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Forward Euler has the closed unit disk centered at −1 as its absolute-stability region

Statement

For forward Euler applied to $y^{'} = λ y$ , let $z = h λ$ . Its stability function is $R_{E} (z) = 1 + z$ , and its absolute-stability region is exactly the closed disk

${z \in C : ∣ z + 1 ∣ \leq 1} .$

On the negative real axis, for $y^{'} = - ρ y$ with $ρ > 0$ , the method is absolutely stable exactly when $0 < h ρ \leq 2$ .

Facts & Assumptions

Given: Forward Euler's scalar test equation, a step size $h > 0$ , and $λ \in C$ .

[F1]

Scalar absolute stability, A-stability, L-stability, and multistep root regions: the one-step region is the set of defined $z$ for which $∣ R (z) ∣ \leq 1$ .

[F2]

Orders and scalar stability functions of Euler, midpoint, and classical RK4: the forward-Euler stability function is $R_{E} (z) = 1 + z$ .

Proof

1.1givenF1F2

By [F2], the forward-Euler multiplier is defined for every $z \in C$ and equals $1 + z$ . Thus [F1] makes its absolute-stability condition $∣ 1 + z ∣ \leq 1$ , which is the closed disk centered at $- 1$ with radius one.

2.1givenstep 1.1algebra∎

For $λ = - ρ$ with $ρ > 0$ , $z = - h ρ$ is real and the disk condition is $∣ 1 - h ρ ∣ \leq 1$ . This is equivalent to $- 1 \leq 1 - h ρ \leq 1$ , hence $0 \leq h ρ \leq 2$ ; since $h ρ > 0$ , it is exactly $0 < h ρ \leq 2$ .

Depends on

Used by

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Sources

Ernst Hairer and Gerhard Wanner, Introduction à l’Analyse Numérique, Chapter III, §III.9 (standard reference, not scraped)
MIT OpenCourseWare, 18.330 Introduction to Numerical Analysis, Chapter 5, §5.2.2: Stability (standard reference, not scraped)