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Polyline interpolation and equal-length redistribution approximate a smooth arc-length path

Statement

Let L>0, let M≥1 be an integer, and let R:[0,L]→Rd be a unit-speed C3 curve with M equal arc-length intervals of length h=L/M. Suppose ∥R′′∥2≤K2 and the coordinate third derivatives satisfy the bound in Second-order centered and unequal-step tangent estimates with constant M3. For sufficiently small h, each chord length ℓi=∥R(ih)−R((i−1)h)∥2 satisfies ∣ℓi−h∣≤C1h3, where C1 depends only on K2,M3. The piecewise-linear interpolant through the samples differs from R in supremum norm by at most C2h2. If this polygon is redistributed by linear interpolation at equally spaced cumulative-length targets, the jth redistributed image is within C3h2 of R(jh), for every j=0,…,M.

Here C3 on [0,L] means that R is the restriction of a C3 curve on an open interval containing [0,L]; derivatives at the endpoints are the continuous one-sided values, and the unit-speed and derivative bounds are required on [0,L].

Facts & Assumptions

Given: The unit-speed C3 curve, equal arc-length mesh, derivative bounds, and small-mesh condition in the Statement.

[F1]

The unit-speed hypothesis [A1] means ∥R′(s)∥2=1 on [0,L]. Thus the normalized tangent convention of Regular minimum-energy paths on a declared collision-free tube gives t(s)=R′(s) on this interval.

[F2]

Second-order centered and unequal-step tangent estimates: the coordinate third-derivative bound used here is ∣Rj′′′∣≤M3/d on the path interval.

[F3]

The Lagrange and Cauchy forms of Taylor's remainder: its one-dimensional Lagrange formula applies to a scalar C3 function on a closed interval when derivatives through order three are continuous to the endpoints; with degree two it bounds the remainder by sup⁡∣g′′′∣ ∣u∣3/6, and with degree one it bounds the remainder by sup⁡∣g′′∣ ∣u∣2/2.

[F4]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): differentiating a composition gives the derivative of the outer function applied to the derivative of the inner function.

[A1]

The curve has the endpoint C3 convention in the Statement, has unit speed on [0,L], and has the stated bounds ∥R′′∥2≤K2 and coordinate third-derivative bound M3/d on the whole interval; the interior C3 convention is Ck maps and multi-index derivative notation in Euclidean space.

Given: The data in the Statement.

Proof

1.1A1F1F2F3F4givenalgebra

On each mesh interval [s,s+h], apply the one-dimensional degree-two formula [F3] to each coordinate of the extended curve at s; the endpoint convention supplies the required derivative continuity, and [A1, F2] gives a remainder vector r with ∥r∥2≤M3h3/6. Thus R(s+h)−R(s)=hT+h22K+r with T=R′(s), K=R′′(s), and ∥K∥2≤K2. Differentiating ∥R′(s)∥22=1 gives TTK=0 by the chain rule [F4]. Hence ∣∥R(s+h)−R(s)∥22−h2∣≤Ch4 for a constant depending only on K2,M3; for sufficiently small h the chord is at least h/2, so division by its sum with h yields ∣ℓi−h∣≤C1h3.

1.2A1F1F3givenalgebra

For s=ih+θh with 0≤θ≤1, apply the one-dimensional degree-one formula [F3] on [ih,s] and [ih,(i+1)h]. The remainder bound in [F3] and ∥R′′∥2≤K2 give R(s)=R(ih)+θhR′(ih)+O(K2h2) and R((i+1)h)=R(ih)+hR′(ih)+O(K2h2). Subtracting the linear interpolant cancels the first-order term, so the pointwise interpolation error is bounded by C2h2, uniformly over all intervals.

2.1givenstep 1.1algebra

Let pi=∑m=1iℓm and Lh=pM. Step 1.1 gives ∣pi−ih∣≤iC1h3≤C1Lh2 and ∣Lh−L∣≤C1Lh2. Thus the redistributed target qj=jLh/M satisfies ∣qj−jh∣≤C1Lh2. Each segment has length at least h/2 for small h. If target qj lies on segment i at fraction θ, then pi−1+θℓi=qj; combining ∣pi−1−(i−1)h∣≤C1Lh2, ∣qj−jh∣≤C1Lh2, and ∣ℓi−h∣≤C1h3 gives ∣(i−1)h+θh−jh∣≤Ch2.

3.1F1step 1.2step 2.1algebra∎

The point on the true curve at parameter (i−1)h+θh is within C2h2 of the redistributed polygon point by step 1.2, and it is within Ch2 of R(jh) by unit speed and step 2.1. The triangle inequality gives the claimed C3h2 bound, including endpoints where redistribution fixes the endpoints.

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