Symplectic time integrators for numerical General Relativity · Symplectic time integrators for...

Introduction Discretisation Constrained evolution schemes Summary and Outlook

Symplectic time integrators for numerical General

Relativity

Ronny Richter, Christian Lubich

Semiannual Meeting of the SFB/TR7Garching

September 25 2007

Ronny Richter University of Tübingen

Symplectic time integrators for numerical General Relativity


Motivation

Symplectic integrators preserve energy and improve stabilitye.g. in simulations of the outer solar system.

Can these integration methods be used to improve simulationsin General Relativity?




Motivation

Symplectic integrators preserve energy and improve stabilitye.g. in simulations of the outer solar system.

Can these integration methods be used to improve simulationsin General Relativity?




Contents

1 IntroductionCanonical Formulation of ODENumerical Integration Methods

2 DiscretisationDerivation of the discrete ADM HamiltonianFree evolution

3 Constrained evolution schemesThe momentum constraints as hidden constraintsNumerical tests




Contents







Canonical Formulation of ODE

Hamiltonian Systems

If a system of ordinary di�erential equations can be written as

pi = −∂qiH(p, q), qi = ∂pi

H(p, q) (1)

then it is a Hamiltonian system.

The function H(p, q) is called Hamiltonian.

If (p(t), q(t)) is a solution of (1) then

H(p(t0), q(t0)) = H(p(t1), q(t1)) ∀t0, t1. (2)

The Hamiltonian is a �rst integral.





Hamiltonian Systems



H(p, q) (1)




H(p(t0), q(t0)) = H(p(t1), q(t1)) ∀t0, t1. (2)






Hamiltonian Systems



H(p, q) (1)




H(p(t0), q(t0)) = H(p(t1), q(t1)) ∀t0, t1. (2)






Canonical and symplectic transformations

Canonical transformations preserve the Hamiltonian character:Let p′ = ψp(p, q) and q′ = ψq(p, q).The map ψ is called canonical transformation if

p′i = −∂q′

iH ′(p′, q′), q′

i = ∂p′iH ′(p′, q′),

with H ′(ψp(p, q), ψq(p, q)) = H(p, q).

Symplectic maps preserve the so-called symplectic 2-form (anotion of area in phase space).

Theorem (Jacobi, 1836):A map ψ is symplectic, if and only if it is canonical.







p′i = −∂q′

iH ′(p′, q′), q′

i = ∂p′iH ′(p′, q′),










p′i = −∂q′

iH ′(p′, q′), q′

i = ∂p′iH ′(p′, q′),







Numerical Integration Methods

Symplectic Integrators

De�nition:A numerical one-step method is called symplectic if theone-step map yn+1 = Φ∆t(yn) is symplectic whenever themethod is applied to a smooth Hamiltonian system.

The value of the Hamiltonian (i.e. the energy) is preserved.

The symplectic 2-form is preserved.





Symplectic Integrators

De�nition:A numerical one-step method is called symplectic if theone-step map yn+1 = Φ∆t(yn) is symplectic whenever themethod is applied to a smooth Hamiltonian system.

The value of the Hamiltonian (i.e. the energy) is preserved.

The symplectic 2-form is preserved.





Examples for symplectic integrators

Often symplectic integrators are implicit methods (dependingon the structure of the Hamiltonian and the method itself).

Integrators for free Hamiltonian systems

Symplectic Euler method

Symplectic Runge-Kutta methods

Störmer-Verlet method

...

Integrators for Hamiltonian systems with holonomicconstraints

Symplectic Euler method with constraints

Partitioned Lobatto IIIA - IIIB pair

Rattle method

...











...




Rattle method

...











...




Rattle method

...




Contents







Derivation of the discrete ADM Hamiltonian

Canonical Formulation of GR

A Hamiltonian that describes General Relativity has beendeveloped since the late �fties.

The ADM Hamiltonian is1

HADM =

∫d3x

(α

(πijπij −

1

2πi iπ

jj

)− αhR

+ 2πijhikDjβk

), (3)

where

πij =√h

(Khij − K ij

). (4)

1We replaced the lapse function N by the slicing density α := N/√h.





Canonical Formulation of GR

A Hamiltonian that describes General Relativity has beendeveloped since the late �fties.

The ADM Hamiltonian is1

HADM =

∫d3x

(α

(πijπij −

1

2πi iπ

jj

)− αhR

+ 2πijhikDjβk

), (3)

where

πij =√h

(Khij − K ij

). (4)

1We replaced the lapse function N by the slicing density α := N/√h.





First simple examples

For �rst tests we consider only those solutions where the 3-metrichij ful�ls the following properties

hij = 0 for i 6= j

hij(x , y , z) ≡ hij(x)

hyy = hzz







hij = 0 for i 6= j


hyy = hzz







hij = 0 for i 6= j


hyy = hzz





A simpli�ed Hamiltonian

These systems are described by a simpli�ed ADM Hamiltonian

H =

∫dx

[α

(1

2πxxπxxhxxhxx − πxxπyyhxxhyy

)− α

(1

2∂xhyy∂xhyy − 2hyy∂

2

xhyy + ∂xhyy∂x log(hxx)

)+ 2πxxhxx∂xβ

x + πxxβx∂xhxx + πyyβx∂xhyy

]. (5)

The 3-metric h provides the position variables (q in (1)).

The corresponding canonical momentum is π (p in (1)).

For the moment we take α and βx to be given in advance.







H =

∫dx

[α

(1


)− α

(1


2


)+ 2πxxhxx∂xβ


]. (5)










H =

∫dx

[α

(1


)− α

(1


2


)+ 2πxxhxx∂xβ


]. (5)










H =

∫dx

[α

(1


)− α

(1


2


)+ 2πxxhxx∂xβ


]. (5)








Method of lines

Derive a discrete Hamiltonian:

Approximation by piecewise linear functions

Centered 2nd order �nite di�erences for the spacial derivatives

Integration of the product of piecewise linear functions

The discrete Hamiltonian only depends on the functionalvalues at the grid points.





Method of lines










Method of lines










Method of lines










Method of lines









Free evolution

Application of free evolution schemes

From the discrete Hamiltonian derive the ordinary di�erentialequations for the grid variables.

The discrete Hamiltonian system is then a system of �rst orderODE in 4N variables (N: number of grid points).

Given initial data it is straight forward to apply well knownintegration methods (symplectic as well as non-symplectic).




Free evolution








Free evolution








Free evolution

Gauge wave testbed

The method was tested by simulating the gauge wave testbed.

The analytical solution is

g = (1 + H)(dt2 − dx2) + dy2 + dz2, (6)

where H = A sin(2π(x − t)).

After the 3 + 1-decomposition w.r.t. t = const slicing oneobtains

hxx = 1 + H, hyy = 1 = hzz , hij = 0 for i 6= j . (7)




Free evolution

Gauge wave testbed



g = (1 + H)(dt2 − dx2) + dy2 + dz2, (6)







Free evolution

Gauge wave testbed



g = (1 + H)(dt2 − dx2) + dy2 + dz2, (6)







Free evolution

Results

Results of simulations with A = 0.01 and A = 0.1 for 50 and 200grid points:




Free evolution

Stability of the Hamiltonian system

In the high amplitude simulations (A = 0.1) the highfrequency modes grow and the code crashes.

It turns out that the linearisation of the system correspondingto (5) is not well-posed (the norm of the solution at a time t

cannot be bounded by the norm of the initial data).

If the continuous problem is not well-posed one cannot expectthat the discrete evolution is stable.




Free evolution









Free evolution









Contents







The momentum constraints as hidden constraints

Type of constraints

There are symplectic constrained evolutionschemes for holonomic constraints, i.e. those thatdepend on the position variables only(e.g. the Rattle method).

To impose these constraints additionalforces of constraints

and corresponding Lagrange multipliers are added:

p = −∂qH(p, q)− Fg (q, λ), q = ∂qH(p, q), g(q) = 0.

The forces of constraints are Fg = λ∂qg(q).





Type of constraints

��

��

��

pfree










Type of constraints

��

��

��

��

��

pfree

Fg










Modi�cation of the Poisson structure

Formally one may impose the momentum constraints in thatway.

Interpret the shift β as the canonical momentum of a positionvariable ρβ that does not appear in the Hamiltonian.

The Hamiltonian equations of motion for β and ρβ are then

β = −∂ρβH = 0, ρβ = ∂βH = −2C, (8)

where C is the momentum constraint.

If the discrete holonomic constraint ρβ = 0 is imposed in theconstrained evolution scheme then automatically a discretemomentum constraint C = 0 is satis�ed.









β = −∂ρβH = 0, ρβ = ∂βH = −2C, (8)











β = −∂ρβH = 0, ρβ = ∂βH = −2C, (8)











β = −∂ρβH = 0, ρβ = ∂βH = −2C, (8)







Invertibility of the forces of constraints

Imposing the constraint ρβ = 0 leads to a force of constraintFβ that in�uences only β:

β = −∂ρβH − Fβ. (9)

There is no β in the equation ρβ = ∂βH = −2C.It is thus not possible to adjust this force of constraint (i.e.the corresponding Lagrange multiplier) such that ρβ = 0 isalways satis�ed.

The way out:

Impose another holonomic constraints for which thecorresponding force has in�uence on πij .







β = −∂ρβH − Fβ. (9)


The way out:








β = −∂ρβH − Fβ. (9)


The way out:








β = −∂ρβH − Fβ. (9)


The way out:






Choosing a gauge

This constraint must depend on the metric hij .

The freedom of choosing a spacial coordinate system was notused yet.

Choose the gauge such that some function g of the metric andits spacial derivatives vanishes.

One obtains

β = −∂ρβH − Fβ = −Fβ, ρβ = ∂βH = −2C,

πij = −∂hijH − F

ijg , hij = ∂πijH,

ρβ = 0, g(hij) = 0. (10)





Choosing a gauge




One obtains

β = −∂ρβH − Fβ = −Fβ, ρβ = ∂βH = −2C,



ρβ = 0, g(hij) = 0. (10)





Choosing a gauge




One obtains

β = −∂ρβH − Fβ = −Fβ, ρβ = ∂βH = −2C,



ρβ = 0, g(hij) = 0. (10)





Choosing a gauge




One obtains

β = −∂ρβH − Fβ = −Fβ, ρβ = ∂βH = −2C,



ρβ = 0, g(hij) = 0. (10)




Numerical tests

Gauge choice for the gauge wave

For the gauge wave testbed the function hxx should havesinusoidal character.

Calculate the Fourier transform of hxx and require that thehigher frequency modes vanish → g(hij).

Use the shift to satisfy the gauge constraint and the force ofconstraint Fijg to satisfy ρβ = 0.

Possible problem:

For yet unknown reasons it is not possible to require thehighest frequency mode of hxx to vanish → we ignore thisconstraint, too.




Numerical tests





Possible problem:





Numerical tests





Possible problem:





Numerical tests





Possible problem:





Numerical tests

Possible problems

The number of forces Fg equals the number of constraintsg(hij) = 0 (i.e. N − 4 here).

The number of constraints ρβ = 0 is N.

It is possible that the freedom in choosing the Lagrangemultipliers in the forces Fijg is not su�cient to satisfy ρβ = 0.

However, the operator that assigns the momentum constraintsto the momenta πij 7→ C clearly has a kernel.

It might be that the missing four forces of constraints onlymove the momentum within this kernel.




Numerical tests

Possible problems









Numerical tests

Possible problems









Numerical tests

Possible problems









Numerical tests

Possible problems









Numerical tests

Results of Rattle

Results of the constrained evolution using Rattle for A = 0.1 and50 grid points:




Summary

For small amplitude in the gauge wave testbed the freeevolution schemes seem to be quite stable.

For higher amplitudes one obtains growing high frequencymodes.

The origin of this instability is presumably the nonwell-posedness of the continuous equations.

Through a modi�cation of the ADM Hamiltonian one mayapply constrained evolution schemes, too.

The scheme investigated here is still unstable.




Summary









Summary









Summary









Summary









Outlook

Try to �nd Hamiltonian formulations of General Relativity thatare well-posed.

Investigate the constrained evolution schemes further.




Outlook

Try to �nd Hamiltonian formulations of General Relativity thatare well-posed.

Investigate the constrained evolution schemes further.



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