LINEARIZATION METHODS FOR VARIATIONAL INTEGRATORS AND EULER-LAGRANGE EQUATIONS

Transcription

1 LINEARIZATION METHODS FOR VARIATIONAL INTEGRATORS AND EULER-LAGRANGE EQUATIONS Tesis by AnilN.Hirani Avisors James R. Arvo, Jerrol E. Marsen In Partial Fulfillment of te Requirements for te Degree of Master of Science in Computer Science California Institute of Tecnology Pasaena, California April 000

2 ii c April 000 Anil N. Hirani All rigts Reserve

3 iii Acknowlegements I like to tank my avisors Prof. Jerrol E. Marsen an Prof. Jim Arvo for teir invaluable elp an avice. Te comments mae by te members of my caniacy committee were also very elpful. Besies my two avisors, te committee members were Professors Peter Scröer, Alan Barr an Tom Hou. I am also tankful to Mattew West an Matieu Desbrun for some very elpful suggestions. In te text I ve cite te specific result were Mattew s suggestions were most important. Matieu was also generous wit is avice an gave me some very goo suggestions. Once, e gave me a table of coefficients for finite ifferences an tat really elpe precipitate te completion of tis work. Discussions wit Antonio Hernanez, Dong-Eui Cang an Sameer Jalnapurkar on some geometric mecanics erivations were also elpful.

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5 v LINEARIZATION METHODS FOR VARIATIONAL INTEGRATORS AND EULER-LAGRANGE EQUATIONS by AnilN.Hirani Abstract In Partial Fulfillment of te Requirements for te Degree of Master of Science in Computer Science Hamiltonian systems arise in a wie variety of iealize pysical systems an Hamilton s equations often must be solve numerically. In general, traitional finite ifference metos for numerically integrating orinary ifferential equations o not take into account te special structure of Hamilton s equations. One woul like te numerical solution to preserve some or all of te properties of te continuous time solution. Variational integrators, also calle iscrete Euler-Lagrange equations in tis tesis, provie a way to o tat. Tis tesis is about ifferentiating, i.e linearizing Euler-Lagrange equations an iscrete Euler-Lagrange equations by ifferentiating te Lagrangian or iscrete Lagrangian. Te original motivation for ifferentiating te Lagrangian was to evelop iger orer variational integrators. However it was foun tat te resulting intgerators are not iger orer. Noneteless, te pursuit of tis goal turne up several interesting results : i. Differentiating a Lagrangian along te acceleration, i.e along ( q, q) results in te Euler- Lagrange equation 0 = 0 (Lemma, page 0). ii. Differentiating a Lagrangian along a general irection (u, u), etermining te Euler- Lagrange equation an ten setting u = q results in an Euler-Lagrange equation tat is te time erivative of te original Euler-Lagrange equation corresponing to te original Lagrangian (Lemma 3, page 1). iii. A similar operation on te iscrete Lagrangian gives a iscrete time erivative of te iscrete Euler-Lagrange equation corresponing to te original iscrete Lagrangian (Lemma 4, page 3). iv. Te operations escribe in te previous two items are part of a commutative iagram (Teorem 1, page 8). v. Numerical measurements an analytical calculation of orer of accuracy for specific examples using tis approac sow tat iger orer metos are not obtaine in general (Capter 5, especially Section 5.4 an 5.5). vi. Te manifol consisting of triples tat satisfy a iscrete Euler Lagrange equation is a smoot manifol (Proposition 4, page 35). vii. Te algoritm on TQ TQ is symplectic an te algoritm tat upates triples of points on Q is presymplectic (Teorem, page 36). I also give a erivation of a symplectic form on TQ TQ were Q is te configuration manifol. Most proofs given in tis tesis are one using local coorinates. A possible irection for furter stuy is to give intrinsic proofs in orer to gain a eeper unerstaning of te geometric meaning of te operations of tis tesis.

6 vi Table of Contents Acknowlegments iii Abstract v 1 Introuction Variational Integrators Results Preliminaries 7.1 Lagrangian Mecanics Geometry Hamiltonian Mecanics Legenre Transform Algoritms Variational Integrators Discrete Mecanics Invariance Properties Linearization Motivation Differentiating Lagrangian along ( q, q) Continuous Time Linearization Discrete Time Linearization Discrete an Continuous Linearization Summary Invariance Properties Symplectic form on TQ TQ Presymplectic algoritm Examples an Conclusion Falling Particle Continuous Discrete Simple Harmonic Oscillator Continuous Discrete Duffing Oscillator Continuous Discrete Numerical Results Accuracy Calculations

7 vii 5.6 Using more accurate u k Conclusion Bibliograpy 53

8 1 Capter 1 Introuction Te traitional meto of numerically solving te initial value problem (IVP) for orinary ifferential equations (ODEs) is to iscretize erivatives by finite ifferences. For example for q R an te equation q = f(q, t), (1.1) t one can coose a step size for stepping te time iscretely an replace te left an sie of (1.1) by q k+1 q k. Muc of te evelopment of numerical metos for IVPs as concerne itself wit wat to o wit te rigt an sie of equation (1.1) in orer to iscretize te equation completely. Te evelopment as resulte in a well evelope teory of truncation errors (accuracy), stability, an convergence. Te books by Hairer et.al [6], Lambert [11], Gear [3] an Iserles [8] are excellent sources of information about numerical metos for ODEs. Te table of coefficients for finite ifference approximations for various orer erivatives on pages 16 an 17 of Fornberg s book [1] is also quite any. But consier now te following simple system of equations, given a scalar function H : R R R suc tat q = H (1.) p ṗ = H q. Tis is an example of a Hamiltonian system of equations, wic is te form tat many iealize equations in pysical sciences take. Some equations tat may not at first appear to be in tis form are also actually Hamiltonian in a sense efine in Capter an in Capter 5 of te book by Marsen an Ratiu [15]. Some examples of equations tat are Hamiltonian are : equation of a particle in a potential, rigi boy equations, rigi boy wit a fixe point in gravitational fiel, Euler equations for an ieal, incompressible flui, Maxwell-Vlasov equations for a collisionless plasma, Maxwell s equations an many oters. Te stuy an solution of Hamiltonian equations is tus a subject wit a long traition. It can be sown tat for suc an equation like (1.) te function H is constant along solutions,i.e Ḣ(q(t),p(t)) = 0 for all time t if (q(t),p(t)) is a solution of te equations 1..

9 If equation (1.) is iscretize using an arbitrary meto tere is no guarantee tat te iscrete solutions will also satisfy te property of keeping H constant along solutions. Even worse, tere may be a systematic biase error introuce wic will make te compute solution rift from te true solution. Since H is often te energy of te system tis is not acceptable for many situations. Anoter important property of Hamiltonian equations is tat te flow is symplectic, i.e te flow of te vector fiel efine by te equations preserves a certain ifferential form (tis will be efine in Capter ). One consequence of symplectic flow is tat areas in pase space are conserve. An arbitrary iscretization of equation (1.) may not yiel an algoritm tat is symplectic. However, one woul like a numerical solution to ave some of tese properties like energy conservation or symplecticness. For example, Figures 1.1 an 1., wic are taken from Iserles an Zanna [7] sow ow initial conitions evolve uner te forwar Euler an implicit mipoint metos. Te smiley face sown is a clou of initial conitions wic are followe over time. Te equation being solve is tat of a simple armonic oscillator, namely q + q = 0 an so te smiley face soul just rotate rigily troug pase space. But in Figure 1.1 te smiley face enlarges. Figure 1.sows te ieal beavior. Te particular implicit meto use in Figure 1.is symplectic, so it conserves pase space area as one steps troug time. Figure 1.1: Pase flow for armonic oscillator as approximate by forwar Euler meto. Horizontal axis is position an vertical axis is momentum. Te uprigt (rigtmost) smiley face is a collection of initial conitions an oters are te location of te collection at various times. From Iserles an Zanna [7]. To see a comparison of te momentum an energy beavior of fourt orer non symplectic Runge-Kutta versus tat of a symplectic algoritm please refer to Figure 13 on page

10 3 Figure 1.: Pase flow for armonic oscillator as approximate by implicit mipoint rule. Horizontal axis is position an vertical axis is momentum. Te uprigt (rigtmost) smiley face is a collection of initial conitions an oters are te location of te collection at various times. From Iserles an Zanna [7]. 18 of Marsen et.al [1]. It is sown tere tat te non symplectic Runge-Kutta meto oes not preserve momentum wile te symplectic meto oes. Furtermore, te energy compute by symplectic metos appears to stay near te true value wile tat compute by te Runge-Kutta meto oes not. Te beavior isplaye by te symplectic integrators in Figure 13 of Marsen et. al [1] is in some sense te best possible because of a teorem prove in Ge an Marsen [], tat a constant time stepping integrator cannot preserve te symplectic form, momentum, an energy simultaneously. However, Kane et.al [9] sow tat by using aaptive time stepping an an appropriate efinition of symplecticness one can inee get symplectic, momentum an energy preserving algoritms. In eveloping symplectic integrators early researcers use generating functions (wic are symplectic transformations) an Hamilton-Jacobi equation. A more recent approac is to create variational integrators using iscrete Euler-Lagrange equations, a meto wic is escribe in te next section. 1.1 Variational Integrators If one works wit te Hamiltonian, symplectic integrators for Hamiltonian equations can be create using generating functions. In general tis is more complicate tan te meto iscovere by Veselov [19]. Tis meto, wic uses te Lagrangian instea of te Hamiltonian, can be caracterize as te creation of integrators base on iscrete variational principles.

11 4 It is escribe an furter evelope by Wenlant an Marsen [0]. Given any Hamiltonian system of equations, uner some general conitions a corresponing Lagrangian exists an te Hamiltonian equations are equivalent to some variational principle. Tis will be mae clearer in Capter 3. Te meto is to iscretize te continuous Lagrangian an use a iscrete variational principle. At least tis was te current wisom wen I starte working on tis problem. In recent work Kane et.al [10] an West [17] ave sown tat te iscrete Lagrangian soul be consiere to be an approximation for te action integral over one time step. But tis point of view will not be explore in tis tesis. Rougly speaking, te meto use in tis tesis is to replace te position variable by some position an te velocity variable in te Lagrangian by some finite ifference. Finally one replaces a variational principle in wic an integral appears by one in wic a sum appears. Extremizing accoring to te principle yiels a iscrete Euler-Lagrange equations wic form te basis for te algoritm. It can be prove tat te algoritm is symplectic an momentum preserving. 1. Results In tis tesis I present some results regaring ifferentiating, i.e linearizing Lagrangians an iscrete Lagrangians, te resulting Euler-Lagrange an iscrete Euler-Lagrange equations an te connections between all tese. Te original motivation for ifferentiating te Lagrangian was to evelop iger orer variational integrators. However it was foun tat te resulting intgerators are not, in general, iger orer. Tus te iscrete sie results escribe below are not computationally useful. Noneteless, te pursuit of tis goal turne up several interesting results : i. Differentiating a Lagrangian along te acceleration, i.e along ( q, q) results in te Euler- Lagrange equation 0 = 0 (Lemma, page 0). ii. Differentiating a Lagrangian along a general irection (u, u), etermining te Euler- Lagrange equation an ten setting u = q results in an Euler-Lagrange equation tat is te time erivative of te original Euler-Lagrange equation corresponing to te original Lagrangian (Lemma 3, page 1). iii. A similar operation on te iscrete Lagrangian gives a iscrete time erivative of te iscrete Euler-Lagrange equation corresponing to te original iscrete Lagrangian (Lemma 4, page 3). iv. Te operations escribe in te previous two items are part of a commutative iagram (Teorem 1, page 8). v. Numerical measurements an analytical calculation of orer of accuracy for specific examples using tis approac sow tat iger orer metos are not obtaine in general (Capter 5, especially Section 5.4 an 5.5). vi. Te manifol consisting of triples tat satisfy a iscrete Euler Lagrange equation is a smoot manifol (Proposition 4, page 35). vii. Te algoritm on TQ TQ is symplectic an te algoritm tat upates triples of points on Q is presymplectic (Teorem, page 36). I also give a erivation of a symplectic form on TQ TQ were Q is te configuration manifol. Te structure of te rest of tis tesis is as follows. Capter is a quick review of

12 5 continuous time Lagrangian an Hamiltonian mecanics. Capter 3 introuces variational integrators an Capter 4 escribes te linearization results mentione above. Capter 5 gives examples of linearization on bot te continuous an te iscrete sie. In tis capter I also o some erivations of orer of accuracy calculations. Tese are accompanie by results from numerical experiments for measuring orer of accuracy. Bot, te numerical experiments an te analytical calculations sow tat our meto oes not, in general, give iger orer symplectic integrators.

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14 7 Capter Preliminaries Before eveloping te teory an escribing te linearization meto I will give some necessary efinitions, stanar results an examples. Tis section is aapte primarily from Lecture 5 of Marsen [13] an from Capters an 5 of te book by Marsen an Ratiu [15]..1 Lagrangian Mecanics Let Q be a configuration manifol of a mecanical system an L : TQ R be a function (calle a Lagrangian). Let q =(q 1,...,q n ) be local coorinates for Q an let c :[a, b] R TQ be a smoot curve on TQ. Using stanar abuse of notation I will write c(t) (q(t), q(t) ) (q(t), q(t)) t for t [a, b]. Te action integral S is efine as S = b a L(q(t), q(t))t. Ten te variational principle calle Hamilton s Principle states tat δs =0 L t q i L =0fori =1,...,n. (.1) qi Definition 1 Te evolution equations L t q i L =0 (.) qi are calle te Euler-Lagrange equations.. Geometry Definition If f : M R is a smoot function on a ifferentiable manifol M ten it can be ifferentiate at any point m M to obtain a map T m f : T m M T f(m) R. Te tangent space of R can be ientifie wit itself since R is a vector space. Tis gives us a linear map f(m) :T m M R. Tus f(m) T mm, te ual of te vector space T m M.

15 8 Tis map f is calle te ifferential of f. Teirectional erivative in te irection v T m M is efine to be f(m) v. Definition 3 A symplectic form Ω on a vector space Z is a nonegenerate skew symmetric, bilinear form on Z. Te pair (Z, Ω) is calle a symplectic vector space. TeformΩ is sai to be (weakly) nonegenerate if Ω(z 1,z ) = 0 for all z Z implies z 1 =0. It can be sown tat tere is a basis for Z in wic Ω(u, v) =u T J v for all u, v Z an were J = ( 0 I I 0 ) were 0 an I are n n zero an ientity matrices for a symplectic vector space Z of imension n. Tus finite imensional symplectic vector spaces always ave an even imension. For an example of a symplectic vector space consier Z = W W were W is a vector space an W is its ual. Te canonical symplectic form ΩonZ is efine as were w 1,w W an α 1,α W. Ω((w 1,α 1 ), (w,α )) = α (w 1 ) α 1 (w ) Definition 4 A symplectic manifol is a pair (P, Ω), were P is a manifol an Ω is a close (weakly) non egenerate two-form on P. An example of a symplectic manifol is te cotangent bunle of te configurations space. Let te manifol Q be te configuration space of a mecanical system. Ten TQ is te isjoint union (over all points of Q) of te tangent spaces an is calle te tangent bunle. At eac point p Q te tangent space T p Q is a vector space an ence tere is an associate ual vector space Tp Q. Te isjoint union of all tese over all te points of te manifol is te cotangent bunle T Q an it is a symplectic manifol. Definition 5 Let (P 1, Ω 1 )an(p, Ω ) be symplectic manifols. A C -map ϕ : P 1 P is calle symplectic or canonical if ϕ (Ω )=Ω 1. Tis pull-back notation means tat for all q P 1,anallv, w T q P 1 Ω 1p (v, w) =Ω ϕ(p) (T p ϕ v, T p ϕ w) (.3) were Ω 1p means Ω 1 evaluate at point p an T p ϕ is te erivative of ϕ at p. Animportant result (see e.g Capter 5 of Marsen an Ratiu [15]) is Proposition 1 A smoot canonical transformation between symplectic manifols of te same imension is volume preserving an is a local iffeomorpism.

16 .3 Hamiltonian Mecanics 9 Te motivation for tese efinitions above concerning symplectic manifols an maps is te following proposition wic is prove by Marsen an Ratiu in Capter of teir book [15]. But first we nee a efinition of Hamiltonian vector fiel on a symplectic manifol. Definition 6 Let (P, Ω) be a symplectic manifol. A vector fiel X on P is calle Hamiltonian if tere is a function H : P R suc tat for all v T z P Ω z (X(z),v)=H(z) v. In tis case one writes X H for X. Hamilton s equations are te evolution equations ż = X H (z). Proposition Te flow ϕ t of a Hamiltonian vector fiel X H consists of canonical (i.e symplectic)transformations. Conversely if te flow of a vector fiel X consists of canonical transformations, ten it is Hamiltonian..4 Legenre Transform A transform tat allows us to move between Hamiltonian an Lagrangian mecanics is te Legenre Transform efine below. Definition 7 Given a Lagrangian L : TQ R efine a map FL : TQ T Q calle te fiber erivative by FL(v) w = s L(v + sw) s=0 were v, w T q Q. Tus FL(v) w is te erivative of L at v along te fiber T q Q in te irection w. In a local cart U E for TQ were U is open in te moel space E for Q, FL(u, e) =(q, D L(u, e)) were D L is te partial erivative of L wit respect to its secon argument. For a finite imensional manifol Q wit (q i ) te coorinate for Q, an(q i, q i ) te inuce coorinates for TQ te fiber erivative as te local expression ( FL(q i, q i )= q i, L ) q i Notation : In te above equation FL(q i, q i ) soul be rea FL(q 1,...,q n, q 1,..., q n ). A similar convention will be use trougout. Tus FL is given by p i = L q i (.4)

17 10 Tis transformation given by is also calle te Legenre transform. j=1 FL :(q i, q i ) (q i,p i ) In te Euler-Lagrange equations (.) te cain rule can be use to to get n ( ) L q j q i qj + L q j q i qj L =0, i =1,...,n (.5) qi Definition 8 To write q j in equation (.5) as a function of q an q it is require tat te matrix L q j q i be invertible. A Lagrangian for wic tis is invertible is calle a regular Lagrangian. If L is regular ten one can o te cange of variables (q i, q i ) (q i,p i ) at least locally ue to te implicit function teorem. Te Legenre transform allows us to go back an fort between Hamiltonian an Lagrangian mecanics ue te following proposition : Proposition 3 Let L(q i, q i ) be a regular Lagrangian. Define a new function ( n ) H(q i,p i )= p i q i L(q i, q i ) i=1 were p i = L q i Wit tis efinition te Euler-Lagrange equations are equivalent to te Hamilton s equations, i.e.5 Algoritms L t q i L q i =0 { q i t = H p i, p i t = H q i Definition 9 An algoritm on a pase space P is a collection of maps F : P P, usually epening smootly on R for small >0. Avancing te solution z k by one time step can be written as z k+1 = F (z k ). Tis means avancing time by wic is also calle te step size. Hereallz k P. Definition 10 An algoritm F is i. a symplectic integrator if eac F is symplectic, ii. an energy integrator if H F = H, iii. a momentum integrator if J F = J, werej is te momentum map for te action of a Lie group G.

18 11 Remark 1 It as been sown by Ge an Marsen [] tat a fixe step size numerical integrator can only be symplectic an momentum preserving or energy an momentum preserving. In te remainer of tis tesis I will refer to symplectic-momentum integrators as simply symplectic integrators. Remark In tis tesis I will not aress te momentum conservation issue an I will not give te efinitions for te terms use in te efinition of a momentum integrator above. Tese can be foun in Wenlant an Marsen [0] an in te references containe terein.

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20 13 Capter 3 Variational Integrators 3.1 Discrete Mecanics As escribe by Veselov [19] an Wenlant an Marsen [0] one can evelop a iscrete picture corresponing to te continuous mecanics escribe in Section.1. Instea of a Lagrangian L : TQ R an te variational principle S = b a L(q(t), q(t))t one efines a iscrete Lagrangian L : Q Q R. Te action integral above is replace by an action sum S : Q N+1 R efine as S = N 1 k=0 L (q k,q k+1 ) were q k Q. Te iscrete variational principle states tat te evolution equations extremize te action sum given fixe en points, q 0 an q N. Extremizing S over q 1,...,q N 1 leas to te iscrete Euler-Lagrange equations : D 1 L (q k,q k+1 )+D L (q k 1,q k )=0 for k =1,...,N 1 (3.1) were D i is te partial erivative wit respect to te i-t argument. Tese equations can also be written as D 1 L Φ+D L =0 (3.) were Φ : Q Q Q Q is efine implicitly by Φ(q k 1,q k )=(q k,q k+1 ). If D 1 L is invertible ten equation (3.) efines te iscrete map Φ wic flows te system forwar in iscrete time. Ten one can sow various properties like symplecticness, momentum map preservation of Φ as one by Wenlant an Marsen [0]. In orer to obtain an algoritm in terms of te original continuous Lagrangian one must relate L wit L an in Wenlant an Marsen [0] tis is one by efining ( y + x L (x, y) =L, y x ) (3.3) were is te step size. For examples of variational integrators constructe accoring to tis sceme please con-

21 14 sult Section 4 of Wenlant an Marsen [0]. See Capter 5 for some aitional examples. Tese examples sow te construction for variational integrators by te sceme escribe above as well as integrators obtaine by ifferentiating te iscrete Lagrangian accoring te main ieas of tis tesis. 3. Invariance Properties To talk about te symplecticness of te algoritm Φ : Q Q Q Q one as to sow tat some symplectic form is being preserve i.e tat Φ is a symplectic map. But wic symplectic form? Te most natural coice is to pull back te canonical symplectic form from te cotangent bunle T Q. Tis is wat was one by Veselov [19] an wic is explaine by Wenlant an Marsen [0]. To o tis, we efine te fiber erivative suc tat FL : Q Q T Q (3.4) FL :(q k,q k+1 ) (q k, D1L (q k,q k+1 )). Now one can efine te symplectic -form on Q Q by pulling back te canonical -form from T Q by FL.TusifΩ CAN is te canonical -form on T Q ten ω on Q Q is ω = FL (Ω CAN) (3.5) Please refer to Sections 3.1 an 3.of Wenlant an Marsen [0] wic give a proof tat te algoritm Φ preserves te symplectic form, i.e Φ ω = ω. Tere one can also fin a local coorinate expression for ω.

22 15 Capter 4 Linearization In tis capter I will escribe our meto for ifferentiating, i.e linearizing Euler-Lagrange equations an variational integrators by manipulating te Lagrangian an te iscrete Lagrangian. Tis manipulation is suc tat in te continuous case, applying te appropriate variational principle yiels a continuous time erivative of te Euler-Lagrange equation corresponing to te original Lagrangian. In te iscrete case, te iscrete Lagrangian is manipulate an te iscrete variational principle yiels a iscrete time erivative of te iscrete Euler-Lagrange equations corresponing to te original iscrete Lagrangian. If te original iscrete Euler-Lagrange equation is linear, te iscrete time erivative is simply a finite ifference. Of course, to linearize continuous (or iscrete) Euler-Lagrange equations one coul just as well start wit te equation an take its erivative (or finite ifference). But as mentione in Section 1., my original aim wic le to te present results, was to moify te Lagrangian an ten examine te resulting Euler-Lagrange equations. Tis is because te iscretization of te Euler-Lagrange equations is generate from te Lagrangian an not from te continuous time equations. 4.1 Motivation Consier te iscrete Euler-Lagrange equation wic gives te symplectic algoritm as escribe in Capter 3. Tis is repeate ere for convenience : D 1 L Φ+D L =0 (4.1) were Φ : Q Q Q Q is efine implicitly by te above equation an Φ(q k 1,q k )= (q k,q k+1 ). Since L : Q Q R, te resulting algoritm map Φ takes a pair of points (q k 1,q k ) Q an prouces a pair (q k,q k+1 ). It is reasonable to conjecture tat an algoritm tat is able to use more points, say 3 points, migt result in a better truncation error since points allow one to approximate erivative an 3 will allow one to approximate acceleration as well. Tis was te motivation for our subsequent ieas tat le to te linearization metos of tis tesis. Tus, wat is esire is an algoritm map Φ () : Q Q Q Q Q Q. 4. Differentiating Lagrangian along ( q, q) Wit a view to introucing accelerations into te framework, consier wat appens wen te Lagrangian is a function of q, q an q, i.e te Lagrangian is some function L () (q, q, q).

23 16 Hamilton s principle (.1) an simple variational calculus (see for example Capter, page 4, equation of Gelfan an Fomin [4]) ensures tat te solutions q(t) satisfytefourt orer Euler-Lagrange equations L () t q i L () t q i + L() q i =0. (4.) Tis is fourt orer because in general te term L () t q i (q, q, q) (4.3) involves a fourt erivative term. Tus in going from L(q, q) tol () (q, q, q) one goes from a secon orer Euler-Lagrange equation to a fourt orer Euler-Lagrange equation. Tis can be represente by Figure 4.1. On te left someting as yet unspecifie is one to introuce L(q, q)? Hamilton s Principle EL(L) is secon orer? L () (q, q, q) Hamilton s Principle EL(L () ) is fourt orer Figure 4.1: On te left q as been introuce into te Lagrangian but on te rigt te equations go from being secon to fourt orer. acceleration into te Lagrangian. On te rigt one sees tat te Euler-Lagrange equations skip an orer an go from being secon orer to fourt orer. Te question marks inicate an as yet unspecifie operation. Wat is te most obvious operation one can perform on L(q, q) to introuce acceleration? Te answer is ifferentiation. Tus one migt expect te question mark on te rigt own arrow to peraps represent ifferentiation also. However, as it turns out, wen te secon orer Euler-Lagrange equations for L(q, q) namely equation (.) are ifferentiate one gets a tir orer equation as sown by te following calculation (using implie summation convention were necessary). First, by ifferentiating te Euler-Lagrange equations for L(q, q) weget: t But te left an sie of te above equation is ( L t q j q i qj + L q j q i qj ( ) L L (q, q) (q, q) =0. (4.4) t q i qi ) L q j q i qj L q j q i qj, (4.5)

24 17 wic is ( = L q j q i qj + q j 3 L q k q j q i qk + ( q j + q j L... q j q i 3 ) L q k q j q i qk + 3 L q k q j q i qk + 3 ) L q k q j q i qk L q j q i qj L q j q i qj = L q j q i qj + 3 L q k q j q i qk q j + L... q j q j q i + 3 L q k q j q i qk q j + 3 L q k q j q i qk q j + 3 L q k q j q i qk q j L q j q i qj L q j q i qj Terefore te ifferentiation of te Euler-Lagrange equation for L(q, q) gives te following equation : L q j q i qj + 3 L q k q j q i qk q j + L... q j q j q i + 3 L q k q j q i qk q j + 3 L q k q j q i qk q j + 3 L q k q j q i qk q j L q j q i qj L q j q i qj =0. (4.6) Tis equation (4.6) is tir orer. Tus te iagram in Fig. 4.1 will not commute if own arrows are tougt of as ifferentiation. On te rigt one nees to go to a tir orer Euler-Lagrange equation. However if L () (q, q, q) was only linear in q ten te term L () q i in equation (4.) will be a function only of q an q as a result of wic L () t q i will only be a tir orer term. Te Euler-Lagrange for suc a Lagrangian will be tir orer. As te left own arrow was being tougt of as ifferentiation tis migt work out. Tis suggests getting L () (q, q, q) froml(q, q) by simply formally ifferentiating L(q, q) wit respect to time. If L(q, q) is simply formally ifferentiate wrt t ten one gets L () (q, q, q) = L q i qi + L q i qi (4.7) Note tat now L () (q, q, q) is linear in q, as esire. Now te next step ougt to be to try an see if te iagram in Figure 4.commutes. But tat will not be wortwile because

25 18 ifferentiate L(q, q) L () (q, q, q) linear in q Hamilton s Principle EL(L) is secon orer ifferentiate Hamilton s Principle EL(L () ) is tir orer Figure 4.: On te left q as been introuce into te Lagrangian by ifferentiating L. Since L () is linear in q, EL(L () ) soul be tir orer. But tere is a major problem as is explaine in te text. of te two Lemmas tat follow. Lemma 1 below sows tat for Lagrangians L of a certain type, wen L () is obtaine as in equation (4.7) an Hamilton s principle is applie using Lagrangian L () one gets te trivial ientity 0 = 0 for te Euler-Lagrange equations. Te next Lemma removes te restriction on te form of Lagrangian. 1 Lemma 1 Let te L be of te form L(q, q) =L T ( q)+l V (q) (4.8) (Te superscripts T an V are reminers tat te term L T ( q) is te kinetic energy an te term L V (q) is minus te potential energy term). Construct a new Lagrangian L () (q, q, q) as L () (q, q, q) = L q i qi + L q i qi (4.9) ForsucaLagrangianL () te Hamilton s principle gives no information about te evolution of state, i.e te Euler-Lagrange equation for L () is 0=0. Proof Wen te Lagrangian of te form (4.8) is ifferentiate to get L () efine by equation (4.9) L () (q, q, q) = LT q i qi + LV q i q i (4.10) is obtaine since te oter terms are 0. Define L (T ) ( q, q) LT q j qj (4.11) L (V ) (q, q) LV q j qj 1 Te more general result in Lemma was suggeste by Mattew West wen I sowe im Lemma 1.

26 19 Now te Euler-Lagrange equation for L () (q, q, q), namely equation (4.) is ( L (T ) q i L (T ) t q i + L (T ) ) t q i + ( L (V ) q i L (V ) t q i + L (V ) ) t q i =0 (4.1) Substitute for L (T ) an L (V ) from equation (4.11) into equation (4.1) an consier te L (T ) an L (V ) parts separately. Te L (T ) part of equation (4.1) becomes 0 L (T ) t q i + L (T ) t q i = ( ) ( ) L T t q i q j qj + L T t q i q j qj ( L T ) + = t q i q j qj = ( L T t q i q j qj =0 ) + t L T t q i ( L T q j q i qj were te last step follows from te equality of mixe partials. Similarly te L (V ) part of equation (4.1) becomes ) L (V ) q i L (V ) t q i +0 = ( ) L V q i q j qj ( L V t q i q j = L V q i q j qj L V t q i = L V q i q j qj LV q j q i qj = 0 qj ) were again te last step follows from te equality of mixe partials. Tus equation (4.1) becomes te trivial ientity 0 = 0. Remark 3 Te type of LagrangianL assume in Lemma 1 arises commonly. Some common examples woul be : a particle in, say, a cubic potential fiel were or a simple penulum of lengt l for wic L(q, q) =m q + q3 L(θ, θ) = l θ + gl cos θ.

27 0 Te restriction on te form of Lagrangian can be remove as is sown by te following more general result. Lemma Let L be any sufficiently smoot function of q an q. Construct a new Lagrangian L () (q, q, q) as L () (q, q, q) = L q i qi + L q i qi (4.13) For tis Lagrangian L () te Hamilton s principle gives no information about te evolution of state, i.e te Euler-Lagrange equation for L () is 0=0. Proof Te Euler-Lagrange equation corresponing to te Lagrangian L () is L () t q i L () t q i + L() q i = 0 (4.14) for i =1,...,n. Te first term of tis equation is (summation over repeate inices) L () t q i = L t q i = ( ) L t q j q i qj + L q j q i qj ( = L ) L q j q i qj + t q j q i q j + ( L... q j q j q i + t Te secon term of equation (4.14) is L () t q i = ( L t q i q j qj + L ) q i + L q i q j qj ( = L ) L q i q j qj + t q i q j q j + ( + L... q j L q i q j + t q i q j an te last term of equation (4.14) is L q j q i qj + L q j q i L q j q i qj ) q j (4.15) ) q j (4.16) L () q i = L q i q j qj + L q i q j qj (4.17) Substituting te expressions (4.15), (4.16) an (4.17) in equation (4.14) an using equality of mixe partials one fins tat te left an sie of equation (4.14) is 0 tus completing te proof. Tus ifferentiating te Lagrangian along (q, q) apparently oes not make te iagram 4.commute. Section 4.3 below sows ow tis can be remeie by a ifferent approac to make it work. Tis results in continuous time linearization of te Euler-Lagrange equations. Section 4.4 sows a similar iscrete time linearization of te iscrete Euler-Lagrange equations.

28 1 4.3 Continuous Time Linearization In Section 4.te Lagrangian L was ifferentiate along te acceleration vector ( q, q) to get a new Lagrangian L () wic was linear in q. I sowe tat tis new Lagrangian L () yiels an Euler-Lagrange equation wic is a trivial ientity. Now I will generalize tis iea of ifferentiating L to get a new Lagrangian but tis time te new Lagrangian oes yiel a nontrivial Euler-Lagrange equation. Ten I will escribe an prove te relationsip between ifferentiating te Lagrangian an te corresponing Euler-Lagrange equations. Given L : TQ R, anq Q te ifferential of L at (q, q) T q Q is te map L(q, q) : T (q, q) TQ T R = R. See Section., Definition for etails. In Section 4., I wrote L () (q, q, q) =L(q, q) ( q, q) = L q i (q, q) qi + L q i (q, q) qi (4.18) an foun tat te Euler-Lagrange equation for L () is 0 = 0. In equation (4.18) te ifferentiation is along te acceleration vector ( q, q) wic is in TTQ at (q, q). (Actually to be correct one soul really write suc an element of TTQ as (q, q, q) or even better, as (q, q, q, q)). Te obvious generalization of tis ifferentiation is to ifferentiate along an arbitrary vector in TTQ. Wit tis view let v q T q Q an let w vq T vq TQ. Define L () by L () (w vq ) L(v q ) w vq (4.19) In coorinates one writes v q =(q, q) anw vq =(q, q,u, u) wereu T q Q an u is not necessarily equal to v q =(q, q). Tus in local coorinates equation (4.19) is L () (q, q,u, u) L q i (q, q) ui + L q i (q, q) ui (4.0) We now introuce te notation EL(L) to represent te Euler-Lagrange equations corresponing to a Lagrangian L. Te main result of tis section is tat ifferentiating EL(L) results in te same equations tat are obtaine by first ifferentiating L to get L (), fining EL(L () ) an setting u = q. Tis is summarize in te following Lemma. Lemma 3 Te following iagram commutes EL(L) EL(L () ) u= q (4.1) L L () Proof First I will escribe ow to rea te commutative iagram above. Te up arrows mean apply Hamilton s principle an te result is te Euler-Lagrange equations corresponing to te Lagrangian one starts wit at te bottom. Te rigt arrows inicate ifferentiation. Te top rigt arrow is ifferentiation wrt t. Te bottom rigt arrow is ifferentiating as in (4.0). Te notation u= q in te top rigt corner only makes sense if one arrives at tat corner from te bottom. To be specific ere is ow te various pats troug te iagram soul be rea. Start at bottom left corner. Going up an ten rigt means starting wit te Lagrangian L fin te Euler-Lagrange equation corresponing to it an ten ifferentiate tat equation wrt t. On te oter an te pat going rigt an ten up means, ifferentiate L as in (4.0) wic gives an expression for L () in terms of L. Ten fin te Euler-Lagrange equation

29 for tis Lagrangian. Finally substitute u = q. Te resulting equations will be te same as tose obtaine by te up an rigt pat. I will now give a local coorinate proof of te commutativity of (4.1). I ave alreay ifferentiate EL(L) an te result is given in equation (4.6). Tus te result of ifferentiating EL(L) can be consiere to be te EL(L) itself an te linearize EL(L) inequation (4.6). Let us now see wat one gets by writing EL(L () ) in terms of te original Lagrangian L.SinceL () (q, q, u, u) as inepenent variables q an u it follows tat EL(L () )consists of te two equations L () t q i L() q i =0 (4.) L () t u i L() u i =0 (4.3) Te left an sie of equation (4.) wen written in terms of L is equivalent to ( ) L ( ) L (4.4) Now = t = t q i q j uj + L q j uj ( L q i q j uj + L q i q j uj q i ) q j uj + L q j uj ( L q i q j uj + L q i q j uj ) (4.5) L q i q j uj + u j L t q i q j + L q i q j üj + u j L t q i q j L q i q j uj L q i q j uj. (4.6) L t q i q j = L t q i q j = 3 L q k q i q j qk + 3 L q k q i q j qk + 3 L q k q i q j qk (4.7) 3 L q k q i q j qk (4.8) Tus one equation of EL(L () ) obtaine by substituting (4.7) an (4.8) in (4.6) is L q i q j uj + L q i q j üj + 3 L q k q i q j uj q k + 3 L q k q i q j uj q k + L q i q j uj L q i q j uj =0 3 L q k q i q j uj q k + (4.9) 3 L q k q i q j uj q k Wen one sets u = q one equation of EL(L () ) u= q in terms of L is obtaine as L q i q j qj + L... q j q i q j + 3 L q k q i q j qj q k + 3 L q k q i q j qj q k + L q i q j qj L q i q j qj =0 3 L q k q i q j qj q k + (4.30) 3 L q k q i q j qj q k

30 3 wic except for a rearrangement of terms is te same as te linearize EL(L) inequation (4.6). Te oter equation of EL(L () ) is equation (4.3) wic in terms of L is equivalent to t u i wic is equivalent to ( L q j uj + L ) q j uj ( L u i q j uj + L ) q j uj =0, (4.31) L t q i L =0, (4.3) qi wic is te same as te original EL(L). Tis completes te proof of commutativity of (4.1). Tus I ave sown tat te Euler-Lagrange equation corresponing to L, namely EL(L) can be ifferentiate by ifferentiating L along (u, u) first to get a new Lagrangian L () an ten eriving te Euler-Lagrange equations EL(L () ) corresponing to tis Lagrangian an finally setting u = q. 4.4 Discrete Time Linearization Wit a view of linearizing te iscrete Euler-Lagrange equations I will mimic on te iscrete sie wat was one on te continuous sie. Tus I first obtain L () from L by ifferentiating. Let L : Q Q R be te iscrete Lagrangian. By analogy wit te continuous case one first ifferentiates tis to get L (q, r) :TQ TQ R were (q, r) Q Q an (v q,v r ) T (q,r) (Q Q) = T q Q T r Q. Ten one efines In local coorinates L () (v q,v r ) L (q, r) (v q,v r ) (4.33) L () (q k,u k,q k+1,u k+1 )=L (q k,q k+1 ) (u k,u k+1 ) (4.34) Te use of subscripts k an k + 1 is suggestive of te fact tat q k+1 is te position after one iscrete time step, starting from q k. However, te above efinition of L () is vali for any q 1,u 1,q,u in te appropriate spaces. Note also tat ere q k etc. are vectors, i.e q k =(qk 1,...,qn k ) etc. I will assume, as one by Wenlant an Marsen [0] tat Q V were V is a vector space. Tis is essential if one is going to be averaging an subtracting points in Q. Now I will prove a result on te iscrete sie analogous to wat was one in Lemma 3 on te continuous sie. Let DEL(L ) represent te Discrete Euler-Lagrange equations corresponing to te iscrete Lagrangian L as in (3.1). Ten te following is true Lemma 4 Te following iagram commutes : L L () DEL(L ) DEL(L () ) uk = q k+1 q k (4.35)

31 4 Proof First I must escribe ow to rea te iagram. Te top rigt arrow represents ifferentiation efine in (4.34). Te own arrows represent using te iscrete variational principle to get te iscrete Euler-Lagrange equations. Te bottom rigt arrow as to be given a meaning. I will efine an operation of iscrete ifferentiation as escribe below. In te case of linear equations tis is te same as finite ifferencing. First let s ceck carefully wat DEL(L () ) is. Tis is obtaine by extremizing te iscrete action sum, i.e by extremizing N 1 S () = L () (q k,u k,q k+1,u k+1 ) (4.36) k=0 One extremizes S () over q k an u k for k =1,...,N 1 keeping q 0,q N,u 0 an u N fixe an by setting S () q j k =0, S () u j k =0 for j =1,...,n. Tis yiels te two sets of equations tat comprise te iscrete Euler- Lagrange equations for L (). Tese equations, wic I will call DEL(L() )are L () q j k L () u j k (q k 1,u k 1,q k,u k )+ L() q j (q k,u k,q k+1,u k+1 ) = 0 (4.37) k (q k 1,u k 1,q k,u k )+ L() u j (q k,u k,q k+1,u k+1 ) = 0 (4.38) k for k =1,...,N 1anj =1,...,n. Now I can start te proof of te commutativity of (4.35). To o tis one must write DEL(L () )intermsofl. Consier te first set of equations represente by (4.37). Substitute for L () in terms of L using (4.34). Wit tis substitution, equation (4.37) becomes ( ) q j u i L k 1 q i (q k 1, q k )+u i L k k k 1 qk i (q k 1,q k ) + ( ) u i L k qk i (q k, q k+1 )+u i L k+1 qk+1 i (q k,q k+1 ) =0, (4.39) wic implies q j k u i k 1 L q j (q k 1,q k )+u i L k qi k k 1 q j (q k 1,q k )+ k qi k u i L k q j (q k,q k+1 )+u i L k qi k+1 k q j (q k,q k+1 )=0. (4.40) k qi k+1 Now let s substitute u i k = qi k+1 qi k

32 5 etc. to get q i k qi k 1 L q j k qi k 1 (q k 1,q k )+ qi k+1 qi k q i k+1 qi k L q j k qi k Te above equation represents one equation of L q j k qi k (q k 1,q k )+ (q k,q k+1 )+ qi k+ qi k+1 DEL(L () ) uk = q k+1 q k In terms of L te oter equation, i.e (4.38) becomes u j k ( wic is L u i k 1 qk 1 i Now DEL(L )is ) (q k 1, q k )+u i L k qk i (q k 1,q k ) + ( u i L k qk i (q k, q k+1 )+u i k+1 L q j k u j k. L q j k qi k+1 L q i k+1 (q k,q k+1 ) = 0 (4.41) ) (q k,q k+1 ) =0 (4.4) (q k 1,q k )+ L q j (q k, q k+1 ) = 0 (4.43) k L q j (q k 1,q k )+ L k q j (q k,q k+1 ) = 0 (4.44) k wic is te same as (4.43). Te iscrete time erivative of tis is same as (4.41). Tis erivative is taken just like te usual erivative except tat instea of q k 1 one writes q k q k 1 If an expression is linear in its variables, its iscrete time erivative is te same as its finite ifference. Consier for example an expression f(q) linear in q. In particular let f(q) =a 1 q + a 0. Ten by te above escription of iscrete time erivative, te iscrete time erivative of f(q) is f q 1 q 0 q q0 wic is Note tat tis is te same as q 1 q 0 a 1. f(q 1 ) f(q 0 ),

33 6 te finite ifference of te expression f(q). Tus we ave sown tat DEL(L () ) qk+1 q k consists of two sets of equations. Te first set, equation (4.37), after substituting u k = q k q k 1 is te iscrete time erivative of DEL(L ). Te secon set, equation (4.38) is te same as DEL(L ) wic sows tat te iagram in equation (4.35) commutes. Tus te situation is exactly te same as te continuous case in wic EL(L () ) u= q consiste of two sets of equations te first of wic was te erivative of EL(L) ante secon of wic was te same as EL(L). Linearize Algoritm Te linearize algoritm is generate by extremizing te action sum corresponing to L (). Tis yiels two iscrete Euler-Lagrange Equations, te linearize an te original, just like for te continuous Lagrangian L (). I call te two iscrete Euler- Lagrange equations collectively as DEL(L () ). It was seen tat te DEL(L() )aregivenby (4.37) an (4.38). After making te substitutions u i k =(qi k+1 qi k )/ in tese equations, one gets wat I calle DEL(L () ) qk+1 q k namely equation (4.41). Notice tat (4.41) can be solve for q k+ assuming invertibility of te matrix L q j (q k,q k+1 ) (4.45) k qi k+1 appearing in te equation. For a regular Lagrangian, by efinition, te matrix L q j q i (4.46) is invertible. It is known [16] tat at least for Lagrangians of te form Kinetic Energy - Potential Energy, te invertibility of (4.46) implies te invertibility of (4.45) for small step size. Presumably te same migt be true for a general Lagrangian. Tus given a triple of points (q k 1,q k,q k+1 ) one can solve (4.41) for q k+. Tus an algoritm tat takes te triple to te triple (q k 1,q k,q k+1 ) (q k,q k+1,q k+ )

34 7 as been efine. One can also write te algoritm in terms of te original continuous Lagrangian but I will not o tat. 4.5 Discrete an Continuous Now te continuous an iscrete Lagrangians must be relate in some way an tis is summarize in te following result. Lemma 5 Te following iagram commutes L L () L L () (4.47) Proof Te vertical arrows represent implementation ecisions, i.e it is up to us to coose a iscretization tat will ave te rigt properties. Starting from te top left corner of (4.47) an going own, i.e L L soul be rea as writing L in terms of L. Tis is provie by equation (3.3) written ere as ( qk+1 + q k L (q k,q k+1 )=L, q ) k+1 q k (4.48) were, in L(q, q) tepositionq as been approximate by (q k+1 + q k )/an q by (q k+1 q k )/. A similar implementation ecision must be mae in going from L () (q, q, u, u) to (q k,u k,q k+1,u k+1 ) i.e on going own te rigt arrow. Tis is one by setting L () in L ().Now q =(q k+1 + q k )/ q =(q k+1 q k )/ u =(u k+1 + u k )/ u =(u k+1 u k )/ L () (q k,u k,q k+1,u k+1 )= L () ( qk+1 + q k, q k+1 q k, u k+1 + u k, u ) k+1 u k (4.49) L () (q k,u k,q k+1,u k+1 )= ui k+1 + ( ui k L qk+1 + q k q i, q ) k+1 q k + ui k+1 ui k L q i ( qk+1 + q k, q ) k+1 q k (4.50) were te last equality follows from (4.0). Equation (4.50) represents starting from top left, going rigt an ten bottom. Tis must be compare wit te oter pat of going

35 8 own from top left an ten going rigt. To o tis one must use equation (4.34). Here, L (q k,q k+1 )= [ 1 L q 1 1 L q 1,...,1 L q n 1 L q n, 1 L q L q 1,...,1 L q n + 1 ] L q n (4.51) Tus L (q k,q k+1 ) (u k,u k+1 )= u i k ( 1 L q i 1 ) ( L 1 q i + u i L k+1 q i + 1 ) L q i (4.5) were i = 1,...,n wic implies tat L () (q k,u k,q k+1,u k+1 )= ui k+1 + ui k L q i + ui k+1 ui k L q i (4.53) were te partials are evaluate at ((q k+1 + q k )/, (q k+1 q k )/). But (4.53) is te same as (4.50) wic completes te proof of commutativity of (4.47). Te previous tree lemmas can be summarize in one iagram as given by te following teorem. Teorem 1 Te following iagram commutes EL(L) EL(L () ) u= q L L () L L () DEL(L ) DEL(L () ) uk = q k+1 q k (4.54) Proof Tis follows from te previous tree lemmas by pasting te corresponing commutative iagrams togeter. 4.6 Linearization Summary I summarize te linearization process by giving te various Euler-Lagrange equations in one place ere. In te following equations of tis section, i, j =1,...,n an k =1,...,N 1, were n is te imension of te configurationmanifol an 0,...,N represents iscrete time.

36 9 Euler Lagrange equations EL(L) for L(q, q) : L t q i L q i = 0 (4.55) Euler Lagrange equations EL(L () ) for L () (q, q,u, u) : L () t q i L() q i = 0 (4.56) L () t u i L() u i = 0 (4.57) EL(L () ) u= q in terms of L : L q i q j qj + L... q j q i q j + 3 L q k q i q j qj q k + 3 L q k q i q j qj q k + 3 L q k q i q j qj q k + (4.58) 3 L q k q i q j qj q k L q i q j qj L q i q j qj =0 L t q i L q i = 0 (4.59) Equation (4.58) is te erivative of equation (4.55). Discrete Euler-Lagrange equations DEL(L ) for L (q k,q k+1 ): L q j k (q k 1,q k )+ L q j (q k,q k+1 ) = 0 (4.60) k Discrete Euler-Lagrange equations DEL(L () ) for L() (q k,u k,q k+1,u k+1 ) L () q j k L () u j k (q k 1,u k 1,q k,u k )+ L() q j (q k,u k,q k+1,u k+1 ) = 0 (4.61) k (q k 1,u k 1,q k,u k )+ L() u j (q k,u k,q k+1,u k+1 ) = 0 (4.6) k

37 30 DEL(L () ) qk+1 q k in terms of L : qk i qi k 1 L q j k qi k 1 qk+1 i qi k L q j k qi k L q j k (q k 1,q k )+ qi k+1 qi k (q k,q k+1 )+ qi k+ qi k+1 L q j (q k 1,q k )+ (4.63) k qi k L q j k qi k+1 (q k,q k+1 )=0 (q k 1,q k )+ L q j (q k,q k+1 ) = 0 (4.64) k Equation (4.63) is te iscrete time erivative of equation (4.60). 4.7 Invariance Properties To see wy te proofs of invariance properties given by Wenlant an Marsen [0] will still work, it is elpful to consier te following picture. L : TQ R L : Q Q R L () : TTQ R L () : TQ TQ R Te left column sows te usual continuous Lagrangian an te iscrete Lagrangians use by Wenlant an Marsen [0]. It was sown by Wenlant an Marsen [0] tat te algoritm obtaine from L is a symplectic-momentum algoritm. On te rigt column are te Lagrangian an iscrete Lagrangian I ave obtaine. Note tat te rigt column is te same as te left one wit Q replace by TQ. Tus all te invariance proofs of te left column still apply. Tis means tat EL(L () ) as symplectic flow an DEL(L () ) in equations (4.37) an (4.38) is a symplectic integrator. I erive te te symplectic form on TQ TQin subsection below. One can ceck explicitly tat tis symplectic form, given as equation (4.68) below, is preserve by te algoritm DEL(L () ) toug tis also follows witout cecking from te proof in Wenlant an Marsen [0] as inicate above. But we must ceck tat similar invariance properties remain true after we set u = q in EL(L () )oru k =(q k+1 q k )/ in DEL(L () ). Tis will be one in subsection 4.7.for DEL(L () ) Symplectic form on TQ TQ I will efine te symplectic form on TQ TQ by pulling back te canonical -form from T TQ. Te pull back will be one uner a fiber erivative of L () were, as efine earlier, L () : TQ TQ R. Tis fiber erivative will be efine by analogy to te fiber erivative for L : Q Q R, efine by Wenlant an Marsen in [0]. Tat fiber erivative can be efine by taking a parameterize curve (q(t),q 1 ) were te secon component is fixe an were q(0) = q 0 an q(0) = u 0. Tis is a curve in Q {q 1 } Q Q an can be ientifie wit te curve q(t). Tus it represents te vector (q 0,u 0 ) T q0 Q. Te fiber erivative FL : Q Q T Q is ten efine as FL (q 0,q 1 ) (q 0,u 0 )= t L (q(t),q 1 ) (4.65) t=0

38 31 I will efine te fiber erivative FL () : TQ TQ T TQ by analogy to tis. But first I note tat te following steps are neee to efine te symplectic form on TQ TQ: i. Define te fiber erivative FL () : TQ TQ T TQ, ii. Compute te tangent map of FL (),namely T ((q0,u 0),(q 1,u 1))FL () : T (q0,u 0)TQ T (q1,u 1)TQ T () FL ((q0,u0),(q1,u1))t TQ iii. Write te local coorinate expression for te canonical one form Θ on T TQ, iv. Define te symplectic form on TQ TQ as Ω = (FL () Θ) were FL () te pullback uner te fiber erivative FL (). means Step 1 :SinceL () maps TQ TQ R, te fiber erivative FL () maps TQ TQ T TQ. Note tat FL () evaluateatapointintq TQis a point in T TQso FL () can be efine by efining ow it acts on a vector in TTQ. By analogy to te efinition of fiber erivative FL given above, let ((q(t),u(t)), (q 1,u 1 )) be a parameterize curve from an open interval of R incluing 0 into TQ {(q 1,u 1 )} TQ TQ suc tat q(0) = q 0 u(0) = u 0 q(0) = e u(0) = f Te curve ((q(t),u(t)), (q 1,u 1 )) can be ientifie wit te curve (q(t),u(t)) an tis curve ten represents te vector ((q 0,u 0 ),e,f) T (q0,u 0)TQ. Te fiber erivative FL () evaluate at ((q 0,u 0 ), (q 1,u 1 )) TQ TQcanbeefinebyowitsactson((q 0,u 0 ),e,f) T (q0,u 0)TQ as follows: FL () ((q 0,u 0 ), (q 1,u 1 )) ((q 0,u 0 ),e,f)= t L () ((q(t),u(t)), (q 1,u 1 )) t=0 Note te analogy wit te efinition in equation(4.65). Wen te above calculation is carrie out, in matrix notation it results in te following efinition FL () ((q 0,u 0 ), (q 1,u 1 )) = ((q 0,u 0 ), D 1 L () ((q 0,u 0 ), (q 1,u 1 )), D L () ((q 0,u 0 ), (q 1,u 1 ))) (4.66) Note tat FL () maps te point ((q 0,u 0 ), (q 1,u 1 )) TQ TQtoacovectorinT(q TQ 0,u 0) base at (q 0,u 0 ) so te first (q 0,u 0 ) tat appears on te rigt an sie of equation(4.66) is te base point of te covector.