TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Backward Euler has a globally defined nonexpansive resolvent (contractive when strongly dissipative)

Statement

Let $d \geq 1$ , let $f : R \times R^{d} \to R^{d}$ be continuous, and suppose for every fixed $t$ and all $x, y \in R^{d}$ ,

$⟨ f (t, x) - f (t, y), x - y ⟩ \leq - μ ∥ x - y ∥_{2}^{2}$

for some $μ \geq 0$ . For every $h > 0$ , define $H_{t, h} (x) = x - h f (t, x)$ . Then $H_{t, h}$ is bijective and its inverse $R_{t, h} = H_{t, h}^{- 1}$ , the backward Euler step map, satisfies

$∥ R_{t, h} (u) - R_{t, h} (v) ∥_{2} \leq \frac{1}{1 + h μ} ∥ u - v ∥_{2} .$

In particular, for each input state $u$ , the equation $z = u + h f (t, z)$ has exactly one solution on all of $R^{d}$ . No $h L < 1$ contraction assumption is needed for this dissipative branch.

Facts & Assumptions

Given: $d \geq 1$ , $h > 0$ , $μ \geq 0$ , a continuous $f$ on the full space, and the dissipativity inequality in the Statement.

[F1]

Backward Euler step equation and algebraic residual: the backward Euler endpoint $z$ solves $z = u + h f (t, z)$ , equivalently $H_{t, h} (z) = u$ .

[F2]

A continuous strongly monotone map of finite-dimensional space is bijective: a continuous map $H : R^{d} \to R^{d}$ satisfying $⟨ H (x) - H (y), x - y ⟩ \geq κ ∥ x - y ∥_{2}^{2}$ for $κ > 0$ is bijective and its inverse is $κ^{- 1}$ -Lipschitz.

[F3]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $∥ x ∥_{2}^{2} = ⟨ x, x ⟩$ for the Euclidean inner product.

[A1]

The vector field is continuous on all of $R \times R^{d}$ , and the same $μ$ in the stated dissipativity inequality works for every time under consideration.

Proof

1.1A1F1F3algebra

For fixed $t, h$ and $x, y \in R^{d}$ , [F3] and the dissipativity assumption give $⟨ H_{t, h} (x) - H_{t, h} (y), x - y ⟩ = ∥ x - y ∥_{2}^{2} - h ⟨ f (t, x) - f (t, y), x - y ⟩ \geq (1 + h μ) ∥ x - y ∥_{2}^{2}$ . The constant $1 + h μ$ is positive, and $H_{t, h}$ is continuous by [A1].

2.1F1F2step 1.1∎

Apply [F2] with $κ = 1 + h μ$ . It gives a unique inverse image $z = R_{t, h} (u)$ for every $u$ , and the inverse Lipschitz bound in the Statement. By [F1], this inverse image is exactly the backward Euler root.

Depends on

Used by

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Sources

Ernst Hairer, 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)