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Constant-step BDF2 is A-stable under the multistep root condition
Statement
For each with , the constant-step BDF2 test polynomial has degree two, all its roots have modulus at most one, and every root of modulus one is simple. Thus its multistep absolute-stability region contains the closed left half-plane, so BDF2 is A-stable.
Facts & Assumptions
Given: Constant-step BDF2 on with and .
Scalar absolute stability, A-stability, L-stability, and multistep root regions: a multistep test polynomial gives absolute stability when its degree is unchanged, every root has modulus at most one, and any unit-modulus root is simple; A-stability means the closed left half-plane lies in this region.
BDF2 roots on or outside the unit disk map to the closed right half-plane: any root of with modulus at least one forces , with equality only at .
Proof
The leading coefficient of is . If it vanished, then , contradicting . Thus has degree two.
By [F2], a root with modulus greater than one would force , which is impossible. A root with modulus one can occur only when and . At , , so the unit root is simple and the other root is .
The degree and root properties in steps 1.1–1.2 satisfy the multistep criterion in [F1] for every . Hence the closed left half-plane lies in the absolute-stability region, proving A-stability.
Depends on
Used by
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Dependency tree · two levels
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Sources
- Ernst Hairer and Gerhard Wanner, Introduction à l’Analyse Numérique, Chapter III, §III.9 (standard reference, not scraped)
- Masayuki Yano, James Douglass Penn, George Konidaris, and Anthony T. Patera, Math, Numerics, and Programming (for Mechanical Engineers), Unit IV, Chapter 21 (standard reference, not scraped)