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Autonomous Rosenbrock–Euler is one Newton correction of backward Euler

Statement

Let $d \geq 1$ and let $f : U \to R^{d}$ be autonomous and $C^{1}$ on an open $U \subseteq R^{d}$ , let $y \in U$ , and let $h > 0$ . Suppose $I_{d} - h D f (y)$ is invertible. Then the exact-arithmetic Rosenbrock–Euler step with $γ = 1$ is the first Newton correction of the backward-Euler residual at the initial guess $y$ . This does not assert that the corrected point solves the full backward-Euler equation.

Facts & Assumptions

Given: The autonomous field $f$ , state $y$ , step $h$ , and the invertible shifted Jacobian.

[F1]

Backward Euler step equation and algebraic residual: for an autonomous field, the backward-Euler residual at candidate $z$ is $G (z) = z - y - h f (z)$ .

[F2]

Continuously differentiable maps, local inverses, and local diffeomorphisms: $C^{1}$ means totally differentiable at every point, with continuous total derivative.

[F3]

The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case: the Jacobian matrix has entries given by the state partial derivatives.

[F4]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : the derivative of a composition is the composition of total derivatives.

[F5]

A total derivative computes every directional derivative, and its matrix is the Jacobian: when a total derivative exists, its matrix is the Jacobian matrix.

[F6]

Newton correction for a square vector residual: for residual vector $r$ and invertible matrix $A$ , the correction is the unique $s$ satisfying $A s = - r$ .

[F7]

The Rosenbrock–Euler step map with an exact or approximate Jacobian: at $γ = 1$ , $k$ solves $(I_{d} - h D f (y)) k = f (y)$ and the update is $y + h k$ .

Proof

Given: The data in the Statement.

1.1F1F2F3F4F5algebra

From [F1], $G (y) = - h f (y)$ . Since [F2] makes $f$ totally differentiable, [F4] differentiates $G$ at $y$ to $D G (y) = I_{d} - h D f (y)$ ; [F3] and [F5] identify the state derivative matrix with the Jacobian.

2.1F6step 1.1F7algebra

By [F6], the Newton correction $s$ solves $(I_{d} - h D f (y)) s = - G (y) = h f (y)$ . Since the matrix is invertible, uniqueness gives $s = h k$ , where $k$ is the stage slope in [F7].

3.1step 2.1F1F7algebra∎

The corrected point is therefore $y + s = y + h k$ , exactly the update in [F7]. It is one correction from $y$ ; the calculation establishes no further Newton correction and makes no claim that this point solves the backward-Euler equation.

Depends on

Used by

Rosenbrock–Euler damps a stiff scalar decay with the backward-Euler factor Example

Dependency tree · two levels

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Sources

WIAS Berlin, Numerical Methods for Stiff Ordinary Differential Equations, §2.4 Rosenbrock Methods (standard reference, not scraped)
Daniel Kressner, Advanced Numerical Analysis, §3.5 Rosenbrock methods (standard reference, not scraped)