Understanding partial derivatives computation issued from an implicit change of variables.

Question

Find $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}$, if $xu-yv=0$ and $yu+xv=1$
Actually, I have an answer but I don't understand anything. Not because it is Chinese but transformations. Could you explain?

Answer 1

The first set of equations:

$$\begin{cases}xu-yv&=&0\\yu+xv&=&1\end{cases} \tag{0}$$

can be differentiated under the form:

$$\begin{cases}(udx+xdu)-(vdy+ydv)&=&0\\(u dy +y du)+(vdx+xdv)&=&0\end{cases},$$

giving the equivalent system:

$$\begin{cases}xdu-ydv&=&-udx+vdy\\ ydu+xdv&=&-vdx-udy\end{cases} $$

as in your book. Now, write it with matrices and vectors:

$$\underbrace{\begin{pmatrix}x&-y\\y&x \end{pmatrix}}_A\begin{pmatrix}du\\dv \end{pmatrix}=\underbrace{\begin{pmatrix}-u&v\\-v&-u\end{pmatrix}}_B \begin{pmatrix}dx\\dy \end{pmatrix}$$

from which you deduce that :

$$\begin{pmatrix}du\\dv \end{pmatrix}=\underbrace{A^{-1}B}_J \begin{pmatrix}dx\\dy \end{pmatrix}$$

It remains to compute matrix $J$:

$$J=\dfrac{1}{x^2+y^2}\begin{pmatrix}x&y\\-y&x \end{pmatrix}\begin{pmatrix}-u&v\\-v&-u\end{pmatrix}$$

$$J=\dfrac{1}{x^2+y^2}\begin{pmatrix}(-ux-vy)&(vx-uy)\\(uy-vx)&(-ux-vy)\end{pmatrix}\tag{1}$$

Remark: We have used the classical formula for the inverse of a $2 \times 2$ matrix.

$J$ is the jacobian matrix of the transformation $(x,y) \to (u,v)$, whose entries are precisely:

$$J=\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y} \end{pmatrix}\tag{2}$$

It remains to identify the corresponding entries in (1) and (2).

Edit : important remark: My attention has been triggered by the fact that:

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \ \ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

which are called "Cauchy-Riemann equations", classical for analytical ("smooth") complex functions.

Nothing surprizing because the initial relationships can be written plainly :

$$(x+iy)(u+iv)=i \ \iff \ u+iv=\dfrac{i}{x+iy}$$

finally giving

$$u+iv=\dfrac{i(x-iy)}{(x+iy)(x-iy)}=\dfrac{y+ix)}{x^2+y^2}$$

Otherwise said, identifying real and imaginary parts, we have explicit expressions:

$$u=\dfrac{y}{x^2+y^2}, \ \ \ v=\dfrac{x}{x^2+y^2},\tag{3}$$

from which the partial derivatives are easily obtained...

(transformation (3) is a parent of the classical "inversion" transform)