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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Cm Math Approximation

1 · Prerequisites

2 · Summary

The approximation layer of the non-relativistic classical mechanics collection, all of it mathematics: the exact solution theory of the linear damped driven oscillator, the convolution (Green's kernel) representation of forced solutions, the steady sinusoidal response with its resonance and amplitude peak, the finite-time linearization error bound, the averaging estimate on the O(1/ε) time scale, and the adiabatic action-drift bound. Every item carries domain: mathematics and depends only on mathematical suppliers; nothing here asserts that any model describes nature, and the physical reading of masses, forces, damping, driving and slow parameter variation belongs to the consuming classical-mechanics pages.

The exact oscillator layer begins with The damped driven oscillator model: second-order equation, phase-plane system and characteristic data, which fixes the second-order model mx′′+cx′+kx=f(t), its phase-plane system and its characteristic data. The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data classifies all solutions for k>0 by the discriminant (distinct real roots, a double root, and the oscillatory family e−γt(Acos⁡ωt+Bsin⁡ωt)) and proves uniqueness for prescribed initial data; The degenerate and repulsive oscillator: solutions for k=0 and the explicit growing solution for k<0 covers k=0 and k<0, including the explicit exponentially growing solution when k<0, and Strictly damped homogeneous oscillator with k>0: every solution tends to zero records the exponential decay of all homogeneous solutions when c>0, k>0.

The forcing layer is Differentiation of a variable-limit convolution integral with a C-one matrix kernel, the exact differentiation rule for variable-limit convolution integrals, and The forced oscillator with k≥0: unique global solution and the convolution (Green's function) formula, which represents the unique solution of the forced model with k≥0 as a homogeneous solution plus the convolution of the forcing with the kernel g(0)=0, g′(0)=1/m. Variation of constants for z′=Az+b(t) via the matrix exponential supplies the corresponding matrix-exponential formula for general linear systems, used again in the linearization estimate. Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at Ω=k/m identifies the steady response Re⁡(f0eiΩt/D) for D=k−mΩ2+icΩ, the resonant solution f0tsin⁡(ω0t)/(2mω0) when c=0 and Ω=ω0, and the decay of the transient when c>0; The amplitude of the sinusoidally forced damped oscillator: location and value of its maximum locates the maximum of the response amplitude at Ω∗=k/m−c2/(2m2) when this is positive and at 0 otherwise. Forced quadratic system Mq′′+Kq=φ(t): decoupled forced modes and the convolution solution decouples the forced quadratic system Mq′′+Kq=φ(t) into the scalar forced modes of the quadratic-mode theorem and writes their solutions by the same kernel.

The approximation layer proper comprises Finite-time linearization error bound: quadratic remainder, exponential growth bound and the variation-of-constants estimate (existence on [0,T] under δeLT<r and the estimate ∥z−y∥≤Cδ2∫0te∥A∥(t−s)e2Lsds), Elementary bounds for the periodic average, the deviation and its time derivative (the explicit bounds B=2MP, B1=2LP for the periodic deviation and its derivatives on the stated compact state ball), Averaging for fast periodic oscillations: order-ε agreement on the O(1/ε) time scale (order-ε agreement of x′=εf(t,x) with the averaged system on [0,A/ε] under the trapping condition εK<ρ), and Adiabatic action drift bound: order-ε change of the action on the O(1/ε) time scale (the order-ε bound for the drift of the action along a one-frequency branch on the same time scale, conditional on a uniform action-angle chart with ∣∂Ih∣≥ω0>0 and on the branch staying in the chart). Scope and limits of the oscillator, linearization, averaging and adiabatic items on this page records the limits of every one of these statements: finite-time validity, the conditional character of the adiabatic chart hypothesis, the absence of any long-time linearization or stability claim, and the absence of empirical content.

The companion page cm-math-approximation-examples records worked checks: explicit oscillator initial value problems, the resonant response and the two regimes of the amplitude maximum, constant forcing with and without linear drag, a pendulum linearization estimate with explicit constants, an averaging estimate compared with its exact solution, an exactly solvable adiabatic branch compared with the drift bound, a nonuniqueness counterexample without a Lipschitz condition, and a finite-time escape from the domain of a bounded field.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableprecheck passOpen item page →

The damped driven oscillator model: second-order equation, phase-plane system and characteristic data

Definition

Let m>0, c≥0 and k∈R be given real numbers, let I⊆R be an interval with nonempty interior (Intervals of R: the nine order-convex forms, nondegeneracy, and length) and let f:I→R be continuous (Continuity of f:A→R at a point of A and on A: the ε-δ condition, its agreement with lim⁡x→cf(x)=f(c) at a limit point, and continuity at an isolated point). The damped driven oscillator model with data (m,c,k,f) is the second-order equation

m x′′(t)+c x′(t)+k x(t)=f(t),t∈I.

A solution of the model on an interval J⊆I with nonempty interior is a twice differentiable x:J→R such that the identity holds for every t∈J, derivatives being ordinary real derivatives (The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral). The model is called homogeneous when f≡0, undamped when c=0, and strictly damped when c>0.

The phase-plane equation of the model has right-hand side

Gf:I×R2⟶R2,Gf(t,x,v):=(v, f(t)−c v−k xm).

Here R2 is the phase plane, with coordinates (x,v), where x is the displacement and v the velocity. For an interval J⊆I with nonempty interior, a differentiable curve z=(x,v):J→R2 solves the phase-plane equation when z′(t)=Gf(t,z(t)) for every t∈J, using the domain-relative derivative at any included endpoint (The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set). If x is a solution of the model on J, then t↦(x(t),x′(t)) solves this equation; conversely, if t↦(x(t),v(t)) solves it, then v=x′ and x is a solution of the model on J. When J is open, this is the first-order system of First-order systems, initial value problems, and solutions on intervals on the open domain J×R2. For the homogeneous model the right-hand side is written G0(x,v)=(v,−(cv+kx)/m).

The characteristic polynomial of the model is

p(λ):=mλ2+cλ+k,

with discriminant Δ:=c2−4mk. When k>0 define the undamped frequency ω0:=k/m, the damping rate γ:=c/(2m) and, when 4mk>c2, the damped frequency ω:=k/m−γ2=4mk−c2/(2m); the corresponding square roots are the nonnegative real ones.

Remarks

  • Equilibrium of the homogeneous phase system. A point (x,v)∈R2 satisfies G0(x,v)=(0,0) if and only if v=0 and kx=0; hence the zero set of G0 is {(0,0)} when k≠0 and the whole line {(x,0):x∈R} when k=0. For the forced model, Gf(t,x,v)=(0,0) holds if and only if v=0 and kx=f(t).

  • Interpretation. This item is a purely mathematical model: m,c,k are real parameters, f an arbitrary continuous real function, and no physical units, force law, constitutive assumption or empirical claim is made. The classical-mechanical reading of m,c,k,f and any identification with measured data belong to the consuming physics item, not to this definition.

  • Regularity. Gf is continuous relative to I×R2 and, for each t, Lipschitz in (x,v) with constant L:=max⁡(1,(c+∣k∣)/m)2, since ∣−cΔv−kΔx∣≤c∣Δv∣+∣k∣∣Δx∣ and ∣a∣+∣b∣≤2a2+b2. At every interior time t∈I∘, its restriction to a sufficiently small open time interval about t satisfies the Picard-Lindelof hypotheses of Picard-Lindelöf local existence and uniqueness for first-order systems. If I includes an endpoint, this item does not assert existence across that endpoint unless f is also defined on an open time neighborhood there.

DefinitionDefinition: AI-adaptedProof: Not applicableprecheck passOpen item page →

Periodic averaging data: the average field, the oscillatory part and the periodic deviation

Definition

Let n≥1, let P>0, let O⊆Rn be open, and let f:R×O→Rn be continuous (Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions). Assume that for each standard coordinate vector ej of Rn, 0≤j<n, and every (t,x)∈R×O, the vector-valued derivative at r=0 of r↦f(t,x+rej) exists, and that each resulting partial derivative (t,x)↦∂xjf(t,x) is continuous on R×O (Directional derivatives and partial derivatives of a map U⊆Rm→Rn). This is the meaning of continuously partially differentiable in the state variable here. Suppose also that f is P-periodic in time, that is

f(t+P,x)=f(t,x)for all (t,x)∈R×O.

The average of such an f is the map

fˉ:O⟶Rn,fˉ(x):=1P∫0Pf(s,x) ds,

where the integral is the componentwise Riemann integral of the continuous curve s↦f(s,x) over the compact interval [0,P] (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of R: the nine order-convex forms, nondegeneracy, and length); it is well defined and finite for every x∈O by componentwise application of A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion. The oscillatory part of f is f−fˉ, and the deviation of f is

g:[0,P]×O⟶Rn,g(t,x):=∫0t(f(s,x)−fˉ(x)) ds.

Because ∫0Pf(s,x)−fˉ(x) ds=Pfˉ(x)−Pfˉ(x)=0 by the linearity clause of The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral and the componentwise constant-integral formula of If m≤f≤M on [a,b] then m(b−a)≤L(f,P)≤∫ab‾f≤∫ab‾f≤U(f,P)≤M(b−a) for every partition P; in particular every constant function is integrable, with ∫abc=c(b−a), the formula g(t+kP,x):=g(t,x) for k∈Z and t∈[0,P) defines a unique extension, again written g, to all of R×O; it satisfies g(t+P,x)=g(t,x) and g(0,x)=0 for all (t,x), and on [0,P] it is the integral displayed above.

The averaged (or reduced) system of f for a parameter ε>0 is the autonomous first-order system on O

y′(t)=ε fˉ(y(t)).

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicableprecheck passOpen item page →

A one-frequency parameter-dependent action-angle branch, its transformed Hamiltonian and its drift generator

Definition

Let ε>0, let J^,S^⊆R be open intervals, and let J⊆J^ and S⊆S^ be intervals with nonempty interiors (Intervals of R: the nine order-convex forms, nondegeneracy, and length). Let h:J^×S^→R and k:J^×R×S^→R be of class C2 on these open Euclidean domains (Ck Euclidean maps and diffeomorphisms), and suppose k is 2π-periodic in its second variable on J×R×S:

k(I,θ+2π,s)=k(I,θ,s)for all (I,θ,s)∈J×R×S.

The transformed Hamiltonian of the datum is

K(I,θ,s):=h(I,s)+ε k(I,θ,s),

its frequency function is

ω(I,s):=∂Ih(I,s),

its angle average is ⟨k⟩(I,s):=12π∫02πk(I,θ,s) dθ, and, on the set where ω≠0, its drift generator is

g(I,θ,s):=k(I,θ,s)−⟨k⟩(I,s)ω(I,s).

The branch vector field of the datum is the restriction to J×R×S of the map Φ^:J^×R×S^→R3,

Φ^(I,θ,s):=(−ε ∂θk(I,θ,s),  ∂Ih(I,s)+ε ∂Ik(I,θ,s),  ε),

and a branch solution on an interval T⊆R with nonempty interior is a differentiable (I,θ,s):T→J×R×S satisfying (I,θ,s)′=Φ^(I,θ,s) at every t∈T, with derivatives relative to T at included endpoints. On open time intervals this is the first-order system of First-order systems, initial value problems, and solutions on intervals on the open domain J^×R×S^ (Directional derivatives and partial derivatives of a map U⊆Rm→Rn). Partial derivatives are those of the ambient Euclidean coordinates.

Remarks

  • Angle interpretation. Since k is 2π-periodic in θ, so are ∂θk and ∂Ik on J×R×S, hence the restriction of Φ^ to J×R×S is 2π-periodic in the second variable: reading the second coordinate in R/2πZ gives a well-defined vector field on J×(R/2πZ)×S, and if (I,θ,s) is a branch solution, so is (I,θ+2πn,s) for every n∈Z. The definition keeps the covering description on R to avoid quotient notation; all statements about θ below are invariant under 2π-translations.

  • Reading. A datum of this definition is the normal form of a one-frequency Hamiltonian problem in action-angle coordinates after the slow parameter is promoted to a state variable s with s′=ε: h is the angle-independent part, εk the angle-dependent remainder including any contribution of the parameter-dependent chart, and ω=∂Ih the fast frequency. The definition asserts no existence theorem: whether a given mechanical family admits such a chart with the stated regularity, on which domain, and with ω≠0, is a separate hypothesis of any consumer, and the adiabatic statements below are conditional on it.

  • Scope. This is pure mathematics: no physical identification of I, θ, s, h, k, ε or of the parameter variation with measured data is made here.

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Differentiation of a variable-limit convolution integral with a C-one matrix kernel

Statement

Let n≥1, equip Rn with the Euclidean inner product and norm of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, and let I⊆R be an interval with nonempty interior (Intervals of R: the nine order-convex forms, nondegeneracy, and length), let t0∈I, let U⊆R be an open interval with I−I:={t−s:t,s∈I}⊆U, let g:U→Mn(R) have C1 entries (The vector space Mm×n(F):=F m×n of m by n matrices over a field, with entrywise operations, Ck Euclidean maps and diffeomorphisms) and let f:I→Rn be continuous (Vector-valued functions f:A→Rm, their limits and continuity, with the dictionary to the metric notions). For B=(bij)∈Mn(R) and v∈Rn, define (Bv)i:=∑j<nbijvj. This is a linear map by the coordinate formula; the Euclidean norm is a norm by Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation, and the standard coordinate vectors form an ordered basis by The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0, so the map is bounded by A linear map from a finite-dimensional normed space is bounded. Let ∥B∥ denote its induced operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum). All matrix actions in the integrals use this convention. Since 0∈I−I⊆U, the value g(0) is defined. Define

F:I⟶Rn,F(t):=∫t0tg(t−s)f(s) ds,

with the oriented integral of The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral and the matrix action just defined; the integrand is continuous in s, hence integrable on every compact subinterval. Then:

  1. F is differentiable at every interior point t of I (The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set) and

F′(t)=g(0)f(t)+∫t0tg′(t−s)f(s) ds.

  1. If moreover g′ is bounded on compact subintervals of U (which holds automatically because g is C1), then F is of class C1 on the interior of I.

Facts & Assumptions

Given: A natural number n≥1, an interval I⊆R with nonempty interior, a point t0∈I, an open interval U⊇I−I, a C1 matrix curve g:U→Mn(R) and a continuous curve f:I→Rn.

[F1]

The vector-valued derivative and the oriented Riemann integral are componentwise: (F′)i=(Fi)′ and (∫abϕ)i=∫abϕi, with ∫ab=−∫ba; the integral is additive over subintervals and linear in the integrand (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, For a<c<b: f is integrable on [a,b] if and only if it is integrable on [a,c] and on [c,b], and then ∫abf=∫acf+∫cbf; with the oriented form for arbitrary a,b,c).

[F2]

For a<b and an integrable ϕ:[a,b]→Rn one has ∥∫abϕ∥2≤∫ab∥ϕ∥2, and for real integrable u≤v one has ∫abu≤∫abv (For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable, If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a)). Consequently, for a continuous ψ on the interval with endpoints a and a+h, h≠0, ∥1h∫aa+hψ∥2≤sup⁡s∥ψ(s)∥2, the supremum over that interval.

[F3]

Leibniz's rule on a compact rectangle: if G,H:[a,b]×[c,d]→R are continuous and for every fixed t∈[c,d] the slice x↦G(x,t) is differentiable on (a,b) with derivative H(x,t), then x↦∫cdG(x,t) dt is differentiable on [a,b] with derivative ∫cdH(x,t) dt (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral). The component maps below satisfy these hypotheses: sums and products of C1 functions are C1 by the chain rule and the algebra of derivatives (The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c), Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(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)≠0).

[F5]

Differentiability at t means that the difference quotients [F(t+h)−F(t)]/h converge to F′(t) as h→0 through nonzero values (The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set).

Proof

technique · direct; split the difference quotient into a parameter-derivative term and a moving-endpoint term, take the two limits separately, then check continuity of the derivative
1.1F1given

Fix t∈I˚ and choose η>0 with [t−η,t+η]⊆I and [−η,η]⊆U. For 0<∣h∣≤η, additivity and linearity of the integral [F1] give F(t+h)−F(t)=∫t0t[g(t+h−s)−g(t−s)]f(s) ds+∫tt+hg(t+h−s)f(s) ds, with the second integral oriented and taken over the interval with endpoints t and t+h; both integrands are continuous on their compact intervals, since every difference of two points of I lies in I−I⊆U.

2.1step 1.1F1F3F6given

For a component index i<n, put ϕi(u,s):=(g(u−s)f(s))i=∑j<ngij(u−s)fj(s) for (u,s)∈[t−η,t+η]×J, where J is the compact interval with endpoints t0 and t. This is continuous, and for fixed s it is differentiable in u with ∂uϕi(u,s)=∑j<ngij′(u−s)fj(s), which is continuous on the rectangle; [F3] therefore gives lim⁡h→01h∫t0t[g(t+h−s)−g(t−s)]f(s) ds=∫t0tg′(t−s)f(s) ds, for t0<t directly, and for t0>t after reversing the orientation and applying [F3] on [t,t0]; [F1] identifies the components.

2.2step 1.1F1F2F4F6given

For the second term put ψh(s):=g(t+h−s)f(s) on the interval with endpoints t and t+h. Since g(0)f(t) is the value of the constant function g(0)f(t) there, [F1] gives 1h∫tt+hg(t+h−s)f(s) ds−g(0)f(t)=1h∫tt+h[ψh(s)−g(0)f(t)]ds, and [F2] bounds its norm by sup⁡s∥ψh(s)−g(0)f(t)∥2=sup⁡s∥g(t+h−s)f(s)−g(0)f(t)∥2 over the interval with endpoints t and t+h. As h→0 the arguments t+h−s lie between 0 and h (in either order) and tend to 0, and s→t, so continuity of g at 0 and of f at t makes this supremum tend to 0; hence the second term tends to g(0)f(t).

3.1step 1.1step 2.1step 2.2F5

Dividing the identity of step 1.1 by h, the two summands are precisely the terms analysed in steps 2.1 and 2.2; their limits exist and sum to g(0)f(t)+∫t0tg′(t−s)f(s) ds. By [F5] this is exactly the differentiability assertion of clause 1 at t, and t∈I˚ was arbitrary.

4.1step 3.1F1F2F4F6∎

For clause 2 write F′(t)=g(0)f(t)+H(t) by step 3.1, with H(t):=∫t0tg′(t−s)f(s) ds. The term g(0)f(t) is continuous. Fix t1∈I˚, let J0 be the closed interval with endpoints t0,t1, and put Q:=t1−J0=[α,β]⊆I−I⊆U. Since U is an open interval containing the compact interval Q, choose η>0 small enough that [t1−η,t1+η]⊆I and K0:=[α−η,β+η]⊆U. Put J1:=[min⁡(t0,t1−η),max⁡(t0,t1+η)]⊆I. For t∈[t1−η/2,t1+η/2], additivity [F1] gives H(t)−H(t1)=∫t1tg′(t−s)f(s) ds+∫t0t1[g′(t−s)−g′(t1−s)]f(s) ds. In the first integral, s lies between t1 and t, so t−s∈K0; in the second, u:=t1−s∈Q and t−s=u+(t−t1)∈K0. By [F2], the first integral has norm at most ∣t−t1∣sup⁡u∈K0∥g′(u)∥sup⁡s∈J1∥f(s)∥2, which tends to 0. The second has norm at most ∣t1−t0∣sup⁡s∈J0∥f(s)∥2sup⁡u∈Q, ∣δ∣≤∣t−t1∣∥g′(u+δ)−g′(u)∥, which tends to 0 by uniform continuity of the matrix-valued g′ on K0. Hence H is continuous at t1, so F′ is continuous on I˚, proving clause 2.

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The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data

Statement

Let m>0, c≥0 and k>0, and consider the homogeneous damped oscillator model m x′′+c x′+k x=0 of The damped driven oscillator model: second-order equation, phase-plane system and characteristic data with time domain I=R and forcing f:R→R identically zero, with characteristic polynomial p(λ)=mλ2+cλ+k, discriminant Δ=c2−4mk, damping rate γ:=c/(2m), and, when applicable, roots λ±:=−c±Δ2m (Δ≥0),λ:=−c2m,ω:=4mk−c22m (Δ<0).

A local solution here may have any interval J⊆R with nonempty interior, with domain-relative derivatives at included endpoints. It has a unique extension to the global family below. A maximal solution means one that has no solution extension to a strictly larger interval; the maximal solutions of this global autonomous equation are precisely the family members on R. For a homogeneous model restricted to a time domain I, its solutions have J⊆I and are restrictions of these same global members; their extensions solve the all-R equation.

  1. Distinct real roots. If Δ>0, then λ+>λ− are exactly the real roots of p, and a function x:J→R on an interval J⊆R with nonempty interior is a solution of the model if and only if x(t)=Aeλ+t+Beλ−t(t∈J) for some A,B∈R.

  2. Double real root. If Δ=0, then λ=−c/(2m) is the only root of p, and x:J→R is a solution if and only if x(t)=(A+Bt)eλt(t∈J) for some A,B∈R.

  3. Complex conjugate roots. If Δ<0, then ω>0, the roots of p are −γ±iω, and x:J→R is a solution if and only if x(t)=e−γt(Acos⁡(ωt)+Bsin⁡(ωt))(t∈J) for some A,B∈R.

  4. Initial data. In each of the three cases and for every (x0,v0)∈R2 there is exactly one solution x:R→R with x(0)=x0 and x′(0)=v0. Its coefficients are A=v0−λ−x0λ+−λ−,B=λ+x0−v0λ+−λ−(Δ>0),A=x0, B=v0−λx0(Δ=0), A=x0,B=v0+γx0ω(Δ<0).

Facts & Assumptions

Given: The parameters m>0, c≥0, k>0, the homogeneous model of The damped driven oscillator model: second-order equation, phase-plane system and characteristic data with I=R and f≡0, its phase-plane field G0(x,v)=(v,−(cv+kx)/m), the characteristic polynomial p(λ)=mλ2+cλ+k, the discriminant Δ=c2−4mk, and, in the three cases, the constants λ±,λ,γ,ω of the statement.

[F1]

The model, its solutions, the phase-plane system and the characteristic data are those of The damped driven oscillator model: second-order equation, phase-plane system and characteristic data with I=R and f≡0: x solves the model on J exactly when t↦(x(t),x′(t)) solves z′=G0(z) on J; G0 is continuous and locally Lipschitz in the state variable, with the global state-Lipschitz constant (1+(c+k)/m)2 on R2. [given]

[F2]

Local uniqueness and Lipschitz dependence: under the hypotheses of Continuous and Lipschitz dependence on the initial condition on a Picard-Lindelof cylinder with field G0, U:=R2, centre a0∈R2, radius R>0, M=sup⁡B‾(a0,R)∥G0∥2, state-Lipschitz constant L, and 0<ρ<R with hM≤R−ρ, Lh<1, for every a∈B‾(a0,ρ) the initial value problem z′=G0(z), z(0)=a has exactly one solution on [−h,h] with graph in [−h,h]×B‾(a0,R). [given]

[F3]

Differentiation rules: for real μ the real exponential satisfies (eμt)′=μeμt; for complex μ and real t the complex exponential satisfies ddtexp⁡(μt)=μexp⁡(μt); sums, products and composites are differentiated by the algebra of derivatives and the chain rule; (cos⁡t)′=−sin⁡t and (sin⁡t)′=cos⁡t. [thm-derivative-of-exponential, thm-complex-exponential-is-entire-with-derivative-itself, thm-algebra-of-derivatives, thm-chain-rule, thm-sine-and-cosine-derivatives]

[F4]

Euler's form: for real γ,ω and t, exp⁡(−γt+iωt)=e−γt(cos⁡(ωt)+isin⁡(ωt)). [cor-complex-exponential-cartesian-form-modulus-and-eulers-identity, given]

Proof

technique · cases; verify the three explicit solution families by differentiation, match initial data by explicit coefficients, and deduce exhaustiveness from the local uniqueness lemma by a connectedness argument
1.1givenalgebra

Algebraic reduction: for all λ, p(λ)=m[(λ+c2m)2−Δ4m2]=m[(λ+γ)2−Δ4m2]. Hence: if Δ>0 the real roots of p are exactly λ±=(−c±Δ)/(2m) with λ+>λ−; if Δ=0 then λ=−c/(2m) is the only root, a double one; and if Δ<0 then p has no real root, Δ=−4m2ω2 for the positive number ω=4mk−c2/(2m), and the complex roots of p are −γ±iω because (±iω)2=Δ/(4m2). In particular p(λ±)=0 in case 1 and p(λ)=0 in case 2.

2.1step 1.1F3

Verification of the real-root families. [assume-case 1, assume-case 2] For a real number μ with p(μ)=0 the function t↦eμt satisfies m(eμt)′′+c(eμt)′+keμt=(mμ2+cμ+k)eμt=0 by [F3]; and t↦teμt satisfies m(teμt)′′+c(teμt)′+kteμt=eμt[(2mμ+c)+t p(μ)], since (teμt)′=eμt(1+μt) and (teμt)′′=eμt(2μ+μ2t) by [F3]; this vanishes exactly when 2mμ+c=0, that is μ=−c/(2m)=λ, and then also p(μ)=0. By step 1.1 the roots λ+,λ− in case 1 and the double root λ in case 2 do satisfy p=0, so by linearity of the differential expression every Aeλ+t+Beλ−t and every (A+Bt)eλt is a solution on every interval.

2.2step 1.1F3F4

Verification of the oscillatory family. [assume-case 3] Let μ:=−γ+iω, so that p(μ)=0 by step 1.1; by [F3] the complex-valued curve z(t):=exp⁡(μt) satisfies mz′′+cz′+kz=p(μ)z=0. Real and imaginary parts of a complex-valued solution of a real linear constant-coefficient equation solve that equation, by componentwise differentiation; by [F4], z(t)=e−γt(cos⁡ωt+isin⁡ωt), so e−γtcos⁡ωt and e−γtsin⁡ωt are real solutions, and every e−γt(Acos⁡ωt+Bsin⁡ωt) is one by linearity. This family too is defined on all of R.

3.1step 2.1step 2.2F3

Matching initial data: in case 1, the two linear conditions A+B=x0 and λ+A+λ−B=v0 are solved uniquely by the coefficients displayed in clause 4 because λ+−λ−≠0, and substitution verifies x(0)=x0, x′(0)=v0. In case 2, x(t)=(A+Bt)eλt has x(0)=A and x′(0)=λA+B by [F3], giving A=x0, B=v0−λx0. In case 3, x(t)=e−γt(Acos⁡ωt+Bsin⁡ωt) has x(0)=A and, by [F3], x′(0)=−γA+ωB, giving A=x0 and B=(v0+γx0)/ω since ω≠0. Thus in each case there is a global solution with the prescribed data.

4.1step 2.1step 2.2step 3.1F1F2F3algebra∎

Exhaustiveness and uniqueness. [cases-exhaustive: Δ>0, Δ=0, Δ<0] Let x solve the model on an interval J with nonempty interior and choose an arbitrary t∗∈J∘. For the ordered fundamental pairs (eλ+t,eλ−t), (eλt,teλt) and (e−γtcos⁡ωt,e−γtsin⁡ωt), the determinant of their value/derivative matrix is respectively (λ−−λ+)e(λ++λ−)t, e2λt and ωe−2γt, by [F3]. Each is nonzero, so there is a unique combination y from the corresponding global family with y(t∗)=x(t∗) and y′(t∗)=x′(t∗). Put S={t∈J∘:(x(t),x′(t))=(y(t),y′(t))}. This set is nonempty and relatively closed by phase-curve continuity. It is relatively open: at t1∈S choose a closed time neighborhood in J∘ and a state ball centered at the common phase value, containing both curves on a smaller time neighborhood. Choose h>0 within that neighborhood with hsup⁡B‾(a0,R)∥G0∥2≤R/2 and Lh<1. Since G0 is autonomous, translate t1 to zero and apply [F2] with ρ=R/2 to the two phase curves; uniqueness gives equality on this neighborhood. The interval J∘ is connected, so S=J∘. Continuity extends the equality to any endpoints belonging to J, proving x=y∣J. Thus every local solution is a restriction of exactly one global member of the appropriate family. In particular the global members matched in step 3.1 are the unique solutions with the prescribed data at zero, and all four clauses follow. A proper interval solution therefore has a strictly larger all-R extension, whereas an all-R solution has no larger real time interval; this proves the maximality assertion. The same classification applies to a restricted homogeneous model on I by restricting these family members to J⊆I.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

The degenerate and repulsive oscillator: solutions for k=0 and the explicit growing solution for k<0

Statement

Let m>0, c≥0, let k≤0, and consider the model m x′′+c x′+k x=0 of The damped driven oscillator model: second-order equation, phase-plane system and characteristic data with time domain I=R and forcing f:R→R identically zero. In the three cases below, x:J→R runs over solutions on intervals J⊆R with nonempty interior.

A local solution here may have any interval J⊆R with nonempty interior, with domain-relative derivatives at included endpoints. It has a unique extension to the global family below. A maximal solution means one that has no solution extension to a strictly larger interval; the maximal solutions of this global autonomous equation are precisely the family members on R. For a homogeneous model restricted to a time domain I, its solutions have J⊆I and are restrictions of these same global members; their extensions solve the all-R equation.

  1. Double zero root (k=0, c=0). Then on every such interval J, x solves the model if and only if x(t)=A+Bt for some A,B∈R; each such local solution is the restriction of a unique global member of this family. For every datum (x0,v0) there is a unique global solution on R with x(0)=x0, x′(0)=v0, namely x(t)=x0+v0t.
  2. Simple zero root (k=0, c>0). Then on every such interval J, x solves the model if and only if x(t)=A+Be−ct/m for some A,B∈R; each such local solution is the restriction of a unique global member of this family. For every datum (x0,v0) there is a unique global solution on R with x(0)=x0, x′(0)=v0, namely the member with A=x0+mv0c and B=−mv0c. This global solution has limit A as t→+∞; any local solution on an interval unbounded above has the same limit along that interval.
  3. One positive and one negative root (k<0). Put λ±:=−c±c2−4mk2m. Then λ−<0<λ+, both are real roots of the characteristic polynomial, and on every such interval J, x solves the model if and only if x(t)=Aeλ+t+Beλ−t for some A,B∈R; each such local solution is the restriction of a global member of this family. In particular the global solution t↦eλ+t has data x(0)=1, x′(0)=λ+ and satisfies eλ+t→+∞ as t→+∞: the model has an unbounded solution growing exponentially on [0,∞).

Facts & Assumptions

Given: The parameters m>0, c≥0, k≤0, the model of The damped driven oscillator model: second-order equation, phase-plane system and characteristic data with I=R, f≡0 and phase field G0(x,v)=(v,−(cv+kx)/m), and the constants λ± of case 3.

[F1]

The model, its solutions and the phase-plane system are those of The damped driven oscillator model: second-order equation, phase-plane system and characteristic data with I=R and f≡0; x is a solution on J exactly when t↦(x(t),x′(t)) solves z′=G0(z) on J, and for k≤0 the field G0(x,v)=(v,−(cv+kx)/m) is linear, hence continuous and globally Lipschitz in the state variable. [given]

[F2]

Local uniqueness: Continuous and Lipschitz dependence on the initial condition on a Picard-Lindelof cylinder in the form used before: for the Lipschitz field G0, any two solutions of z′=G0(z) defined on a common interval and taking the same value at one point agree on that interval (this is its "exactly one solution through given data with graph in a Picard-Lindelof cylinder" clause together with the cylinder-local uniqueness, applied along the interval). [given]

[F3]

Derivative rules: (eμt)′=μeμt for real μ; sums and products are differentiated by the algebra of derivatives and the chain rule. [thm-derivative-of-exponential, thm-algebra-of-derivatives, thm-chain-rule]

[F5]

Fundamental theorem of calculus for continuous integrands: if G is a primitive of a continuous u on an interval containing a and t, then for a<t the evaluation clause of Every continuous function on an interval has a primitive; two primitives differ by a constant; and ∫abf=G(b)−G(a) for any primitive G gives ∫atu=G(t)−G(a). For t<a, The integral with oriented limits: ∫aaf:=0 and ∫baf:=−∫abf gives ∫atu=−∫tau=−(G(a)−G(t)) by the same clause on [t,a]; for t=a both sides vanish. Thus the formula holds for every ordering of a,t.

[F6]

The real exponential is strictly increasing, and its image is (0,∞). [thm-exponential-is-strictly-increasing, cor-exponential-is-a-bijection-onto-positive-reals]

Proof

technique · cases; identify the solution families in the three degenerate/repulsive cases, then obtain uniqueness from the local uniqueness lemma
1.1F1F3F4

Case k=0, c=0. [assume-case 1] The model is mx′′=0. By [F4], x′ is a constant B on J, and x(t)−Bt has zero derivative, hence is a constant A on J. Thus every solution on an arbitrary interval is A+Bt, and conversely each such function solves by [F3]. For a global solution the data at zero give A=x0, B=v0.

1.2F1F3F4F5F6algebra

Case k=0, c>0. [assume-case 2] Put u=x′. By [F3], (ect/mu)′=ect/m(u′+(c/m)u)=0, so [F4] gives u(t)=De−ct/m for a constant D on J. Fix a∈J∘. Applying [F5] to the primitives x and −(mD/c)e−ct/m of u gives x(t)−x(a)=−(mD/c)(e−ct/m−e−ca/m) for every t∈J. Hence (x(t)+(mD/c)e−ct/m)′=0 and x(t)=A+Be−ct/m for constants A,B with B=−mD/c, without requiring 0∈J. Conversely this family solves by [F3] and extends globally. Its extension is unique: if two such global members agree on J, their difference A1−A2+(B1−B2)e−ct/m vanishes on the nonempty open interval J∘; differentiating there and using c>0 and e−ct/m>0 gives B1=B2, then A1=A2. On the global member, data at zero give B=−mv0/c and A=x0+mv0/c. The exponential term tends to zero at +∞ by monotonicity and surjectivity of the exponential in [F6]: for any η>0, choose r∈R with er=η; whenever t>max⁡{0,−mr/c}, −ct/m<r and 0<e−ct/m<η. Thus the global extension has limit A; when J is unbounded above, the original solution agrees with this extension on its positive-time tail and has the same limit.

1.3givenF3F6algebra

Case k<0. [assume-case 3] Here c2−4mk>c2≥0, so c2−4mk>c and λ+>0>λ−; the completing-square identity mλ2+cλ+k=m[(λ+c2m)2−c2−4mk4m2] shows that λ± are exactly the two real roots of the characteristic polynomial. For a real root μ, t↦eμt solves the model because mμ2+cμ+k=0 and (eμt)′′=μ2eμt by [F3]; by linearity every Aeλ+t+Beλ−t solves it. Finally eλ+t→+∞ as t→+∞: given R>0, surjectivity in [F6] gives u with eu>R, and monotonicity gives eλ+t≥eu>R for all t≥u/λ+.

2.1step 1.1step 1.2step 1.3F1F2F3algebra∎

Uniqueness and exhaustiveness. [cases-exhaustive: k=0,c=0; k=0,c>0; k<0] Cases 1 and 2 exhaust the arbitrary-interval solution sets by steps 1.1 and 1.2. In case 3 choose any t0∈J∘. The value/derivative matrix of eλ+t and eλ−t at t0 has determinant (λ−−λ+)e(λ++λ−)t0≠0, so a unique global family member y matches (x(t0),x′(t0)). By [F1] the full phase curves solve the same system, and [F2] gives equality on J: cylinder-local uniqueness propagates on the connected interior using the closed-and-open set where both phase components agree, then extends to included endpoints by continuity. Thus x=y∣J. Each explicit family is defined on R, its coefficient/data map at zero is invertible, and these exhaust all solutions, proving global existence and uniqueness for prescribed data. The repulsive member with data (1,λ+) grows without bound by step 1.3. Because every interval solution has a unique all-R extension and R has no larger real time interval, these global family members are exactly the maximal solutions. Restricting them to J⊆I gives the solutions of a restricted homogeneous model on I.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Variation of constants for z′=Az+b(t) via the matrix exponential

Statement

Let n≥1, equip Rn with the Euclidean inner product and norm of The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, let A∈Mn(R) (The vector space Mm×n(F):=F m×n of m by n matrices over a field, with entrywise operations), let t0∈R and z0∈Rn, and let b:R→Rn be continuous (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral). The Euclidean norm is a norm by Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation, and the standard coordinate vectors form an ordered basis by The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0. For A=(aij) and z∈Rn, define (Az)i:=∑j<naijzj; this is a linear map by Linear map between vector spaces over the same field, hence bounded by A linear map from a finite-dimensional normed space is bounded, and ∥A∥ denotes its induced Euclidean operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Matrix powers and products use Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes; their action agrees with composition of the corresponding linear maps by distributing the finite sums in the displayed coordinate formula. Define the matrix exponential

E:R⟶Mn(R),E(t):=∑k=0∞tkk!Ak.

Then:

  1. E is well defined and of class C1, with E(0)=In, E(u+v)=E(u)E(v), dduE(u)=AE(u)=E(u)A for all u,v∈R, and ∥E(u)∥≤e∥A∥∣u∣ for the operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum) subordinate to the Euclidean norm.
  2. The map z:R⟶Rn,z(t):=E(t−t0)z0+∫t0tE(t−s)b(s) ds, satisfies z(t0)=z0 and z′(t)=A z(t)+b(t) for every t∈R.
  3. Uniqueness: if J⊆R is an interval with nonempty interior containing t0 (Intervals of R: the nine order-convex forms, nondegeneracy, and length) and w:J→Rn is differentiable with w′=Aw+b on J and w(t0)=z0, then w=z∣J.

Facts & Assumptions

Given: n≥1, A∈Mn(R) identified with a linear endomorphism of Rn, z0∈Rn, a continuous b:R→Rn, and the matrix exponential E displayed above.

[F1]

The matrix action defined in the Statement is a bounded linear endomorphism A^ of normed Euclidean Rn: the Euclidean norm is a norm, and the standard coordinate vectors form a finite ordered basis, so the finite-dimensional boundedness theorem applies (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation, The standard list e:n→Fn with ei(i)=1F and ei(j)=0F for j≠i is an ordered basis of Fn; hence dim⁡FFn=n, and F0 is the zero space with basis ∅ and dimension 0, A linear map from a finite-dimensional normed space is bounded). Its induced metric agrees with d2 (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn), which is complete (R and Rn for n≥1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R); hence this normed space is Banach. For matrices B=(bij) and C=(cjℓ), ((BC)z)i=∑ℓ<n(∑j<nbijcjℓ)zℓ=(B(Cz))i, so matrix multiplication represents composition (Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes). The exponential series lemma applies to A^ and gives absolute convergence in operator norm, E∈C1 with E′=AE=EA, E(0)=In, E(u+v)=E(u)E(v) and ∥E(u)∥≤e∥A∥∣u∣ (The exponential series of a bounded operator).

[F2]

The convolution differentiation lemma (Differentiation of a variable-limit convolution integral with a C-one matrix kernel) applies with interval I=R, open kernel domain U=R⊇I−I, g:=E:U→Mn(R) and f:=b:I→Rn: for every t, the map t↦∫t0tE(t−s)b(s) ds is differentiable with derivative E(0)b(t)+∫t0tE′(t−s)b(s) ds.

[F3]

Differentiation rules: the chain rule and the algebra of derivatives differentiate products and compositions; for a differentiable matrix curve M and constant vector v, t↦M(t)v is differentiable with derivative M′(t)v. [thm-chain-rule, thm-algebra-of-derivatives]

[F4]

Fundamental theorem for continuous derivatives: if u:J→Rn is differentiable with continuous derivative on an interval J containing t0, then ∫t0tu′=u(t)−u(t0) (If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz); and ∥∫t0tϕ∥2≤∣∫t0t∥ϕ∥2∣ (For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[F5]

Gronwall: if a continuous real u≥0 satisfies u(t)≤A+∫t0tB u for t≥t0 with constants A,B≥0, then u(t)≤AeB(t−t0); the time-reflected form with the integral from t to t0 applies for t≤t0 (Gronwall's integral inequality with variable and constant coefficients).

Proof

technique · direct; verify the explicit solution by the convolution lemma and the exponential identities, then prove uniqueness by Gronwall
1.1F1F3given

By [F1] the exponential E is C1 with E′=AE=EA, E(0)=In, E(u+v)=E(u)E(v) and ∥E(u)∥≤e∥A∥∣u∣ for all u,v; in particular t↦E(t−t0) is differentiable with derivative AE(t−t0), by the chain rule for the affine inner map t↦t−t0.

2.1step 1.1F2F3given

By step 1.1 and [F3], t↦E(t−t0)z0 has derivative AE(t−t0)z0. By [F2] the integral term has derivative E(0)b(t)+∫t0tE′(t−s)b(s) ds=b(t)+∫t0tAE(t−s)b(s) ds, where the second identity uses E′=AE of step 1.1 and linearity of the integral. Hence z′(t)=AE(t−t0)z0+b(t)+A∫t0tE(t−s)b(s) ds=Az(t)+b(t), by linearity of matrix-vector multiplication. Moreover z(t0)=E(0)z0+∫t0t0=z0, so clause 2 holds.

3.1step 2.1F4F5∎

For clause 3 let w solve w′=Aw+b on an interval J containing t0 with w(t0)=z0, and put u:=w−z∣J. By step 2.1, u′=Au and u(t0)=0, so by [F4] u(t)=∫t0tA u(s) ds(t∈J), and therefore ∥u(t)∥2≤∣∫t0t∥A∥ ∥u(s)∥2 ds∣≤∥A∥ ∣∫t0t∥u(s)∥2 ds∣ for t≥t0 and the reflected inequality for t≤t0. Applying [F5] with A0=0 and B=∥A∥ to the nonnegative continuous function ∥u∥2 gives ∥u(t)∥2≤0 on both sides of t0, hence u=0 and w=z∣J.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Strictly damped homogeneous oscillator with k>0: every solution tends to zero

Statement

Let m>0, c>0 and k>0, let x:R→R be a solution of the homogeneous model m x′′+c x′+k x=0 with time domain I=R and f≡0 (The damped driven oscillator model: second-order equation, phase-plane system and characteristic data, The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data), and write Δ=c2−4mk, γ=c/(2m), λ±=(−c±Δ)/(2m) in case Δ>0 and λ=−c/(2m) in case Δ=0. Then x(t)→0 as t→+∞ (Limits at +∞ and −∞, and infinite limits at a point); more precisely, with the coefficients A,B of the solution in the notation of The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data:

  1. if Δ>0: λ+<0 and ∣x(t)∣≤(∣A∣+∣B∣)eλ+t for all t≥0;
  2. if Δ=0: λ<0 and ∣x(t)∣≤(∣A∣+∣B∣t)eλt for all t≥0;
  3. if Δ<0: γ>0 and ∣x(t)∣≤(∣A∣+∣B∣)e−γt for all t≥0.

Facts & Assumptions

Given: m>0, c>0, k>0, a solution x of the homogeneous model, and the notation Δ,γ,λ±,λ of the statement.

[F1]

The solution is one of the three explicit families of The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data, on all of R, with the coefficients A,B of the relevant case. [given]

[F2]

Signs: if Δ>0 then λ+=−c+Δ2m<−c+c2m=0 because k>0 gives Δ<c2 and hence Δ<c; if Δ=0 then λ=−c/(2m)<0; and γ=c/(2m)>0. [F1, algebra]

[F3]

Ranges: ∣cos⁡u∣≤1 and ∣sin⁡u∣≤1 for every real u (Signs, monotonicity intervals, and ranges of sine and cosine).

[F4]

Exponential limits: exp⁡(u)→0 as u→−∞ and exp⁡(u)→+∞ as u→+∞ (The exponential tends to +∞ at +∞ and to 0 at −∞); consequently for λ<0, exp⁡(λt)→0 as t→+∞. The reciprocal identity eλt=1/e−λt and positivity hold by The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x). For t>0 and λ<0, the exponential series has nonnegative terms, so exp⁡(−λt)≥(−λt)2/2 and hence 0≤t eλt≤2λ2t (The real exponential function and the number e by a power series, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[F5]

Convergence to 0 at +∞ is the quantified assertion of Limits at +∞ and −∞, and infinite limits at a point: for every ε>0 there is T with ∣u(t)∣<ε for all t>T.

[F6]

The real exponential is strictly increasing (The exponential function is strictly increasing), so u≤v implies eu≤ev.

Proof

technique · cases; read the three explicit solution families off the classification theorem, bound them with the sign and range facts, and take the limits
1.1F1F3F6

Bounds for the three families. [assume-case 1, assume-case 2, assume-case 3] By [F1], in case 1 the solution is x(t)=Aeλ+t+Beλ−t with λ−<λ+; since t≥0 and λ−<λ+ give λ−t≤λ+t, [F6] gives eλ−t≤eλ+t, the triangle inequality yields ∣x(t)∣≤(∣A∣+∣B∣)eλ+t. In case 2, ∣x(t)∣=∣A+Bt∣eλt≤(∣A∣+∣B∣t)eλt. In case 3, ∣x(t)∣=e−γt∣Acos⁡(ωt)+Bsin⁡(ωt)∣≤(∣A∣+∣B∣)e−γt by [F3].

2.1step 1.1F2F4F5

The three majorants tend to zero at +∞. By [F2] the exponents λ+,λ,−γ are negative, so eλ+t→0 and e−γt→0 by [F4]. For the middle case, [F4] gives 0≤teλt≤2/(λ2t)→0: given ε>0, every t>2/(λ2ε) has teλt<ε; hence teλt→0 and therefore (∣A∣+∣B∣t)eλt=∣A∣eλt+∣B∣ teλt→0.

3.1cases-exhaustive: the three discriminant casesstep 1.1step 2.1F5∎

Since ∣x(t)∣ is bounded above in each case by the corresponding majorant, and all three majorants tend to 0, [F5] gives x(t)→0 as t→+∞ in each case; the estimates are those displayed in clause 1.

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The forced oscillator with k≥0: unique global solution and the convolution (Green's function) formula

Statement

Let m>0, c≥0, k≥0, let I⊆R be an interval with nonempty interior (Intervals of R: the nine order-convex forms, nondegeneracy, and length) and let f:I→R be continuous. Fix t0∈I and data (x0,v0)∈R2. Let g:R→R be the (unique) solution of the associated homogeneous model with time domain R and zero forcing, with g(0)=0 and g′(0)=1/m.

Then:

  1. There is exactly one solution x:I→R of the forced model m x′′+c x′+k x=f with x(t0)=x0 and x′(t0)=v0. It is twice differentiable on I, with derivatives at included endpoints taken relative to I, and x′′ is continuous on I in its relative topology.
  2. If xh:R→R is any homogeneous solution with xh(t0)=x0 and xh′(t0)=v0 (it exists and is unique by The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data and The degenerate and repulsive oscillator: solutions for k=0 and the explicit growing solution for k<0), then x(t)=xh(t)+∫t0tg(t−s)f(s) ds(t∈I).

Remarks

Facts & Assumptions

Given: The parameters m>0, c≥0, k≥0, the interval I with nonempty interior, the continuous f:I→R, the point t0∈I and the data (x0,v0). The homogeneous kernel g and member xh solve the associated all-R homogeneous model; the forced model uses the given time domain I. No extension of f outside I is used.

[F1]

The model and its phase-plane system are those of The damped driven oscillator model: second-order equation, phase-plane system and characteristic data; x solves the forced model on I exactly when t↦(x(t),x′(t)) solves z′=Gf(t,z) on I. [given]

[F2]

For k>0 the homogeneous model has exactly one global solution for every initial datum, and its three explicit solution families are those of The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data clauses 1-3; in particular the solution g with data (0,1/m) exists, is unique and is defined on all of R. [given]

[F3]

For k=0 the homogeneous model has exactly the solution families of The degenerate and repulsive oscillator: solutions for k=0 and the explicit growing solution for k<0 clauses 1-2, and every interval solution extends to a global member; in particular g with data (0,1/m) exists, is unique and is defined on all of R. [given]

[F4]

Convolution differentiation (Differentiation of a variable-limit convolution integral with a C-one matrix kernel): if I is an interval, t0∈I, U is open with I−I⊆U, G:U→Mn(R) is C1 and b:I→Rn is continuous, then t↦∫t0tG(t−s)b(s) ds is differentiable with derivative G(0)b(t)+∫t0tG′(t−s)b(s) ds. Here n=1, U=R, G=g; when g is C2, the same application with G=g′ is legitimate. [given]

[F5]

Homogeneous uniqueness from Variation of constants for z′=Az+b(t) via the matrix exponential: for a fixed A∈M2(R) and the globally continuous forcing b0(t)≡0, the global solution of z′=Az+b0 with zero datum is z≡0, and every differentiable solution on an interval containing its initial time with that datum is its restriction. [given]

Proof

technique · constructive; build the solution as homogeneous solution plus convolution with the kernel $g$, verify the equation by two applications of the convolution lemma, and invoke the variation-of-constants uniqueness
1.1F2F3

The kernel g exists, is unique and is defined on all of R by [F2] (case k>0) and [F3] (case k=0), with g(0)=0 and g′(0)=1/m. Since g solves m g′′+c g′+k g=0, we have g′′=−(c g′+k g)/m; the right-hand side is continuous because g and g′ are, so g′′ is continuous and g is of class C2 on R.

2.1step 1.1F4

Put F(t):=∫t0tg(t−s)f(s) ds for t∈I. Applying [F4] with kernel g and b:=f gives, at every interior point, F′(t)=g(0)f(t)+∫t0tg′(t−s)f(s) ds=∫t0tg′(t−s)f(s) ds, and applying [F4] once more with the C1 kernel g′ (legitimate by step 1.1) gives F′′(t)=g′(0)f(t)+∫t0tg′′(t−s)f(s) ds=f(t)m+∫t0tg′′(t−s)f(s) ds. If I includes an endpoint, the same split-difference-quotient argument as in the proof of Differentiation of a variable-limit convolution integral with a C-one matrix kernel applies with increments h restricted to the side lying in I; it gives these formulas for the domain-relative endpoint derivatives, first for F with kernel g and then for F′ with kernel g′. The displayed right-hand sides are continuous on I, since f,g,g′,g′′ are continuous and the integrals vary continuously on compact subintervals. In particular F(t0)=0 and F′(t0)=0.

3.1step 2.1F1F2F3construct

Let xh be a homogeneous solution with the data (x0,v0) (existing by [F2], [F3]) and set x:=xh+F on I. [construct] Then x is twice differentiable on I with continuous second derivative, using domain-relative derivatives at included endpoints, and, by step 2.1 and the linearity of the differential expression, m x′′+c x′+k x=(m xh′′+c xh′+k xh)+m F′′+c F′+k F=f(t)+∫t0t(m g′′+c g′+k g)(t−s)f(s) ds=f(t), because the bracket vanishes identically by step 1.1. Moreover x(t0)=xh(t0)+F(t0)=x0 and x′(t0)=xh′(t0)+F′(t0)=v0.

4.1step 3.1F1F5discharge-construct: the constructed convolution is the unique solution∎

Uniqueness: let x~:I→R be any other solution with the prescribed data and put u:=x~−(xh+F), where xh+F is the constructed solution of step 3.1. By linearity, u solves the homogeneous equation on I with u(t0)=u′(t0)=0. Its phase curve U(t):=(u(t),u′(t)) therefore solves U′=AU on I with U(t0)=0, where A=(01−k/m−c/m). Apply [F5] with the globally continuous forcing b0≡0: the unique global solution from zero data is zero, and clause 3 of the matrix variation-of-constants result makes every interval solution with the same data its restriction. Thus U=0 and x~=xh+F. This proves uniqueness using only a forcing defined on its actual domain I for the constructed solution.

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Forced quadratic system Mq′′+Kq=φ(t): decoupled forced modes and the convolution solution

Statement

Let n≥1, let M∈Mn(R) be symmetric positive definite, let K∈Mn(R) be symmetric and non-negative, let S:=M−1/2KM−1/2=OΛOT with O orthogonal and Λ=diag⁡(λ1,…,λn), λi≥0 (Decoupling of the constant-coefficient quadratic system Mq′′+Kq=0 into normal modes, with the zero-mode and energy identities), let I⊆R be an interval with nonempty interior (Intervals of R: the nine order-convex forms, nondegeneracy, and length), let φ:I→Rn be continuous and let t0∈I. Put

ψ(t):=OTM−1/2φ(t),ωi:=λi (λi>0),

and for a curve q:I→Rn whose first and second derivatives exist on I (relative to I at included endpoints) and whose second derivative is continuous on I, put z:=OTM1/2q. Then:

  1. q solves the forced quadratic system Mq′′+Kq=φ on I if and only if z solves the decoupled forced system zi′′+λizi=ψi(1≤i≤n) on I.
  2. For each i, the unique solution zi:I→R of the scalar equation in clause 1 with data zi(t0)=ξi, zi′(t0)=ηi is zi(t)=ξicos⁡(ωi(t−t0))+ηiωisin⁡(ωi(t−t0))+1ωi∫t0tsin⁡(ωi(t−s))ψi(s) ds(λi>0), zi(t)=ξi+ηi(t−t0)+∫t0t(t−s)ψi(s) ds(λi=0).
  3. Consequently, for every initial datum (q(t0),q′(t0))∈R2n there is exactly one solution q:I→Rn of Mq′′+Kq=φ with that datum, and it is q=M−1/2Oz for the z of clause 2.

Facts & Assumptions

Given: The symmetric positive definite M, symmetric non-negative K, the orthogonal diagonalisation S=OΛOT, the interval I, the continuous φ:I→Rn, the point t0∈I, and the transformed forcing ψ=OTM−1/2φ.

[F1]

The quadratic-mode theorem (Decoupling of the constant-coefficient quadratic system Mq′′+Kq=0 into normal modes, with the zero-mode and energy identities) supplies the positive-definite square-root factors and an orthogonal diagonalisation S=OΛOT with λi≥0. Its scalar solution classification is stated only for intervals containing 0; the present interval I is arbitrary, so the proof below uses only this algebraic diagonalisation and verifies the forced scalar responses at the prescribed t0.

[F2]

The forced oscillator theorem (The forced oscillator with k≥0: unique global solution and the convolution (Green's function) formula) applies on this arbitrary interval I at the specified t0: with m=1, c=0, k=λi≥0 and continuous ψi, it gives exactly one solution and its homogeneous-plus-convolution formula. The homogeneous classification (The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data) and degenerate proposition (The degenerate and repulsive oscillator: solutions for k=0 and the explicit growing solution for k<0) classify homogeneous solutions on every interval. For λi>0, translating an interval by −t0 gives one containing 0, where the classification determines the homogeneous member from its data; for λi=0 the proposition directly determines the affine member from its value and derivative at t0. Trigonometric derivatives and values at zero are (sin⁡)′=cos⁡, (cos⁡)′=−sin⁡, sin⁡0=0, and cos⁡0=1 (The derivatives of sine and cosine are cosine and minus sine).

[F3]

M1/2, M−1/2, O and OT=O−1 are the matrices of the modal theorem; M−1/2MM−1/2=In and OTO=In. [given]

Proof

technique · direct; transport the forced system through the modal transformation and apply the scalar forced-oscillator theorem mode by mode
1.1F1F3algebra

Transport: differentiating q=M−1/2Oz twice gives q′′=M−1/2Oz′′, hence Mq′′+Kq=M1/2O(z′′+OTM−1/2KM−1/2Oz)=M1/2O(z′′+Λz) by the definition of S and [F1], and M1/2O is invertible. Therefore Mq′′+Kq=φ is equivalent to z′′+Λz=OTM−1/2φ=ψ, which is the decoupled system of clause 1.

2.1step 1.1F2F4

The i-th scalar equation is an oscillator equation with m=1, c=0, k=λi≥0 and continuous forcing ψi; by [F2] its unique solution on I with data (ξi,ηi) at t0 is the homogeneous solution with those data plus ∫t0tgi(t−s)ψi(s) ds. If λi>0, set hi(t)=ξicos⁡(ωi(t−t0))+ηiωisin⁡(ωi(t−t0)). By the chain rule and trigonometric derivatives in [F2], hi′′+λihi=0, hi(t0)=ξi, and hi′(t0)=ηi; translating any competing homogeneous solution to an interval containing 0 and applying the classification in [F2] proves uniqueness. If λi=0, hi(t)=ξi+ηi(t−t0) directly solves the homogeneous equation with the required data and is the unique affine family member. Likewise gi(u)=sin⁡(ωiu)/ωi for λi>0 and gi(u)=u for λi=0 solve the homogeneous equation and have gi(0)=0, gi′(0)=1; the classified families give uniqueness of the kernel. Substitution in the forced theorem's formula gives exactly the two expressions in clause 2.

3.1step 1.1step 2.1F1F3∎

Clauses 1 and 2 exhibit z as the unique solution of the decoupled forced system with prescribed data; since q=M−1/2Oz is a bijection with (q(t0),q′(t0))↦(z(t0),z′(t0)) linear and invertible (q′(t0)=M−1/2Oz′(t0)), clause 3 follows.

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Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at Ω=k/m

Conventions

For a bounded real-valued curve u:R→R, its amplitude about zero means sup⁡t∈R∣u(t)∣.

Statement

Let m>0, k>0, c≥0, let f0∈R and Ω≥0, and consider the forced model m x′′+c x′+k x=f with time domain I=R and forcing f:R→R, f(t)=f0cos⁡(Ωt) (The damped driven oscillator model: second-order equation, phase-plane system and characteristic data). Put

D:=k−mΩ2+i c Ω∈C,ω0:=k/m.

  1. Non-resonant response. If D≠0, then the function xp(t):=Re⁡ ⁣(f0eiΩtD)=f0 (k−mΩ2)cos⁡(Ωt)+c Ωsin⁡(Ωt)(k−mΩ2)2+c2Ω2 is a solution of the forced model on all of R. Its amplitude about zero is ∣f0∣/(k−mΩ2)2+c2Ω2. For Ω>0 and f0≠0, a phase-lag angle relative to the drive means a real ϕ such that xp(t)=(f0/∣D∣)cos⁡(Ωt−ϕ) for every t, with the signed factor f0 retained in both drive and response. Such an angle exists: any ϕ∈R satisfying cos⁡ϕ=k−mΩ2∣D∣,sin⁡ϕ=cΩ∣D∣ has this property. No temporal phase is assigned when Ω=0 or f0=0.

  2. Resonance. If c=0 and Ω=ω0, then D=0 and xp(t):=f0 tsin⁡(ω0t)2mω0 is a solution of the forced model on all of R; if f0≠0 then xp is unbounded on [0,∞).

  3. Transient decay. If c>0, then D≠0 for every Ω≥0, and every solution x:[0,∞)→R of the forced model satisfies x(t)=xp(t)+xh(t) where xh is a homogeneous solution; consequently x(t)−xp(t)→0 as t→+∞ (Limits at +∞ and −∞, and infinite limits at a point) and x is bounded on [0,∞).

Facts & Assumptions

Given: The parameters m>0, k>0, c≥0, the number f0∈R and frequency Ω≥0, the all-R forcing f(t)=f0cos⁡(Ωt) and D=k−mΩ2+icΩ, and ω0=k/m.

[F1]

The model and its phase system are those of The damped driven oscillator model: second-order equation, phase-plane system and characteristic data; the differential expression x↦mx′′+cx′+kx is linear over R, so sums of solutions of the forced and homogeneous equations are solutions of the corresponding forced equation. [given]

[F2]

Complex exponential: ddtexp⁡(μt)=μexp⁡(μt) for complex μ and real t, and exp⁡(−γt+iΩt)=e−γt(cos⁡(Ωt)+isin⁡(Ωt)), so Re⁡exp⁡(iΩt)=cos⁡(Ωt); complex conjugation and modulus are those of Real and imaginary parts, complex conjugation, and modulus, with ww‾=∣w∣2 and 1/w=w‾/∣w∣2 for w≠0. [thm-complex-exponential-is-entire-with-derivative-itself, cor-complex-exponential-cartesian-form-modulus-and-eulers-identity, def-complex-conjugate-real-imaginary-part-and-modulus]

[F3]

Real derivatives: (sin⁡t)′=cos⁡t, (cos⁡t)′=−sin⁡t; sums and products are differentiated by the algebra of derivatives and the chain rule; sin⁡(π/2)=1 and sine has period 2π. [thm-sine-and-cosine-derivatives, thm-algebra-of-derivatives, thm-chain-rule, thm-quarter-turn-values-and-shift-formulas, thm-sine-cosine-zero-sets-and-fundamental-period]

[F4]

For k>0, every homogeneous solution is one of the three families of The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data, and if c>0 every homogeneous solution satisfies xh(t)→0 and is bounded on [0,∞) by Strictly damped homogeneous oscillator with k>0: every solution tends to zero. [given]

[F5]

Sine and cosine have range [−1,1] (Signs, monotonicity intervals, and ranges of sine and cosine); cos⁡(−u)=cos⁡u, sin⁡(−u)=−sin⁡u and cos⁡2u+sin⁡2u=1 (Parity and the Pythagorean identity for sine and cosine). The addition formulas (The addition formulas for sine and cosine) and parity give cos⁡(u−v)=cos⁡ucos⁡v+sin⁡usin⁡v for all real u,v.

Proof

technique · cases; verify the complex steady-state solution and its real part, verify the resonant solution directly, and obtain the transient statement from the homogeneous decay
1.1F2F5givenalgebra

Non-resonant case. [assume-case 1] Suppose D≠0 and put z(t):=f0exp⁡(iΩt)/D. By [F2], z′=iΩz and z′′=−Ω2z, hence mz′′+cz′+kz=(−mΩ2+icΩ+k)z=Dz=f0exp⁡(iΩt). The differential expression has real coefficients and the real and imaginary parts of a complex-valued C2 function may be differentiated separately, so xp=Re⁡z satisfies m xp′′+c xp′+k xp=Re⁡(f0exp⁡(iΩt))=f0cos⁡(Ωt) on R. The displayed real form follows from [F2] and 1/D=D‾/∣D∣2, D‾=k−mΩ2−icΩ, ∣D∣2=(k−mΩ2)2+c2Ω2. For the amplitude, put a=k−mΩ2, b=cΩ and r=∣D∣=a2+b2>0. By [F5], (acos⁡u+bsin⁡u)2+(asin⁡u−bcos⁡u)2=r2, so ∣xp(t)∣≤∣f0∣/r. If Ω=0, then D=k and xp≡f0/k, attaining that bound. If Ω>0, the cosine range in [F5] supplies a real u with cos⁡u=a/r; the Pythagorean identity gives ∣sin⁡u∣=b/r since b≥0, and replacing u by −u if necessary makes sin⁡u=b/r while preserving its cosine. At t=u/Ω the numerator acos⁡u+bsin⁡u equals r, attaining ∣xp(t)∣=∣f0∣/r. This also covers f0=0, proving the stated amplitude. When Ω>0 and f0≠0, take ϕ=u from that construction. For this angle, and for any other angle with the stated sine and cosine, the difference formula in [F5] gives cos⁡(Ωt−ϕ)=(acos⁡(Ωt)+bsin⁡(Ωt))/r; therefore xp(t)=(f0/r)cos⁡(Ωt−ϕ) for every t, proving the phase relation with the declared signed convention.

1.2F3givenalgebra

Resonance. [assume-case 2] Suppose c=0 and Ω=ω0, and put xp(t)=f0tsin⁡(ω0t)/(2mω0), so that f(t)=f0cos⁡(ω0t) and k=mω02. By [F3], xp′(t)=f0(sin⁡(ω0t)+ω0tcos⁡(ω0t))2mω0,xp′′(t)=f0(2ω0cos⁡(ω0t)−ω02tsin⁡(ω0t))2mω0, so m xp′′+k xp=f0cos⁡(ω0t)−12f0ω0tsin⁡(ω0t)+12f0ω0tsin⁡(ω0t)=f0cos⁡(ω0t): it is a solution on R. For each j∈N, put tj=(π/2+2πj)/ω0. These times are nonnegative and tend to +∞ because ω0>0 and π>0. Integer periodicity and the quarter-turn value in [F3] give sin⁡(ω0tj)=1, so ∣xp(tj)∣=∣f0∣tj/(2mω0)→+∞ whenever f0≠0; hence xp is unbounded on [0,∞).

2.1step 1.1step 1.2F1F4F5∎

Transient decay. [assume-case 3, cases-exhaustive: D≠0; c=0, Ω=ω0; c>0] Suppose c>0; then D=0 would force cΩ=0, hence Ω=0 and D=k≠0, a contradiction, so D≠0 for every Ω and step 1.1 supplies the solution xp. Let x:[0,∞)→R be any solution of the forced model. Then xh:=x−xp satisfies the homogeneous model by linearity [F1]; by [F4] it is one of the three families of the classification theorem, and since c>0, Strictly damped homogeneous oscillator with k>0: every solution tends to zero gives xh(t)→0 as t→+∞ and boundedness of xh on [0,∞); xp is bounded by [F5] because ∣cos⁡∣≤1, ∣sin⁡∣≤1 and D≠0 in the real form of step 1.1. Hence x(t)−xp(t)=xh(t)→0 and x=xp+xh is bounded.

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The amplitude of the sinusoidally forced damped oscillator: location and value of its maximum

Statement

Let m>0, k>0, c>0 and f0≠0. For Ω≥0 let

A(Ω):=∣f0∣(k−mΩ2)2+c2Ω2

be the amplitude about zero of the steady response of Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at Ω=k/m clause 1, and put

u∗:=km−c22m2,Ω∗:=u∗  when u∗>0.

Then:

  1. If u∗>0 (equivalently c2<2mk), A is strictly increasing on [0,Ω∗], strictly decreasing on [Ω∗,∞), and its maximum on [0,∞) is attained at Ω∗ and at no other point, with A(Ω∗)=2m∣f0∣c4mk−c2.
  2. If u∗≤0 then A is strictly decreasing on [0,∞) and its maximum on [0,∞) is attained at Ω=0 and at no other point.

Facts & Assumptions

Given: The parameters m>0, k>0, c>0, f0≠0, the amplitude function A of the statement and the number u∗.

[F1]

For every Ω≥0 the denominator (k−mΩ2)2+c2Ω2 is strictly positive, and A(Ω)=∣f0∣/q(Ω2) with q(u):=(k−mu)2+c2u; this is the amplitude computed in Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at Ω=k/m clause 1. [given]

[F2]

Derivatives: for a differentiable F with F′>0 on an interval, F is strictly increasing there, and with F′<0 strictly decreasing, by the mean value theorem (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)); polynomials are differentiated by the algebra of derivatives and the power rule, and composites by the chain rule. [thm-algebra-of-derivatives, lem-derivative-of-a-power, thm-chain-rule, def-interval]

[F3]

For 0<a<b one has 0<b−1<a−1 and 0<a<b, and x↦x is strictly increasing on [0,∞) (Inverses of positives are positive, and reciprocation reverses order, Squaring is monotone on the nonnegatives, with nonnegative roots supplied by Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}). [def-interval]

Proof

technique · cases on the sign of the minimiser; compute the quadratic exactly, locate its vertex, and convert the monotonicity through the strictly increasing square-root map followed by the strictly decreasing reciprocal map on positive arguments
1.1givenalgebra

Write q(u)=(k−mu)2+c2u=m2u2+(c2−2mk)u+k2 and u∗=km−c22m2. [assume-case 1, assume-case 2] Expanding m2(u−u∗)2 and using −2m2u∗=c2−2mk gives q(u)=m2(u−u∗)2+(k2−m2u∗2),q′(u)=2m2(u−u∗), and the constant is (k2−m2u∗2)=(k−mu∗)(k+mu∗)=c22m⋅4mk−c22m=c2(4mk−c2)4m2, because k−mu∗=c22m and k+mu∗=4mk−c22m; in particular q is a quadratic with positive leading coefficient and unique global minimiser u∗ and q(u∗)=c2(4mk−c2)4m2. Also u∗>0 if and only if c2<2mk; in this branch 4mk−c2>0, so q(u∗)>0. For every u≥0, q(u)=(k−mu)2+c2u>0: at u=0 it equals k2>0, and at u>0 the second summand is positive.

2.1step 1.1F2

Monotonicity of Ω↦q(Ω2): by [F2], its derivative is 2Ω q′(Ω2)=4m2Ω(Ω2−u∗). Hence if u∗>0 this derivative is negative on (0,Ω∗) and positive on (Ω∗,∞), so q(Ω2) is strictly decreasing on [0,Ω∗] and strictly increasing on [Ω∗,∞) (strictness on the closed intervals follows from [F2] on every subinterval and continuity); if u∗≤0 the derivative is positive on (0,∞), so q(Ω2) is strictly increasing on [0,∞).

3.1step 2.1F1F3algebra∎

Conclusion. [cases-exhaustive: u∗>0; u∗≤0] If u∗>0, then on [0,Ω∗] the value q(Ω2) strictly decreases to q(Ω∗2)=q(u∗) and on [Ω∗,∞) it strictly increases from q(u∗); since x↦1/x is strictly decreasing on (0,∞) and x↦x strictly increasing on [0,∞) by [F3], A=∣f0∣/q(Ω2) is strictly increasing on [0,Ω∗] and strictly decreasing on [Ω∗,∞), so its maximum is attained exactly at Ω∗, with A(Ω∗)=∣f0∣/q(u∗)=2m∣f0∣c4mk−c2. If u∗≤0, then q(Ω2) is strictly increasing on [0,∞), so A is strictly decreasing there and its maximum is attained exactly at Ω=0.

Remarks

  • The undamped case. For c=0 the same computation gives q(u)=(k−mu)2, whose zero is u=k/m, i.e. Ω=ω0=k/m; there the denominator of the steady-response formula vanishes and A (the modulus of the response function) is unbounded above in every neighbourhood of ω0 meeting [0,∞)∖{ω0}. Hence for c=0 the amplitude has no maximum on [0,∞); it is strictly increasing on [0,ω0), which is the increasing branch that persists in the c>0 case above. This matches the resonant case of Sinusoidal forcing: the steady response, its amplitude and phase, and resonance at Ω=k/m clause 2, where the response has absolute values tending to infinity along the quarter-turn sequence when f0≠0.
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Elementary bounds for the periodic average, the deviation and its time derivative

Statement

Let n≥1, P>0, O⊆Rn open, and let f:R×O→Rn be continuous, continuously partially differentiable in the second variable and P-periodic in time, with average fˉ, oscillatory part f−fˉ and periodic deviation g as in Periodic averaging data: the average field, the oscillatory part and the periodic deviation (Intervals of R: the nine order-convex forms, nondegeneracy, and length). Here Dxf(t,x):Rn→Rn denotes the total derivative in the state variable, with coordinate columns ∂xjf(t,x); it exists and is continuous by If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative. The same notation is used for Dxfˉ and Dxg when those total derivatives exist. Let x0∈O, R>0 with B‾(x0,R)⊆O (Open ball, closed ball and sphere in a metric space) and let M,L≥0 satisfy

∥f(t,x)∥2≤M,∥f(t,x)−f(t,y)∥2≤L∥x−y∥2(t∈R, x,y∈B‾(x0,R)).

Then:

  1. ∥fˉ(x)∥2≤M and ∥fˉ(x)−fˉ(y)∥2≤L∥x−y∥2 for all x,y∈B‾(x0,R).
  2. g is P-periodic and ∥g(t,x)∥2≤2MP for every t∈R and x∈B‾(x0,R); moreover ∥g(t,x)−g(t,y)∥2≤2LP∥x−y∥2 for all t∈R and x,y∈B‾(x0,R).
  3. ∂tg(t,x)=f(t,x)−fˉ(x) for every (t,x); consequently ∥∂tg(t,x)∥2≤2M for x∈B‾(x0,R) and ∥∂tg(t,x)−∂tg(t,y)∥2≤2L∥x−y∥2 for all t∈R and x,y∈B‾(x0,R).
  4. g has continuous first partial derivatives on R×O; its partial derivative in the second variable is Dxg(t,x)=∫0t(Dxf(s,x)−Dxfˉ(x))ds, and ∥Dxg(t,x)∥≤2LP for t∈R, x∈B‾(x0,R), with ∥⋅∥ the operator norm on Mn(R) (The operator norm as the least bound and as the unit-sphere or unit-ball supremum). In particular g is totally differentiable at every point of R×O (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

Facts & Assumptions

Given: The data n,P,O,f of Periodic averaging data: the average field, the oscillatory part and the periodic deviation, the ball centre x0, radius R, constants M,L, and the derived objects fˉ and g.

[F2]

For integrable vector-valued ϕ and a<b: ∥∫abϕ∥2≤∫ab∥ϕ∥2, and for real integrable u≤v: ∫abu≤∫abv; for constant c and h>0: ∫aa+hc=c h (For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable, If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a)).

[F4]

Fundamental theorem of calculus: if u is continuous on an interval I and a∈I, then F(t):=∫atu is a primitive of u on I (Every continuous function on an interval has a primitive; two primitives differ by a constant; and ∫abf=G(b)−G(a) for any primitive G); in particular F has derivative u at every point of I, and a periodic function is determined by its values on [0,P].

[F5]

Leibniz's rule on a compact rectangle (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral): if G,H:[a,b]×[c,d]→R are continuous and for every fixed t∈[c,d] the slice x↦G(x,t) is differentiable with derivative H(x,t), then x↦∫cdG(x,t) dt has derivative ∫cdH(x,t) dt. The componentwise application of this rule to f (whose partial derivatives in the state variable are continuous by hypothesis) supplies the identities of step 3.1 (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative for the passage from partial derivatives to the total derivative).

[F6]

A continuous scalar function on an interval with zero derivative at every interior point is constant (A function continuous on an interval I whose derivative vanishes at every interior point of I is constant on I; consequently two such functions with the same derivative differ by a constant); applying this to every coordinate gives the same assertion for vector-valued functions.

Proof

technique · direct; estimate the averages and the deviation by the integral norm inequality and monotonicity, and identify the time derivative by the fundamental theorem
1.1F1F2given

Bounds for fˉ: for x∈B‾(x0,R), ∥fˉ(x)∥2=1P∥∫0Pf(s,x) ds∥2≤1P∫0P∥f(s,x)∥2 ds≤1P∫0PM ds=M by [F2]; and for x,y in the ball, ∥fˉ(x)−fˉ(y)∥2=1P∥∫0P(f(s,x)−f(s,y))ds∥2≤1P∫0P∥f(s,x)−f(s,y)∥2 ds≤L∥x−y∥2, again by [F2] and the assumed Lipschitz bound; this is clause 1.

2.1step 1.1F1F2F3given

Bounds for g: for t∈[0,P] and x in the ball, step 1.1 and [F2] give ∥g(t,x)∥2=∥∫0t(f(s,x)−fˉ(x))ds∥2≤∫0t(∥f(s,x)∥2+∥fˉ(x)∥2)ds≤2MP, and for x,y in the ball, ∥g(t,x)−g(t,y)∥2≤∫0t(∥f(s,x)−f(s,y)∥2+∥fˉ(x)−fˉ(y)∥2)ds≤2LP∥x−y∥2. Since g is P-periodic by its definition (it satisfies g(t+P,x)=g(t,x) and g(P,x)=0), the same bounds hold for all t∈R; this is clause 2.

2.2step 1.1F1F3F4F6given

Time derivative: fix x and put u(s):=f(s,x)−fˉ(x), a continuous Rn-valued function; on (0,P) the fundamental theorem [F4] gives ∂∂tg(t,x)=u(t)=f(t,x)−fˉ(x). At t=0 the right difference quotient converges to u(0)=f(0,x)−fˉ(x), while for t→0− periodicity reduces the quotient to the left derivative of g(⋅,x) at P, namely u(P)=f(P,x)−fˉ(x)=f(0,x)−fˉ(x) by periodicity of f; the same argument at every t=kP shows the derivative exists there with the same value, and away from these points it is u(t). Hence ∂tg(t,x)=f(t,x)−fˉ(x) everywhere. The oriented integral H(t):=∫0tu(s) ds has the same derivative by [F4], so H−g(⋅,x) has zero derivative and vanishes at zero; [F6] gives g(t,x)=H(t) for every real t. The bounds on f−fˉ follow from the triangle inequality and step 1.1: ∥f(t,x)−fˉ(x)∥2≤M+M=2M and ∥f(t,x)−fˉ(x)−f(t,y)+fˉ(y)∥2≤2L∥x−y∥2; this is clause 3.

3.1step 1.1step 2.2F1F2F5given∎

Second-variable derivative and its norm bound: first fix t∈R. For x in the open ball B(x0,R) and any unit vector v∈Rn, the assumed state-Lipschitz estimate gives ∥(f(t,x+hv)−f(t,x))/h∥2≤L for every sufficiently small nonzero h; taking h→0 yields ∥Dxf(t,x)v∥2≤L. For a boundary point x∈B‾(x0,R), the sequence xj=x0+(1−1/(j+1))(x−x0), j∈N lies in the open ball and tends to x, so continuity of Dxf(t,⋅) gives the same bound at x. Taking the supremum over unit v proves ∥Dxf(t,x)∥≤L throughout the closed ball. Next fix any x∈O and a component index i. The map (s,u)↦fi(s,u), with u near x, is continuous and has continuous partial derivatives in u; [F5] applied componentwise at fixed s and t gives ∂xj∫0tfi(s,x) ds=∫0t∂xjfi(s,x) ds (first for t in a compact interval, then by additivity for all t∈R), and likewise Dxfˉ(x)=1P∫0PDxf(s,x) ds. For x∈B‾(x0,R) and every unit v, the vector integral inequality [F2] now gives ∥Dxfˉ(x)v∥2≤1P∫0PL ds=L, hence ∥Dxfˉ(x)∥≤L. Using the all-time integral identity of step 2.2, Dxg(t,x)=∫0t(Dxf(s,x)−Dxfˉ(x))ds, continuously in (t,x). For x∈B‾(x0,R), t∈[0,P] and every unit v, another application of [F2] gives ∥Dxg(t,x)v∥2≤∫0t∥(Dxf(s,x)−Dxfˉ(x))v∥2 ds≤2Lt≤2LP. Taking the supremum over unit v proves ∥Dxg(t,x)∥≤2LP on one period; periodicity gives the same bound for all t. This proves clause 4.

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Finite-time linearization error bound: quadratic remainder, exponential growth bound and the variation-of-constants estimate

Statement

Let n≥1, let U⊆Rn be open with 0∈U, let F:U→Rn be continuous and locally Lipschitz in the state variable, and let r>0, C≥0, L≥0 and A∈Mn(R) satisfy B‾(0,r)⊆U (Open ball, closed ball and sphere in a metric space) and

∥F(z)−Az∥2≤C∥z∥22,∥F(z)∥2≤L∥z∥2(∥z∥2≤r), where Az and ∥A∥ use the matrix-action and induced Euclidean operator-norm convention of Variation of constants for z′=Az+b(t) via the matrix exponential.

Let z0∈Rn with δ:=∥z0∥2>0, let T>0 with δeLT<r, and let y:R→Rn be the solution of the linear system y′=Ay, y(0)=z0 (Variation of constants for z′=Az+b(t) via the matrix exponential). Then:

  1. The maximal solution z of z′=F(z), z(0)=z0 is defined on [0,T] and satisfies ∥z(t)∥2≤δeLt for t∈[0,T] (Extensions, maximal solutions, maximal intervals, and global solutions).
  2. For every t∈[0,T], ∥z(t)−y(t)∥2≤Cδ2∫0te∥A∥(t−s)e2Ls ds≤Cδ2 t e(∥A∥+2L)t, where ∥A∥ is the operator norm of A for the Euclidean norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Remarks

Facts & Assumptions

Given: The field F and constants A,r,C,L as in the statement, the datum z0 with δ=∥z0∥2>0, the time T>0 with δeLT<r, and the linear solution y(t)=E(t)z0 with E(t) the matrix exponential of A.

[F1]
[F2]

Local existence on a Picard-Lindelof cylinder and the a priori Lipschitz estimate are provided by Continuous and Lipschitz dependence on the initial condition on a Picard-Lindelof cylinder; every initial value problem has a unique maximal solution, and any solution is a restriction of it (Every Picard–Lindelöf initial value problem has one maximal solution on an open interval, Extensions, maximal solutions, maximal intervals, and global solutions). [given]

[F3]

Compact continuation: a solution whose graph stays in a compact subset of U near a finite endpoint extends past that endpoint (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). [given]

[F5]

Gronwall: if continuous u≥0 satisfies u(t)≤A0+K∫0tu on [0,T], then u(t)≤A0eKt there (Gronwall's integral inequality with variable and constant coefficients). [given]

[F6]

Matrix exponential data (Variation of constants for z′=Az+b(t) via the matrix exponential, The operator norm as the least bound and as the unit-sphere or unit-ball supremum): E is C1, E′(t)=AE(t)=E(t)A, E(t)E(s)=E(t+s), and ∥E(u)∥≤e∥A∥∣u∣; in particular y(t)=E(t)z0 is the global solution of y′=Ay, y(0)=z0, obtained by taking the globally continuous forcing b≡0.

Proof

technique · direct; bound the maximal solution by Gronwall while it stays in the ball, extend it to $[0,T]$ by compact continuation, and derive the finite-interval comparison identity with an integrating factor
1.1F1F4F5given

Let z be the maximal solution through z0 and let J⊆[0,∞) be that part of its interval on which ∥z(t)∥2≤r; while t∈J, [F4] and the bound ∥F(z(s))∥2≤L∥z(s)∥2 give ∥z(t)∥2≤δ+∫0tL∥z(s)∥2 ds, so [F5] yields ∥z(t)∥2≤δeLt for all t∈J. In particular, if t∈J and t≤T then ∥z(t)∥2≤δeLT<r.

2.1step 1.1F2F3given

The maximal solution is defined on all of [0,T]. Indeed, if its forward interval ended at some τ≤T, then the graph of z on [0,τ) would lie in the compact set [0,T]×B‾(0,r) by step 1.1 applied on J∩[0,τ) — more precisely, should z reach ∥z∥2=r at some first time τ0, step 1.1 would give ∥z(τ0)∥2≤δeLτ0≤δeLT<r at that time, a contradiction; so ∥z(t)∥2<r for all t in the maximal forward interval that are ≤T, and the graph stays in the compact subset [0,T]×B‾(0,r)⊆[0,T]×U. By [F3] the solution extends past every finite endpoint below or at T, so it is defined on [0,T] and step 1.1 gives ∥z(t)∥2≤δeLt there; this is clause 1.

3.1step 2.1F6given

Finite-interval comparison identity: on [0,T] put q:=z−y and b(t):=F(z(t))−Az(t). The curve b is continuous on [0,T], and q′=Aq+b, q(0)=0. Define r(t):=E(−t)q(t). By [F6], the chain rule and product rule, r′(t)=−AE(−t)q(t)+E(−t)(Aq(t)+b(t))=E(−t)b(t), since AE(−t)=E(−t)A. The vector-valued fundamental theorem of calculus, applied on the actual interval [0,T], gives r(t)=∫0tE(−s)b(s) ds; multiplying by E(t) and using [F6]'s group law yields q(t)=∫0tE(t−s)b(s) ds. This derivation uses only the continuous values of b on [0,T]. Consequently, using ∥E(t−s)∥≤e∥A∥(t−s) for 0≤s≤t, the quadratic remainder hypothesis and clause 1, ∥z(t)−y(t)∥2≤∫0te∥A∥(t−s)∥F(z(s))−Az(s)∥2 ds≤Cδ2∫0te∥A∥(t−s)e2Ls ds, for every t∈[0,T].

4.1step 3.1algebra∎

For the final estimate, e∥A∥(t−s)≤e∥A∥t and e2Ls≤e2Lt on 0≤s≤t, so ∫0te∥A∥(t−s)e2Lsds≤t e(∥A∥+2L)t; substituting into step 3.1 gives the displayed product bound. This proves clause 2.

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Averaging for fast periodic oscillations: order-ε agreement on the O(1/ε) time scale

Statement

Let n≥1, P>0, let O⊆Rn be open, and let f:R×O→Rn be continuous, continuously partially differentiable in the second variable and P-periodic in time, with average fˉ and deviation g as in Periodic averaging data: the average field, the oscillatory part and the periodic deviation. Here Dxf(t,x) is the total derivative in the state variable defined in Elementary bounds for the periodic average, the deviation and its time derivative. Let x0∈O, let 0<ρ<R with B‾(x0,R)⊆O (Open ball, closed ball and sphere in a metric space), and let M,L≥0 satisfy

∥f(t,x)∥2≤M,∥Dxf(t,x)∥≤L(t∈R, x∈B‾(x0,R)),

∥⋅∥ being the operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Put

B:=2MP,B1:=2LP,K:=eLA[B+(LB+B1M)A],

where ε>0 and A>0 are parameters. Suppose y:[0,A/ε]→B‾(x0,R−ρ) is a solution of the averaged system

y′(t)=εfˉ(y(t)),y(0)=x0.

If εK<ρ, then the solution x of x′(t)=εf(t,x(t)), x(0)=x0 exists on all of [0,A/ε] and satisfies, for every t∈[0,A/ε],

∥x(t)−y(t)∥2≤εK=ε eLA[B+(LB+B1M)A].

Remarks

  • Scaling. The two curves agree to order ε on the time interval of length A/ε, For a fixed period P, this horizon contains A/(εP) periods. The constant is explicit and grows like AeLA; no claim is made for times longer than A/ε or beyond the trapping radius ρ.
  • Scope. f, y, g, ε, A are mathematical objects; no physical identification or empirical claim is made. The theorem compares two explicitly specified ODE solutions; the averaged system is a different model, not an exact description.

Facts & Assumptions

Given: The data n,P,O,f,x0,ρ,R,M,L, the derived fˉ,g,B,B1, and the parameters ε>0, A>0, with a solution y:[0,A/ε]→B‾(x0,R−ρ) of the averaged system.

[F1]

Derivative-to-Lipschitz bound and periodic-deviation bounds: for each fixed t, the map x↦f(t,x) is L-Lipschitz on the convex ball B‾(x0,R). Indeed, for x,y in the ball the segment x+s(y−x) lies in it, the composite s↦f(t,x+s(y−x)) is differentiable with derivative Dxf(t,x+s(y−x))(y−x), and the vector-valued fundamental theorem estimate gives ∥f(t,x)−f(t,y)∥2≤L∥x−y∥2 (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a), If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz). Thus Elementary bounds for the periodic average, the deviation and its time derivative applies and gives ∥fˉ(x)∥2≤M and ∥fˉ(x)−fˉ(y)∥2≤L∥x−y∥2 on the ball; ∥g(t,x)∥2≤B=2MP for t∈R, x∈B‾(x0,R); ∥Dxg(t,x)∥≤B1=2LP there; ∂tg(t,x)=f(t,x)−fˉ(x); and g is totally differentiable on R×O. [given]

[F2]

The field Fε(t,x):=εf(t,x) is continuous. Near any (t∗,x∗)∈R×O, choose a time interval and a convex state ball contained in R×O on which continuity of Dxf bounds its operator norm by some finite C. For each fixed time in that interval, the line-segment argument of [F1], with C in place of L, gives ∥Fε(t,x)−Fε(t,y)∥2≤εC∥x−y∥2 for states in that ball. This is local Lipschitz continuity in the state variable as defined in Local Lipschitz continuity in the state variable, locally uniform in time and parameters, so the Picard-Lindelof theorem gives a unique maximal solution of the initial-value problem (First-order systems, initial value problems, and solutions on intervals, Every Picard–Lindelöf initial value problem has one maximal solution on an open interval, Extensions, maximal solutions, maximal intervals, and global solutions). Every such solution satisfies the Volterra equation x(t)=x0+∫0tεf(s,x(s)) ds (A first-order initial value problem is equivalent to its Volterra integral equation), and compact continuation extends a maximal solution whose graph stays in a compact subset of [0,∞)×O near a finite endpoint (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). [given]

[F3]

Chain rule: for a totally differentiable curve G on R×O and a differentiable curve t↦x(t), the composite t↦G(t,x(t)) is differentiable with derivative ∂tG+DxG⋅x′ (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)). [given]

[F4]

Integral inequalities: ∥∫abϕ∥2≤∣∫ab∥ϕ∥2∣ and monotonicity for real integrands (For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable, If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a), The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); Gronwall: if continuous u≥0 satisfies u(t)≤L0+M0t+K0∫0tu on [0,T] with constants ≥0, then u(t)≤(L0+M0t)eK0t (apply the nondecreasing-a clause of Gronwall's integral inequality with variable and constant coefficients with a(t)=L0+M0t and b(t)=K0). [given]

Proof

technique · direct; conjugate the fast curve by the periodic deviation, apply Gronwall to the difference with the averaged curve, and control the trapping radius by the final constant
1.1F1F2F3given

By [F2] let x:(α,β)→O be the unique maximal solution of x′=εf(t,x) with x(0)=x0, where α<0<β. Define τR:=inf⁡{t∈[0,β):∥x(t)−x0∥2=R}, with τR:=+∞ if this set is empty, and T0:=min⁡{A/ε,β,τR}. For 0≤t<T0 one has x(t)∈B(x0,R), so the bounds of [F1] apply; also g is totally differentiable near (t,x(t)), as required by [F3].

2.1step 1.1F1F3F4algebra

Estimate on [0,T0). Put d(t):=x(t)−εg(t,x(t)) and w:=d−y. Since g(0,x0)=0, w(0)=0. By [F3] and ∂tg=f−fˉ, d′=εf−ε(∂tg+Dxg⋅x′)=εf−ε(f−fˉ)−ε2Dxg⋅f=εfˉ(x)−ε2Dxg(t,x)f(t,x), so w′=ε(fˉ(x)−fˉ(y))−ε2Dxg(t,x)f(t,x) for 0≤t<T0. Using [F1], ∥fˉ(x)−fˉ(y)∥2≤L∥x−y∥2≤L∥w∥2+εLB because ∥x−y∥2≤∥x−d∥2+∥d−y∥2=ε∥g(t,x(t))∥2+∥w∥2≤εB+∥w∥2, and ∥ε2Dxg⋅f∥2≤ε2B1M. Therefore, by [F4], ∥w(t)∥2≤∫0t(εL∥w(s)∥2+ε2(LB+B1M))ds, and the Gronwall inequality of [F4] with L0=0, M0=ε2(LB+B1M), K0=εL gives ∥w(t)∥2≤ε2(LB+B1M) t eεLt≤ε(LB+B1M)A eLA for t<T0≤A/ε. Hence ∥x(t)−y(t)∥2≤∥w(t)∥2+εB≤ε[B+(LB+B1M)AeLA]≤εK.

3.1step 2.1F1F2given∎

No exit and continuation. If τR<min⁡{β,A/ε}, continuity lets the estimate of step 2.1 pass to t=τR; then R=∥x(τR)−x0∥2≤∥x(τR)−y(τR)∥2+∥y(τR)−x0∥2≤εK+(R−ρ)<R, a contradiction. Hence for every 0≤t<min⁡{β,A/ε}, step 2.1 and the hypothesis on y give ∥x(t)−x0∥2≤εK+(R−ρ)<R. If β≤A/ε were finite, the graph of x near β would stay in the compact subset [0,A/ε]×B‾(x0,R) of [0,∞)×O, so compact continuation [F2] would extend x past β, contradicting maximality. Thus β>A/ε, and x is defined on all of [0,A/ε]; continuity extends the estimate of step 2.1 to the endpoint, proving the claim.

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Adiabatic action drift bound: order-ε change of the action on the O(1/ε) time scale

Statement

Let ε>0, A>0, ω0>0, let J^,S^⊆R be open intervals, and let J⊆J^ and S⊆S^ be intervals with nonempty interiors (A one-frequency parameter-dependent action-angle branch, its transformed Hamiltonian and its drift generator, Intervals of R: the nine order-convex forms, nondegeneracy, and length). Let h:J^×S^→R and k:J^×R×S^→R be C2, with k 2π-periodic in θ on J×R×S. Let K⊆J and S0⊆S be compact intervals, let s0∈S0, and suppose

∣∂Ih(I,s)∣≥ω0((I,s)∈K×S0).

Let (I,θ,s):[0,A/ε]→J×R×S be a branch solution of the branch vector field of the datum (A one-frequency parameter-dependent action-angle branch, its transformed Hamiltonian and its drift generator) with s(0)=s0, and suppose that for every t∈[0,A/ε] one has I(t)∈K and s(t)∈S0. Then for every t∈[0,A/ε],

∣I(t)−I(0)∣≤ε[2sup⁡K×[0,2π]×S0∣g∣+Asup⁡K×[0,2π]×S0∣∂Ig ∂θk−∂θg ∂Ik−∂sg∣],

where g=(k−⟨k⟩)/∂Ih is the drift generator of the datum.

Remarks

  • Conditional character. The bound is conditional on the branch solution staying in the compact sets K×S0, i.e. on the action coordinate not leaving the chart along the motion; the trapping is an explicit hypothesis, not a consequence. The chart itself — the existence of a smooth parameter-dependent action-angle family with ∣ω∣≥ω0 — is likewise a hypothesis of the definition, not proved here. No claim of exact invariance is made: the right-hand side is a bound of order ε and grows linearly in the rescaled time A.
  • Comparison with the sources. Both cited treatments derive the order-ε (averaged) statement ⟨I˙⟩=0; Tong's Section 4.6 explicitly remarks that stricter proofs are deferred to the literature, and Likharev's averaged identity likewise treats the parameter as frozen over one period. The present theorem replaces averaging by the exact chain-rule identity (I+εg)′=−ε2(…) and hence yields a uniform-in-time bound on the stated horizon under the explicit trapping and frequency hypotheses. No exact conservation is asserted.

Facts & Assumptions

Given: The intervals J,S, the compact intervals K,S0, the constants ε,A,ω0,s0, the C2 data h,k with k 2π-periodic in θ, the drift generator g, and a branch solution (I,θ,s) as in the statement.

[F1]

The branch vector field is (I,θ,s)′=Φ^ with Φ^(I,θ,s)=(−ε∂θk, ∂Ih+ε∂Ik, ε); consequently s(t)=s0+εt, I′=−ε∂θk(I,θ,s) and θ′=ω+ε∂Ik(I,θ,s) with ω=∂Ih; see A one-frequency parameter-dependent action-angle branch, its transformed Hamiltonian and its drift generator. [given]

[F2]

The angle average ⟨k⟩(I,s)=12π∫02πk(I,θ,s) dθ is of class C1 on the open parameter domain J^×S^, with ∂I⟨k⟩=⟨∂Ik⟩ and ∂s⟨k⟩=⟨∂sk⟩, by differentiation under the integral sign on compact rectangles (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral). Since ∣∂Ih∣≥ω0>0 on K×S0, the quotient g=(k−⟨k⟩)/∂Ih is C1 on a neighborhood of K×[0,2π]×S0; in particular ∂θg=∂θk/∂Ih there, because ⟨k⟩ does not depend on θ. [given]

[F3]

The suprema in the statement are finite: g and its first partial derivatives and the partial derivatives ∂θk,∂Ik,∂sg are continuous on the compact set K×[0,2π]×S0 (periodicity of k and of its derivatives in θ makes [0,2π] a full period; If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative for the differentiable structure used below). [given]

[F4]

Chain rule for a composite t↦g(I(t),θ(t),s(t)): if g is totally differentiable at the relevant points and t↦(I,θ,s) is differentiable, then ddtg(I,θ,s)=∂Ig I′+∂θg θ′+∂sg s′ (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)). [given]

[F5]

Fundamental theorem and norm bound: a differentiable curve on an interval with continuous derivative satisfies u(t)−u(0)=∫0tu′ (If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral), and ∣∫0tϕ∣≤∫0t∣ϕ∣ for real continuous ϕ (For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable for the one-dimensional case, or monotonicity of the integral). [given]

Proof

technique · direct; conjugate the action by the drift generator so that the fast angle cancels exactly, then integrate the remaining order-$\varepsilon^2$ identity and bound the errors
1.1F1F2F4algebra

Along the branch solution the composite t↦g(t):=g(I(t),θ(t),s(t)) is differentiable, and by [F1], [F2], [F4], g′=∂Ig (−ε∂θk)+∂θg (ω+ε∂Ik)+∂sg ε. Since ∂θg ω=∂θkωω=∂θk by [F2], this gives g′=−ε∂Ig ∂θk+∂θk+ε∂θg ∂Ik+ε∂sg, hence I′+εg′=−ε∂θk+εg′=−ε2(∂Ig ∂θk−∂θg ∂Ik−∂sg), that is, ddt(I+εg(I,θ,s))=−ε2(∂Ig ∂θk−∂θg ∂Ik−∂sg) at every t∈[0,A/ε].

2.1step 1.1F3F5

Integrating: by [F5] applied to the curve u(t):=I(t)+εg(I(t),θ(t),s(t)), whose derivative is continuous by step 1.1 and [F3], u(t)−u(0)=−ε2∫0t(∂Ig ∂θk−∂θg ∂Ik−∂sg)(I(r),θ(r),s(r)) dr, so by [F3] and the norm bound of [F5], ∣u(t)−u(0)∣≤ε2t C≤εAC, where C:=sup⁡K×[0,2π]×S0∣∂Ig ∂θk−∂θg ∂Ik−∂sg∣ and t≤A/ε.

3.1step 2.1F3algebra∎

Conclusion: writing u=I+εg and using the triangle inequality, ∣I(t)−I(0)∣≤∣u(t)−u(0)∣+ε∣g(I(t),θ(t),s(t))∣+ε∣g(I(0),θ(0),s(0))∣≤εAC+2εsup⁡K×[0,2π]×S0∣g∣, which is the displayed bound.

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