Inverse of Jacobian matrices

[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
$$\begin{pmatrix} du \\ dv \\ \end{pmatrix} = A \begin{pmatrix} dx \\ dy \\ \end{pmatrix}$$

$$\begin{pmatrix} dx \\ dy \\ \end{pmatrix} = B \begin{pmatrix} du \\ dv \\ \end{pmatrix}$$

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

 

[pasmith]

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

 

[parsesnip]

By the chain rule $$1 = \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial x}$$ and $$0 = \frac{\partial y}{\partial x} = \frac{\partial y}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x}$$ and similarly $\frac{\partial y}{\partial y} = 1$ and $\frac{\partial x}{\partial y} = 0$. Now I suspect that if you expand $$1 = \frac{\partial x}{\partial x}\frac{\partial y}{\partial y} - \frac{\partial y}{\partial x}\frac{\partial x}{\partial y}$$ 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

$$A= \begin{pmatrix} u_x & u_y \\ v_x & v_y \\ \end{pmatrix}$$
$$B= \begin{pmatrix} x_u & x_v \\ y_u & y_v \\ \end{pmatrix}$$
$$AB= \begin{pmatrix} u_x & u_y \\ v_x & v_y \\ \end{pmatrix} \begin{pmatrix} x_u & x_v \\ y_u & y_v \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$

It meets with post #3.

$$A=B^{-1}$$
$$\begin{pmatrix} u_x & u_y \\ v_x & v_y \\ \end{pmatrix} =\frac{1}{x_uy_v-x_vy_u} \begin{pmatrix} y_v & -x_v \\ -y_u & x_u \\ \end{pmatrix}$$

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.