Poincaré-Bendixson theorem in $\mathbb{R}^1$

Question

I have a question about a problem in Teschl's ODE&DS book.

Consider a continuous dynamical system induced by an autonomous ODE $\dot{x} = f(x)$, $f \in C^1(M, \mathbb{R}^2)$ where $M$ is an open subset of $\mathbb{R}^2$.

Poincaré-Bendixson theorem says that if the $\omega$-limit set $\omega_{\sigma}(x)$ of $x \in M$ is nonempty compact connected and contains only finitely many fixed points, then $\omega_{\sigma}(x)$ is a fixed orbit, or a regular periodic orbit, or consists of finitely many fixed points and non-closed orbits.

Problem 7.9 asks me to prove a one-dimensional Poincare-Bendixson theorem. I have no idea about this. I just guess that

• There cannot be any regular periodic orbit since the orbit must pass a fixed point by the mean value theorem, which seems absurd.

• If there are finitely many fixed points of $f$, then I guess that $\omega_{\sigma}(x)$ is a subset of the collection of fixed points, with size $1$ (i.e. a singleton), and non-closed orbits $\gamma (y)$ with $\omega_{\sigma}(y)$ is a subset of the collection of fixed points, with size $1$.

• If there is not a fixed point of $f$, then I guess that $\omega_{\sigma}(x)$ is an empty set $\gamma_{\sigma}(x)$.

I'm not sure about what the author wants in this question. What does a 'Poincaré-Bendixson theorem' in $\mathbb{R}^1$ mean?

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edited:

Answer 1

Your last question asks what the author means by a Poincaré–Bendixson theorem on the real line—but you're on the right track: it's asking you to categorise $\omega$-limit sets, whenever they reasonably exist.

Notice how, given your $\mathcal C^1$ ODE on some open interval $U \subset \mathbb R$,

$$\dot x = f(x),$$

treating $\mathbb R = \mathbb R \times \{0\}$, you can always enlarge this to an ODE on the plane: i.e.,

$$\ \dot x = f(x), \quad \dot y = 0.$$

So P-B's theorem in 2D certainly applies here (you can project down to 1D again as you like): an omega limit set must be a homoclinic orbit, a fixed point, or a periodic loop.

Further, as you almost-correctly hinted at, only one of these three are possible on the real line (e.g. periodic orbits are ruled out by the MVT and the uniqueness-of-solutions, as you rightly pointed out). Indeed, some of the restrictions to the 2D P.-B. can be relaxed.

(Ruling out homoclinic tangencies is a little trickier. To start: how can your orbit get close to all of the points on the homoclinic tangency? Only by being on it. . .)

However, it is impossible to have more than one fixed point in the omega limit set without there being a homoclinic tangency, since the statement of P-B does not have this as an option (in the 2D version).

In sum, your statement would read something like this: given a $\mathcal C^1$ ODE on an open set $U$ in $\mathbb R$, then if $x \in U$ and $\omega(x)$ is non-empty, then it consists of only one fixed point.

(Yes, that may have been what you were trying to say in the above.)

Answer 2

I suppose that the idea in the question is that you should digest the definition of what it means to be an $\omega$-limit and conclude that in one dimension the $\omega$-limit must be trivial, either empty or a point. Hint: Let $x(t), t\geq 0$ be the trajectory of $x$. If $y,z \in \omega(x)$ then there are time sequences $(t_n)_n, (t'_n)_n$ with $t_n, t'_n\rightarrow +\infty$ as $n\rightarrow +\infty$ such that $$ \lim_n x(t_n) = y \ \ \ \mbox{and} \ \ \ \lim_n x(t'_n) = z. $$ Show that in one dimension this is impossible unless $y=z$. It is very simple, and using the 2D-Poincare-Bendixson in this context would be like using a sledgehammer to crack a nut.