06 - Jacobian Matrix

Jacobian Matrix

Suppose $x=x(u,v)$ and $y=y(u,v)$
We have a transformation o coordinates

$$(x,y)\to (u,v)$$

By Chain Rule then

$$\left(\begin{array}{cc}\frac{\partial f}{\partial u}& \frac{\partial f}{\partial v}\end{array}\right)=\left(\begin{array}{cc}\frac{\partial f}{\partial x}& \frac{\partial f}{\partial y}\end{array}\right)\left(\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ & \\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right)$$

where the Jacobian Matrix is

$$\left(\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ & \\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right)$$

Jacobian

$$\frac{\partial (x,y)}{\partial (u,v)}=\left|\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right|=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}$$

Note that the Jacobian is equivalent to an area scale factor

Property of Jacobians

Let $r,s$ be functions of variables $u,v$ which are functions of $x,y$
Then

$$\frac{\partial (r,s)}{\partial (x,y)}=\frac{\partial (r,s)}{\partial (u,v)}\cdot \frac{\partial (u,v)}{\partial (x,y)}$$

Proof

$$\begin{gathered} \begin{array}{rl}\frac{\partial (r,s)}{\partial (x,y)}& =\left| \\ \begin{array}{cc}\frac{\partial r}{\partial x}& \frac{\partial r}{\partial y}\\ \frac{\partial s}{\partial x}& \frac{\partial s}{\partial y}\end{array}\right|\\ & \\ & =\left|\begin{array}{cc}\frac{\partial r}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial r}{\partial v}\frac{\partial v}{\partial x}& \frac{\partial r}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial r}{\partial v}\frac{\partial v}{\partial y}\\ & \\ \frac{\partial s}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial s}{\partial v}\frac{\partial v}{\partial x}& \frac{\partial s}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial s}{\partial v}\frac{\partial v}{\partial y}\end{array}\right|\\ & =\left|\left(\begin{array}{cc}\frac{\partial r}{\partial u}& \frac{\partial r}{\partial v}\\ \frac{\partial s}{\partial u}& \frac{\partial s}{\partial v}\end{array}\right)\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right)\right|\\ & =\left|\begin{array}{cc}\frac{\partial r}{\partial u}& \frac{\partial r}{\partial v}\\ \frac{\partial s}{\partial u}& \frac{\partial s}{\partial v}\end{array}\right|\left|\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right|\\ & =\frac{\partial (r,s)}{\partial (u,v)}\cdot \frac{\partial (u,v)}{\partial (x,y)}\end{array} \end{gathered}$$

"Inverse"-like result of Jacobians corollary

$$\frac{\partial (x,y)}{\partial (u,v)}\frac{\partial (u,v)}{\partial (x,y)}=1$$

Proof

Take $r=x$ and $s=y$ in the proof above

Note that the stronger result holds as

$$\left(\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right)\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$