How to determine the existence of limit cycle?

Question

I'm student learning about dynamical system. I understood well how to find fixed points and determine stability thanks to eigenvalues of the jacobian matrix, but not how to find limit cycle...

I heard about Poincaré–Bendixson theorem, but it remain totally unclear how to applied it in "pratical" way...

Thanks

Answer 1

The Poincare-Bendixson theorem, states that :

Theorem (Poincare-Bendixson) : Given a differentiable real dynamical system defined on an open subset of the plane, then every non-empty compact $\omega-$limit set of an orbit, which contains only finitely many fixed points, is either :

• a fixed point
• a periodic orbit
• or a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting these.

Moreover, there is at most one orbit connecting different fixed points in the same direction. However, there could be countably many homoclinic orbits connecting one fixed point.

Essentialy, the theorem tells us that if there exists an orbit that "bounds" the derivative of a corresponding dimesnion of a system and includes finitely many fixed points, it has some certain characteristics regarding the properties yielded.

I think the best way to get an understanding over the existence of $\omega$ limit sets and limit cycles on real dynamical systems, the best way is to elaborate some examples. Read carefully through the following :

Exercise :

Consider the following differential system, given in polar coordinates: $$r' = r(μ-r^2)$$ $$θ' = ρ(r^2)$$ where $μ>0$ is a constant and $p: \mathbb R_+ \to \mathbb R$ smooth function with $ρ(μ) > 0$. Show that there exists a unique limit cycle.

Solution :

The interesting part of the given system, is the expression $r(μ-r^2)$. Recall that since we are working over polar coordinates, it is $r>0$. Thus, for the given expression, it is :

$$r(μ-r^2) = 0 \Rightarrow r=\sqrt{μ}$$

Now, to derive a conclusion about the existence of an $\omega-$limit set, all you have to do is draw an one-dimensional phase portrait, where an $\longrightarrow$ arrow deonotes the area where $r'>0$ and on the other hand, an $\longleftarrow$ deones the area where $r'<0$. Thus, the following portrait is yielded :

$$\longrightarrow \quad \sqrt{\mu} \quad \longleftarrow$$

From this one-dimension portrait, we can see that $r'$ gets "bounded" around $\sqrt{\mu}$. Thus we have proved the existence of an $\omega$ limit set, which proves to be a clockwise limit-cycle, since $θ'(\sqrt{\mu}) = ρ(μ) > 0$.

Exercise :

Given the dynamical system : $$x_1' = x_2+2μx_1(5-x_1^2-x_2^2)$$ $$x_2' = -x_1+2μx_2(5-x_1^2-x_2^2)$$ where $(x_1,x_2) \in \mathbb R^2$ and $μ>0$ a constant. Applying polar coordinates, determine the omega limit set $ω(x_0)$ for any given vector $(x_1,x_2)$.

Discussion :

I used the polar coordinates substitution :

$$x_1 = r\cosθ$$ $$x_2 = r\sinθ$$

and via the expressions :

$$rr' = x_1x_1' + x_2x_2'$$

$$θ' = \frac{x_1x_2' - x_2x_1'}{r^2}$$

Using a polar coordinate substitution correctly for $x_1$ and $x_2$, one yields :

$$r' = 2μr(5-r^2)$$ $$θ' = -1$$

Now, what we can see is that the sign of $r'$ entirely depends on the factor $5-r^2$, since by it's definition $r>0$ and $μ>0$. This means that until $r=\sqrt5$ $r'>0$ and after that $r'<0$. Also, noting that $θ'=-1$ means that the direction of the phase portrait flow follows an anticlockwise flow.

Other than that, defining an one dimension phase portrait for $r'$ and noting the arrows just like the previous exercise, we get :

$$ \longrightarrow \sqrt{5} \longleftarrow$$

Again, we see that $r'$ is bounded around $\sqrt{5}$, thus this means that there exists a limit cycle defined with $r=\sqrt{5}$.

This implies that the omega limit set for any given values, will be :

$$\text{For} \space x_0 \neq 0 \space \rightarrow \space ω(x_0) = S_{\sqrt5}$$

$$\text{For} \space x_0 = 0 \space \rightarrow \space ω(x_0) = \{0\}$$

Here's a phase portrait of the initial system (the non-polar one) where I used $x$ as $x_1$ and $y$ as $x_2$ with $\mu =1$, where you can see the radius of the limit cycle $r = \sqrt{5}$ and that it evolves around the origin $(0,0)$ :

$\qquad \qquad \qquad \quad$