Can a spiral be represented as $F(x)+G(y)=0$

Question

Can you characterise a spiral in $\mathbb R^2$ by the equation $F(x)+G(y)=0$, where $F$ and $G$ are some continuous real-valued functions.

A spiral is any continuous parametric curve $(u(t),v(t)),t\in \mathbb R_+$ starting in the origin $(0,0)$ and such that the "angle" of $(u(t),v(t))$ relative to the origin is continuous and increasing. (I omit using polar coordinate, which makes the formulation of the angle somewhat vague. My intention is to allow for consideration of a spiral consisting of straight segments.)

Related question discussing characterisation of closed curves as $F(x)+G(y)=0$: Closed curves of the form $F(x)+G(y)=0$

Answer 1

No.

The complement of the spiral is connected, hence $F(x)+G(x)$ cannot change sign there. So wlog. $F(x)+G(x)\ge 0$ for all $x,y$. Then for every $y\in \Bbb R$, the $x\in \Bbb R$ with $(x,y)$ on the spiral are precisely the global minima of $F$. These global minima of $F$ do not depent on $y$, of course. But then the figure consists only if straight vertical lines and is not a spiral.

Answer 2

You can still have spiraling curves in a certain sense of the word, for example with

$$x+x \sin(x)=y+y \cos(y)$$

as you can see on the following graphical representation (Desmos plotting):