Flows on the Circle

Module 2: Synchronization and Collective Behaviour

A periodic rhythm can be described by a single angle, the phase $\theta$. The phase lives on the circle: $\theta$ and $\theta + 2\pi$ are the same state. A one-dimensional flow on the circle is

\[ \dot\theta = f(\theta), \qquad f(\theta + 2\pi) = f(\theta). \tag{1}\]

Unlike a flow on the line, a flow on the circle can return to its starting point. This is the simplest model of an oscillator (Strogatz 2024, sec. 4.1).

Uniform oscillator

The simplest choice is a constant speed, $\dot\theta = \omega$. The solution is $\theta(t) = \omega t + \theta_0$ and the period is $T = 2\pi/\omega$ (Strogatz 2024, sec. 4.2).

Nonuniform oscillator

A phase that speeds up and slows down along the cycle obeys

\[ \dot\theta = \omega - a \sin\theta, \qquad \omega > 0,\; a \ge 0. \tag{2}\]

This equation appears in the firefly model, in overdamped pendulums and in Josephson junctions (Strogatz 2024, sec. 4.3 to 4.6). Its behaviour depends on the ratio $a/\omega$.

For $a < \omega$ there are no fixed points. The phase circulates, slowest near $\theta = \pi/2$.
At $a = \omega$ a half-stable fixed point appears at $\theta = \pi/2$. This is a saddle-node bifurcation on the circle (see Saddle-Node Bifurcation).
For $a > \omega$ there are two fixed points with $\sin\theta^* = \omega/a$. One is stable and one is unstable. The oscillation stops.

For $a < \omega$ the period is

\[ T = \int_0^{2\pi} \frac{d\theta}{\omega - a\sin\theta} = \frac{2\pi}{\sqrt{\omega^2 - a^2}}. \tag{3}\]

As $a \to \omega^-$ the period diverges like $T \approx \dfrac{\sqrt{2}\,\pi}{\sqrt{\omega}}\,(\omega - a)^{-1/2}$. The trajectory spends almost all its time in the bottleneck left behind by the fixed points that are about to appear (Strogatz 2024, sec. 4.3).

Figure 1: Nonuniform oscillator with $\omega = 1$. Left: the vector field $f(\theta)$ for three values of $a$. Right: the period from Equation 3 and the square-root scaling near $a = \omega$.

Exercise

Take $a > \omega$. Find the two fixed points of Equation 2 in $[0, 2\pi)$ and decide which one is stable.

Solution

The fixed points satisfy $\sin\theta^* = \omega/a < 1$, so $\theta_1^* = \arcsin(\omega/a) \in (0, \pi/2)$ and $\theta_2^* = \pi - \theta_1^*$. The derivative of the vector field is $f'(\theta) = -a\cos\theta$. At $\theta_1^*$ the cosine is positive, so $f' < 0$ and the point is stable. At $\theta_2^*$ the cosine is negative and the point is unstable.

Exercise

Check Equation 3 numerically for $\omega = 1$ and $a = 0.8$. Integrate the ODE with solve_ivp and measure the time it takes $\theta$ to grow by $2\pi$.

Solution

The formula gives $T = 2\pi/\sqrt{1 - 0.64} = 2\pi/0.6 \approx 10.47$.

import numpy as np
from scipy.integrate import solve_ivp

def rhs(t, theta, omega, a):
    return omega - a * np.sin(theta)

def one_turn(t, theta, omega, a):
    return theta[0] - 2 * np.pi

one_turn.terminal = True
sol = solve_ivp(rhs, (0, 100), [0.0], args=(1.0, 0.8), events=one_turn, rtol=1e-10)
print(sol.t_events[0][0])  # about 10.472

References

Strogatz, Steven H. 2024. Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering. Chapman; Hall/CRC.

--- title: "Flows on the Circle" subtitle: "Module 2: Synchronization and Collective Behaviour" format: html --- A periodic rhythm can be described by a single angle, the **phase** $\theta$. The phase lives on the circle: $\theta$ and $\theta + 2\pi$ are the same state. A one-dimensional flow on the circle is $$ \dot\theta = f(\theta), \qquad f(\theta + 2\pi) = f(\theta). $$ {#eq-circle-flow} Unlike a flow on the line, a flow on the circle can return to its starting point. This is the simplest model of an oscillator [@strogatz2024nonlinear, Section 4.1]. ## Uniform oscillator The simplest choice is a constant speed, $\dot\theta = \omega$. The solution is $\theta(t) = \omega t + \theta_0$ and the period is $T = 2\pi/\omega$ [@strogatz2024nonlinear, Section 4.2]. ## Nonuniform oscillator A phase that speeds up and slows down along the cycle obeys $$ \dot\theta = \omega - a \sin\theta, \qquad \omega > 0,\; a \ge 0. $$ {#eq-nonuniform} This equation appears in the firefly model, in overdamped pendulums and in Josephson junctions [@strogatz2024nonlinear, Sections 4.3 to 4.6]. Its behaviour depends on the ratio $a/\omega$. - For $a < \omega$ there are no fixed points. The phase circulates, slowest near $\theta = \pi/2$. - At $a = \omega$ a half-stable fixed point appears at $\theta = \pi/2$. This is a saddle-node bifurcation on the circle (see [Saddle-Node Bifurcation](../low-dimensional-dynamics/saddle-node-bifurcation.qmd)). - For $a > \omega$ there are two fixed points with $\sin\theta^* = \omega/a$. One is stable and one is unstable. The oscillation stops. For $a < \omega$ the period is $$ T = \int_0^{2\pi} \frac{d\theta}{\omega - a\sin\theta} = \frac{2\pi}{\sqrt{\omega^2 - a^2}}. $$ {#eq-nonuniform-period} As $a \to \omega^-$ the period diverges like $T \approx \dfrac{\sqrt{2}\,\pi}{\sqrt{\omega}}\,(\omega - a)^{-1/2}$. The trajectory spends almost all its time in the bottleneck left behind by the fixed points that are about to appear [@strogatz2024nonlinear, Section 4.3]. ```{python} #| label: fig-nonuniform #| fig-cap: "Nonuniform oscillator with $\\omega = 1$. Left: the vector field $f(\\theta)$ for three values of $a$. Right: the period from @eq-nonuniform-period and the square-root scaling near $a = \\omega$." #| fig-width: 8 #| fig-height: 3.5 #| echo: false import numpy as np import matplotlib.pyplot as plt from amlab.synchronization.phase_models import nonuniform_oscillator_period omega = 1.0 theta = np.linspace(0, 2 * np.pi, 400) fig, axs = plt.subplots(1, 2, figsize=(8, 3.5)) for a, style in [(0.5, "-"), (1.0, "--"), (1.5, ":")]: axs[0].plot(theta, omega - a * np.sin(theta), style, label=f"a = {a}") axs[0].axhline(0, color="black", linewidth=0.8) axs[0].set_xlabel(r"$\theta$") axs[0].set_ylabel(r"$\dot\theta$") axs[0].set_xticks([0, np.pi / 2, np.pi, 3 * np.pi / 2, 2 * np.pi]) axs[0].set_xticklabels(["0", r"$\pi/2$", r"$\pi$", r"$3\pi/2$", r"$2\pi$"]) axs[0].legend() a_values = np.linspace(0, 0.999, 400) axs[1].plot(a_values, nonuniform_oscillator_period(omega, a_values), label="Exact period") axs[1].plot( a_values[200:], np.sqrt(2) * np.pi / np.sqrt(omega) / np.sqrt(omega - a_values[200:]), "--", label="Square-root scaling", ) axs[1].set_xlabel("a") axs[1].set_ylabel("T") axs[1].set_ylim(0, 60) axs[1].legend() plt.tight_layout() plt.show() ``` ::: {.callout-note} ## Exercise Take $a > \omega$. Find the two fixed points of @eq-nonuniform in $[0, 2\pi)$ and decide which one is stable. ::: ::: {.callout-tip collapse="true"} ## Solution The fixed points satisfy $\sin\theta^* = \omega/a < 1$, so $\theta_1^* = \arcsin(\omega/a) \in (0, \pi/2)$ and $\theta_2^* = \pi - \theta_1^*$. The derivative of the vector field is $f'(\theta) = -a\cos\theta$. At $\theta_1^*$ the cosine is positive, so $f' < 0$ and the point is stable. At $\theta_2^*$ the cosine is negative and the point is unstable. ::: ::: {.callout-note} ## Exercise Check @eq-nonuniform-period numerically for $\omega = 1$ and $a = 0.8$. Integrate the ODE with `solve_ivp` and measure the time it takes $\theta$ to grow by $2\pi$. ::: ::: {.callout-tip collapse="true"} ## Solution The formula gives $T = 2\pi/\sqrt{1 - 0.64} = 2\pi/0.6 \approx 10.47$. ```python import numpy as np from scipy.integrate import solve_ivp def rhs(t, theta, omega, a): return omega - a * np.sin(theta) def one_turn(t, theta, omega, a): return theta[0] - 2 * np.pi one_turn.terminal = True sol = solve_ivp(rhs, (0, 100), [0.0], args=(1.0, 0.8), events=one_turn, rtol=1e-10) print(sol.t_events[0][0]) # about 10.472 ``` :::