Limit cycles and nonlinear oscillators

A state can return without standing still

Imagine a model whose state has two coordinates, x and y. Both are normalized numbers with no physical unit. Start at (2, 0). After a quarter of one revolution the state is (0, 2); after a complete revolution it returns to (2, 0). Its radius stays 2 while its coordinates keep changing.

One exact motion with these properties is x(t) = 2 cos(2t/T), y(t) = 2 sin(2t/T), where t is elapsed time and T = 1 s. The angle 2t/T is measured in radians, increasing counterclockwise. A complete turn has angle 2π, so the least positive return period is πT = π s. T is a time scale, not the period itself.

A periodic solution repeats its entire state motion after a positive period, not merely at two isolated recorded times. A fixed point stays at the same state. A limit cycle is a nonconstant periodic orbit isolated from other periodic orbits nearby. The orbit is the closed set of states traced out; the solution also tells us when each state is visited.

Substitution and graph reading are the main preparation; Algebra, functions, graphs, and logarithms provides a return path. The coordinate and radian ideas in Geometry, trigonometry, and coordinate systems are optional preparation. On a first reading, follow the states and rate signs below; return to the exact formula later with Derivatives, Taylor series, and local approximation or Ordinary differential equations if needed.

Specify how the whole state changes

A circular drawing alone does not determine a model. We supply the following autonomous rate rule: dx/dt = [(1 − (x² + y²)/4)x − 2y]/T and dy/dt = [(1 − (x² + y²)/4)y + 2x]/T, with T = 1 s. Autonomous means that the right sides depend on the current state and fixed parameters, with no explicit time dependence.

The symbols dx/dt and dy/dt mean instantaneous rates of change. Because x and y are dimensionless, these rates have units s⁻¹. They form the complete rate vector (dx/dt, dy/dt). At (1, 0), substitution gives (3/4, 2) s⁻¹: the motion changes both coordinates.

These polynomial right sides are continuously differentiable. The local existence and uniqueness result for a smooth rate field therefore gives a unique local solution for each finite initial state. That local statement by itself does not guarantee a solution for all time. We will check the future-time behavior of this particular model explicitly.

All values here are stipulated teaching values. The coordinates are not distances in metres, and the rule has not been fitted to measurements of a mechanical or biological oscillator.

Separate radial change from rotation

Define r = √(x² + y²), the nonnegative distance from the origin in the normalized state plane. For r > 0, write x = r cos θ and y = r sin θ. The radial rate describes how that distance changes; the angular rate describes rotation around the origin.

For this supplied rule, dr/dt = r(1 − r²/4)/T and dθ/dt = 2/T. To check the radial equation, use r dr/dt = x dx/dt + y dy/dt; the two rotation terms cancel. To check the angular equation, use dθ/dt = (x dy/dt − y dx/dt)/r²; the radial terms cancel. The angular rate is 2 rad/s when T = 1 s.

At r = 0 the angle is undefined, so the formulas involving division by r cannot be used there. Direct substitution in the original Cartesian rule gives both rates zero at (0, 0). The origin is a fixed point.

The table uses T = 1 s. A zero radial rate does not necessarily mean a zero complete rate vector.

Read the circulating orbit and radial-rate graph

For 0 < r < 2 the radial rate is positive; for r > 2 it is negative. At r = 2 it is zero, but the angular rate remains 2/T. Thus the circle r = 2 is a moving periodic orbit, not a circle of fixed points. The exact trigonometric motion at the start is a solution of the complete rule.

The upper panel uses equal scales for x and y. Its four arrows show the instantaneous complete rate at four states on the cycle. Their tips are not subsequent states. The lower panel instead plots the scalar radial rate against radius. Its horizontal axis is radius, not time, and its vertical rate scale differs from the arrow scale.

Any other positive radius changes monotonically toward 2, as the exact formula below will confirm. Such a motion cannot return to its initial radius after a positive period. Consequently there are no other periodic orbits in a sufficiently small neighborhood of this circle: it is a limit cycle.

Approaching an orbit does not fix the phase

Distance to the circular orbit is |r − 2|. Orbital stability means that an initial state sufficiently close to the orbit stays arbitrarily close in the future. Asymptotic orbital stability additionally requires that this distance tend to zero for sufficiently close initial states.

Here the radial rate points toward 2 on both sides. A positive initial radius stays on its side of 2 and approaches it; its distance |r − 2| never grows. This gives orbital stability, and the exact future-time formula in the next section verifies convergence. In fact every positive initial radius approaches this orbit. The fixed origin is an exception.

This compares distance to the whole orbit, not distance to one chosen solution at the same time. Two states already on the circle can keep a constant phase difference. If that difference is π/2, their state-plane distance is always 2√2 even though each has zero distance to the orbit.

The angular rule advances both phases at the same rate. It does not make their phases agree. Attraction to a limit cycle and synchronization of two motions are different conclusions.

Check an exact formula for the amplitude

For a positive initial radius r₀, set z = r² and z₀ = r₀². Then dz/dt = 2r dr/dt = 2z(1 − z/4)/T. Squaring the radius is only a useful change of variable here; z is not a mechanical energy.

An exact solution is z(t) = 4/[1 + (4/z₀ − 1) exp(−2t/T)], for z₀ > 0 and t ≥ 0. The function exp is the exponential function; its input −2t/T is dimensionless. At t = 0 this formula gives z₀. Differentiating the formula gives dz/dt = 2z(1 − z/4)/T, so it passes both the rate-rule and initial-value checks.

Write c = 4/z₀ − 1. Since z₀ > 0, c > −1. For future times, 0 < exp(−2t/T) ≤ 1, so 1 + c exp(−2t/T) stays positive. The formula therefore stays finite and positive for every finite t ≥ 0. As t increases without bound the exponential tends to zero, z tends to 4, and r = √z tends to 2.

For z₀ ≠ 4 the radius approaches 2 without reaching it at any finite future time. To recover the full state, combine this radius with θ(t) = θ₀ + 2t/T. The positive-radius formula excludes z₀ = 0; the separate Cartesian solution at the origin stays fixed.

A family of closed orbits is a different case

Now compare a separate linear model: dx/dt = −2y/T and dy/dt = 2x/T. Its radial rate is zero at every positive radius, while its angular rate is 2/T. Each positive radius therefore gives a different circular periodic orbit with period πT.

No one of these circles is isolated: circles with slightly larger or smaller radii lie arbitrarily close to it. They do not attract one another. A closed orbit can be periodic and still fail to be a limit cycle.

The linear comparison is homogeneous: multiplying any solution by a constant gives another solution of the same rule. In a finite-dimensional autonomous homogeneous linear system, scaling a nonconstant periodic solution by nearby positive factors similarly prevents its orbit from being isolated. This statement does not cover an externally forced linear system with an explicit time-dependent input.

Our nonlinear model contains the cubic amplitude term −r³/(4T), which changes the radial rate with amplitude and selects r = 2. This explains this model’s limit cycle; nonlinearity alone does not guarantee that an arbitrary system has one.

What would a physical oscillator require?

The supplied autonomous model produces sustained periodic motion without an explicit periodic input in its rate rule. That mathematical fact does not identify a physical energy source or prove that a real device obeys this rule. Physical interpretation requires measured state variables, their normalization, a justified rate model and evidence about its behavior under relevant perturbations.

A finite record with repeated sampled states can suggest a period. By itself it does not establish repetition at all intervening and future times, isolation from neighboring periodic orbits, or orbital stability. These conclusions were established here from a complete stipulated rule and exact reasoning.

The circular orbit and constant angular rate make this example especially transparent. Other limit cycles need not be circular, and their speed along the orbit need not be constant. Keep the general ideas—nonconstant periodic motion, isolation and specified stability—separate from the special geometry of this teaching model.

Worked examples

Try it yourself

1. For T = 1 s and r = 3/2, calculate dr/dt. Does knowing r alone also determine the complete Cartesian rate vector?

   Show the answer and reasoning

   dr/dt = 21/32 s⁻¹, and the complete Cartesian vector is not determined by r alone.

   The radial factor is 1 − (3/2)²/4 = 7/16; multiplying by 3/2 gives 21/32. Different angles at this radius give different x and y, so the original two-component rule needs that state information too.

2. At the cycle state (0, −2), with T = 1 s, what is the full rate vector? Is the state stationary because its radial rate is zero?

   Show the answer and reasoning

   The full rate is (4, 0) s⁻¹. The state is moving to the right at that instant.

   The amplitude factor is zero on r = 2, leaving dx/dt = −2y/T = 4 s⁻¹ and dy/dt = 2x/T = 0. Zero radial change means constant radius here, not a fixed point.

3. In a separate case set T = 2 s and start on the cycle at (2, 0). Find the least positive period and the state after a quarter of that period.

   Show the answer and reasoning

   The period is 2π s. After π/2 s the state is (0, 2).

   The angular rate is 2/T = 1 rad/s. A full 2π-radian turn takes 2π s, and a quarter turn takes π/2 s. Changing T changes the timing, not the cycle radius.

4. A motion starts at (0, 0). May we insert z₀ = 0 into the positive-radius formula? Does this motion become the limit cycle?

   Show the answer and reasoning

   No. That formula divides by z₀; the original Cartesian rule instead gives the fixed solution (0, 0).

   Both coordinate rates vanish at the origin. Smooth local uniqueness and the directly checked constant solution keep this initial state there. A constant state is not a nonconstant periodic orbit, and its distance to the cycle stays 2.

5. The separate linear comparison has circular periodic motions at every positive radius. Is its r = 2 circle a limit cycle?

   Show the answer and reasoning

   No. It is periodic but not isolated from other periodic orbits.

   For any small positive change in radius there is another circular periodic orbit nearby. These neighboring radii stay constant, so their distance to r = 2 does not decay. Neither isolation nor attraction follows from the closed shape alone.

6. A record shows the same state at two sampled times. Without a complete rate rule or other information, does this prove a stable limit cycle?

   Show the answer and reasoning

   No. The two recorded values alone prove neither a complete periodic solution nor an isolated, stable periodic orbit.

   The intervening and future states are unspecified, and neighboring initial states were not tested. The supplied model in this lesson supports stronger conclusions because its complete rule, exact solution and neighboring-radius behavior can be checked.

What to carry forward

• A limit cycle is an isolated nonconstant periodic orbit; a fixed state or an arbitrary closed orbit is insufficient.
• Radial and angular rates describe different parts of a two-coordinate motion.
• In the supplied model, every positive initial radius approaches r = 2, while the origin stays fixed.
• Orbital attraction compares distance to a whole orbit and does not imply phase synchronization.
• Exact model conclusions and finite experimental records require different kinds of evidence.