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Stability, perturbations and chaos — Examples

1 · Prerequisites

2 · Summary

These examples exercise the stability, linearisation, chaos and recurrence items of cm16-stability-perturbations-and-chaos on declared model data and probe the boundaries of the page's statements.

The ideal pendulum on declared data: strict minimum and Lyapunov stability at the downward equilibrium, negative Hessian and instability at the upward equilibrium, the exact linearised eigenvalues and oscillation frequency, and failure of attraction works the ideal pendulum in the angle chart with $m = 1 k g$ , $ℓ = 1 m$ and $g = 9.81 m s^{- 2}$ : the downward equilibrium is a strict minimum, hence Lyapunov stable and not attracting, with the purely imaginary linearisation frequencies $\pm i 3 \sqrt{109} / 10 s^{- 1}$ ; the upward equilibrium has the negative Hessian $- m g ℓ$ and is unstable, with the real rates $\pm 3 \sqrt{109} / 10 s^{- 1}$ . Two declared linear models worked through: the harmonic flow has zero finite-time growth in every direction, while the repulsive linear model has exact rates $\pm 1$ and the positive Lyapunov exponent $1$ at its equilibrium computes the flows and growth rates of two dimensionless one-label Cartesian models in $R^{3}$ : the harmonic force $F (q) = - q$ has an orthogonal rotation flow, so every finite-time growth rate vanishes and every existing Lyapunov exponent is $0$ ; the repulsive force $F (q) = q$ has exact rates $\pm 1$ along its eigendirections and positive Lyapunov exponent $1$ at equilibrium. The three-dimensional isotropic oscillator: the stroboscopic map at half period is minus the identity, every state is periodic, the periodic points are dense, and the map is neither mixing nor sensitive shows the three-dimensional isotropic force model's stroboscopic map at half period to be minus the identity on its six-dimensional phase space: every state is periodic, the periodic points are dense, but the map is neither topologically mixing nor sensitive, so continuous maps with dense periodic points need not be chaotic in the page's stated sense. The pulsed horseshoe on declared data: $a = 100$ , the radicand box $[74, 126]$ , the contraction factor $1 / \sqrt{74}$ , the limiting quadratic fixed points $1 \pm \sqrt{101}$ , and the physical bounds of the coded states at $ℓ_{0} = 1 m$ , $m = 1 k g$ , $ω_{0} = 2 π s^{- 1}$ evaluates the constants of the pulsed horseshoe on its invariant axis at $a = 100$ : the coding box $R = 13$ , the radicands in $[74, 126]$ , the contraction factor $1 / \sqrt{74}$ , the limiting quadratic fixed points $1 \pm \sqrt{101}$ , the limiting kick composition, and the axial physical bounds $13 m$ and $26 π k g m s^{- 1}$ at $ℓ_{0} = 1 m$ , $m = 1 k g$ , $ω_{0} = 2 π s^{- 1}$ .

The counterexamples delimit the statements. A positive Lyapunov exponent at a hyperbolic equilibrium does not produce a chaotic invariant set of the time-one map: the three-dimensional repulsive model has rate $1$ and no compact invariant set with more than one point exhibits the repulsive linear model, whose equilibrium has a positive Lyapunov exponent and whose time-one map has no nonempty compact invariant set beyond ${0}$ ; that singleton fails sensitivity, so a positive exponent at an equilibrium does not produce a chaotic invariant set of that map. Lyapunov stability does not imply attraction: the harmonic model is stable at its equilibrium while every nonzero datum keeps its distance and never converges exhibits the harmonic model, Lyapunov stable at its equilibrium while the datum $(δ, 0)$ keeps its distance: stability is not attraction. A trapped one-label model: the compact energy sublevel of the radial quartic potential is invariant with finite phase volume, so the recurrence theorem applies, while the free model's energy sublevels are unbounded and the hypothesis fails verifies the recurrence theorem on the trapped one-label, three-dimensional radial quartic model of the flat Cartesian presentation, whose compact invariant energy sublevels have finite six-volume, and contrasts it with the free model, whose unbounded energy sublevels violate the finite-volume hypothesis. Unit restoration and scale checks for the two-rotor KAM statement: the scaling identities at $I = E_{0} = 1$ , the physical angular frequencies, the exclusion of rational frequency labels, and the equal-rescaling invariance restores units for the two-rotor KAM statement at $I = 1 k g m^{2}$ , $E_{0} = 1 J$ , checks the exact scaling identities, the physical angular frequencies $ω_{0} ν_{i}$ , the exclusion of rational frequency labels and the invariance under equal rescaling of inertia and energy; it does not evaluate the KAM constant $c$ . All data on this page are declared model data; no measurement or empirical comparison is asserted anywhere.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

The ideal pendulum on declared data: strict minimum and Lyapunov stability at the downward equilibrium, negative Hessian and instability at the upward equilibrium, the exact linearised eigenvalues and oscillation frequency, and failure of attraction

Physical scope

A declared-data check of the page's energy-stability theorem on the ideal planar pendulum: with mass 1 kg, length 1 m and g=9.81 m s^-2 in the angle chart, the downward equilibrium is a strict minimum, hence Lyapunov stable but not attracting, with purely imaginary linearised eigenvalues and a positive angular oscillation frequency; the upward equilibrium has a strictly negative Hessian, hence is unstable, with real linearised eigenvalues. All data are declared model data; no measurement and no empirical comparison is made.

Example

Declared model data. Take one label on the real line as the angle chart of the ideal planar pendulum: the chart domain is $Ω : = R$ , the configuration coordinate is the angle $q$ (radians, identified modulo $2 π$ at the end), the kinetic metric is the constant matrix $M : = m ℓ^{2}$ with $m : = 1 k g$ , $ℓ : = 1 m$ , and the potential is $U (q) : = m g ℓ (1 - \cos q), g : = 9.81 m s^{- 2}, m g ℓ = 9.81 J .$ This is the mechanical chart datum of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart with $Ω = R$ , $M$ constant and $U$ of class $C^{\infty}$ . Its equations are $m ℓ^{2} \ddot{q} = - m g ℓ \sin q$ , equivalently $\ddot{q} = - (g / ℓ) \sin q$ .

Assertions to check. (i) $q = 0$ and $q = π$ are mechanical equilibria; (ii) at $q = 0$ the potential is a strict local minimum and Energy stability and instability of the mechanical model at an equilibrium: the strict minimum is Lyapunov stable, a negative Hessian eigenvalue gives linear instability, a negative definite potential Hessian gives nonlinear instability, and exact energy conservation rules out attraction clause 1 makes $(0, 0)$ Lyapunov stable, while the linearised eigenvalues are $\pm i \sqrt{g / ℓ} = \pm i 3 \sqrt{109} / 10 s^{- 1}$ (so the angular oscillation frequency is $3 \sqrt{109} / 10 s^{- 1}$ ) and the equilibrium is not asymptotically stable; (iii) at $q = π$ the Hessian is $- m g ℓ < 0$ and clause 3 makes $(π, 0)$ unstable, with linearised eigenvalues $\pm 3 \sqrt{109} / 10 s^{- 1}$ .

Facts & Assumptions

Given: the declared chart data $M = 1 k g m^{2}$ , $U (q) = 9.81 (1 - \cos q) J$ on $Ω = R$ and the two equilibria $q = 0$ , $q = π$ .

[L1]

The energy-stability theorem (Energy stability and instability of the mechanical model at an equilibrium: the strict minimum is Lyapunov stable, a negative Hessian eigenvalue gives linear instability, a negative definite potential Hessian gives nonlinear instability, and exact energy conservation rules out attraction): under the chart hypotheses, a strict local minimum of $U$ makes $(q_{0}, 0)$ Lyapunov stable; a negative Hessian eigenvalue gives an unbounded solution $e^{κ t} (w, κ w)$ of the associated vector equation $Y^{'} = D G (q_{0}, 0) Y$ , the action on an initial deviation of the normalized matrix equation (linear instability); a negative definite Hessian $D^{2} U (q_{0})$ makes the equilibrium unstable; and exact energy conservation prevents asymptotic stability whenever states of different energy accumulate at the equilibrium.

[L2]

Trigonometric facts: $(\sin)^{'} = \cos$ , $(\cos)^{'} = - \sin$ , $\sin 0 = 0$ and $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine); $\sin π = 0$ , $\cos π = - 1$ and the half-turn shifts are those of Quarter-turn values and shifts by pi/2 and pi. Both functions are $2 π$ -periodic, and $\sin q = 0$ exactly when $q \in π Z$ (The zero sets of sine and cosine and the least positive common period 2 pi). Cosine is strictly decreasing on $[2 j π, (2 j + 1) π]$ and strictly increasing on $[(2 j + 1) π, (2 j + 2) π]$ for every integer $j$ (Signs, monotonicity intervals, and ranges of sine and cosine); with its value $1$ at $- 2 π, 0, 2 π$ , these inequalities give $\cos q < 1$ whenever $0 < ∣ q ∣ < 2 π$ .

Verification

technique · direct

1.1givenL2algebra

Equilibria and second derivative: by [L2], $U^{'} (q) = m g ℓ \sin q$ and $U^{''} (q) = m g ℓ \cos q$ ; hence $U^{'} (0) = U^{'} (π) = 0$ , so $q = 0$ and $q = π$ are mechanical equilibria (The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart), with $U^{''} (0) = m g ℓ = 981 / 100 J > 0$ and $U^{''} (π) = - m g ℓ < 0$ .

2.1step 1.1givenL1L2algebra

Downward equilibrium: $U (q) - U (0) = m g ℓ (1 - \cos q) > 0$ for $0 < ∣ q ∣ < 2 π$ , because $\cos q < 1$ there; thus $q = 0$ is a strict local minimum (take $r_{0} : = π$ ), and [L1] first clause gives Lyapunov stability of $(0, 0)$ in the chart. Since $g / ℓ = 981 / 100 s^{- 2}$ , the linearised first-order matrix is $A_{0} = (\begin{array}{cc} 0 & 1 \\ - g / ℓ & 0 \end{array})$ , whose characteristic polynomial is $λ^{2} + g / ℓ$ and whose eigenvalues are $\pm i \sqrt{g / ℓ} = \pm i 3 \sqrt{109} / 10 s^{- 1}$ . The coefficient $ω : = \sqrt{g / ℓ}$ in $\ddot{η} + ω^{2} η = 0$ is the angular oscillation frequency of this linearised model, so the equilibrium is not hyperbolic and no exponential conclusion follows.

2.2step 1.1givenL1algebra

Upward equilibrium: $U^{''} (π) = - 981 / 100 J < 0$ , and since $n = 1$ the Hessian is negative definite at $q = π$ ; [L1] third clause therefore makes $(π, 0)$ unstable. The linearised first-order matrix is $A_{π} = (\begin{array}{cc} 0 & 1 \\ g / ℓ & 0 \end{array})$ , with characteristic polynomial $λ^{2} - g / ℓ$ and real eigenvalues $\pm \sqrt{g / ℓ} = \pm 3 \sqrt{109} / 10 s^{- 1}$ , one positive, matching the linear-instability statement of [L1] second clause.

3.1step 2.1givenL1algebra

Failure of attraction: for every $0 < ε < π$ the datum $z_{0} : = (ε, 0)$ has energy $E (z_{0}) = m g ℓ (1 - \cos ε) > 0 = E (0, 0)$ , so letting $ε ↓ 0$ supplies states of different energy accumulating at the equilibrium; by [L1] fourth clause (and the strict-minimum clause of step 2.1) $(0, 0)$ is Lyapunov stable but not asymptotically stable. In particular no datum of positive energy can converge to $(0, 0)$ , because the energy is conserved.

4.1step 2.1step 2.2givenalgebra∎

All quantities above are declared model data substituted into the chart formulas; $M$ has dimension $k g m^{2}$ , the potential and the energies share the dimension $J = k g m^{2} s^{- 2}$ , and the frequencies have dimension $s^{- 1}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). No pendulum, material or measurement is involved.

Remarks

Periodic identification. The chart coordinate $q$ is the lift of the circle angle; the potential is $2 π$ -periodic and the equation is invariant under $q \mapsto q + 2 π$ , so the equilibria at $q = 2 π k$ (stable) and $q = π + 2 π k$ (unstable) are the same physical configurations as those checked. The two equilibria are the only mechanical equilibria, by step 1.1 and the zero set of the sine.
What the check does not show. The check verifies the hypotheses of Energy stability and instability of the mechanical model at an equilibrium: the strict minimum is Lyapunov stable, a negative Hessian eigenvalue gives linear instability, a negative definite potential Hessian gives nonlinear instability, and exact energy conservation rules out attraction on declared data; it does not produce a basin of attraction, a decay rate (none exists at the stable equilibrium) or any statement about the pendulum as an apparatus.

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

Two declared linear models worked through: the harmonic flow has zero finite-time growth in every direction, while the repulsive linear model has exact rates $\pm 1$ and the positive Lyapunov exponent $1$ at its equilibrium

Physical scope

A declared-data check of two physical one-label Cartesian force models with m0=1 kg, ell0=1 m and t0=1 s: F_1,phys=-(m0/t0^2)X and F_2,phys=(m0/t0^2)X, in newtons. The formulas use normalized q=X/ell0, v=t0 V/ell0 and t=T/t0, mass 1 and potentials U_1(q)=||q||^2/2 and U_2(q)=-||q||^2/2 with energy reference E0=m0 ell0^2/t0^2. In this numerical state metric the harmonic derivative is orthogonal, so all finite-time rates and Lyapunov exponents are zero; the repulsive derivative has exact rates plus and minus 1 along its eigendirections, with positive exponent 1 at equilibrium. The corresponding physical rates divide by t0. All data are declared, with no measurement and no claim of chaos.

Example

Declared model data and normalization. Take one label with $Q : = R^{3}$ and declare physical reference data $m_{0} : = 1 k g$ , $ℓ_{0} : = 1 m$ , $t_{0} : = 1 s$ . For $σ_{1} = - 1$ , $σ_{2} = 1$ define physical force maps $F_{i, p h y s} : R \times Q \times R^{3} \to R^{3}$ by $F_{i, p h y s} (T, X, V) = σ_{i} (m_{0} / t_{0}^{2}) X$ , with $T$ in seconds, $X$ in metres, $V$ in metres per second, and outputs in newtons. These are smooth globally state-Lipschitz force maps of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with physical mass $m_{0}$ . Put $t = T / t_{0}$ , $q = X / ℓ_{0}$ , $v = t_{0} V / ℓ_{0}$ and $E_{0} = m_{0} ℓ_{0}^{2} / t_{0}^{2}$ . Then $m_{0} \ddot{X} = F_{i, p h y s}$ becomes $q^{''} = σ_{i} q$ , with $q^{'} = v$ , since $\ddot{X} = (ℓ_{0} / t_{0}^{2}) q^{''}$ . The normalized force maps are $F_{i} (t, q, v) = F_{i, p h y s} (t_{0} t, ℓ_{0} q, (ℓ_{0} / t_{0}) v) / (m_{0} ℓ_{0} / t_{0}^{2}) = σ_{i} q$ , with normalized mass $m = 1$ . The physical potentials are $U_{i, p h y s} (X) = - σ_{i} m_{0} ∥ X ∥_{2}^{2} / (2 t_{0}^{2})$ ; normalizing their energies by $E_{0}$ gives, in the constant-metric chart $Ω : = R^{3}$ with $M : = I_{3}$ , they have the potentials $U_{1} (q) : = \frac{1}{2} ∥ q ∥_{2}^{2}$ and $U_{2} (q) : = - \frac{1}{2} ∥ q ∥_{2}^{2}$ of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart. The associated fields and complete motion maps are listed below; $e_{1}, e_{2}, e_{3}$ denote the coordinate basis, explicitly $e_{j} (k) = 1$ for $k = j - 1$ and $0$ otherwise, for $k \in {0, 1, 2}$ and $j \in {1, 2, 3}$ . All state and tangent norms below apply to the dimensionless numerical coordinates $(q, v)$ .

The harmonic model has field $G_{1} (q, v) = (v, - q)$ and motion map $Φ_{t}^{1} (q_{0}, v_{0}) = (q_{0} \cos t + v_{0} \sin t, - q_{0} \sin t + v_{0} \cos t) .$
The repulsive linear model has field $G_{2} (q, v) = (v, q)$ and motion map $Φ_{t}^{2} (q_{0}, v_{0}) = (\frac{1}{2} (q_{0} + v_{0}) e^{t} + \frac{1}{2} (q_{0} - v_{0}) e^{- t}, \frac{1}{2} (q_{0} + v_{0}) e^{t} - \frac{1}{2} (q_{0} - v_{0}) e^{- t}) .$

For a state $a$ and a nonzero tangent vector $u$ with $D_{a} Φ_{t} (a) u \neq 0$ , the finite-time tangent growth rate is $t^{- 1} \log (∥ D_{a} Φ_{t} (a) u ∥_{2} / ∥ u ∥_{2})$ ; when the forward lifetime is unbounded and its finite limit as $t \to \infty$ exists, that limit is the Lyapunov exponent (Finite-time tangent growth rates and Lyapunov exponents of the model's flow: the rate per unit time of a transported deviation, with its existence convention and its limits clauses (i)-(ii)). Both displayed flows are complete, so this lifetime condition holds for each state.

Assertions to check. (i) Both displayed maps are the local flows of their fields; (ii) for the harmonic model $D_{a} Φ_{t}^{1} (a)$ is the orthogonal block rotation $(\begin{array}{cc} \cos t I_{3} & \sin t I_{3} \\ - \sin t I_{3} & \cos t I_{3} \end{array}),$ so every finite-time tangent growth rate vanishes and every existing Lyapunov exponent is $0$ ; (iii) for the repulsive model the rates along the directions $(e_{1}, - e_{1})$ and $(e_{1}, e_{1})$ are $- 1$ and $+ 1$ , so the Lyapunov exponent of the equilibrium along the unstable eigenvector is $1$ .

Facts & Assumptions

Given: the two force maps and their equivalent chart data with $M = I_{3}$ , the displayed candidate flows and the finite-time growth-rate definition of Finite-time tangent growth rates and Lyapunov exponents of a C¹ flow as recorded in Finite-time tangent growth rates and Lyapunov exponents of the model's flow: the rate per unit time of a transported deviation, with its existence convention and its limits.

[L1]

Uniqueness of force-model motions (Local existence and uniqueness for the Newtonian initial-value problem, with the quantitative Picard-Lindelöf cylinder): for the adopted physical force maps, every initial datum has a unique maximal model motion; consequently a globally defined curve that solves the equation with the given initial data is the model's maximal motion. The invertible state and time normalization transfers this uniqueness and maximality to the numerical systems.

[L2]

calculus (The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivatives of sine and cosine are cosine and minus sine, The natural logarithm as the inverse of the exponential function, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): the exponential and trigonometric functions have their standard derivatives, sums and constant multiples differentiate termwise, and $\log$ is the natural logarithm, so that $\log e^{s} = s$ for real $s$ .

[L3]

Pythagorean identity (Parity and the Pythagorean identity for sine and cosine): for every real $t$ , $\sin^{2} t + \cos^{2} t = 1$ .

Verification

technique · direct

1.1givenalgebra

The two fields: for the constant metric $M = I_{3}$ the mechanical field of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart is $G (q, v) = (v, - \nabla U (q))$ , giving $G_{1} (q, v) = (v, - q)$ and $G_{2} (q, v) = (v, q)$ . These are also the normalized force-model fields of the declared physical $F_{1, p h y s}, F_{2, p h y s}$ .

2.1givenstep 1.1L1L2algebra

Harmonic flow: differentiating the displayed $Φ_{t}^{1}$ with respect to $t$ by [L2] gives $(\dot{q}, \dot{v}) = (v, - q)$ ; at $t = 0$ the map is the identity. By [L1] the displayed map is the force model's maximal flow.

2.2givenstep 1.1L1L2algebra

Repulsive flow: differentiating the displayed $Φ_{t}^{2}$ gives $\dot{q} = v (t)$ and $\dot{v} = q (t)$ ; at $t = 0$ the map is the identity, so by [L1] it is the force model's maximal flow of $G_{2}$ .

3.1step 2.1givenL2L3algebra

Harmonic growth rates: $Φ_{t}^{1}$ is linear in $a = (q_{0}, v_{0})$ , with derivative $D_{a} Φ_{t}^{1} (a) = (\begin{array}{cc} \cos t I_{3} & \sin t I_{3} \\ - \sin t I_{3} & \cos t I_{3} \end{array});$ writing $c = \cos t$ , $s = \sin t$ , its transpose times itself has diagonal blocks $(c^{2} + s^{2}) I_{3} = I_{3}$ by [L3] and off-diagonal blocks $c s I_{3} - s c I_{3} = 0$ , so it is orthogonal, whence $∥ D_{a} Φ_{t}^{1} (a) u ∥_{2} = ∥ u ∥_{2}$ for every $u \neq 0$ . The explicit flow exists for every real time, so the forward lifetime is unbounded, as required by the exponent definition. Therefore $r (t, a, u) = t^{- 1} \log 1 = 0$ for every $t > 0$ , and the limit defining the Lyapunov exponent exists and equals $0$ ; the variational equation of Linearisation along a motion of the mechanical model: the flow derivative solves the variational equation, with Lipschitz dependence, the constant-metric second-order form, and the conditional finite-time error bound of the linear approximation is satisfied by this derivative matrix, whose columns solve $\dot{Y} = D G_{1} Y$ by step 2.1.

3.2step 2.2givenL2algebra

Repulsive growth rates: for an initial tangent vector $(δ q, δ v)$ put $u : = δ q + δ v$ and $w : = δ q - δ v$ . The derivative flow satisfies $δ q (t) + δ v (t) = e^{t} u$ and $δ q (t) - δ v (t) = e^{- t} w$ . For each $j \in {1, 2, 3}$ , input $(e_{j}, e_{j})$ gives output $e^{t} (e_{j}, e_{j})$ and rate $+ 1$ , while input $(e_{j}, - e_{j})$ gives output $e^{- t} (e_{j}, - e_{j})$ and rate $- 1$ . The explicit flow exists for every real time, so the lifetime hypothesis holds and these Lyapunov exponents exist. At the equilibrium $a = (0, 0)$ this gives a positive exponent along each unstable coordinate eigenvector.

4.1step 3.1step 3.2givenL1L2algebra∎

Scope check: the chain rule transfers each numerical motion to $X (T) = ℓ_{0} q (T / t_{0})$ satisfying its physical Newton equation. A rate $r$ per numerical time $t$ becomes $r / t_{0}$ per physical time $T$ ; thus the physical rates are $0$ and $\pm 1 / t_{0}$ , and the positive physical exponent is $1 / t_{0}$ . All steps substitute declared normalized data into the chart formulas and the finite-time-rate definition; no step produces a chaotic invariant set, a mixing property or a measurement, and the positive exponent of step 3.2 concerns the linear model at its equilibrium, as the page's counterexample discusses.

Remarks

What is genuinely checked. The example verifies that the variational equation holds with the displayed derivative matrices and that the growth rate is invariant under multiplication of the deviation by nonzero scalars (the logarithm of a positive factor cancels against the norm ratio), so the rate is a function of the direction, not of the length of $u$ . Both flows are complete on the declared chart; no domain-escape phenomenon occurs.
The unstable direction. For the repulsive model the directions $(e_{j}, e_{j})$ and $(e_{j}, - e_{j})$ are exactly the eigendirections of $A = D G_{2} (0, 0) = (\begin{array}{cc} 0 & I_{3} \\ I_{3} & 0 \end{array})$ , with eigenvalues $+ 1$ and $- 1$ , respectively, so the rates $\pm 1$ are the eigenvalues of the linearisation, in agreement with Hyperbolic linearisation of an equilibrium of the model: exponential dichotomy of the linearised flow, local exponential asymptotic stability in the stable case, and instability in the saddle case clause 1 with $c = 1$ .

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

The three-dimensional isotropic oscillator: the stroboscopic map at half period is minus the identity, every state is periodic, the periodic points are dense, and the map is neither mixing nor sensitive

Physical scope

A declared-data check on the physical one-label force model with mass m0=1 kg, force F_phys(T,X,V)=-(m0/t0^2)X and potential U_phys(X)=m0||X||^2/(2t0^2), using ell0=1 m and t0=1 s. The formulas use numerical coordinates q=X/ell0, v=t0 V/ell0 and t=T/t0, normalized mass 1 and potential ||q||^2/2. The stroboscopic map at numerical half-period pi is minus the identity and at 2pi is the identity; the first-return map to q_y=0, v_y>0 is the identity. All states return, including the constant zero state. Periodic points of the map are dense, but the map is neither mixing nor sensitive and is not chaotic in the stated sense. Physical periods multiply by t0. All parameters are declared model data; no measurement is reported.

Example

Declared model data and normalization. Fix one label, $Q : = R^{3}$ , $m_{0} : = 1 k g$ , $ℓ_{0} : = 1 m$ and $t_{0} : = 1 s$ . The physical force map is $F_{p h y s} : R \times Q \times R^{3} \to R^{3}$ , $F_{p h y s} (T, X, V) = - (m_{0} / t_{0}^{2}) X$ , in newtons, with $T$ in seconds, $X$ in metres and $V$ in metres per second. This smooth globally state-Lipschitz map and mass $m_{0}$ specify a force model of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance. Put $t = T / t_{0}$ , $q = X / ℓ_{0}$ and $v = t_{0} V / ℓ_{0}$ . Newton's equation becomes $q^{''} = - q$ since $\ddot{X} = (ℓ_{0} / t_{0}^{2}) q^{''}$ . The normalized force is $F (t, q, v) : = F_{p h y s} (t_{0} t, ℓ_{0} q, (ℓ_{0} / t_{0}) v) / (m_{0} ℓ_{0} / t_{0}^{2}) = - q$ ; $m = 1$ is its normalized mass. With $U_{p h y s} (X) = m_{0} ∥ X ∥_{2}^{2} / (2 t_{0}^{2})$ and $E_{0} = m_{0} ℓ_{0}^{2} / t_{0}^{2}$ , $U = U_{p h y s} / E_{0}$ gives, in the chart $Ω : = R^{3}$ the constant kinetic metric is $M : = I_{3}$ and the potential is $U (q) : = \frac{1}{2} ∥ q ∥_{2}^{2}$ of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart. The field and motion through $(q_{0}, v_{0})$ are $G (q, v) = (v, - q), q (t) = q_{0} \cos t + v_{0} \sin t, v (t) = - q_{0} \sin t + v_{0} \cos t .$ Regard the autonomous force model as $P$ -periodic for any $P > 0$ , and let $Ψ_{P}$ be its stroboscopic map of The stroboscopic map of a periodic model and the Poincare return map of a transversal section: domains, iterates, fixed points and the local C1 diffeomorphism property clause (ii). The complete flow makes each $Ψ_{P}$ defined on all of $R^{6}$ . Numerical period $P$ means physical period $t_{0} P$ ; all state norms below use the dimensionless coordinates $(q, v)$ . Use the section $Σ_{+} : = {(q, v) : q_{y} = 0, v_{y} > 0}$ ; its first-return map sends each section state to the state at its least positive return time, with the local $C^{1}$ construction supplied by The stroboscopic map of a periodic model and the Poincare return map of a transversal section: domains, iterates, fixed points and the local C1 diffeomorphism property clause (iii).

Assertions to check. (i) The displayed curves are the model's motions; (ii) $Ψ_{π} = - i d$ and $Ψ_{2 π} = i d$ , so every state of the model is periodic under the flow with period $2 π$ and every point is a periodic point of $Ψ_{π}$ with period dividing $2$ (the zero state has a constant motion; only nonzero states give nonconstant periodic motions); (iii) the first-return map of $Σ_{+}$ is the identity; (iv) with the Euclidean metric of the phase chart, $Ψ_{π}$ on $X : = R^{6}$ is continuous and has dense periodic points, but is not topologically mixing and has no sensitivity constant, hence is not chaotic in the stated sense of Chaos in the stated sense for maps of the model: topological mixing, dense periodic points and sensitive dependence, as one precise list among several.

Facts & Assumptions

Given: the declared one-label force model $Q = R^{3}$ , $m = 1$ , $F (T, q, v) = - q$ , its equivalent chart data $M = I_{3}$ , $U (q) = ∥ q ∥_{2}^{2} / 2$ , the field $G$ and the displayed motion formulas.

[L1]

Unique force-model motions (Local existence and uniqueness for the Newtonian initial-value problem, with the quantitative Picard-Lindelöf cylinder): for the adopted physical smooth force map, every initial datum has a unique maximal motion, and every other solution through that datum is its restriction. The invertible state and time normalization above transfers existence, uniqueness and maximality to the displayed numerical system.

[L2]

Calculus (The derivatives of sine and cosine are cosine and minus sine, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ): $(\sin)^{'} = \cos$ , $(\cos)^{'} = - \sin$ , $\sin 0 = 0$ and $\cos 0 = 1$ , with the sum, constant-multiple and chain rules.

[L3]

The chaos list (Topological mixing, dense periodic points, sensitive dependence, and the stated chaos list): a map is chaotic in the stated sense exactly when it is continuous, topologically mixing, has dense periodic points and has sensitive dependence with a positive constant; the negation of topological mixing or of sensitivity therefore suffices to fail the list.

[L4]

The local first-return proposition (A transversal return of a C¹ flow yields a local C¹ Poincaré map, with inverse given by the reversed crossing): for a $C^{1}$ autonomous field and a $C^{1}$ section, two transversal crossings define a local return map, and exclusion of intervening hits from the section closure makes it a first-return map.

[L5]

Trigonometric values and zeros (Quarter-turn values and shifts by pi/2 and pi, The zero sets of sine and cosine and the least positive common period 2 pi): $\sin π = 0$ , $\cos π = - 1$ , and sine and cosine have period $2 π$ , so $\sin (2 π) = \sin 0 = 0$ , $\cos (2 π) = \cos 0 = 1$ using [L2]. For every real $t$ , $\sin t = 0$ exactly when $t = k π$ for an integer $k$ . In particular, the only zero in $0 < t < 2 π$ is $t = π$ .

Verification

technique · direct

1.1givenL1L2algebra

Field and motion: with $M = I_{3}$ and $U (q) = ∥ q ∥_{2}^{2} / 2$ the mechanical field is $G (q, v) = (v, - q)$ (The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart), which is also the normalized field of the declared physical force map. Differentiating the displayed vector formulas gives $\dot{q} = v$ and $\dot{v} = - q$ , and at $t = 0$ they reproduce the initial datum; by [L1] these are the unique maximal model motions.

2.1step 1.1givenL2L5algebra

Stroboscopic maps: evaluating step 1.1 at $t = π$ uses $\cos π = - 1$ , $\sin π = 0$ and gives $Ψ_{π} (q, v) = (- q, - v)$ ; at $t = 2 π$ it gives the identity. Thus every state returns after $2 π$ , and $Ψ_{π}$ is an involution, so every point of $X = R^{6}$ is a periodic point of $Ψ_{π}$ with period $1$ or $2$ ; in particular the periodic points are all of $X$ and are dense.

2.2step 1.1givenL2L4L5algebra

First return to $Σ_{+}$ : take the open state neighbourhood $W : = {(q, v) : v_{y} > 0}$ and the section function $h : W \to R$ , $h (q, v) : = q_{y}$ , so $h^{- 1} (0) = Σ_{+}$ within $W$ . At every point of $Σ_{+}$ , $D h G = v_{y} > 0$ . A state with $q_{y 0} = 0$ and $v_{y 0} > 0$ has $q_{y} (t) = v_{y 0} \sin t$ and $v_{y} (t) = v_{y 0} \cos t$ , so the next hit of $Σ_{+}$ is at $t = 2 π$ ; at that time the full six-dimensional state returns to its initial value by the formulas of step 1.1. The open segment $0 < t < 2 π$ avoids $\overline{Σ_{+}} = {q_{y} = 0, v_{y} \geq 0}$ : by [L5] its only intermediate zero is at $t = π$ , where $v_{y} = - v_{y 0} < 0$ . Applying [L4] at each reference point gives the local first-return map, and the explicit calculation identifies it with the identity on all of $Σ_{+}$ .

3.1step 2.1L3algebra

Failure of mixing: let $Z \in X = R^{6}$ be a state with $∥ Z ∥_{2} > 1$ and let $U : = B (Z, 1)$ be the open Euclidean ball (Open ball, closed ball and sphere in a metric space); then $U \cap (- U) = \emptyset$ , and step 2.1 gives $Ψ_{π}^{n} (U) = U$ for even $n$ and $Ψ_{π}^{n} (U) = - U$ for odd $n$ . Taking $V : = U$ in the mixing condition of [L3], the intersection $Ψ_{π}^{n} (U) \cap V$ is empty for every odd $n$ ; consequently no integer $N$ has $Ψ_{π}^{n} (U) \cap V \neq \emptyset$ for all $n \geq N$ , and $Ψ_{π}$ is not topologically mixing on $X$ .

4.1step 2.1step 3.1L3algebra

Failure of sensitivity: for all states $Z, Z^{'} \in R^{6}$ and all $n \geq 1$ , step 2.1 gives $∥ Ψ_{π}^{n} (Z) - Ψ_{π}^{n} (Z^{'}) ∥_{2} = ∥ Z - Z^{'} ∥_{2}$ because each iterate is $\pm i d$ , an isometry; hence for no $δ > 0$ can one find, for every $Z$ and every ball around $Z$ , a nearby state whose iterate is $δ$ -separated. The sensitivity condition of [L3] fails, and with it the chaos list; the periodic-point and continuity clauses alone do not make $Ψ_{π}$ chaotic in the stated sense.

5.1step 2.2step 4.1givenL1L2algebra∎

Reading: the maps use the dimensionless numerical state and time variables defined above. By the chain rule, $X (T) = ℓ_{0} q (T / t_{0})$ obeys $m_{0} \ddot{X} = - (m_{0} / t_{0}^{2}) X$ ; uniqueness transfers through this invertible normalization. The physical first-return time is $2 π t_{0}$ and the physical half-period is $π t_{0}$ . No measurement, sampling or empirical comparison is performed.

Remarks

Why the check matters. It exhibits a model whose stroboscopic map is continuous with dense periodic points yet fails two of the four clauses of the page's chaos list, showing that the list is a genuine restriction rather than a set of automatic consequences of periodicity. It is the discrete-time companion of the finite-time growth checks: rate zero everywhere (Two declared linear models worked through: the harmonic flow has zero finite-time growth in every direction, while the repulsive linear model has exact rates $\pm 1$ and the positive Lyapunov exponent $1$ at its equilibrium) and no chaos.
The return map. On the regular section ${q_{y} = 0, v_{y} \neq 0}$ the first-return time is $π$ , and its return map sends $(q, v)$ to $(- q, - v)$ ; composing this return map twice gives the identity, including on $Σ_{+}$ . The section is transversal because $D h G = v_{y} \neq 0$ on it, in agreement with the hypotheses of The stroboscopic map of a periodic model and the Poincare return map of a transversal section: domains, iterates, fixed points and the local C1 diffeomorphism property clause (iii). For $Σ_{+}$ , its closure is ${q_{y} = 0, v_{y} \geq 0}$ ; the only intermediate zero at time $π$ has $v_{y} = - v_{y 0} < 0$ , so the open reference segment avoids that closure. For the regular two-sided section its closure is ${q_{y} = 0}$ , and no intermediate zero occurs before $π$ . These observations also verify the proposition's extra first-return condition.

ExampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

The pulsed horseshoe on declared data: $a = 100$ , the radicand box $[74, 126]$ , the contraction factor $1 / \sqrt{74}$ , the limiting quadratic fixed points $1 \pm \sqrt{101}$ , and the physical bounds of the coded states at $ℓ_{0} = 1 m$ , $m = 1 k g$ , $ω_{0} = 2 π s^{- 1}$

Physical scope

A declared-data check of the constants and bounds in the page's pulsed-horseshoe theorem: at amplitude a=100 the coding box is R=13, the radicands lie in [74,126], the contraction factor is 1/sqrt(74), the limiting quadratic map has fixed points 1 plus or minus sqrt(101), the limiting kick satisfies K composed with the rotation = F, and with length scale 1 m, mass 1 kg and frequency scale 2 pi per second the coded states on the invariant axis are bounded by 13 m in axial position and 26 pi kg m per second in axial momentum, with period (pi/2+delta)/(2 pi) seconds. All quantities are declared model data; the pulse-width threshold is not numerically evaluated and no measurement is involved.

Example

Declared model data. Take the full physical pulsed oscillator model of A smooth periodically forced one-label Cartesian model with a transverse homoclinic horseshoe: the physical stroboscopic map has a compact invariant set conjugate to the shift, chaotic in the stated sense clause (i), and restrict its coded states to the invariant axis, where transverse coordinates and momenta vanish. Use the declared data $a : = 100, ℓ_{0} : = 1 m, m : = 1 k g, ω_{0} : = 2 π s^{- 1},$ and a pulse width $0 < δ < δ_{0}$ with $δ_{0}$ the supplier threshold (not evaluated numerically here). Let $R : = \sqrt{a} + 3$ , $L : = \sqrt{a - 2 R}$ and let $r_{\pm}$ be the fixed points of the limiting quadratic map.

Assertions to check. (i) $R = 13$ and the coding radicands $a + x_{n - 1} + x_{n + 1}$ lie in $[74, 126] = [a - 2 R, a + 2 R]$ , with $\sqrt{126} < 13 = R$ ; (ii) the contraction factor of the coding map is $1 / L = 1 / \sqrt{74}$ and $r_{\pm} = 1 \pm \sqrt{101}$ ; (iii) for the limiting pulse, $K \circ R_{0} = F$ with $R_{0} (q, p) = (p, - q)$ , $K (q, p) = (q, p + q^{2} - a)$ and $F (x, y) = (y, y^{2} - a - x)$ ; (iv) every coded state satisfies $∣ Q ∣ \leq 13 m$ and $∣ Π ∣ \leq 26 π k g m s^{- 1}$ , and the forcing period is $T_{p} = (π / 2 + δ) / (2 π) s$ .

Facts & Assumptions

Given: the constants $a = 100$ , $R = \sqrt{a} + 3$ , $L = \sqrt{a - 2 R}$ and the scales $ℓ_{0}, m, ω_{0}$ above; the dimensionless model, its rotation $R_{0}$ , pulse endpoint maps $K_{δ}$ and stroboscopic map $G_{δ} = K_{δ} \circ R_{0}$ of The smooth $P$ -periodic pulsed oscillator: the normalized pulse, its Hamiltonian and its stroboscopic (period) map; the limiting kick $K$ .

[L1]

The horseshoe theorem and its bounds (A smooth periodically forced one-label Cartesian model with a transverse homoclinic horseshoe: the physical stroboscopic map has a compact invariant set conjugate to the shift, chaotic in the stated sense clauses (ii)-(iv)): for $0 < δ < δ_{0}$ the full physical stroboscopic map $Ψ_{δ}$ has a compact invariant coded set on the invariant axis, with transverse coordinates and momenta zero; each state satisfies $∣ Q ∣ \leq R ℓ_{0}$ and $∣ Π ∣ \leq m ℓ_{0} ω_{0} R$ , and its coded trajectories exist for all times.

[L2]

The coding contraction (The square-root recurrence $T_{s} (x)_{n} = s_{n} \sqrt{a + x_{n - 1} + x_{n + 1}}$ is a contraction of $B_{R}$ with a unique coded solution) and the limiting pulse (The rescaled pulse endpoint map converges to the kick $K (q, p) = (q, p + q^{2} - a)$ in $C^{1}$ , and the period map is smooth, symplectic and $C^{1}$ -close to $F$ ): the coding map $T_{σ}$ on the ball of bounded sequences with $∣ x ∣_{\infty} \leq R$ is a contraction with factor $L^{- 1}$ when the radicands stay positive and the derivative bound holds, and $K_{δ} \to K$ in $C^{1}$ as $δ \to 0$ with $K (q, p) = (q, p + q^{2} - a)$ .

[L3]

Elementary arithmetic: $R^{2} - a - 2 R = 4 \sqrt{a} + 3 > 0$ for $a = 100$ , so $R = \sqrt{a} + 3$ satisfies $\sqrt{a + 2 R} < R$ ; the fixed points of $x \mapsto$ the constant-code recurrence $x^{2} - a = 2 x$ solve $x^{2} - 2 x - a = 0$ .

Verification

technique · direct

1.1givenL2L3algebra

Arithmetic of the coding box: for $a = 100$ , $R = \sqrt{100} + 3 = 13$ ; hence $a - 2 R = 100 - 26 = 74$ and $a + 2 R = 100 + 26 = 126$ , so the radicands of [L2] lie in $[74, 126]$ whenever the coded sequence satisfies $∣ x_{n} ∣ \leq R$ . Moreover $\sqrt{126} < 13$ because $126 < 169$ , and $\sqrt{74} > 0$ , so all square roots in the coding recurrence are defined with the derivative bound required by [L2]. Also $R^{2} - a - 2 R = 169 - 100 - 26 = 43 = 4 \sqrt{100} + 3 > 0$ , which is the supplier's inequality ensuring $T_{σ}$ maps the ball into itself.

1.2givenL2algebra

Limiting pulse composition: with $R_{0} (q, p) = (p, - q)$ and $K (q, p) = (q, p + q^{2} - a)$ one computes $K (R_{0} (q, p)) = K (p, - q) = (p, - q + p^{2} - a)$ and $F (q, p) = (p, p^{2} - a - q)$ , which agree; this is the algebraic core of the $C^{1}$ -closeness statement $G_{δ} \to F$ of [L2].

2.1step 1.1givenL2L3algebra

Contraction and fixed points: the radicand bound of step 1.1 gives the derivative bound $\frac{1}{2 \sqrt{74}}$ for the square root, hence the contraction factor $(\frac{1}{2 \sqrt{74}}) \cdot 2 = \frac{1}{\sqrt{74}} = L^{- 1}$ for the coding map, as in [L2] with $L = \sqrt{74}$ . Constant codes satisfy $x_{n + 1} + x_{n - 1} = x_{n}^{2} - a$ with $x_{n} \equiv r$ , that is $2 r = r^{2} - a$ or $r^{2} - 2 r - a = 0$ , whose solutions are $r = 1 \pm \sqrt{1 + a} = 1 \pm \sqrt{101}$ . The branch conditions hold with exact inequalities: $\sqrt{101} > 8$ because $101 > 64$ , so $∣ r_{-} ∣ = \sqrt{101} - 1 > 7$ ; $\sqrt{101} > 1 + \sqrt{74}$ because $101 > 75 + 2 \sqrt{74}$ reduces to $13 > \sqrt{74}$ , that is $169 > 74$ ; so $∣ r_{\pm} ∣ > L$ ; and $r_{+} \leq 13$ , $∣ r_{-} ∣ \leq 13$ because $\sqrt{101} \leq 12$ and $\sqrt{101} - 1 \leq 13$ , both from $101 \leq 144$ .

2.2step 1.1givenL1algebra

Physical bounds: by [L1] clause (iv) every coded physical state satisfies $∣ Q ∣ \leq R ℓ_{0} = 13 m$ and $∣ Π ∣ \leq m ℓ_{0} ω_{0} R = 13 \cdot 2 π k g m s^{- 1} = 26 π k g m s^{- 1}$ ; the forcing period is $T_{p} = P / ω_{0} = (π / 2 + δ) / (2 π) s$ and the pulse duration is $δ / (2 π) s$ .

3.1step 2.1step 1.2step 2.2givenalgebra∎

Reading: every number above is declared model data or an exact surd; the threshold $δ_{0}$ is not numerically evaluated, no decimal approximation is substituted into a proof step, and no measurement or apparatus is involved (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).

Remarks

Status of the numerical values. The values $13$ , $74$ , $126$ , $1 / \sqrt{74}$ , $1 \pm \sqrt{101}$ and $26 π$ are exact consequences of the declared data; they are quoted as a check of the supplier's constants, not as an empirical estimate of any oscillation. In particular the threshold $δ$ below which the horseshoe exists is left symbolic because the supplier does not compute it.
Unit consistency. $[ℓ_{0}] = ℓ$ , $[Π] = m ℓ_{0} ω_{0} = μ ℓ τ^{- 1}$ , $[T_{p}] = τ$ , and the coded box is bounded in both position and canonical momentum (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

A positive Lyapunov exponent at a hyperbolic equilibrium does not produce a chaotic invariant set of the time-one map: the three-dimensional repulsive model has rate $1$ and no compact invariant set with more than one point

Physical scope

A declared-model counterexample in the one-label three-dimensional force branch, refuting the implication from a positive Lyapunov exponent at an equilibrium to a chaotic compact invariant set of the time-one map: the dimensionless repulsive model with potential U(q)=-||q||^2/2 has a complete flow, exponent 1 at its equilibrium along each unstable direction, and a nonlinearly unstable equilibrium, yet every nonempty compact invariant set of its time-one map is {0}; this singleton fails the sensitivity clause of the page's chaos list. The counterexample is a model computation; it reports no measurement.

Statement refuted

False statement. Let $G : S_{0} \to R^{6 N}$ be the $C^{1}$ associated field of a complete autonomous stability model in the sense of The model vocabulary of stability and chaos: the associated flow of a declared force model, equilibria, periodic and homoclinic orbits, and invariant sets, with flow $Φ : R \times S_{0} \to S_{0}$ . If an equilibrium has a positive real eigenvalue $λ$ of $D G$ and Lyapunov exponent $λ$ along a corresponding eigenvector, then the time-one map $f : = Φ_{1} : S_{0} \to S_{0}$ possesses a compact invariant set with more than one point on which its restriction is chaotic in the stated sense of Chaos in the stated sense for maps of the model: topological mixing, dense periodic points and sensitive dependence, as one precise list among several (continuous, topologically mixing, dense periodic points, sensitive dependence).

Facts & Assumptions

Given: the dimensionless one-label Cartesian force model with $Q = R^{3}$ , mass $m = 1$ , kinetic matrix $M = I_{3}$ , potential $U (q) = - \frac{1}{2} ∥ q ∥_{2}^{2}$ , force $F (q, v) = q$ , field $G (q, v) = (v, q)$ , equilibrium $(0, 0)$ , and complete flow $Φ_{t} (q, v) = (\frac{1}{2} (q + v) e^{t} + \frac{1}{2} (q - v) e^{- t}, \frac{1}{2} (q + v) e^{t} - \frac{1}{2} (q - v) e^{- t}) .$

[L1]

The growth computation (Two declared linear models worked through: the harmonic flow has zero finite-time growth in every direction, while the repulsive linear model has exact rates $\pm 1$ and the positive Lyapunov exponent $1$ at its equilibrium): for $u : = q + v$ and $w : = q - v$ , the derivative flow acts as $(u, w) \mapsto (e^{t} u, e^{- t} w)$ ; along $(e_{1}, e_{1})$ the finite-time rate at the equilibrium is $1$ for every $t > 0$ , so the Lyapunov exponent is $1$ , and the linearisation has positive eigenvalue $1$ .

[L2]

The chaos list (Topological mixing, dense periodic points, sensitive dependence, and the stated chaos list): sensitivity requires $δ > 0$ such that for every $x$ and every $ε > 0$ there are $y$ and $n \geq 1$ with $d (x, y) < ε$ and $d (f^{n} (x), f^{n} (y)) \geq δ$ ; on a one-point space the only admissible $y$ is $x$ , so no positive $δ$ satisfies the condition.

[L3]

Time-one invariance and instability (The model vocabulary of stability and chaos: the associated flow of a declared force model, equilibria, periodic and homoclinic orbits, and invariant sets clause (iv), Hyperbolic linearisation of an equilibrium of the model: exponential dichotomy of the linearised flow, local exponential asymptotic stability in the stable case, and instability in the saddle case clause 3): for the complete flow, $f = Φ_{1}$ is a bijection with inverse $Φ_{- 1}$ ; a compact invariant set of $f$ satisfies $f (K) = K$ , and a hyperbolic equilibrium with a positive-real-part eigenvalue is unstable.

[L4]

Norms (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Open ball, closed ball and sphere in a metric space): $∥ (q, v) ∥_{2}^{2} = \frac{1}{2} (∥ u ∥_{2}^{2} + ∥ w ∥_{2}^{2})$ , so boundedness in the state chart implies boundedness of both transformed vectors.

Counterexample

technique · direct

1.1givenL1

The antecedent holds: by [L1] the model has equilibrium $(0, 0)$ , the linearisation has eigenvector $(e_{1}, e_{1})$ with eigenvalue $1 > 0$ , and the Lyapunov exponent along that direction exists and equals $1$ . The explicit flow is defined for every real time. This satisfies the antecedent of the refuted statement.

1.2givenL1L3L4algebra

Classification of compact invariant sets of $f$ : let $K$ be compact with $f (K) = K$ , and let a state of $K$ have transformed coordinate $u \neq 0$ . By [L1], $∥ f^{n} (q, v) ∥_{2} \geq \frac{1}{\sqrt{2}} ∥ u ∥_{2} e^{n}$ for $n \geq 0$ , which is unbounded; this contradicts boundedness of $K$ . Hence $u = 0$ on $K$ . Since $f$ is a bijection and $f (K) = K$ , applying the same argument to $f^{- n} (K) = K$ shows $w = 0$ on $K$ . Therefore every nonempty compact invariant set of $f$ is the singleton ${(0, 0)}$ .

2.1step 1.2L2algebra

The one-point invariant set fails the chaos list: for $X = {(0, 0)}$ every admissible $y$ in the sensitivity condition of [L2] coincides with the base point, so $d (f^{n} (x), f^{n} (y)) = 0 < δ$ for every $n$ and every $δ > 0$ ; the sensitivity clause fails, and with it the four-clause list of Chaos in the stated sense for maps of the model: topological mixing, dense periodic points and sensitive dependence, as one precise list among several.

3.1step 1.1step 1.2step 2.1L3∎

Conclusion: the antecedent of the refuted statement holds for the declared model (step 1.1), while its time-one map has no compact invariant set with more than one point (step 1.2) and its only compact invariant set is not chaotic in the stated sense (step 2.1); the implication is therefore false. Nonlinear instability of the equilibrium from [L3] does not repair it: instability of a single state and chaos of a compact invariant set are different properties.

Remarks

Scope of the refutation. The counterexample refutes the stated implication for the time-one map and the page's precise chaos list; it does not argue that positive exponents are irrelevant to chaotic behaviour, only that the value of an exponent at a single equilibrium does not by itself produce the listed properties. The horseshoe theorem A smooth periodically forced one-label Cartesian model with a transverse homoclinic horseshoe: the physical stroboscopic map has a compact invariant set conjugate to the shift, chaotic in the stated sense establishes those properties by an explicit conjugacy, not from an exponent.
No empirical content. The computation is a substitution in a dimensionless linear model; it involves no measurement, no numerical integration and no claim about physical systems exhibiting exponential separation.

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedOpen item page →

Lyapunov stability does not imply attraction: the harmonic model is stable at its equilibrium while every nonzero datum keeps its distance and never converges

Physical scope

A declared-model counterexample refuting the implication from Lyapunov stability to asymptotic stability for the page's mechanical models: the one-dimensional harmonic model with unit mass and unit stiffness has the equilibrium 0, which is Lyapunov stable by the strict-minimum theorem, while the datum (delta,0) has the periodic motion delta cos(t) whose distance from the equilibrium returns to delta at every period, so it never converges. The counterexample is a model computation; it reports no measurement.

Statement refuted

False statement. Every Lyapunov stable equilibrium of a one-dimensional mechanical chart model attracts every sufficiently close initial state; equivalently, Lyapunov stability implies asymptotic stability.

Facts & Assumptions

Given: the dimensionless harmonic chart data: the one-dimensional model of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart with constant kinetic metric $M = 1$ , potential $U (q) = \frac{1}{2} q^{2}$ , field $G (q, v) = (v, - q)$ and equilibrium $(0, 0)$ ; the comparison data $(δ, 0)$ for real $δ > 0$ .

[L1]

Stability of the strict minimum (Energy stability and instability of the mechanical model at an equilibrium: the strict minimum is Lyapunov stable, a negative Hessian eigenvalue gives linear instability, a negative definite potential Hessian gives nonlinear instability, and exact energy conservation rules out attraction clause 1): a strict local minimum of the potential makes the equilibrium Lyapunov stable, with the exact quantifiers of Stability vocabulary of the model: Lyapunov, asymptotic and orbital stability, linear stability of equilibria and map fixed points, and hyperbolicity of equilibria and fixed points clause (i).

[L2]

The harmonic motions (The motions of the harmonic model are the sinusoids $A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0}))$ : amplitude and phase, the least period, the conserved energy $k R^{2} / 2$ and the elliptical phase curves clauses 1-3): for the harmonic model with $m = k = 1$ and $ω = \sqrt{k / m} = 1$ , every motion is $x (t) = A \cos (t - t_{0}) + B \sin (t - t_{0})$ with $A = x (t_{0})$ , $B = v (t_{0})$ ; if $R = \sqrt{A^{2} + B^{2}} > 0$ then $x (t) = R \cos ((t - t_{0}) - φ)$ and the least positive period is $2 π$ ; the datum $(x (t_{0}), v (t_{0})) = (δ, 0)$ has $A = δ$ , $B = 0$ .

Counterexample

technique · direct

1.1givenL1

The equilibrium is Lyapunov stable: $U (q) = \frac{1}{2} q^{2}$ has the strict local minimum $q = 0$ ( $U (q) > U (0)$ for $q \neq 0$ ), so [L1] makes $(0, 0)$ Lyapunov stable in the chart model.

1.2givenL2algebra

The nonzero datum does not converge: by [L2] the datum $(δ, 0)$ at time $t_{0} = 0$ has the maximal motion $x (t) = δ \cos t$ , with $A = δ$ , $B = 0$ , $R = δ > 0$ ; the least positive period is $2 π$ and $x (2 π n) = δ \cos (2 π n) = x (0) = δ$ for every integer $n$ . Hence the sequence $t_{n} = 2 π n$ tends to infinity while $∥ (x (2 π n), x^{'} (2 π n)) - (0, 0) ∥_{2} \geq δ$ ; the state does not converge to the equilibrium.

2.1step 1.1step 1.2given∎

Conclusion: the equilibrium is Lyapunov stable (step 1.1) but for every $δ > 0$ the datum $(δ, 0)$ , arbitrarily close to the equilibrium, fails to converge (step 1.2); the equilibrium is therefore not asymptotically stable, and the implication of the refuted statement is false for the declared model. The stronger statement of Energy stability and instability of the mechanical model at an equilibrium: the strict minimum is Lyapunov stable, a negative Hessian eigenvalue gives linear instability, a negative definite potential Hessian gives nonlinear instability, and exact energy conservation rules out attraction clause 4 explains the failure: the energy $\frac{1}{2} (v^{2} + q^{2})$ is conserved, equals $δ^{2} / 2$ on this datum, and no state of different energy can converge to the equilibrium.

Remarks

Scope. The refutation concerns the page's exact definition of asymptotic stability (attraction for all data in a neighbourhood, after Lyapunov stability). It does not deny that this particular datum has bounded distance from the equilibrium, nor that a damped model would converge; damping is a different model with its own postulates (The damped oscillator model: a harmonic restoring force with a phenomenological linear drag $- c \dot{x}$ , $c \geq 0$ ).
No empirical content. The computation uses dimensionless declared data; no measurement, no friction and no apparatus are involved.

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A trapped one-label model: the compact energy sublevel of the radial quartic potential is invariant with finite phase volume, so the recurrence theorem applies, while the free model's energy sublevels are unbounded and the hypothesis fails

Physical scope

A declared-data check in dimensionless normalized coordinates of the page's recurrence theorem on a trapped one-label, three-dimensional Cartesian model: for the unit-mass radial quartic potential U(x)=||x||^4/4 the six-dimensional phase-space energy sublevel {E<=E1}, E1>0, is compact with finite dimensionless coordinate volume bounded by 512 E1^(9/4), invariant under the flow by energy conservation and compact continuation, so the recurrence theorem gives almost-everywhere infinite returns; in contrast the dimensionless free model U=0 has unbounded energy sublevels and each nonzero-velocity trajectory eventually exits every bounded set, showing the finite-volume hypothesis is not automatic. No SI quantity, measurement or empirical comparison is asserted.

Example

Declared model data. Use dimensionless normalized position, velocity, mass, time and energy coordinates; this is a dimensionless model computation and no SI units are assigned. Take one label in the flat Cartesian presentation of Poincare recurrence for the model's Hamiltonian flow on a bounded invariant set of finite phase volume: almost every state of any Borel subset returns to it infinitely often with $N = 1$ : the configuration domain is $Q : = R^{3}$ , the mass is $m : = 1$ so that $M = I_{3}$ and $W = Q \times R^{3}$ with coordinates $(x, v)$ , and the radial quartic potential is $U (x) : = \frac{1}{4} ∥ x ∥_{2}^{4},$ a $C^{\infty}$ datum of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart with field $G (x, v) = (v, - ∥ x ∥_{2}^{2} x)$ and energy $E (x, v) = \frac{1}{2} ∥ v ∥_{2}^{2} + \frac{1}{4} ∥ x ∥_{2}^{4}$ . For $E_{1} > 0$ put $A_{E_{1}} : = {(x, v) \in W : E (x, v) \leq E_{1}} .$

Assertions to check. (i) $A_{E_{1}}$ is a nonempty compact Borel subset of $W$ with $λ_{6} (A_{E_{1}}) \leq (2 (4 E_{1})^{1 / 4})^{3} (2 \sqrt{2 E_{1}})^{3} = 512 E_{1}^{9 / 4} < \infty$ ; (ii) the flow of the model maps $A_{E_{1}}$ onto itself for every real time, so the hypotheses of Poincare recurrence for the model's Hamiltonian flow on a bounded invariant set of finite phase volume: almost every state of any Borel subset returns to it infinitely often are met and for every Borel $E \subseteq A_{E_{1}}$ almost every state of $E$ returns to $E$ infinitely often; (iii) for the flat model $U \equiv 0$ the energy sublevels are unbounded, the phase measure of $W$ is infinite, and each nonzero-velocity free motion eventually exits every bounded set (hence visits a fixed bounded set at only finitely many integer times), so the finite-volume hypothesis is not automatic.

Facts & Assumptions

Given: the data $N = 1$ , $M = I_{3}$ , $U (x) = ∥ x ∥_{2}^{4} / 4$ , its field and energy; the sublevel $A_{E_{1}}$ ; the flat comparison data $U \equiv 0$ with field $G_{0} (x, v) = (v, 0)$ .

[L1]

Recurrence (Poincare recurrence for the model's Hamiltonian flow on a bounded invariant set of finite phase volume: almost every state of any Borel subset returns to it infinitely often): for an autonomous potential presentation with a bounded invariant Borel set $A$ whose compact closure lies in the phase domain and whose flow is defined for all real times on $A$ , phase volume is preserved on its Borel subsets by finite local covers and time subdivision; almost every state of every Borel $E \subseteq A$ returns to $E$ infinitely often at integer times. The theorem assumes ${A C}_{ω}$ , inherited from the phase-volume result.

[L2]

Energy conservation and compact continuation (Energy is conserved along solutions of the mechanical system with a configuration-dependent kinetic metric, Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it): the energy of the mechanical system is constant along every maximal solution, and a solution whose graph stays in a compact subset of the state domain cannot reach a finite forward or backward endpoint.

[L3]

Borel and Lebesgue vocabulary (The Borel sigma-algebra of a topological space, Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ): closed sets are Borel, and the Lebesgue measure of a box in $R^{6}$ is the product of its side lengths.

Verification

technique · direct

1.1givenL3algebra

Compactness and phase volume: $E$ is continuous and $A_{E_{1}} = E^{- 1} ((- \infty, E_{1}])$ is closed; if $(x, v) \in A_{E_{1}}$ then $\frac{1}{4} ∥ x ∥_{2}^{4} \leq E_{1}$ and $\frac{1}{2} ∥ v ∥_{2}^{2} \leq E_{1}$ , so $∥ x ∥_{2} \leq (4 E_{1})^{1 / 4}$ and $∥ v ∥_{2} \leq \sqrt{2 E_{1}}$ ; hence $A_{E_{1}}$ is contained in the box of side lengths $2 (4 E_{1})^{1 / 4}$ (three position coordinates) and $2 \sqrt{2 E_{1}}$ (three velocity coordinates), is bounded and closed, therefore compact, and by [L3] has $λ_{6} (A_{E_{1}}) \leq (2 (4 E_{1})^{1 / 4})^{3} (2 \sqrt{2 E_{1}})^{3} = 512 E_{1}^{9 / 4} < \infty$ ; it is nonempty because $(0, 0) \in A_{E_{1}}$ .

2.1step 1.1givenL2algebra

Invariance: along a maximal solution of the quartic model the energy is constant by [L2], so the solution stays in $A_{E_{1}}$ ; because $A_{E_{1}}$ is compact and contained in the state domain $W$ , [L2] also shows the solution is defined for all times, and applying the same argument to the reversed motion gives $Φ_{t} (A_{E_{1}}) \supseteq A_{E_{1}}$ as well; hence $Φ_{t} (A_{E_{1}}) = A_{E_{1}}$ for every $t \in R$ .

3.1step 1.1step 2.1L1

Recurrence conclusion: $A_{E_{1}}$ is a bounded Borel set of finite measure with closure in the flow domain and is invariant by step 2.1; [L1] therefore gives, for every Borel $E \subseteq A_{E_{1}}$ , that almost every state of $E$ returns to $E$ at infinitely many positive integer times.

4.1step 1.1givenL1algebra∎

Failure of the hypothesis for the flat model: for $U \equiv 0$ , $E_{0} (x, v) = \frac{1}{2} ∥ v ∥_{2}^{2}$ and the sublevel ${E_{0} \leq E_{1}} = R^{3} \times {v : ∥ v ∥_{2} \leq \sqrt{2 E_{1}}}$ is unbounded, indeed it meets every translate of the velocity ball in the position directions; the phase domain $W = R^{6}$ has infinite Lebesgue measure, so [L1]'s boundedness and finite-measure hypotheses fail. For the motion through $(0, e_{1})$ , $x (t) = t e_{1}$ : if a fixed bounded phase set has $∥ x ∥_{2} \leq R$ on it, this trajectory can be in the set only for $0 \leq t \leq R$ . It eventually exits and makes only finitely many integer-time visits; this does not claim that no finite return occurs. The finite-volume hypothesis is a genuine restriction.

Remarks

What the check establishes. The radial quartic model is trapped: every energy sublevel is compact and invariant, so recurrence applies to every positive energy level in the flat $N = 1$ Cartesian presentation for which the theorem is stated. This realizes the hypothesis of Poincare recurrence for the model's Hamiltonian flow on a bounded invariant set of finite phase volume: almost every state of any Borel subset returns to it infinitely often on declared data and shows that the theorem is not vacuous. The conclusion is almost-everywhere in phase volume, gives no return time and says nothing about the flat model beyond the failure of its hypotheses.
Choice assumption. The check inherits the countable-choice assumption of [L1] through the cited theorem; no further choice is used.
Units. The coordinates and measure here are dimensionless. To give the model a physical-unit interpretation one would have to specify compatible length, mass, time and energy scales and restore them in both the potential and phase-volume bound; this example does not make that identification (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).

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Unit restoration and scale checks for the two-rotor KAM statement: the scaling identities at $I = E_{0} = 1$ , the physical angular frequencies, the exclusion of rational frequency labels, and the equal-rescaling invariance

Physical scope

A declared-data check of the units and labels in the page's two-rotor KAM theorem: with moment of inertia 1 kg m^2 and energy scale 1 J the scaling constants are J0=1 kg m^2 s^-1, t0=1 s and omega0=1 s^-1; the physical and dimensionless Hamilton equations correspond exactly; a surviving torus with dimensionless label nu carries physical angular frequencies omega0 nu_i; every rational label of the frequency cube is excluded from the Diophantine set; and equal rescaling of inertia and energy leaves the dimensionless model and the threshold unchanged. The constant c in the smallness condition is not numerically evaluated and no measured value is asserted.

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as in the cited two-rotor KAM theorem.

Declared model data. Take the two-rotor model of The two-rotor model: a positive-measure family of invariant Lagrangian tori with Diophantine frequencies for sufficiently small coupling, with the explicit units and the stated scope with $I : = 1 k g m^{2}, E_{0} : = 1 J, ε dimensionless,$ so that $J_{0} = \sqrt{I E_{0}} = 1 k g m^{2} s^{- 1}$ , $t_{0} = J_{0} / E_{0} = 1 s$ and $ω_{0} = \sqrt{E_{0} / I} = 1 s^{- 1}$ .

Assertions to check. (i) The dimensions $[H_{p h y s}] = J$ , $[Π_{i}] = k g m^{2} s^{- 1}$ , $[I] = k g m^{2}$ , $[ε] = 1$ and $[ω_{0}] = s^{- 1}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). (ii) With $I_{i} = Π_{i} / J_{0}$ , $s = t / t_{0}$ the physical Hamilton equations ${\dot{θ}}_{i} = Π_{i} / I$ , ${\dot{Π}}_{i} = E_{0} ε \sin θ_{i} \cos θ_{3 - i}$ are equivalent to $θ_{i}^{'} = I_{i}$ , $I_{i}^{'} = ε \sin ϑ_{i} \cos ϑ_{3 - i}$ . (iii) A surviving torus with dimensionless label $ν$ carries the physical angular frequencies $ω_{0} ν_{i}$ , here $ν_{i} s^{- 1}$ . (iv) Every rational label in the cube $C^{*}$ is excluded from $Ω_{α, τ}$ for every $α > 0$ and $τ > 1$ . (v) Replacing $(I, E_{0})$ by $(λ I, λ E_{0})$ with $λ > 0$ leaves $ω_{0}$ , the dimensionless model and the smallness condition unchanged, while rescaling $J_{0}$ by $λ$ .

Facts & Assumptions

Given: the declared data $I = 1 k g m^{2}$ , $E_{0} = 1 J$ and the scaling constants $J_{0}, t_{0}, ω_{0}$ ; the frequency cube $C^{*}$ and the Diophantine set $Ω_{α, τ}$ of The two-rotor model: a positive-measure family of invariant Lagrangian tori with Diophantine frequencies for sufficiently small coupling, with the explicit units and the stated scope.

[L1]

The two-rotor KAM statement and its scaling (The two-rotor model: a positive-measure family of invariant Lagrangian tori with Diophantine frequencies for sufficiently small coupling, with the explicit units and the stated scope): the physical Hamiltonian is $Π_{1}^{2} / (2 I) + Π_{2}^{2} / (2 I) + E_{0} ε \cos θ_{1} \cos θ_{2}$ with $J_{0}^{2} = I E_{0}$ and $t_{0} = J_{0} / E_{0}$ ; under $I_{i} = Π_{i} / J_{0}$ , $s = t / t_{0}$ it becomes $\frac{1}{2} (I_{1}^{2} + I_{2}^{2}) + ε \cos ϑ_{1} \cos ϑ_{2}$ , and each surviving torus carries $φ (t) = φ_{0} + ω_{0} ν t$ in its torus parameter, with the physical angle given by the embedding.

[L2]

Diophantine labels (Analytic KAM data: angular tori, action-angle neighbourhoods, Fourier cutoffs and Diophantine frequency labels): a frequency $ν$ is $(α, τ)$ -Diophantine exactly when $∣ k \cdot ν ∣ \geq α ∣ k ∣_{1}^{- τ}$ for every nonzero integer vector $k$ .

Verification

technique · direct

1.1givenL1algebra

Dimensions: the physical Hamiltonian has $[Π_{i}^{2} / I] = k g^{2} m^{4} s^{- 2} / (k g m^{2}) = k g m^{2} s^{- 2} = J$ and $[E_{0} ε] = 1 \cdot J$ with $ε$ dimensionless; $ω_{0} = \sqrt{E_{0} / I}$ has $[ω_{0}] = \sqrt{k g m^{2} s^{- 2} / (k g m^{2})} = s^{- 1}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). With the declared values, $J_{0} = 1 k g m^{2} s^{- 1}$ , $t_{0} = 1 s$ , $ω_{0} = 1 s^{- 1}$ .

1.2givenL1algebra

Scaling of the equations, first equation: with $Π_{i} = J_{0} I_{i}$ and $t = t_{0} s$ the chain rule gives ${\dot{θ}}_{i} = t_{0}^{- 1} θ_{i}^{'}$ and ${\dot{Π}}_{i} = t_{0}^{- 1} J_{0} I_{i}^{'}$ ; substituting into ${\dot{θ}}_{i} = Π_{i} / I$ and using $J_{0} t_{0} = I$ yields $t_{0}^{- 1} θ_{i}^{'} = J_{0} I_{i} / I = I_{i} / t_{0}$ , that is the dimensionless equation $θ_{i}^{'} = I_{i}$ .

1.3givenL2algebra

Rational labels are non-Diophantine: let $ν = (p_{1} / q, p_{2} / q) \in C^{*}$ with integers $p_{1}, p_{2}$ and $q \geq 1$ . If $p_{1} = p_{2} = 0$ , choose $k = (1, 0)$ ; otherwise set $k : = (p_{2}, - p_{1}) \neq 0$ ; in either case $k \cdot ν = 0 < α ∣ k ∣_{1}^{- τ}$ , so $ν \notin Ω_{α, τ}$ by [L2]. Hence the Diophantine set of the theorem contains no rational point of the cube; the positive measure it retains is carried by irrational labels.

2.1step 1.2givenL1algebra

Scaling of the equations, second equation: substituting into ${\dot{Π}}_{i} = E_{0} ε \sin θ_{i} \cos θ_{3 - i}$ gives $t_{0}^{- 1} J_{0} I_{i}^{'} = E_{0} ε \sin θ_{i} \cos θ_{3 - i}$ ; since $J_{0} / t_{0} = J_{0} E_{0} / J_{0} = E_{0}$ , this is equivalent to $I_{i}^{'} = ε \sin θ_{i} \cos θ_{3 - i}$ , the dimensionless Hamilton equation of the potential $ε \cos ϑ_{1} \cos ϑ_{2}$ . With the declared values $J_{0} / t_{0} = 1 J$ and $E_{0} = 1 J$ the substitution is an identity.

2.2step 1.1givenL1algebra

Physical frequencies: by [L1] a surviving torus with label $ν$ carries $φ (t) = φ_{0} + ω_{0} ν t$ in its torus parameter; the physical angle is its embedded angular component; with $ω_{0} = 1 s^{- 1}$ the components $ω_{0} ν_{i} = ν_{i} s^{- 1}$ are the physical angular frequencies, of dimension $s^{- 1}$ as in step 1.1.

3.1step 1.1step 1.2givenL1algebra∎

Equal rescaling: replace $(I, E_{0})$ by $(λ I, λ E_{0})$ , $λ > 0$ ; then $ω_{0} = \sqrt{E_{0} / I}$ is unchanged, $J_{0} = \sqrt{I E_{0}}$ is multiplied by $λ$ and $t_{0} = J_{0} / E_{0}$ is unchanged, so the dimensionless Hamiltonian, the frequency cube, the Diophantine set and the smallness condition are all unchanged, while the physical momenta $Π_{i} = J_{0} I_{i}$ are rescaled by $λ$ . The constant $c$ of the theorem is not numerically evaluated in this check; only the exact scaling behaviour is asserted.

Remarks

What this check does not do. It verifies dimensions, the scaling correspondence and the exclusion of rational labels; it does not evaluate the KAM constant $c$ , does not certify any particular pair $(α, ε)$ and does not assert that any laboratory rotor realizes the model. Those would require the numerical computation of the theorem's constant and a separate apparatus model.
Why rational labels are worth recording. The frequency cube is full of rational labels, so the positive-measure conclusion of The two-rotor model: a positive-measure family of invariant Lagrangian tori with Diophantine frequencies for sufficiently small coupling, with the explicit units and the stated scope concerns an open-ended irrational set; the check makes the Diophantine condition concrete by exhibiting the exact obstruction for rational labels.

Stability, perturbations and chaos — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

The ideal pendulum on declared data: strict minimum and Lyapunov stability at the downward equilibrium, negative Hessian and instability at the upward equilibrium, the exact linearised eigenvalues and oscillation frequency, and failure of attraction

Example

Facts & Assumptions

Verification

Remarks

Two declared linear models worked through: the harmonic flow has zero finite-time growth in every direction, while the repulsive linear model has exact rates ±1±1 and the positive Lyapunov exponent 11 at its equilibrium

Example

Facts & Assumptions

Verification

Remarks

The three-dimensional isotropic oscillator: the stroboscopic map at half period is minus the identity, every state is periodic, the periodic points are dense, and the map is neither mixing nor sensitive

Example

Facts & Assumptions

Verification

Remarks

The pulsed horseshoe on declared data: a=100a=100, the radicand box [74,126][74,126], the contraction factor 1/741/74​, the limiting quadratic fixed points 1±1011±101​, and the physical bounds of the coded states at ℓ0=1 mℓ0​=1m, m=1 kgm=1kg, ω0=2π s−1ω0​=2πs−1

Example

Facts & Assumptions

Verification

Remarks

A positive Lyapunov exponent at a hyperbolic equilibrium does not produce a chaotic invariant set of the time-one map: the three-dimensional repulsive model has rate 11 and no compact invariant set with more than one point

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Lyapunov stability does not imply attraction: the harmonic model is stable at its equilibrium while every nonzero datum keeps its distance and never converges

Statement refuted

Facts & Assumptions

Counterexample

Remarks

A trapped one-label model: the compact energy sublevel of the radial quartic potential is invariant with finite phase volume, so the recurrence theorem applies, while the free model's energy sublevels are unbounded and the hypothesis fails

Example

Facts & Assumptions

Verification

Remarks

Unit restoration and scale checks for the two-rotor KAM statement: the scaling identities at I=E0=1I=E0​=1, the physical angular frequencies, the exclusion of rational frequency labels, and the equal-rescaling invariance

Example

Facts & Assumptions

Verification

Remarks

Sources

Two declared linear models worked through: the harmonic flow has zero finite-time growth in every direction, while the repulsive linear model has exact rates $\pm 1$ and the positive Lyapunov exponent $1$ at its equilibrium

The pulsed horseshoe on declared data: $a = 100$ , the radicand box $[74, 126]$ , the contraction factor $1 / \sqrt{74}$ , the limiting quadratic fixed points $1 \pm \sqrt{101}$ , and the physical bounds of the coded states at $ℓ_{0} = 1 m$ , $m = 1 k g$ , $ω_{0} = 2 π s^{- 1}$

A positive Lyapunov exponent at a hyperbolic equilibrium does not produce a chaotic invariant set of the time-one map: the three-dimensional repulsive model has rate $1$ and no compact invariant set with more than one point

Unit restoration and scale checks for the two-rotor KAM statement: the scaling identities at $I = E_{0} = 1$ , the physical angular frequencies, the exclusion of rational frequency labels, and the equal-rescaling invariance