Synchronization and coupled systems

Start with a shared-rate question

Imagine marking where each of two repeating cycles has reached. Give each cycle a phase angle: 0 rad marks a chosen reference point, and an increase of 2π rad completes one phase cycle. Two phases can advance at the same rate while one stays ahead of the other. Same rate does not require same phase.

We will start with phases θ₁ = 0 rad and θ₂ = (π/6) rad. The second cycle is one twelfth of a full cycle ahead. The question is whether their interaction can keep that offset constant even when their uncoupled rates differ.

Prepare with Geometry, trigonometry, and coordinate systems for radians and sine. Derivatives, Taylor series, and local approximation supplies the instantaneous-rate and slope ideas used later. A first reading can follow substitutions and rate signs; the slope test is an optional return after those ideas are clear.

State the two phase equations

Let t be time in seconds. The phases θ₁(t) and θ₂(t) are unwrapped angle quantities: they may increase past 2π rad instead of resetting to zero. Define their difference as δ = θ₂ − θ₁. A phase wheel will display these angles modulo 2π, but the equations retain the unwrapped values.

Supply the exact model dtheta_1/dt = omega_1 + K*sin(delta/(1 rad)) and dtheta_2/dt = omega_2 - K*sin(delta/(1 rad)). Here ω₁ = 1 rad/s and ω₂ = 2 rad/s are the rates each phase would have without coupling; K = 1 rad/s is the equal coupling coefficient in both equations. The sine argument δ/(1 rad) is dimensionless.

The derivative dθⱼ/dt is the instantaneous phase rate. It need not equal the natural rate ωⱼ once coupling is present. K in this convention is an angular rate, not a force or an energy; the model contains no amplitude or spatial displacement.

Calculate how the gap changes

Subtract the first equation from the second. The natural-rate difference is Δω = ω₂ − ω₁, and the two coupling contributions add in the difference: delta_dot = (omega_2 - omega_1) - 2*K*sin(delta/(1 rad)). For the supplied values this becomes delta_dot = (1 - 2*sin(delta/(1 rad))) rad/s.

At δ = 0 rad the gap rate is +1 rad/s, so equal phases do not remain equal. At δ = (π/6) rad, sin(π/6) = 1/2: the first phase rate becomes 1 + 1/2 = 3/2 rad/s and the second becomes 2 − 1/2 = 3/2 rad/s. Their gap then has zero rate.

Adding the equations gives a different relation. The mean phase Φ = (θ₁ + θ₂)/2 always has rate (ω₁ + ω₂)/2 = 3/2 rad/s in this symmetric model. A constant mean rate alone does not show locking; locking concerns whether δ stays constant.

Read the gap rate and two locked snapshots

The upper graph places phase difference δ on the horizontal axis and its instantaneous rate on the vertical axis. A positive rate increases the unwrapped gap; a negative rate decreases it. The filled circle and hollow square mark zero gap rates. This is a rate-versus-gap graph, rather than a time history.

The lower wheels show simultaneous pairs of phases at t = 0 s and t = (π/3) s for the stated initial values. Blue circles mark phase 1; hollow orange diamonds mark phase 2. Phase increases counterclockwise from the right-hand reference. These are schematic phase positions, not the spatial paths or amplitudes of physical oscillators.

The initial difference already equals (π/6) rad, so this selected solution is locked from the start. Both phases advance by (π/2) rad between the two snapshots, keeping their nonzero difference. These pictures illustrate an already locked solution; they do not show a transient approaching it.

The table gives the exact rates at all seven selected gaps. A zero gap rate identifies a possible constant gap, but its response to a small disturbance still needs a stability check.

Follow the already locked solution

Because both rates are exactly 3/2 rad/s at the initial gap, an exact solution is theta_1(t) = (3/2 rad/s)*t and theta_2(t) = (pi/6) rad + (3/2 rad/s)*t for t ≥ 0 s. Substitution leaves the gap constant, so the two coupled equations remain satisfied.

At t = (π/3) s the first unwrapped phase is (π/2) rad and the second is (2π/3) rad. They have different values while their rates agree. It takes (4π/3) s for each phase to increase by 2π rad and return to its own initial position on the phase wheel.

Adding the same initial angle to both phases gives another solution with the same relative phase. Relative locking does not choose a unique common phase origin.

Check relative stability

Near (π/6) rad, a slightly smaller gap has a positive rate and a slightly larger gap has a negative rate. Both changes move the gap back toward that value. This fixed lag is locally attracting for the relative phase.

There is another zero at (5π/6) rad. A slightly smaller gap decreases, while a slightly larger gap increases: small disturbances move away from this lag. Starting exactly there still gives a constant-gap solution, but that solution is relatively unstable.

For a formal slope check, write the gap-rate function as f(δ). Its derivative is df/ddelta = -(2*K/(1 rad))*cos(delta/(1 rad)). At K = 1 rad/s the slopes are −√3 s⁻¹ at (π/6) rad and +√3 s⁻¹ at (5π/6) rad. The sign tests agree with the rate directions.

The fixed lags repeat after adding 2π rad. Stability refers to small changes of the relative phase near a selected copy, not convergence of every unwrapped initial gap to the same literal angle. A common shift of both phases is not pulled back by these equations.

Compare coupling with the natural-rate difference

Keep the same equation form but allow K > 0. A constant gap requires sin(delta/(1 rad)) = (omega_2 - omega_1)/(2*K). In this sinusoidal pair, the strict condition |ω₂ − ω₁| < 2K gives one attracting and one repelling lag per phase cycle. The factor 2 follows from the equal per-link coupling in these two equations; it is not a universal threshold for every coupled system.

For the natural-rate difference 1 rad/s, take K = 1/4 rad/s instead. The gap rate is between 1/2 and 3/2 rad/s and is always positive. There is no constant gap for any initial phase in this case: the unwrapped gap continues to grow, even though the wheel angles stay bounded by wrapping.

At the boundary K = 1/2 rad/s the gap rate is (1 − sin(δ/(1 rad))) rad/s. It vanishes at (π/2) rad but has zero slope there. Nearby smaller gaps move up toward this value; nearby larger gaps move farther above it. The exact constant-gap state and this one-sided approach do not establish two-sided local stability.

Separate the model from physical synchronization

The supplied equations are an exact mathematical model for this lesson. A phase-only approximation derived from a real limit-cycle oscillator normally needs justified assumptions about weak perturbations, frequency differences and the retained phase description. The chosen numbers alone do not establish those assumptions for a device.

Constant phase difference is the locking criterion used here. It does not state that amplitudes, full physical states or energy transfers become identical. Nor does it give a theorem about a large network, a noisy system or a different coupling function.

Keep the evidence question separate from the equation check. Our two selected snapshots come from an exact already locked solution. To claim that a physical pair synchronizes, one would need an appropriate physical model, observations and a stated test of the proposed synchronization behavior.

Worked examples

Try it yourself

1. In the main model, set δ = 0 rad at one instant. Find both phase rates and decide whether that gap can remain zero.

   Show the answer and reasoning

   The rates are 1 and 2 rad/s, so the gap rate is +1 rad/s; equal phases do not remain equal.

   The sine term vanishes at zero. Subtract the first instantaneous rate from the second instead of assuming that equal phase values imply locking.

2. In the selected locked solution, what are θ₁, θ₂ and their difference at t = (π/3) s?

   Show the answer and reasoning

   θ₁ = (π/2) rad, θ₂ = (2π/3) rad and δ = (π/6) rad.

   Multiply 3/2 rad/s by (π/3) s to get a (π/2) rad increment for each phase, then add each initial phase separately.

3. In the main model, does the zero gap rate at δ = (5π/6) rad describe an attracting relative state?

   Show the answer and reasoning

   No. It is a constant-gap solution if initialized exactly, but it is relatively unstable.

   Its gap-rate slope is +√3 s⁻¹. A small decrease in the gap makes it decrease further, and a small increase makes it increase further.

4. For natural rates 1 and 2 rad/s with K = 1/4 rad/s, what is the smallest possible gap rate? Can the gap be constant?

   Show the answer and reasoning

   The smallest gap rate is 1/2 rad/s, so no constant gap is possible.

   The expression is 1 − (1/2)sin(δ/(1 rad)) in rad/s; its minimum occurs when the sine is 1.

5. For natural rates 1 and 2 rad/s with K = 1/2 rad/s, the gap rate vanishes at δ = (π/2) rad and its slope is zero. Does that prove attraction from both sides?

   Show the answer and reasoning

   No. This exact gap can remain constant, but nearby smaller gaps approach it and nearby larger gaps move away.

   The gap rate is nonnegative on both sides. A zero derivative alone cannot replace the two-sided sign check.

6. Set K = 0 while keeping natural rates 1 and 2 rad/s and initial phases 0 rad and (π/6) rad. What is the unwrapped gap after 1 s?

   Show the answer and reasoning

   The gap is (π/6 + 1) rad.

   The uncoupled gap grows at 1 rad/s. Add its 1 rad increase to the initial (π/6) rad offset; no coupling is available to hold that offset fixed.

What to carry forward

• Phase locking can keep a nonzero offset constant; matching rates does not require identical phases.
• Subtract the coupled phase equations to find the relative dynamics; the mean phase describes a different quantity.
• A zero gap rate needs a stability check, and the coupling threshold belongs to the specified equation convention.
• An exact phase model and two locked snapshots do not by themselves establish synchronization of a physical system.