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Unrestricted variable-step ratios amplify a vanishing BDF2 starter error

Statement refuted

For every family of variable-step grids with maximum step tending to zero, applying BDF2 to $y^{'} = 0$ with any starter errors tending to zero produces maximum grid error tending to zero, without any bound on adjacent step ratios.

Facts & Assumptions

Given: The variable-step BDF2 procedure Variable-step BDF2 with an explicit local implicit-branch rule applied to $y^{'} = 0$ , and its homogeneous increment recurrence The variable-step BDF2 parasitic increment contracts below the sharp ratio threshold.

[L1]

The variable-step BDF2 parasitic increment contracts below the sharp ratio threshold gives $Δ_{n + 1} = \frac{r_{n}^{2}}{1 + 2 r_{n}} Δ_{n}$ for $f = 0$ and every positive ratio.

Counterexample

For each integer $m \geq 2$ , take $T = 1$ and grid increments

$h_{0} = m^{- 2}, h_{1} = m^{- 1}, h_{2} = \dots = h_{m - 1} = m^{- 1}, h_{m} = m^{- 1} - m^{- 2} .$

They sum to one and have maximum $H_{m} = 1 / m \to 0$ , but the first adjacent ratio is $r_{1} = m$ . For the exact solution $y = 0$ , use starters $Y_{0} = 0$ , $Y_{1} = 1 / m$ , whose error tends to zero.

For the branch input in Variable-step BDF2 with an explicit local implicit-branch rule, prescribe every ball to be the closed radius-two ball centered at zero. Supply the returning exact branch oracle that computes $Z = ((1 + r)^{2} Y_{n} - r^{2} Y_{n - 1}) / (1 + 2 r)$ and returns unique-root with $Z$ if $Z$ is in the prescribed ball, and no-root otherwise. For $f = 0$ the field is defined on every ball and the positive coefficient $1 + 2 r$ excludes multiple roots.

1.1givenalgebra

Summing the increments gives $m^{- 2} + m^{- 1} + (m - 2) m^{- 1} + m^{- 1} - m^{- 2} = 1$ , and every increment is positive for $m \geq 2$ . The maximum is $H_{m} = m^{- 1}$ .

1.2givenalgebra

The first ratio is $r_{1} = h_{1} / h_{0} = m$ and the starter increment is $Δ_{1} = 1 / m \to 0$ .

2.1L1step 1.1step 1.2algebra

For $n \geq 2$ the grid has $r_{n} \leq 1$ , so [L1] gives $0 < q (r_{n}) = r_{n}^{2} / (1 + 2 r_{n}) \leq r_{n} / 2 \leq 1 / 2$ . The increments from $Δ_{2}$ onward are positive and their total is at most $Δ_{2} \sum_{j \geq 0} 2^{- j} = 2 Δ_{2} < 1$ . Since $Y_{1} = 1 / m \leq 1 / 2$ , every computed value is less than $3 / 2$ . For $f = 0$ the step equation has exactly one root because its coefficient $1 + 2 r_{n}$ is positive, so all roots lie in the prescribed balls and the algorithm completes.

2.2L1step 1.2algebra

By [L1], $Δ_{2} = \frac{m^{2}}{1 + 2 m} \frac{1}{m} = \frac{m}{1 + 2 m} \to \frac{1}{2}$ . Therefore $Y_{2} = Y_{1} + Δ_{2} \to \frac{1}{2}$ , while $t_{2} = h_{0} + h_{1} \to 0$ and the exact solution remains zero.

3.1step 1.1step 2.1step 2.2∎

Thus a vanishing starter perturbation produces an error bounded away from zero even though $H_{m} \to 0$ . The claimed ratio-free uniform zero-stability fails.

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