Chaos theory and weather prediction / personal lessonsLesson 2 of 14
Chaos and weather · Lesson 2

Explain nonlinear feedback

See why the same input change need not have the same effect everywhere.

By the end: Distinguish nonlinearity, feedback, and chaos.

State-dependent response

A linear rule multiplies an input by a fixed number; an affine rule also adds a constant. The logistic map contains x squared when expanded. Its response to a small change in x depends on where x starts. This is nonlinearity. The rising and falling halves also fold different inputs onto the same output: x and one minus x give equal next states.

For r = 4, inputs 0.2 and 0.3 give 0.64 and 0.84, a positive change of 0.20. Inputs 0.7 and 0.8 give 0.84 and 0.64, a negative change of 0.20. The same increase in input can raise or lower the next state.

Measure how the response depends on the starting input

To test linearity, compare responses to changes in input, not whether a time trace looks straight. A rule y equal to three times x gives an output increase of 0.3 whenever the input increases by 0.1. Adding a fixed offset changes output levels but leaves that response unchanged. Such an offset gives an affine rule. For the logistic rule, expanding the product gives r times x minus r times x squared. The squared term makes equal input increments produce different output increments at different positions on the curve.

Hold r at three and increase the input by 0.1 in several separate calculations. Going from 0.1 to 0.2 changes the output from 0.27 to 0.48, an increase of 0.21. Going from 0.4 to 0.5 changes it from 0.72 to 0.75, only 0.03. Going from 0.6 to 0.7 changes it from 0.72 to 0.63, a decrease of 0.09. These are comparisons across inputs to one update, not three episodes in a single trajectory. The same positive input increment can produce a smaller positive response or a negative one.

Equal input increments at r = 3

Input intervalOutput intervalOutput change / 0.1
0.1 to 0.20.27 to 0.482.1
0.4 to 0.50.72 to 0.750.3
0.6 to 0.70.72 to 0.63−0.9
Each row compares independent inputs separated by 0.1 under one fixed rule. The final column divides the output change by that same 0.1 denominator. It is an average one-update response, not a long-run growth rate.

High-school algebra gives an exact expression for this change. Call the input increment d. Expand the next output at x plus d, then subtract the output at x. After canceling equal terms, the difference is r times d times the quantity one minus twice x minus d. Dividing by d, when d is nonzero, gives the average response per unit input over that interval. At r equal to three and d equal to 0.1, this average is 2.1 near x equal to 0.1 but only 0.3 near x equal to 0.4.

Expand the changed input
f(x+d)=r(x+d−x2−2×x×d−d2)

f(x) = rx(1 − x); d is the signed input increment. States, parameters, and response ratios here are dimensionless.

Subtract the unchanged output
f(x+d)−f(x)=r×d×(1−2×x−d)
Divide by the nonzero increment
f(x+d)−f(x)d=r(1−2×x−d)

For r = 3 and d = 0.1, this ratio is 2.1 at x = 0.1 and 0.3 at x = 0.4.

The finite increment matters near the peak. Starting at x equal to 0.49 and adding 0.02 ends at 0.51. Symmetry makes their outputs identical, so the average response over that interval is zero. Nevertheless, the output rises between 0.49 and 0.5 and falls between 0.5 and 0.51. A zero net change over an interval does not mean the rule is constant throughout it. To describe response near one starting point, you must specify a sufficiently small change or inspect the rule over the entire interval under discussion.

The same algebra explains what a local slope means without requiring calculus notation. Divide the exact difference by d and let the tested increments become smaller. The remaining expression approaches r times one minus twice x. It is positive on the left half of the unit interval, zero at the middle, and negative on the right half. Its magnitude measures the approximate one-update amplification of a sufficiently small input difference. The word approximate matters because the discarded part is minus r times d squared in the output difference, not identically zero for finite d.

Samples of the nonlinear response

Output after one update

Samples of the nonlinear response00.40.800.51

Input x, dimensionless

  • Solid: r = 3 rule
These are sampled input-output values at r equal to three, not a trajectory indexed by time. Straight segments interpolate between samples and are not the exact quadratic rule. Symmetric inputs on either side of 0.5 share an output.

Response per unit input is dimensionless here because both input and output are dimensionless states. In a physical application, the units would depend on what each axis measured. Comparing slopes from graphs with different scales can be misleading even when the underlying numbers are the same. Use the numerical ratio of output change to input change rather than the apparent angle on the screen. This lesson's finite-change comparisons establish nonlinearity directly. They do not establish that nearby trajectories diverge indefinitely, since that requires applying the rule repeatedly along the changing states.

Amplification and restraint in one update

The factor r x increases with x, while one minus x decreases. Their product first rises, then falls. The largest output occurs at x = 0.5 and equals r divided by four. For r between zero and four, inputs in the unit interval stay there.

Feedback means a result changes a later input. It can stabilize, oscillate, or destabilize a system depending on its strength and timing. Calling a feedback positive is not the same as calling it beneficial, and calling it negative does not mean it is always stable.

Derive the peak and the permitted interval

The two factors x and one minus x compete in a precise algebraic sense. Their sum is always one. Near zero, x limits their product; near one, the complementary factor limits it. At the middle they are equal. Rewrite their product as one quarter minus the square of x minus one half. Expanding that square verifies the identity. A square cannot be negative, so this expression cannot exceed one quarter. Multiplying by nonnegative r places the largest output at r divided by four, with no graph or calculus required.

Complete a square
x×(1−x)=14−(x−12)2

x is a state in [0,1] and r is nonnegative; both are dimensionless.

Bound the output
0≤f(x)=r4−r×(x−12)2≤r4
Locate the peak
f(12)=r4

The squared distance vanishes at x = 1/2. For r ≤ 4, the output remains in [0,1].

For example, use r equal to 3.6. The largest output is 0.9, reached at input 0.5. Inputs 0.3 and 0.7 both give 0.756 because their factors are 0.3 and 0.7 in opposite order. Inputs 0.1 and 0.9 both give 0.324. This symmetry is called folding because two distinct inputs become indistinguishable after one update. Once two trajectories have the same state under the same future rule, their later states agree too. Determinism does not require retaining enough information to reconstruct which of the earlier inputs occurred.

Distance from the peak at r = 3.6

Input xSquared distance from 0.5Output x next
0.1 or 0.90.160.9 − 3.6 × 0.16 = 0.324
0.3 or 0.70.040.9 − 3.6 × 0.04 = 0.756
0.500.9
Each row is one independent input to the same quadratic rule. The squared distance from 0.5 determines the shortfall below the peak output 0.9. Equal distances on opposite sides give equal outputs, so the update cannot recover the side.

Nonnegativity follows separately from the peak argument. For x between zero and one, both x and one minus x are nonnegative. If r is also nonnegative, their product cannot be negative. Combining this lower bound with the maximum gives an output interval from zero to r divided by four. When r is at most four, that entire output interval lies inside the allowed state interval. Repeated updates then remain valid by the same argument at every step. This is an algebraic all-update bound, unlike checking a finite sample of trajectories.

The bound depends on the stated domain. If an input exceeds one, the complementary factor becomes negative and the output need not remain a valid unit-interval state. If r were negative, the nonnegativity argument would also fail. A physical interpretation cannot repair an invalid mathematical input by silently replacing it with a boundary value. Replacing negative outputs by zero, or values above one by one, defines a clipped rule. Its future behavior may differ because clipping changes both the graph and which starting states become identical after an update.

A maximum output is not automatically an equilibrium. At r equal to 3.6, the peak output 0.9 becomes the next input and then produces 0.324. It does not remain at 0.9. The word maximum concerns the largest possible output across all current inputs. A fixed state instead must equal its own output. Likewise, a bounded sequence need not approach its largest possible value. The domain bound tells you where every trajectory is allowed to go; it does not determine which parts of that interval a particular trajectory actually visits.

A bound is not a fixed state

CalculationResultMeaning
All inputs in [0,1]Outputs in [0,0.9]Domain bound
Input 0.5Next 0.9Peak output
Input 0.9Next 0.324Leaves the peak
Input 0.324Next 0.7884864Another update, not a fixed state
All entries concern r equal to 3.6. The first row bounds every one-update output for permitted inputs. The other rows follow the peak input forward, demonstrating why reaching the largest possible output does not make that output permanent.

This distinction also limits explanations based only on amplification and restraint. Those factors show why the single-update graph bends and why outputs stay bounded for the supported parameter range. They do not by themselves classify repeated motion. Two nearby points can separate on one part of the curve and approach each other on another. To understand a trajectory, follow which parts it visits and how the response changes there. Boundedness prevents unlimited growth of the state, but it does not force the difference between two states to decrease after every update.

Inspect the rules and their result

The logistic update
xn+1=rxn(1−xn)

x is dimensionless and n counts updates. r is fixed within a run. We restrict 0 ≤ r ≤ 4 and 0 ≤ x ≤ 1.

Changing r restarts the run from x = 0.213; it does not continue the old run. The first sixty updates remain separate from the late sample.

Logistic state over sixty updatesState x10.5003060Update n
Solid blue: r = 2.8.

Nonzero fixed point: 0.6429. Local error multiplier: -0.80. Its magnitude is below one.

UpdateLate value
8930.642857
8940.642857
8950.642857
8960.642857
8970.642857
8980.642857
8990.642857
9000.642857
Last eight values after 900 updates. Rounded near-repeats suggest a pattern; this is not a proof of an exact period.
InputsOutputs at r = 4Output change
0.2 to 0.30.64 to 0.84+0.20
0.7 to 0.80.84 to 0.64−0.20
Equal input differences can have opposite effects.

Compare a rule with the trajectory it generates

The map experiment plots state against update number, rather than output against input. Its sixty-update curve therefore cannot by itself show whether the defining rule is linear. The same nonlinear formula produces every preset in this experiment. What changes is r, held fixed during each run. Choosing a preset restarts from 0.213. The slider also restarts from that same initial state when it changes r. There is no playback clock in this experiment, so moving the slider compares fresh calculations rather than speeding, slowing, or advancing one continuing trajectory.

A useful additional setting is r equal to 2.6, which the slider supports even though it has no named preset button. From 0.213, the first outputs are 0.4358406, 0.639297286, and 0.599550292 after rounding for display. The nonzero fixed state is eight thirteenths, about 0.615384615. The sequence crosses that value, then returns from the other side. A time plot makes these crossings visible. The nonlinear response graph would instead show that the same output difference can have different signs depending on the current input's position.

A supported slider setting at r = 2.6

State x, dimensionless

A supported slider setting at r = 2.60.20.450.7036

Update n, count

  • Solid: From 0.213
  • Dashed: Fixed state 8/13
These are the first six updates from the experiment's initial state 0.213 at fixed r equal to 2.6. The second series marks the exact fixed-state level. Connections indicate update order, not continuous motion between integer model updates.

The initial rise at r equal to 2.6 is not inconsistent with later settling. Far from the fixed state, a trajectory can make a substantial change even though small errors near that state tend to shrink. The local multiplier printed by the experiment is minus 0.6 at this setting. That is a near-fixed-state approximation, not a claim that every input difference anywhere in the unit interval shrinks by exactly 0.6. For example, the first step from 0.213 is governed by the full product, not by multiplying the state itself by minus 0.6.

You can check the approximation with a paper perturbation that the ordinary map controls do not set directly. Start 0.01 above eight thirteenths at r equal to 2.6. If e denotes displacement from the fixed state, the exact next displacement is minus 0.6 times e minus 2.6 times e squared. Substituting 0.01 gives minus 0.00626, while the local linear estimate gives minus 0.006. Their difference is minus 0.00026. This separates an approximation error from a disagreement with the nonlinear recurrence, which still determines the example exactly.

Compare this slider setting with the existing Two-cycle preset at r equal to 3.2. Both runs use the same initial state and the same formula family, but the late values have different patterns. The experiment's last-eight table comes from a separate calculation through update 900. It is not simply the final eight entries of the sixty-update plot. Rounded near-repeats in that late sample can suggest alternating behavior. They cannot identify the exact cycle values or establish that a trajectory has no longer transient without further analysis of the rule.

Exact and approximate one-step displacement

ContributionCalculationDisplacement
Linear estimate−0.6 × 0.01−0.006
Quadratic correction−2.6 × 0.01²−0.00026
Exact sum−0.006 − 0.00026−0.00626
This paper calculation uses r equal to 2.6 and starts 0.01 above the nonzero fixed state. Displacements are dimensionless. The linear term is a local estimate; adding the quadratic correction recovers the exact one-update displacement for this example.

The fixed-state readout remains present even when the observed trajectory does not settle there. That is intentional. Solving the equation for an unchanging state and testing whether nearby states approach it are different tasks. Nor does attraction near a fixed state guarantee attraction from every possible state. The boundary input zero remains zero under every supported r, for instance, and input one maps to zero immediately. Those exceptional cases are enough to reject a statement that all unit-interval starting states approach the nonzero fixed state, regardless of what the displayed interior trajectory does.

Nonlinear is not a synonym for chaotic

At r = 2.8, most starting values in the interior settle toward one value. At r = 3.2 they approach alternating values. Other parameter settings produce irregular traces. All use the same nonlinear form.

A complicated graph can come from external forcing, random inputs, hidden variables, or numerical mistakes. First identify the rule and its state. A finite irregular record alone does not establish deterministic chaos.

Feedback timing can change the behavior

Nonlinearity and feedback refer to different relations. A single evaluation of a curved input-output rule shows nonlinearity without repeating anything. Feedback occurs when the result affects a later input. Even an affine rule can participate in feedback and produce repeated changes. To isolate timing, consider a hypothetical correction model for a dimensionless deviation u from a target. Positive u means above target and negative u means below. A correction of opposite sign tries to reduce that deviation. Its effect depends on when the correction acts and how strongly it changes the next state.

First stipulate that the next deviation equals minus one half of the current deviation. Starting at one gives minus 0.5, then 0.25, then minus 0.125. The sign alternates, so every update crosses the target. The magnitude halves each time, so the sequence approaches zero despite those crossings. This rule is linear and deterministic. Oscillation therefore does not establish either nonlinearity or chaos. Its stability follows directly from successive multiplication by a number whose magnitude is less than one, not from whether the plotted line appears smooth or complicated.

Now delay the correction by one update. Stipulate that the next deviation equals minus one half of the previous deviation, rather than the current one. This is a different recurrence. Give it initial values u at update zero equal to one and u at update one equal to zero. It then produces minus 0.5, zero, 0.25, zero, and minus 0.125. The same coefficient now produces a slower alternating pattern with zeros between nonzero values. The model needs two consecutive deviations as its state, because the current deviation alone no longer determines the next one.

The same coefficient with different timing

Deviation u, dimensionless

The same coefficient with different timing-0.601.1036

Update n, count

  • Solid: Immediate half-strength
  • Dashed: Delayed half-strength
These invented dimensionless correction models run through six updates. Immediate feedback starts at one; delayed feedback also needs the stipulated value zero at update one. Segments connect discrete trajectory samples only and do not model any physical motion between updates.

The delayed example shows why calling a feedback negative does not settle a stability question. The coefficient's sign says the response opposes the deviation it reads, but it does not say whether that deviation is current or old. Its magnitude matters too. If the immediate rule used minus 1.2 instead of minus 0.5, a starting deviation of one would become minus 1.2, then 1.44, then minus 1.728. These alternating magnitudes grow. The response still has opposite sign, yet it overshoots by more than the previous deviation instead of reducing it.

In the delayed half-strength example, the zeros are not evidence that the system has permanently reached the target. At update one the current deviation is zero, but the stored previous deviation is one. The next correction therefore produces minus 0.5. A displayed scalar can hide that memory. Compare a second history with both initial deviations zero. It has the same current value at update one but remains zero afterward. This is a concrete state-completeness test. A single plotted number may not contain the information needed to determine how a feedback model continues.

Zero current deviation can hide a later correction

Previous deviationCurrent deviationNext deviation
10−0.5
000
−0.50+0.25
00.250
Every row uses the delayed rule, next deviation equals minus half the previous deviation. The current scalar alone is insufficient. Equal current values can produce different next values because the rule reads a stored earlier value from the full state.

These correction rules are stipulated comparisons, not experiments on a thermostat or measured weather. They isolate timing and gain so that their effects can be calculated without extra physical assumptions. Returning to the logistic map, the whole nonlinear product defines the feedback, rather than a free-standing label such as positive or negative. Some parameters support settling and others support persistent changes, but a finite irregular trace is insufficient to prove chaos. A classification needs evidence about repeated dynamics under a specified rule, including what constitutes the complete state and which initial conditions the claim covers.

Exercises

Check the allowed range

If r = 4.4 and x = 0.5, is the next value still a valid unit-interval state?

Hint

Find the maximum output.

Worked answer
rxNext valueWithin [0,1]?
40.51Yes
4.40.51.1No
The assumed state space matters.

It is 1.1, outside the stipulated interval. The lab rejects r above four rather than clipping the value. Clipping would define a different model and could hide a modeling error.

Reject a shortcut

A nonlinear model settles to one value. Does that contradict its nonlinearity?

Hint

Nonlinearity describes the rule, not one required outcome.

Worked answer
ClaimRelevant check
Nonlinear ruleResponse depends on input location
Stable behaviorNearby trajectories approach a fixed state
ChaosSensitive, bounded, nonperiodic dynamics under suitable definitions
Different questions need different evidence.

No. The rule can be nonlinear and still have an attracting fixed point. A linearity test compares how the rule responds to inputs; a stability test concerns what trajectories do near a state.

Sources and further reading

Selected reference passages support the definitions and modeling distinctions below. The examples, explanations, and implementations are original. Terminology and assumptions are stated within each lesson; source review is not an independent audit of this course. Optional links are not runtime dependencies.

  • MathWorld: Logistic map

    Recurrence, fixed points, period doubling, and periodic windows. Selected explanatory passages reviewed; the plots here are generated independently.

  • Chaos theory, selected definitions

    Secondary overview of deterministic dynamics and sensitive dependence. Not a proof that every irregular numerical trace is chaotic.