In deriving continuity equations using Lagrangian .
We consider the element of fluid which occupied a rectangular parallelopiped having its centre at the point $(a,b,c)$ and its edges $\delta a$ , $\delta b$ ,$\delta c $ parallel to the axes . At the time $t$ the same element for an oblique parallelepiped . The centre now has for its co-ordinates $x$ , $y$ , $z$; and the projections of the edges on the co-ordinate axes are respectively $$ \frac{\partial x}{\partial a} \delta a \ , \ \frac{\partial y}{\partial a} \delta a \ , \ \frac{\partial z}{\partial c} \delta a$$ $$\frac{\partial x}{\partial b} \delta b \ , \ \frac{\partial y}{\partial b} \delta b \ , \ \frac{\partial z}{\partial b} \delta b$$ $$\frac{\partial x}{\partial c} \delta c \ , \ \frac{\partial y}{\partial c} \delta c \ , \ \frac{\partial z}{\partial c} \delta c$$
How can i get these projections ? The volume of the parallelepiped is therefore $$\begin{vmatrix} \frac{\partial x}{\partial a} & \frac{\partial y}{\partial a} & \frac{\partial z}{\partial a} \\ \frac{\partial x}{\partial b} & \frac{\partial y}{\partial b} & \frac{\partial z}{\partial b} \\ \frac{\partial x}{\partial c} & \frac{\partial y}{\partial c} & \frac{\partial z}{\partial c} \end{vmatrix} \delta a \delta b \delta c$$ or as its often written $$\frac{D(x,y,z)}{D(a,b,c)} \delta a \delta b \delta c$$
since the fluid mass is unchanged and the fluid is incompressible we have $$\frac{D(x,y,z)}{D(a,b,c)} =1$$
Is there a way to prove that $$\frac{D(a,b,c)}{D(x,y,z)}= 1$$ without expanding the determinant ?
The projections can be obtained by using the chain rule: $$ \delta x = \frac{\partial x}{\partial a}\delta a + \frac{\partial x}{\partial b}\delta b + \frac{\partial x}{\partial c}\delta c, $$ et cetera.
If I understand the situation, the (infinitesimally small) oblique parallelogram with coordinates $x$, $y$, $z$ gives the volume element at a time $t$. Consider the map $F_t \colon \mathbb{R}^3\to \mathbb{R}^3: (a,b,c)\mapsto(x,y,z)$ that gives the future position $(x,y,z)$ at time $t$ for a point with initial position $(a,b,c)$. Let $G_t$ be the inverse map. Then $F_t \circ G_t= id$.
Note that the matrices of the differentials $dF_t$ and $dG_t$ are $$ \begin{bmatrix} \frac{\partial x}{\partial a} & \frac{\partial y}{\partial a} & \frac{\partial z}{\partial a} \\ \frac{\partial x}{\partial b} & \frac{\partial y}{\partial b} & \frac{\partial z}{\partial b} \\ \frac{\partial x}{\partial c} & \frac{\partial y}{\partial c} & \frac{\partial z}{\partial c} \end{bmatrix} \quad \text{resp.} \quad \begin{bmatrix} \frac{\partial a}{\partial x} & \frac{\partial b}{\partial x} & \frac{\partial c}{\partial x} \\ \frac{\partial a}{\partial y} & \frac{\partial b}{\partial y} & \frac{\partial c}{\partial y} \\ \frac{\partial a}{\partial z} & \frac{\partial b}{\partial z} & \frac{\partial c}{\partial z} \end{bmatrix}. $$ Since $F_t$ and $G_t$ are inverse, these matrices are also each others inverses. Consequently their determinants $$ \frac{D(x,y,z)}{D(a,b,c)} \quad \text{and} \quad \frac{D(a,b,c)}{D(x,y,z)} $$ are inverses, so $\frac{D(a,b,c)}{D(x,y,z)}$ is $1$ as well.