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One-pass PECE and fully implicit Adams have different stability

Example

For one known derivative sample, PECE uses the explicit Euler predictor and one trapezoidal Adams corrector evaluation. On the scalar test equation $y^{'} = λ y$ , its one-step multiplier is $R_{P E C E} (z) = 1 + z + z^{2} / 2$ , whereas the fully implicit trapezoidal Adams method has multiplier $R_{A M} (z) = (1 + z / 2) / (1 - z / 2)$ , where $z = h λ$ . At $z = - 3$ , one pass is unstable ( $∣ R_{P E C E} ∣ = 5 / 2$ ) while the fully implicit method is stable ( $∣ R_{A M} ∣ = 1 / 5$ ).

Facts & Assumptions

Given: The one-pass procedure Variable-step Adams predictor-evaluate-correct-evaluate procedure and its Lagrange predictor/corrector weights Variable-step Adams weights are integrals of the actual-node Lagrange bases.

[F1]

Variable-step Adams predictor-evaluate-correct-evaluate procedure performs one explicit predictor, evaluates at the predicted endpoint, applies one corrector, then evaluates at the corrected endpoint.

[L1]

Variable-step Adams weights are integrals of the actual-node Lagrange bases gives the $k = 1$ predictor weight $h$ and the two-node corrector weights $h / 2, h / 2$ on a uniform grid.

Verification

1.1F1L1algebra

Set $z = h λ$ and $Y_{n} = 1$ . By [F1] and [L1], the Euler predictor is $P = 1 + z$ . The PECE correction is $Y_{n + 1} = 1 + \frac{z}{2} (1 + P) = 1 + z + z^{2} / 2$ ; the final evaluation in PECE stores $λ Y_{n + 1}$ and does not alter this value.

1.2L1algebra

The fully implicit trapezoidal corrector instead solves $Y_{n + 1} = 1 + \frac{z}{2} (1 + Y_{n + 1})$ , so $(1 - z / 2) Y_{n + 1} = 1 + z / 2$ and its multiplier is $(1 + z / 2) / (1 - z / 2)$ when $z \neq 2$ .

2.1step 1.1step 1.2algebra∎

At $z = - 3$ , the multipliers are $5 / 2$ and $- 1 / 5$ , respectively. Define scalar test stability at $z$ by the condition that the multiplier's modulus is at most one. Hence PECE is unstable there and the fully implicit Adams method is stable there, so their stability differs.

Depends on

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