e^{- im θ}

and summation over

m \in Z

give

\sum_{m} e^{- im θ} u_{m + k}^{n} = e^{ik θ} u^{n} (θ) .

Thus the recurrence becomes

A (θ) u^{n + 1} (θ) = B (θ) u^{n} (θ),

where

A (θ) = \sum_{k = r}^{s} a_{k} e^{ik θ}, B (θ) = \sum_{k = r}^{s} b_{k} e^{ik θ} .

Assuming

A (θ) \neq = 0

H (θ) = \frac{B ( θ )}{A ( θ )} = \frac{\sum _{k = r}^{s} b _{k} e ^{ik θ}}{\sum _{k = r}^{s} a _{k} e ^{ik θ}} .

ii

Solution

u^{n} (θ) = H (θ)^{n} u^{0} (θ) .

∣ H (θ) ∣ \leq 1

everywhere, the Parseval identity gives

∥ u^{n} ∥_{ℓ^{2}}^{2} = \frac{1}{2 π} \int_{- π}^{π} ∣ H (θ) ∣^{2 n} ∣ u^{0} (θ) ∣^{2} d θ \leq ∥ u^{0} ∥_{ℓ^{2}}^{2} .

Conversely, if

∣ H (θ_{0}) ∣ > 1

, continuity supplies a neighbourhood on which

∣ H ∣ \geq 1 + ϵ

. Choose nonzero square-integrable Fourier data supported there. Its norm then grows at least as

(1 + ϵ)^{n}

, contradicting boundedness. Hence

{u^{n}} is bounded for every u^{0} ⟺ ∣ H (θ) ∣ \leq 1 for all θ .

This is the one-step Von Neumann stability analysis criterion.

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