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The system of ODE is

$$\begin{cases} \dot{x}=-y(1-x^{2}) \\ \dot{y}=x+y(1-x^{2}) \end{cases} \tag{$\ast$}$$

Claim: $\forall p\in\left\{ (x,y)\in\mathbb{R}^{2} : |x|<1,\ x^{2}+y^{2}>0\right\} $, $$\omega(p)=\left\{ (-1,y) : y\in\mathbb{R}\right\} \cup\left\{ (1,y) : y\in\mathbb{R}\right\}.$$

I understand that the following are solutions to the system $(\ast)$:

$$x=1 \text{ and } y=t$$

$$x=-1 \text{ and } y=-t$$

When we talk about omega limit set, aren't we supposed to make $t$ go to infinity? Why do we have $(-1,y)$? As $t\rightarrow\infty$, shouldn't $y\rightarrow \infty$?

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1 Answer 1

up vote 1 down vote accepted

Yes, the $\omega$-limit set is the set of accumulation points of the orbit as $t\to\infty$, so for $x=\pm 1$ and $y$ arbitrary, this accumulation set is empty because all orbits diverge to $\infty$. However, the points $(\pm 1, y)$ are not in the set of allowed initial values for $p$ that you are given, since $|x|<1$ is violated.

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Yes. I agree. But does it mean the claim is wrong. If not, how do they find that omega limit set? –  user38335 Dec 4 '12 at 12:04
    
No, it means nothing about the claim, since the claim does not say anything about solutions with $x=\pm 1$. How to prove or disprove the claim is a different question. –  Lukas Geyer Dec 4 '12 at 14:22

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