Physics theoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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First-order reduction of order with an explicit residual and error bound

Physical scope

Exact residual and an absolute acceleration-error bound for the first-order reduction of order of the adopted prescribed-force ODE, on the unique bounded-future branch and for bounded twice continuously differentiable forces with bounded second derivative. The result is conditional on the effective model; it is not a bound for nonlinear self-consistent Maxwell–Lorentz trajectories, and it gives no experimental validation.

Statement

Assume the equation and bounded-force hypotheses of the preceding theorem: $m, τ > 0$ , $f : R \to R^{3}$ is bounded and twice continuously differentiable, and the unique bounded-future solution $a_{*}$ is used. Suppose

$M_{2} : = \sup_{t \in R} ∣ f^{''} (t) ∣ < \infty .$

Since $f$ has units $N$ and $t$ has units $s$ , $M_{2}$ has units $N s^{- 2}$ .

Define the first-order reduction-of-order approximation by

$a_{R O} (t) : = \frac{f (t) + τ f^{'} (t)}{m} .$

Its residual when substituted into the adopted equation is exactly

$m (a_{R O} (t) - τ a_{R O}^{'} (t)) - f (t) = - τ^{2} f^{''} (t) .$

On the bounded-future branch the acceleration error obeys

$∣ a_{*} (t) - a_{R O} (t) ∣ \leq \frac{τ^{2} M_{2}}{m} (t \in R) .$

The residual and bound are exact conditional statements about the prescribed-force ODE; they do not assert that a nonlinear self-consistent particle trajectory is approximated by $a_{R O}$ .

Statement references

Runaway and bounded-future branches of the Abraham–Lorentz equation

Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$

The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation

Facts & Assumptions

Given: The adopted equation, constants and bounded-future branch of the preceding theorem, with $f \in C^{2} (R, R^{3})$ bounded and $M_{2} < \infty$ .

[F1]

For every bounded continuous $f$ , the preceding theorem gives $a_{*} (t) = (m τ)^{- 1} \int_{0}^{\infty} \exp (- v / τ) f (t + v) d v$ ; the integral converges componentwise and this solution is the unique one bounded on a future half-line. (Runaway and bounded-future branches of the Abraham–Lorentz equation)

[A1]

The force and constants satisfy the preceding theorem's bounded-force hypotheses; additionally $f$ is twice continuously differentiable and $∣ f^{''} (t) ∣ \leq M_{2}$ for all $t$ . (Runaway and bounded-future branches of the Abraham–Lorentz equation)

[F2]

The chain and product rules apply to $\exp (- v / τ) f (t + v)$ ; the real exponential has derivative $\exp$ ; the first fundamental theorem evaluates integrals of continuous functions on finite intervals. (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The exponential function is smooth and $(\exp)^{'} = \exp$ , The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive)

[F3]

The mean value theorem bounds each component of $f^{'}$ by its value at one point plus $M_{2}$ times the time increment. For $x \geq 0$ , the exponential power series gives $\exp (x) \geq 1 + x + x^{2} / 2$ , and $\exp (- x) = 1 / \exp (x)$ , so both $\exp (- x)$ and $x \exp (- x)$ tend to zero. (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 $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ , The real exponential function and the number $e$ by a power series, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ )

[F4]

Riemann integration on finite intervals is linear and monotone; $\int_{0}^{\infty} \exp (- v / τ) d v = τ$ . For a scalar integrand $h$ with $∣ h (v) ∣ \leq M \exp (- v / τ)$ , monotonicity applied to $- M \exp (- v / τ) \leq h (v) \leq M \exp (- v / τ)$ yields $∣ \int h ∣ \leq M \int \exp (- v / τ)$ . (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive)

[F5]

The dot product is the finite coordinate sum defining the Euclidean inner product, and $∣ u \cdot w ∣ \leq ∣ u ∣ ∣ w ∣$ . (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation)

Proof

technique · direct; differentiate the proposed reduced-order expression, then integrate the exact future kernel twice by parts and estimate its remainder

1.1givenA1F2algebra

The product and chain rules give $a_{R O}^{'} = (f^{'} + τ f^{''}) / m$ . Substitution therefore yields $m (a_{R O} - τ a_{R O}^{'}) = f + τ f^{'} - τ f^{'} - τ^{2} f^{''} = f - τ^{2} f^{''}$ , proving the stated residual.

1.2A1F3F4algebra

Fix $t \in R$ . The boundedness of $f$ makes $\exp (- R / τ) f (t + R) \to 0$ as $R \to \infty$ . For each component $j$ , [A1] and [F3] give $∣ f_{j}^{'} (t + R) ∣ \leq ∣ f_{j}^{'} (t) ∣ + M_{2} R$ for $R \geq 0$ by the mean value theorem. Since $\exp (- R / τ) (1 + R / τ) \to 0$ , also $\exp (- R / τ) f^{'} (t + R) \to 0$ . All finite-interval integrals below exist because their integrands are continuous.

2.1A1F2F4algebrastep 1.2

On $[0, R]$ , differentiate $\exp (- v / τ) f (t + v)$ and integrate the product-rule identity using [F2]. After rearranging, this gives $\int_{0}^{R} \exp (- v / τ) f (t + v) d v = τ f (t) - τ \exp (- R / τ) f (t + R) + τ \int_{0}^{R} \exp (- v / τ) f^{'} (t + v) d v .$ Applying the same calculation to $\exp (- v / τ) f^{'} (t + v)$ gives $\int_{0}^{R} \exp (- v / τ) f^{'} (t + v) d v = τ f^{'} (t) - τ \exp (- R / τ) f^{'} (t + R) + τ \int_{0}^{R} \exp (- v / τ) f^{''} (t + v) d v .$ These identities hold componentwise.

3.1F1F4F5step 1.2step 2.1algebra∎

Let $R \to \infty$ in the identities of step 2.1; the boundary terms vanish by step 1.2 and the last integral converges componentwise since $∣ f^{''} ∣ \leq M_{2}$ and $\int_{0}^{\infty} \exp (- v / τ) d v = τ$ by [F4]. Substitution into [F1] yields $a_{*} (t) - a_{R O} (t) = \frac{τ}{m} \int_{0}^{\infty} \exp (- v / τ) f^{''} (t + v) d v .$ If this vector integral is $z \neq 0$ , set $w = z / ∣ z ∣$ . By [F5], $∣ w ∣ = 1$ and $∣ w \cdot f^{''} (t + v) ∣ \leq M_{2}$ . Coordinatewise linearity gives $w \cdot z = \int_{0}^{\infty} \exp (- v / τ) w \cdot f^{''} (t + v) d v$ , and [F4] gives $∣ w \cdot z ∣ \leq M_{2} \int_{0}^{\infty} \exp (- v / τ) d v = M_{2} τ$ . Thus $∣ z ∣ = w \cdot z \leq M_{2} τ$ . The same bound is trivial for $z = 0$ , proving $∣ a_{*} (t) - a_{R O} (t) ∣ \leq τ^{2} M_{2} / m$ .

First-order reduction of order with an explicit residual and error bound

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources