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Coupled oscillations and approximation — Examples

1 · Prerequisites

2 · Summary

Worked checks for cm09-coupled-oscillations-and-approximation. Modal vectors and coefficients use the model's fixed-unit numerical convention: physical mass-normalized vector components have unit $s_{i} / (\sqrt{e_{*}} t_{*})$ and modal amplitudes have unit $\sqrt{e_{*}} t_{*}$ . Two coupled masses on a line: the in-phase and anti-phase normal modes, their frequencies, mode-energies and the zero-stiffness translation mode verifies the free-mode theorem on two equal masses joined by a coupling spring: the modal vectors $(1, \pm 1) / \sqrt{2 m}$ , the frequencies $\sqrt{k / m}$ and $\sqrt{(k + 2 c) / m}$ , the single-mode data and energies, and the zero-stiffness case in which the pair's translation is an affine zero mode. A linear triatomic chain: the translation zero mode, the antisymmetric stretch and the symmetric stretch, with the corrected frequency comparison computes the three modal patterns of a linear chain with masses $m, m_{c}, m$ — translation, antisymmetric stretch and symmetric stretch — with frequencies $\sqrt{k / m}$ and $\sqrt{(k / m) (1 + 2 m / m_{c})}$ , and corrects a frequency comparison of the cited treatment whose printed matrices give the opposite ordering. The equal-arm double pendulum in the linear approximation: the slow same-side mode, the fast opposite-side mode and the bounded modal motions treats the two-angle small-oscillation data of the equal-arm double pendulum, with mass metric $m l^{2}$ times the matrix with rows $(2, 1)$ and $(1, 1)$ and stiffness $m g l d i a g (2, 1)$ , with the slow same-side mode $\sqrt{(g / l) (2 - \sqrt{2})}$ and the fast opposite-side mode $\sqrt{(g / l) (2 + \sqrt{2})}$ . A doubly degenerate quadratic model: every direction is a mode, the modal basis is not intrinsic, and coupling splits the double frequency exhibits a doubly degenerate model in which every direction is a mode, the modal basis is not intrinsic, and the coupling of the two-mass example splits the double frequency. The inverted pendulum about its upward equilibrium: the negative modal stiffness, the exponential quadratic mode and the nonlinear instability computes the inverted pendulum's negative stiffness $- m g l$ , the exponential quadratic motions and, through the hyperbolic-saddle theorem, the nonlinear instability of the upward equilibrium.

The approximation statements are checked on explicit data. A drive along the in-phase pattern of the two-mass chain: resonance exactly at that mode's frequency and a bounded steady response off resonance drives the two-mass chain along its in-phase pattern: resonance occurs exactly at $ω_{+} = \sqrt{k / m}$ with a linearly growing mode-shaped motion, while at $ω_{-} = \sqrt{(k + 2 c) / m}$ the drive is non-resonant with steady amplitude $∣ F ∣ / (2 c)$ . An explicit weakly damped oscillator: the averaging constants and the energy-decay bound over five thousand seconds (about 796 undamped periods) evaluates the averaging constants $M = 3 \sqrt{2} / 2$ , $L = \sqrt{2}$ , $K = 2 π (3 \sqrt{2} + 6) e^{\sqrt{2} / 2}$ and verifies the smallness and trapping hypotheses for $m = k = γ = 1$ , $ε = 10^{- 4}$ , concluding that the mechanical energy tracks $e^{- t / 10^{4}}$ within $2.63 \times 10^{- 2}$ J on $[0, 5000 s]$ , with amplitude unit $1 m s^{- 1 / 2}$ and $A = 1 / 2 s$ . A cubic-potential chart system: the exact quadratic remainder, the explicit error constants and a checked finite-time bound computes the exact quadratic remainder of a cubic potential and the explicit constants $C = ∣ λ ∣ / (2 m)$ , $L = \max (1, (k + ∣ λ ∣ r / 2) / m)$ in the explicitly normalized metre/second phase coordinates, checks the finite-time bound on a numerical instance and records the exactly linear case $λ = 0$ .

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

Two coupled masses on a line: the in-phase and anti-phase normal modes, their frequencies, mode-energies and the zero-stiffness translation mode

Physical scope

A worked instance of the coupled quadratic model: two equal masses m on a line joined by a coupling spring c with outer springs k to fixed points, the explicit normal modes, frequencies sqrt(k/m) and sqrt((k+2c)/m), the single-mode initial data, the conserved mode energies, and the zero-stiffness case k=0 in which the pair's translation is an affine zero mode and only the relative motion oscillates. The masses and spring constants are model parameters with declared units; no measured value, no realized apparatus and no empirical claim is present.

Example

The algebra below uses the fixed-unit numerical convention of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium, with coordinate units metres, time unit $1 s$ and energy unit $1 J$ . Physical model parameters are given their stated units; in matrix products, initial-vector identities and coefficient formulas their numerical coefficients in these fixed units are used. Modal amplitudes have common unit $a_{*} = (1 J)^{1 / 2} (1 s)$ ; the $j$ -th component of a mass-normalized modal vector has unit $s_{j} / a_{*}$ . Modal oscillatory and hyperbolic coefficients have unit $a_{*}$ , and zero-mode slopes have unit $a_{*} / (1 s)$ . Physical coordinates and energies are reconstructed with the coordinate units and $1 J$ , respectively. In particular a numerical single-mode datum $q (0) = u_{i}$ means a modal amplitude of one in this common amplitude unit.

Global-time, growth and boundedness claims below refer to the unique maximal extension of the autonomous free equation on $R$ ; local motions remain its restrictions. Values at $0$ in coefficient formulas are evaluated on that extension when the original interval omits $0$ .

Consider two particle labels of equal mass $m > 0$ constrained to a line, with coordinates $x_{1}, x_{2}$ in metres read in a fixed inertial frame (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). Each label is attached to a fixed point by a spring of stiffness $k > 0$ and the labels are coupled to each other by a spring of stiffness $c > 0$ ; both constants are read in kilograms per second squared. The model is the coupled quadratic model of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium with $n = 2$ , coordinates $q = (x_{1}, x_{2})$ , mass metric matrix $M = m I_{2}$ and stiffness

$K = (\begin{array}{cc} k + c & - c \\ - c & k + c \end{array}),$

so that the equation $M q^{''} + K q = 0$ is the linear chain $m x_{1}^{''} = - (k + c) x_{1} + c x_{2}$ , $m x_{2}^{''} = c x_{1} - (k + c) x_{2}$ . Then:

the vectors $u_{\pm} : = (1, \pm 1) / \sqrt{2 m}$ satisfy $u_{+}^{T} M u_{+} = u_{-}^{T} M u_{-} = 1$ , $u_{+}^{T} M u_{-} = 0$ and $K u_{\pm} = λ_{\pm} M u_{\pm}$ with $λ_{+} = \frac{k}{m}, λ_{-} = \frac{k + 2 c}{m};$
the modal coordinates are $z_{\pm} = e_{\pm}^{T} M^{1 / 2} q$ with $e_{\pm} = (1, \pm 1) / \sqrt{2}$ , that is $z_{+} = \sqrt{m / 2} (x_{1} + x_{2})$ and $z_{-} = \sqrt{m / 2} (x_{1} - x_{2})$ , and every motion is $q (t) = (A \cos (ω_{+} t) + B \sin (ω_{+} t)) u_{+} + (C \cos (ω_{-} t) + D \sin (ω_{-} t)) u_{-}$ with $ω_{+} = \sqrt{k / m}$ and $ω_{-} = \sqrt{(k + 2 c) / m}$ ;
the initial data $q (0) = u_{+}$ , $q^{'} (0) = 0$ excite exactly the in-phase mode $q (t) = \cos (ω_{+} t) u_{+}$ , whose mechanical energy is the constant $E = k / (2 m)$ ; the data $q (0) = u_{-}$ , $q^{'} (0) = 0$ excite exactly the anti-phase mode $q (t) = \cos (ω_{-} t) u_{-}$ , with $E = (k + 2 c) / (2 m)$ ;
the in-phase frequency $ω_{+}$ is independent of the coupling $c$ and the anti-phase frequency $ω_{-}$ is strictly increasing in $c$ ;
if $k = 0$ while $c > 0$ , then $λ_{+} = 0$ and $u_{+}$ spans a zero mode: the general motion is $q (t) = (A + B t) u_{+} + (C \cos (ω_{-} t) + D \sin (ω_{-} t)) u_{-}$ , the pair translates freely as a whole and only the relative coordinate oscillates, at frequency $\sqrt{2 c / m}$ .

Facts & Assumptions

Given: The masses $m > 0$ , the stiffnesses $k > 0$ (and in clause 5, $k = 0$ ), $c > 0$ , the matrices $M = m I_{2}$ and $K$ of the Example, and the vectors $u_{\pm}, e_{\pm}$ .

[F1]

Model dictionary (The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium): the data $(M, K, 0)$ declare a free coupled quadratic model with two degrees of freedom, kinetic energy $\frac{1}{2} m (x_{1}^{' 2} + x_{2}^{' 2})$ , potential energy $\frac{1}{2} q^{T} K q$ and phase system $z^{'} = A z$ ; the modal vocabulary is that of clause (e) with mass metric $M$ and stiffness $K$ .

[F2]

Free-mode theory (The free coupled quadratic model: global existence, the complete modal classification, mode energies, intrinsic degenerate mode subspaces, boundedness and the unstable direction clauses 2, 5 and 6): motions of the free model are the modal superpositions of that theorem, with oscillatory families for $λ_{i} > 0$ , affine families for $λ_{i} = 0$ and exponential families for $λ_{i} < 0$ ; these are complete; a positive definite stiffness gives bounded motions.

[F3]

Direct linear algebra (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): matrix-vector products and inner products in $R^{2}$ are computed componentwise; $M^{1 / 2} = \sqrt{m} I_{2}$ , so $u_{\pm} = M^{- 1 / 2} e_{\pm}$ and $z = O^{T} M^{1 / 2} q$ with $O = [e_{+} e_{-}]$ ; the displayed products below are direct computations.

[F4]

Scalar oscillator calculus (The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , $C^{k}$ Euclidean maps and diffeomorphisms): the scalar equations $z_{\pm}^{''} + λ_{\pm} z_{\pm} = 0$ with $λ_{\pm} > 0$ have exactly the families $A \cos (ω_{\pm} t) + B \sin (ω_{\pm} t)$ with $ω_{\pm} = \sqrt{λ_{\pm}}$ , and the displayed curves are differentiated by the chain rule.

Verification

Proof technique: direct; verify the modal identities by explicit matrix multiplication and invoke the free-mode theorem for the motion and energy statements.

1.1F1F3algebra

Modal identities: $M$ is symmetric positive definite and $K$ is symmetric with $q^{T} K q = (k + c) (x_{1}^{2} + x_{2}^{2}) - 2 c x_{1} x_{2} = k (x_{1}^{2} + x_{2}^{2}) + c (x_{1} - x_{2})^{2} \geq 0$ for $k > 0, c > 0$ (indeed positive definite), so the model is an oscillatory coupled quadratic model of [F1]. Multiplying, $K u_{+} = (k, k) / \sqrt{2 m} = k (1, 1) / \sqrt{2 m} = \frac{k}{m} M u_{+}$ and $K u_{-} = (k + 2 c, - (k + 2 c)) / \sqrt{2 m} = \frac{k + 2 c}{m} M u_{-}$ ; and $u_{\pm}^{T} M u_{\pm} = m ∣ u_{\pm} ∣_{2}^{2} = m \cdot \frac{2}{2 m} = 1$ , $u_{+}^{T} M u_{-} = m (1 - 1) / (2 m) = 0$ . This proves clause 1 with $λ_{\pm}$ as displayed.

2.1F1F3step 1.1algebra

Modal coordinates: with $e_{\pm} = (1, \pm 1) / \sqrt{2}$ the matrix $O = [e_{+} e_{-}]$ is orthogonal and $M^{- 1 / 2} K M^{- 1 / 2} = \frac{1}{m} K$ has the orthonormal eigenbasis $e_{\pm}$ with eigenvalues $λ_{\pm}$ by clause 1; hence $z = O^{T} M^{1 / 2} q$ has $z_{\pm} = e_{\pm}^{T} \sqrt{m} q = \sqrt{m / 2} (x_{1} \pm x_{2})$ and $u_{\pm} = M^{- 1 / 2} e_{\pm}$ , so the modal data of clause (e) of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium are exactly those displayed.

2.2step 1.1algebra

Frequency comparison: $ω_{+} = \sqrt{k / m}$ contains no $c$ , while $c \mapsto ω_{-} = \sqrt{(k + 2 c) / m}$ is strictly increasing on $c > 0$ with $ω_{-} > ω_{+}$ : the in-phase oscillation is softer than the anti-phase one. This is clause 4.

2.3F2step 1.1algebra

Zero stiffness: if $k = 0$ then $K = c (\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array})$ , $K u_{+} = 0$ and $K u_{-} = \frac{2 c}{m} M u_{-}$ , so $λ_{+} = 0$ and $λ_{-} = 2 c / m$ ; by [F2] the general motion is the affine-plus-oscillatory family displayed in clause 5, in which the pair translates as a whole (arbitrary constant velocity $B$ ) and only the relative coordinate $x_{1} - x_{2}$ oscillates, with $ω_{-} = \sqrt{2 c / m}$ . This is clause 5.

3.1F2F4step 1.1step 2.1

Motions: by [F2] and [F4], every motion is the superposition of clause 2 with $λ_{\pm} > 0$ , i.e. the displayed two-frequency family, and the coefficients $(A, B, C, D)$ are bijective onto the data $(q (0), q^{'} (0)) = (A u_{+} + C u_{-}, B ω_{+} u_{+} + D ω_{-} u_{-})$ . This is clause 2.

4.1F2step 1.1step 3.1algebra

Single-mode data: for $q (0) = u_{+}$ , $q^{'} (0) = 0$ the coefficient system gives $A = 1$ , $B = C = D = 0$ , so $q (t) = \cos (ω_{+} t) u_{+}$ ; its energy is $E = \frac{1}{2} q^{' T} M q^{'} + \frac{1}{2} q^{T} K q = \frac{1}{2} ω_{+}^{2} \sin^{2} (ω_{+} t) + \frac{1}{2} \frac{k}{m} \cos^{2} (ω_{+} t) = \frac{k}{2 m}$ , constant because $ω_{+}^{2} = k / m$ ; the same computation with $λ_{-} = (k + 2 c) / m$ gives $E = (k + 2 c) / (2 m)$ for the anti-phase datum. This is clause 3.

5.1step 1.1step 3.1step 4.1step 2.2step 2.3∎

Clauses 1-5 are verified on the explicit data; all parameters are model data with declared units, and no measured frequency, damping or apparatus behaviour is asserted.

Remarks

The coupling acts only on the relative coordinate. The change of coordinates $(x_{1}, x_{2}) \mapsto (x_{1} + x_{2}, x_{1} - x_{2})$ decouples the chain: the sum is governed by the outer springs alone and the difference by the outer springs plus twice the coupling. This is the physical reading of clauses 1-4 and the reason the in-phase frequency is independent of $c$ .
The zero mode is a translation, not an oscillation. Clause 5 is the two-mass instance of the zero-mode warning recorded in Scope of the coupled quadratic model: zero modes, degenerate eigenspaces, unstable quadratic directions and the limits of the approximation statements: the mode of zero frequency is the uniform motion of the pair, its family is affine, and it is unbounded whenever the pair has nonzero total momentum.

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A linear triatomic chain: the translation zero mode, the antisymmetric stretch and the symmetric stretch, with the corrected frequency comparison

Physical scope

A worked instance of the coupled quadratic model: a linear chain of three labels with masses m, m_c, m joined by identical nearest-neighbour springs of stiffness k, modelled by the quadratic stiffness matrix k[[1,-1,0],[-1,2,-1],[0,-1,1]], with its three modal patterns (translation at zero modal stiffness, antisymmetric stretch at sqrt(k/m) and symmetric stretch at sqrt((k/m)(1+2m/m_c))), the general motion, the affine translation zero mode and the corrected frequency ordering omega_3>omega_2. The parameters are model data with declared units; the example asserts no measured molecular frequency and no empirical identification of the chain with a real molecule.

Example

Consider three particle labels constrained to a line, with coordinates $x_{1}, x_{2}, x_{3}$ in metres and masses $m, m_{c}, m > 0$ read in kilograms (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework); the outer labels have mass $m$ and the middle label mass $m_{c}$ . Adjacent labels interact through identical springs of stiffness $k > 0$ in kilograms per second squared, and the potential is approximated by its quadratic part $U = \frac{k}{2} ((x_{1} - x_{2})^{2} + (x_{2} - x_{3})^{2})$ about the equilibrium. The model is the coupled quadratic model of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium with $n = 3$ , $q = (x_{1}, x_{2}, x_{3})$ ,

$M = diag (m, m_{c}, m), K = k (\begin{array}{ccc} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array}) .$

Then:

$M$ is symmetric positive definite and $K$ is symmetric with $\ker K = span {(1, 1, 1)}$ ;
the modal data are as follows, with $M$ -orthonormal modal vectors: (a) $λ_{1} = 0$ with $u_{1} = (1, 1, 1) / \sqrt{2 m + m_{c}}$ : the translation mode, in which the three labels move together; (b) $λ_{2} = k / m$ with $u_{2} = (1, 0, - 1) / \sqrt{2 m}$ : the antisymmetric stretch, in which the outer labels move oppositely and the middle label is at rest; (c) $λ_{3} = \frac{k}{m} (1 + \frac{2 m}{m_{c}})$ with $u_{3} = (1, - 2 m / m_{c}, 1) / \sqrt{2 m (1 + 2 m / m_{c})}$ : the symmetric stretch, in which the outer labels move together and the middle label moves oppositely;
every motion is $q (t) = (a + b t) u_{1} + (A \cos (ω_{2} t) + B \sin (ω_{2} t)) u_{2} + (C \cos (ω_{3} t) + D \sin (ω_{3} t)) u_{3}$ with $ω_{2} = \sqrt{k / m}$ and $ω_{3} = \sqrt{\frac{k}{m} (1 + \frac{2 m}{m_{c}})}$ , and the pure-zero-mode motions are exactly the affine curves $q (t) = (a + b t) u_{1}$ ; for $b \neq 0$ they are unbounded on $[0, \infty)$ ;
$ω_{3} > ω_{2}$ for all $m, m_{c} > 0$ : in this model the symmetric stretch is the stiffer vibration.

Facts & Assumptions

Given: The data $m, m_{c}, k > 0$ , the matrices $M, K$ of the Example, the vectors $u_{1}, u_{2}, u_{3}$ and the numbers $λ_{1}, λ_{2}, λ_{3}$ .

[F1]

Model dictionary (The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium): the data $(M, K, 0)$ declare a free coupled quadratic model with three degrees of freedom, kinetic energy $\frac{1}{2} \sum_{i} m_{i} x_{i}^{' 2}$ and potential energy $\frac{1}{2} q^{T} K q$ ; the modal vocabulary is that of clause (e).

[F2]

Free-mode theory (The free coupled quadratic model: global existence, the complete modal classification, mode energies, intrinsic degenerate mode subspaces, boundedness and the unstable direction clauses 2, 4 and 5): every free motion is a modal superposition with oscillatory, affine or exponential families according to the sign of $λ_{i}$ ; the pure- $λ$ motions are the solutions of $q^{''} + λ q = 0$ with values in the intrinsic subspace $N_{λ}$ , which for $λ = 0$ are the affine curves in $N_{0} = \ker K$ ; an affine component with nonzero slope is unbounded on $[0, \infty)$ .

[F3]

Two-mode check (Two equal masses with two outer springs and a coupling spring: eigenmodes, decoupling and the explicit normal frequencies): the two-label symmetric case with outer stiffness $k$ and coupling $c$ has frequencies $\sqrt{k / m}$ and $\sqrt{(k + 2 c) / m}$ ; it is the $2 \times 2$ instance of the same eigenproblem and is used below only as a consistency check on the antisymmetric sector.

[F4]

Direct linear algebra (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): matrix-vector products and $M$ -inner products are computed componentwise; the products displayed in the Verification are direct computations.

[F5]

Scalar oscillator calculus: for $λ_{i} > 0$ , The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data with scalar mass $1$ , damping $0$ and stiffness $λ_{i}$ gives exactly $z_{i} (t) = A_{i} \cos (\sqrt{λ_{i}} t) + B_{i} \sin (\sqrt{λ_{i}} t)$ . Its hypothesis $k > 0$ excludes the zero-stiffness case; the affine family for $λ_{i} = 0$ is supplied by [F2]. The chain rule The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ governs differentiation of composites, and $C^{k}$ Euclidean maps and diffeomorphisms supplies the $C^{2}$ regularity convention.

Verification

technique · direct; verify the three eigenpairs by explicit matrix multiplication and read the motion statement from the free-mode theorem

1.1F1F4algebra

Kernel and positivity: $q^{T} K q = k ((x_{1} - x_{2})^{2} + (x_{2} - x_{3})^{2}) \geq 0$ with equality if and only if $x_{1} = x_{2} = x_{3}$ , so $K$ is symmetric non-negative with the stated kernel; $M$ is diagonal with positive entries.

1.2F1F4algebra

Eigenpairs: multiplying, $K u_{1} = 0, K u_{2} = k (1, 0, - 1) / \sqrt{2 m} = \frac{k}{m} M u_{2}, K u_{3} = k (1 + \frac{2 m}{m_{c}}) (1, - 2, 1) / \sqrt{2 m (1 + 2 m / m_{c})} = \frac{k}{m} (1 + \frac{2 m}{m_{c}}) M u_{3},$ using $M u_{3} = m (1, - 2, 1) / \sqrt{2 m (1 + 2 m / m_{c})}$ . Direct products give $u_{1}^{T} M u_{1} = (2 m + m_{c}) / (2 m + m_{c}) = 1$ , $u_{2}^{T} M u_{2} = 2 m / (2 m) = 1$ , and $u_{3}^{T} M u_{3} = (2 m + 4 m^{2} / m_{c}) / (2 m (1 + 2 m / m_{c})) = 1$ ; the three mixed numerators are $m - m = 0$ , $m - 2 m + m = 0$ , and $m - m = 0$ , respectively, so the modal vectors are $M$ -orthonormal. This proves clauses 1 and 2.

2.1F3step 1.2

Consistency check: on the antisymmetric sector $x_{1} = - x_{3}$ , $x_{2} = 0$ , the displayed equations give $m x_{1}^{''} = - k x_{1}$ and $m x_{3}^{''} = - k x_{3}$ , with the middle equation identically zero. Thus the frequency is $\sqrt{k / m}$ . Equivalently, this is the two-label model of [F3] with outer stiffness $k$ and zero coupling, whose two equal frequencies are $\sqrt{k / m}$ ; no spurious coupling term is used.

2.2F2F5step 1.2

Motions: since $λ_{1} = 0 < λ_{2} < λ_{3}$ , [F2] gives that every motion is the displayed superposition with the affine family for $λ_{1}$ and oscillatory families for $λ_{2}, λ_{3}$ ; the pure-zero-mode motions form the affine curves in $N_{0} = \ker K = span {u_{1}}$ , i.e. $q (t) = (a + b t) u_{1}$ , which are unbounded on $[0, \infty)$ for $b \neq 0$ because $∣ a + b t ∣ \to \infty$ . This is clause 3.

2.3step 1.2algebra

Frequency ordering: $ω_{3}^{2} / ω_{2}^{2} = 1 + 2 m / m_{c} > 1$ , so $ω_{3} > ω_{2}$ for every $m_{c} > 0$ ; this is clause 4.

3.1step 1.2step 2.2step 2.3given∎

Clauses 1-4 are verified on the explicit data. The comparison in clause 4 is read off the eigenvalues computed from the displayed matrices; the cited treatment's sentence comparing the two frequencies is not used, because with the matrices it displays the computed eigenvalues give the opposite ordering, and the discrepancy is recorded as a source error in the reader record for that treatment. All parameters are model data; no molecular frequency, force constant or apparatus behaviour is asserted.

Remarks

The zero mode is the centre-of-mass motion. $u_{1}$ moves all three labels together; the quadratic potential is blind to it, and the model has no restoring force in that direction. This is the exact mechanism of a translation mode: a linearization about a non-isolated minimum (the equilibrium manifold here is the line $x_{1} = x_{2} = x_{3}$ ) acquires a zero modal stiffness. If the physical chain is confined, or an external restoring force acts on the centre of mass, the data change (an additional stiffness term appears) and the zero mode disappears.
What the example does not establish. The quadratic potential is the second-order truncation of the real interatomic potential; the example computes the normal modes of the truncated model and asserts nothing about the accuracy of that truncation, the anharmonic behaviour at larger amplitudes, or the vibrational spectrum of any real molecule.

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The equal-arm double pendulum in the linear approximation: the slow same-side mode, the fast opposite-side mode and the bounded modal motions

Physical scope

A worked instance of the coupled quadratic model: the two-angle small-oscillation data M = m l^2 [[2,1],[1,1]], K = m g l diag(2,1) attributed to the linearisation of an equal-mass equal-length planar double pendulum, with its two modal data (the slower same-side mode of frequency sqrt((g/l)(2-sqrt2)) and the faster opposite-side mode of frequency sqrt((g/l)(2+sqrt2))), the general bounded modal motion and the pure-mode data. The example verifies the quadratic model on the declared matrices; it does not derive the pendulum Lagrangians, does not control the error of the linear approximation at finite amplitude and asserts no measured period.

Example

The algebra below uses the fixed-unit numerical convention of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium, with coordinate units radians, time unit $1 s$ and energy unit $1 J$ . Physical model parameters are given their stated units; in matrix products, initial-vector identities and coefficient formulas their numerical coefficients in these fixed units are used. Modal amplitudes have common unit $a_{*} = (1 J)^{1 / 2} (1 s)$ ; the $j$ -th component of a mass-normalized modal vector has unit $s_{j} / a_{*}$ . Modal oscillatory and hyperbolic coefficients have unit $a_{*}$ , and zero-mode slopes have unit $a_{*} / (1 s)$ . Physical coordinates and energies are reconstructed with the coordinate units and $1 J$ , respectively. In particular a numerical single-mode datum $q (0) = u_{i}$ means a modal amplitude of one in this common amplitude unit.

Let a planar double pendulum consist of two equal particle labels of mass $m > 0$ connected by two equal massless rods of length $l > 0$ , pivoted at a fixed point; let $θ_{1}, θ_{2}$ be the angles of the rods from the downward vertical, read in radians (dimensionless under the angle convention of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework), and let $g > 0$ be the uniform gravitational acceleration of the model. In the linear approximation about the downward equilibrium of the cited treatment, the small-oscillation data are the coupled quadratic model of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium with $n = 2$ ,

$M = m l^{2} (\begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array}), K = m g l (\begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array}), J \equiv 0.$

Write $γ_{s} : = 2 - \sqrt{2}$ , $γ_{f} : = 2 + \sqrt{2}$ . Then:

$M$ is symmetric positive definite (with $M^{1 / 2}$ its unique positive definite square root) and $K$ is symmetric positive definite, so the model is oscillatory with both modal stiffnesses positive;
the modal data are $λ_{s} = \frac{g}{l} γ_{s} with w_{s} = (\begin{array}{c} 1 \\ \sqrt{2} \end{array}), λ_{f} = \frac{g}{l} γ_{f} with w_{f} = (\begin{array}{c} 1 \\ - \sqrt{2} \end{array}),$ that is $K w_{s} = λ_{s} M w_{s}$ and $K w_{f} = λ_{f} M w_{f}$ ; the $M$ -orthonormal modal vectors are $u_{s} = w_{s} / \sqrt{2 m l^{2} γ_{f}}$ and $u_{f} = w_{f} / \sqrt{2 m l^{2} γ_{s}}$ ;
the mode frequencies are $ω_{s} = \sqrt{\frac{g}{l} (2 - \sqrt{2})} < ω_{f} = \sqrt{\frac{g}{l} (2 + \sqrt{2})},$ and every motion is bounded on $R$ and has the form $q (t) = (A \cos (ω_{s} t) + B \sin (ω_{s} t)) u_{s} + (C \cos (ω_{f} t) + D \sin (ω_{f} t)) u_{f};$
in the slow mode $u_{s}$ the two angles have the same sign ( $θ_{2} = \sqrt{2} θ_{1}$ ) and in the fast mode $u_{f}$ they have opposite signs ( $θ_{2} = - \sqrt{2} θ_{1}$ ); the datum $q (0) = u_{s}$ , $q^{'} (0) = 0$ excites exactly the slow mode and the datum $q (0) = u_{f}$ , $q^{'} (0) = 0$ exactly the fast mode.

Facts & Assumptions

Given: The parameters $m, l, g > 0$ , the matrices $M, K$ of the Example, the vectors $w_{s}, w_{f}$ and the numbers $γ_{s}, γ_{f}$ .

[F1]

Model dictionary (The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium): the data $(M, K, 0)$ declare a free coupled quadratic model with two degrees of freedom, kinetic energy $\frac{1}{2} q^{' T} M q^{'}$ , potential energy $\frac{1}{2} q^{T} K q$ and the modal vocabulary of clause (e).

[F2]

Free-mode theory (The free coupled quadratic model: global existence, the complete modal classification, mode energies, intrinsic degenerate mode subspaces, boundedness and the unstable direction clauses 1, 2, 4 and 5): every free motion is the modal superposition of that theorem; positive definite stiffness gives positive modal stiffnesses and bounded motions; the pure- $λ$ motions are the solutions of $q^{''} + λ q = 0$ with values in the intrinsic subspace $N_{λ}$ .

[F3]

Direct linear algebra (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): $M$ -inner products, matrix products and positive definiteness are checked by the displayed quadratic forms and products. The existence of the positive definite square root is supplied by [F1], not by these arithmetic definitions.

[F4]

Scalar oscillator calculus (The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , $C^{k}$ Euclidean maps and diffeomorphisms): for $λ > 0$ the scalar equation $z^{''} + λ z = 0$ has exactly the family $A \cos (ω t) + B \sin (ω t)$ with $ω = \sqrt{λ}$ .

Verification

Proof technique: direct; check positive definiteness, verify the two eigenpairs by explicit multiplication, and read the motion and boundedness statements from the free-mode theorem.

1.1F1F3algebra

Positive definiteness: $M$ is symmetric and, for $w = (x, y) \neq 0$ , $w^{T} M w = m l^{2} (2 x^{2} + 2 x y + y^{2}) = m l^{2} (x^{2} + (x + y)^{2}) > 0$ , since both squares can vanish only when $x = y = 0$ . Likewise $w^{T} K w = m g l (2 x^{2} + y^{2}) > 0$ , so $K$ is symmetric positive definite; hence both modal stiffnesses are positive and the model is oscillatory.

1.2F1F3algebra

Eigenpairs: multiplying, $K w_{s} = m g l (\begin{array}{c} 2 \\ \sqrt{2} \end{array}), M w_{s} = m l^{2} (\begin{array}{c} 2 + \sqrt{2} \\ 1 + \sqrt{2} \end{array}),$ and $2 / (2 + \sqrt{2}) = \sqrt{2} / (1 + \sqrt{2}) = 2 - \sqrt{2}$ , so $K w_{s} = \frac{g}{l} (2 - \sqrt{2}) M w_{s}$ ; similarly $K w_{f} = m g l (\begin{array}{c} 2 \\ - \sqrt{2} \end{array}), M w_{f} = m l^{2} (\begin{array}{c} 2 - \sqrt{2} \\ 1 - \sqrt{2} \end{array}),$ and $2 / (2 - \sqrt{2}) = - \sqrt{2} / (1 - \sqrt{2}) = 2 + \sqrt{2}$ , so $K w_{f} = \frac{g}{l} (2 + \sqrt{2}) M w_{f}$ . The normalizations give $u_{s}^{T} M u_{s} = u_{f}^{T} M u_{f} = 1$ and $u_{s}^{T} M u_{f} = 0$ , because $w_{s}^{T} M w_{s} = 2 m l^{2} γ_{f}$ , $w_{f}^{T} M w_{f} = 2 m l^{2} γ_{s}$ and $w_{s}^{T} M w_{f} = m l^{2} (2 - 2) = 0$ ; this proves clauses 1 and 2.

2.1F2F4step 1.2

Motions: by [F2] with $λ_{s}, λ_{f} > 0$ every motion is the displayed two-frequency superposition, with coefficients bijective onto the data, and it is bounded on $R$ because both modal families are bounded; the frequencies are $\sqrt{λ_{s}}$ and $\sqrt{λ_{f}}$ by [F4]. This is clause 3.

2.2F2step 1.2algebra

Mode patterns and pure-mode data: reading the coordinates of $w_{s} = (1, \sqrt{2})$ and $w_{f} = (1, - \sqrt{2})$ gives $θ_{2} = \sqrt{2} θ_{1}$ in the slow mode and $θ_{2} = - \sqrt{2} θ_{1}$ in the fast mode; the data $q (0) = u_{s}$ , $q^{'} (0) = 0$ correspond to $A = 1$ , $B = C = D = 0$ , hence to $q (t) = \cos (ω_{s} t) u_{s}$ , and analogously for $u_{f}$ . This is clause 4.

2.3step 1.2algebra

Frequency comparison: $γ_{f} - γ_{s} = 2 \sqrt{2} > 0$ gives $ω_{f} > ω_{s}$ : the mode in which the rods swing in opposite directions is the faster one, consistent with the cited treatment's conclusion.

3.1givenstep 1.1step 2.1step 2.2∎

Clauses 1-4 are verified on the declared matrices, which are the small-oscillation data attributed to the double pendulum in the cited linearization. The verification does not derive those matrices from the pendulum Lagrangian, makes no claim about the accuracy of the linear approximation at finite angle, and asserts no measured period; the gravitational acceleration $g$ is a model parameter.

Remarks

Why the slow mode is the same-side mode. The potential energy $\frac{1}{2} q^{T} K q = m g l (θ_{1}^{2} + \frac{1}{2} θ_{2}^{2})$ penalizes $θ_{1}$ more strongly than $θ_{2}$ , while the kinetic metric couples the two angles; the resulting slow direction is the one in which both rods move together and the fast direction the one in which they move oppositely. The computation above verifies the frequencies and patterns; the physical reading is the interpretation of the model data.
The linear model is not the pendulum. The matrices are the coefficients of the second-order Taylor expansion of the full Lagrangian in displacement and velocity at the downward rest point: $M$ is the zeroth-order value of the kinetic metric at that equilibrium, whereas $K$ is the Hessian of the potential. Derivatives of the kinetic metric multiply terms of total degree at least three in displacement and velocity, so they do not enter this quadratic Lagrangian. The exact pendulum is not a quadratic model, and the finite-time accuracy of the linear approximation is controlled only by Finite-time error control for the small-oscillation approximation: the quadratic remainder of the mechanical field and the variation-of-constants bound on small data, not by this example.

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A doubly degenerate quadratic model: every direction is a mode, the modal basis is not intrinsic, and coupling splits the double frequency

Physical scope

A worked instance of degeneracy in the coupled quadratic model: two identical uncoupled oscillators, M = m I, K = k I, with the double modal stiffness k/m, the full two-dimensional eigenspace, the four-parameter family of motions, the non-intrinsic status of every modal basis and the splitting of the double modal stiffness into k/m and (k+2c)/m, with corresponding frequencies sqrt(k/m) and sqrt((k+2c)/m), when a coupling c>0 of the two-mass model is switched on. The parameters are model data; the example asserts no measured frequency and no general stability-of-degeneracy theorem.

Example

Consider two uncoupled oscillators of equal mass and stiffness, modelled by the coupled quadratic model of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium with $n = 2$ ,

$M = m I_{2}, K = k I_{2}, J \equiv 0, m > 0, k > 0.$

The configuration-vector coefficients $a, b$ in clause 2 are not modal coefficients: their physical components have the coordinate unit $1 m$ ; converting them to modal amplitudes multiplies by $M^{1 / 2}$ in the numerical convention.

Then:

$M$ and $K$ are symmetric positive definite, the matrix $M^{- 1 / 2} K M^{- 1 / 2} = (k / m) I_{2}$ has the double eigenvalue $λ = k / m$ , the modal eigenspace is $E_{λ} = R^{2}$ and $N_{λ} : = R^{2}$ ; every orthonormal basis ${e_{1}, e_{2}}$ of $R^{2}$ is an orthonormal eigenbasis of that matrix, so the decoupling data are not unique;
every motion is $q (t) = \cos (ω t) a + \sin (ω t) b, a, b \in R^{2}, ω = \sqrt{k / m},$ and every nonzero vector $a \in R^{2}$ spans a pure mode: the datum $q (0) = a$ , $q^{'} (0) = 0$ produces the single-frequency oscillation $q (t) = \cos (ω t) a$ ;
the frequency $ω$ and the subspace $N_{λ} = R^{2}$ are intrinsic, while individual modal vectors are not: there is no basis-independent meaning to "the first mode" and no separation of the motions into two distinguished one-parameter families;
splitting by coupling. The two-mass model of Two coupled masses on a line: the in-phase and anti-phase normal modes, their frequencies, mode-energies and the zero-stiffness translation mode with coupling $c > 0$ has stiffness $K_{c} = (\begin{array}{cc} k + c & - c \\ - c & k + c \end{array})$ , which equals $k I_{2}$ at $c = 0$ and coincides with the data of this example when $c = 0$ ; for $c > 0$ its modal stiffnesses are the distinct numbers $k / m$ and $(k + 2 c) / m$ , so an explicit change of the model data separates the double eigenvalue into the unchanged in-phase frequency and the raised relative frequency.

Facts & Assumptions

Given: The data $m, k > 0$ , $M = m I_{2}$ , $K = k I_{2}$ , the frequency $ω = \sqrt{k / m}$ and, in clause 4, the mass metric and stiffness $K_{c}$ of Two coupled masses on a line: the in-phase and anti-phase normal modes, their frequencies, mode-energies and the zero-stiffness translation mode with coupling $c > 0$ .

[F1]

[F2]

Free-mode theory (The free coupled quadratic model: global existence, the complete modal classification, mode energies, intrinsic degenerate mode subspaces, boundedness and the unstable direction clauses 2, 4 and 5): motions are the modal superpositions; for a modal stiffness $λ$ the pure- $λ$ motions are the solutions of $q^{''} + λ q = 0$ with values in the intrinsic subspace $N_{λ}$ ; a positive definite stiffness gives bounded motions.

[F3]

Coupled two-mass data and modes (Two coupled masses on a line: the in-phase and anti-phase normal modes, their frequencies, mode-energies and the zero-stiffness translation mode clauses 1, 2 and 4): for $M = m I_{2}$ and $K_{c} = (\begin{array}{cc} k + c & - c \\ - c & k + c \end{array})$ with $c > 0$ , the modal stiffnesses are $λ_{+} = k / m$ and $λ_{-} = (k + 2 c) / m$ with the stated modal vectors.

[F4]

Direct linear algebra (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): $M^{- 1 / 2} = \frac{1}{\sqrt{m}} I_{2}$ , so $M^{- 1 / 2} K M^{- 1 / 2} = \frac{k}{m} I_{2}$ ; a scalar matrix commutes with every orthogonal matrix, so every orthonormal basis diagonalizes it.

[F5]

Scalar oscillator calculus (The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data): for $λ > 0$ the scalar equation $z^{''} + λ z = 0$ has exactly the family $A \cos (ω t) + B \sin (ω t)$ with $ω = \sqrt{λ}$ .

Verification

Proof technique: direct; identify the scalar multiple of the identity, read the double eigenvalue and the flexibility of the modal basis, and compare the data with the coupled two-mass instance.

1.1F1F4algebra

Degeneracy: by [F4] the matrix $M^{- 1 / 2} K M^{- 1 / 2}$ is $(k / m) I_{2}$ , whose only eigenvalue is $k / m$ with eigenspace $R^{2}$ ; any orthonormal basis diagonalizes it because a scalar matrix commutes with every matrix. Hence $λ_{1} = λ_{2} = k / m$ , $E_{λ} = R^{2}$ , $N_{λ} = R^{2}$ and the decoupling data of clause (e) of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium are non-unique. This is clause 1.

2.1F1F2F5step 1.1

Motions: by [F2] and [F5] the motions are the superpositions of the scalar families with the single frequency $ω = \sqrt{k / m}$ in every direction, i.e. the displayed four-parameter family $q (t) = \cos (ω t) a + \sin (ω t) b$ ; the datum $q (0) = a$ , $q^{'} (0) = 0$ corresponds to $b = 0$ and gives $q (t) = \cos (ω t) a$ , a motion whose values lie in the line spanned by $a$ for all $t$ : by clause 4 of [F2] this is a pure- $λ$ motion, and every nonzero $a$ gives one. This is clause 2.

2.2F3step 1.1

Splitting by coupling: the stiffness of [F3] reduces to $k I_{2}$ at $c = 0$ , so the data of this example and of the coupled two-mass model agree at $c = 0$ ; for $c > 0$ , [F3] gives the distinct modal stiffnesses $k / m$ (the in-phase mode, unchanged by the coupling) and $(k + 2 c) / m$ (the relative motion, raised by the coupling). This is clause 4.

3.1F2step 1.1step 2.1

Intrinsic versus non-intrinsic data: the number $λ = k / m$ and the subspace $N_{λ} = R^{2}$ are determined by $M$ and $K$ alone, whereas an individual modal vector is a member of a chosen orthonormal basis and different admissible decouplings assign the same motion different coordinate labels; hence no basis-independent decomposition of the four-parameter family into two distinguished one-parameter families exists here. This is clause 3.

4.1step 1.1step 3.1step 2.2∎

Clauses 1-4 are verified on the explicit data; the splitting statement is a comparison of two explicitly specified models at different parameter values and is not a general assertion that degeneracies are always removed by perturbations.

Remarks

Degeneracy is a property of the data. The double eigenvalue exists because the two oscillators are identical and uncoupled; the identity of the oscillators is a model assumption, and a model with slightly different masses or stiffnesses need not have any degeneracy. Nothing in this example asserts how degeneracies behave under general parameter changes.
What clause 4 does show. In the specific family of the two-mass model, switching on the coupling from $c = 0$ to $c > 0$ splits the double eigenvalue $k / m$ into $k / m$ and $(k + 2 c) / m$ : the in-phase frequency is unchanged and the relative frequency is raised. The two limiting descriptions are consistent, as they must be, since the data agree at $c = 0$ .

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The inverted pendulum about its upward equilibrium: the negative modal stiffness, the exponential quadratic mode and the nonlinear instability

Physical scope

A worked instance of an unstable quadratic direction: the planar inverted pendulum of mass m and length l about its upward equilibrium, with chart data M = m l^2 and U(theta) = m g l cos(theta) measured from the upward vertical, negative Hessian -m g l, the exact exponential motions of the small-oscillation model and the exponential escaping family of the quadratic model, together with the nonlinear Lyapunov instability of the upward equilibrium obtained from the hyperbolic-saddle theorem. The example is a conditional statement about the declared mathematical model; g, m and l are model parameters and no measured period or escape time is asserted.

Example

The algebra below uses the fixed-unit numerical convention of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium, with coordinate units radians, time unit $1 s$ and energy unit $1 J$ . Physical model parameters are given their stated units; in matrix products, initial-vector identities and coefficient formulas their numerical coefficients in these fixed units are used. Modal amplitudes have common unit $a_{*} = (1 J)^{1 / 2} (1 s)$ ; the $j$ -th component of a mass-normalized modal vector has unit $s_{j} / a_{*}$ . Modal oscillatory and hyperbolic coefficients have unit $a_{*}$ , and zero-mode slopes have unit $a_{*} / (1 s)$ . Physical coordinates and energies are reconstructed with the coordinate units and $1 J$ , respectively. In particular a numerical single-mode datum $q (0) = u_{i}$ means a modal amplitude of one in this common amplitude unit.

In the formulas below, $\cosh u = (e^{u} + e^{- u}) / 2$ and $\sinh u = (e^{u} - e^{- u}) / 2$ , as in the hyperbolic family of The free coupled quadratic model: global existence, the complete modal classification, mode energies, intrinsic degenerate mode subspaces, boundedness and the unstable direction.

Let a planar pendulum consist of one particle label of mass $m > 0$ attached to a fixed pivot by a massless rod of length $l > 0$ , and let $θ$ be its angle from the upward vertical, read in radians (dimensionless under Dimensions and SI unit conventions of the non-relativistic classical mechanics framework); let $g > 0$ be the uniform gravitational acceleration of the model. In the chart with kinetic metric field $M (θ) = m l^{2}$ and potential $U (θ) = m g l \cos θ$ , the mechanical system of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart has energy $E = \frac{1}{2} m l^{2} θ^{' 2} + m g l \cos θ$ and equation $m l^{2} θ^{''} = m g l \sin θ$ , that is $θ^{''} = (g / l) \sin θ$ . Then:

$θ = 0$ is a mechanical equilibrium and a strict local maximum of $U$ on the chart, with $K : = D^{2} U (0) = - m g l < 0$ ; the small-oscillation model of clause (d) of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium is the one-dimensional quadratic model $m l^{2} ξ^{''} - m g l ξ = 0$ , whose modal stiffness is $λ = - g / l < 0$ ;
with $κ : = \sqrt{g / l} > 0$ , physical frequency unit $s^{- 1}$ , every motion of that quadratic model on an interval with nonempty interior is $ξ (t) = a \cosh (κ t) + d \sinh (κ t), a, d \in R,$ Here $a, d$ are coefficients of the angle coordinate, read in radians; $b$ below denotes its initial angular velocity in radians per second. The datum $ξ (0) = 0$ , $ξ^{'} (0) = b \neq 0$ gives $ξ (t) = (b / κ) \sinh (κ t)$ , unbounded on $[0, \infty)$ ;
the rest point of the quadratic model is not Lyapunov stable (Equilibria, Lyapunov stability and asymptotic stability of an autonomous first-order system): with $ε_{0} = 1$ and every $δ > 0$ , a datum of size $< δ$ leads to a motion reaching norm $1$ ;
the nonlinear upward equilibrium $(θ, θ^{'}) = (0, 0)$ of the pendulum chart system is not Lyapunov stable either: the field $F (θ, v) = (v, (g / l) \sin θ)$ is $C^{1}$ with $F (0, 0) = 0$ and $D F (0, 0) = (\begin{array}{cc} 0 & 1 \\ g / l & 0 \end{array})$ , whose eigenvalues $\pm \sqrt{g / l}$ are real, nonzero and include the positive value $\sqrt{g / l}$ , so the hyperbolic-saddle theorem applies.

Facts & Assumptions

Given: The parameters $m, l, g > 0$ , the chart metric field and potential $M (θ) = m l^{2}$ , $U (θ) = m g l \cos θ$ , the field $F (θ, v) = (v, (g / l) \sin θ)$ , and the numbers $κ = \sqrt{g / l}$ and $ε_{0} = 1$ .

[F1]

Mechanical chart model (The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart): $θ = 0$ satisfies $\nabla U (0) = U^{'} (0) = 0$ and is a mechanical equilibrium; $U^{''} (0) = - m g l$ , so $K = D^{2} U (0)$ is the symmetric matrix $- m g l < 0$ ; the mechanical vector field is $G (θ, v) = (v, - U^{'} (θ) / m l^{2}) = (v, (g / l) \sin θ)$ .

[F2]

Small-oscillation model (The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium clause (d)): the model at the equilibrium is the coupled quadratic model with mass metric $m l^{2}$ , stiffness $K = - m g l$ and zero generalized force, in the displacement coordinate $ξ$ ; its modal stiffness is $λ = K / M = - g / l$ .

[F3]

Free-mode theory (The free coupled quadratic model: global existence, the complete modal classification, mode energies, intrinsic degenerate mode subspaces, boundedness and the unstable direction clauses 2 and 6): for $λ < 0$ the free motions are $a \cosh (κ t) + d \sinh (κ t)$ with $κ = \sqrt{- λ}$ , and the rest point of the free model is not Lyapunov stable, for every requested initial size, with an escaping datum of smaller size and an explicit growing family. Stable decaying families are not asserted to escape.

[F4]

Hyperbolic instability (A hyperbolic equilibrium whose linearisation has an eigenvalue with positive real part is unstable): for a $C^{1}$ field $F$ on an open set with $F (0) = 0$ , if no complex eigenvalue of $D F (0)$ has real part $0$ and at least one eigenvalue has positive real part, then $0$ is not a Lyapunov stable equilibrium (Equilibria, Lyapunov stability and asymptotic stability of an autonomous first-order system).

[F5]

Hyperbolic solutions and growth (The degenerate and repulsive oscillator: solutions for k=0 and the explicit growing solution for k<0, The exponential function is strictly increasing): the scalar proposition gives $e^{κ t}$ as a solution of $ξ^{''} - κ^{2} ξ = 0$ with initial data $(1, κ)$ and proves $e^{κ t} \to + \infty$ ; for $t \geq 0$ the strictly increasing exponential also gives $0 < e^{- κ t} \leq 1$ . By the definitions in the Statement, $\sinh (κ t) = (e^{κ t} - e^{- κ t}) / 2$ is therefore unbounded as $t \to + \infty$ .

[F6]

Trigonometric derivative and continuity (The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$ , $C^{k}$ Euclidean maps and diffeomorphisms): $D F (θ, v) = (\begin{array}{cc} 0 & 1 \\ (g / l) \cos θ & 0 \end{array}),$ so $D F (0, 0) = (\begin{array}{cc} 0 & 1 \\ g / l & 0 \end{array})$ and its eigenvalues are $\pm \sqrt{g / l}$ . The sine derivative is cosine, and cosine is continuous because it is differentiable; hence $F$ is $C^{1}$ . The square root is positive since $g / l > 0$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[F7]

Intermediate value theorem (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ): a continuous real-valued function on a closed interval takes every value between its endpoint values.

Verification

technique · direct; read the equilibrium data from the chart, solve the repulsive scalar equation, and apply the free-mode instability clause and the hyperbolic-saddle theorem

1.1F1F2algebra

Equilibrium and stiffness: $U^{'} (θ) = - m g l \sin θ$ vanishes at $θ = 0$ , and $U (θ) < U (0)$ for $0 < ∣ θ ∣ < π$ on the chart, so $θ = 0$ is a strict local maximum and $K = U^{''} (0) = - m g l < 0$ ; the small-oscillation model has data $M = m l^{2}$ , $K = - m g l$ , i.e. modal stiffness $λ = - g / l$ . This is clause 1.

2.1F3F5step 1.1

Exponential families: by [F3] and [F5] the motions of the quadratic model are exactly $ξ (t) = a \cosh (κ t) + d \sinh (κ t)$ with $κ = \sqrt{g / l}$ ; for the datum $ξ (0) = 0$ , $ξ^{'} (0) = b \neq 0$ the hyperbolic coefficients are $a = 0$ and $d = b / κ$ , so the motion is $(b / κ) \sinh (κ t)$ , which is unbounded on $[0, \infty)$ . This is clause 2.

2.2F4F6step 1.1

Nonlinear instability: the field $F (θ, v) = (v, (g / l) \sin θ)$ is $C^{1}$ on $R^{2}$ with $F (0, 0) = 0$ ; by [F6] its derivative at the origin has the real eigenvalues $\pm \sqrt{g / l}$ , none on the imaginary axis and one positive, so the hypotheses of [F4] hold and $(0, 0)$ is not a Lyapunov stable equilibrium of the pendulum chart system. This is clause 4.

3.1F3F5F7step 2.1algebra

Quadratic instability: by clause 6 of [F3] applied to the one-dimensional free model with negative modal stiffness, the rest point is not Lyapunov stable, with the displayed escaping family; explicitly, for every $δ > 0$ choose $0 < ∣ b ∣ < \min (δ, 1)$ . The motion $(b / κ) \sinh (κ t)$ has initial phase norm $∣ b ∣ < 1$ and its position norm is unbounded by [F5]. Thus its continuous phase norm starts below $1$ and exceeds $1$ at some finite time; [F7] gives a time $t > 0$ at which the phase norm is exactly $1 = ε_{0}$ . This is clause 3.

4.1step 1.1step 2.1step 3.1step 2.2∎

Clauses 1-4 are verified on the explicit chart data; clause 4 is the nonlinear conclusion supplied by the hyperbolic-saddle theorem, whereas clause 3 concerns the quadratic model alone. The parameters $m, l, g$ are model data, and no escape time, measured period or apparatus behaviour is asserted.

Remarks

Why the linear and nonlinear conclusions agree here. The linearization at the upward equilibrium is hyperbolic — its spectrum $\pm \sqrt{g / l}$ is disjoint from the imaginary axis — so the saddle theorem converts the positive eigenvalue into a genuine nonlinear instability. This is exactly the conditional route recorded in Scope of the coupled quadratic model: zero modes, degenerate eigenspaces, unstable quadratic directions and the limits of the approximation statements; for a nonhyperbolic critical point (in particular one with a zero mode) the linear calculation would not be sufficient.
The energy picture. The mechanical energy $E = \frac{1}{2} m l^{2} θ^{' 2} + m g l \cos θ$ has a saddle at the upward equilibrium: its potential part has a strict local maximum while its velocity part is positive quadratic; energy conservation therefore cannot trap nearby data in a small neighbourhood, in contrast with the strict-minimum case where the same energy argument gives stability (Small oscillations at a strict minimum: the linearized model, its normal modes and the Lyapunov stability of the nonlinear equilibrium clause 4).

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A drive along the in-phase pattern of the two-mass chain: resonance exactly at that mode's frequency and a bounded steady response off resonance

Physical scope

A worked instance of the forced coupled quadratic model: the two equal-mass chain of the companion example with a prescribed generalized force J(t)=F cos(Omega t) u_+ along the in-phase pattern (a sinusoidal drive for Omega>0 and a constant-force specialization for Omega=0), the modal force components (F/m,0), the unbounded resonance at Omega = sqrt(k/m) with the explicit linearly growing modal motion, the bounded steady response of amplitude |F|/(2c) at Omega = sqrt((k+2c)/m), and the boundedness of every motion for Omega^2 not equal to k/m. The force is prescribed external data; no back-reaction, measured response or empirical value is asserted.

Example

Adopt the two-mass chain of Two coupled masses on a line: the in-phase and anti-phase normal modes, their frequencies, mode-energies and the zero-stiffness translation mode: mass metric $M = m I_{2}$ , stiffness $K = (\begin{array}{cc} k + c & - c \\ - c & k + c \end{array})$ with $m > 0$ , $k > 0$ , $c > 0$ , and modal vectors $u_{\pm} = (1, \pm 1) / \sqrt{2 m}$ with modal stiffnesses $λ_{+} = k / m$ , $λ_{-} = (k + 2 c) / m$ and frequencies $ω_{+} = \sqrt{k / m}$ , $ω_{-} = \sqrt{(k + 2 c) / m}$ . Choose a real drive coefficient $F \neq 0$ . In the physical equation $J = F u_{+} \cos (Ω t)$ its unit is ${k g}^{3 / 2} m s^{- 2}$ , since $u_{+}$ has unit ${k g}^{- 1 / 2}$ and $J$ is a force. Thus $F / m$ is a modal force amplitude with unit $a_{*} / s^{2}$ , and $∣ F ∣ / (2 c)$ below is a modal coordinate amplitude with unit $a_{*}$ ; multiplying by $u_{+}$ gives each particle's displacement amplitude. Choose a frequency $Ω \geq 0$ and prescribe $J (t) = F \cos (Ω t) u_{+} .$ For $Ω > 0$ this is the sinusoidal drive of The sinusoidally driven oscillator model: a prescribed external force $f_{0} \cos (Ω t)$ added to the damped oscillator; for $Ω = 0$ it is the constant force allowed by clause (b) of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium. In either case it is prescribed external data, so the model is the forced coupled quadratic model $M q^{''} + K q = J$ . The all-time claims below concern maximal motions on $R$ ; local solutions are their restrictions. Then:

the modal force components of $J (t) = J_{0} \cos (Ω t)$ with $J_{0} = F u_{+}$ are $ψ_{0} = O^{T} M^{- 1 / 2} J_{0} = (F / m, 0)$ : the drive excites the in-phase modal coordinate alone, with modal amplitude $F / m$ , and leaves the relative coordinate unforced;
if $Ω = ω_{+} = \sqrt{k / m}$ , then $q^{p} (t) = \frac{F}{2 m ω_{+}} t \sin (ω_{+} t) u_{+}$ is a motion of the model, of in-phase pattern, with linearly growing amplitude $∣ F ∣ t / (2 m ω_{+})$ , and every motion of the model is unbounded on $[0, \infty)$ ;
if $Ω^{2} \neq k / m$ then every motion of the model is bounded on $[0, \infty)$ ; in particular at $Ω = ω_{-} = \sqrt{(k + 2 c) / m}$ — the frequency of the unforced relative mode — the drive is non-resonant and the steady modal response has amplitude $\frac{∣ F ∣ / m}{∣ k / m - (k + 2 c) / m ∣} = \frac{∣ F ∣}{2 c};$
the resonance condition is exactly that the drive frequency equal the frequency of a mode whose modal force component does not vanish: here the in-phase mode at $ω_{+}$ , not the relative mode at $ω_{-}$ .

Facts & Assumptions

Given: The data $m, k, c > 0$ of the chain, the modal vectors $u_{\pm}$ and modal stiffnesses $λ_{\pm}$ , the modal data $O$ , $M^{1 / 2}$ with $e_{\pm} = (1, \pm 1) / \sqrt{2}$ , real $F \neq 0$ and $Ω \geq 0$ .

[F1]

Chain data and modes (Two coupled masses on a line: the in-phase and anti-phase normal modes, their frequencies, mode-energies and the zero-stiffness translation mode clauses 1, 2 and 4): $M = m I_{2}$ , $K$ as displayed, $O = [e_{+} e_{-}]$ , $u_{\pm} = M^{- 1 / 2} e_{\pm}$ , $K u_{\pm} = λ_{\pm} M u_{\pm}$ with the displayed $λ_{\pm}$ , and $K$ is positive definite.

[F2]

Forced modal response (The forced coupled quadratic model: modal responses, resonance, and boundedness for positive definite stiffness off resonance clauses 1-3): the unique motion with a datum has the modal convolution form with $ψ = O^{T} M^{- 1 / 2} J$ ; for $J (t) = J_{0} \cos (Ω t)$ the modal force is $ψ (t) = ψ_{0} \cos (Ω t)$ with $ψ_{0} = O^{T} M^{- 1 / 2} J_{0}$ ; the sinusoidal particular functions of clauses 2(a)-(c) hold; every motion is $q^{p} + q^{h}$ with $q^{h}$ free; in the resonant cases (some $ψ_{0, i} \neq 0$ with $λ_{i} > 0$ , $Ω^{2} = λ_{i}$ , or $λ_{i} = 0$ , $Ω = 0$ ) every motion is unbounded on $[0, \infty)$ ; and for positive definite $K$ with no resonant index of nonzero modal force, every motion is bounded.

[F3]

Prescribed generalized force (The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium clause (b)): $J$ is continuous prescribed model data. For $Ω > 0$ the cosine term has the restricted sinusoidal-drive status of The sinusoidally driven oscillator model: a prescribed external force $f_{0} \cos (Ω t)$ added to the damped oscillator clause (ii); for $Ω = 0$ it is the constant-force specialization of the general $J$ data. In both cases it depends on time alone and has no modelled back-reaction.

[F4]

Direct linear algebra (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): $M^{- 1 / 2} = \frac{1}{\sqrt{m}} I_{2}$ , $J_{0} = F (1, 1) / \sqrt{2 m}$ , $e_{+}^{T} (1, 1) = \sqrt{2}$ , $e_{-}^{T} (1, 1) = 0$ ; the products below are direct computations. The particular motions displayed are differentiated by the chain rule (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

Verification

technique · direct; compute the modal force, instantiate the resonant and non-resonant cases of the forced modal theorem, and evaluate the steady amplitude

1.1F1F4algebra

Modal force: with $J_{0} = F u_{+} = F (1, 1) / \sqrt{2 m}$ , $ψ_{0} = O^{T} M^{- 1 / 2} J_{0} = \frac{F}{\sqrt{m}} (\begin{array}{c} e_{+}^{T} \\ e_{-}^{T} \end{array}) \frac{(1, 1)}{\sqrt{2 m}} = \frac{F}{m} (\begin{array}{c} 1 \\ 0 \end{array}),$ because $e_{+}^{T} (1, 1) = \sqrt{2}$ and $e_{-}^{T} (1, 1) = 0$ . This is clause 1.

2.1F1F2F4step 1.1algebra

Resonant case: for $Ω = ω_{+}$ , $λ_{+} = Ω^{2}$ and $ψ_{0, +} = F / m \neq 0$ ; by [F2] clause 2(b) the function $z_{+}^{p} (t) = \frac{F}{m} \frac{t \sin (ω_{+} t)}{2 ω_{+}}$ solves the in-phase modal equation and $q^{p} = M^{- 1 / 2} O z^{p} = \frac{F}{2 m ω_{+}} t \sin (ω_{+} t) u_{+}$ is a motion of the model, with the stated linearly growing amplitude; and by [F2] clause 3(a) every motion of the model is unbounded on $[0, \infty)$ . This is clause 2.

2.2F1F2F4step 1.1algebra

Non-resonant case: if $Ω^{2} \neq k / m = λ_{+}$ , then the only index with $ψ_{0, i} \neq 0$ is $i = +$ (since $ψ_{0, -} = 0$ ), and it is not resonant; $K$ is positive definite by [F1], so [F2] clause 3(b) gives that every motion is bounded on $[0, \infty)$ . At $Ω = ω_{-}$ the steady in-phase response of [F2] clause 2(a) has amplitude $∣ ψ_{0, +} ∣ / ∣ λ_{+} - Ω^{2} ∣ = \frac{∣ F ∣}{m} / \frac{2 c}{m} = ∣ F ∣ / (2 c)$ , and the relative mode has zero modal force. This is clause 3.

3.1F2step 1.1step 2.1step 2.2

Resonance reading: by [F2] the resonant frequencies are exactly the frequencies of the modal directions with nonzero modal force component; here $ψ_{0}$ is supported on the in-phase direction alone, so the resonance is at $ω_{+}$ and the equally available frequency $ω_{-}$ of the relative mode does not produce resonance for this drive. This is clause 4.

4.1F3step 1.1step 2.1step 2.2step 3.1∎

Clauses 1-4 are verified on the explicit data under the prescribed drive [F3]; no damping, back-reaction, measured response or apparatus detail is asserted, and the parameters are model data.

Remarks

The unbounded motion is a mode-shaped oscillation. The resonant motion of clause 2 keeps the in-phase pattern $u_{+}$ fixed and grows linearly in amplitude; it is not a runaway in an arbitrary direction, and the prescribed particular response has zero relative coordinate. An arbitrary motion can additionally contain the bounded homogeneous relative oscillation of the free model.
Why the amplitude at $Ω = ω_{-}$ grows as $c \to 0^{+}$ . $∣ F ∣ / (2 c)$ is the steady amplitude at the relative-mode frequency; as the coupling tends to zero the two frequencies merge ( $ω_{-} \to ω_{+}$ ) and the drive approaches resonance, so the bounded steady amplitude grows without bound. This is the limiting consistency of clauses 2 and 3 and not an assertion about any particular apparatus.

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An explicit weakly damped oscillator: the averaging constants and the energy-decay bound over five thousand seconds (about 796 undamped periods)

Physical scope

A worked numerical instance of the weak-damping averaging estimate: the model m x'' + epsilon gamma x' + k x = 0 with m = 1 kg, k = 1 kg/s^2, epsilon = 10^-4 and gamma = 1 kg/s, initial rotating-frame amplitude zeta_0 = 1, ball radii R = 1/2 and rho = 1/4 and horizon parameter A = 1/2, verifying the smallness and trapping hypotheses with the exact constants M = 3 sqrt2/2, L = sqrt2, K = 2 pi (3 sqrt2 + 6) exp(sqrt2/2) and concluding that the mechanical energy tracks the averaged exponential exp(-t/10^4) within 2.63 x 10^-2 on the interval [0,5000]. The numbers are exact-arithmetic evaluations of the model formulas, not measurements; no apparatus, no observed decay and no empirical statement is present.

Example

Adopt the damped oscillator model of The damped oscillator model: a harmonic restoring force with a phenomenological linear drag $- c \dot{x}$ , $c \geq 0$ with data

$m = 1 kg, k = 1 {kg/s}^{2}, ε = 10^{- 4}, γ = 1 kg/s,$

so that $c = ε γ = 10^{- 4}$ kg/s and $ω_{0} = \sqrt{k / m} = 1$ rad/s; the angles and frequencies are read under the convention of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework. Fix amplitude unit $a_{*} = 1 m s^{- 1 / 2}$ and time unit $1 s$ . Take $ζ_{0} = a_{*}$ , $R = a_{*} / 2$ , $ρ = a_{*} / 4$ and $A = (1 s) / 2$ . Below these symbols, $t$ and the constants in coefficient calculations mean their numerical values in these fixed units. In particular $M$ is in $a_{*} / s$ , $L$ in $s^{- 1}$ , $B, K$ in $a_{*}$ and $B_{1}$ is dimensionless. Every energy exponential below is multiplied by $E_{0} = 1 J$ when read as a physical quantity, and the horizon $[0, 5000]$ means $[0, 5000 s]$ . For $x > 0$ , write $\ln x : = \log x$ , where $\log : (0, \infty) \to R$ is the inverse of the real exponential (The natural logarithm as the inverse of the exponential function). Then:

the averaging data of Averaging for a weakly damped oscillator: the rotating-frame amplitude, the averaged exponential energy decay and the order- $ε$ bound on the $O (1 / ε)$ time scale specialize to $P = 2 π$ , $F (t, ζ) = - i e^{i t} (\cos t Im ζ - \sin t Re ζ)$ and $\overset{ˉ}{F} (ζ) = - ζ / 2$ , with averaged solution $\overset{ˉ}{ζ} (t) = e^{- ε t / 2} = e^{- t / (2 \cdot 10^{4})}$ ;
the constants of the averaging estimate are $M = \frac{3 \sqrt{2}}{2}, L = \sqrt{2}, B = 6 \sqrt{2} π, B_{1} = 4 \sqrt{2} π, K = 2 π (3 \sqrt{2} + 6) e^{\sqrt{2} / 2},$ so that $K \approx 1.30522 \times 10^{2}, ε K \approx 1.30522 \times 10^{- 2} < 0.25 = ρ;$
the trapping hypothesis holds: $∣ ζ_{0} ∣ (1 - e^{- γ A / (2 m)}) = 1 - e^{- 1 / 4} \approx 0.22120 < R - ρ = 0.25$ , and also $1 / 4 = γ A / (2 m) \leq \ln (4 / 3) \approx 0.28768$ ;
consequently the motion with datum $x (0) = \sqrt{2}$ m, $x^{'} (0) = 0$ — which has $ζ (0) = 1$ and mechanical energy $E_{0} = m ω_{0} ∣ ζ_{0} ∣^{2} = 1$ J — satisfies $∣ E (t) - e^{- t / 10^{4}} ∣ \leq m ω_{0} ε K (2 ∣ ζ_{0} ∣ + ε K) \approx 2.62748 \times 10^{- 2} J (0 \leq t \leq 5000),$ the interval $[0, 5000]$ being $[0, A / ε]$ and containing $5000 / (2 π) \approx 795.8$ oscillation periods of the undamped frequency;
at the end of the horizon the averaged energy is $e^{- 1 / 2} \approx 0.60653$ J, so the estimate bounds the deviation of the exact energy from the averaged exponential decay by about $4.3 %$ of its value there.

Facts & Assumptions

Given: The model data $m = 1$ , $k = 1$ , $ε = 10^{- 4}$ , $γ = 1$ ; the numbers $ζ_{0} = 1$ , $R = 1 / 2$ , $ρ = 1 / 4$ , $A = 1 / 2$ ; the constants $M, L, B, B_{1}, K$ of the Example.

[F1]

Averaging theorem (Averaging for a weakly damped oscillator: the rotating-frame amplitude, the averaged exponential energy decay and the order- $ε$ bound on the $O (1 / ε)$ time scale clauses 1-4): for the model with data $(m, ε γ, k)$ the rotating-frame amplitude maps motions bijectively onto solutions of $ζ^{'} = ε F (t, ζ)$ ; $F$ is $P = 2 π / ω_{0}$ -periodic with average $\overset{ˉ}{F} (ζ) = - γ ζ / (2 m)$ ; under the hypotheses $ε K < ρ$ and the trapping condition, the estimate $∥ ζ (t) - \overset{ˉ}{ζ} (t) ∥_{2} \leq ε K$ and the energy estimate $∣ E (t) - E_{0} e^{- ε γ t / m} ∣ \leq m ω_{0} ε K (2 ∣ ζ_{0} ∣ + ε K)$ hold on $[0, A / ε]$ , with $M = (γ / m) \sqrt{2} (∣ ζ_{0} ∣ + R)$ , $L = (γ / m) \sqrt{2}$ , $B = 2 M P$ , $B_{1} = 2 L P$ and $K = e^{L A} [B + (L B + B_{1} M) A]$ .

[F2]

Data specialization (Periodic averaging data: the average field, the oscillatory part and the periodic deviation, Real and imaginary parts, complex conjugation, and modulus): with $ω_{0} = 1$ and $γ / m = 1$ the field of the theorem reads $F (t, ζ) = - i e^{i t} (\cos t Im ζ - \sin t Re ζ)$ and $P = 2 π$ ; $∣ ζ_{0} ∣ = 1$ .

[F3]

Elementary arithmetic (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The derivatives of sine and cosine are cosine and minus sine): the displayed decimal values are evaluations of the exact expressions, obtained by direct arithmetic from $\sqrt{2} \approx 1.41421356$ , $π \approx 3.14159265$ and $e^{\sqrt{2} / 2} \approx 2.02811498$ ; in particular the datum with $ζ_{0} = 1$ corresponds to $x (0) = \sqrt{2}$ m, $x^{'} (0) = 0$ and $E_{0} = m ω_{0} ∣ ζ_{0} ∣^{2} = 1$ J.

[F4]

Exact numerical inequalities: $\sqrt{2} < 3 / 2$ by squaring and $2 < e < 3$ by The elementary numerical bound $2 < e < 3$ . For $x \in [1 / 4, 3 / 4]$ , $x (1 - x) \geq 3 / 16$ and $1 + x^{2} \leq 25 / 16$ , so $\frac{x^{4} (1 - x)^{4}}{1 + x^{2}} \geq \frac{16}{25} (\frac{3}{16})^{4} > 0.$ Polynomial division gives $\frac{x^{4} (1 - x)^{4}}{1 + x^{2}} = x^{6} - 4 x^{5} + 5 x^{4} - 4 x^{2} + 4 - \frac{4}{1 + x^{2}} .$ By Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series, $\int_{0}^{1} d t / (1 + t^{2}) = \arctan 1$ . Also $\arctan 1 = π / 4$ : the principal inverse tangent of The principal inverse tangent $\arctan : R \to (- π / 2, π / 2)$ inverts $\tan$ on $(- π / 2, π / 2)$ ; the cofunction identity of Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions at $x = π / 4$ gives $\sin (π / 4) = \cos (π / 4)$ , hence $\tan (π / 4) = 1$ by Tangent, cotangent, secant, and cosecant on their exact natural domains. Termwise integration of the polynomial using For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term and Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ now gives $\int_{0}^{1} \frac{x^{4} (1 - x)^{4}}{1 + x^{2}} d x = \frac{22}{7} - 4 \arctan 1 = \frac{22}{7} - π .$ The integrand is nonnegative on $[0, 1]$ , and its integral on $[1 / 4, 3 / 4]$ is at least $\frac{1}{2} \cdot \frac{16}{25} (\frac{3}{16})^{4} > 0$ by If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , so the full integral is positive and $π < 22 / 7$ . Also, for $x > 0$ , write $\ln x : = \log x$ as in The natural logarithm as the inverse of the exponential function. Then $\ln (4 / 3) = \int_{1}^{4 / 3} d t / t \geq \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$ by The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t and integral monotonicity. For $u > 0$ , the fundamental theorem and exponential derivative give $1 - e^{- u} = \int_{0}^{u} e^{- s} d s < u$ : on $[0, u / 2]$ the integrand is at most $1$ , and on $[u / 2, u]$ it is at most $e^{- u / 2} < 1$ by strict increase of the exponential (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , The exponential function is smooth and $(\exp)^{'} = \exp$ , The exponential function is strictly increasing, If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ). Finally, $e^{- 1 / 2} > 1 / 3$ follows from $e^{1 / 2} < e < 3$ and $e^{- x} = 1 / e^{x}$ (The elementary numerical bound $2 < e < 3$ , The exponential function is strictly increasing, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

Verification

technique · direct; specialize the theorem's data, evaluate the constants, check the two explicit hypotheses, and state the conclusion

1.1F1F2algebra

Specialization: with $m = k = γ = 1$ and $ε = 10^{- 4}$ the quantities of [F1] become $ω_{0} = 1$ , $P = 2 π$ , $\overset{ˉ}{F} (ζ) = - ζ / 2$ and $\overset{ˉ}{ζ} (t) = e^{- ε t / 2} = e^{- t / (2 \cdot 10^{4})}$ ; the field is the one displayed in clause 1.

1.2F1F4givenalgebra

Constants: substituting $∣ ζ_{0} ∣ = 1$ , $R = 1 / 2$ , $γ / m = 1$ gives $M = \sqrt{2} \cdot \frac{3}{2} = \frac{3 \sqrt{2}}{2}$ and $L = \sqrt{2}$ , hence $B = 2 M P = 6 \sqrt{2} π$ and $B_{1} = 2 L P = 4 \sqrt{2} π$ . Then $L B = \sqrt{2} \cdot 6 \sqrt{2} π = 12 π$ and $B_{1} M = 4 \sqrt{2} π \cdot \frac{3 \sqrt{2}}{2} = 12 π$ , so $K = e^{\sqrt{2} / 2} [6 \sqrt{2} π + \frac{1}{2} (12 π + 12 π)] = 2 π (3 \sqrt{2} + 6) e^{\sqrt{2} / 2},$ and by [F4] $K < 198 < 200$ : $\sqrt{2} < 3 / 2$ and $e^{\sqrt{2} / 2} < e < 3$ give $K < 2 π \cdot (3 \cdot \frac{3}{2} + 6) \cdot 3 = 63 π < 63 \cdot 22 / 7 = 198$ . This is clause 2.

2.1F1F4step 1.2givenalgebra

Hypotheses: $ε K < 198 / 10^{4} < 1 / 50 < 1 / 4 = ρ$ by step 1.2, so the smallness condition holds; at the full horizon $A / ε = 5000$ the trapping quantity is $∣ ζ_{0} ∣ (1 - e^{- γ A / (2 m)}) = 1 - e^{- 1 / 4} < 1 / 4 = R - ρ$ by [F4]. The sufficient logarithmic condition also holds, since $γ A / (2 m) = 1 / 4 \leq \ln (4 / 3)$ by [F4]. Hence the averaged solution stays in the required ball. This is clause 3.

3.1F1F3F4step 1.1step 1.2step 2.1algebra

Conclusion: applying [F1] with the verified hypotheses and the horizon $A / ε = 5000$ gives $∥ ζ (t) - \overset{ˉ}{ζ} (t) ∥_{2} \leq ε K < 1 / 50$ for $0 \leq t \leq 5000$ , hence the energy bound $∣ E (t) - e^{- t / 10^{4}} ∣ \leq ε K (2 + ε K) < \frac{101}{2500} < \frac{1}{24}$ J; the datum corresponds to $x (0) = \sqrt{2}$ m, $x^{'} (0) = 0$ , $E_{0} = 1$ J, and the horizon contains more than $795$ periods of the undamped frequency, because $5000 > 1590 π$ and [F4] gives $1590 π < 1590 \cdot \frac{22}{7} = \frac{34980}{7} < 5000$ ; at $t = 5000$ the averaged energy is $e^{- 1 / 2} > 1 / 3$ by [F4], so the bound is below one eighth of it. This is clauses 4 and 5.

4.1step 1.1step 1.2step 2.1step 3.1∎

Clauses 1-5 are verified by exact computation from the formulas of the averaging theorem, with the rational and elementary estimates above in place of decimal arithmetic; no measured quantity, apparatus or empirical claim is present.

Remarks

Decimal orientation values. On the stated instance, $K \approx 1.30522 \times 10^{2}$ , $ε K \approx 1.30522 \times 10^{- 2}$ and the energy bound is approximately $2.63 \times 10^{- 2}$ J; the averaged energy at $t = 5000$ is approximately $0.60653$ J. These decimals are evaluations of the exact expressions of the proof and play no role in it.
Reading the constant. The dominating factor in $K$ is $e^{L A} = e^{\sqrt{2} / 2}$ , below $3$ , because the horizon parameter $A = 1 / 2$ is small: the estimate is useful for horizons of order one in the slow time $A$ , and $K$ grows like $A e^{L A}$ with $A$ . Taking $A$ larger lengthens the interval $[0, A / ε]$ but increases both $K$ and the required smallness margin.
The finite check is arithmetic, not empirical. Every number above is computed from the model formulas; nothing in the example depends on an experiment, and the estimate is a conditional bound for the declared model.

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A cubic-potential chart system: the exact quadratic remainder, the explicit error constants and a checked finite-time bound

Physical scope

A worked instance of the finite-time linearization error bound: the one-dimensional chart system with constant mass m, potential U(q) = k q^2/2 + lambda q^3/6 and field G(q,v) = (v,-(kq+lambda q^2/2)/m), whose translated field has the exact quadratic remainder -(lambda/(2m)) delta q^2, giving the exact constants C = |lambda|/(2m) and L = max(1,(k+|lambda|r/2)/m) on the ball of radius r; a numerical instance with m=k=lambda=1, r=1, delta=10^-2 and T=1 checks the smallness condition delta exp(LT) = 0.04482 < 1 and the bound 2.73 x 10^-3, and the case lambda=0 reduces the field to its linearization with C=0. All quantities are exact-arithmetic evaluations of model formulas; no measurement or empirical claim is present.

Example

Let $m > 0$ , $k > 0$ and $λ \in R$ be constants, read with the dimensions $μ$ , $μ τ^{- 2}$ and $μ ℓ^{- 1} τ^{- 2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework), and consider the one-dimensional mechanical chart system of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart with constant metric field $M (q) = m$ and potential

$U (q) = \frac{1}{2} k q^{2} + \frac{λ}{6} q^{3},$

For all field, norm and constant formulas below fix $s_{1} = 1 m$ , $t_{*} = 1 s$ and $e_{*} = 1 J$ , and use the dimensionless numerical coordinates $q = Q / s_{1}$ , $t = t_{p h y s} / t_{*}$ , $v = (t_{*} / s_{1}) \dot{Q}$ . In these formulas $m, k, λ$ mean the numerical coefficients $m_{p h y s} s_{1}^{2} / (e_{*} t_{*}^{2})$ , $k_{p h y s} s_{1}^{2} / e_{*}$ and $λ_{p h y s} s_{1}^{3} / e_{*}$ , respectively. Consequently $r, δ, T, C, L, ∥ A ∥$ are numerical normalized quantities, and $\max (1, (k + ∣ λ ∣ r / 2) / m)$ compares dimensionless numbers. The physical horizon is $t_{*} T$ , and the numerical error bound $b$ implies $∣ Q - Q_{l i n} ∣ \leq s_{1} b$ and $∣ \dot{Q} - {\dot{Q}}_{l i n} ∣ \leq (s_{1} / t_{*}) b$ .

so that the mechanical field is $G (q, v) = (v, - (k q + \frac{λ}{2} q^{2}) / m)$ . Then:

$q_{0} = 0$ is a mechanical equilibrium with $K : = D^{2} U (0) = k > 0$ ; the small-oscillation model of clause (d) of The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium is the one-dimensional quadratic model $m ξ^{''} + k ξ = 0$ with frequency $ω = \sqrt{k / m}$ ;
for the translated field $F (δ q, δ v) = (δ v, - (k δ q + \frac{λ}{2} δ q^{2}) / m)$ one has the exact remainder identity $F (ζ) - A ζ = (0, - \frac{λ}{2 m} δ q^{2}), A = (\begin{array}{cc} 0 & 1 \\ - k / m & 0 \end{array}), ζ = (δ q, δ v),$ so the constants of Finite-time error control for the small-oscillation approximation: the quadratic remainder of the mechanical field and the variation-of-constants bound can be taken as $C = \frac{∣ λ ∣}{2 m}, L = \max (1, \frac{k + ∣ λ ∣ r / 2}{m})$ on every ball $∥ ζ ∥_{2} \leq r$ , and $∥ A ∥ = \max (1, k / m)$ ;
the numerical instance $m = 1$ kg, $k = 1$ kg/s $^{2}$ , $λ = 1$ kg/(m $\cdot$ s $^{2}$ ), $r = 1$ , $δ = 10^{- 2}$ , $T = 1$ in the normalized variables (a one-second horizon), for any initial normalized phase datum of norm $δ$ , satisfies $δ e^{L T} = 10^{- 2} e^{3 / 2} \approx 0.04482 < 1$ and the resulting bound is $C δ^{2} T e^{(∥ A ∥ + 2 L) T} \approx 2.73 \times 10^{- 3};$
for $λ = 0$ the potential is exactly quadratic, $C = 0$ , and the nonlinear model coincides with its linearization: the bound of the theorem degenerates to the identity $z = y$ .

Facts & Assumptions

Given: The parameters $m, k > 0$ , $λ$ , the field $G$ and translated field $F$ , the matrix $A$ , and in clause 3 the numerical values $m = k = λ = r = 1$ , $δ = 10^{- 2}$ , $T = 1$ .

[F1]

Chart model and linearization (The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart, Linearisation of the mechanical system at an equilibrium: M(q_0) xi'' + D²U(q_0) xi = 0, and the small-oscillation quadratic model): $\nabla U (0) = 0$ and $D G (0, 0) (δ q, δ v) = (δ v, - \frac{k}{m} δ q)$ ; the second-order Taylor data at the equilibrium are $M = m$ , $K = k$ , so the small-oscillation model is $m ξ^{''} + k ξ = 0$ with $ω = \sqrt{k / m}$ (The coupled quadratic oscillator model: mass metric, stiffness, generalized force, modal vocabulary and the small-oscillation model at a mechanical equilibrium clause (d)).

[F2]

Finite-time error bound (Finite-time error control for the small-oscillation approximation: the quadratic remainder of the mechanical field and the variation-of-constants bound): for a $C^{3}$ metric field and potential with vanishing gradient at $q_{0}$ and constants $C, L, r_{0}$ with $∥ F (ζ) - A ζ ∥_{2} \leq C ∥ ζ ∥_{2}^{2}$ and $∥ F (ζ) ∥_{2} \leq L ∥ ζ ∥_{2}$ on $∥ ζ ∥_{2} \leq r_{0}$ , every datum with $δ = ∥ ζ_{0} ∥_{2} > 0$ and $δ e^{L T} < r_{0}$ has its motion defined on $[0, T]$ with $∥ z (t) - y (t) ∥_{2} \leq C δ^{2} t e^{(∥ A ∥ + 2 L) t}$ there.

[F3]

Direct computation (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The operator norm as the least bound and as the unit-sphere or unit-ball supremum, The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ ): differentiating $U$ gives $U^{'} (q) = k q + \frac{λ}{2} q^{2}$ and $U^{''} (q) = k + λ q$ ; the remainder identity of clause 2 is read off the components of $F$ ; the norm of $∥ F (ζ) ∥_{2}$ on the ball is bounded by $\max (1, (k + ∣ λ ∣ r / 2) / m) ∥ ζ ∥_{2}$ because $∣ k δ q + \frac{λ}{2} δ q^{2} ∣ \leq (k + ∣ λ ∣ r / 2) ∣ δ q ∣$ ; and $∥ A ∥ = \max (1, k / m)$ from $A^{T} A = diag ((k / m)^{2}, 1)$ .

Verification

Proof technique: direct; compute the exact remainder of the cubic potential, bound the field on a ball, and evaluate the theorem's hypotheses and conclusion on explicit numbers.

1.1F1algebra

Equilibrium and linearization: $U^{'} (0) = 0$ gives the equilibrium; $U^{''} (0) = k$ ; by [F1] the small-oscillation model is $m ξ^{''} + k ξ = 0$ with frequency $\sqrt{k / m}$ , and the derivative of $G$ at the origin is the displayed $A$ . This is clause 1.

1.2F1F3algebra

Exact remainder: the translated field is $F (ζ) = (δ v, - (k δ q + \frac{λ}{2} δ q^{2}) / m)$ and $A ζ = (δ v, - \frac{k}{m} δ q)$ , so subtracting gives $F (ζ) - A ζ = (0, - \frac{λ}{2 m} δ q^{2})$ ; its norm is $\frac{∣ λ ∣}{2 m} ∣ δ q ∣^{2} \leq \frac{∣ λ ∣}{2 m} ∥ ζ ∥_{2}^{2}$ , so $C = ∣ λ ∣ / (2 m)$ is exact on every ball; the growth bound with $L$ and the norm of $A$ are [F3]. This is clause 2.

2.1F2step 1.2algebra

Numerical instance: with $m = k = λ = r = 1$ one has $C = 1 / 2$ , $L = \max (1, 3 / 2) = 3 / 2$ and $∥ A ∥ = 1$ ; for $δ = 10^{- 2}$ and $T = 1$ the smallness condition $δ e^{L T} = e^{3 / 2} / 100 < 1 = r$ holds because $e^{3 / 2} < e^{2} < 8$ , so the bound applies and gives $C δ^{2} T e^{(∥ A ∥ + 2 L) T} = e^{4} / 20000$ ; this is clause 3.

2.2F2step 1.2algebra

Linear case: for $λ = 0$ the potential is exactly quadratic and $F (ζ) = A ζ$ for all $ζ$ ; hence $C = 0$ , the theorem's bound reads $∥ z (t) - y (t) ∥_{2} \leq 0$ , and the nonlinear motion and its linearization coincide on the whole line. This is clause 4.

3.1step 1.1step 1.2step 2.1step 2.2∎

Clauses 1-4 are verified by exact computation from the model formulas; the numbers in clause 3 are arithmetic evaluations of the displayed expressions and are not measured quantities.

Remarks

The cubic term is the first genuinely nonlinear term. Its remainder is exactly quadratic in the displacement, which is why the error-bound theorem's hypothesis holds globally here with the exact constant $C = ∣ λ ∣ / (2 m)$ ; for a general $U$ the constant comes from compactness and is not explicit, as in Finite-time error control for the small-oscillation approximation: the quadratic remainder of the mechanical field and the variation-of-constants bound.
Decimal values of the instance. On the instance of clause 3 the smallness quantity is approximately $0.04482$ and the bound is approximately $2.73 \times 10^{- 3}$ ; these decimals are evaluations of the exact expressions $e^{3 / 2} / 100$ and $e^{4} / 20000$ and play no role in the verification.
What the bound does not say. The estimate compares the nonlinear motion with the harmonic one only on the stated finite horizon and for the stated small datum. It gives no conclusion about the approximation error beyond that horizon. Likharev Section 5.2 discusses slowly accumulating perturbative effects under its own weak-nonlinearity assumptions, including a cubic force arising from a quartic potential. That discussion does not supply an all-time estimate or a deduction of secular behaviour for every motion of the cubic-potential model here; when $λ = 0$ the equality in clause 4 holds at all times.

Coupled oscillations and approximation — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Two coupled masses on a line: the in-phase and anti-phase normal modes, their frequencies, mode-energies and the zero-stiffness translation mode

Example

Facts & Assumptions

Verification

Remarks

A linear triatomic chain: the translation zero mode, the antisymmetric stretch and the symmetric stretch, with the corrected frequency comparison

Example

Facts & Assumptions

Verification

Remarks

The equal-arm double pendulum in the linear approximation: the slow same-side mode, the fast opposite-side mode and the bounded modal motions

Example

Facts & Assumptions

Verification

Remarks

A doubly degenerate quadratic model: every direction is a mode, the modal basis is not intrinsic, and coupling splits the double frequency

Example

Facts & Assumptions

Verification

Remarks

The inverted pendulum about its upward equilibrium: the negative modal stiffness, the exponential quadratic mode and the nonlinear instability

Example

Facts & Assumptions

Verification

Remarks

A drive along the in-phase pattern of the two-mass chain: resonance exactly at that mode's frequency and a bounded steady response off resonance

Example

Facts & Assumptions

Verification

Remarks

An explicit weakly damped oscillator: the averaging constants and the energy-decay bound over five thousand seconds (about 796 undamped periods)

Example

Facts & Assumptions

Verification

Remarks

A cubic-potential chart system: the exact quadratic remainder, the explicit error constants and a checked finite-time bound

Example

Facts & Assumptions

Verification

Remarks

Sources