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Newton can exhaust its budget although the dissipative backward Euler root is unique

Facts & Assumptions

Given: The scalar field $f (x) = - x^{3}$ , step $h = 1$ , input $y_{0} = 1$ , initial Newton iterate $x_{0} = 0$ , residual tolerance $ε = 0.1$ , and correction budget $N = 0$ .

[F1]

Backward Euler has a globally defined nonexpansive resolvent (contractive when strongly dissipative): if $f : R \to R$ is continuous and $⟨ f (x) - f (y), x - y ⟩ \leq 0$ on $R$ , the BE equation has exactly one root for every input and every $h > 0$ .

[F2]

Finite-budget Newton and chord solves with scheduled factorization reuse: the finite solve tests the current residual; if it exceeds tolerance when the correction budget is exhausted, it returns iteration-budget-exhausted with the last valid iterate and residual.

[F3]

Backward Euler step equation and algebraic residual: the BE residual is $G (z) = z - y_{0} - h f (z)$ .

Proof

1.1F1algebra

For all $x, y \in R$ , $(f (x) - f (y)) (x - y) = - (x - y)^{2} (x^{2} + x y + y^{2}) \leq 0,$ because $x^{2} + x y + y^{2} = (x + y / 2)^{2} + 3 y^{2} / 4 \geq 0$ . This polynomial field is continuous on $R$ and is globally dissipative with $μ = 0$ .

2.1F1F3step 1.1algebra

By [F1], the equation $z = 1 - z^{3}$ , equivalently $z + z^{3} = 1$ , has exactly one real root. At the supplied initial iterate $x_{0} = 0$ , [F3] gives $G (0) = 0 - 1 - (- 0) = - 1$ , so $∥ G (0) ∥ = 1 > 0.1$ .

3.1F2step 2.1cases∎

Since the current residual exceeds tolerance and the correction budget is already zero, the finite solver returns iteration-budget-exhausted, despite the unique BE root proved in step 2.1. No forward-error statement follows from this status.

Depends on

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Sources

Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, §4.5.2 Newton for nonlinear systems (standard reference, not scraped)
Ernst Hairer, Introduction à l’Analyse Numérique, Chapter III (standard reference, not scraped)