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Simple unit-circle roots give bounded oscillatory modes

Example

The consistent three-step formula with ρ(ζ)=(ζ−1)(ζ2+1) and σ(ζ)=2 has simple roots 1,i,−i on the unit circle. Its homogeneous modes include 1,in,(−i)n; their real and imaginary parts are bounded periodic real solutions.

Facts & Assumptions

Given: The root-condition and order conventions in Linear multistep methods, consistency, order, and the root condition.

[F1]

Taylor moment conditions characterize the order of a linear multistep method characterizes consistency by the first two moment conditions.

[L1]

The root condition is equivalent to bounded powers of the multistep companion matrix proves that the root condition implies bounded companion powers.

[F4]

Quarter-turn values and shifts by pi/2 and pi gives cos⁡(π/2)=0 and sin⁡(π/2)=1.

Verification

1.1F1F5algebra

Expanding gives ρ(ζ)=ζ3−ζ2+ζ−1. Thus ρ(1)=0, ρ′(1)=2=σ(1), and [F1] proves consistency. Since i2=−1 by [F5], the factorization displays three distinct roots 1,i,−i, all of modulus one, so the root condition holds.

1.2F2F3F4algebra

By [F2] and [F4], i=exp⁡(iπ/2) and ∣i∣=∣−i∣=1. Repeated use of [F3] gives in=exp⁡(inπ/2), so in has modulus one by [F2]; also (−i)n=(−1)nin has modulus one. Each is a recurrence mode because substituting Yn=λn multiplies the homogeneous recurrence by ρ(λ).

2.1L1step 1.1step 1.2∎

Apply [L1] to the root condition in step 1.1. The full homogeneous companion recurrence is bounded, including these periodic modes; no decay of unit-circle modes is asserted.

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