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Science Forum Index » Nonlinear Science Forum » AUTO07p and continuation of periodic solutions
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| Sven |
 Posted: Fri Mar 02, 2007 3:31 pm |
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Guest
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Hi,
is here anybody who has some experience using AUTO for the
continuation of periodic solutions in Hamiltonian
systems?
Thanks,
Sven Schmidt |
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| Sven |
| Posted: Fri Mar 02, 2007 7:48 pm |
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Guest
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Hi,
I would like to continue a periodic solution from the Lyapunov family
of L1 in the restricted three body problem for varying mu.
AUTO found the Lyapunov family with the help of a detuning parameter
because of the conserved quantity.
If I try to continue a periodic solution with varying mu, AUTO either
aborts with MX or it just doesn't vary mu and stops immediately.
The constant file I use looks like
4 2 5 0 NDIM,IPS,IRS,ILP
3 2 1 11 NICP,(ICP(I),I=1 NICP)
100 4 3 3 -1 15 0 0 NTST,NCOL,IAD,ISP,ISW,IPLT,NBC,NINT
60 0.01215 0.1 -3 3 NMX,RL0,RL1,A0,A1
3 -5 2 15 5 3 0 NPR,MXBF,IID,ITMX,ITNW,NWTN,JAC
1e-6 1e-6 1e-4 EPSL,EPSU,EPSS
2e-2 1e-5 5e-2 1 DS,DSMIN,DSMAX,IADS
1 NTHL,((I,THL(I)),I=1,NTHL)
11 0
0 NTHU,((I,THU(I)),I=1,NTHU)
0 NUZR,((I,UZR(I)),I=1,NUZR)
Since the solution was found previously by varying the detuning
parameter it appears as first index (2) in ICP, then mu and last the
period.
Does anybody know what is wrong?
I successfully could continue periodic orbits in the Lorenz system so
I guess it must have something to do with the detuning.
Thanks and kind regards,
Sven Schmidt |
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| Alois Steindl |
| Posted: Mon Mar 05, 2007 5:29 am |
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Guest
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Hello,
I have never applied AUTO for this kind of problem, but are you aware,
that in Hamiltonian systems periodic solutions generically appear as
family of periodic solutions, not as isolated orbits?
So if you fix your parameter (e.g. mu) at some value, you will have a
manifold of periodic solutions, which might display special behaviour
(shrink to a point, bifurcate) for special parameter values.
Alois |
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| Sven |
| Posted: Mon Mar 05, 2007 4:15 pm |
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Guest
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On Mar 5, 2:29 am, Alois Steindl <Alois.Stei...@tuwien.ac.at> wrote:
Quote: Hello,
I have never applied AUTO for this kind of problem, but are you aware,
that in Hamiltonian systems periodic solutions generically appear as
family of periodic solutions, not as isolated orbits?
So if you fix your parameter (e.g. mu) at some value, you will have a
manifold of periodic solutions, which might display special behaviour
(shrink to a point, bifurcate) for special parameter values.
Alois
Hi Alois,
yes, I know what you mean (cylinder theorem). That's why I introduced
an unfolding parameter lambda that destroys this family
of periodic solutions for lambda != 0.
I was able to continue a periodic orbit in the Lorenz system for
varying r and I can continue a family of periodic solutions
in a system with integral with the help of a detuning parameter. Now I
try to glue these two things together, but don't know how
because the obvious solution doesn't work.
The way I tried to do this is the following. I extracted the numerical
data for this orbits using a software that integrates the
equations of motion. Then I converted this file (using fcon) into a
form AUTO can work with (a solution file you can access with
load(e="r3b",c="r3b.1",s=solution file)).
I guess I haven't really understood how AUTO continues in more than
one free parameter since you can only specify the continuation
range for the principal parameter.
Thanks for you help and kind regards,
Sven |
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| Alois Steindl |
| Posted: Tue Mar 06, 2007 7:56 am |
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Guest
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"Sven" <s.schmidt@lboro.ac.uk> writes:
Quote: yes, I know what you mean (cylinder theorem). That's why I introduced
an unfolding parameter lambda that destroys this family
of periodic solutions for lambda != 0.
I would be afraid that this causes additional troubles. You would have
to design your unfolding such that it admits a unique periodic
solution for lambda=0. (If I understand your method correctly.)
Quote: I was able to continue a periodic orbit in the Lorenz system for
varying r and I can continue a family of periodic solutions
in a system with integral with the help of a detuning parameter. Now I
try to glue these two things together, but don't know how
because the obvious solution doesn't work.
The way I tried to do this is the following. I extracted the numerical
data for this orbits using a software that integrates the
equations of motion. Then I converted this file (using fcon) into a
form AUTO can work with (a solution file you can access with
load(e="r3b",c="r3b.1",s=solution file)).
I guess I haven't really understood how AUTO continues in more than
one free parameter since you can only specify the continuation
range for the principal parameter.
I would suggest to choose one of the following strategies:
a) Keep the parameter mu fixed and calculate all periodic solutions
(at least the connected family, you are interested in.)
b) For varying parameter mu calculate special periodic solutions, like
resonant ones.
Of course there could be interesting combinations of these methods:
Find an interesting solution with a) and continue it with method b.
BTW, I think that you know already that you can leave out one periodic
boundary condition, because that one is usually automatically
fulfilled (if you start on the proper branch).
Good luck
Alois
--
Alois Steindl, Tel.: +43 (1) 58801 / 32558
Inst. for Mechanics and Mechatronics Fax.: +43 (1) 58801 / 32598
Vienna University of Technology, A-1040 Wiedner Hauptstr. 8-10 |
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| Sven |
| Posted: Tue Mar 06, 2007 12:35 pm |
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Guest
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On Mar 6, 4:56 am, Alois Steindl <Alois.Stei...@tuwien.ac.at> wrote:
Quote: "Sven" <s.schm...@lboro.ac.uk> writes:
yes, I know what you mean (cylinder theorem). That's why I introduced
an unfolding parameter lambda that destroys this family
of periodic solutions for lambda != 0.
I would be afraid that this causes additional troubles. You would have
to design your unfolding such that it admits a unique periodic
solution for lambda=0. (If I understand your method correctly.)> I was able to continue a periodic orbit in the Lorenz system for
varying r and I can continue a family of periodic solutions
in a system with integral with the help of a detuning parameter. Now I
try to glue these two things together, but don't know how
because the obvious solution doesn't work.
The way I tried to do this is the following. I extracted the numerical
data for this orbits using a software that integrates the
equations of motion. Then I converted this file (using fcon) into a
form AUTO can work with (a solution file you can access with
load(e="r3b",c="r3b.1",s=solution file)).
I guess I haven't really understood how AUTO continues in more than
one free parameter since you can only specify the continuation
range for the principal parameter.
I would suggest to choose one of the following strategies:
a) Keep the parameter mu fixed and calculate all periodic solutions
(at least the connected family, you are interested in.)
b) For varying parameter mu calculate special periodic solutions, like
resonant ones.
Of course there could be interesting combinations of these methods:
Find an interesting solution with a) and continue it with method b.
BTW, I think that you know already that you can leave out one periodic
boundary condition, because that one is usually automatically
fulfilled (if you start on the proper branch).
Good luck
Alois
--
Alois Steindl, Tel.: +43 (1) 58801 / 32558
Inst. for Mechanics and Mechatronics Fax.: +43 (1) 58801 / 32598
Vienna University of Technology, A-1040 Wiedner Hauptstr. 8-10
Hi Alois,
thank you very much for your answer, I appreciate this very much!
The method with the detuning parameter is prety much standard, see
recent papers by Doedel et. al., especially
"Continuation of periodic orbits in conservative and Hamiltonian
systems", Physica D 181 (2003) 1-38 by
F.J. Munoz-Almaraz.
Your suggested approach sounds good and I pretty much did what you
said. But as I said, the connection of both continuations (in mu and
the detuning parameter) won't work. So I figured I use an even simpler
Hamiltonian system with natural parameter (lambda). I found one in
Guckenheimer&Holmes (page 45-47). For lambda>0 it has an elliptic
equilibrium at the origin so I detuned, AUTO found the Hopf
bifurcation and calculated the family of periodic orbits. I took one
of them and continued that one in three parameters (lambda, detuning,
period) and it worked!, although I dont't know how AUTO decides how to
vary the parameters.
So, what I think the problem is, is that if I try to feed AUTO with
the numerical data of one complete periodic solution, something goes
wrong because I tried this with the same system and it failed. Now I
try to find out what is going on here.
The other issue is that I want AUTO to continue the periodic orbits
for fixed energy. I'd like to use the user-supplied routine pvls to
add the energy as a parameter but I can't get it working with C. I
think it is broken in AUTO07P.
Kind regards,
Sven Schmidt |
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| Sven |
| Posted: Tue Mar 06, 2007 4:32 pm |
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Guest
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So, some good news.
If I let AUTO compute the Ljaounov family around L1 in R3BP and take
one periodic solution, say S1, and continue it in all three
parameters,
it works. But I want to give AUTO an extern numerical starting
solution to continue. So what I did was compute the exact same
solution S1 with Mathematica, wrote it to an external file and
converted it into an AUTO readable file using fcon.
But AUTO just doesn't want to continue it.
This is the Ljapunov family:
Starting r3b ...
BR PT TY LAB PAR(2) L2-NORM MAX U(1) MAX
U(2) MAX U(3) MAX U(4) PERIOD
2 3 4 2.89740E-14 8.38429E-01 8.43244E-01
2.14889E-02 1.41894E-02 4.95911E-02 2.69903E+00
2 6 5 1.77247E-14 8.46342E-01 8.54133E-01
5.41004E-02 3.77104E-02 1.22275E-01 2.73956E+00
2 9 6 7.61080E-15 8.60917E-01 8.66911E-01
8.77537E-02 6.48178E-02 1.93103E-01 2.82186E+00
2 12 7 2.47572E-15 8.81845E-01 8.83582E-01
1.23358E-01 9.49996E-02 2.61984E-01 2.96096E+00
2 15 8 1.28509E-15 9.08315E-01 9.06929E-01
1.62323E-01 1.26583E-01 3.28348E-01 3.18381E+00
2 18 9 6.67108E-14 9.38311E-01 9.36744E-01
2.06730E-01 1.56427E-01 3.90969E-01 3.52641E+00
2 21 10 3.81608E-13 9.68262E-01 9.69965E-01
2.58436E-01 1.81236E-01 4.48662E-01 4.00200E+00
2 24 11 3.03833E-13 9.95219E-01 1.00155E+00
3.17113E-01 2.24969E-01 5.02495E-01 4.55862E+00
2 27 12 6.73898E-13 1.01906E+00 1.02868E+00
3.80226E-01 2.94125E-01 5.55303E-01 5.11374E+00
2 30 13 9.10726E-13 1.04111E+00 1.05130E+00
4.45186E-01 3.57710E-01 6.09633E-01 5.61330E+00
2 33 14 5.04378E-12 1.06257E+00 1.07032E+00
5.09980E-01 4.16182E-01 6.66840E-01 6.03754E+00
2 36 15 1.44079E-12 1.08420E+00 1.08664E+00
5.73371E-01 4.70898E-01 7.27860E-01 6.38580E+00
2 39 16 2.91691E-12 1.10647E+00 1.10088E+00
6.34929E-01 5.22887E-01 7.93183E-01 6.66537E+00
2 42 17 2.35866E-12 1.12959E+00 1.11352E+00
6.94121E-01 5.73018E-01 8.62853E-01 6.88596E+00
2 45 18 1.31161E-11 1.15367E+00 1.12485E+00
7.50696E-01 6.21574E-01 9.37228E-01 7.05727E+00
2 48 19 4.20597E-12 1.17874E+00 1.13506E+00
8.04498E-01 6.68668E-01 1.01643E+00 7.18796E+00
2 51 20 1.90910E-11 1.20478E+00 1.14431E+00
8.55843E-01 7.14223E-01 1.10047E+00 7.28540E+00
2 54 21 3.65348E-13 1.23171E+00 1.15268E+00
9.04391E-01 7.58111E-01 1.18988E+00 7.35567E+00
2 57 22 -9.48457E-12 1.25945E+00 1.16024E+00
9.50354E-01 8.00027E-01 1.28469E+00 7.40369E+00
2 60 EP 23 1.04773E-11 1.28791E+00 1.16701E+00
9.93588E-01 8.39608E-01 1.38522E+00 7.43343E+00
Total Time 0.154E+01
I chose the solution with label 23 to continue with varying mu:
Starting r3b ...
BR PT TY LAB PAR(2) L2-NORM MAX U(1) MAX
U(2) MAX U(3) MAX U(4) PAR(1) PERIOD
2 100 24 -7.28198E-09 1.24250E+00 1.15916E+00
9.23556E-01 7.86534E-01 1.22278E+00 1.04799E-02 7.43343E+00
2 200 25 -8.10595E-09 1.19917E+00 1.15068E+00
8.48235E-01 7.30994E-01 1.07187E+00 8.53341E-03 7.43343E+00
2 300 26 -7.40446E-10 1.15874E+00 1.14120E+00
7.68726E-01 6.71933E-01 9.31457E-01 6.59989E-03 7.43343E+00
2 400 27 -2.23981E-09 1.12165E+00 1.13026E+00
6.84462E-01 6.09284E-01 7.99411E-01 4.83242E-03 7.43343E+00
2 500 28 -3.53121E-09 1.08843E+00 1.11791E+00
5.95969E-01 5.41839E-01 6.73825E-01 3.31499E-03 7.43343E+00
2 600 29 8.83879E-10 1.05959E+00 1.10390E+00
5.04055E-01 4.69175E-01 5.54087E-01 2.08973E-03 7.43343E+00
2 700 30 -5.81186E-09 1.03568E+00 1.08803E+00
4.08938E-01 3.91421E-01 4.39099E-01 1.17001E-03 7.43343E+00
2 800 31 6.22663E-09 1.01724E+00 1.07021E+00
3.11821E-01 3.07487E-01 3.27897E-01 5.45754E-04 7.43343E+00
2 900 32 -3.33477E-09 1.00476E+00 1.05032E+00
2.13539E-01 2.17379E-01 2.20807E-01 1.84441E-04 7.43343E+00
2 1000 33 -7.26866E-09 9.98677E-01 1.02823E+00
1.14087E-01 1.20389E-01 1.16490E-01 2.98367E-05 7.43343E+00
2 1100 34 -1.36822E-08 9.99321E-01 1.00380E+00
1.46148E-02 1.60265E-02 1.47772E-02 6.66593E-08 7.43343E+00
2 1200 35 -3.48002E-09 1.00690E+00 1.04414E+00
8.42005E-02 9.63178E-02 2.02237E-01 1.36219E-05 7.43343E+00
2 1300 36 -3.38797E-09 1.02130E+00 1.09610E+00
1.81905E-01 2.17357E-01 4.54741E-01 1.46581E-04 7.43343E+00
2 1400 37 -2.28709E-09 1.04220E+00 1.14813E+00
2.77791E-01 3.47800E-01 7.24814E-01 5.60671E-04 7.43343E+00
2 1500 EP 38 -9.43391E-10 1.06918E+00 1.20005E+00
3.72259E-01 4.88255E-01 1.01424E+00 1.44723E-03 7.43343E+00
Total Time 0.110E+02
So something goes wrong when I try to use an external numerical
solution. Any thoughts?
Thanks, Sven |
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