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

Statement

Consider $y^{'} = - 100 y$ with $y_{0} = 1$ and one step $h = 1$ . The named Rosenbrock–Euler method ( $γ = 1$ , exact Jacobian) returns $y_{1} = 1 / 101$ , exactly the backward-Euler value. Its scalar factor is $1 / (1 + 100)$ ; it lies in $(0, 1)$ and tends to zero as the decay parameter times the step tends to infinity. This is a scalar exact-arithmetic calculation and makes no finite-precision forward-error claim.

Facts & Assumptions

Given: $f (y) = - 100 y$ , $y_{0} = 1$ , $h = 1$ , and exact arithmetic.

[F1]

The Rosenbrock–Euler step map with an exact or approximate Jacobian: at $γ = 1$ , $(1 - h J) k = f (y)$ and $y_{+} = y + h k$ .

[F2]

The Rosenbrock–Euler scalar stability function is A-stable exactly for gamma at least one half: $R_{1} (z) = 1 / (1 - z)$ and the $γ = 1$ method is L-stable.

[F3]

Autonomous Rosenbrock–Euler is one Newton correction of backward Euler: for an autonomous field, the $γ = 1$ step is one Newton correction of the backward-Euler residual from the starting value.

[F4]

Backward Euler step equation and algebraic residual: the scalar endpoint equation is $z = y + h f (z)$ .

Proof

Given: The data in the Statement.

1.1givenF1algebra

Here $J = - 100$ and $z = h (- 100) = - 100$ . By [F1], $k = (- 100) / (1 + 100) = - 100 / 101$ , hence $y_{1} = 1 - 100 / 101 = 1 / 101$ .

2.1F3F4step 1.1algebra

The backward-Euler equation [F4] is $z = 1 - 100 z$ , so $101 z = 1$ and its unique scalar root is $z = 1 / 101$ . This agrees with step 1.1; [F3] also identifies the Rosenbrock update as one Newton correction, which is exact here because the residual is affine.

3.1F2givenalgebra∎

Since $z = - 100 < 0$ , the factor in [F2] is $1 / 101 \in (0, 1)$ . For general decay parameter $λ > 0$ , the factor is $1 / (1 + h λ)$ , which tends to zero as $h λ \to \infty$ .

Finite check

For $h λ = 100$ , the method factor and backward-Euler factor both equal $1 / 101$ ; explicit Euler would have factor $1 - 100 = - 99$ , which grows in magnitude rather than damping this mode.

Depends on

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Sources

WIAS Berlin, Numerical Methods for Stiff Ordinary Differential Equations, §2.3 (standard reference, not scraped)
Daniel Kressner, Advanced Numerical Analysis, §3.5 Rosenbrock methods (standard reference, not scraped)