# Confusing result about stability of a limit cycle

I want to understand a confusing result about this ODE system from Arnold's book on dynamical systems.

$$\dot{r}=[r^2-1][2 r \cos \phi - 1]$$ $$\dot{\phi}=1$$

The limit cycle is given for $r=1$ and $\phi = t$.

My question: The limit cycle seems to be asymptotically stable (if I didn't do any mistakes; see 1. Determining stability by Floquet analysis) but the phase portrait seems to suggest, that there are trajectories that start close to the limit cycle but drift away from it (see: 2. Understanding stability in the phase portrait). I would interpret the phase portrait of the original system as evidence that the limit cycle is not stable because there are trajectories that start close to it but then drift away on the $x$-axis $r=1$. I would be glad if someone could help me in resolving this confusion.

1. Determining stability by Floquet analysis

One way of dealing with the stability of this limit cycle is by first rewriting the ODE such that the origin is the trivial solution to the ODE. In order to do this use the following substitution $r = 1 + x_1$ and $\phi = t + x_2$.

The system then can be written as $$\dot{x}_1=[x_1^2+2x_1]\left[2(x_1+1)\cos(t+x_2)-1 \right]$$

$$\dot{x}_2=0.$$

It is obvious that $x_1=0$ and $x_2=0$ is now the trivial solution, which corresponds to the limit cycle in the original equation.

Now we can linearize the system at the origin to obtain:

$$\Delta \dot{x}_1=-2\Delta x_1 +4\Delta x_1\cos(t)$$ $$\Delta \dot{x}_2=0.$$

Note, that the system in question is a linear time-variant system but has periodic coefficients with period $T = 2\pi$. Such systems can be investigated using Floquet theory.

One will have to simulate the system for two pairs of initial conditions $x_{1,1}(0)=1\, \wedge \,x_{2,1}(0)=0$ and $x_{1,2}(0)=0 \,\wedge \, x_{2,2}(0)=1$ form $t=0$ to $t=2\pi$. One will obtain the states $x_{1,1}(2\pi),x_{2,1}(2\pi),x_{1,2}(2\pi),x_{2,2}(2\pi)$ which can be put together into the so called monodromy matrix

$$C=\begin{bmatrix} x_{1,1}(2\pi) & x_{1,2}(2\pi)\\ x_{2,1}(2\pi) & x_{2,2}(2\pi) \end{bmatrix}.$$

This can be as previously told by numerical integration or analytically like in this case.

Note that the linearized equation is a decoupled first order linear equation. The general solution is given by

$$\Delta x_1(t) = c_1\exp(4\sin(t)-2t)$$ $$\Delta x_2(t) = c_2.$$

Using the initial conditions and setting $t=2\pi$ we obtain:

$$C = \begin{bmatrix}\exp(-4\pi) & 0 \\ 0 &1 \end{bmatrix}$$ The eigenvalues of this matrix, which can also be complex, are called Floquet multipliers $\lambda_i$. The eigenvalues are given by the entry on the diagonal, as we have an upper triangular matrix. The first eigenvalue is $\lambda_1= \exp(-4\pi)<1$ and the second one is given as $\lambda_2=1$. It can be shown that linearization of a limit cycle will always lead to one Floquet multiplier $\lambda = 1$, which can be excluded from the analysis. This eigenvalue corresponds to a disturbance along the limit cycle.

If the remaining Floquet multipliers fulfill the following inequality $|\lambda_i| < 1$, then the limit cycle is asymptotically stable. If there is at least one remaining Floquet multiplier with $|\lambda_i|>1$ then the limit cycle is unstable. If there exist at least one remaining Floquet multipliers for which $|\lambda_i|=1$ but the absolute value of all other Floquet multipliers is smaller or equal to $1$ then this method is indecisive.

As we can see that the only remaining eigenvalue is $0<\lambda_1=\exp(-4\pi)<1$, we can conclude that the limit cycle is asymptotically stable.

2. Understanding stability in the phase portrait

In order to get a better picture of the limit cycle, I plotted some trajectories of the ODE system. Which you see in the following figure ($x$-axis: $r$ and $y$-axis: $\phi$). • When you determine stability of trajectory, you are most likely will use Lyapunov stability definition. Just have in mind that it's not important that trajectory drifts away at first: how far it drifts is the only thing that matters. That's what these $\varepsilon$ and $\delta$ in definition are responsible for. And even visually you can estimate these $\varepsilon$-$\delta$ on your phase portrait. – Evgeny Sep 27 '17 at 10:20
• @MrYouMath Sure, but who doesn't there? Can you highlight a trajectory which you think doesn't tend to trajectory that corresponds to limit cycle? – Evgeny Sep 27 '17 at 13:26
• @MrYouMath The focus now is to track that trajectory further. Can you do it? You will notice that in the next strip $\pi \leqslant \varphi \leqslant 3\pi$ the trajectory will go closer to your "limit cycle" though it will also have that "drift away segment". The distance between "limit cycle" and nearby trajectory doesn't have to go to $0$ monotonously: I think that might be a source of your confusion here (correct me if I'm wrong). – Evgeny Sep 28 '17 at 8:24
• @MrYouMath Well not so arbitrarily large, just enough to satisfy the existence of limit. It's like with $f_1(x) = e^{-x}$ and $f_2(x) = e^{-x} (2 + \sin{x} )$: both go to zero as $x \rightarrow + \infty$, but the first one does it monotonously and the second one drifts away from zero just like the trajectory in your example. – Evgeny Sep 28 '17 at 13:08
• @MrYouMath Don't worry about that :) Maybe I'll try to post this as an answer later, but I think it might require at least a nice illustration as an addition and I can't do this right now) – Evgeny Sep 30 '17 at 14:35