$\alpha$-limit set in $A=\{x\in \mathbb{R}^2|1<|x|<2\}$

Question

Consider the system $$ \dot{x}=-y+x(r^4-3r^2+1)\\ \dot{y}=x+y(r^4-3r^2+1) $$ where $r^2=x^2+y^2$.

Let $A=\{x\in \mathbb{R}^2|1<|x|<2\}$.

Affirmation (fait accompli): $\dot{r}<0$ on the circle $r=1$ and $\dot{r}>0$ on the circle $r=2$.

My question (doubt) is: Why the affirmation above implies that the $\alpha$-limit set any trajectory that starts in $A$ is in $A$?

Thank You!

Answer 1

Considering

$$ \left\{ \begin{array}{rcl} \dot{x}&=&-y+x(r^4-3r^2+1)\\ \dot{y}&=&x+y(r^4-3r^2+1) \end{array} \right. \Rightarrow \left\{ \begin{array}{rcl} x\dot{x}&=&-x y+x^2(r^4-3r^2+1)\\ y\dot{y}&=&xy+y^2(r^4-3r^2+1) \end{array} \right. \Rightarrow\frac 12\left(r^2\right)' = r^2(r^4-3r^2+1) $$

Now analyzing

$$ \frac 12\left(r^2\right)' = r^2(r^4-3r^2+1) $$

we conclude that for $r^2 = \frac 12\left(3\pm \sqrt 5\right)\;$ we have $\frac 12\left(r^2\right)' = 0\Rightarrow r_1 = \sqrt{\frac 12\left(3-\sqrt 5\right)}$ and $r_2 = \sqrt{\frac 12\left(3+ \sqrt 5\right)}\;$ are orbits.

We know that $r_1 < 1 < r_2 < 2\;$ and also we know that orbits in the phase plane do not cross. Those arguments suffice. Attached a stream plot for the system.

NOTE

We can also conclude from the signal of $(r^4-3r^2+1)$ the kind of attractiveness involved.