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BDF2 roots on or outside the unit disk map to the closed right half-plane

Statement

On a uniform grid, the variable-step BDF2 algorithm of Variable-step BDF2 with an explicit local implicit-branch rule has the scalar test-equation polynomial

$P_{z} (ζ) = (3 - 2 z) ζ^{2} - 4 ζ + 1, z = h λ .$

If $ζ$ is a root and $∣ ζ ∣ \geq 1$ , then $Re z \geq 0$ . Equality is possible only for $z = 0$ and $ζ = 1$ .

Facts & Assumptions

Given: A uniform step $h > 0$ , the scalar test equation $y^{'} = λ y$ , and a root $ζ$ of the BDF2 test polynomial.

[F1]

Variable-step BDF2 with an explicit local implicit-branch rule: when adjacent step ratios equal one, the update is $(3 Y_{n + 1} - 4 Y_{n} + Y_{n - 1}) / (2 h) = f (t_{n + 1}, Y_{n + 1})$ .

[F2]

Linear multistep methods, consistency, order, and the root condition: a linear multistep method has characteristic polynomials $ρ$ and $σ$ , and substitution of $f (t, y) = λ y$ gives the characteristic polynomial $ρ (ζ) - z σ (ζ)$ .

[F3]

Scalar absolute stability, A-stability, L-stability, and multistep root regions: the test variable is $z = h λ$ , and the multistep test polynomial is $π_{z} (ζ) = ρ (ζ) - z σ (ζ)$ .

[F4]

Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive: complex modulus is nonnegative and definite, $∣ u v ∣ = ∣ u ∣ ∣ v$ , and $u \overline{u} = ∣ u ∣^{2}$ .

[F5]

Complex modulus restricts to real absolute value: the complex modulus of the embedded real number $1$ is $1$ .

Proof

1.1F1F2F3algebra

Setting $r_{n} = 1$ in [F1] gives $3 Y_{n + 1} - 4 Y_{n} + Y_{n - 1} = 2 h f (t_{n + 1}, Y_{n + 1})$ . Substituting $f (t, y) = λ y$ and a trial mode $Y_{n} = ζ^{n}$ , the polynomial convention in [F2] gives $(3 - 2 z) ζ^{2} - 4 ζ + 1 = 0$ .

2.1F4F5step 1.1algebra

The constant term is one, so $ζ \neq 0$ . Put $w = 1 / ζ$ , $a = Re w$ , and $r = ∣ w ∣$ . By [F4] and [F5], $1 = ∣ 1 ∣ = ∣ ζ w ∣ = ∣ ζ ∣ ∣ w ∣$ , so $r = 1 / ∣ ζ ∣$ . Solving the equation in step 1.1 for $z$ gives $z = 3 / 2 - 2 w + w^{2} / 2$ . Writing $w = a + b i$ , [F4] gives $r^{2} = a^{2} + b^{2}$ and hence $Re (w^{2}) = a^{2} - b^{2} = 2 a^{2} - r^{2}$ ; therefore $Re z = (1 - a)^{2} + (1 - r^{2}) / 2$ .

3.1F4step 2.1algebra∎

If $∣ ζ ∣ > 1$ , then step 2.1 gives $r < 1$ , hence $Re z > 0$ . If $∣ ζ ∣ = 1$ , then $r = 1$ , so $Re z = (1 - a)^{2} \geq 0$ ; equality forces $a = 1$ . Writing $w = 1 + b i$ , [F4] gives $1 = r^{2} = 1 + b^{2}$ , so $b = 0$ and $w = 1$ , hence $ζ = 1$ and $z = 0$ . This proves both assertions.

Depends on

Used by

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)