Understanding the Inverse of Jacobian Matrices: A Brief Overview

[parsesnip]

Homework Statement

I want to prove that $\left| \frac{\partial(x,y)}{\partial(u,v)} \right|=\frac{1}{\left|\frac{\partial(u,v)}{\partial(x,y)}\right|}$ (If this is true)

Relevant Equations

$\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix} x_u & x_v\\ y_u&y_v \end {vmatrix}$
$\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix} u_x & u_y\\ v_x&v_y \end {vmatrix}$

I got that ${x_u}{y_v}-{x_y}{y_u}=$$\frac{1}{\frac{1}{{x_u}{y_v}}-\frac{1}{{y_u}{x_v}}}$. But this implies that ${x_u}{x_v}{y_u}{y_v}=-1$ and I don't see how that is true?

 

[anuttarasammyak]

Very simple case of u=x, v=y obviously the statement stands. but
x_u x_v y_u y_v = 1*0*0*1=0
So your result seems wrong.

Find 2X2 matrices A, B by partial differentiation
<br /> \begin{pmatrix}<br /> du \\<br /> dv \\<br /> \end{pmatrix}<br /> <br /> = A<br /> <br /> \begin{pmatrix}<br /> dx \\<br /> dy \\<br /> \end{pmatrix}

<br /> \begin{pmatrix}<br /> dx \\<br /> dy \\<br /> \end{pmatrix}<br /> <br /> = B<br /> <br /> \begin{pmatrix}<br /> du \\<br /> dv \\<br /> \end{pmatrix}

So
AB=BA=E
det\ A\ \ det\ B = 1

 

[pasmith]

By the chain rule <br /> 1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial x} and <br /> 0 = \frac{\partial y}{\partial x} = \frac{\partial y}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x} and similarly \frac{\partial y}{\partial y} = 1 and \frac{\partial x}{\partial y} = 0. Now I suspect that if you expand <br /> 1 = \frac{\partial x}{\partial x}\frac{\partial y}{\partial y} - \frac{\partial y}{\partial x}\frac{\partial x}{\partial y} and rearrange it then you will obtain your result.

 

[parsesnip]

By the chain rule <br /> 1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial x} and <br /> 0 = \frac{\partial y}{\partial x} = \frac{\partial y}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x} and similarly \frac{\partial y}{\partial y} = 1 and \frac{\partial x}{\partial y} = 0. Now I suspect that if you expand <br /> 1 = \frac{\partial x}{\partial x}\frac{\partial y}{\partial y} - \frac{\partial y}{\partial x}\frac{\partial x}{\partial y} and rearrange it then you will obtain your result.

I think I understand. But if $\frac {\partial x}{\partial u}=\frac{1}{\frac {\partial u}{\partial x}}$, then doesn't $\frac {\partial x}{\partial u}\frac {\partial u}{\partial x}=1$? Then how do you resolve the contradiction that $1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial x} = 2$?

 

[pasmith]

I think I understand. But if $\frac {\partial x}{\partial u}=\frac{1}{\frac {\partial u}{\partial x}}$, then doesn't $\frac {\partial x}{\partial u}\frac {\partial u}{\partial x}=1$? Then how do you resolve the contradiction that $1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial x} = 2$?

The relation \frac{\partial x}{\partial u} = \left(\frac{\partial u}{\partial x}\right)^{-1} does not hold in general. For example, with plane polar coordinates we have \frac{\partial r}{\partial x} = \frac xr and \frac{\partial x}{\partial r} = \cos \theta = \frac xr \neq \frac rx.

Rather, if that relation does hold then it implies that x is independent of v so that \frac{\partial x}{\partial v} = 0.

 

[anuttarasammyak]

Further to my post #2

A=<br /> \begin{pmatrix}<br /> u_x &amp; u_y \\<br /> v_x &amp; v_y \\<br /> \end{pmatrix}
B=<br /> \begin{pmatrix}<br /> x_u &amp; x_v \\<br /> y_u &amp; y_v \\<br /> \end{pmatrix}
AB=<br /> \begin{pmatrix}<br /> u_x &amp; u_y \\<br /> v_x &amp; v_y \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> x_u &amp; x_v \\<br /> y_u &amp; y_v \\<br /> \end{pmatrix}<br /> =<br /> \begin{pmatrix}<br /> 1 &amp; 0 \\<br /> 0 &amp; 1 \\<br /> \end{pmatrix}

It meets with post #3.

A=B^{-1}
<br /> \begin{pmatrix}<br /> u_x &amp; u_y \\<br /> v_x &amp; v_y \\<br /> \end{pmatrix}<br /> =\frac{1}{x_uy_v-x_vy_u}<br /> \begin{pmatrix}<br /> y_v &amp; -x_v \\<br /> -y_u &amp; x_u \\<br /> \end{pmatrix}

For an example (1,1) component says

u_x=\frac{1}{x_uy_v-x_vy_u}y_v

You see in order $u_x =\frac{1}{x_u}$ as you expect, $x_vy_u=0$ and $y_v \neq 0$ is required.