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Distributed-delay midpoint quadrature separates two history errors

Example

Take a scalar delay window of length $τ$ , constant kernel $K (θ) = 1$ , and continuous path $y (t) = t^{2}$ on $[- τ, T]$ , so its history is $ϕ (s) = s^{2}$ . At query time $t = 0$ , let $h = τ / m$ and refine the composite midpoint rule with $r = q m$ cells, where $m, q \geq 1$ are integers and $Δ = τ / r = h / q$ . The exact integral is $τ^{3} / 3$ . The exact-history midpoint remainder is $τ Δ^{2} / 12$ , while using the piecewise-linear stored history adds a distinct node-history term. The stored-history rule returns an error of $τ h^{2} / 6$ in this example; the two terms are separated in Distributed-delay midpoint error separates history and quadrature errors.

Facts & Assumptions

Given: A scalar history $ϕ (s) = s^{2}$ on $[- τ, 0]$ , the constant kernel on $[0, τ]$ , an aligned state mesh $h = τ / m$ , and midpoint refinement $Δ = h / q$ .

[F1]

The distributed-delay integral is $I_{K} [y] (t) = \int_{0}^{τ} K (θ) y (t - θ) d θ$ (Distributed delays through a finite-window history integral).

[F2]

The composite midpoint rule samples the stored piecewise-linear history at $θ_{j} = (j + 1 / 2) Δ$ (Composite midpoint quadrature for a distributed delay history).

[F3]

Under its uniform stored-path error bound on $[- τ, T]$ , the error bound splits into a stored-history node term and an exact-history quadrature remainder (Distributed-delay midpoint error separates history and quadrature errors).

Proof

1.1F1givenalgebra

At $t = 0$ , [F1] gives $I = \int_{0}^{τ} θ^{2} d θ = τ^{3} / 3$ . Put $Q_{e x} = Δ \sum_{j = 0}^{r - 1} θ_{j}^{2}$ , the midpoint sum using the exact history. On a cell centered at $c_{j}$ , write $θ = c_{j} + u$ with $- Δ / 2 \leq u \leq Δ / 2$ . Then $\int_{- Δ / 2}^{Δ / 2} (c_{j} + u)^{2} d u = c_{j}^{2} Δ + Δ^{3} / 12$ . Summing the cells gives $I - Q_{e x} = r Δ^{3} / 12 = τ Δ^{2} / 12$ .

1.2F2givenalgebra

On the history cell $[ℓ h, (ℓ + 1) h]$ in the variable $θ = - s$ , the stored chord of $ϕ$ satisfies ${\hat{ϕ}}_{h} (- θ) - θ^{2} = (θ - ℓ h) ((ℓ + 1) h - θ) .$ Its maximum is $h^{2} / 4$ , so the uniform interpolation error on $[- τ, 0]$ is $h^{2} / 4$ . Since $Δ = h / q$ , every midpoint subcell is contained in one such chord cell.

2.1step 1.1step 1.2algebra

The stored chord is affine on each width- $h$ cell, and the midpoint rule is exact for an affine function on each of its $q$ equal subcells: the centered linear term integrates to zero. Therefore the stored-history midpoint sum $Q_{s t}$ equals the integral of the chord. The area between the chord and $θ^{2}$ on one width- $h$ cell is $\int_{0}^{h} v (h - v) d v = h^{3} / 6$ . Summing $m$ cells gives $Q_{s t} - I = m h^{3} / 6 = τ h^{2} / 6$ . Combining with step 1.1 yields $Q_{s t} - Q_{e x} = τ h^{2} / 6 + τ Δ^{2} / 12$ .

3.1F2F3givenstep 1.1step 1.2step 2.1algebra∎

At $t = 0$ every sampled time $- θ_{j}$ lies in $[- τ, 0]$ , so [F2] and step 1.2 give $∣ Q_{s t} - Q_{e x} ∣ \leq Δ \sum_{j = 0}^{r - 1} ∣ {\hat{ϕ}}_{h} (- θ_{j}) - θ_{j}^{2} ∣ \leq τ h^{2} / 4$ . Combining this with step 1.1 directly gives $∣ I - Q_{s t} ∣ \leq τ h^{2} / 4 + τ Δ^{2} / 12$ , the same history/quadrature split as [F3]; this calculation needs no error bound for positive-time stored values. The exact node-history term from step 2.1 is $τ h^{2} / 6 + τ Δ^{2} / 12 \leq τ h^{2} / 4$ because $Δ \leq h$ . Its positive sign cancels the midpoint underestimate $Q_{e x} - I = - τ Δ^{2} / 12$ , leaving $∣ Q_{s t} - I ∣ = τ h^{2} / 6$ .

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Sources

Nicola Guglielmi and Ernst Hairer, Solving Delay Differential Equations, Chapter 1 (standard reference, not scraped)
Roland Pulch, Equivalent Systems for Differential Equations with Polynomially Distributed Delay, version 3 (standard reference, not scraped)