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Finite execution and storage counts for compatible neutral Euler

Statement

For Explicit Euler for a scalar neutral delay on a compatible grid, set N=Km and h=τ/m. With no budget let M=N; with a budget B∈N0 let M=min⁡{B,N}. Assuming each prescribed history, derivative-history and field evaluation returns, the procedure terminates after exactly M calls to f, returning completed exactly when M=N and budget-exhausted otherwise. Every requested index and evaluation is admitted. If M≥1, compatibility gives V0=ϕ′(0).

With the initialized history retained, the state and derivative arrays contain 2m+2M+1 scalar values. In particular completion uses N field calls and O(N+m) retained scalar state and derivative values, with no memory bound constant uniformly in N+m. Counts exclude the internal costs of evaluating ϕ, ϕ′ and f.

Facts & Assumptions

Given: The admitted compatible scalar neutral problem and evaluation convention in the Statement.

[F1]

Explicit Euler for a scalar neutral delay on a compatible grid stores Yj,Vj for −m≤j<0 and Y0, then at each accepted index n forms Dn=ϕ′(tn−τ) for n<m or Vn−m otherwise, computes Vn=aDn+f(tn,Yn,Yn−m), and appends Yn+1.

[F2]

Compatibility is ϕ′(0)=aϕ′(−τ)+f(0,ϕ(0),ϕ(−τ)) (Scalar neutral delay equations and the derivative-segment shift).

Proof

1.1givenF1F2algebra

At accepted index n, induction gives states with indices −m,…,n and derivative values −m,…,n−1. The state index n−m is in range. For n<m, tn−τ∈[−τ,0), so the history derivative is defined; for n≥m, 0≤n−m≤n−1, so the computed derivative is available. Since n<N and Nh=Kτ=T, tn<T, and the field accepts all scalar states at that time. Thus each accepted update is defined. At n=0, substituting the initialized states and D0=ϕ′(−τ) into the recurrence gives V0=ϕ′(0) by [F2]. This assertion requires an accepted first update.

2.1givenF1step 1.1algebra∎

There are finitely many loop indices 0,…,N−1, each with one field call if accepted. The budget test rejects the first index B when B<N, while no index is rejected when there is no budget or B≥N. Thus exactly M updates occur and the status is as asserted; all prescribed evaluations return, so the procedure terminates. Initialization stores 2m+1 scalar values; each accepted update adds Vn and Yn+1, for a total 2m+2M+1. At completion this is O(N+m) and grows with the retained array lengths, yielding the claimed memory conclusion.

Depends on

Used by

Nothing in the library uses this result yet.

Cited to discharge well-definedness by Explicit Euler for a scalar neutral delay on a compatible grid.

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