Exercise :
Given the dynamical system : $$x_1' = x_2+2μx_1(5-x_1^2-x_2^2)$$ $$x_2' = -x_1+2μx_2(5-x_1^2-x_2^2)$$ where $(x_1,x_2) \in \mathbb R^2$ and $μ>0$ a constant. Applying polar coordinates, determine the omega limit set $ω(x_0)$ for any given vector $(x_1,x_2)$.
Discussion :
I used the polar coordinates substitution :
$$x_1 = r\cosθ$$ $$x_2 = r\sinθ$$
and via the expressions :
$$rr' = x_1x_1' + x_2x_2'$$
$$θ' = \frac{x_1x_2' - x_2x_1'}{r^2}$$
I derived the system in polar coordinates :
$$r' = 2μr(5-r^2)$$ $$θ' = -1$$
Now, what we can see is that the sign of $r'$ entirely depends on the factor $5-r^2$, since by it's definition $r>0$ and $μ>0$. This means that until $r=\sqrt5$ $r'>0$ and after that $r'<0$. Also, noting that $θ'=-1$ means that the direction of the phase portrait flow follows an anticlockwise flow.
Other than that, defining an one dimension phase portrait for $r'$ and noting as a right arrow for where $r'>0$ and left for $r'<0$, we get :
$$--- \quad 0 \quad \rightarrow \sqrt{5} \leftarrow$$
This means that there is a circle defined with $r=\sqrt{5}$.
This implies that the omega limit set for any given values, will be :
$$\text{For} \space x_0 \neq 0 \space \rightarrow \space ω(x_0) = S_{\sqrt5}$$
$$\text{For} \space x_0 = 0 \space \rightarrow \space ω(x_0) = \{0\}$$
Question : Is the above approach correct ?