Omega and alpha limit set of unstable spiral

Question

I have a set of nonlinear equations in polar coordiantes:

$$\dot{r}=r+r^2$$ $$\dot{\theta }=1/2 $$

What I know: the equilibrium is at $(0, 0)$, and it is an unstable spiral (anticlockwise). What would the $\omega$-limit and $\alpha$-limit sets be in this instance? I think $\alpha$ is the point $(0,0)$ itself, but is $\omega$ 'infinity' as it is spiralling outward, or is it too $(0,0)$ as it will stay there for all negative time? Any insight appreciated.

Answer 1

The $\omega$-limit and $\alpha$-limit are defined individually for each point.

• As $(0,0)$ is a fixed point, $\omega$ and $\alpha$ limits will be both $(0,0)$.

• The points $\neq (0,0)$ will have $\alpha$ limit $(0,0)$ and $\omega$ limit $\infty$.

Justification: Notice that, as you said, it's spiralling outward because $\dot{r} = r + r^2$ is always positive (because $r$ is positive) and $\dot{\theta}$ is contant with value $1/2$.