# Understanding the Poincare-Bendixson theorem.

I'm trying to understand applications of the Poincaré-Bendixson theorem. The Poincaré-Bendixson theorem goes as follows:

Poincaré-Bendixson Theorem: Consider the equation $$\dot{x} = f(x)$$ in $$\mathbb{R}^2$$ and assume that $$\gamma^+$$ is a bounded, positive orbit and that $$\omega(\gamma^+)$$ contains ordinary points only. Then $$\omega(\gamma^+)$$ is a periodic orbit. If $$\omega(\gamma^+)\neq \gamma^+$$ the periodic orbit is called a limit cycle. An analogous result is valid for a bounded, negative orbit.

(Here $$\omega(\gamma^+)$$ is called the $$\omega$$-limitset and denotes the set of positive limitpoints of $$\gamma^+$$.)

In the example below the Bendixson criterium is used as well:

Bendixson's Criterium: Suppose that the domain $$D\subset \mathbb{R}^2$$ is simply connected; $$(f,g)$$ is continuously differentiable in $$D$$. The equation $$\dot{x} = f(x)$$ can only have periodic solutions if $$\nabla .(f,g)$$ changes sign in $$D$$ or if $$\nabla . (f,g) = 0$$ in $$D$$.

The Poincaré-Bendixson theorem is illustrated in the following example:

Example: Consider the following system \begin{align} \dot{x} &= x(x^2 + y^2 -2x - 3) - y\\ \dot{y} & = y(x^2 + y^2 - 2x -3) + x \end{align} The only critical point is $$(0,0)$$; this is a spiral point with positive attraction. To see whether closed orbits are possible apply the Bendixson criterium. We find for the divergence of vector function on the righthand-side $$4x^2 + 4y^2 -6x - 6 = 4[(x - \frac{3}{4})^2 + y^2 -\frac{33}{16}]$$ Inside the Bendixson circle with centre $$(\frac{3}{4}, 0)$$ and radius $$\sqrt{33}/4$$ the expression is sign definite and no closed orbits can be contained in the interior of this circle.

Closed orbits are possible which enclose or which intersect the Bendixson-circle. We transform the system to polar coordinates by $$x = r\cos\theta$$ and $$y = r\sin\theta$$ to find \begin{align} \dot{r} &= r(r^2 - 2r\cos\theta -3)\\ \dot{\theta} &= 1 \end{align} If $$r<1$$ we have $$\dot{r} < 0$$, if $$r > 3$$ we have $$\dot{r} > 0$$. According to the Poincaré-Bendixson theorem the annulus $$1 must contain one or more limit cycles.

Question:

Why does the Poincaré-Bendixson theorem show that there have to be limit cycles inside the annulus $$1 < r< 3$$? I understand how the Bendixson criterium is used and why there cannot be any closed orbits inside the circle with centre $$(\frac{3}{4}, 0)$$ and radius $$\sqrt{33}/4$$, but I don't get why the fact that $$\dot{r} < 0$$ when $$r > 1$$ and $$\dot{r} > 0$$ when $$r<3$$ insinuates that the are limit cycles inside the annulus.

After applying the time change $$t\mapsto -t$$, the conditions on the sign of $$\overset{\cdot}{r}$$ imply that the annulus $$S=\{1\leq r\leq 3\}$$ is a "trapping region" for the flow, i.e. the flow stays in this compact set (in other words, this ensures that the negative orbits in this region are bounded). By observing that there are no critical points in $$S$$, one can invoke the Poincare-Bendixson theorem to deduce the existence of a limit cycle in the given annulus. Specifically, the absence of fixed points implies that the (negative) omega limit set of any orbit in this region contains a periodic orbit.
• Thanks for your answer! Could you elaborate a bit more though on why the conditions on the sign of $\dot{r}$ imply that the annulus is a trapping region for the flow? Commented Apr 2, 2019 at 7:35