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Symplectic Isospectral Runge--Kutta Methods as Lie group methods

Paolo Cifani, Klas Modin, Cecilia Pagliantini, Milo Viviani

Abstract

We compare three approaches for structure preserving numerical integration of isospectral flows on quadratic Lie algebras. Such flows originate from Hamiltonian dynamics on the cotangent bundle of the Lie group. It is known, via discrete reduction theory, that symplectic Runge--Kutta methods applied to the cotangent bundle formulation induce isospectral symplectic Runge--Kutta (ISOSYRK) schemes on the Lie algebra. Here, we show that the same symplectic Runge--Kutta method, but applied to the transport formulation of the flow on the Lie group, is equivalent to the corresponding ISOSYRK scheme. We also give numerical results suggesting that the formulation on the Lie group is more efficient for schemes with two or more intermediate stages.

Symplectic Isospectral Runge--Kutta Methods as Lie group methods

Abstract

We compare three approaches for structure preserving numerical integration of isospectral flows on quadratic Lie algebras. Such flows originate from Hamiltonian dynamics on the cotangent bundle of the Lie group. It is known, via discrete reduction theory, that symplectic Runge--Kutta methods applied to the cotangent bundle formulation induce isospectral symplectic Runge--Kutta (ISOSYRK) schemes on the Lie algebra. Here, we show that the same symplectic Runge--Kutta method, but applied to the transport formulation of the flow on the Lie group, is equivalent to the corresponding ISOSYRK scheme. We also give numerical results suggesting that the formulation on the Lie group is more efficient for schemes with two or more intermediate stages.

Paper Structure

This paper contains 3 sections, 2 theorems, 10 equations, 1 table.

Table of Contents

  1. Equivalent formulations of isospectral flows on quadratic Lie algebras
  2. Isospectral numerical integration
  3. Numerical tests

Key Result

Lemma 1.1

Let $\mathfrak{g}$ be a Lie subalgebra of $\mathfrak{gl}(n,\mathbb{C})$ such that there exists a matrix $J$ for which Then $\mathfrak{g} = \mathfrak{g}^\dagger$ implies $J^2\in\mathfrak{c}(\mathfrak{g}):=\lbrace A\mid[A,\mathfrak{g}]=0\rbrace$. Moreover, if $J$ is invertible and $J^2\in\mathfrak{c}(\mathfrak{g})$, then $\mathfrak{g} = \mathfrak{g}^\dagger$. $\blacktriangleleft$$\blacktriangleleft

Theorems & Definitions (3)

  • Lemma 1.1
  • Theorem 2.1
  • Remark 2.2