jitse niesen (university of leeds) per christian moan ...jitse/cas11.pdf · the numerical method...
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![Page 1: Jitse Niesen (University of Leeds) Per Christian Moan ...jitse/cas11.pdf · The numerical method reduces to y n+1 = y n + h y n = (1 + h )y n, so y n = (1 + h )ny 0. The function](https://reader034.vdocuments.mx/reader034/viewer/2022042915/5f51799528897f0eab257298/html5/thumbnails/1.jpg)
Modified equations for Hamiltonian systems
Jitse Niesen (University of Leeds)
Per Christian Moan
Institute of Computational Mathematics andScientific/Engineering Computing
Chinese Academy of Sciences, Beijing
22 August 2011
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Feng Kang
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Numerical integratorsA numerical integrator applied to the initial value problem (IVP)
y ′ = f (y), y(0) = y0,
yields a sequence y1, y2, y3, . . . that approximates the solution tothe IVP: yn ≈ y(nh).
For instance, the (forward) Euler method is:
yn+1 = yn + hf (yn).
Apply this to y ′ = 0.2 y , y(0) = 1 with step size h = 0.5:
0 2 4 6 8t
012345
y
y(t)
y0y1
y2y3
y4
y5
y6
y7
y8
![Page 4: Jitse Niesen (University of Leeds) Per Christian Moan ...jitse/cas11.pdf · The numerical method reduces to y n+1 = y n + h y n = (1 + h )y n, so y n = (1 + h )ny 0. The function](https://reader034.vdocuments.mx/reader034/viewer/2022042915/5f51799528897f0eab257298/html5/thumbnails/4.jpg)
Modified equations
The modified equation is the initial value problem
y ′ = f (y), y(0) = y0,
such that its solution interpolates the numerical solution:yn = y(nh) (modulo technicalities).
The numerical method solves the original equation approximately,but it solves the modified equation exactly (backward erroranalysis).
0 2 4 6 8t
012345
y
y(t)
y(t)
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Modified equations: An exampleConsider the Euler method yn+1 = yn + hf (yn) applied to thelinear test equation
y ′ = λy , y(0) = y0,
The numerical method reduces to yn+1 = yn + hλyn = (1 + hλ)yn,so yn = (1 + hλ)ny0.
The function
y(t) = (1 + hλ)t/h = exp
(log(1 + hλ)
ht
)interpolates the numerical solution: yn = y(nh).
The function y satisfies the differential equation
y ′ =log(1 + hλ)
hy .
This is the modified equation.
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Outline
I Modified equations
I Hamiltonian equations and symplectic integrators
I The modified energy
I The work of Skeel et al.
I Richardson extrapolation
I Example: The pendulum
I Exponentially small remainder for modified energy
I Error when computing the modified energy
I Example: The Kepler problem
I Conclusion
![Page 7: Jitse Niesen (University of Leeds) Per Christian Moan ...jitse/cas11.pdf · The numerical method reduces to y n+1 = y n + h y n = (1 + h )y n, so y n = (1 + h )ny 0. The function](https://reader034.vdocuments.mx/reader034/viewer/2022042915/5f51799528897f0eab257298/html5/thumbnails/7.jpg)
Outline
I Modified equations
I Hamiltonian equations and symplectic integrators
I The modified energy
I The work of Skeel et al.
I Richardson extrapolation
I Example: The pendulum
I Exponentially small remainder for modified energy
I Error when computing the modified energy
I Example: The Kepler problem
I Conclusion
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Hamiltonian equations
A Hamiltonian equation is of the form
q′ =∂H
∂p(q, p), p′ = −∂H
∂q(q, p)
for some function H.
We can write this as y ′ = J∇H(y) where J =[
0 I−I 0
]and y =
[ qp
].
Solutions preserve energy H, meaning that H(y(t)) = H(y(0)).
In this talk, we only consider separable Hamiltonians, meaning thatH is of the form H(q, p) = T (p) + V (q) (kinetic plus potentialenergy), yielding the equations q′ = ∇T (p), p′ = −∇V (q).
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The pendulum
Hamiltonian equation: q′ =∂H
∂p(q, p), p′ = −∂H
∂q(q, p)
For example, the mathematical pendulum is described by
q′ = p, p′ = − sin q,
which is the Hamiltonian equation for H = 12p
2 − cos q.
0 2 4 6 8 101.51.00.50.00.51.01.5
q(t)
p(t)
H(t)
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The pendulum: Numerical solution
Apply Euler method with h = 0.05.
{qn+1 = qn + hpnpn+1 = pn − h sin qn
0 81.5
0.0
1.5
0 100
Apply leap-frog method with h = 1.
qn+ 1
2= qn + 1
2hpnpn+1 = pn − h sin qn+ 1
2
qn+1 = qn+ 12
+ 12hpn+1
0 81.5
0.0
1.5
0 100
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Symplectic integrators
Let Φt be the flow map of the Hamiltonian equation y ′ = J∇H(y);that is, Φt maps y(0) to y(t) where y is a solution of the equation.
Then, Φt is a symplectic map: it satisfies (DΦt)T J DΦt = J
where DΦt is the Jacobian matrix.
Let Ψf be the time stepping function of a numerical integratorapplied to y ′ = f (y); that is, Ψf maps yk to yk+1. For forwardEuler, Ψf (y) = y + hf (y).
The method is a symplectic integrator if Ψf is a symplectic mapwhenever the method is applied to a Hamiltonian equation. Theleap-frog method is symplectic; Euler’s method is not.
Thus, symplectic integrators preserve the symplectic structure ofthe ODE. They do not preserve energy, but energy drift isbounded. Modified equations can explain this.
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Modified energy
If a non-symplectic method is used on a Hamiltoian equation, themodified equation will not be Hamiltonian.
However, if a symplectic integrator is used, then the modifiedequation is Hamiltonian: y ′ = J∇H(y). Modified equation reflectthe structure-preserving properties of the integrator.
We call H the modified energy.
The main topic of this talk is how to compute the modified energy.
Motivation: The modified energy can be used to detect numericalinstabilities.1 Computing the modified energy is also useful toillustrate and guide backward error analysis.
1 Skeel & Hardy, SIAM J. Sci. Comp., 23:1172–1188 (2001).
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Outline
I Modified equations
I Hamiltonian equations and symplectic integrators
I The modified energy
I The work of Skeel et al.
I Richardson extrapolation
I Example: The pendulum
I Exponentially small remainder for modified energy
I Error when computing the modified energy
I Example: The Kepler problem
I Conclusion
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The extended Hamiltonian
Given Hamiltonian H(q, p), equations are q′ = ∂H∂p and p′ = −∂H
∂q .
Define homogeneous extension of H by
He(q, α, p, β) = α2H(1αq,
1αp).1
Hamilton’s equations for He are:
q′ =∂He
∂p= α
∂H
∂p, p′ =
∂He
∂q= −α∂H
∂q,
α′ =∂He
∂β= 0, β′ = −∂H
e
∂α= −2αH + qT
∂H
∂q+ pT
∂H
∂p.
If α(0) = 1, then α(t) = 1 for all t, so (q, p) evolve under He asunder H. Furthermore,
β′ = −2H + qTp′ − pTq′.
1 Skeel & Hardy, SIAM J. Sci. Comp., 23:1172–1188 (2001).
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Numerical integrators and the extended Hamiltonian
Given Hamiltonian H(q, p), equations are q′ = ∂H∂p and p′ = −∂H
∂q .
Modified eq’n is q′ = ∂H∂p , p′ = −∂H
∂q where H is modified energy.
Extended Hamiltonian He = α2H(1αq,
1αp)
yields equations:
q′ =∂H
∂p, p′ = −∂H
∂q, α′ = 0, β′ = −2H + qTp′ − pTq′.
If a splitting method is used, then modified energy for He ishomogeneous extension of modified energy for H: He = (H)e .1
Thus modified equation for He is
q′ =∂H
∂p, p′ = −∂H
∂q, α′ = 0, β′ = −2H + qT p′ − pT q′.
Last equation implies H = 12(qT p′ − pT q′ − β′).
1 Skeel & Hardy, SIAM J. Sci. Comp., 23:1172–1188 (2001).
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Computing the modified energy
We can use numerical solution {qn, pn, βn} to estimateH = 1
2(qT p′ − pT q′ − β′). For instance, a second-orderapproximation is
1
2
(qTn
pn+1 − pn−12h
− pTnqn+1 − qn−1
2h− βn+1 − βn−1
2h
)Higher-order approximations can be constructed:2
2 Engle, Skeel & Drees, J. Comput. Phys., 206:432–452 (2005).
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Richardson extrapolation
We propose the use of Richardson extrapolation to approximateH = 1
2(qT p′ − pT q′ − β′).
Richardson extrapolation combines two approximations of the sameorder into one higher-order approximation.
Suppose we want to approximate z ′(0) for some function z .Construct two second-order approximations:
T1 =z(12h)− z(−1
2h)
hand T2 =
z(h)− z(−h)
2h
Then T ∗ = T2 + 13(T2 − T1) is a fourth-order approximation.
This process can be repeated to get approximants of higher andhigher orders. It is used in Romberg’s method to compute integralsand in the Stoer–Bulirsch method to solve ODEs.
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Computing modified energy with Richardson extrapolation
We want to approximate H = 12(qT p′ − pT q′ − β′).
First, compute second-order approximations using centraldifferences:
T nj ,1 =
1
2
(qTn
pn+j − pn−j2jh
− qTnpn+j − qn−j
2jh−βn+j − βn−j
2jh
).
Then, apply Richardson extrapolation recursively:
T nj ,k+1 = T n
j ,k +T nj ,k − T n
j−1,k(1− k/j)2 − 1
(for k < j).
After j iterations we get a (2j)-th order approximation to themodified energy: T n
j ,j = H(qn, pn) +O(h2j).
The error in Tj ,j can be estimated as Tj+1,j+1 − Tj ,j .
We use Arprec multiple-precision library with 200 digits.
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The pendulum: H = 12p
2 − cos(q)Solve pendulum with leap-frog rule over t = [0, 100] and computeenergy drift ∆H = maxn H(qn, pn)−minn H(qn, pn) using Tj ,j forj = 2, 5, 10, 15, 20, 30, 40 and various values of step size h.
0 20 40 60 80 1001/h
10-50
10-40
10-30
10-20
10-10
100
drift
usi
ng Tj,j
2
5
10
15
20
Red line, corresponding to j = 200, shows that ∆H ≈ C1e−C2/h.
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The pendulum: H = 12p
2 − cos(q)
Set q(0) = 0 and vary initial condition for p(0).
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4p(0)
10-100
10-80
10-60
10-40
10-20
100dr
ift in
mod
ified
ene
rgy
h=1/5
h=1/10
h=1/15
h=1/20
The drift in modified energy decreases as we approach theequilibrium point (q, p) = (0, 0).
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Outline
I Modified equations
I Hamiltonian equations and symplectic integrators
I The modified energy
I The work of Skeel et al.
I Richardson extrapolation
I Example: The pendulum
I Exponentially small remainder for modified energy
I Error when computing the modified energy
I Example: The Kepler problem
I Conclusion
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Modified equations done rigorously
Actually, there is no f such that solution of y ′ = f (y) interpolatesnumerical solution, yn = y(nh).
The standard approach is to use a divergent but asymptotic series:
f ∼ f + hp+1fp+1 + hp+2fp+2 + hp+3fp+3 + · · ·
If f[k] is truncation of order k and y[k] solves y ′[k] = f[k](y[k]) then
yn = y[k](nh) +O(hk+1).
If f is analytic and truncation index k is chosen suitably (varyingwith h), then the remainder is exponentially small:
|y1 − y[k](h)| ≤ C1he−C2/h.
In Hamiltonian setting, y[k] preserves exactly the modified
energy H[k], so numerical drift in modified energy is exponentially
small over exponentially long intervals.3
3 Tang Yifa (1994), Benettin & Giorgilli (1994), Hairer & Lubich (1997), . . .
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An alternative approach
It is possible to find a modified equation if we allow it to benon-autonomous.4
The modified equation is y ′ = f1(y) + f2(y , t) where thetime-dependent part f2 is h-periodic and exponentially small:
‖f2‖ ≤ C1h−1‖f ‖ exp
(− C2
h‖f ‖
),
where ‖ · ‖ denotes max-norm in complex neighbourhood of y(t).
Solution of modified equation interpolates numerical solutionexactly: yn = y(nh).
In Hamiltonian setting, modified energy H = H1(y) + H2(y , t) istime-dependent, so not conserved.
However, energy drift ddt H(y(t)) = ∂H2
∂t is exponentially small.
4 Moan, J. Phys. A, 39:5545–5561 (2006).
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Theorem on time-dependent modified equationsSuppose that f is analytic in
D =⋃t>0
{z ∈ Cd : |=(y(t)− z)|∞ < R}.
Let ‖f ‖D = supz∈D |f (z)|∞.
If one-step method Ψf is analytic, then there exists a modifiedvector field f (y , t) = f1(y) + f2(y , t) analytic in
D′ =
{(z , τ) ∈ Cd × C : |=(z − y(t))‖∞ < R − δ, |=(τ − t)| < δη
e‖f ‖D
}for all δ ∈ (0,R) and η ∈ (0, 1).The time-dependent part f2 is h-periodic: f2(y , t + h) = f2(y , t).The modified equation satisfies the estimates:
‖f ‖D′ ≤2
1− η‖f ‖D
‖f2‖D′ = O(‖f ‖Dh
exp
(− 2πδη
eh‖f ‖D
)).
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Error in computing the modified energy
Drift in modified energy is exponentially small.Can we recover it numerically?
Yes! If j∗ is chosen suitably, then error in j∗-th approximation inRichardson tableau is exponentially small:
|H(qn, pn, ty )− T nj∗,j∗ | ≤ C ′1h
−1‖f ‖ exp
(− C ′2h‖f ‖
).
This bound is of same form as bound on time-dependent part
‖f2‖ ≤ C1h−1‖f ‖ exp
(− C2
h‖f ‖
).
Thus, we should be able to track (accumulated) drift in modifiedenergy numerically.
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Kepler problem: Estimated error in Tj ,j against j
0 20 40 60 80 100j
10-50
10-40
10-30
10-20
10-10
100
|Tj+
1,j
+1−Tj,j|
We use ad hoc criterion to find truncation point: smoothen curveby taking maximum over 10 values and find minimum ofsmoothened curve (see stars in plot).
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Kepler problem: Energy drift against time
0 20 40 60 80 100time
10-50
10-40
10-30
10-20
10-10
100
drift
in m
odifi
ed e
nerg
y
h=1/10
h=1/20
h=1/40
h=1/80
Modified energy shows drift when approaching point closest tosingularity but recovers afterwards.
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Kepler problem: Energy drift against h
0 50 100 150 2001/h
10-80
10-60
10-40
10-20
100
drift
in m
odifi
ed e
nerg
y
Again, ∆H ≈ C1e−C2/h, confirming theory.
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Kepler problem: Other methods
0 20 40 60 80 1001/h
10-40
10-30
10-20
10-10
100
drift
in m
odifi
ed e
nerg
y
Leap-frogYoshidaBlanes/Moan
Also include Yoshida’s fourth-order method5 and highly optimized6-stage fourth order method of Blanes and Moan.6
5 Yoshida, Phys. Lett. A, 150:262–268 (1990).6 Blanes & Moan, J. Comput. Appl. Math, 142:313–330 (2002).
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Conclusion
I Richardson extrapolation can be used to track drift inmodified energy.
I Error estimates show that error is of similar magnitude as driftin modified energy, so we can capture the drift numerically.
I Experiments confirm exponentially small drift, also forHenon–Heiles system which Skeel could not do.
I If preserving modified energy over long times is priority, thenthere seems no reason to use higher-order methods.
I Can we use modified energy to study chaotic islands inpendulum (numerically induced chaos)?
I Need better stopping criterion for Richardson extrapolation; orperhaps use convergence acceleration?