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Geometric numerical integration — Examples

1 · Prerequisites

2 · Summary

Summary

These computations verify the fixed kick-then-drift convention for symplectic Euler, exhibit the exactly preserved positive modified quadratic form for Störmer–Verlet when $∣ h ∣ < 2$ , and give backward Euler as an A-stable but non-symplectic method. The identities are derived from the actual step matrices. The Verlet estimate is specific to the harmonic oscillator and does not substitute for the finite-order theorem on the A page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Kick-then-drift symplectic Euler on the harmonic oscillator

Example

For $H (q, p) = \frac{1}{2} (q^{2} + p^{2})$ on $R^{2}$ and $h > 0$ , the kick-then-drift symplectic Euler step is $(\begin{array}{c} q^{+} \\ p^{+} \end{array}) = (\begin{array}{cc} 1 - h^{2} & h \\ - h & 1 \end{array}) (\begin{array}{c} q \\ p \end{array}) .$ The same matrix formula is its algebraic signed-step extension, and that extension is symplectic for every real $h$ . The original Hamiltonian need not be exactly conserved.

Facts & Assumptions

Given: $V (q) = q^{2} / 2$ , $M = 1$ , the local kick-then-drift algorithm, and $J = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ .

[F1]

The method applies $p^{+} = p - h V^{'} (q)$ and then $q^{+} = q + h p^{+}$ . Local kick-then-drift symplectic Euler step

[F2]

For $S = (\begin{array}{cc} a & b \\ c & d \end{array})$ , multiplication gives $S^{T} J S = (\begin{array}{cc} 0 & a d - b c \\ b c - a d & 0 \end{array}) = (\det S) J$ .

Verification

1.1givenF1algebra

Since $V^{'} (q) = q$ , [F1] gives $p^{+} = p - h q$ and $q^{+} = q + h (p - h q) = (1 - h^{2}) q + h p$ , which is the displayed matrix.

2.1F2step 1.1algebra

Its determinant is $(1 - h^{2}) \cdot 1 - h (- h) = 1$ . By [F2], $S^{T} J S = J$ , verifying symplecticity.

3.1step 1.1givenalgebra∎

For any $h > 0$ , starting from $(q, p) = (0, 1)$ gives $(q^{+}, p^{+}) = (h, 1)$ by step 1.1. Thus $H (q^{+}, p^{+}) - H (q, p) = h^{2} / 2 > 0$ , so the original Hamiltonian is not conserved by every step.

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Störmer–Verlet has bounded oscillatory energy error on the harmonic oscillator

Example

For $H (q, p) = \frac{1}{2} (q^{2} + p^{2})$ , use the real-parameter kick and drift shears of One-step leapfrog map as force and drift shears and define the signed kick-drift-kick step $S_{h} : = K_{h / 2} \circ D_{h} \circ K_{h / 2}$ for real $h$ ; for $h > 0$ this is the published one-step map, and for $h < 0$ it is its same-form signed-step extension. If $0 < ∣ h ∣ < 2$ , this step preserves the positive quadratic form ${\tilde{H}}_{h} (q, p) = \frac{1}{2} ((1 - h^{2} / 4) q^{2} + p^{2}) .$ Consequently its original-energy error is bounded uniformly over all step counts by $∣ H (q_{n}, p_{n}) - H (q_{0}, p_{0}) ∣ \leq \frac{h^{2}}{8} (\frac{2 {\tilde{H}}_{h} (q_{0}, p_{0})}{1 - h^{2} / 4} + q_{0}^{2}) .$

Facts & Assumptions

Given: The published kick-drift-kick map, $M = 1$ , $V (q) = q^{2} / 2$ , and $0 < ∣ h ∣ < 2$ .

[F1]

The force kicks and drifts are defined for real parameters, with formulas $K_{a} (q, p) = (q, p - a q)$ and $D_{b} (q, p) = (q + b p, p)$ ; their signed composition is $S_{h} = K_{h / 2} \circ D_{h} \circ K_{h / 2}$ . For $h > 0$ it is the one-step map in One-step leapfrog map as force and drift shears.

Verification

1.1F1givenalgebra

The first kick gives $p_{1 / 2} = p - h q / 2$ , the drift gives $q_{1} = q + h p_{1 / 2} = (1 - h^{2} / 2) q + h p$ , and the last kick gives $p_{1} = p_{1 / 2} - h q_{1} / 2 = - h (1 - h^{2} / 4) q + (1 - h^{2} / 2) p$ . Thus $S_{h} = (\begin{array}{cc} a & h \\ - h s & a \end{array}), a = 1 - h^{2} / 2, s = 1 - h^{2} / 4.$

2.1step 1.1algebra

Direct multiplication gives $S_{h}^{T} (\begin{array}{cc} s & 0 \\ 0 & 1 \end{array}) S_{h} = (\begin{array}{cc} s (a^{2} + h^{2} s) & 0 \\ 0 & a^{2} + h^{2} s \end{array}) = (\begin{array}{cc} s & 0 \\ 0 & 1 \end{array}),$ because $a^{2} + h^{2} s = 1$ . Hence ${\tilde{H}}_{h}$ is exactly invariant. Since $s > 0$ , invariance implies $q_{n}^{2} \leq 2 {\tilde{H}}_{h} (q_{0}, p_{0}) / s$ for every $n$ .

3.1step 2.1algebra∎

The identity ${\tilde{H}}_{h} = H - (h^{2} / 8) q^{2}$ gives $H (q_{n}, p_{n}) - H (q_{0}, p_{0}) = \frac{h^{2}}{8} (q_{n}^{2} - q_{0}^{2}) .$ Apply the bound from step 2.1 and the triangle inequality to obtain the stated uniform error estimate.

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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.

Geometric numerical integration — Examples

1 · Prerequisites

2 · Summary

Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Kick-then-drift symplectic Euler on the harmonic oscillator

Example

Facts & Assumptions

Verification

Störmer–Verlet has bounded oscillatory energy error on the harmonic oscillator

Example

Facts & Assumptions

Verification

Backward Euler is A-stable but not symplectic

Example

Facts & Assumptions

Verification

Sources