📂Dynamics

Proof of Poincaré bendixson Theorem

Theorem

$2$ Consider a manifold $\mathcal{P}$ and a function $f,g \in C^{r} \left( \mathcal{P} \right)$ such that the following vector field is given as a differential equation: $$\dot{x} = f(x,y) \\ \dot{y} = g(x,y)$$ If $\mathcal{M}$ represents an invariant set with a finite number of fixed points, then the omega limit set $\omega (p)$ of $p \in \mathcal{M}$ satisfies one of the following three conditions:

• (1): $\omega (p)$ is a singleton set, meaning it contains only one fixed point.
• (2): $\omega (p)$ is a closed orbit.
• (3): $\omega (p)$ consists of orbits $\gamma$ that satisfy the following for some fixed points $p_{1} , \cdots , p_{n}$ of $i,j \in [1,n]$: $$\alpha ( \gamma ) = \left\{ p_{i} \right\} \\ \omega ( \gamma ) = \left\{ p_{j} \right\}$$

Explanation

A metric space is naturally a $T_{1}$ space, and in $T_{1}$, singleton sets are closed sets, so it can be said that $\omega (p) = \left\{ p \right\}$ is naturally a closed orbit. However, in the context of the statement, let’s differentiate the case that includes only one fixed point as something else.

In fact, in the Poincaré-Bendixson theorem, chaos doesn’t even need to be defined, and the statement that chaos does not occur is close to a corollary. What the theorem tells us is just a classification of omega limit sets, and it is deduced that chaos cannot occur because it is precisely made up of what we know. However, due to such theorems, the interest in chaos theory is able to definitively move beyond the dimension $2$.

The intuitive understanding of the theorem is not very difficult. If $\mathcal{M}$ is not bounded, it cannot be chaotic in the first place, and if it is bounded, it cannot extend indefinitely, so the flow must either narrow down and rotate or spread out and rotate. However, unlike in the dimension $3$, in the dimension $2$, a line divides the plane into two regions, so one of the regions must always be abandoned. This is metaphorically like abandoning the remaining part of the bounded space $\mathcal{M}$ as time passes. If you try to pass through the flow you have already passed to use an area you have not passed yet, at that moment, it becomes a closed orbit, and eventually, it converges to a fixed point or a closed orbit, making it impossible to cause chaos.

Proof ¹

Strategy: As befitting a theorem named after Poincaré, it’s topologically oriented. Let $\Sigma$ be a continuous, connected arc in the interior $\mathcal{P}$.

If the normal vector at all points of $\Sigma$ and the vector field do not have an inner product of $0$ and do not change sign, then $\Sigma$ is said to transverse the vector field of $\mathcal{P}$. This concept can also be thought of for just one point, where the vector field and $\Sigma$ would not be tangent at that point. In terms of the flow’s sense, it not only meets at a point but also penetrates $\Sigma$.

Let’s represent the flow created by the given vector field as $\phi_{t}$, and represent the orbit of a point $p \in \mathcal{P}$ under the flow $\phi_{t}$ for positive time as $O_{+}(p)$. Let’s represent the orbit that a point $p_{i}$ reaches $p_{j}$ following the flow of time $t$ under the flow $\phi_{t}$ as $\widehat{p_{i} p_{j}} \subset O_{+} (p)$. The omega limit set denoted as $\omega ( \cdot )$ was originally defined for a given point, but the $\omega \left( X \right)$ for some set $X$ can be thought of as follows: $$\omega (X) := \bigcup_{x \in X} \omega (x)$$ The same applies to the alpha limit set denoted as $\alpha ( \cdot )$.

In addition, the following auxiliary theorems will be used continuously.

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

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

First, the omega limit sets that occur in dimension $2$ will not be in the form of having an area, so the omega limit sets mentioned below can be thought of in the form of some curve.

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Part 1.

If $\Sigma \subset \mathcal{M}$ is an arc that crosses the vector field, and since $\mathcal{M}$ is an invariant set in the dimension $2$ vector field, $\Sigma$ cannot go out of $\mathcal{M}$ against the flow of the vector field. Therefore, for any $p \in \mathcal{M}$, if we call the $k$th point where $O_{+} (p)$ and $\Sigma$ meet as $p_{k}$, then it must be $p_{k}\subset \widehat{p_{k-1} p_{k+1}} \subset O_{+} (p)$. In other words, the flow is converging to some inner core while meeting $\Sigma$, and it doesn’t happen that the intersection points get closer and then move away again.

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Part 2. The omega limit set $\omega (p)$ of $p \in \mathcal{M}$ intersects $\Sigma$ at most at one point.

This will be shown by contradiction. Assume that $\omega (p)$ and $\Sigma$ intersect at two different points $q , \overline{q}$.

Then, by the definition of the omega limit set, when $n \to \infty$, a sequence $\left\{ q_n \right\}_{n \in \mathbb{N}} , \left\{ \overline{q}_n \right\}_{n \in \mathbb{N}} \subset O_{+} (p)$ that satisfies $$q_{n} \to q \\ \overline{q}_{n} \to \overline{q}$$ exists. However, according to Part 1, these intersection points are arranged in a sequence $p_{1} , p_{2} , \cdots$, so there is a contradiction to the assumption. Therefore, $\omega (p)$ and $\Sigma$ either do not meet at all or meet at only one point if they do. [ NOTE: In the case of a torus, this logic cannot be applied directly, but the same conclusion can be reached by dividing it into pieces so that it becomes a shape like $\mathcal{M}$. ]

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Part 3. If $\omega (p)$ does not include fixed points, it is a closed orbit.

After showing that the orbit $O_{+}(q)$ of $q \in \omega (p)$ is a closed orbit, it needs to be shown that it is $\omega (p) = O_{+} (q)$.

• Part 3-1. Orbit $O_{+}(q)$ is closed.
  • If we pick a point $x \in \omega (q)$, according to auxiliary theorem [2], since $\omega (p)$ is closed and a union of orbits without fixed points, $x$ must also not be a fixed point. Be careful not to get confused with $p,q$, the assumption is that $\omega (p)$ does not have fixed points, and since $x$ is said to be $x \in \omega (q)$, there is no guarantee that it is necessarily $x \in \omega (p)$, but it can be concluded that it is not a fixed point anyway. Now, let’s take one arc $\Sigma_{x}$ that crosses the vector field of this non-fixed point $x$. According to Part 1., the sequence of intersections $\left\{ q_{n} \right\}_{n \in \mathbb{N}}$ of $\Sigma_{x}$ and $O_{+} (q)$ is $q_{n} \to x$ when $n \to \infty$, and since $x \in \mathcal{M}$, according to Part 2., for $\forall n \in \mathbb{N}$, it must be $q_{n} = x$. Since $x$ is not a fixed point, if $O_{+} (q)$ intersects with $x$, it must cross and then return to intersect again repeatedly. Since it is said here that $x \in \omega (q)$, $O_{+}(q)$ actually intersects with $x$ without stopping or pausing as it approaches $x$, and therefore, $O_{+}(q)$ becomes a closed orbit.
• Part 3-2. $O_{+}(q) = \omega (p)$
  • If we take one arc $\Sigma_{q}$ that crosses the vector field from point $q \in \omega (p)$, according to Part 2, $\omega (p)$ and $\Sigma_{q}$ meet only at $q$. According to auxiliary theorem [3], since $\omega (p)$ is a union of orbits, if $q \in \omega (p)$ then $O_{+} (q) \subset \omega (p)$, but since $\omega (p)$ does not contain fixed points and is a connected space, it must be exactly $O_{+}(q) = \omega (p)$.

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Part 4. For $p \in \mathcal{M}$, if $p_{1} , p_{2} \in \omega (p)$ are different fixed points of the vector field, there exists at most one orbit $\gamma \subset \omega (p)$ that satisfies $\alpha (\gamma) = \left\{ p_{1} \right\}$ and $\omega (\gamma) = \left\{ p_{2} \right\}$.

This will be shown by contradiction. If there are two different orbits connecting two points, there would be some area $\mathcal{K}$ with an area between the two orbits, and a contradiction will be derived from there. Assume that there exist two different orbits $\gamma_{1} , \gamma_{2} \subset \omega (p)$ satisfying the following conditions. $$\alpha \left( \gamma_{i} \right) = \left\{ p_{1} \right\} \\ \omega \left( \gamma_{i} \right) = \left\{ p_{2} \right\}$$ Let’s take one point each from these orbits, $q_{1} \in \gamma_{1}$, $q_{2} \in \gamma_{2}$, and take arcs that cross the vector field from $q_{1}$ and $q_{2}$ as $\Sigma_{1}, \Sigma_{2}$.

Since $\gamma_{1} , \gamma_{2} \subset \omega (p)$, according to Part 2, let’s say $O_{+} (p)$ intersects with $\Sigma_{1}$ at one point $a$ and then $\Sigma_{2}$ intersects at one point $b$. Then, in the dimension $2$ manifold, there will be a subregion $\color{red}{\mathcal{K}}$ surrounded by the following path.

$$q_{1} \overset{\Sigma_{1}}{\to} a \overset{ O_{+} (p) }{ \to } b \overset{\Sigma_{2}}{\to} q_{2} \overset{ \omega (\gamma) }{ \to } p_{2} \overset{ \gamma_{1} }{ \gets } q_{1}$$ The notation $\displaystyle x \overset{\mathcal{C}}{\to} y$ was used to mean that point $x,y$ was connected to curve $\mathcal{C}$. The flow starting from $\color{red}{\mathcal{K}}$ cannot go over $\gamma_{1} , \gamma_{2}$, so $\color{red}{\mathcal{K}}$ becomes an invariant set. However, the fact that the orbit $O_{+}(p)$ starting from $p$ cannot exit once it enters $\color{red}{\mathcal{K}}$ means that neither $\gamma_{1}$ nor $\gamma_{2}$ can belong to $\omega (p)$. For example, if you think about $\gamma_{2}$, $q_{2} \overset{\gamma_{2}}{\to} p_{2}$ may belong to $\omega (p)$, but it cannot go to the front part, $p_{1} \overset{\gamma_{2}}{\to} q_{2}$. Therefore, the claim that the entire $\gamma_{2}$ belongs to $\omega (p)$ cannot be made, and it contradicts $\gamma_{1} , \gamma_{2} \subset \omega (p)$.

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Part 5.

In this part only, let’s call a point that is not a fixed point a Regular Point. There’s no need to limit this to just this part, but since the expression Regular is often used in academia regardless of the field, using it without caution or warning can cause great confusion.

• Case 1. If $\omega (p)$ only has fixed points
  • Since $\mathcal{M}$ has a finite number of fixed points and $\omega (p)$ is a connected space, it must have only one fixed point.
• Case 2. If $\omega (p)$ only has regular points
  • According to Part 3, $\omega (p)$ is a closed orbit.
• Case 3. If $\omega (p)$ has both fixed and regular points
  • Consider the orbit $\gamma \subset \omega (p)$ consisting only of regular points.

Since $\gamma$ consists only of regular points, according to Part 3, $\omega ( \gamma )$ and $\alpha (\gamma)$ are closed orbits, but they must also have fixed points. However, according to auxiliary theorem [4], since $\omega ( \gamma )$ is a connected space, the closed orbit and the fixed point cannot be separated, and the fixed point must be located at one point on the closed orbit, which means $\omega ( \gamma )$ is a singleton set containing only fixed points. The same discussion can be repeated for $\alpha ( \gamma )$, so all regular points of $\omega (p)$ have fixed points as their omega and alpha limit points.

$\omega (p)$ must fall into one of these three cases. This concludes the proof.

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