On Yoshida's Method for Constructing Explicit Symplectic Integrators For

faculty of science mathematics and applied and engineering mathematics

On Yoshida’s Method For Constructing Explicit Symplectic Integrators For Separable Hamiltonian Systems

Bachelor’s Project Mathematics

July 2019

Student: J.C. Pim

First supervisor: Dr. M. Seri

Second assessor: Dr. A.E. Sterk

Abstract We describe Yoshida’s method for separable Hamiltonians H = T (p)+ V (q). Hamiltonians of this form occur frequently in classical mechanics, for example the n-body problem, as well as in other ﬁelds. The class of symplectic integrators constructed are explicit, reversible, of arbitrary even order, and have bounded energy growth. We give an introduction to Hamiltonian mechanics, symplectic geometry, and Lie theory. We compare the performance of these integrators to more commonly used methods such as Runge-Kutta, using the ideal pendulum and Kepler problem as examples.

2 Contents

1 Introduction 4

2 Preliminaries And Prerequisites 5 2.1 Manifolds, Vector Fields, And Diﬀerential Forms ...... 5 2.2 The Matrix Exponential ...... 6 2.3 The Vector Field Exponential ...... 7

3 Hamiltonian Mechanics 9 3.1 Motivating Examples ...... 10

4 Symplectic Geometry 15 4.1 Symplectic Vector Spaces ...... 15 4.2 Symplectic Manifolds ...... 17

5 Lie Theory And BCH Formula 21

6 Symplectic Integrators 28 6.1 Separable Hamiltonians ...... 28 6.2 Reversible Hamiltonians ...... 29 6.3 Yoshida’s Method For Separable Hamiltonians ...... 30 6.4 Properties Of The Yoshida Symplectic Integrators ...... 38 6.5 Backward Error Analysis ...... 38

7 Numerical Simulations 43 7.1 The Ideal Pendulum ...... 43 7.2 The Kepler Problem ...... 46

8 Conclusion 50

References 51

3 1 Introduction

Many interesting phenomena in science can be modelled by Hamiltonian systems, for example the n-body problem, oscillators, problems in molecular dynamics, and models of electric circuits [11]. In general these are defined by systems of differential equations that cannot be solved analytically and very often numerical methods are needed in order to study their behaviour. This gives rise to some issues: many of these systems have invariants such as conservation of energy or angular momentum which numerical methods do not necessarily preserve, sometimes with drastic consequences on the evolution of the computed solution. In the case of chaotic systems, like the n-body problem, this is wors- ened by the fact that the perturbations introduced by numerical methods may lead to wildly different solutions over long time scales. This is also undesirable in many applications such as modelling the trajectory of spacecraft or other small bodies, Hamiltonian Monte-Carlo methods, and physics-based animation used in the production of video games, and movies [2]. A way to mitigate these problems is to use high order numerical methods with a sufficiently small step size, but this can be extremely computationally expensive. Also, techniques such as adaptive time-stepping may not preserve important features of the system’s dynamics. For example, in neuronal dynamics [3] the model’s limit cycles have an important role in neuronal spiking. However, these are not well preserved by commonly used “Euler-like” methods. Hence, we would like to have numerical methods which preserve at least some of the system’s invariants. We can do this by exploiting the underling geometry of Hamiltonian systems, called symplectic geometry. The natural space for the Hamiltonian dynamics, the so-called phase space, is endowed with a canonical differential 2-form, called the symplectic form, which the flow of the Hamilto- nian system preserves. This can be used to derive numerical methods, called symplectic integrators, which themselves preserve the symplectic form. In this bachelor project, we aim to describe Yoshida’s method for constructing explicit symplectic integrators for separable Hamiltonian systems. The class of integrators we construct are reversible, of arbitrary even order and have bounded energy growth. The construction makes use of Lie algebras and the Baker-Campbell-Hausdorff (BCH) formula. However, we will first describe the basic theory of Hamiltonian systems and their geometric properties using symplectic geometry. We will study some examples of Hamiltonian systems includ- ing the ideal pendulum and the Kepler problem, which will act as our main motivation for developing symplectic integrators. Then, we will give an introduction to Lie theory, before explaining Yoshida’s construction, and showing some of its important properties. This will include a backward error analysis to show that this class of integrators has bounded energy growth. Finally, we will implement and compare some integrators from this class to more commonly used methods, such as Runge-Kutta, by applying them to our motivating examples.

4 2 Preliminaries And Prerequisites

We will briefly describe and recall some of the theory used in later sections. We discuss some concepts from differential geometry such as manifolds, vector fields and differential forms, which will be useful later. We then review some facts about the matrix exponential, and give some intuition for the vector field exponential.

2.1 Manifolds, Vector Fields, And Differential Forms In later sections of this thesis we will make use of many concepts from differential geometry, especially in Section 4 and Section 5. The details about any of what follows, can be found in any book on differential geometry such as [14]. We will mainly use, though not discuss in detail: manifolds, smooth maps between manifolds, vector fields and differential forms on manifolds. In general, it will be sufficient to consider manifolds as open subsets of, or as surfaces in Rn. When we say smooth it is safe to read this as: of class C∞, though many results also hold for Ck with k ≥ 1. We will write x1, . . . , xm for the local coordinates on a smooth manifold M ∂ ∂ of dimension m. We use ∂x1 ,..., ∂xm as the basis of the tangent space TpM 1 m ∗ for p ∈ M, and dx ,..., dx as the basis of the cotangent space Tp M. We denote the differential of a smooth map F : M → N between M and another manifold N, as F∗ : TpM → TF (p)N at p ∈ M. For a tangent vector Xp ∈ TpM we define it pointwise as F∗(Xp)f = Xp(f ◦ F ) where f is a smooth function on M. It can also be computed using a smooth curve c(t) on M with c(0) = p 0 d and c (0) = Xp as F∗(Xp) = dt t=0(F ◦ c)(t). Pm ∂ A vector field X = i=1 ai ∂xi is called smooth if its coefficients ai are smooth functions on M. Smooth vector fields have a flow Φ: I × M → M, where I is an interval. Definition 1 A smooth vector field on a manifold is called complete if its flow is defined for all time.

That is, if the vector field’s flow Φt has I = R. We will often make use of complete vector fields as they simplify the theory in places. We interpret vector fields as differential operators, and thus they can be applied to smooth functions. Pm i Similarly a differential 1-form or covector field on M is given by η = i=1 aidx and is again called smooth if the ai are smooth. Differential k-forms for k > 1 can be written as a sum of wedge products of the basis 1-forms: dx1,..., dxm, 1 2 2 3 for example dx ∧dx +5dx ∧dx . The 0-forms are functions on M. We use ηp ∗ for p ∈ M to denote it as a covector in Tp M. We can compute the application 1 k of a k-form dx ∧ · · · ∧ dx to k tangent vectors v1, . . . , vk as  1 1  dx (v1) ··· dx (vk) 2 2 dx (v1) ··· dx (vk) dx1 ∧ · · · ∧ dxk(v , . . . , v ) = det   . 1 k  .   .  k k dx (v1) ··· dx (vk)

5 The exterior derivative d transforms k-forms into (k+1)-forms. The exterior ∂f 1 ∂f m derivative of a smooth function f is given by df = ∂x1 dx + ··· + ∂xm dx . Since covectors “consume” tangent vectors, we can see that the diﬀerential f∗ and exterior derivative df of a function f are in fact the same. A k-form η is called closed if dη = 0, while it is called exact if there exists a (k − 1)-form µ s.t. dµ = η. Finally, the pullback of a k-form η at p ∈ M by a smooth function ∗ F is given by F ηp(v1, . . . , vk) = ηF (p)(F (v1),...,F (vn)) for v1, . . . , vn ∈ TpM. We will also make use of the following results. Theorem 1 (Regular Level Set Theorem[14, Theorem 9.9, pg. 105]) Let N and M be smooth manifolds of dimension n and m respectively. Let F : N → M be a smooth map. If c ∈ M is a regular value of F s.t. F −1(c) 6= ∅, then F −1(c) is a regular submanifold of N of dimension n − m.

Theorem 2 ([14, Theorem 11.15, pg. 124]) Let f : N → M be a smooth map of manifolds, and let f(N) ⊂ S ⊂ M. If S is a regular submanifold of M, then the map i ◦ f = f˜: N → S induced by the inclusion map i: S → M, is smooth.

Theorem 3 ([14, Proposition 14.3, pg. 151]) A vector ﬁeld X on a manifold M is smooth if and only Xf ∈ C∞(M) for all f ∈ C∞(M).

2.2 The Matrix Exponential We will brieﬂy review some facts about the matrix exponential which will be useful when we discuss Lie theory and the BCH formula. Proofs of these facts can be found in books on ordinary diﬀerential equations such as [1, Chapter 3] or [14, Section 15.3 and 15.4].

Deﬁnition 2 For X ∈ Rn×n the matrix exponential eX or exp(X) is deﬁned by: 1 1 eX = I + X + X2 + X3 + .... 2! 3!

It is well defined as the series converges absolutely for all X ∈ Rn×n. Theorem 4 For X,Y ∈ Rn×n, if X and Y commute, then eX eY = eX+Y . It’s natural to wonder whether the reverse of this statement holds, that is can we find W s.t. eX eY = eW . This is answered by the BCH formula (Theorem 17). Theorem 5 The trace of a matrix satisfies:

• For X,Y ∈ Rn×n, tr(XY ) = tr(YX). n×n −1 • For X,Y ∈ R s.t. det Y 6= 0, tr(X) = tr YXY . • For X ∈ Rn×n, det eX = etr X .

n×n d tX tX tX Theorem 6 For X ∈ R , dt e = Xe = e X.

6 This is interesting as for real matrices X, tr X ∈ R, and ∀x ∈ R, ex ≥ 0, hence by the above theorem det eX 6= 0. We fruitfully exploit this method of constructing non-singular matrices from arbitrary real matrices, as it allows us to explicitly construct curves in GL(n, R). Using these we can more easily compute the diﬀerentials of smooth maps on GL(n, R), which will be use in Section 5. Consider the curve

tX n×n c: R → GL(n, R); t 7→ Ae , for X ∈ R , and A ∈ GL(n, R).

It has initial point c(0) = A ∈ GL(n, R), and initial “velocity” d c0(0) = etX A| = AX ∈ n×n = T GL(n, ). dt t=0 R A R To show the utility of such curves, we compute the the diﬀerential of the determinant map, det: GL(n, R) → R at A ∈ GL(n, R). GL(n, R) is an open n×n n×n subset of R , thus we can identify TAGL(n, R) with R , see Example 5. −1 Consider the map c(t) = AetA X , for X ∈ Rn×n. By the above c(0) = A and c0(0) = AA−1X = X. Then

d tA−1X d tA−1X det∗(X) = det Ae = det(A) det e dt t=0 dt t=0 d −1 = det(A) et tr(A X) dt t=0 = det(A) trA−1X.

2.3 The Vector Field Exponential We will give a brief introduction to the vector field exponential. We will show that it is a generalization of the matrix exponential and solves the differential equation induced by the vector field. Recall that the flow of the scalar equation ta y˙(t) = ay(t), a ∈ R is given by Φt(u) = e u. Notice that, Theorem 6 implies the matrix exponential etA satisfies the equationy ˙(t) = Ay(t) where A ∈ Rn×n, tA thus the flow of such a system is given by Φt(u) = e u. Consider a more general case, let U ⊂ Rn be open, and let X¯ : U → Rn be a complete smooth vector field. We can interpret X¯ as the derivation given

Pn ¯ ∂ ¯ pointwise by Xp = i=1 Xi(p) ∂xi where the Xi are smooth functions on U. p X also induces a linear operator on the space of smooth functions on U. Of course, X¯ and X are the same object, and we will use X for both in later sections, but distinguishing them will be somewhat helpful here. Now, X¯ also induces the diﬀerential equationy ˙(t) = X(y(t)). Comparing both X¯ and X we see n n X X ∂ X(y(t)) = X¯ (y(t))e ↔ X¯ = X, i i i ∂xi i=1 i=1 n where { e1, . . . , en } is a basis of R .

7 By assumption the vector field X¯ is complete, thus it has a flow Φt which is diffeomorphism of U for each t ∈ R, and is smooth since X is smooth. Recall that d ¯ the flow has the properties: Φ0 = idU ,Φt ◦Φs = Φt+s, and dt Φt(u) = X(Φt(u)). ∞ Assume that X and hence X¯ are C , and that for each p ∈ U,Φt(p) is real analytic in t, in a neighbourhood of 0. Hence, there is some r > 0 s.t.

∞ k X t (k) Φ (p) = Φ (p) for all t ∈ (−r, r). t k! 0 k=0 By the chain rule we have that d Φ(0)(p) = id (Φ (p)) = Φ (p) and Φ(1)(p) = Φ (p) = X¯(Φ (p)), t U t t t dt t t i ¯ ¯ where Φt and Xi are the i-th components of Φt and X respectively. Now assume (k) k−1 ¯ for k ≥ 1 that Φt (p) = (X X)(Φt(p)) with this being the composition of the linear operator X applied to the function X¯, then d d Φ(k+1)(p) = Φ(k)(p) = Xk−1X¯(Φ (p)) t dt t dt t n X ∂ k−1 ¯ d i = X X Φt(p) ∂xi Φ (p) dt i=1 t n X ∂ k−1 = X¯i(Φt(p)) (X X¯) ∂xi Φ (p) i=1 t k = (X X¯)(Φt(p)).

(n+1) n ¯ Hence by induction, we have that Φt (p) = X X(Φt(p)), and in particular (n+1) n ¯ that Φ0 (p) = (X X)(p). Therefore,

∞ X tk Φ (p) = p + (Xk−1X¯)(p) := etX p. t k! k=1 This has the same form as the series form of the matrix and scalar exponentials since X and X¯ are interpretations of the same vector field. Thus we call etX := Φt the vector field exponential. When Φt is not analytic we can no longer expand it as an infinite series. We can however still approximate it with a Taylor polynomial. Assuming Φt(p) is of class C`, this approximation along with the above computations gives that n for each integer ` ≥ n ≥ 1, there exists some function Rn : R → R s.t.

n k n k X t (k) X t Φ (p) = Φ (p) + R (t) = p + (Xk−1X¯)(u) + R (t) t k! 0 n k! n k=0 k=1

n+1 n ¯ with limt→0 Rn(t) = 0. This expansion along with Φt (p) = (X X)(Φt(u)), tX is enough for us to write e := Φt.

8 3 Hamiltonian Mechanics

In this section we aim to give an introduction to Hamiltonian mechanics in the familiar setting of R2n. We will consider only autonomous Hamiltonians, i.e. models of energy conserving systems. We will study some examples of Hamiltonian systems such as the ideal pendulum and the Kepler problem. These example will act, both as motivation for the abstraction to symplectic manifolds, which we discuss in Section 4, and as test cases for the numerical integrators we construct in Section 6. We will not yet prove any properties of the flows of Hamiltonian systems, such as the conservation of energy. We instead choose to leave them for later when we can employ the Lie derivative and other tools of differential geometry. It is well known that energy is an important concept in physics. The idea of Hamiltonian mechanics is to use the total energy of a system to define the dynamics of that system. For such energy conservative systems, Hamiltonian mechanics is an equivalent reformulation of classical Newtonian mechanics where the dynamics are directly derived from Newton’s formula F = ma. It is in some sense “dual” to the Lagrangian formulation through conjugate momentum and the Legendre transform. However, for dissipative systems the situation is more complicated. We will neither discuss these reformulations nor show how they are equivalent (the details can be found in books on classical mechanics, or in [6, Chapter VI.]).

Definition 3 Let M ⊂ R2n be open, and H ∈ C2(M, R). 0 −In 2n×2n x˙ = XH (x) := J∇H(x) where J = ∈ R In 0 is called the Hamiltonian differential equation, while XH is called the Hamiltonian vector field, and H is the Hamiltonian.

2n n n Giving R = R × R the coorindates (p, q) = (p1, . . . , pn, q1, . . . , qn), the Hamiltonian diﬀerential equations take the form ∂H ∂H p˙ = − andq ˙ = , (1) ∂q ∂p h iT wherep ˙ = dp and ∂H = ∂H , ··· , ∂H . This “twisting”, where the time dt ∂p ∂p1 ∂pn derivatives of one variable depend on the special derivatives of H w.r.t. another variable is where the special structure of Hamiltonian dynamics comes from. We call p the momentum, and q the position. The Hamiltonian is also called the total energy of the system, while n is the number of degrees of freedom, and M is the phase space. Hamiltonians are often of the form H(p, q) = T (p)+V (q), where T is called the kinetic energy and V the potential energy. Such Hamiltonians are called a separable Hamiltonian. The Hamiltonian vector ﬁeld for these splits as XH = XT + XV with XT depending only on p and XV only on q. This is precisely the property we will exploit in Section 6 to construct symplectic integrators.

9 The Hamiltonian vector ﬁeld (1) can be written as derivation

n n X ∂H ∂ X ∂H ∂ X = − + . H ∂q ∂p ∂p ∂q i=1 i i i=1 i i

We see that the flow of a Hamiltonian vector field XH is given formally by the vector field exponential exp(tXH ). With the initial condition (p0, q0) ∈ M, the formal solution of the Hamiltonian equations becomes exp{tXH }(p0, q0). Given a vector field X how do we determine whether it is a Hamiltonian vector field? We know that for a Hamiltonian H, XH = J∇H, thus ∇H = −JXH . Determining whether a vector field is the gradient of some function is relatively easy, at least for open and simply connected subsets of Rn. Theorem 7 Let M ⊂ Rn be open and simply connected, and f ∈ C1(M, Rn). Then, f is a gradient vector field if and only if the Jacobian of f is symmetric.

Proof n To the vector field f = (f1, . . . , fn): M → R we associate the differential 1- 1 n form ωf = f1dx +···+fndx . Notice that under this association f is a gradient vector field if and only if ωf is exact. Since M is open and simply connected, Poincaré’sLemma [14, Corollary 27.13, pg. 300] implies that ωf is exact if and only if ωf is closed. We compute dωf :

n n  n  X X X ∂fi dω = df ∧ dxi = dxj ∧ dxi f i  ∂xj  i=1 i=1 j=1 n n X X ∂fi = dxj ∧ dxi ∂xj i=1 j=1 X ∂fj ∂fi = − dxi ∧ dxj ∂xi ∂xj 1≤i

∂fj ∂fi It is clear that dωf = 0 ⇐⇒ ∂xi = ∂xj for i, j = 1, . . . , n. Therefore f is a gradient vector ﬁeld ⇐⇒ ωf is closed ⇐⇒ the Jacobian of f is symmetric.

3.1 Motivating Examples We study some examples of Hamiltonians systems.

Example 1 (Linear Hamiltonian Systems) A linear vector field on R2n is given by f(x) = Bx for B ∈ R2n×2n. f will be a Hamiltonian vector field if −JBx is a gradient vector field. By Theorem 7, this is precisely when its Jacobian −JB is symmetric. Hence, we want to find a function H s.t. ∇H = −JBx. Recall, for A ∈ Rn×n we have that ∇xT Ax = AT x + Ax = (A + AT )x,

10 where A + AT is symmetric. Thus, if JB is symmetric, f is the Hamiltonian vector ﬁeld generated by 1 H : 2n → ; x 7→ H + xT (− B)x, R R 0 2 J where H0 ∈ R. Therefore, all linear Hamiltonian systems are generated by Hamiltonians of the form 1 H : 2n → ; x 7→ H + xT Ax, (2) R R 0 2 2n×2n T where A ∈ Sym(2n, R) := { X ∈ R | X = X }, and H0 ∈ R. However, these Hamiltonians will not, in general, be separable. An easy computation shows that for a linear Hamiltonian to be separable, A must be of the form TD n×n −DT V where T,V,D ∈ R and D is skew-symmetric. Notice (2) generates Hamiltonian systems of the form

x˙ = JAx, (3) who’s system matrices JA have the property T T T (JA) J + J(JA) = A J J − I2nA = A(−I2n) − A = A − A = 0. We known the flow of (3) is given by the vector field exponential which reduces to the matrix exponential since the vector field is linear. Thus the flow is

2n 2n t A R × R → R ;(t, x) 7→ e J x. Example 2 (Ideal Pendulum) A non-linear Hamiltonian system who’s dynamics are still relatively easy to describe is the ideal pendulum (see Figure 1). Here, q is the angular displacment and p is the momentum which is the velocity q˙ in this case since m = 1. Its Hamiltonian is given by 1 H(p, q) = p2 − cos q, 2 which is clearly separable and generates the dynamics ∂H ∂H p˙ = − = − sin q andq ˙ = = p. ∂q ∂p We can see that these dynamics are the same as those derived from Newton’s equation F = ma which in this case givesq ¨ = − sin q. The Hamiltonian is constant along solution curves of the system, thus they lie in the energy shells H−1(c) for c ∈ R. Examining the energy surface in Figure 1, we can clearly see that the equilibrium points are given by (0, kπ) for k ∈ Z. The stable and unstable equilibria are the basins and saddle points respectively of this surface (i.e. (0, 2kπ) and (0, 2kπ+1) for k ∈ Z). These can be analytically determined by computing the eigenvalues of the linearized system around these points. We see that the dynamics change at H = 1, between periodic motion of q when H < 1 and monotonic motion of q when H > 1.

11 Figure 1: Ideal pendulum of length l = 1, and mass m = 1 (left). Energy surface of the ideal pendulum with solution curves for E = 2, E = 1, and E = 0 (right).

Example 3 (The Kepler Problem) Consider two bodies which are attracted to each other gravitationally, for example a sun and planet. If we denote their positions with qs and q, and their momentums with ps and p, their Hamiltonian is 2 2 kpsk kpk GmM H(p, ps, q, qs) = + − , 2M 2m kq − qsk where M and m are the masses of the sun and planet, while G is the gravitational 3 constant, and p, ps, q, qs ∈ R . The norm is Euclidean 2-norm. This Hamiltonian generates the dynamics

∂H q − qs ∂H p p˙ = − = −GmM 3 , q˙ = = ∂q kq − qsk ∂p m and ∂H q − qs ∂H ps p˙s = − = GmM 3 , q˙s = = . ∂qs kq − qsk ∂p s m

p p Using Heliocentric coordinates Q = q − qs we get that Q˙ = /m − s/M, and p˙ p˙ Q Q¨ = − s = −G(M + m) . m M kQk3

We also have d G(M + m) Q × Q˙ = Q˙ × Q˙ + Q × Q¨ = − (Q × Q) = 0. dt kQk3

12 T 3 ˙ This implies Q(t) is constrained to the plane x ∈ R | Q(0) × Q(0) x = 0 for all t ∈ R. Therefore, we can choose a basis of the plane, consider Q ∈ R2 and take a suitable rescalling of units, to get that the simpliﬁed dynamics are given by Q P˙ = Q¨ = − , Q˙ =: P. (4) kQk3 These dynamics are generated by the separable Hamiltonian

kP k2 1 H(P,Q) = − . (5) 2 kQk

d A quick computation shows that dt L = 0 where L is the angular momentum

L(p1, p2, q1, q2) = q1p2 − q2p1, with P = (p1, p2) and Q = (q1, q2). Hence both H and L are conserved along solutions curves of (4). Thus, a solution (P,Q) of (4) with initial point (P (0),Q(0)) = (P0,Q0) has

H(P (t),Q(t)) = H(P0,Q0) = H0 ∈ R and L(P (t),Q(t)) = L(P0,Q0) = L0 ∈ R.

In polar coordinates (q1, q2) = (r cos θ, r sin θ), we have

˙ ˙ (p1, p2) = (q ˙1, q˙2) = r˙ cos θ − rθ sin θ, r˙ sin θ + rθ cos θ , which gives the expressions 1 1 H = r˙2 + r2θ˙2 − , and L = r2θ.˙ 0 2 r 0 ∂r ˙ Considering r as a function of θ, we seer ˙ = ∂θ θ by the chain rule , and ! L2 ∂r 2 1 H = 0 + r2 − . 0 2r4 ∂θ r

From this we get

2 4 3 2 2 p 2 2 ∂r 2r H0 2r r L0 ∂r r 2r H0 + 2r − L0 = 2 + 2 − 2 =⇒ = ± . ∂θ L0 L0 L0 ∂θ L0 By separation of variables, and other computations it possible to see

Z Z Z 2 L0 L0 θ − θ0 = dθ = dr = dr p 2 2 p 2 2 2 r 2r H0 + 2r − L0 r e r − (L0 − r) 1L2 = arccos 0 − 1 , e r

13 p 2 where e := 1 + 2H0L0 is called the eccentricity. Therefore, L2 r(θ) = 0 , 1 + e cos(θ − θ0) where θ0 is determined by the initial conditions r(θ(0)) and θ(0). These are in turn speciﬁed by q1(0) and q2(0) using

q1(0) = r(θ(0)) cos(θ(0)), and q2(0) = r(θ(0)) sin(θ(0)).

Therefore we can explicitly compute

1 L2 θ = θ(0) − arccos 0 − 1 . 0 e r(θ(0))

For L0 6= 0, r(θ) is a conic section, which depending on H0 gives the following types of trajectories for the Kepler problem:

• For H0 < 0, we have e ∈ [0, 1) which are ellipses in general and a circle if e = 0.

• For H0 = 0, we have e = 1 which is a parabola.

• For H0 = 0, we have e = (1, ∞) which are hyperbola.

For example, the trajectory of P0 = (0, 1.35), Q0 = (1, 0) is an ellipse, while that of P0 = (0, 1.5), Q0 = (1, 0) is a hyperbola (see Figure 2).

Figure 2: Ellipse generated with r(θ) using H0 = −0.08875, L0 = 1.35, e = 0.8225, and θ0 = 0 (left). Hyperbola generated with r(θ) using H0 = 0.125, L0 = 1.5, e = 1.25, and θ0 = 0 (right).

14 4 Symplectic Geometry

To gain greater insight into the behaviour and properties of Hamiltonian systems, we will take a more abstract perspective by considering their phase spaces as manifolds rather than just as open subsets of R2n. This allows us to employ the tools of differential geometry. These more clearly display the geometric structure of these properties. We equip the phases space with an additional structure, a differential 2-form which defines the dynamics that a Hamiltonian generates. Further explanation of why symplectic geometry is the natural setting of classical mechanics can be found in Cohn’s essay [5]. However, to begin we will study symplectic vector space which will make the jump to abstract symplectic manifolds easier.

4.1 Symplectic Vector Spaces Let V be a real vector space of dimension n.

Deﬁnition 4 A function ω : V × V → R is called bilinear or a 2-form if ω is linear in both of its arguments.

• ω is anti-symmetric if for all a, b ∈ V, ω(a, b) = −ω(b, a). • ω is non-degenerate if for all non-zero a ∈ V , ∃b ∈ V s.t. ω(a, b) 6= 0. • ω is a symplectic form if it anti-symmetric and non-degenerate. Then the pair (V, ω) is called a symplectic vector space.

A symmetric and positive deﬁnite 2-form on V is an inner product on V . We can think of a symplectic form as an anti-symmetric analogue to an inner product. We can also see that an anti-symmetric 2-form is a diﬀerential 2-form on V since ∼ TpV = V for p ∈ V .

Deﬁnition 5 Let { v1, . . . , vn } be a basis of V . The matrix representation n×n J ∈ R of a 2-form ω w.r.t. that basis, is given by (J)ij = ω(vi, vj) for i, j = 1, . . . , n. The rank of ω is the rank of its matrix representation J. The matrix representation of an anti-symmetric 2-form can be put in a standard form by constructing a suitable basis of the vector space.

Theorem 8 (Darboux Theorem: Linear Case) Let ω be an anti-symmetric 2-form of rank r. Then r = 2m for some m ≥ 0, and there exists a basis of V such that the matrix representation of ω has the form   0 −Im 0 J = Im 0 0 . 0 0 0

15 Proof Assume that ω 6= 0, otherwise we’re done. So there exists vb1, vbm+1 ∈ V v s.t. c1 := ω(vbm+1, vb1) 6= 0. Let v1 := b1/c1 and vm+1 = vbm+1. Since ω is anti-symmetric

ω(v1, v1) = ω(vm+1, vm+1) = 0 and ω(v1, vm+1) = −ω(vm+1, v1) = −1.

Let U1 = span {v1, vm+1} ⊂ V and V2 := { p ∈ V | ω(p, q) = 0 ∀q ∈ U1 }. Notice that U1 ∩ V2 = {0} and deﬁne y := x + ω(v1, x)vm+1 − ω(vm+1, x)v1 for x ∈ V . Then

ω(v1, y) = ω(v1, x + ω(v1, x)vm+1 − ω(vm+1, x)v1)

= ω(v1, x) + ω(v1, x)ω(v1, vm+1) − ω(vm+1, x)ω(v1, v1)

= ω(v1, x) − ω(v1, x) = 0, and similarly ω(vm+1, y) = 0. Thus, y ∈ V2 which implies U1 + V2 = V . We can now consider ω|V2 and repeat this process using V2 instead of V . By induction 0 this gives us a basis { v1, . . . , v2m } for a subspace V of V of dimension 2m. If 0 dim V = n > 2m, we can extended this basis for V with e1, . . . , e2m−n ∈ V to get a basis of V . Since ω has rank 2m we have that ω(ek, v) = 0 for all v ∈ V . With respect to the basis { v1, . . . , v2m, e1 . . . , e2m−n } the matrix representation of ω is given by   1, if i = m + j (J)ij = −1, if j = m + i  0, otherwise for i, j = 1,..., 2m and (J)ij = 0 for i, j > 2m. We can use this theorem to derive two useful facts about symplectic vector spaces. Corollary 1 Symplectic vector spaces are even dimensional.

Proof Let (V, ω) be a symplectic vector space. ω is non-degenerative, hence by Theorem 8 dim V = rank ω = 2m for some m ≥ 0.

Corollary 2 A symplectic vector space (V, ω) has a basis {p1, . . . , pm, q1, . . . , qm} Pm i i s.t. ω has the standard form i=1 dp ∧ dq .

Proof From dim V = 2m and the proof of Theorem 8, setting pi := vi and qi := vm+i gives us the required basis. Looking at J, the matrix representation Pm i i of ω w.r.t. this basis, shows us that ω is of the form i=1 dp ∧ dq . We can use this corollary to construct a volume form on a symplectic vector space. By the above results it suﬃces to show that on a symplectic vector space (V, ω) of dimension 2m, the 2-form ω deﬁnes a volume form ωm = ω ∧ ... ∧ ω

16 i.e. ωm is a 2m-form s.t. ωm 6= 0. Notice that

m !  m  m X i i X j j ω = dp ∧ dq ∧ ...  dp ∧ dq  i=1 j=1 = n!dp1 ∧ dq1 ∧ ... ∧ dpm ∧ dqm.

Hence, ωm 6= 0 so it is a volume form on V . Just as isometries between inner product spaces preserve the inner product, we define maps which preserve the symplectic structure of a vector space. Let W be another finite dimensional R-vector space. Definition 6 Let η be a 2-form on W , and f ∈ L(V,W ) a linear map form V to W . The function f is called a symplectic if f preserves the symplectic form i.e. if for all a, b ∈ V

(f ∗η)(a, b) = η(f(a), f(b)) = ω(a, b).

Thus the symplectic maps f ∈ L(V,V ) are those which satisfy f ∗ω = ω. If we 0 −I choose a basis of V s.t. the matrix representation of ω has the form J = I 0 and let A be the matrix representation of f, this condition becomes AT JA = J. The set of all symplectic maps on V becomes a group under composition, called the symplectic group. It is shown to be a Lie group in Example 6.

4.2 Symplectic Manifolds We will now generalise the concept of a symplectic vector space to that of a symplectic manifold. We will see how the phase spaces of Hamiltonians systems form symplectic manifolds, and are thus the natural space for Hamiltonian dynamics. Using the Lie derivative, an operation intrinsic to manifolds, we will show two important properties of the Hamiltonian flow: it preserves the Hamil- tonian function, and it preserves the symplectic form on the manifold. We begin by generalising symplectic forms from vector spaces to manifolds. Definition 7 A differential 2-form ω on a smooth manifold M is called a symplectic form if ω is closed, and for each p ∈ M, ωp is a symplectic form on TpM. The pair (M, ω) is called a symplectic manifold.

A symplectic manifold (M, ω) is even dimensional. This is due to ωp being a symplectic form on TpM for p ∈ M. Recall dim TpM = dim M, and that TpM is even dimension by Corollary 3. A symplectic form ω on manifold M also deﬁnes a volume form on M by m similar argument. Let dim M = 2m. For p ∈ M, ωp is a volume form on TpM by Corollary 2. Therefore, ωm 6= 0, implying ωm is a volume form and that symplectic manifolds are orientable. Consider a smooth manifold N of dimension n. The tangent and cotangent bundles of N, TN and T ∗N respectively are both manifolds of dimension 2n.

17 Thus, both satisfy the even dimensionality requirement, and could be equipped with a symplectic form. The cotangent bundle is used as the phase space for the Hamiltonian formalism while the Lagrangian formalism uses the tangent bundle. Therefore, in the rest of this section we will consider symplectic manifolds of the form M := T ∗N. We write q = (q1, . . . , qn) for the local coordinates of N. This induces local coordinates (p, q) = (p1, . . . , pn, q1, . . . , qn) on T ∗M giving ∗ 1 1 n n p ∈ Tq M which implies p = p dq + ··· + p dq . With these coordinates we can equip M with the symplectic form n X i i ω0 = dp ∧ dq . i=1 This can be derived in a coordinate free way as the exterior derivative of the Liouville form (see [14, Example 17.4, pg. 193] or [9, Section 10.1, pg. 218]). Hence, it is called the canonical symplectic form. We define interior multiplication of a differential form and use it to define the dynamics generated by a Hamiltonian, as well as the Lie derivative. Definition 8 Let X be a smooth vector field on a manifold N. For k ≥ 2, the interior multiplication or contraction of a k-form ω with X is a (k−1)-form defined by ιX ω(X2,...,Xk) := ω(X,X2,...,Xk).

We deﬁne ιX ω = ω(X) for a 1-form ω and ιX ω = 0 for a 0-form ω i.e. a function.

Lemma 1 Let X be a smooth vector ﬁeld on a manifold N. Then ιX ◦ ιX = 0. Proof Let ω be a k-form on N.

ιX ◦ ιX ω(X3,...,Xk) = ιX ω(X,X3,...,Xk) = ω(X,X,X3,...,Xk) = 0, as ω is alternating.

Definition 9 The vector field XH defined by ιXH ω := ω(XH , ·) = dH for a smooth function H : M → R on a symplectic manifold (M, ω), is called the Hamiltonian vector field generated by H. The triple (M, ω, H) is called a Hamiltonian system.

This definition extends our previous definition of Hamiltonian systems on Rn. ∗ n Example 4 Consider the symplectic manifold (M, ω) = (T R , ω0) with coordinates (p, q), and the vector field

n n X ∂ X ∂ X = X i + X i . p ∂pi q ∂qi i=1 i=1 Given a Hamiltonian H : M → R, we determine the coeﬃcients of X s.t. it is the Hamiltonian vector ﬁeld generated by H. Now, n n X i i X i i ιX ω0 = dp ∧ dq (X, ·) = Xqi dp − Xpi dq , i=1 i=1

18 and n X ∂H ∂H dH = dpi + dqi . ∂pi ∂qi i=1

We can now solve ιX ω0 = dH be comparing coefficients to see that ∂H ∂H X i = − and X i = p ∂qi q ∂pi for i = 1, . . . , n. This is agrees with our previous definition of the Hamiltonian vector field (see Definition 3). Another operation intrinsic to manifolds and related to interior multiplication is the Lie derivative of a differential form. Geometrically it is how the form changes along the vector field. Definition 10 Let X be a smooth vector field on a manifold N. The Lie derivative of a differential k-form ω along X is the k-form defined by the Cartan homotopy formula

LX ω := (ιX d + dιX )ω. It is well deﬁned since for a k-form ω the exterior derivative raises it to a k + 1- form while interior multiplication lowers it again to a k-form. A smooth function f : N → R is a 0-form, thus its Lie derivative along X is

LX f = ιX df + dιX f = df(X) = Xf, which is the direction derivative of f along X. Furthermore, the Lie derivative along X interacts nicely with time derivative of the flow of X. Theorem 9 ([9, Theorem B.34, pg. 521]) Let X be a smooth vector field on a manifold N, and let Φ: R × M → M be the flow generated by X. Then for t ∈ R, each differential form ω on N satisfies d (Φ∗ω) = Φ∗L ω. dt t t X For more details about the Lie derivative or interior multiplication see [14, Section 20] or [9, Section B.5]. We have now built up enough machinery to prove two properties of the Hamiltonian flow.

Theorem 10 The ﬂow Φ: R × M → M of a Hamiltonian system (M, ω, H) preserves H.

Proof d d H(Φ ) = Φ∗H = Φ∗L H = 0 dt t dt t t XH since

LXH H = dιXH (H) + ιXH dH = ddH(H) + ιXH ◦ ιXH = 0.

19 Just as we have symplectic maps between symplectic vector spaces we also define structure preserving maps between symplectic manifolds. Definition 11 Let (M, ω) and (N, η) be symplectic manifolds of the same dimension. A smooth map F : M → N is called a symplectic map if F ∗η = ω. If F is also a diffeomorphism, it is called a symplectomorphism.

∗ k k In fact, F η = ω for each k ∈ {1,..., 1/2 dim M}, since the pullback dis- tributes over the wedge product i.e. g∗(α ∧ β) = g∗α ∧ F ∗β.

Theorem 11 Let (M, ω, H) be a Hamiltonian system, and let Φ: R × M → M be the ﬂow generated by the Hamiltonian vector ﬁeld XH . Then for each t ∈ R, Φt is symplectic.

∗ ∗ Proof Φ0 = idM which implies Φ0ω = idM ω = ω. Further using Theorem 9 d Φ∗ω = Φ∗L ω = Φ∗(ι dω + dι ω) = 0 dt t t XH t XH XH since ω is closed and dιXH ω = ddH = 0. This means that the ﬂow of Hamiltonian system not only preserves the sym- k plectic form, but also its exterior products ω for k ∈ {1,..., 1/2 dim M}. In particular it preserves the volume form, thus preserving phase space volume.

20 5 Lie Theory And BCH Formula

Lie groups are manifolds with a group structure on them. Such manifolds are homogeneous in the sense that locally the manifold looks the same around any point. Thus we can study the manifold by examining a neighbourhood of the identity element. The tangent space at identity of the Lie group has a canonical bracket operation transferred from the space of vector ﬁelds on the Lie group which are “compatible” with the group structure. This tangent space and bracket is the Lie algebra of that Lie group. Much of the information about the group is encoded in its Lie algebra which is easier to study as it is a vector space. Some of this encoding of the group’s structure in its Lie algebra can be seen in the BCH formula. In this section, we will describe the basics of Lie groups and Lie algebras before discussing the BCH formula. We will use M to denote a plain manifold, while G will be used for Lie groups. We will use lower case Gothic letters for the Lie algebra of a Lie group, that is we write g for the Lie algebra of the Lie group G. More about Lie theory can be found in books such as [14, Chapter 4] or [8]. Deﬁnition 12 A Lie group is a smooth manifold G which is also a group s.t. the group operations

µ: G × G → G;(a, b) 7→ ab and ι: G → G; a 7→ a−1 are smooth.

Example 5 (GL(n, R) is a Lie group) b By deﬁnition,

n×n −1 GL(n, R) = { X ∈ R | det{X}= 6 0 } = det (R \{0}).

2 As vector spaces n×n is isomorphic to n , hence we give n×n the topology of 2 R R R Rn . This allows us to investigate the smoothness of the determinant function det: Rn×n → R. n×n Given A ∈ R with entries aij for i, j = 1, . . . , n, let Mij(A) be the (n − 1) × (n − 1) matrix obtained from A by deleting the i-th row and j-th column of A. From linear algebra we know that the determinant of A can be deﬁned in terms of the determinants of these submatrices, i.e.

n X i+j det A = (−1) aij det Mij(A), for any i = 1, . . . , n. j=1

From this it is clear that det A is a polynomial in terms of the entries aij of A. Polynomials are smooth functions of their coeﬃcients, thus det is smooth. −1 The set R \{0} is an open subset of R, thus det (R \{0}) is an open subset of Rn×n. Therefore GL(n, R) is an n2-manifold. Pn For A, B ∈ GL(n, R) the entries of AB are given by (AB)ij = k=1 aikbkj. Hence the multiplication map

µ: GL(n, R) × GL(n, R) → GL(n, R); (A, B) 7→ AB

21 is smooth since each entry of a AB is a polynomial in terms of aij and bij. Similarly, Cramer’s rule gives that the entries of A−1 are 1 (A−1) = · (−1)i+j det M (A). ij det A ji The determinant is smooth, thus the inverse map

−1 ι: GL(n, R) → GL(n, R); A 7→ A is smooth. Recall that the determinant satisﬁes det(AB) = det(A) det(B) and detA−1 = −1 det(A) . Thus GL(n, R) is closed under µ and ι. Therefore, GL(n, R), with matrix multiplication and inversion is a Lie Group of dimension n2.

Example 6 (Sp(2n, R) is a Lie group) The real symplectic group Sp(2n) is given by T Sp(2n) = Sp(2n, R) = { A ∈ GL(2n, R) | A JA = J },

0 −In where J = In 0 . Let Alt(2n) be the real vector space of skew-symmetric 2 matrices, A = −AT . Note dim Alt(2n) = 2n2 − n, thus Alt(2n) =∼ R2n −n. Consider the map

T F : GL(2n, R) → Alt(2n); A 7→ A JA.

T T T T For A ∈ GL(2n, R), A JA = A (−J)A = −A JA. Hence F (GL(2n, R)) ⊂ Alt(2n), and F is smooth since it deﬁned in terms of matrix multiplication. We will use the regular level set theorem (Theorem 1) to show Sp(2n) is a submanifold of GL(2n, R). To apply this theorem, J must be a regular value of F . We will show that, since F∗ is surjective for all A ∈ GL(2n, R), every A and in particular J, is a regular value of F . ∼ 2n×2n ∼ For A ∈ GL(2n, R), we know TAGL(2n, R) = R , and TF (A)Alt(2n) = Alt(2n). There exists a curve c(t) on GL(2n, R) s.t. for X ∈ R2n×2n, c(0) = A and c0(0) = X. Hence, the diﬀerential of F at A is

d d T F∗(X) = F (c(t)) = c(t) Jc(t) dt t=0 dt t=0 0 T T 0 = c (t) Jc(t) + c(t) Jc (t) t=0 T T = X JA + A JX.

Notice (XT JA)T = −AT JX, so for B ∈ Alt(2n), we can solve the matrix equation XT JA + AT JX = B by ﬁnding X s.t.

1 −1 XT A = B ⇐⇒ X = − AT B. J 2 J

This implies F∗ is surjective at each A ∈ GL(2n, R). Thus J is a regular value of F , and Sp(2n) = F −1(J) is a regular submanifold of GL(2n, R) of dimension 4n2 − (2n2 − n) = 2n2 + n.

22 For A, B ∈ Sp(2n), we have

T T T (AB) JAB = B A JAB = J =⇒ AB ∈ Sp(2n) and T −1T −1 −1 A JA = J ⇐⇒ J = A JA =⇒ A ∈ Sp(2n). Sp(2n) is closed under the multiplication and inversion maps induced by those on GL(2n, R). They are smooth by Theorem 2. Therefore Sp(2n) with matrix multiplication and inversion is a Lie Group of dimension 2n2 + n. In the following section we assume G is a Lie group, and we use the notation X(M) to denote the R-vector space of smooth vector fields on M. By Theorem 3 X(M) = L(C∞(M), C∞(M)). This is reasonable as vector fields are derivations. They can be interpreted as linear operators on the space of smooth functions on M. This also implies X(M) is infinite dimensional.

Deﬁnition 13 Let G be a Lie group. For g ∈ G, lg : G → G; x 7→ µ(g, x) is called left multiplication or translation. Left translation by g ∈ G is a diﬀeomorphism. G is a group, thus g−1 ∈ G. Hence for x ∈ G,

−1 −1 lg ◦ lg−1 (x) = µ(g, µ(g , x)) = µ(µ(g, g ), x) = x, and similarly lg−1 ◦ lg(x) = x. Therefore lg−1 is the inverse of lg, and lg is smooth, since µ is smooth. This means that lg maps a neighbourhood of identity e, to a neighbourhood of g, for any g ∈ G. It implies that all of the local properties of G can studied by focusing on a neighbourhood of e. An example of this, is that it allows us to describe TgG by describing TeG and using that the diﬀerential lg∗ : TeG → TgG of lg is an isomorphism of vector spaces.

Example 7 (The Tangent space TI GL(n, R)) In Example 5, we saw that n×n we could identify the tangent space TgGL(n, R) with R for g ∈ GL(2n, R). We explicitly compute the isomorphism lg∗ : TI GL(n, R) → TgGL(n, R). For X ∈ Rn×n, there exists a curve in GL(n, R) with c(0) = I, and c0(0) = X, which is given by c(t) = etX . Then,

d tX d tX lg∗(X) = lg(e ) = ge = gX. dt t=0 dt t=0

Example 8 (The Tangent space TI Sp(2n)) Take X ∈ TI Sp(2n), there exists a curve c(t) in Sp(2n) with c(0) = I and c0(0) = X so take c(t) = etX . We have the condition that c(t) ∈ Sp(2n) ∀t ∈ R, hence c(t) must satisfy c(t)T Jc(t) = J.

T tX T tX tXT −t X c(t) Jc(t) = J ⇐⇒ (e ) Je = J ⇐⇒ e e J J = I T ⇐⇒ etX = etJXJ T T ⇐⇒ X = JXJ ⇐⇒ X J + JX = 0

23 n×n T Therefore TI Sp(2n) consists of the X ∈ R s.t. X J+JX = 0. Such matrices are called inﬁnitesimally symplectic. Note the connection to Example 1, i.e. the system matrices of linear Hamiltonians are inﬁnitesimally symplectic. Sp(2n) is a Lie group, and is closed under lg. This map is a restriction of left translation on GL(2n, R) by g. Hence, lg∗ : TI Sp(2n) → TgSp(2n) is an isomorphism between the tangent spaces of Sp(2n), which is explicitly given by lg∗(X) = gX, for g ∈ Sp(2n).

Deﬁnition 14 A vector ﬁeld X on a Lie group G is called left-invariant if for g ∈ G, lg∗(X) = X.

If X is a left-invariant vector field on G, ∀g, h ∈ G we have that lg∗(Xh) = Xgh. Hence, h = e implies lg∗(Xe) = Xg. Thus X is completely determined by Xe ∈ TeG. This suggests that we can generate a left-invariant vector field on G, given a tangent vector at identity. Definition 15 On a Lie group G, the left-invariant vector field A˜ generated by A ∈ TeG is given by (A˜)g = lg∗(A) for g ∈ G. We use L(G) to denote the R-vector space of left-invariant vector fields on G.

In fact, TeG is isomorphic to L(G) via the maps A 7→ A˜ and Xe ← X. This is useful as both spaces have a natural structure that can be transfer[ to the other using this isomorphism. We will make use of only one direction of this isomorphism to deﬁne a Lie bracket on TeG. Theorem 12 ([14, Proposition 16.8, pg. 181]) Let G be a Lie group. Then L(G) is a subspace of X(G) i.e. left-invariant vector ﬁelds are smooth.

Example 9 (Left-Invariant Vector Fields On GL(n, R)) For g ∈ GL(n, R) n×n we can identify TgGL(n, R) with R via X ∂ aij ↔ [aij], ∂xij g where [aij] is the n × n matrix with entries aij. Hence, we can use B to denote P ∂ n×n ˜ both bij ∂xij I ∈ TI GL(n, R) and [bij] ∈ R . Thus B the left-invariant vector field generated by B is given by B˜g = lg∗(B) ↔ gB. In local coordinates ˜ P ∂ Bg = (gB)ij ∂xij g. We introduce a kind of “product” operation on vector fields, giving them additional structure which we will need for the BCH formula. Definition 16 Let M be a smooth manifold, and U ⊂ M be open. Given X,Y ∈ X(U), the Lie bracket [X,Y ] at p ∈ U is

[X,Y ]pf = (XpY − YpX)f, where f is a smooth function on M.

24 For the Lie bracket to a useful operation, we would like it to be closed on X(M), or put more succinctly: we would like the Lie bracket of smooth vector ﬁelds to be a smooth vector ﬁeld. It turns out that this is the case: Theorem 13 For a smooth manifold M,

X,Y ∈ X(M) =⇒ [X,Y ] ∈ X(M).

Proof Let f be a smooth function on M. Both X(Y f) and Y (Xf) are also smooth functions. Therefore [X,Y ]f = (XY − YX)f = X(Y f) − Y (Xf) is also smooth by linearity of vector ﬁelds as operators. Theorem 14 Let G be a Lie group. For X,Y ∈ L(G), [X,Y ] ∈ L(G).

For a proof see [14, Proposition 16.9, pg. 182]. Hence, L(G) is a subspace of X(G) which is closed under the Lie bracket. Vector spaces can be equipped with a product a product operation satisfying properties similar to the Lie bracket. This deﬁnes a useful class of spaces.

Deﬁnition 17 A Lie algebra is an R-vector space V with a product [·, ·]: V × V → V called the bracket which is bilinear, anti-commutative, and satisﬁes the Jacobi identity:

[X, [Y,Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0, ∀X,Y,Z ∈ V.

Gothic lowercase letters such as g, h, etc. are often used to denote Lie algebras.

Example 10 For a manifold M, X(M) is an R-vector space. When equipped with the Lie bracket it becomes a Lie Algebra.

From the name we might expect there to be a connection between Lie groups and Lie algebras. This is in fact the case, as we can make TeG into a Lie algebra by defining a bracket on it. We do this by using that L(G) is naturally a Lie algebra when equipped with the Lie bracket, and the isomorphism between L(G) and TeG. Consider the bracket [A, B] := [A,˜ B˜]e on TeG. The left-invariant vector field generated by this bracket should be equal to the vector field [A,˜ B˜]. Theorem 15 Given left-invariant vector fields A˜ and B˜ on a Lie group G, generated by A, B ∈ TeG respectively. Then [A,˜ B˜] = [Â, B].

Proof On TeG, the Lie bracket of A and B is [A, B] = [A,˜ B˜]e. Thus

[^A, B] = [A,^˜ B˜]e = [A,˜ B˜] since the isomorphism between TeG and L(G) is given via the maps A 7→ A˜ and Xe ← X. [ Deﬁnition 18 The Lie algebra of Lie group G, denoted g is the vector space TeG with the bracket [A, B] := [A,˜ B˜]e.

25 We computed TI GL(n, R) in Example 7 and TI Sp(2n) in Example 8. We can ﬁnd Lie brackets on them, to form their Lie algebras gl(n, R) and sp(2n). Example 11 (Lie Bracket on gl(n, R)) Take

X ∂ X ∂ A = aij ,B = bij ∈ TI GL(n, ). ∂xij I ∂xij I R

Then, making use of the coordinate function xij, we get

[A, B]xij = [A,˜ B˜]I xij = A˜I Bx˜ ij − B˜I Ax˜ ij = ABx˜ ij − BAx˜ ij ˜ ˜ ˜ since AI = A and BI = B. We previously computed Bg for g ∈ GL(n, R) as ˜ P ∂ Bg = (gB)ij ∂xij g. Hence,

˜ X X Bgxij = (gB)ij = gikbkj = bkjxik(g) k k ˜ P which holds for all g. Therefore, Bxij = k bkjxik. This implies ! ˜ X ∂ X X ABxij = alm bkjxik = aikbkj = (AB) . ∂xlm I ij l,m k k

˜ Interchanging A and B gives BAxij = (BA)ij. Therefore,

˜ ˜ X ∂ X ∂ [A, B] = [A, B]I = (AB) − (BA) = (AB − BA) . ij ij ∂xij I ij ∂xij I

P ∂ If we use the identiﬁcation aij ∂xij ↔ [aij] as in Example 9, between deriva- g tions and matrices, then we have [A, B] = AB − BA.

Example 12 (Lie Bracket on sp(2n)) Similar computations show that the Lie Bracket on sp(2n) is just a restriction of the Lie Bracket on gl(2n, R). This is somewhat comparable to how lg∗, the isomorphism between tangent spaces of Sp(2n), is a restrict of the isomorphism between tangent spaces of GL(2n, R). We now have a enough theory to discuss the Baker-Campbell-Hausdorﬀ For- mula (BCH) formula. Theorem 16 (The BCH Formula For Matrices[8, Theorem 5.3, pg. 114]) n×n For all X,Y ∈ C with kXk + kY k ≤ 1/2 · log 2, we have 1 1 1 logeX eY = X + Y + [X,Y ] + [X, [X,Y ]] − [Y, [X,Y ]] + ··· . 2 12 12 It implies that all of the information about the product operation a the Lie group is encoded in its Lie algebra. This can be generalised, to elements of a Banach algebra.

26 Theorem 17 (The BCH Formula[10, Theorem 3.17, pg. 12]) Let A be a Banach algebra over R or C. If X,Y ∈ A s.t. kXk + kY k ≤ 1/2 · log 2, then 1 1 1 logeX eY = X + Y + [X,Y ] + [X, [X,Y ]] − [Y, [X,Y ]] + ··· . 2 12 12 A Banach algebra is a complete normed vector space with a product which is compatible with both the structure of the vector space and the norm. The exponential and logarithm are defined using series, similarly to how the matrix exponential is defined, and [X,Y ] = XY − YX. For example, the space of smooth vector fields on a manifold becomes a Banach algebra if equipped with an appropriate norm. We will not concern ourselves here with the choice of norm. It does not play a large role in our use of the BCH formula in Yoshida’s construction. For more details about this approach and for many other proof of different versions of the BCH formula see [10].

27 6 Symplectic Integrators

In this section, we will derive a class of explicit symplectic integrators for separable Hamiltonian systems which are of arbitrary even order. We will do this using a technique described by Yoshida in [15] and [16] based on repeated application of the BCH formula. We will describe some their properties, and perform a backward error analysis. We will test these integrators by applying them to our motivating examples in the following section. Before this, we will discuss separable Hamiltonians, a general problem that leads to symplectic integrators, and some theory and properties of numerical integrators.

6.1 Separable Hamiltonians Consider a Hamiltonian of the form H(p, q) = T (p)+V (q) deﬁned on U an open subset of R2n. Both T (p) and V (q) are also Hamiltonians, having Hamiltonian vector ﬁelds XT and XV , given by " # " # − ∂T − ∂V ∂V ∂q 0 ∂q − ∂q XT = ∂T = ∂T and XV = ∂V = . (6) ∂p ∂p ∂p 0

Notice that the Hamiltonian vector field of H splits, that is XH = XT + XV . Due to this property, we will call Hamiltonians with such a form separable. The flows of XT and XV can be easily computed explicitly. For XT we have ∂T ∂T p˙ = 0 =⇒ p(t) = p , q˙ = =⇒ q(t) = q + t (p ), 0 ∂p 0 ∂p 0 where (p0, q0) ∈ U. The computations for XV are similar. Thus, we find that their flows are p p − t ∂V (q) tXT tXV ∂q e (p, q) = ∂T and e (p, q) = . q + t ∂p (p) q Both of these maps are symplectic by Theorem 11, as they are the flows of Hamiltonian vector fields. Thus, their composition is also symplectic, Based on the separable structure of XH it appears reasonable to try to approximate its flow etXH = et(XT +XV ) with either the composition etXT etXV tXV tXT TXH or e e . For t = T , approximating e (p0, q0) results in the numerical integration schemes

∂V ∂T pn+1 = pn − h (qn) qn+1 = qn + h (pn) ∂q and ∂p ∂T ∂V qn+1 = qn + h ∂p (pn+1) pn+1 = pn − h ∂q (qn+1), where we divide the interval [0,T ] into N intervals of length h = T/N. These methods are both called Symplectic Euler, and have the same properties. They are studied by Niiranen in [11].

28 Assuming for a given n that we can ﬁnd coeﬃcients c1, . . . , ck, d1, . . . , dk ∈ R s.t. k Y etXH = et(XT +XV ) = ecitXT editXV + O(tn+1), i=1

Qk citXT ditXV we would have an integrator of order n. The composition i=1 e e is a symplectic map as it is a composition of symplectic maps. We deﬁne this question more precisely, and show that it has solutions in Section 6.3. Since the ﬂows ecitXT , editXV are explicitly computable, we approximate etXH with their composition. This results in the numerical method

(i) (i−1) ∂T (i−1) qn+1 = qn+1 + cih ∂p pn+1 (i) (i−1) ∂V (i) pn+1 = pn+1 − dih ∂q qn+1 ,

(k) (k) (0) (0) for i = 1, . . . , k where qn+1 = qn+1, pn+1 = pn+1, qn+1 = qn, and pn+1 = pn. We call the map

k Y cihXT dihXV Φh :(pn, qn) 7→ (pn+1, qn+1) = e e (pn, qn) (7) i=1 the numerical flow of the vector field XH . This numerical flow is a symplectic map. Hence we call the numerical method that generates it a symplectic integrator. Of course, in general there exist symplectic integrators which are not given by a composition of flows. For example, the implicit midpoint rule is a symplectic integrator [6, Theorem 3.4]. Definition 19 We call a one-step numerical method a symplectic integrator if the numerical flow generated by it is a symplectic map when applied to a smooth Hamiltonian system.

6.2 Reversible Hamiltonians

Let U be an open subset of R2n and let T be the map T : U → U;(p, q) 7→ (−p, q).

Deﬁnition 20 A Hamiltonian H : U → R is called reversible if H ◦ T = H. A bijective map Φ: U → U is called reversible if T ◦ Φ = Φ−1 ◦ T .

Clearly a separable Hamiltonian H(p, q) = T (p) + V (q) is reversible if T (p) = T (−p), and the ﬂow generated by a reversible Hamiltonian is reversible. Re- versible Hamiltonians occur frequently in classical mechanics, in fact both of our motivating examples are reversible. They encode the intuition that reversing the initial velocity vectors of a system, should result in the same trajectory as for the backwards evolution of the original system. They are discussed in [9, Chapter 11]. It is desirable to ﬁnd numerical methods which have reversible

29 numerical flows when applied to a reversible Hamiltonian system. It increases our confidence that the long term behaviour of the numerical flow is similar to that of the exact flow. −1 The exact flow Φt of a vector field satisfies Φt = Φ−t, which implies that −1 Φt = Φ−t and Φt ◦ Φ−t = Φ−t ◦ Φt = id.

Definition 21 We say that the numerical flow Φh of a vector field is symmet- −1 ric if Φh = Φ−h and time-reversible if Φh ◦ Φ−h = Φ−h ◦ Φh = id.

Notice that these definitions are equivalent. A numerical flow Φh is symmetric −1 if and only if Φ−h = Φh , which holds if and only if Φh is time-reversible. It is not only desirable that a numerical flow has the same properties as the exact flow, but symmetric methods have a criterion for determining whether their numerical flows are reversible.

Theorem 18 If a numerical ﬂow of a reversible Hamiltonian system satisﬁes T ◦ Φh = Φ−h ◦ T , then Φh is reversible if and only if it is symmetric.

−1 −1 −1 Proof Φ−h ◦ T = T ◦ Φh = Φh ◦ T ⇐⇒ Φ−h = Φh ⇐⇒ Φh = Φ−h.

6.3 Yoshida’s Method For Separable Hamiltonians We will show how the problem of finding coefficients can be solved. Once a solution for n = 2 has been found, this can be used to find solutions for any even n ∈ N, and these solutions define reversible symmetric symplectic integrators. Following the approach described by Yoshida in [15]. Consider the problem: given n ∈ N, A and B complete, non-commutative vector fields of class Cn+1, and t sufficiently small. Find c1, . . . , ck, d1, . . . , dk ∈ R and k ∈ N s.t.

k Y et(A+B) = ecitAeditB + O(tn+1). (8) i=1

To make this precise, by t sufficiently small, and f(t) = O(tn+1) we mean that there exists some δ > 0 and M > 0 s.t. |f(t)| ≤ M|t|n+1 for all t ∈ (−δ, δ). For Qk citA ditB notational convenience we define Φ(t) := i=1 e e . Theorem 19 Let A and B be complete, non-commutative vector fields of class Cn+1, and t sufficiently small.

• For n = 1, any solution to the algebraic equations

k k X X 1 = cj and 1 = dj, j=1 j=1

is a solution of Equation (8).

30 • For n = 2, any solution to the algebraic equations k k k k k−1 j X X 1 X X 1 X X 1 = c , 1 = d , = c d and = c d , j j 2 j i 2 j+1 i j=1 j=1 j=1 i=j j=1 i=1 is a solution of Equation (8).

t(A+B) Qk citA ditB Proof We expand e and i=1 e e as finite Taylor series of order n in terms of t, equate the coefficients of each power of t, and solve the resulting algebraic equations for the coefficients ci and di. By examining the Lagrange form of the remainder, we see that it is clearly O(tn+1) for t in some small interval (−δ, δ). We begin with the left hand side of (8) which is easy to expand in powers of t n X tk tn et(A+B) = (A + B)k + O(tn+1) = I + t(A + B) + ··· + (A + B)n + O(tn+1). k! n! k=0 The difficulties with extending this approach to compute solutions for n > 2 begin to appear when we expand the right hand side of (8). However, we need a solution for n = 2 to apply Yoshida’s method. Thus we proceed with this u Qu citA ditB direct approach. Define ϕl (t) = i=l e e where u ≥ l, u, l ∈ N, and k Qk citA ditB (n) Φ(t) = ϕ1 (t) = i=1 e e . We use the notation f to denote the n-th time derivative of f, i.e. f˙ := f (1), f¨ := f (2),... . We expand Φ(t) as n X tk Φ(t) = Φ(k)(t) + O(tn+1) k! t=0 k=0 t2 tn = Φ(0) + tΦ(0)˙ + Φ(0)¨ + ··· + Φ(n)(0) + O(tn+1), 2! n! ˙ ¨ u where Φ(0) = I. Next, we explicitly compute Φ(0) and Φ(0) fromϕ ˙ l (0) and u ϕ¨l (0) respectively. Thus u d Y ϕ˙ u(t) = ecj tAedj tB l dt j=1

cltA dltB cutA dutB cltA dltB dutB = clAe e ··· e e + e dlBe ··· e

cltA dltB cutA dutB + ··· + e e ··· e duBe u−1 u X j u = clAϕl (t) + cj+1(ϕl (t)Aϕj+1(t)) j=l u−1 cltA dltB u X j cj+1tA dj+1tB u + dl(e Be ϕl+1(t)) + dj+1(ϕl (t)e Be ϕj+2(t)), j=l u and since ϕl (0) = I, u u u X X ϕ˙ l (0) = cj A + dj B. j=l j=l

31 Therefore, for n = 1 equating the coeﬃcients of t gives

k k ˙ k X X A + B = Φ(0) =ϕ ˙ 1 (0) = cj A + dj B, j=1 j=1

Pk Pk which results in the algebraic equations 1 = j=1 cj and 1 = j=1 dj. u Diﬀerentiatingϕ ˙ l (t) termwise and rearranging the resulting expression gives

u−1 u−1 u u X j u X j u ϕ¨l (t) = clAϕ˙ l (t) + cj+1ϕ˙ l (t)Aϕj+1(t) + cj+1ϕl (t)Aϕ˙ j+1(t) j=l j=l

cltA dltB 2 cltA 2 dltB u cltA dltB u + (cldlAe Be + dl e B e )ϕl+1(t) + dle Be ϕ˙ l+1(t) u−1 X j cj+1tA j cj+1tA dj+1tB u + dj+1(ϕ ˙ l (t)e + cj+1ϕl (t)Ae )Be ϕj+2(t) j=l u−1 X j cj+1tA dj+1tB u dj+1tB u + dj+1ϕl (t)e B(dj+1Be ϕj+2(t) + e ϕ˙ j+2(t)). j=l

u Using the expression forϕ ˙ l (0) and the following identities

u u−1 j u u u−1 u u−1 j X X X X X X X X X cjdj + dj+1 ci = cj di, dj ci = cj+1 dj, j=l j=l i=l j=l i=j j=l i=j+1 j=l i=l and j u u−1 u u−1 u 2 X 2 X X X X X dj + dj di + dj+1 di = di , j=l j=l i=j+1 j=l i=l i=l we get that

u−1 u−1 u u X j X u ϕ¨l (0) = clAϕ˙ l (0) + cj+1ϕ˙ l (0)A + cj+1Aϕ˙ j+1(0) j=l j=l u−1 2 2 u X j + (cldlAB + dl B ) + dlBϕ˙ l+1(0) + dj+1(ϕ ˙ l (0) + cj+1A)B j=l u−1 X u + dj+1B(dj+1B +ϕ ˙ j+2(0)) j=l j u 2 u u u−1 X 2 X X X X = ci A + 2 cj di AB + 2 cj+1 dj BA i=l j=l i=j j=l i=l u 2 X 2 + di B . i=l

32 Therefore, for n = 2 comparing the coeﬃcients of t2 gives us

(A + B)2 = A2 + AB + BA + B2

j k 2 k k k−1 k 2 X 2 X X X X X 2 = ci A + 2 cj di AB + 2 cj+1 dj BA + di B , i=1 j=1 i=j j=1 i=1 i=1

Pk Pk 1 which results in the algebraic equations 1 = j=1 cj, 1 = j=1 dj, /2 = Pk Pk 1 Pk−1 Pj j=1 cj i=j di, and /2 = j=1 cj+1 i=1 di. Notice that for n = 1 this theorem gives us the simple solutions k = 1 with c1 = d1 = 1, and k = 2 with c1 = d2 = 0, c2 = d1 = 1. When applied to the vector ﬁelds XT and XV of a separable Hamiltonian these two solutions correspond to the two version of the Symplectic Euler method. 1 Also notice that for n = 2 we get the solutions k = 2 with c1 = c2 = /2, 1 d1 = 1 and d2 = 0, and k = 2 with d1 = d2 = /2, c1 = 0 and c2 = 1. These coeﬃcients lead to the two numerical methods (pn, qn) 7→ (pn+1, qn+1) given by

0 h ∂T qn+1 = qn + 2 ∂p (pn) ∂V 0 pn+1 = pn − h ∂q (qn+1) (9) 0 h ∂T qn+1 = qn+1 + 2 ∂p (pn+1), and 0 h ∂V pn+1 = pn − 2 ∂q (qn) ∂T 0 qn+1 = qn + h ∂p (pn+1) (10) 0 h ∂V pn+1 = qn+1 − 2 ∂q (qn+1). These methods are called the St¨ormer-Verlet or leapfrog methods (compare with [7, Equation (1.5) and (1.6)]). By writing one and a third iterations of either method, we see that the last step of one iteration can be concatenated with 0 0 the ﬁrst step of the next iteration to get the methods (pn, qn+1) 7→ (pn+1, qn+2) 0 0 and (pn, qn) 7→ (pn+1, qn+1), by

∂V 0 0 0 ∂V pn = pn−1 − h qn pn+1 = pn − h qn ∂q and ∂q (11) 0 0 ∂T ∂T 0 qn+1 = qn + h ∂p (pn) qn+1 = qn + h ∂p (pn).

0 0 Under the identification (pn, qn+1) ↔ (pn+1, qn+1) these are the same method, up to requiring different transformations of the initial point (p0, q0). Many properties of the Störmer-Verlet method such as which geometric structures it preserves are described by Hairer et al. in [7]. We hope from the above proof of Theorem 19, it is clear why trying to use this direct approach to find solutions of (8) will quickly become difficult for n > 2. Thus we require a better technique. We will use the BCH formula to construct higher order solutions using a solution for n = 2. However, first we observe that we can reformulate Equation (8).

33 Theorem 20 Given n ∈ N, A and B complete, non-commutative vector fields of class Cn+1, and t sufficiently small. Finding coefficients which satisfy

k Y et(A+B) = ecitAeditB + O(tn+1) i=1 is equivalent to ﬁnding c1, . . . , ck, d1, . . . , dk ∈ R and k ∈ N s.t.

k Y Φ(t) = ecitAeditB = expt(A + B) + O(tn+1) . (12) i=1

Proof Use the BCH formula repeatedly on Φ to combine the exponentials, and compare it to (8). We can apply the BCH formula, as taking t suﬃciently small rescales the norms. This gives

n j n j 2 3 X t X t Φ(t) = eα1t+α2t +α3t +··· = C(j)(0) + O(tn+1) = (A + B)j + O(tn+1) j! j! j=0 j=0 = et(A+B) + O(tn+1),

2 3 where C(t) = eα1t+α2t +α3t +···. Computing C(j)(0) and equating the coeﬃ- cients of each power of t, we get that α1 = (A+B) and α2 = α3 = ··· = αn = 0. Yoshida’s method is built upon a useful property of symmetric methods of Qk citA ditB the form i=1 e e .

Qk citA ditB Lemma 2 Given a symmetric method of the form Φ(t) = i=1 e e . If we can expand Φ(t) as

2 3 4 Φ(t) = exp α1t + α2t + α3t + α4t + ··· , then 0 = α2 = α4 = α6 = ··· .

Proof Φ(t) is symmetric, thus Φ(t) = Φ−1(−t), and

2 3 4 Φ(t) = exp α1t + α2t + α3t + α4t + ··· = Φ−1(−t) 2 3 4 −1 = exp −α1t + α2t − α3t + α4t + ··· 2 3 4 = exp α1t − α2t + α3t − α4t + ··· .

Hence, by equating coeﬃcients we see that α1 = α1, α2 = −α2, α3 = α3, α4 = −α4, ... and therefore 0 = α2 = α4 = α6 = ··· . We use this property to show in detail, the construction for a special case of the general method. This special case deﬁnes a 4-th order symplectic integrator.

34 Theorem 21 Given A and B complete, non-commutative vector ﬁelds of class n+1 t/2A tB t/2A C , and t suﬃciently small. Let Φ2(t) = e e e . Then there exists x0, x1 ∈ R s.t. the composition

Φ4(t) := Φ2(x1t)Φ2(x0t)Φ2(x1t)

5 satisﬁes Φ4(t) = exp t(A + B) + O(t ) . Moreover √ − 3 2 1 x0 = √ , and x1 = √ . 2 − 3 2 2 − 3 2 Proof Observe that the BCH formula, when repeatedly applied to a triple composition eX eY eX , gives 1 eX eY eX = eW =⇒ W = 2X + Y + ([Y, [Y,X]] − [X, [X,Y ]]) 6 7 1 + [X, [X, [X, [X,Y ]]]] − [Y, [Y, [Y, [Y,X]]]] + ··· . 360 360

Now, Φ2 is clearly symmetric, so we apply the BCH formula and Lemma 2 to expand it as

t/2A tB t/2A 3 5 Φ2(t) = e e e = exp α1t + α3t + α5t + ··· ,

1 1 7 where α1 = A+B, α3 = /12[B, [B,A]]− /24[A, [A, B]], α5 = /5760[A, [A, [A, [A, B]]]]+ ··· , etc. Consider the symmetric composition of Φ2’s

Φ4(t) := Φ2(x1t)Φ2(x0t)Φ2(x1t).

t(A+B)+O(t5) If we can ﬁnd x0, x1 ∈ R s.t. Φ4(t) = e ,Φ4 will be 4th order integrator. We can apply BCH formula to Φ2(x0t) and Φ2(x1t) to get

3 3 5 5 Φ2(x0t) = exp α1x0t + α3x0t + α5x0t + ··· 3 3 5 5 Φ2(x1t) = exp α1x1t + α3x1t + α5x1t + ··· . While applying it to their composition, gives

3 3 3 5 5 5 Φ4(t) = exp α1t(2x1 + x0) + α3t (2x1 + x0) + α5t (2x1 + x0) + ··· .

P 2k+1 P 2k+1 To see this, set X = k=0 α2k+1x1t , Y = k=0 α2k+1x0t . Computing the following terms results in

3 3 3 5 5 5 2X + Y = α1t(2x1 + x0) + α3t (2x1 + x0) + α5t (2x1 + x0) + ··· ,

5 5 and as [α1, α1] = 0, we get [Y [Y,X]] = t (··· ). Similarly [X[X,Y ]] = t (··· ). 5 Since Φ4 should be of 4th order, we can ignore terms with t and higher powers. After the iterated Lie brackets with three of X or Y , the brackets have at least ﬁve of X or Y . By the bilinearity of the Lie brackets, these terms are of order 5 or higher. Hence we ignore them.

35 Therefore, we have that

3 3 3 5 5 5 Φ4(t) = exp α1t(2x1 + x0) + α3t (2x1 + x0) + α5t (2x1 + x0) + ··· = expt(A + B) + O(t5) .

By equating coeﬃcients of t and t3, we get the algebraic equations

3 3 2x1 + x0 = 1 and 2x1 + x0 = 0, which have the unique real solution √ − 3 2 1 x0 = √ , and x1 = √ . 2 − 3 2 2 − 3 2

Qk citA ditB We can determine the coeﬃcients in Φ4(t) = i=1 e e from the con- A A 2A stants x0 and x1 in Theorem 21 by remembering that e e = e . This gives

Φ4(t) = Φ2(x1t)Φ2(x0t)Φ2(x1t) x t t = exp 1 A exp{x tB} exp (x + x )A exp{x tB} 2 1 2 1 0 0 t x t exp (x + x )A exp{x tB} exp 1 . 2 1 0 1 2

Hence, a solution to problem (8) for n = 4 is given by the coefficients √ 1 1 1 1 − 3 2 c1 = c4 = x1 = √ , c2 = c3 = (x0 + x1) = √ , 2 2(2 − 3 2) 2 2(2 − 3 2) √ 1 − 3 2 d1 = d3 = x1 = √ , d2 = x0 = √ , and d4 = 0. 2 − 3 2 2 − 3 2 We now treat the general case of Yoshida’s method for recursively constructing an integrator of arbitrary even order. Theorem 22 Given A and B complete, non-commutative vector fields of class Cn+1, and t sufficiently small. Then the integrators defined recursively by

t t Φ2(t) = exp{ /2A} exp{tB} exp{ /2A} (13) (n) (n) (n) Φ2n+2(t) =: Φ2n x1 t Φ2n x0 t Φ2n x1 t ,

2n+3 (n) (n) for n ≥ 1 statisfy Φ2n+2(t) = exp t(A + B) + O(t ) for some x0 , x1 ∈ R. Moreover √ − 2n+1 2 1 x(n) = √ , and x(n) = √ . 0 2 − 2n+1 2 1 2 − 2n+1 2

36 Proof For this general case we will repeat the procedure from the proof of Theorem 21. We will skip most of the details, due to the arguments being similar to that special case. It is clear that Φ2n(t) is a symmetric method for n ≥ 1 by deﬁnition as Φ2(t) is symmetric. Thus, we assume Φ2n is of order 2n. We expand it using the BCH formula, and Lemma 2 as

2n+1 2n+3 Φ2n(t) = exp t(A + B) + t α˜2n+1 + t α˜2n+3 + ··· .

(n) (n) We can now try to ﬁnd x0 , x1 ∈ R s.t. (n) (n) (n) Φ2n+2(t) := Φ2n x1 t Φ2n x0 t Φ2n x1 t is of order 2n+2. Using the BCH formula with similar arguments to the special case, we get that

2n+1 2n+1 (n) (n) 2n+1 (n) (n) Φ2n+2(t) = exp t 2x1 + x0 (A + B) + t 2 x1 + x0 α˜2n+1 + ···

= expt(A + B) + O(t2n+3) .

Hence x0 and x1 are solutions of the algebraic equations 2n+1 2n+1 (n) (n) (n) (n) 2x1 + x0 = 1 and 2 x1 + x0 = 0, which have the unique real solution √ − 2n+1 2 1 x(n) = √ , and x(n) = √ . 0 2 − 2n+1 2 1 2 − 2n+1 2

Qk citA ditB We can determine the coefficients in Φ2n+2(t) = i=1 e e recursively (n) (n) from the constants x0 and x1 in Theorem 22. Writing the coefficients explicitly while straight forward for any particular n, appears to requires a combina- torial argument for the general case. We will not try that here, however we can determine k. Our construction of Φ2n+2 uses Φ2n, 3 times and Φ4 required Φ2, 3 times. If we construct the intermediate Φ2j’s in the same manner, this implies that n Φ2n+2 requires Φ2, 3 times. Each Φ2 uses 3 exponentials, but in a composition of Φ2’s the last exponential of one Φ2 can be combined with the first exponential n n n of the next. Hence Φ2n+2 requires 3 · 3 − 3 + 1 = 2 · 3 + 1 exponentials, and 2·3n+1 n our k for this solution to (8) is k = ceiling( 2 ) = 3 + 1. When applied to a separable Hamiltonian, each exponential uses one function ∂T ∂V evaluation of either ∂p or ∂q . Therefore, higher order integrators constructed with this method quickly become computationally expensive, requiring 2·3n +1 functions evaluations for an integrator of order 2n + 2. However, methods for deriving more economical integrators based on this construction have been found by Yoshida in [15] and Suzuk in [12].

37 6.4 Properties Of The Yoshida Symplectic Integrators We summarise some of the important properties of the Yoshida symplectic integrators. We have already seen that they are explicit and symmetric by construction. We will show they are symplectic and reversible, before using backward error analysis to show that they preserve a perturbed Hamiltonian. Theorem 23 The Yoshida symplectic integrators constructed in Theorem 22 are symplectic integrators.

t t Proof Φ2(t) = exp{ /2A} exp{tB} exp{ /2A} is a symplectic integrator, as it is a composition of symplectic maps. Assume for some n ≥ 1 that Φ2n(t) is (n) (n) (n) a symplectic map. Then Φ2n+2(t) = Φ2n x1 t Φ2n x0 t Φ2n x1 t is again (n) (n) just a composition of symplectic maps, where x0 , x1 are the constants from Theorem 22. By induction, for each n ≥ 1, Φ2n(t) is a symplectic integrator. As a consequence of Deﬁnition 11, a symplectic map on a 2n dimensional space with symplectic form ω preserves ωk for k = 1, . . . , n. Thus, a symplectic integrator applied on that space actually preserves n invariants of the system, i.e. it preserves all ωk. In particular, symplectic integrators preserve phase space volume since they preserve ωn. Theorem 24 The Yoshida symplectic integrators constructed in Theorem 22 are reversible when applied to a reversible Hamiltonian system.

t t Proof When Φ2(t) = exp{ /2A} exp{tB} exp{ /2A} is applied to a reversible Hamiltonian system, it satisﬁes T ◦ Φ2(t) = Φ2(−t) ◦ T . To see this, notice that (pn+1, qn+1) = Φ2(t)(pn, qn) implies that T ◦ Φ2(t)(pn, qn) = (−pn+1, qn+1). ∂T ∂T Since the Hamiltonian is reversible, T (p) = T (−p), thus ∂p (p) = − ∂p (−p). From this and the iteration formula for Φ2(−t) derived from 9, we see that Φ2(−t) ◦ T (pn, qn) = (−pn+1, qn+1). Hence via Theorem 18 Φ2(t) is reversible since it is symmetric. Assume for some n ≥ 1 that Φ2n(t) is symmetric and satisﬁes T ◦ Φ2n(t) = (n) (n) (n) (n) (n) Φ2n(−t)◦T . Then Φ2n+2(t) = Φ2n x1 t Φ2n x0 t Φ2n x1 t where x0 , x1 are the constants from Theorem 22, is clearly symmetric. Notice

(n) (n) (n) Φ2n+2(−t) ◦ T = Φ2n − x1 t Φ2n − x0 t Φ2n − x1 t ◦ T (n) (n) (n) = Φ2n − x1 t Φ2n − x0 t ◦ T ◦ Φ2n x1 t = T ◦ Φ2n+2(t),

Thus Φ2n+2(t) is reversible. By induction, for each n ≥ 1, Φ2n(t) is reversible.

6.5 Backward Error Analysis The idea behind backward error analysis is to interpret the numerical solution yn+1 = Φh(yn) of a diﬀerential equationy ˙ = f(y) as the exact solution of a

38 modified differential equation, y˜˙ = fh(˜y). We assume that the modified equation is given by a power series in terms of the step-size h

2 fh(˜y) = f(˜y) + hf2(˜y) + h f3(˜y) + ··· , (14) and that yn =y ˜(nh). In general, we do not expect this series to converge, hence suitable truncations must be taken. By studying the difference between f(y) and fh(y), we can gain some insight into the qualitative behaviour of numerical solutions toy ˙ = f(y). In order for the modified equation (14) to be useful, we need to determine the coefficient functions fj for j ≥ 2. Assume that Φh can be expanded as a series 2 3 Φh(˜y) =y ˜ + hf(˜y) + h d2(˜y) + h d3(˜y) + ··· , where the functions dj for j ≥ 2 are know. For example, the expansion of the Forward Euler method is Φh(y) = y + hf(y) with dj = 0 for all j ≥ 2. We ˙ expandy ˜, the solution of y˜ = fh(˜y), as a Taylor series around a fixed t ∈ R. 1 y˜(s) =y ˜(t) + y˜˙(t)(s − t) + y˜¨(t)(s − t)2 + ··· 2 1 =y ˜(t) + f (˜y)(s − t) + f 0 (˜y)f (˜y)(s − t)2 + ··· h 2 h h Set s = t + h. 2 =⇒ y˜(t + h) =y ˜(t) + h f(˜y) + hf2(˜y) + h f3(˜y) + ··· (15) h2 + (f 0(˜y) + hf 0 (˜y)) + ··· )(f(˜y) + hf (˜y) + ··· ) + ··· 2 2 2

Since yn =y ˜(nh), we havey ˜(t + h) = Φh(˜y). Comparing the coeﬃcients of powers of h, gives 1 f (˜y) = d (˜y) − f 0(˜y)f(˜y), 2 2 2 1 1 f (˜y) = d (˜y) − f 00(˜y)f(˜y)2 + f 0(˜y)2f(˜y) − (f 0 (˜y)f(˜y) + f 0(˜y)f (˜y)), 3 3 3! 2 2 2 . . where the fj for j > 3 are computed similarly. We would like use properties of a numerical method to study its associated modiﬁed equation. We would like to understand what powers of h appear in it, to know how it deviates from the original equation. To begin, we use the order of the numerical integrator.

Theorem 25 Suppose that a numerical method yn+1 = Φh(yn) is of order n, i.e. n+1 n+2 Φh(y) = ϕh(y) + h δn+1(y) + O h , n+1 where ϕt is the exact ﬂow of y˙ = f(y) and h δn+1(y) is the leading term of the local truncation error. Then the modiﬁed equation is of the form

n n+1 y˜˙ = f(˜y) + h fn+1(˜y) + h fn+2(˜y) + ··· , with fn+1(˜y) = δn+1(˜y).

39 Proof Our computation of the fj (see (15)) shows that fj = 0 for j = 2, . . . , n n+1 if and only if Φh(y) − ϕh(y) = O(h ). We can use the symmetry of a numerical method to say more about the coeﬃ- cient functions. Lemma 2 makes similar use of symmetry.

Theorem 26 The coefficient functions fj(y) with j ≥ 2 of a symmetric method Φh satisfy f2k = 0 for k ≥ 1. −1 Proof Φh is symmetric, thus Φh = Φ−h. Hence, the solutiony ˜ of the modified equation for Φh must satisfyy ˜ = Φ−h(˜y(t + h)). This is equivalent toy ˜(t−h) = Φ−hy˜(t). If we replace h with −h in the above series (beginning with (14)) we j+1 used to compute fj, we find fj(˜y) = (−1) fj(˜y), which implies f2k = 0. Therefore, the modified equation 14 of a symmetric method is an expansion in even powers of h. We can fruitfully apply this since the Yoshida symplectic integrators are symmetric. We will now specialise to separable Hamiltonian systemsx ˙ = J∇H(x) with H(p, q) = T (p)+V (q). We will again make use of the BCH formula to show that the modified equations of the numerical integrators constructed in Section 6.3, are Hamiltonian. For these, we introduce the Poisson bracket and its relation to the Lie bracket of Hamiltonian vector fields. For the coordinate free definition and more details see [9, Section 10.2]. Definition 22 Let U ⊂ R2n be open with coordinates (p1, . . . , pn, q1, . . . , qn), and let f, g : U → R be smooth. The Poisson bracket of f and g is n X ∂f ∂g ∂f ∂g {f, g} := − . ∂qi ∂pi ∂pi ∂qi i=1 Theorem 27 ([9, Remark 10.23]) Let U ⊂ R2n be open, and let f, g : U → R be smooth Hamiltonians. Then the relationship between the Hamiltonian vector fields Xf , Xg and X{f,g} is

X{f,g} = [Xg,Xf ]. We show how these techniques work using the Symplectic Euler and the Leapfrog method as examples, then state a general theorem about the compo- sitional integrators from Section 6.3. Theorem 28 For a separable Hamiltonian H(p, q) = T (p) + V (q), the Sym- hXT hXV plectic Euler method Φh = e e is the exact solution to the Hamiltonian system generated by the perturbed Hamiltonian H˜ given by the formal power series 2 3 H˜ = H + hH2 + h H3 + h H4 + ··· .

The coefficient functions Hj are written in terms of the iterated Poisson brackets of T and V : 1 1 H = {T,V },H = ({{T,V },V } + {{V,T },T }),... 1 2 2 12 and the coefficients fj of the modified equation are fj = J∇Hj for j ≥ 2.

40 Proof By the BCH formula, 1 X˜ = log(exp{hX } exp{hX }) H h T V h h2 = X + X + [X ,X ] + ([X , [X ,X ]] + [X , [X ,X ]]) + ··· T V 2 T V 12 T T V V V T Using X{f,g} = [Xg,Xf ]. h h2 = X + X + X + X + ··· , H 2 {T,V } 12 {{T,V },V } {{V,T },T } which is a Hamiltonian vector field since it is a sum of Hamiltonian vector fields. As J∇ is a linear operator, we can see this Hamiltonian vector field is generated ˜ ˜ by H. Thus, XH = XH˜ and fj = J∇Hj. Theorem 29 For a separable Hamiltonian H(p, q) = T (p)+V (q), the Leapfrog method h h Φh = exp{ /2XT } exp{hXV } exp{ /2XT } is the exact solution to the Hamiltonian system generated by the perturbed Hamil- tonian H˜ given by the formal power series

1 1 H˜ = H + h2 {{T,V },V } − {{V,T },T } + O(h4). 12 24

Proof By the BCH formula

1 h h X˜ = log exp X exp{hX } exp X H h 2 T V 2 T 1 1 = X + X + h2 [X , [X ,X ]] − [X , [X ,X ]] + O(h4). T T 12 V V T 24 T T V

˜ Following the proof of Theorem 28, we get H of the correct form. Theorem 30 Given a separable Hamiltonian H(p, q) = T (p) + V (q), and an Qk citXT ditXB n-th order method of the form Φh = i=1 e e , where the coeﬃcients ci and di are a solution of equation (8). Then Φh is the exact solution to the Hamiltonian system generated by a perturbed Hamiltonian of the form

n n+1 H˜ = H + h Hn+1 + O(h ).

n Proof By Theorem 20, Φh = exp{h(XT + XV + O(h ))}. From its proof, we see that the higher order terms are constructed by repeated application of the BCH formula. Thus, they consist of iterated Lie brackets of XT and XV . Using the relationship between the Poisson bracket and the Lie bracket, we see ˜ n n+1 that XH = XH + h XHn + O(h ) is a Hamiltonian vector ﬁeld. It is clearly ˜ n n+1 generated by the perturbed Hamiltonian H = H + h Hn+1 + O(h ).

41 Notice that, since the Yoshida symplectic integrators Φ2n are symmetric, their associated Hamiltonians are of the form 2n 2n+2 H˜ = H + h H2n+1 + O(h ). We can use this to derive the error in the energy. Given a separable Hamiltonian H(p, q) = T (p) + V (q) with initial point (p0, q0) and exact solution (p(t), q(t)). Let (pn+1, qn+1) = Φh(pn, qn) be its numerical solution, with Φh being a method of order n as in Theorem 30. We have that H˜ is preserved exactly by Φh, thus H(p(nh), q(nh)) − H˜ (pn, qn) = H(p0, q0) − H˜ (p0, q0) := H0 − H˜0. The error in the energy in the n-th time step is

En := H(p(nh), q(nh)) − H(pn, qn) (16)

= H˜ (pn, qn) − H˜ (pn, qn) + H(p(nh), q(nh)) − H(pn, qn) n n+1 = H0 − H˜0 + h Hn+1(pn, qn) + O(h ). (17)

We see that the error in the energy is of order n. When the series for H˜ is convergent, the error remains constant for all time. Otherwise, if h is too large, the series for H˜ may diverge quickly, thus having a reasonably small error in the energy, is no longer guaranteed. There are two other interesting consequences of Theorem 30. First, assume a method Φh preserves H exactly, i.e. H˜ = H. Hence, the solution of modified equation (i.e. the numerical solution) is the same as the solution of the original Hamiltonian up to a reparametrization of time. A more general version is proved in [17] for the case of Hamiltonian systems who’s invariants are all functions of the Hamiltonian. This implies that we cannot hope to find symplectic integrators which preserve the energy exactly. Second, the proof of Theorem 30 uses the BCH formula which relies on the step size being fixed. This suggests that trying to use those integrators with a variable step size may be problem- atic, as it is not clear how to compute the modified Hamiltonian of the resulting method. Moreover, it seems possible that the modified equation of the resulting method may not be Hamiltonian. Stronger statements can be made about the local error in the solution and the long-time behaviour of the error in the energy. This requires appropriate analyticity assumptions on the differential equation and the numerical method. We will not describe this theory here, see [6, Chapter IX.7 and IX.8] for details. As an example, using it gives that energy growth of symplectic integrators is bounded. Theorem 31 ([6, Theorem 8.1, pg. 312]) Consider a Hamiltonian system with analytic H : D → R (where D ⊂ R2n), and apply a symplectic numerical method Φh(y) of order r with step size h. If the numerical solution stays in the compact set K ⊂ D, then there exists h0 and N = N(h) (appropriately chosen [6, Theorem 7.6]) such that

−h0/2h r H˜ (yn) = H˜ (y0) + O(e ) and H(yn) = H(y0) + O(h ) over exponentially long time intervals nh ≤ eh0/2h.

42 7 Numerical Simulations

We will test the performance of the Yoshida symplectic integrators by applying them to our motivating examples, and comparing the results to more commonly used methods. We will compare the symplectic methods: Symplectic Euler, Leapfrog and 4th order Yoshida (Theorem 21), to the well known methods: Forward and Backward Euler, and the explicit 4th order Runge-Kutta method, RK4. We use these methods with a constant step size for a “fair” comparison. We will apply them ﬁrst to the ideal pendulum (Example 2), before consider the more interesting case of the Kepler problem (Example 3).

7.1 The Ideal Pendulum Recall that the ideal pendulum has Hamiltonian 1 H(p, q) = p2 − cos q, 2 which is separable and generates the dynamics ∂H ∂H p˙ = − = − sin q andq ˙ = = p. ∂q ∂p

We use the initial points (p0, q0) = (0, 2.7),(0, 2.96) and (5, 27) for the solution curves with energy H0 = 0.9, 9.8 and 13.4 respectively. We plot the trajectories and energies of the numerical solutions for a variety of step sizes and total times.

Figure 3: H0 = 0.9, with step size h = 0.01, and total time T = 30.

The Forward and Backward Euler methods diverge from the exact solution relatively quickly, even for this small step size. Their energies change monotonically while Symplectic Euler’s energy oscillates around H0 despite being of the same order.

43 Figure 4: H0 = 0.9, with step size h = 0.5, and total time T = 300.

Figure 5: H0 = 0.98, with step size h = 0.7, and total time T = 3000.

Using a much larger step size and longer time, it is clear that RK4’s energy also decreases monotonically, though only very slowly due to being of 4th order. We see that the oscillations in Symplectic Euler’s energy cause its trajectory to be distorted from that of the exact solution. Its qualitative behaviour is the same as for an exact solution with H0 < 1. However, for a slightly larger step size and energy, this is no longer the case. The oscillations in its energy are large enough to cause its trajectory to leave the basin of H0 < 1. The higher order symplectic methods still exhibit the correct behaviour, due to their energies deviating less from H0.

44 Figure 6: H0 = 0.9, with step size h = 1, and total time T = 150.

Figure 7: H0 = 0.9 (left) and H0 = 13.4 (right), with step size h = 1, and total time T = 3000.

For h = 1, none of the symplectic methods have the correct qualitative behaviour for very long. However, their energies remain bounded even over a long time period, though the oscillations increase in size as energy of the exact solution increases. This is due to the fact that for H0 > 1 the exact solution for the momentum p(t) no longer oscillates around 0, this eﬀects the error in the energy En (Equation (16)). Truncating the modiﬁed Hamiltonians of Symplectic Euler and Leapfrog gives

h 2 h H˜ = H + {p /2, − cos q} = H + (p sin q) 2 2

45 and 2 1 2 1 2 2 H˜ = H + h {{p /2, − cos q}, − cos q} − {{− cos q, p /2}, p /2} 12 24 sin2 q p2 cos q = H + h2 − . 12 24

We see that p can have a large eﬀect on En when h is large. Therefore larger oscillations can occur at higher energies. From the next plots we can see that the truncations of En are quite close to the error in the energy along the numerical solutions.

Figure 8: Error in the energy of the numerical solution vs the truncation of the error in the energy (Equation (17)), with H0 = 0.9, step size h = 0.1 (left) and h = 0.5 (right), and total time T = 10.

7.2 The Kepler Problem We now test our symplectic integrators on the Kepler problem. Its Hamiltonian

kP k2 1 H(P,Q) = − , 2 kQk is separable and generate the dynamics Q P˙ = Q¨ = − , Q˙ =: P, kQk3

2 where P,Q ∈ R . We use the initial point (p0, q0) = (0, 1.35, 1, 0) for the exact orbit with energy H0 = −0.09, eccentricity e = 0.8, and angular momentum L0 = 1.35. We plot the trajectories in conﬁguration space, and the energies and angular momentums of the numerical solutions for a variety of step sizes and total times.

46 Figure 9: h = 0.1 (left) and h = 0.5 (right) with total time T = 300. 47 Again, the trajectories produced by Forward Euler and RK4, deviate from the exact solution since the energy and angular momentum increase or decrease along their computed solutions. We see that the symplectic methods conserve the angular momentum exactly. This is due to its relationship to the symplectic 4 form on R , and interpretation as the area of the parallelogram spanned by p0 and q0.

Figure 10: h = 0.1 with total time T = 3000.

Figure 11: h = 0.5 with total time T = 3000.

The symplectic integrators preserve energy relatively well with the oscillations remaining bounded even over a long time period, using the larger step size h = 0.5. However, Symplectic Euler’s “average” energy is shifted upwards with the shift increasing in size as the step size increases. A similar eﬀect is seen in

48 Figure 7 (right). This is probably due to the first order term in its modified Hamiltonian. For the larger step size this shifting is enough to significantly distort the shape of the numerical orbit. The oscillations in the energies of the Leapfrog and 4th order Yoshida methods cause their numerical orbits to process around (0, 0), though they retain the shape of the exact orbit. We see that the spike in energy occurs at the beginning of each orbit. As expected, the procession is slower for the 4th order method. It occurs because the numerical orbits are the exact solutions to perturbations of the Kepler problem. They can cause relatively large (but periodic) errors in the numerical solution. However, the processions appear regular enough to be counteractable with a clever technique. From our numerical experiments, we find that the symplectic integrators that we have considered here, perform well compared to more commonly used methods, at least for large step sizes and long time intervals. We see that they have bounded energy growth, and that their qualitative behaviour mimics well the behaviour of the analytic system. In some cases, theses advantages can likely be nullified by employing methods using an adaptive step size or other techniques. In other cases, for example integrating over extremely long time periods, symplectic methods will still be preferable.

49 8 Conclusion

In this thesis, we have seen how Hamiltonian systems have an underling geometry and how this geometry can be exploited to construct integrators for separable Hamiltonian systems. We described Yoshida’s method for constructing this class of integrators, and showed that they have nice properties. These include being explicit, symmetric, reversible, and of arbitrary even order. We saw that when applied to our motivating examples, they exhibited favourable long term behaviour for relatively large step sizes. The work here could be extended in many diﬀerent directions. We could study the papers of Yoshida [15] or Suzuk [12] to derive more economical integrators similar to this class. We could look at how these techniques can be extended to non-separable Hamiltonians, for example [13] or [4]. We could study variational integrators [6, Chapter VI.6] and see how they can be used to counteract the procession in the numerical orbits computed by the Yoshida symplectic integrators. Furthermore, we could test these symplectic integrators performance on larger examples, such as n-body simulation for n > 2, or described how symplectic integrators are used in video games and movie production.

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