dealing with a function $f$ that is a radial distribution.

Question

How do you show that if the function $f$ is a radial distribution then

$x\frac{\partial f}{\partial y}-y\frac{\partial f}{\partial x}=0$?

Answer 1

Consider polar coordinates $x=r\cos(\theta)$, $y=r\sin(\theta)$. Now $f$ is radial if and only if $f_\theta=0$; the chain rule says $$f_\theta = f_x\frac{\partial x}{\partial\theta}+f_y\frac{\partial y}{\partial\theta}=-y f_x+xf_y.$$

So the correct version of the assertion is that $f$ is radial if and only if $yf_x-xf_y=0$.

Answer 2

Computing the partial derivatives to polar coordinates we get $(x^{2}-y^{2})/r f_r=0$. So what is true is $f(r,\theta)$ depends only on $\theta$. A radial function is one which depends only on $r$ so the statement is wrong.

On the other hand $f$ is radial iff the partial derivative w.r.t. $\theta $ is $0$ and this transaltes to $xf_y-yf_x=0$.