Daily Archives: August 9, 2018

The Jacobian of the inverse of a transformation

Posted on by

The Jacobian of the inverse of a transformation is the inverse of the Jacobian of that transformation

.

In this post, we would like to illustrate the meaning of

the Jacobian of the inverse of a transformation = the inverse of the Jacobian of that transformation

by proving a special case.

.

Consider a transformation \mathscr{T}: \bar{x}^i=\bar{x}^i (x^1,x^2), which is an one-to-one mapping from unbarred x^i‘s to barred \bar{x}^i coordinates, where i=1, 2.

By definition, the Jacobian matrix J of \mathscr{T} is

J= \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{pmatrix}

.

Now we consider the the inverse of the transformation \mathscr{T}:

\mathscr{T}^{-1}: x^i=x^i(\bar{x}^1,\bar{x}^2)

By definition, the Jacobian matrix \bar{J} of this inverse transformation, \mathscr{T}^{-1}, is

\bar{J}= \begin{pmatrix} \displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^1}{\partial \bar{x}^2}} \\ \displaystyle{\frac{\partial x^2}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^2}{\partial \bar{x}^2}} \end{pmatrix}

.

On the other hand, the inverse of Jacobian J of the original transformation \mathscr{T} is

J^{-1}=\displaystyle{\frac{1}{ \begin{vmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{vmatrix} }} \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} & \displaystyle{-\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{-\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} \end{pmatrix}

.

If \bar{J} = J^{-1}, their (1, 1)-elementd should be equation:

\displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}}\stackrel{?}{=}\displaystyle{\frac{1}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}-\displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} }} \bigg( \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \bigg)

Let’s try to prove that.

.

Consider equations

\bar{x}^1 = \bar{x}^1(x^1,x^2)

\bar{x}^2 = \bar{x}^2(x^1,x^2)

Differentiate both sides of each equation with respect to \bar{x}^1, we have:

A := 1=\displaystyle{\frac{\partial \bar{x}^1}{\partial \bar{x}^1}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}

B := 0 = \displaystyle{\frac{\partial \bar{x}^2}{\partial \bar{x}^1}=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}

.

A \times \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}:~~~~~C := \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}}

B \times \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}:~~~~~D := \displaystyle{0=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}}

.

D-C:

\displaystyle{ \frac{\partial \bar{x}^2}{\partial x^2}= \bigg( \frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial \bar{x}^1}{\partial x^2}\bigg) \frac{\partial x^1}{\partial \bar{x}^1}},

results

\displaystyle{ \frac{\partial x^1}{\partial \bar{x}^1}}=\frac{\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial \bar{x}^2}{\partial x^1}}}

…

— Me@2018-08-09 09:49:51 PM

.

.

2018.08.09 Thursday (c) All rights reserved by ACHK