1. Synchronisation — Two oscillators
Naoki Masuda
Department of Engineering Mathematics
naoki.masuda@bristol.ac.uk
http://www.naokimasuda.net
Modified from a lecture I gave in Bristol
(EMATM001 Advanced Nonlinear Dynamics and Chaos)
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3. Christian Huygens (1629–1695)
• Dutch physicist, mathematician, astronomer and inventor,
• Pendulum clock (1656)
• ‘An odd sympathy’, an unexpected discovery he made at home (1665)
Left figure: public domain; right figure: original drawing by Huygens
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4. Sync, but not oscillatory or dynamic in the end
Deffuant model of collective opinion dynamics
(Deffuant, et al., Advances in Complex Systems, 3, 87–98, 2000):
Interact if and only if |xi (t) − xj (t)| < ϵ,
{
xi (t + 1) = xi (t) + κ [xj (t) − xi (t)]
xj (t + 1) = xj (t) + κ [xi (t) − xj (t)]
0 20 40 60 80 100
time
0.0
0.2
0.4
0.6
0.8
1.0
agent'sopinion
Dynamics of the Deffuant model.
N = 100 agents, ϵ − 0.25, κ = 0.2.
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5. Two ways to synchronise
Left: figure in the public domain. Right: clip from the video: https://www.youtube.com/watch?v=oJ2ZLr87lLY
Q: Which of the two sync mechanisms is at work in the following
examples?
• Fireflies?
• Clapping?
• Millennium bridge?
• Metronomes?
• Candle frames?
• Students’ sync
walking?
• Heart?
• Circadian clock?
• Dancing robots?
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6. Phase dynamics of two coupled phase oscillators
{
˙ϕ1 = ω1 + κ sin(ϕ2 − ϕ1)
˙ϕ2 = ω2 + κ sin(ϕ1 − ϕ2)
where ϕi (i = 1, 2) is the phase variable, ∈ [0, 2π), rotating, ωi is the
angular velocity, and κ is the coupling strength.
Q:
1 What happens if κ = 0?
2 Taylor expand the sin term and tell its role when ϕ1 and ϕ2 are not
too far.
3 What do you expect as κ(> 0) increases?
4 What do you expect as κ goes negative large?
5 Synchronisation easier or harder as |ω2 − ω1| becomes larger?
6 Why sin?
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7. Analysis of a two-oscillator system
˙ϕ1 = ω1 + κ sin(ϕ2 − ϕ1)
˙ϕ2 = ω2 + κ sin(ϕ1 − ϕ2)
• Let ψ ≡ ϕ2 − ϕ1 and ∆ω = ω2 − ω1.
What dynamics does ψ obey?
˙ψ = ∆ω − 2κ sin ψ
Worked example 11.1
Show that this system have a solution (i.e. ˙ψ = 0) when
∆ω
2κ
≤ 1
Is this condition intuitive?
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8. Analysis of a two-oscillator system
Worked example 11.2
Analyse
˙ψ = ∆ω − 2κ sin ψ
by drawing a bifurcation diagram in terms of κ.
Which bifurcation happens where?
Perfect synchrony (i.e. ψ = 0) happens?
For small positive κ, what is happening?
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9. Analysis of a two-oscillator system
Worked example 11.3
Do a linear stability analysis of phase-locked solutions (why are they so
called?) of
˙ψ = ∆ω − 2κ sin ψ
when κ > ∆ω/2.
A: By setting ˙ψ = 0, we get sin ψ∗ = ∆ω/2κ.
Set ψ = ψ∗ + ϵ, where ϵ is small, to obtain
˙ϵ = ∆ω − 2κ sin(ψ∗
+ ϵ)
= ∆ω − 2κ(sin ψ∗
+ ϵ cos ψ∗
)
= −2κ cos ψ∗
· ϵ
So the in-phase solution (0 < ψ∗ < π/2, assuming ω1 < ω2) is linearly
stable, whereas the anti-phase solution (π/2 < ψ∗ < π) is linearly
unstable.
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10. Analysis of a two-oscillator system
Worked example 11.4
What is the oscillation frequency when the phase locking is happening?
˙ϕ1 = ω1 + κ sin(ϕ2 − ϕ1)
˙ϕ2 = ω2 + κ sin(ϕ1 − ϕ2)
˙ψ = ∆ω − 2κ sin ψ
Is the solution intuitive?
A: Under phase locking, sin ψ∗ = ∆ω/2κ. So,
˙ϕ1 = ω1 + κ sin ψ∗
=
ω1 + ω2
2
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11. Back to Huygens
Oliveira & Melo, Scientific Reports, 5, 11548 (2015)
https://doi.org/10.1038/srep11548
• Andronov clock model (1966)
¨θ + µ · sign( ˙θ) + ω2
θ = 0
• Plus kicking in a constant
energy to compensate the loss
of kinetic energy due to dry
friction
• µ(> 0): dry friction
coefficient, at θ ≈ 0 in each
cycle
• ω: natural angular frequency
of the pendulum
This and the following figures are from the Oliveira & Melo paper, which has been published under CC BY license.
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12. Two clocks
• Assumption: When one clock receives a kick, the impact propagates
in the wall to instantaneously perturb the other clock slightly.
• Sound travels fast.
{
¨θ1 + µ1 · sign( ˙θ1) + ω2
1θ1 = −α1F(θ2),
¨θ2 + µ2 · sign( ˙θ2) + ω2
2θ2 = −α2F(θ1).
plus kicking, with ω1 = ω + ϵ and ω2 = ω − ϵ
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13. Flavour of analysis
• ϕn: The phase of clock 2 when the phase of clock 1 is 2nπ.
• Derive the Poincar´e map: ϕn+1 = T(ϕn)
• Can show that T has a stable fixed point near π.
• What does this mean physically?
• Consistent with Huygens’ observation.
Simulations
Red: ϵ = 1.5 × 10−4 rad/s
Black: ϵ = 3 × 10−3 rad/s
ω = 4.4879 rad/s
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14. Experiments
In the bottom panel, the free clock
freq of the two clocks are closer than
in the top panel.
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