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Backward Euler propagates unscaled algebraic update residuals

Statement

Assume the hypotheses of Backward Euler has a globally defined nonexpansive resolvent (contractive when strongly dissipative) with a common $μ \geq 0$ , let $N \in N_{0}$ , and use a constant $h > 0$ grid $t_{n} = t_{0} + n h$ for $0 \leq n \leq N$ . Let $u_{0} \in R^{d}$ and let $u_{n + 1}$ be the exact backward Euler sequence, $u_{n + 1} = R_{t_{n + 1}, h} (u_{n})$ . For any approximate sequence ${\tilde{u}}_{n} \in R^{d}$ , define its update residual $η_{n} = {\tilde{u}}_{n + 1} - {\tilde{u}}_{n} - h f (t_{n + 1}, {\tilde{u}}_{n + 1}) .$ With $q = (1 + h μ)^{- 1} \leq 1$ , for $0 \leq n \leq N$ , $∥ {\tilde{u}}_{N} - u_{N} ∥_{2} \leq q^{N} ∥ {\tilde{u}}_{0} - u_{0} ∥_{2} + \sum_{j = 0}^{N - 1} q^{N - j} ∥ η_{j} ∥_{2} .$ This bound uses the unscaled update residual. If $r_{n} = η_{n} / h$ is the derivative residual, its contribution is $h q^{N - j} ∥ r_{j} ∥_{2}$ . The result compares to the exact discrete BE sequence, not directly to the continuous solution.

Facts & Assumptions

Given: A continuous globally dissipative field with constant $μ \geq 0$ , a constant positive step $h$ , the exact BE sequence, and approximate endpoints as in the statement.

[F1]

Backward Euler has a globally defined nonexpansive resolvent (contractive when strongly dissipative): the BE resolvent $R_{t, h}$ is defined on all $R^{d}$ and satisfies $∥ R_{t, h} (a) - R_{t, h} (b) ∥_{2} \leq q ∥ a - b ∥_{2}$ , where $q = (1 + h μ)^{- 1}$ .

[F2]

Implicit residuals and their update and derivative units: the BE update residual is endpoint minus input minus $h f$ ; the derivative residual is this vector divided by $h$ .

[F3]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: the Euclidean norm satisfies the triangle inequality.

[A1]

The same dissipativity constant works at every grid time and $f$ is continuous on the full state space, so each exact BE endpoint in the recurrence exists uniquely.

Proof

1.1A1F1F2algebra

Write $H_{n + 1} (z) = z - h f (t_{n + 1}, z)$ . By [F2], $H_{n + 1} ({\tilde{u}}_{n + 1}) = {\tilde{u}}_{n} + η_{n}$ , while the exact update satisfies $H_{n + 1} (u_{n + 1}) = u_{n}$ .

2.1F1F3step 1.1algebra

Apply the inverse Lipschitz estimate [F1] to these two right-hand sides. With $e_{n} = ∥ {\tilde{u}}_{n} - u_{n} ∥_{2}$ , it gives $e_{n + 1} \leq q ∥ {\tilde{u}}_{n} + η_{n} - u_{n} ∥_{2} \leq q e_{n} + q ∥ η_{n} ∥_{2}$ by the Euclidean triangle inequality [F3].

3.1F2step 2.1inductionalgebra∎

Induction in the recurrence of step 2.1 gives $e_{N} \leq q^{N} e_{0} + \sum_{j = 0}^{N - 1} q^{N - j} ∥ η_{j} ∥_{2}$ : for $N = 0$ this is equality, and multiplying the bound at $N$ by $q$ and appending $q ∥ η_{N} ∥_{2}$ proves the next case. The derivative-residual form follows from $η_{j} = h r_{j}$ in [F2].

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Sources

Ernst Hairer and Gerhard Wanner, Introduction à l’Analyse Numérique, Chapter III: Équations Différentielles Ordinaires (standard reference, not scraped)
Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, Chapter 11: Stiffness (standard reference, not scraped)