📂Dynamics

Omega Limit Sets of Autonomous Systems

Definition

Given a metric space $X$ and a function $f : X \to X$, let’s assume that we have the following vector field represented as a differential equation: $$ \dot{x} = f(x) $$ Considering the flow $\phi ( t, x )$ of this vector field and a point $x_{0} \in X$, if there exists a sequence of times $\left\{ t_{i} \right\} \subset \mathbb{R}$ satisfying $$ \phi \left( t_{i} , x_{0} \right) \to x $$ when $t_{i} \to \infty$, then $ x \in X$ is called the omega limit point of $x_{0}$. The set of omega limit points of $x_{0}$ is called the omega limit set, denoted as $\omega \left( x_{0} \right)$.

Explanation

With the definition above unchanged except for replacing $t_{i} \to - \infty$, it becomes the alpha limit point, alpha limit set, denoted as $\alpha \left( x_{0} \right)$. Alpha and Omega, being the first and last letters of the Greek alphabet, intuitively make sense for the naming since alpha deals with the infinite past (beginning) with $\alpha \left( x_{0} \right)$ and omega with the infinite future (end) with $\omega \left( x_{0} \right)$. Limit points translate to accumulation points and differ from topological mathematics in that the flow given requires attention to whether time flows into the past or future. Naturally, most interests will lie with the future, namely the omega limit.

Given that our system is defined by ordinary differential equations, the omega limit set of $x_{0}$ starting from an initial position is likely to be a curve or a single point. Although it’s not impossible for the path made by a moving point to form an area with volume according to the definition of the vector field, it’s practically considered impossible.

Properties

Properties of the Omega Limit Set: ¹Assuming the whole space is a Euclidean space $X = \mathbb{R}^{n}$ and a point $p \in \mathcal{M}$ in a compact invariant set $\mathcal{M}$ is given in the flow $\phi_{t} ( \cdot )$:

• [1]: $\omega (p) \ne \emptyset$
• [2]: $\omega (p)$ is a closed set.
• [3]: $\omega (p)$ is invariant under the flow. That is, $\omega (p)$ is the union of orbits.
• [4]: $\omega (p)$ is a connected space.

These properties hold true for the alpha limit as well.

Proof of Property [1]

Let’s define the set $\Phi (p) \subset \mathcal{M}$ with respect to the sequence of times $\left\{ t_{k} \right\} \subset \mathbb{R}$ where $p \in \mathcal{M}$ and $\displaystyle \lim_{k \to \infty} t_{k} = \infty$. $$ \Phi (p) := \left\{ p_{k} : p_{k} = \phi_{t_{k}} (p) \right\} $$ Since $\mathcal{M}$ is compact, according to the Bolzano-Weierstrass Theorem, there exist a subsequence of $\Phi (p)$ converging to a point in $\omega (p)$, thus proving $\omega (p) \ne \emptyset$.

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Proof of Property [2]

It suffices to show that $\omega (p)^{c}$ is an open set. Taking an arbitrary point $q \notin \omega (p)$ outside of $\omega (p)$, there must exist a neighborhood $\mathcal{N} (q)$ that satisfies $$ \left\{ \phi_{t} (p) : t \ge T \right\} \cap \mathcal{N} (q) = \emptyset $$ for some $T > 0$, meaning that $q$ must be included in some open set that is disjoint with $\omega (p)$. Since $q$ was arbitrarily chosen from outside $\omega (p)$, $\omega (p)^{c}$ must be an open set.

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Proof of Property [3]

First, let’s assume that for all $q \in \omega (p)$ and $s \in \mathbb{R}$, there exists a $\phi_{s} ( q )$. Although this assumption requires proof and is possible, it seems to hold negligible significance for the discussion, thus it is omitted. $$ q \in \omega (p) \\ \widetilde{q} := \phi_{s} (q) $$ Taking a sequence of times $\left\{ t_{k} \right\} \subset \mathbb{R}$ that makes $\phi_{t_{k}} (p) \to q$ when $\lim_{k \to \infty } t_{k} = \infty$, we find that $$ \phi_{t_{k} + s} (p) = \phi_{s} \left( \phi_{t_{k} } (p) \right) \to \widetilde{q} $$ hence $\widetilde{q} \in \omega (p)$ and, therefore, $\omega (p)$ is invariant.

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Proof of Property [4]

Assuming $\omega (p)$ is not a connected space, there exist open sets $V_{1} , V_{2} \subset \mathbb{R}^{n}$ satisfying the following: $$ \omega (p) \subset V_{1} \cap V_{2} \\ \omega (p) \cap V_{1} \ne \emptyset \\ \omega (p) \cap V_{2} \ne \emptyset \\ V_{1} \cap V_{2} = \emptyset $$ Let’s denote by $K := \mathcal{M} \setminus \left( V_{1} \cup V_{2} \right)$. Given $V_{1}, V_{2}$, the orbit of $p \in \mathcal{M}$ must intersect both sides of $V_{1}, V_{2}$, and for all $T > 0$, there exists a $t > T$ satisfying $\phi_{t} (p) \in K$, so when $k \to \infty$, a sequence of times $\left\{ t_{k} \right\}$ that holds $t_{k} \to \infty$ while being $\phi_{t_{k}} (p) \in K$ can be chosen. This is because $V_{1}$ and $V_{2}$ cannot always stay separate and must cross through $K$ as they aren’t disjoint with $\omega (p)$. Since $V_{1} , V_{2}$ is an open set, and $K$ is compact, applying the Bolzano-Weierstrass Theorem warrants a subsequence of $\left\{ t_{k} \right\}$ converging to $q \in K$, which leads to $q \notin V_{1} \cup V_{2}$. However, this is contradictory to the definition of the omega limit set which requires $q \in \omega (p)$, thus $\omega (p)$ must be a connected space.

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