Let's say, $$f(x,y)=x^2+y^2=1$$ gives the unit circle.
Now I would like to get a smooth distortion on the circle with a function $g(x,y)$.
my guess is to consider the perimeter as one dimension, so I should found a function like :
$F(X)$ : $0 \le X \le 2 \pi$, $F(0)=F(\pi)$, $F'(0)=F'(2 \pi)$ (IRC I should take in account the first derivative, maybe the second derivative to be sure to have something smooth)
Let's say, $$F(X) = \cos(5 \cdot X) $$ so I will get no problem at the borders. I imagine I will get something like a flower when applied to $f(x,y)$.
Now my problem is to link the cartesian system with my (?) parametric system (?).
This question is for helping me to solve a problem in blender.
It seems you use $X$ to mean the polar angle. It would be less confusing to pick another letter, like $t$ or $s$. More importantly, the range should be $0$ to $2\pi$, not to $\pi$.
In terms of angle $t$, the parametric equations of the circle are: $$x = \cos t,\quad y=\sin t,\quad 0\le t\le 2\pi \tag1$$
Apparently, you want to distort the circle by pushing its point by something like $\cos 5t$. Most natural direction to push is away from the center, that is along the vector $(\cos t, \sin t)$ This means adding $d\cos 5t (\cos t,\sin t)$ to the parametric equations of the circle, where $d$ is the maximal distance by which to push. Result:
$$x = \cos t + d \cos 5t \cos t,\quad y=\sin t+d\cos 5t\sin t,\quad 0\le t\le 2\pi \tag2$$
With $d=0.2$, I get this (in FooPlot):
It would be extremely difficult and impractical to eliminate $t$ from equations (2). Hopefully, Blender can accept parametric equations: they are actually easier to draw than Cartesian equations such as $x^2+y^2=1$.