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Backward Euler is A-stable but not symplectic

Example

Backward Euler on the oscillator $y^{'} = A y$ , $A = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ , is A-stable but not symplectic for any $h > 0$ . Here A-stable means its scalar stability function has modulus at most one for every $z$ with $Re z \leq 0$ .

Facts & Assumptions

Given: The backward Euler equation $y^{+} = y + h A y^{+}$ , $h > 0$ , and the canonical matrix $J = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ .

[F1]

A one-step matrix on this scalar test equation has stability function $R (z)$ defined by $y^{+} = R (h λ) y$ when $y^{'} = λ y$ ; A-stability is the modulus condition in the Example.

[F2]

In dimension two, $S^{T} J S = (\det S) J$ , by direct multiplication.

Verification

1.1givenF1algebra

For $y^{'} = λ y$ , backward Euler gives $(1 - h λ) y^{+} = y$ , so $R (z) = 1 / (1 - z)$ wherever defined. If $Re z \leq 0$ , then $∣ 1 - z ∣^{2} = 1 - 2 Re z + ∣ z ∣^{2} \geq 1$ , whence $∣ R (z) ∣ \leq 1$ and the method is A-stable.

2.1F2algebra∎

For the oscillator, $I - h A = (\begin{array}{cc} 1 & - h \\ h & 1 \end{array})$ , so $S_{B E} = (I - h A)^{- 1} = \frac{1}{1 + h^{2}} (\begin{array}{cc} 1 & h \\ - h & 1 \end{array}), \det S_{B E} = \frac{1}{1 + h^{2}} < 1.$ By [F2], $S_{B E}^{T} J S_{B E} = (\det S_{B E}) J \neq J$ . It is therefore not symplectic.

Depends on

Symplectic form and symplectic manifold

Used by

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Dependency tree · two levels

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Sources

Ernst Hairer, Geometric Numerical Integration, TU München Lecture 2: Symplectic integrators (standard reference, not scraped)
Nikolaj Nordkvist and Poul G. Hjorth, Classical Mechanics and Symplectic Integration (standard reference, not scraped)