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A repeated interior root preserves bounded companion powers

Example

Consider the three-step method with $ρ (ζ) = (ζ - 1) (ζ - \frac{1}{2})^{2}$ and $σ (ζ) = \frac{1}{4}$ . It is consistent and satisfies the root condition even though the root $1 / 2$ is repeated; its homogeneous companion powers are bounded.

Facts & Assumptions

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

[F1]

Taylor moment conditions characterize the order of a linear multistep method identifies consistency from $ρ (1) = 0$ and $ρ^{'} (1) = σ (1)$ .

[F2]

The root condition is equivalent to bounded powers of the multistep companion matrix proves that repeated roots strictly inside the unit disk are permitted and still give bounded companion powers.

Verification

1.1F1algebra

Expanding gives $ρ (ζ) = ζ^{3} - 2 ζ^{2} + \frac{5}{4} ζ - \frac{1}{4}$ , so $ρ (1) = 0$ and $ρ^{'} (1) = \frac{1}{4} = σ (1)$ . Hence [F1] gives consistency.

1.2givenalgebra

The factorization displays roots $1$ and $1 / 2$ with multiplicities one and two, respectively. The unit root is simple; the repeated root has modulus strictly below one. Thus the root condition holds.

2.1F2step 1.2algebra∎

Apply [F2] to the root condition in step 1.2. The companion powers are uniformly bounded. In the scalar recurrence, the repeated-root mode $n 2^{- n}$ is bounded as well, since $n 2^{- n} \to 0$ .

Depends on

Used by

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