Limit cycles
1. Periodic orbits and the Poincaré map
As usual, we will consider
\dot{\bx} = \bfF(\bx),
whose flow is denoted \phi(t; \bx).
Definition 1.1.
The orbit of an initial condition \bx_0 is called periodic if there exists T > 0 such that:
- \phi(T; \bx_0) = \bx_0.
- \phi(t; \bx_0) \ne \bx_0 for 0 < t < T.
T is called the period (or minimum period) of the orbit.
Example 1.2.
All orbits of the Predator-Prey model
\begin{aligned} \dot{x} & = x(a - by) \\
\dot{y} & = y(- c + dx)
\end{aligned}
in the first quadrant except a single fixed point are periodic.
Example 1.3.
The damped pendulum
\begin{aligned} \dot{x} & = y \\
\dot{y} & = - \sin(x) - by,
\end{aligned}
where b > 0, has no periodic orbits.
Example 1.4.
The polar system
\begin{aligned} \dot{r} & = r (1 - r^2) (4 - r^2) \\
\dot{\theta} & = - 1.
\end{aligned}
We observe that the system has two periodic oribts given by r = 1 and r = 2. And observing from the phase portrait, the periodic orbit r = 1 seems to be attracting, while the periodic orbit r = 2 seems to be repelling. We need a method to measure whether orbits close to a periodic orbit is getting closer or father away from that oribt.
For the rest of this section, let's focus only on systems in dimension 2.
Definition 1.5.
Let \gamma = \{\phi(t; \bx_0) \st t \in [0, T]\} be a periodic orbit. A (possibly) curved open segment \Sigma is called a Poincaré section at \bx_0 if \bx_0 \in \Sigma, and at every point \bx \in \Sigma, the vector field \bfF(\bx) is nonzero and not tangent to \Sigma.
We often say that \Sigma is transversal to the vector field \bfF, and call \Sigma a cross section.
Definition 1.6.
Let \gamma be a periodic orbit, and let \Sigma be a Poincaré section. For any \bx \in \Sigma, the firs return time is
\tau(x) = \min\{ t > 0 \st \phi(t; \bx) \in \Sigma\}.
In other words, this is the first time that the forward orbit of \bx has returned to the section \Sigma. If the orbits of \bx never returns to \Sigma, we set \tau(x) = \infty.
If the orbit of \bx \in \Sigma returns to \Sigma, i.e. \tau(x) < \infty, define the Poincaré return map P(\bx) to be \phi(\tau(\bx); \bx). The map is considered undefined if \tau(x) = \infty. This maps \bx to the point where its orbit first returns to \Sigma, hence also called the first return map.
Figure 1. The Poincaré map
Let us return to Example 1.4. Consider the section \Sigma given by r \in (0, \infty) and \theta = 0. It's clear that \Sigma is transveral to the vector field since we always have \dot{\theta} = - 1. Consider an initial condition r = r_0, \theta = 0 on \Sigma, since \dot{\theta} = -1 , the return time \tau(r_0, 0) = 2\pi for any r_0. This implies the Poincaré map is
P(r_0, 0) = (r_{r_0}(2\pi), 0),
where r_{r_0}(t) is the solution to the one dimensional ODE
\dot{r} = r(1 - r^2)(4 - r^2), \quad r(0) = r_0.
Since all points on \Sigma has \theta component 0, we can write P as a function of r only. In other words,
P(r_0) = r_{r_0}(2\pi).
By inspecting the phase portrait of the r system, we obtain the following property of the Poincaré map P:
- P(1) = 1, P(2) = 2.
- P maps the intervals (0, 1), (1, 2), and (2, \infty) to themselves.
- P is increasing on (0, 1), decreasing on (1, 2), and increasing on (2, \infty).
Starting with r_0, the first return of r_0 to the section will be P(r_0). The next return will be P(P(r_0)) which we denote by P^2(r_0). Note that in dynamical systems, the notation P^n means P \circ P \cdots \circ P (n times composition), not the nth power of P.
Proposition 1.7.
Let P be the Poincaré map as described. Then
\lim_{n \to \infty} P^n(r_0) =
\begin{cases} 1 & x \in (0, 1) \cup (1, 2) \\
\infty & x \in (2, \infty).
\end{cases}
Consider an initial condition starting close to the point r = 1 on \Sigma. Then its orbit will be crosing closer and closer to the point r = 1 on \Sigma. This means r = 1 is attracting. For r = 2, we note the inverse of the Poincaré map is attracting, therefore r = 2 should be repelling.
We have seen that Poincaré maps can be useful if defined. The following theorem ensures it is at least defined near a periodic orbit.
Theorem 1.8.
Suppose \gamma = \{\phi(t; \bx_0) \st t \in [0, T]\} is a periodic orbit and \Sigma is a cross section at \bx_0. Then both the Poincaré map P and its inverse P^{-1} are defined for points on \Sigma sufficiently close to \bx_0.
Nonrigorous proof.
While the rigorous proof of this theorem requires some set up, it's relatively easy to explain why any orbit starting close to \bx_0 on \Sigma should return to \Sigma. First note that since \phi(T; \bx_0) = \bx_0, we always have \tau(\bx_0) = T and P(\bx_0) = \bx_0.
Due to continous dependence of the solution on initial conditions, if \bx \in \Sigma is close enough to \bx_0, the point \phi(T; \bx) should be very close to \phi(T; \bx_0) = \bx_0, but not necessarily on \Sigma. However, since the flow goes across the section \Sigma, by flowing \phi(T; \bx_0) forward or backward a bit, we can always land on the section \Sigma.
Figure 2.
Definition 1.9.
Let \gamma = \{\phi(t; \bx_0) \st t \in [0, T]\} be a periodic orbit. We say that \gamma is attracting if there exists a Poincaré section \Sigma on which the return map P satisfies \lim_{n \to \infty} P^n(\bx) = \bx_0 for any \bx \in \Sigma sufficiently close to \bx_0. \gamma is repelling if the same holds with P replaced with P^{-1}.
Example 1.10.
In the Predator-Prey model, take any small section that that does not contain the unique fixed point in the first quadrant. Then the Poincaré map is always identity.
The following notion is quite commonly used:
Definition 1.11.
A limit cycle is an isolated periodic orbit. In other words, no nearby orbit is periodic.
2. Poincaré-Bendixon Theorem
For systems that do not admit a nice polar form, it is often quite hard to determine if there is a periodic orbit. We will show that one can often find periodic orbit by finding proper positively invariant regions.
Example 2.1.
Show that for the equation
\begin{aligned} \dot{x} & = y \\
\dot{y} & = -x + y(4 - x^2 - 4y^2),
\end{aligned}
the region
\cA = \{(x, y) \st 1 \le x^2 + y^2 \le 4\}
is positively invariant.Proof.
Set
L(x, y) = x^2 + y^2,
then
\frac{d}{dt} L = 2 x \dot{x} + 2y \dot{y}
= 2x y + 2y (-2x + y(4 - x^2 - 4y^2))
= 2y^2(4 - x^2 - 4y^2).
On one hand,
\frac{dL}{dt} = 2y^2(4 - x^2 - 4 y^2) \le 2y^2(4 - x^2 - y^2),
therefore \frac{dL}{dt} \le 0 for x^2 + y^2 = 4. On the other hand,
\frac{dL}{dt} = 2y^2 (4 - x^2 - 4y^2) \ge 2y^2 (4 - 4x^2 - 4y^2),
therefore \frac{dL}{dt} \ge 0 for x^2 + y^2 = 1. The conclusion follows from the theory of bounding regions.
The region is denoted \cA because its an annulus.
The following theorem shows how a positively invariant region can be useful.
Theorem 2.2. (Poincaré-Bendixon Theorem).
Let \dot{\bx} = \bfF(\bx) be a differential equation on \R^2.
- Assume that \bfF is defined on all of \R^2 and a forward orbit \phi(t; \bx_0), t \ge 0 is bounded. Then \omega(\bx_0) is either a fixed point or a periodic orbit.
- Assume that \cA is a closed, bounded region in \R^2. Suppose \bF is defined on \cA, \cA is positively invariant, and contains no fixed point. Then there is a periodic orbit in \cA.
Remark 2.3.
There is a general theorem in topology that shows if a positively invariant bounded region in \R^2 contains no holes (simply connected), then there must be a fixed point inside. Therefore the second item of Theorem 2.2 can never be applied to a region without holes. Since a region with a single hole is an annulus, this theorem is often applied on annuli.
Idea of the proof.
1. We need to use a theorem from topology that every infinite sequence of points in a closed, bounded set much accumulate somewhere. Essentially, this is due to the fact that a bounded set cannot accommodate infinitely many points that keep a fixed distance from each other.
This implies any bounded orbit must have an non-empty \omega limit set. Suppose \by \in \omega(\bx_0), and suppose s_n is a sequence of times such that
\phi(s_n; \bx_0) \to \by, \quad n \to \infty.
We now suppose \by is not a fixed point, and will try to show that the orbit of \by is periodic. This argument works for both part 1 and part 2 of the theorem.
Pick a transversal section \Sigma at \by. Because \phi(s_n; \bx_0) gets arbitrarily close to \by , for n sufficiently large, the orbit of \phi(s_n; \bx_0) crosses the section \Sigma. In the same way as in the proof of Theorem 1.8, we can either flow the time forward or backward a little bit, such that for the newly chosen times t_n, the points \phi(t_n; \bx_0) lie on the section \Sigma. What we have shown is: for n large enough, the forward orbit of \phi(t_n; \bx_0) returns infinitely many times to \Sigma. Pick one such point and call it \by_1, and let \by_2 = P(\by_1), \by_3 = P(\by_2), etc.
We will next show that all the \by_n's must accumulate to \by from one side. Consider the closed loop formed by the orbit of the flow from \by_1 to \by_2, and the segment from \by_1 to \by_2. The region bounded by this loop is forward invariant. This means y_3 cannot be on the other side of y_2.
Figure 3. Proof of Poincaré-Bendixon Theorem
To show that \by is periodic, we argue that P(\by) must be equal to \by. \by splits the section into two parts, \Sigma_1 and \Sigma_2. We just showed that all by_n must accumulate to \by on one side, say all \by_n's are in \Sigma_1. P(\by) cannot be on \Sigma_1 since doing so will cause the orbit of some \by_n to intersect with orbit of \by. On the other hand, P(\by) cannot be on \Sigma_2 either, because \by_n \to \by, P(\by_n) \to P(\by), this means P(\by_n) = \by_{n+1} must be on \Sigma_2 for n large enough. But this is not possible since all \by_n's are on \Sigma_1. We are left with the only possibility: P(\by)= \by.
The proof of Theorem 2.2 actually shows more. We summarize it in the following corollary.
Corollary 2.4.
Let \gamma be an isolated periodic orbit of a two-dimensional system.
- If \bx_0 is not on \gamma and \omega(\bx_0) = \gamma. Then all points \bx sufficiently close to \gamma on the same side as \bx_0 also has \omega(\bx) = \gamma. \gamma is attracting from that side.
- If there exists two points \bx_1 and \bx_2 on each side of \gamma such that \omega(\bx_1) = \omega(\bx_2) = \gamma. Then \gamma is attracting.
Coming back to Example 2.1, we have shown that there is a positively invariant annular region \cA = \{1 \le x^2 + y^2 \le 4\}. Moreover, the only fixed point for the system is x = y = 0, therefore there is no fixed point in \cA. By Theorem 2.2, there exists a periodic orbit in \cA.
Example 2.5.
\begin{aligned} \dot{x} & = y \\
\dot{y} & = - x^3 + y (4 - x^2 - 4y^2).
\end{aligned}
Use the test function
L(x, y) = \frac12 y^2 + \frac14 x^4.