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ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Harmonic forcing gives an exact error criterion for reduction of order

Physical scope

An exact harmonic-forcing calculation for the adopted prescribed-force ODE. It quantifies the model's absolute phasor-amplitude error for first-order reduction of order and shows that a fixed small tolerance requires small ∣ωτ∣. It is not an empirical frequency-response measurement or a validation of a self-consistent radiation-reaction law.

Example

Use the adopted equation m(a−τa˙)=f with positive constants m>0 and τ>0, of units kg and s, respectively, and the laboratory conventions of the bounded-future branch theorem.

A phasor coefficient at frequency ω∈R is a complex vector g0=gR+igI∈C3, with gR,gI∈R3, representing a real map g:R→R3 by the real-part representation

g(t)=Re⁡ ⁣(g0exp⁡(−iωt))=gRcos⁡(ωt)+gIsin⁡(ωt).

At ω=0 this representation determines only gR. Adopt the canonical static convention g0=G+i0 for the constant function g(t)≡G∈R3; in particular the static force below has v=0 and f0=u+i0. Static acceleration phasors use the same convention.

Let ω∈R and f0=u+iv∈C3, with u,v∈R3. For any z=u+iv∈C3, define ∣z∣:=(∑j=02∣uj∣2+∑j=02∣vj∣2)1/2, the Euclidean norm of its six real coordinates. The force amplitude f0 has units N, ω has units s−1, and acceleration phasors have units m s−2. Prescribe the real force

f(t):=Re⁡ ⁣(f0exp⁡(−iωt)),

where real part is taken componentwise. The bounded-future acceleration has complex phasor amplitude

a∗,0=f0m(1+iωτ).

The first-order reduction-of-order acceleration aRO=(f+τf′)/m has phasor amplitude aRO,0=f0(1−iωτ)/m. For f0≠0, the absolute amplitude error relative to the force acceleration scale ∣f0∣/m is

∣a∗,0−aRO,0∣∣f0∣/m=(ωτ)21+(ωτ)2.

For any chosen tolerance 0<ϵ<1, this relative amplitude error is at most ϵ exactly when

∣ωτ∣≤ϵ2+ϵ4+4ϵ22.

Hence small error at a small prescribed tolerance requires small ∣ωτ∣; no universal numerical cutoff or experimental frequency response is asserted.

Facts & Assumptions

Given: The adopted ODE with m,τ>0, ω∈R, a complex vector amplitude f0=u+iv with u,v∈R3 and v=0 if ω=0 by the canonical static convention, and the real harmonic force f(t)=ucos⁡(ωt)+vsin⁡(ωt).

[F1]

For m,τ>0 and bounded continuous f:R→R3, the preceding theorem gives the unique bounded-future solution of m(a−τa˙)=f. (Runaway and bounded-future branches of the Abraham–Lorentz equation)

[F2]

For m,τ>0 and bounded C2 forces f:R→R3 with bounded f′′, the first-order reduction is aRO=(f+τf′)/m. (First-order reduction of order with an explicit residual and error bound)

[F3]

The Euclidean norm on R3 is induced by its inner product; real and imaginary parts and complex modulus are componentwise, and exp⁡(iθ)=cos⁡θ+isin⁡θ for real θ. (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Real and imaginary parts, complex conjugation, and modulus, exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0)

Verification

1.1F1F3F4algebra

Write f0=u+iv. The candidate a∗(t)=Re⁡ ⁣(f0m(1+iωτ)exp⁡(−iωt)) is a bounded real harmonic vector function. Put x=ωτ; its cosine and sine coefficient vectors are A=(u+xv)/(m(1+x2)) and B=(v−xu)/(m(1+x2)). By [F4], a∗′=−ωAsin⁡(ωt)+ωBcos⁡(ωt), so a∗−τa∗′ has cosine coefficient A−xB=u/m and sine coefficient B+xA=v/m. Thus m(a∗−τa∗′)=f, and uniqueness in [F1] identifies this candidate with the bounded-future branch.

2.1F2F3step 1.1algebra

Differentiating aRO=(f+τf′)/m or using its phasor gives aRO,0=f0(1−ix)/m. Therefore a∗,0−aRO,0=f0m(11+ix−(1−ix))=−f0mx21+ix. Since ∣1+ix∣=1+x2, the displayed error ratio follows.

3.1step 2.1algebra∎

For z=x2≥0, the tolerance inequality z/1+z≤ϵ is equivalent, by squaring nonnegative sides, to z2−ϵ2z−ϵ2≤0. The nonnegative solutions satisfy z≤(ϵ2+ϵ4+4ϵ2)/2, which is exactly the stated bound on ∣x∣. This is a conditional model calculation, not an empirical claim.

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