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Constant-step BDF2 is A-stable under the multistep root condition

Statement

For each z=hλ with Re⁡z≤0, the constant-step BDF2 test polynomial Pz(ζ)=(3−2z)ζ2−4ζ+1 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 y′=λy with h>0 and Re⁡(hλ)≤0.

[F1]

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.

[F2]

BDF2 roots on or outside the unit disk map to the closed right half-plane: any root of Pz with modulus at least one forces Re⁡z≥0, with equality only at z=0,ζ=1.

Proof

1.1givenF1algebra

The leading coefficient of Pz is 3−2z. If it vanished, then z=3/2, contradicting Re⁡z≤0. Thus Pz has degree two.

1.2givenF2algebra

By [F2], a root with modulus greater than one would force Re⁡z>0, which is impossible. A root with modulus one can occur only when z=0 and ζ=1. At z=0, P0(ζ)=3ζ2−4ζ+1=(3ζ−1)(ζ−1), so the unit root is simple and the other root is 1/3.

2.1F1step 1.1step 1.2∎

The degree and root properties in steps 1.1–1.2 satisfy the multistep criterion in [F1] for every Re⁡z≤0. Hence the closed left half-plane lies in the absolute-stability region, proving A-stability.

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Sources