidean space where there is a fluid whose velocity at any point $\mathbf {\vec x}\in D$ at any time $t$ is given by the vector field $\mathbf{\vec u}(\mathbf{\vec x}, t)$. Let us write $\mathbf{\vec \varphi}(\mathbf{\vec x}, t)$ for the trajectory followed by the particle that is at point $\mathbf{\vec x}$ at time $t = 0$. We will assume $\varphi$ is smooth enough so the following manipulations are legitimate and for fixed $t$, $\mathbf{\vec \varphi}$ is an invertible mapping. Let $\varphi_t$ denote the map $\mathbf{\vec x} \rightarrow \mathbf{\vec \varphi}(\mathbf{\vec x}, t)$; that is, with fixed $t$, this map advances each fluid particle from its position at time $t = 0$ to its position at time $t$. Here, of course, the subscript does not denote differentiation. We call $\mathbf{\vec \varphi}$ the fluid flow map. If $W$ is a region in $D$, then $\varphi_t(W) = W_t$ is the volume $W$ moving with the fluid as shown in figure.

Now let's say that $\mathbf{\vec x}=x\hat i+y\hat j+z\hat k$; $\mathbf{\vec \varphi}=\epsilon \hat i+\eta\hat j+\zeta \hat k$ and $\mathbf{\vec u}=u\hat i+v\hat j+w\hat k$. Since $\mathbf{\vec \varphi}$ is the displacement field and $\mathbf{\vec u}$ is the velocity field so we have $$\frac{\delta}{\delta t}(\mathbf{\vec \varphi}(\mathbf{\vec x}, t))=\mathbf{\vec u}(\mathbf{\vec \varphi}(\mathbf{\vec x}, t), t)$$ So $$\frac{\delta\epsilon}{\delta t}=u;\ \frac{\delta\eta}{\delta t}=v;\ \frac{\delta\zeta}{\delta t}=w\ \ \ \ \ \cdots(i)$$ The Jacobian of $\mathbf{\vec x}$ w.r.t $\mathbf{\vec \varphi}(\mathbf{\vec x}, t)$ is given as:

$$J(\mathbf {\vec x},t)=\left[\begin{matrix} \displaystyle{\frac{\delta\epsilon}{\delta x}}&\displaystyle\frac{\delta\eta}{\delta x}&\displaystyle\frac{\delta\zeta}{\delta x}\\ \displaystyle{\frac{\delta\epsilon}{\delta y}}&\displaystyle\frac{\delta\eta}{\delta y}&\displaystyle\frac{\delta\zeta}{\delta y}\\ \displaystyle{\frac{\delta\epsilon}{\delta z}}&\displaystyle\frac{\delta\eta}{\delta z}&\displaystyle\frac{\delta\zeta}{\delta z} \end{matrix}\right]$$ Now the task is to prove that

$$\displaystyle\frac{\delta}{\delta t}J(\mathbf{\vec x}, t) = J(\mathbf{\vec x}, t)\left[\text{div }\mathbf{\vec u}(\mathbf{\vec \varphi}(\mathbf{\vec x}, t), t)\right]$$

What I did:

$$\begin{align} \displaystyle\frac{\delta}{\delta t}J&=\left[\begin{matrix} \displaystyle{\frac{\delta}{\delta t}\frac{\delta\epsilon}{\delta x}}&\displaystyle\frac{\delta\eta}{\delta x}&\displaystyle\frac{\delta\zeta}{\delta x}\\ \displaystyle{\frac{\delta}{\delta t}\frac{\delta\epsilon}{\delta y}}&\displaystyle\frac{\delta\eta}{\delta y}&\displaystyle\frac{\delta\zeta}{\delta y}\\ \displaystyle{\frac{\delta}{\delta t}\frac{\delta\epsilon}{\delta z}}&\displaystyle\frac{\delta\eta}{\delta z}&\displaystyle\frac{\delta\zeta}{\delta z} \end{matrix}\right]+\left[\begin{matrix} \displaystyle{\frac{\delta\epsilon}{\delta x}}&\displaystyle{\frac{\delta}{\delta t}\frac{\delta\eta}{\delta x}}&\displaystyle\frac{\delta\zeta}{\delta x}\\ \displaystyle{\frac{\delta\epsilon}{\delta y}}&\displaystyle{\frac{\delta}{\delta t}\frac{\delta\eta}{\delta y}}&\displaystyle\frac{\delta\zeta}{\delta y}\\ \displaystyle{\frac{\delta\epsilon}{\delta z}}&\displaystyle{\frac{\delta}{\delta t}\frac{\delta\eta}{\delta z}}&\displaystyle\frac{\delta\zeta}{\delta z} \end{matrix}\right]+\left[\begin{matrix} \displaystyle{\frac{\delta\epsilon}{\delta x}}&\displaystyle\frac{\delta\eta}{\delta x}&\displaystyle{\frac{\delta}{\delta t}\frac{\delta\zeta}{\delta x}}\\ \displaystyle{\frac{\delta\epsilon}{\delta y}}&\displaystyle\frac{\delta\eta}{\delta y}&\displaystyle{\frac{\delta}{\delta t}\frac{\delta\zeta}{\delta y}}\\ \displaystyle{\frac{\delta\epsilon}{\delta z}}&\displaystyle\frac{\delta\eta}{\delta z}&\displaystyle{\frac{\delta}{\delta t}\frac{\delta\zeta}{\delta z}} \end{matrix}\right]\\ &=\left[\begin{matrix} \displaystyle{\frac{\delta u}{\delta x}}&\displaystyle\frac{\delta\eta}{\delta x}&\displaystyle\frac{\delta\zeta}{\delta x}\\ \displaystyle{\frac{\delta u}{\delta y}}&\displaystyle\frac{\delta\eta}{\delta y}&\displaystyle\frac{\delta\zeta}{\delta y}\\ \displaystyle{\frac{\delta u}{\delta z}}&\displaystyle\frac{\delta\eta}{\delta z}&\displaystyle\frac{\delta\zeta}{\delta z} \end{matrix}\right]+\left[\begin{matrix} \displaystyle{\frac{\delta\epsilon}{\delta x}}&\displaystyle\frac{\delta v}{\delta x}&\displaystyle\frac{\delta\zeta}{\delta x}\\ \displaystyle{\frac{\delta\epsilon}{\delta y}}&\displaystyle\frac{\delta v}{\delta y}&\displaystyle\frac{\delta\zeta}{\delta y}\\ \displaystyle{\frac{\delta\epsilon}{\delta z}}&\displaystyle\frac{\delta v}{\delta z}&\displaystyle\frac{\delta\zeta}{\delta z} \end{matrix}\right]+\left[\begin{matrix} \displaystyle{\frac{\delta\epsilon}{\delta x}}&\displaystyle\frac{\delta\eta}{\delta x}&\displaystyle\frac{\delta z}{\delta x}\\ \displaystyle{\frac{\delta\epsilon}{\delta y}}&\displaystyle\frac{\delta\eta}{\delta y}&\displaystyle\frac{\delta z}{\delta y}\\ \displaystyle{\frac{\delta\epsilon}{\delta z}}&\displaystyle\frac{\delta\eta}{\delta z}&\displaystyle\frac{\delta z}{\delta z} \end{matrix}\right]\ \ \ \ [\text{By }(i)] \end{align}$$ At this point the book says to use the fact that $$\frac{\delta u}{\delta x}=\frac{\delta u}{\delta \epsilon}\frac{\delta \epsilon}{\delta x}+\frac{\delta u}{\delta \eta}\frac{\delta \eta}{\delta x}+\frac{\delta u}{\delta \zeta}\frac{\delta \zeta}{\delta x}$$ $$\frac{\delta u}{\delta y}=\frac{\delta u}{\delta \epsilon}\frac{\delta \epsilon}{\delta y}+\frac{\delta u}{\delta \eta}\frac{\delta \eta}{\delta y}+\frac{\delta u}{\delta \zeta}\frac{\delta \zeta}{\delta y}$$ $$\vdots$$ $$\frac{\delta w}{\delta z}=\frac{\delta w}{\delta \epsilon}\frac{\delta \epsilon}{\delta z}+\frac{\delta w}{\delta \eta}\frac{\delta \eta}{\delta z}+\frac{\delta w}{\delta \zeta}\frac{\delta \zeta}{\delta z}$$ However I can't quite understand how to use these equations to continue the determinant. Please help me in this problem.

Thanks for the attention.

Answer 1

https://www.owlnet.rice.edu/~ceng501/Chap4.pdf

This link might be helpful to answer your question.

There are cancellations when you expand the determinant.

Answer 2

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}

The key observation is $$ (\dot{\mathbf{J}})_{ij} = \frac{\partial}{\partial t} \left( \frac{\partial x_i}{\partial \xi_j} \right) = \frac{\partial}{\partial \xi_j} \left( \frac{\partial x_i}{\partial t} \right) = \frac{\partial v_i}{\partial \xi_j} = \sum_k \frac{\partial v_i}{\partial x_k} \frac{\partial x_k}{\partial \xi_j} = \sum_k \frac{\partial v_i}{\partial x_k} J_{kj} $$ In matrix form, this writes $$ \dot{\mathbf{J}} = \begin{pmatrix} \frac{\partial v_1}{\partial x_1} & \frac{\partial v_1}{\partial x_2} & \frac{\partial v_1}{\partial x_3} \\ \frac{\partial v_2}{\partial x_1} & \frac{\partial v_2}{\partial x_2} & \frac{\partial v_2}{\partial x_3} \\ \frac{\partial v_3}{\partial x_1} & \frac{\partial v_3}{\partial x_2} & \frac{\partial v_3}{\partial x_3} \end{pmatrix} \mathbf{J} =\mathbf{A}\mathbf{J} $$

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}