Limit Cycles and Closed Orbits

Limit cycles are the mathematical backbone of every self-sustained oscillator: the neuron's action potential (FitzHugh–Nagumo and Hodgkin–Huxley models), the cardiac pacemaker cycle, circadian rhythm models, and predator–prey population cycles beyond the neutrally stable Lotka–Volterra case. In electrical engineering, the Van der Pol equation directly models the vacuum-tube and transistor oscillators used in radio and clock circuits, where the limit cycle's amplitude sets the output signal level regardless of small perturbations in the power supply. In chemistry, the Belousov–Zhabotinsky reaction exhibits a genuine stable limit cycle in concentration space, observable as periodic color oscillations. In control engineering, semistable and unstable limit cycles are diagnosed via describing function analysis to predict unwanted self-oscillation ("hunting") in feedback systems. In conservative systems no limit cycles occur at all, which is exactly why introducing dissipation (as in a driven, damped pendulum) is the standard route to generating them; the tools built here — the Poincaré map and its derivative — are also the direct technical engine behind the Hopf Bifurcation theorem, which identifies the precise parameter threshold at which a stable limit cycle is born from an equilibrium losing stability, and behind the Poincaré–Bendixson theorem, which guarantees existence of a limit cycle inside any trapping region of the plane free of equilibria.

Ask the tutor: How does the Poincaré map's derivative formula connect to Floquet theory for periodic linear systems, and how is it used to prove the existence of the Van der Pol limit cycle rigorously?

Picture the phase plane near a closed curve $\Gamma$. Because $\Gamma$ is a trajectory, no other trajectory can cross it (uniqueness of solutions to the ODE). So every nearby trajectory is confined to spiral in an annular region either just inside or just outside $\Gamma$, forever trapped between $\Gamma$ and its neighbors. The only question is whether, as $t\to\infty$, that spiral winds in toward $\Gamma$, unwinds away from it, or does one on each side. Isolation is what makes this question meaningful. In a conservative system such as the pendulum $\ddot\theta + \sin\theta = 0$, every energy level $E = \tfrac12\dot\theta^2 - \cos\theta \in (-1,1)$ gives a distinct closed orbit, and these orbits fill an entire annular region continuously — there is no room for a neighborhood free of other closed orbits. Limit cycles require dissipation: a mechanism that removes energy from large-amplitude motion and injects it into small-amplitude motion, so that trajectories are funneled toward one specific amplitude rather than preserving whatever energy they started with.

To make "spirals in" and "spirals out" rigorous, we use the Poincaré (first-return) map. Fix a point $p_0 \in \Gamma$ and a smooth transversal segment $\Sigma$ through $p_0$ that is nowhere tangent to the vector field $(f,g)$ (a cross-section). By continuous dependence on initial conditions and the fact that $\varphi_T(p_0) = p_0$, trajectories starting at points of $\Sigma$ near $p_0$ return to $\Sigma$ after a time close to $T$, defining a map

on a sub-neighborhood $\Sigma_0 \subset \Sigma$ of $p_0$, where $\tau(p)$ is the first return time. Parametrize $\Sigma$ near $p_0$ by signed arc length $s$, with $s=0$ at $p_0$, so $P$ becomes a scalar map $s \mapsto P(s)$ with $P(0)=0$.

Example 1 — Van der Pol oscillator with explicit stability check. Consider the Van der Pol system in Liénard-type polar form via the substitution to polar coordinates $x = r\cos\theta,\, y = r\sin\theta$ applied to

Instead, to get full explicit control (Van der Pol itself resists closed-form solution), use the exactly solvable polar model

which is a standard device for constructing a system with a prescribed, provable set of limit cycles (this is not artificial — such normal forms arise directly from Hopf bifurcation analysis). Finding the periodic orbits. Equilibria of the radial equation with $r>0$: $\dot r = 0$ when $r(1-r^2)(4-r^2)=0$, i.e. $r=0,1,2$ (discarding $r=-1,-2$ as radii must be nonnegative). So $r=1$ and $r=2$ are constant solutions of the radial equation; combined with $\dot\theta=1$, the circles $r=1$ and $r=2$ are periodic orbits of period $T=2\pi$. Are they isolated? The map $r \mapsto r(1-r^2)(4-r^2)$ has exactly the zeros $r=0,1,2$ for $r\ge 0$ (a cubic in $r^2$ with three real roots $r^2 = 0,1,4$, each simple), so no other value of $r$ gives $\dot r=0$. Hence no other circle is a closed orbit, and $r=1$, $r=2$ are each isolated: both are limit cycles. Stability via direct sign analysis (equivalent to the Poincaré map $s\mapsto r$, since here $\theta(t)=\theta_0+t$ makes the transversal $\theta=\theta_0$ literally the $r$-axis, and the return time is exactly $T=2\pi$ independent of $r$, so $P$ is just the time-$2\pi$ flow of the radial ODE alone). Determine the sign of $g(r):=(1-r^2)(4-r^2)$ on each side of each root:

• For $0<r<1$: $1-r^2>0$, $4-r^2>0$, so $g(r)>0$: $\dot r>0$, i.e. $r$ increases toward $1$.
• For $1<r<2$: $1-r^2<0$, $4-r^2>0$, so $g(r)<0$: $\dot r<0$, i.e. $r$ decreases toward $1$.
• For $r>2$: $1-r^2<0$, $4-r^2<0$, so $g(r)>0$: $\dot r>0$, i.e. $r$ increases, moving away from $2$.

So near $r=1$: trajectories with $r$ slightly less than $1$ increase toward $1$; trajectories with $r$ slightly more than $1$ decrease toward $1$. By Theorem part 3 (with the identification $P(s)-s$ having the sign of $g$ near the fixed point, since $\dot r>0 \Leftrightarrow$ the return value exceeds the start), $r=1$ is a stable limit cycle. Near $r=2$: trajectories with $r$ slightly less than $2$ decrease (moving away, down toward $1$); trajectories with $r$ slightly more than $2$ increase (moving away, up toward $\infty$). Both sides move away from $r=2$: $r=2$ is an unstable limit cycle. This example — two isolated periodic orbits with opposite stability nested around the origin — is impossible in any conservative system, since a conservative system's closed orbits form a continuum, never two isolated circles with an empty annulus in between except for a spiral pattern. Sanity check via explicit solution. Separating variables in $\dot r = r(1-r^2)(4-r^2)$ with partial fractions confirms these are the only interior equilibria of the radial flow and that convergence to $r=1$ from both sides is monotone and asymptotic (never overshooting, since $g$ does not change sign again between $0$ and $1$, nor between $1$ and $2$), consistent with the sign analysis above.

Example 2 — a semistable limit cycle, with asymmetric coefficients. Consider

for $r \ge 0$. Zeros of the radial factor at $r=0,3,5$; $r=3$ is a double root, $r=5$ a simple root, an intentionally asymmetric setup (unequal multiplicities, unequal spacing) to avoid masking behavior with symmetry. Isolation. For $r>0$, the equation $r(r-3)^2(5-r)=0$ has exactly the roots $r=3,5$ (the factor $r$ contributes only the non-periodic equilibrium $r=0$, a degenerate point, not a closed orbit since $r=0$ is the origin itself, an equilibrium of the full planar system, not a periodic orbit). So $r=3$ and $r=5$ are each isolated closed orbits (circles traversed with angular speed $2$, period $\pi$). Stability of $r=3$. Examine $h(r) = r(r-3)^2(5-r)$ near $r=3$:

• For $r$ slightly less than $3$ (say $r=3-\epsilon$, $0<\epsilon$ small): $r>0$, $(r-3)^2 = \epsilon^2 >0$, and $5-r = 2+\epsilon>0$. So $h(r) > 0$: $\dot r > 0$, meaning $r$ increases toward $3$ — attraction from below.
• For $r$ slightly more than $3$ (say $r = 3+\epsilon$): $r>0$, $(r-3)^2 = \epsilon^2>0$, $5-r = 2-\epsilon>0$ for small $\epsilon$. So $h(r)>0$ again: $\dot r>0$, meaning $r$ increases, moving away from $3$ — repulsion from above.

So $r=3$ attracts from below and repels from above: by Theorem part 3, this is a semistable limit cycle (the sign of $h$ does not change across $r=3$, since $(r-3)^2\ge0$ never changes sign — the hallmark algebraic signature of semistability at a double root). Stability of $r=5$. For $r$ slightly less than $5$: $r>0$ and $(r-3)^2>0$ are both positive and $5-r = \epsilon>0$, so $h(r)>0$: $r$ increases toward $5$. For $r$ slightly more than $5$: $5-r = -\epsilon <0$ while $r>0,(r-3)^2>0$, so $h(r)<0$: $\dot r<0$, so $r$ decreases, moving back toward $5$. Both sides converge: $r=5$ is a stable limit cycle. This example shows algebraically why double roots of the radial equation are the generic mechanism producing semistable cycles — exactly the mechanism seen in Hopf Bifurcation, where a limit cycle is born or annihilated precisely at a parameter value where such a double root first appears (a saddle-node bifurcation of cycles).