Classical Mechanics Numerical Example Discrete Mechanics Taylor …j2schmit/TALK.pdf · 2017-01-29 · Classical Mechanics Numerical Example Discrete Mechanics Taylor Variational

Classical MechanicsNumerical ExampleDiscrete Mechanics

Taylor Variational IntegratorDiscrete Hamiltonian Variational Integrators

Advancement Talk

Jeremy Schmitt

June 17, 2015

Jeremy Schmitt Advancement Talk



Outline

1 Classical Mechanics

2 Numerical Example

3 Discrete Mechanics

4 Taylor Variational Integrator

5 Discrete Hamiltonian Variational Integrators




Outline


2 Numerical Example







Lagrangian Dynamical System

Lagrangian System

The Configuration Space is a differentiable manifold, Q.The State Space is the corresponding tangent bundle, TQ, withlocal coordinates (q, q).The Lagrangian function, L : TQ → R is a differentiable function.Together Q and L define a Lagrangian system.

Dynamics of the Lagrangian system

C(Q) = {q : [0,T ]→ Q|q is a C 2 curve } is the path space.q ∈ C(Q) where q(t) ∈ Tq(t)Q, is called a motion in theLagrangian system, if q extremizes the action functional,

S(q) =

∫ T

0L(q(t), q(t))dt




Lagrangian Dynamical System

Lagrangian System

The Configuration Space is a differentiable manifold, Q.The State Space is the corresponding tangent bundle, TQ, withlocal coordinates (q, q).The Lagrangian function, L : TQ → R is a differentiable function.Together Q and L define a Lagrangian system.

Dynamics of the Lagrangian system

C(Q) = {q : [0,T ]→ Q|q is a C 2 curve } is the path space.q ∈ C(Q) where q(t) ∈ Tq(t)Q, is called a motion in theLagrangian system, if q extremizes the action functional,

S(q) =

∫ T

0L(q(t), q(t))dt




Theorem

A motion, q, satisfies the Euler − Lagrange equations,

d

dt

∂L

∂q− ∂L

∂q= 0.

Proof

0 = δS

= δ

∫ T

0L(q, q)dt

=

∫ T

0δL(q, q)dt

=

∫ T

0

∂L

∂qδq +

∂L

∂qδqdt




Theorem

A motion, q, satisfies the Euler − Lagrange equations,

d

dt

∂L

∂q− ∂L

∂q= 0.

Proof

0 = δS

= δ

∫ T

0L(q, q)dt

=

∫ T

0δL(q, q)dt

=

∫ T

0

∂L

∂qδq +

∂L

∂qδqdt




Proof, cont’d

=

∫ T

0(∂L

∂q− d

dt

∂L

∂q)δqdt

The last line comes from integration by parts, where we haveconstrained our variations to vanish on the boundary. Finally, thefudamental theorem of the calculus of variations implies,

d

dt

∂L

∂q− ∂L

∂q= 0.




Legendre Transform

Definition

Given a Lagrangian system, define the fiber derivativeFL : TQ → T ∗Q by,

FL(v) · w =d

ds|s=0L(v + sw),

where v ,w ∈ TqQ.

Legendre Transform

FL is called the Legendre Transform. For a finite dimensionalconfiguration manifold,

FL : (q, q)→ (q,∂L

∂q) = (q, p).




Legendre Transform

Definition

Given a Lagrangian system, define the fiber derivativeFL : TQ → T ∗Q by,

FL(v) · w =d

ds|s=0L(v + sw),

where v ,w ∈ TqQ.

Legendre Transform

FL is called the Legendre Transform. For a finite dimensionalconfiguration manifold,

FL : (q, q)→ (q,∂L

∂q) = (q, p).




Definition

The generalized momentum coordinates, p ∈ T ∗Q, are defined by,

p =∂L

∂q

Hamiltonian

Assuming FL is a diffeomorphism, define the HamiltonianH : T ∗Q → R as,

H(q, p) = pq − L(q, q)|p= ∂L∂q.




Definition

The generalized momentum coordinates, p ∈ T ∗Q, are defined by,

p =∂L

∂q

Hamiltonian

Assuming FL is a diffeomorphism, define the HamiltonianH : T ∗Q → R as,

H(q, p) = pq − L(q, q)|p= ∂L∂q.




Hamilton’s Equations

Definition

The system of first-order differential equations,

q =∂H

∂p

p = −∂H

∂q

is called Hamilton′s Equations.

Theorem

Hamilton’s equations are equivalent to the Euler-Lagrangeequations.




Hamilton’s Equations

Definition

The system of first-order differential equations,

q =∂H

∂p

p = −∂H

∂q

is called Hamilton′s Equations.

Theorem

Hamilton’s equations are equivalent to the Euler-Lagrangeequations.




Sketch of the Proof

0 = p +∂H

∂q

=d

dtp +

∂H

∂q

Noting p = ∂L∂q by defintion, and ∂H

∂q = −∂L∂q implies,

0 =d

dt

∂L

∂q− ∂L

∂q.




Conservation of Energy

Theorem

Given a Lagrangian, L(q, q), that is not explicitly a function oftime, then the Hamiltonian is an invariant of the flow.

Proof

dH

dt=∂H

∂q

dq

dt+∂H

∂p

dp

dt

= −pq + qp

= 0




Conservation of Energy

Theorem

Given a Lagrangian, L(q, q), that is not explicitly a function oftime, then the Hamiltonian is an invariant of the flow.

Proof

dH

dt=∂H

∂q

dq

dt+∂H

∂p

dp

dt

= −pq + qp

= 0




Can we build a numerical method that takes into account theunderlying (geometric)structure of the dynamical system?




Outline


2 Numerical Example







Simple Pendulum

Consider a pendulum consisting of a ball with unit mass attachedto a rod of unit length, where for simplicity we assume the rod hasnegligible weight. The pendulum’s configuration manifold is S1.The Lagrangian will be of the form L=T-U, where T is the kineticenergy and U is the potential energy of the system. We willparametrize the manifold by θ the angle between the rod and thenegative y-axis.

L(θ, θ) =1

2θ2 + 9.8sin(θ)

The corresponding Hamiltonian will then be the total energy of thesystem, which will remain constant.




Simple Pendulum

The corresponding Euler-Lagrange equation is,

θ = −9.8sin(θ)




Simple Pendulum

Taylor’s Method 2nd Order h=0.1




Simple Pendulum

Taylor Variational Integrator 2nd Order(TVI2) h=0.1




Simple Pendulum

Total Energy for Taylor’s Method 2nd Order and TVI2 h=0.1




Taylor’s Method 6th Order h=0.05 and TVI6




Total Energy of Taylor’s Method 6th Order h=0.05 and TVI




Taylor’s Method 6th order h=0.5 and TVI6




Taylor’s Method 6th order h=0.5 and TVI6




Outline


2 Numerical Example







Discrete Mechanics

Discrete Lagrangian System

Q configuration manifold.Q × Q is the discrete State Space.Ld : Q × Q → R is the discrete Lagrangian function.




Definition

Given an increasing sequence of times {tk = kh|k = 0, . . . ,N} ⊂ Rdefine the discrete Path Space to be,

Cd(Q) = {qd : {tk}Nk=0 → Q}

Definition

The discrete Action Sum, Sd : Cd(Q)→ R is defined as,

Sd(q0, q1, . . . , qN) =N−1∑i=0

Ld(qi , qi+1).




Definition

Given an increasing sequence of times {tk = kh|k = 0, . . . ,N} ⊂ Rdefine the discrete Path Space to be,

Cd(Q) = {qd : {tk}Nk=0 → Q}

Definition

The discrete Action Sum, Sd : Cd(Q)→ R is defined as,

Sd(q0, q1, . . . , qN) =N−1∑i=0

Ld(qi , qi+1).




discrete Hamilton’s Principle

q ∈ Cd(Q) is called a Discrete Motion if it satisfies,

δSd = 0

Theorem

Discrete motion’s satisfy the discrete Euler − Lagrangeequations,

D2Ld(qk−1, qk) + D1Ld(qk , qk+1) = 0.




discrete Hamilton’s Principle

q ∈ Cd(Q) is called a Discrete Motion if it satisfies,

δSd = 0

Theorem

Discrete motion’s satisfy the discrete Euler − Lagrangeequations,

D2Ld(qk−1, qk) + D1Ld(qk , qk+1) = 0.




discrete Hamiltonian map

The discrete Euler-Lagrange equations are equivalent to theimplicit discrete Euler − Lagrange equations,

pk = −D1Ld(qk , qk+1),

pk+1 = D2Ld(qk , qk+1).

This defines the discrete Hamiltonian map,

FLd : (qk , pk) 7→ (qk+1, pk+1)




Discrete Legendre Tranforms

Define the discrete Legendre transforms,F±Ld : Q × Q → T ∗Q, as,

F+Ld : (q0, q1) 7→ (q1, p1) = (q1,D2Ld(q0, q1)),

F−Ld : (q0, q1) 7→ (q0, p0) = (q0,−D1Ld(q0, q1)).

The discrete Hamiltonian map is given by, FLd = F+Ld ◦ (F−Ld)−1.The discrete Lagrangian map is given by, FLd = (F−Ld)−1 ◦ F+Ld .This leads to the following commutative diagram.




(qk , pk)FLd // (qk+1, pk+1)

(qk−1, qk)

F+Ld

DD

FLd

// (qk , qk+1)FLd

//

F+Ld

BB��

F−Ld

ZZ4444444444444

(qk+1, qk+2)

F−Ld

]]<<<<<<<<<<<<<<<




Connecting the discrete to the continuous

Now that we have both a discrete and continuous mechanicalsystem, how do we connect the two?

Exact Discrete Lagrangian

LEd (q0, q1; h) =

∫ h

0L(q01(t), q01(t))dt,

where q01(0) = q0, q01(h) = q1, and q01 satisfies theEuler-Lagrange equations on the interval (0, h).




Connecting the discrete to the continuous

Now that we have both a discrete and continuous mechanicalsystem, how do we connect the two?

Exact Discrete Lagrangian

LEd (q0, q1; h) =

∫ h

0L(q01(t), q01(t))dt,

where q01(0) = q0, q01(h) = q1, and q01 satisfies theEuler-Lagrange equations on the interval (0, h).




Variational Error Analysis

Theorem (Marsden and West, 2001)

If a discrete Lagrangian, Ld : Q × Q → R, approximates the exactdiscrete Lagrangian, LE

d : Q × Q → R to order r , i.e.,

Ld(q0, q1; h) = LEd (q0, q1; h) +O(hr+1),

then the discrete Hamiltonian map, FLd : (qk , pk) 7→ (qk+1, pk+1),viewed as a one-step method, is order r accurate.




How do we construct the discrete Lagrangian?

The variational form of the exact discrete Lagrangian isLEd (q0, q1; h) = ext

q∈C2([0,h],Q),q(0)=q0,q(h)=q1

∫ h0 L(q(t), q(t))dt,

which leads naturally to Galerkin methods. Pick a finitedimensional subspace and a numerical quadrature method tobuild the discrete Lagrangian.(Leok, Shingel)

The type-I generating function form of the exact discreteLagrangian is LE

d (q0, q1; h) =∫ h

0 L(q01(t), q01(t))dt, whereq01(0) = q0, q01(h) = q1, and q01 satisfies the Euler-Lagrangeequation in (0, h). Pick a one-step method, apply the shootingmethod to the Euler-Lagrange BVP, and use a quadrature ruleto generate the discrete Lagrangian.(Leok, Shingel)




How do we construct the discrete Lagrangian?

The variational form of the exact discrete Lagrangian isLEd (q0, q1; h) = ext

q∈C2([0,h],Q),q(0)=q0,q(h)=q1

∫ h0 L(q(t), q(t))dt,

which leads naturally to Galerkin methods. Pick a finitedimensional subspace and a numerical quadrature method tobuild the discrete Lagrangian.(Leok, Shingel)

The type-I generating function form of the exact discreteLagrangian is LE

d (q0, q1; h) =∫ h

0 L(q01(t), q01(t))dt, whereq01(0) = q0, q01(h) = q1, and q01 satisfies the Euler-Lagrangeequation in (0, h). Pick a one-step method, apply the shootingmethod to the Euler-Lagrange BVP, and use a quadrature ruleto generate the discrete Lagrangian.(Leok, Shingel)




Prolongate the Euler-Lagrange vector field, and then apply acollocation method using Hermite interpolating polynomialsto approximate values of q and q satisfying theEuler-Lagrange BVP. Combing these with a quadrature ruleyields a discrete Lagrangian.(Leok, Shingel)

Prolongate the Euler-Lagrange vector field, and use this toconstruct Taylor’s method. Find the initial velocity by solvingan inverse problem, and then apply a quadrature rule to createa discrete Lagrangian.




Prolongate the Euler-Lagrange vector field, and then apply acollocation method using Hermite interpolating polynomialsto approximate values of q and q satisfying theEuler-Lagrange BVP. Combing these with a quadrature ruleyields a discrete Lagrangian.(Leok, Shingel)

Prolongate the Euler-Lagrange vector field, and use this toconstruct Taylor’s method. Find the initial velocity by solvingan inverse problem, and then apply a quadrature rule to createa discrete Lagrangian.




Outline


2 Numerical Example







Goal: Construct a one-step method to solve an IVP fort ∈ [0,T ].

Step 1: Given the Euer-Lagrange BVP, use a r − th orderTaylor’s method on TQ to solve for an approximation to v0

implicitly defined by q1 = πQ ◦ Φh(q0, v0).

Step 2: Combine v0 with the r − th order Taylor’s method togenerate the quadrature points to define our discreteLagrangian.

Step 3: Apply the implicit Euler-Lagrange equations to thediscrete Lagrangian to generate the one-step method.


pk+1 = D2Ld(qk , qk+1).










pk+1 = D2Ld(qk , qk+1).










pk+1 = D2Ld(qk , qk+1).










pk+1 = D2Ld(qk , qk+1).




The discrete Hamiltonian map, FLd : (qk , pk) 7→ (qk+1, pk+1),viewed as a one-step method, will be initialized with some(q0, p0) ∈ T ∗Q.

(q0, q1) = (F−Ld)−1(q0, p0) yields the initial values for thediscrete Lagrangian map, whose values also provide theboundary conditions for the Euler-lagrange BVP.

Combining the discrete Legendre transform with thecontinuous Legendre transform provides a connection betweenthe Euler-Lagrange BVP and the Euler-Lagrange IVP via,

(q0, q1) = (F−Ld)−1 ◦ FL(q0, v0).







(q0, q1) = (F−Ld)−1 ◦ FL(q0, v0).







(q0, q1) = (F−Ld)−1 ◦ FL(q0, v0).




The flow of the Euler-Lagrange IVP,Φt : TQ → TQ combined withthe canonical projection onto the configurationmanifold,πQ : TQ → Q, gives an explicit connection to theEuler-Lagrange BVP.

(q0, q1) = (q0, πQ ◦ Φh(q0, v0)),




(q0, v0)

FL

��

Φh // (q1, v1)

FL

��(q0, p0)

FLEd // (q1, p1)

(q0, q1)

F−LEd

]]<<<<<<<<<<<<<<<

F+LEd

AA��




Euler-Lagrange Boudary Value Problem

Given an Euler–Lagrange equation of the form,

q(t) = f (q(t), q(t), t),

we denote the exact solution of the Euler–Lagrange boundary-valueproblem with boundary conditions (q0, q1) by (q(t), v(t)).

We seek an estimate of the true initial velocity, v0, for thecorresponding Euler–Lagrange initial-value problem, withorder of accuracy r . Let us denote this estimate by v0.

Combining v0 with the prolongation of the Euler-Lagrangevector field will allow us to construct Taylor’s method up tothe desired order.






q(t) = f (q(t), q(t), t),









q(t) = f (q(t), q(t), t),







Theorem

A r -order Taylor’s method,

Ψh(q0, v0) =

(∑r

k=0

hk

k!q(k)(0),

∑r+1

k=1

hk−1

(k − 1)!q(k)(0)

),

with initial conditions (q0, v0), where v0 is an order rapproximation to the true intial velocity v0, has order of accuracy rfor the Euler–Lagrange boundary-value problem with boundaryconditions (q0, q1).




Proof

Denote the solution to the Euler-Lagrange BVP withboundary conditions (q0, q1) by (q(t), v(t)) for t ∈ [0, h].

Then (q(t), v(t)) is also a solution to the Euler-Lagrange IVPwith intial conditions (q0, v0), where v0 satisfiesπQ ◦ Φh(q0, v0) = q1.

Denote the solution of the Euler-Lagrange IVP with intialconditions (q0, v0) by (q(t), v(t)).

Let (qd(t), vd(t)) denote the values generated by the order rTaylor’s method with initial conditions (q0, v0).




Proof








Proof








Proof








Proof

Let M be the Lipschitz constant with respect to initial velocity ofthe well-posed Euler-Lagrange IVP. Then,

‖(q(t), v(t))− (q(t), v(t))‖ ≤ M‖v0 − v0‖ ≤ O(hr+1).

Combining this inequality with our r -order method yields,

‖(q(t), v(t))− (qd(t), vd(t))‖= ‖(q(t), v(t))− (q(t), v(t)) + (q(t), v(t))− (qd(t), vd(t))‖≤ ‖(q(t), v(t))− (q(t), v(t))‖+ ‖(q(t), v(t))− (qd(t), vd(t))‖≤ O(hr+1).




We need an order r approximation to v0.

Given (q0, q1) we must approximately solve the inverseproblem,

πQ ◦ Φh(q0, v0) = q1.

What if we used our r-order Taylor’s method to solve for v0

πQ ◦Ψh(q0, v0) = q1






πQ ◦ Φh(q0, v0) = q1.


πQ ◦Ψh(q0, v0) = q1






πQ ◦ Φh(q0, v0) = q1.


πQ ◦Ψh(q0, v0) = q1




2nd order Taylor’s method

q1 = q0 + hv0 +h2

2f (q0) +O(h3)

v0 =q1 − q0

h− h

2f (q0) +O(h2)

Setting v0 = q1−q0h − h

2 f (q0) only provides a 1st orderapproximation to the intitial velocity!




2nd order Taylor’s method

q1 = q0 + hv0 +h2

2f (q0) +O(h3)

v0 =q1 − q0

h− h

2f (q0) +O(h2)

Setting v0 = q1−q0h − h

2 f (q0) only provides a 1st orderapproximation to the intitial velocity!




Recall the Taylor’s method is approximating elements in TQ,thus the full 2nd order Taylor’s method is,[

qd(h)vd(h)

]=

[q0

v0

]+

[v0

f (q0)

]h +

[f (q0)∇f (q0)v0

]h2

2.

Since we already must calculate ∇f (q0) for the 2nd orderTaylor’s method, we can use it on our inverse problem tomake v0 a 2nd order approximation to v0, at no additionalcomputational cost.




Recall the Taylor’s method is approximating elements in TQ,thus the full 2nd order Taylor’s method is,[

qd(h)vd(h)

]=

[q0

v0

]+

[v0

f (q0)

]h +

[f (q0)∇f (q0)v0

]h2

2.

Since we already must calculate ∇f (q0) for the 2nd orderTaylor’s method, we can use it on our inverse problem tomake v0 a 2nd order approximation to v0, at no additionalcomputational cost.




Theorem

v0 as defined by, πQ ◦ Ψh(q0, v0) = q1, where Ψh is a r + 1-orderTaylor’s method, approximates v0 with order of accuracy r .

Taylor discrete Lagrangian

Given a r -order Taylor method and a s-order accurate numericalquadrature formula with quadrature weights bi and quadraturenodes ci , we construct the Taylor discrete Lagrangian,

Ld(q0, q1; h) = h∑n

i=0biL(Ψcih(q0, v0)),

where πQ ◦ Ψh(q0, v0) = q1.




Theorem

v0 as defined by, πQ ◦ Ψh(q0, v0) = q1, where Ψh is a r + 1-orderTaylor’s method, approximates v0 with order of accuracy r .

Taylor discrete Lagrangian

Given a r -order Taylor method and a s-order accurate numericalquadrature formula with quadrature weights bi and quadraturenodes ci , we construct the Taylor discrete Lagrangian,

Ld(q0, q1; h) = h∑n

i=0biL(Ψcih(q0, v0)),

where πQ ◦ Ψh(q0, v0) = q1.




Theorem

Given a r -order Taylor’s method Ψh, a s-order accurate quadratureformula, and a Lagrangian L that is Lipschitz continuous in bothvariables, the associated Taylor discrete Lagrangian approximatesthe exact discrete Lagrangian with order of accuracy min(r + 1, s).




Proof

q01(cih) = qd(cih) +O(hr+1),

v01(cih) = vd(cih) +O(hr+1).

LEd (q0, q1; h) =

∫ h

0L(q01(t), v01(t))dt

=[h∑m

i=1biL(q01(cih), v01(cih))

]+O(hs+1)

=[h∑m

i=1biL(qd(cih) +O(hr+1), vd(cih) +O(hr+1))

]+O(hs+1)




Proof cont’d

=[h∑m

i=1biL(qd(cih), vd(cih))

]+O(hr+2) +O(hs+1)

= Ld(q0, q1; h) +O(hr+2) +O(hs+1)

= Ld(q0, q1; h) +O(hmin(r+1,s)+1)




2nd order Taylor Variational Integrator

Assume the Lagrangian has the form,

L(q(t), q(t)) =1

2qT (t)Mq(t)− V (q(t)),

where M is symmetric positive definite and V is sufficientlysmooth.


q = −M−1OV (q).

We construct the 2nd order Taylor variational integrator usinga 1st order Taylor’s method combined with the trapezoid rule.






L(q(t), q(t)) =1

2qT (t)Mq(t)− V (q(t)),



q = −M−1OV (q).







L(q(t), q(t)) =1

2qT (t)Mq(t)− V (q(t)),



q = −M−1OV (q).






The 1st order Taylor’s method yields,

q1 = q0 + v0h

v1 = v0 −M−1OV (q0)h

To find v0 we solve,

q1 = q0 + v0h −M−1OV (q0)h2

2

v0 =q1 − q0

h+ M−1OV (q0)

h

2






q1 = q0 + v0h

v1 = v0 −M−1OV (q0)h


q1 = q0 + v0h −M−1OV (q0)h2

2

v0 =q1 − q0

h+ M−1OV (q0)

h

2






q1 = q0 + v0h

v1 = v0 −M−1OV (q0)h


q1 = q0 + v0h −M−1OV (q0)h2

2

v0 =q1 − q0

h+ M−1OV (q0)

h

2




Using the trapezoidal quadrature rule our discrete Lagrangianbecomes,

Ld(q0, q1) =h

2(vT

0 Mv0 − V (q0) + vT1 Mv1 − V (q1)).

Given q0 and p0 our one-step method is defined implicitly by,

p0 = −D1Ld(q0, q1)

p1 = D2Ld(q0, q1)

Working out these equations gives the following explicitone-step method,

q1 = q0+hM−1p0−h2

2M−1OV (q0)+

h4

4M−1OOV (q0)M−1OV (q0)

p1 = p0 −h

2(OV (q0) + OV (q1)) +

h3

4OOV (q0)M−1OV (q0)





Ld(q0, q1) =h

2(vT

0 Mv0 − V (q0) + vT1 Mv1 − V (q1)).


p0 = −D1Ld(q0, q1)

p1 = D2Ld(q0, q1)


q1 = q0+hM−1p0−h2

2M−1OV (q0)+

h4


p1 = p0 −h

2(OV (q0) + OV (q1)) +

h3






Ld(q0, q1) =h

2(vT

0 Mv0 − V (q0) + vT1 Mv1 − V (q1)).


p0 = −D1Ld(q0, q1)

p1 = D2Ld(q0, q1)


q1 = q0+hM−1p0−h2

2M−1OV (q0)+

h4


p1 = p0 −h

2(OV (q0) + OV (q1)) +

h3





Higher-Order Taylor Variational Integrators

The 2nd and 3rd order TVI have can have explicit forms dependingon the Euler-Lagrange equations, but all higher-order methods willbe implicit and require non-linear solvers.

A good initial guess to v0 can be found by taking theLegendre Transform of the initial momentum.

A good initial guess for q1 and other values can be found byutilizing Taylor’s method.


















Numerical Considerations

Taylor’s method is well-suited to provide additional quadraturenodes without the need for additional function evaluations.

The derivatives required for Taylor’s method are computedaccurately and at the cost of a function evaluation by usingAutomatic Differentiation.

Any one-step method could be used in place of Taylor’smethod, but it would require both a r order and r+1 order todevelop the variational integrator.


















Outline


2 Numerical Example







Exact Discrete Hamiltonian

H+,Ed (q0, p1; h) = p(h)q(h)−

∫ h

0[p(t)v(t)−H(q(t), p(t))]v= ∂H

∂pdt,

where (q(t), p(t)) is a solution of Hamilton’s equations satisfyingthe boundary conditions q(0) = q0 and p(h) = p1.

Discrete Right Hamilton’s Equations

q1 = D2H+d (q0, p1)

p0 = D1H+d (q0, p1)




Exact Discrete Hamiltonian

H+,Ed (q0, p1; h) = p(h)q(h)−

∫ h

0[p(t)v(t)−H(q(t), p(t))]v= ∂H

∂pdt,

where (q(t), p(t)) is a solution of Hamilton’s equations satisfyingthe boundary conditions q(0) = q0 and p(h) = p1.

Discrete Right Hamilton’s Equations

q1 = D2H+d (q0, p1)

p0 = D1H+d (q0, p1)




If the Lagrangian is degenerate, then the Legendre transformis not invertible, and it may make more sense to approach theproblem from the Hamiltonian side.

Letting h→ 0 causes a breakdown in the well-posedness of aBVP, with Dirichlet boundary conditions q(0) = q0 andq(h) = q1, as the initial velocity becomes unbounded.

Letting h→ 0 does not cause this singularity when theboundary conditions are q(0) = q0 and p(h) = p1.
















Harmonic Oscillator

Consider the harmonic oscillator given by the Euler-Lagrange BVP,

q(t) = −kx ,

where q(0) = 0 and q( π√k

) = 0. This BVP does not have a unique

solution!




Harmonic Oscillator




Harmonic Oscillator Period=0.025




Discrete variational Hamitonian mechanics has been developedin papers by (Lall, West 2006) and (Leok, Zhang 2011).

The exact discrete Hamiltonian is a type 2 generating functionof the exact flow of Hamilton’s equations.(Leok, Zhang 2011)

The map defined by constructing a discrete right Hamitonianand applying the discrete right Hamilton’s equations issymplectic and preserves momentum maps. (Leok, Zhang2011)
















Definition

Given the discrete right Hamilton’s equation, q1 = D2H+d (q0, p1)

and p0 = D1H+d (q0, p1) we define the following maps:

FH+d

: (qk , pk) 7→ (qk+1, pk+1)

FH+d

: (qk , pk+1) 7→ (qk+1, pk+2)




Definition

Define the discrete left and right Legendre Transforms by,

F+H+d (q0, p1) · δp1 = D2H+

d (q0, p1) · δp1

F−H+d (q0, p1) · δq0 = D1H+

d (q0, p1) · δq0.

Thus we have the maps,

F+H+d : (qk , pk+1) 7→ (qk+1, pk+1) = (D2H+

d (q0, p1), pk+1)

F−H+d : (qk , pk+1) 7→ (qk , pk) = (q0,D1H+

d (q0, p1))

Note that this implies, F+H+d = F−H+

d ◦ FH+d

and

F−H+d (q1, p2) = F+H+

d (q0, p1).




Definition

Define the discrete left and right Legendre Transforms by,

F+H+d (q0, p1) · δp1 = D2H+

d (q0, p1) · δp1

F−H+d (q0, p1) · δq0 = D1H+

d (q0, p1) · δq0.

Thus we have the maps,

F+H+d : (qk , pk+1) 7→ (qk+1, pk+1) = (D2H+

d (q0, p1), pk+1)

F−H+d : (qk , pk+1) 7→ (qk , pk) = (q0,D1H+

d (q0, p1))

Note that this implies, F+H+d = F−H+

d ◦ FH+d

and

F−H+d (q1, p2) = F+H+

d (q0, p1).




(qk , pk)FH+d // (qk+1, pk+1)

(qk−1, pk)

F+H+d

DD

FH+d

// (qk , pk+1)FH+d

//

F+H+d

BB��

F−H+d

ZZ4444444444444

(qk+1, pk+2)

F−H+d

]]<<<<<<<<<<<<<<<




The previous commutative diagram implies the discreteHamiltonian map is given by,

FH+,Ed

(q(0), p(h), h) = F+H+d ◦ (F−H+

d )−1.

We say the discrete right Hamiltonian is of order r if,

‖H+d (q(0), p(h), h)− H+,E

d (q(0), p(h), h)‖ ≤ Chr+1.

Equivalently, this may be expressed as,

H+d (q(0), p(h), h) = H+,E

d (q(0), p(h), h)+hr+1e(q(0), p(h), h),

where e(q(0), p(h), h) is bounded on compact sets.





FH+,Ed

(q(0), p(h), h) = F+H+d ◦ (F−H+

d )−1.


‖H+d (q(0), p(h), h)− H+,E

d (q(0), p(h), h)‖ ≤ Chr+1.


H+d (q(0), p(h), h) = H+,E

d (q(0), p(h), h)+hr+1e(q(0), p(h), h),






FH+,Ed

(q(0), p(h), h) = F+H+d ◦ (F−H+

d )−1.


‖H+d (q(0), p(h), h)− H+,E

d (q(0), p(h), h)‖ ≤ Chr+1.


H+d (q(0), p(h), h) = H+,E

d (q(0), p(h), h)+hr+1e(q(0), p(h), h),





Theorem

If a discrete right Hamiltonian, H+d : T ∗Q → R, approximates the

exact discrete Hamiltonian, H+,Ed : T ∗Q → R to order r , i.e.,

H+d (q0, p1; h) = H+,E

d (q0, p1; h) +O(hr+1),

then the discrete map, FH+d

: (qk , pk) 7→ (qk+1, pk+1), viewed as a

one-step method, is order r accurate.




proof

By assumption we have,

H+d (q(0), p(h), h) = H+,E

d (q(0), p(h), h)+hr+1e(q(0), p(h), h).

Differentiating then yields,

D1H+d (q(0), p(h), h) = D1H+,E

d (q(0), p(h), h)+hr+1D1e(q(0), p(h), h),

where ‖D1e(q(0), p(h), h)‖ ≤ C .

‖F−H+d (q(0), p(h), h)− F−H+,E

d (q(0), p(h), h)‖The same can be shown for F+.




proof


H+d (q(0), p(h), h) = H+,E

d (q(0), p(h), h)+hr+1e(q(0), p(h), h).


D1H+d (q(0), p(h), h) = D1H+,E

d (q(0), p(h), h)+hr+1D1e(q(0), p(h), h),

where ‖D1e(q(0), p(h), h)‖ ≤ C .

‖F−H+d (q(0), p(h), h)− F−H+,E





proof


H+d (q(0), p(h), h) = H+,E

d (q(0), p(h), h)+hr+1e(q(0), p(h), h).


D1H+d (q(0), p(h), h) = D1H+,E

d (q(0), p(h), h)+hr+1D1e(q(0), p(h), h),

where ‖D1e(q(0), p(h), h)‖ ≤ C .

‖F−H+d (q(0), p(h), h)− F−H+,E

d (q(0), p(h), h)‖

The same can be shown for F+.




proof


H+d (q(0), p(h), h) = H+,E

d (q(0), p(h), h)+hr+1e(q(0), p(h), h).


D1H+d (q(0), p(h), h) = D1H+,E

d (q(0), p(h), h)+hr+1D1e(q(0), p(h), h),

where ‖D1e(q(0), p(h), h)‖ ≤ C .

‖F−H+d (q(0), p(h), h)− F−H+,E





proof

Given smooth functions related by,

f1(x , h) = g1(x , h) + hr+1e1(x , h)

f2(x , h) = g2(x , h) + hr+1e2(x , h),

the following can be shown.

f2(f1(x , h), h) = g2(g1(x , h), h) + hr+1e12(x , h)

f −11 (y , h) = g−1

1 (y , h) + hr+1e1(y , h)




proof

Given smooth functions related by,

f1(x , h) = g1(x , h) + hr+1e1(x , h)

f2(x , h) = g2(x , h) + hr+1e2(x , h),

the following can be shown.

f2(f1(x , h), h) = g2(g1(x , h), h) + hr+1e12(x , h)

f −11 (y , h) = g−1

1 (y , h) + hr+1e1(y , h)




proof

Combining this with the fact that FH+d

= F+H+d ◦ (F−H+

d )−1

yields,

FH+d

(q(0), p(h), h) = FH+,Ed

(q(0), p(h), h)+hr+1e3(q(0), p(h), h).




TVI on the Hamiltonian side

All Taylor Variational Integrators on the Hamiltonian side areimplicit.

Using an order r Taylor’s method results in an order r TVI.

H+d (q0, p1; h) = p1q1−

h

2(pT

0 q0−H(q0, p0)+pT1 q1−H(q1, p1))







H+d (q0, p1; h) = p1q1−

h

2(pT

0 q0−H(q0, p0)+pT1 q1−H(q1, p1))







H+d (q0, p1; h) = p1q1−

h

2(pT

0 q0−H(q0, p0)+pT1 q1−H(q1, p1))





The inverse problem for p0 is,

p1 = πT∗Q ◦Ψh(q0, p1) = p0 −∇V (q0) + . . . .

Upshot: No need to construct a higher order-method for theinverse problem. Thus, this generalizes easily to any one-stepmethod.





The inverse problem for p0 is,

p1 = πT∗Q ◦Ψh(q0, p1) = p0 −∇V (q0) + . . . .

Upshot: No need to construct a higher order-method for theinverse problem. Thus, this generalizes easily to any one-stepmethod.




Future Work

Analyze further the stability and other qualitative propertiesof the discrete Hamiltonian versus discrete Lagrangianvariational integrator.

Examine the stability of the discrete Hamiltonian variationalintegrator for high oscillatory mechanical systems.

Further develop a symplectic method for dealing withmechanical systems involving a low and high frequencycomponent via time averaging.

Investigate the discretization of the boundary Lagrangian andpossibly a boundary Hamiltonian for multi-symplectic fieldtheories.




Future Work








Future Work








Future Work






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Classical Mechanics Numerical Example Discrete Mechanics Taylor …j2schmit/TALK.pdf · 2017-01-29 · Classical Mechanics Numerical Example Discrete Mechanics Taylor Variational