Show that the unit circle centered at the origin is an attractor of the system $\dot{x}=-y +(1-x^2-y^2)$, $\dot{y}=x+(1-x^2-y^2)$ Given:


$\dot{x}=-y +(1-x^2-y^2)$


$\dot{y}=x+(1-x^2-y^2)$

Definition given: 
A closed invariant set A ⊂ X is an attracting set under the flow if there
exists N(A) such that $\phi(x)_{t} ∈ N(A)$ for all $x ∈ N(A)$ and $t ≥ 0$, and

inf$_{y ∈ A} ∣ \phi_{t}(x) − y∣ → 0$ as $t → ∞, ∀x ∈ N(A)$.

If in addition, for some x ∈ A, the forward orbit O
+
(x) is a dense subset
of A, then we say that A is an attractor.

My Question:
Is there a way to use this definition of an attractor to show that the unit circle centered at the origin is an attractor for the above system? Or is my attempt sufficient?

My Attempt: Transformed to polar co-ordinates and found that:

$\dot{r}=r^2(1-r^2)$

$\dot{\theta}=1$

I then substituted $r=1+\delta$ (with $|\delta| &lt;&lt;1 $) into the equation for $\dot{r}$ to show that $\dot{r}&lt;0$ for $\delta&gt;0$ and $\dot{r}&gt;0$ for $\delta&lt;0$.
You might as well solve ( separation of variables) the diffeq in $r$ and see that $r(t) \to 1$ as $t \to \infty$ for any solution. That will guarantee that the circle is an attracting set.  Moreover, it's easy to see that if you start on the circle you get the solution covering the circle. Thus, all the checks are done.
If one sets

$R(r) = r^2 - r^4, \tag 1$

then

$\dot r = R(r) = r^2 - r^4, \tag 2$

and one sees that

$R(1) = 1^2 - 1^4 = 0 \tag 3$

and

$\displaystyle \dfrac{dR}{dr} = 2r - 4r^3; \tag 4$

thus

$\dfrac{dR(1)}{dr} = -2  &lt; 0; \tag 5$

thus one sees that $r = 1$ is a hypberbolic equilibrium of the differential equation (2) and thus by (5), since $dR(1)/dr &lt; 0$, it is attractive; thus $r(t) \to 1$  as $t \to \infty$ provided $r(t_0) \in (1 - \epsilon, 1 + \epsilon)$ for some $t_0 \in \Bbb R$ and $\epsilon$ sufficiently small; furthermore, since

$R(r) &lt; 0 \; \text{for} \; r &gt; 1, \tag 6$

and 

$R(r) &gt; 0 \; \text{for} \; r &lt; 1, \tag 7$

we see that the convergence of $r(t) \to 1$ is strictly monotonic.  One also notes that 

$\dot \theta = 1 \tag 8$

implies

$\theta(t) = t - t_0 + \theta(t_0); \tag 9$

then it is easily seen that the point $(r(t), \theta(t)) = (r(t), t - t_0 + \theta(t_0)) \to (1, t  - t_0 + \theta(t_0))$ as $t \to \infty$; indeed the Euclidean distance $d(t)$ 'twixt $ (r(t), t  - t_0 + \theta(t_0))$ and $(1, t - t_0 + \theta(t_0))$ is

$d(t) = \sqrt{(r(t) \cos \theta(t) - 1 \cdot \cos \theta(t))^2 + (r(t) \sin \theta(t) - 1 \cdot \sin \theta(t))^2}$
$= \sqrt{((r(t) - 1) \cos \theta(t))^2 + ((r(t) - 1) \sin \theta(t))^2} = \sqrt{(r(t) - 1)^2 (\cos^2 \theta(t) + \sin^2 \theta(t))}$
$= \sqrt{(r(t) -1)^2} = \vert r(t) - 1 \vert \to 0 \tag{10}$

as $t \to \infty$; it is a reasonably straightforward matter to see that such $d(t)$ is the minimum possible distance twixt $(r(t), \theta(t))$ and the circle $r = 1$, since it is their separation along a radial line.  Thus it may be concluded that the circle is an attracting set for the flow of the system (2), (8); and, as indicated by orangeskid in his answer, any trajectory originating in the circle $r = 1$ passes through every point in said circle; this is very easy to see, geometrically, since the point $(1, t  - t_0 + \theta(t_0))$ returns to the point $(1, \theta(t_0))$ for $t = t_0 + 2\pi$, covering the entire circle $r = 1$ in transit.

Thus one sees that the circle $r = 1$ is an attractor for the system (2), (8).