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A dissipative backward Euler solve with a controlled update residual

Facts & Assumptions

Given: The scalar field $f (x) = - x - x^{3}$ , input ${\tilde{u}}_{0} = u_{0} = 1$ , step $h = 2$ , and approximate endpoint ${\tilde{u}}_{1} = 1 / 3$ .

[F1]

Backward Euler has a globally defined nonexpansive resolvent (contractive when strongly dissipative): strong dissipativity with constant $μ$ makes the BE resolvent globally defined and $1 / (1 + h μ)$ -Lipschitz.

[F2]

Backward Euler propagates unscaled algebraic update residuals: for one BE update from the exact input, endpoint error from an approximate update is at most $q ∣ η ∣$ , where $q = (1 + h μ)^{- 1}$ and $η$ is its unscaled update residual.

[F3]

Backward Euler step equation and algebraic residual: the BE root satisfies $z = u_{0} + h f (z)$ and its update residual is $z - u_{0} - h f (z)$ .

Proof

1.1F1F3algebra

The scalar field is strongly dissipative with $μ = 1$ , since for all $x, y$ , $(f (x) - f (y)) (x - y) = - (x - y)^{2} (1 + x^{2} + x y + y^{2}) \leq - (x - y)^{2},$ where $x^{2} + x y + y^{2} \geq 0$ . Hence [F1] gives a unique BE root on all of $R$ . The root equation is $z + 2 z + 2 z^{3} = 1$ , or $3 z + 2 z^{3} = 1$ .

2.1F2step 1.1algebra∎

At ${\tilde{u}}_{1} = 1 / 3$ , the left side of the root equation is $1 + 2 / 27$ , so the update residual is $η_{0} = 2 / 27$ and the derivative residual is $η_{0} / h = 1 / 27$ . Here $q = 1 / (1 + 2) = 1 / 3$ . Applying [F2] gives $∣ {\tilde{u}}_{1} - u_{1} ∣ \leq q ∣ η_{0} ∣ = \frac{1}{3} \frac{2}{27} = \frac{2}{81} .$

Depends on

Used by

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Sources

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