Re: Poincare section for double pendulum

• To: mathgroup at smc.vnet.net
• Subject: [mg120677] Re: Poincare section for double pendulum
• From: WetBlanket <wyvern864 at gmail.com>
• Date: Wed, 3 Aug 2011 07:07:46 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <j0u7jr\$fbg\$1@smc.vnet.net> <j10l7m\$ou9\$1@smc.vnet.net>

```On Jul 30, 4:08 am, JUN <noec... at gmail.com> wrote:
> On Jul 29, 5:04 am, gal bevc <gal.b... at gmail.com> wrote:
>
>
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>
> > Hello,
>
> > I'm a relatively new user of Mathematica, who doesn't have much of
> > programming skills. For my undergraduate assingment I must analyze chaotic
> > motion of double pendulum.
>
> > Until now i have got system of differential equations for equations of
> > motion for double pendulum(i have x''[t]=function(t) and
> > y''[t]=function(t)). System of differential equations can be solved for 4
> > inital conditions, x[0],y[0],x'[0] and y'[0]. With using function NDSolvei
> > got functions of angles and angular velocities for upper and lower pendulum
> > with respect to time, x[t],x'[t],y[t] and y'[t].
>
> > To get a poincare section of double pendulum, i have to record position of
> > y[t] and y'[t] whenever x[t] is equal to zero and the velocity of x'[t] is a
> > positive number. In the end I must get some sort of phase diagram y[t] and
> > y'[t].
> > Because this is a Hamilton non-dissipative system, inital energy of the
> > system is a constant of time and initial energy is a function of initial
> > conditions. To get a real poincare diagram i must repeat the procedure
> > described above for different initial conditions, but for the same energy
> > level. I need mathematica to use some random numbers for initial conditions
> > in a way that the initial energy of the system stays the same. So i must
> > repeat procedure for poincare section(surface of section) for let's say 50
> > different initial conditions and then display all results in one y[t],y'[t]
> > diagram.
> > Hope that someone can help me.
>
> > Thank you,
> > Gal Bevc
>
> Hi,
> one way of doing this is by using StepMonitor during the numerical
> solution to look for zero-crossings of the variable x[t]:
>
> (a)
> Define an empty list, say
> zeros = {};
> This will be used to collect the zeros in your time interval.
>
> (b)
> Define an auxiliary variable "lastX" and set it equal to the initial
> value of x (say xInitial),
> lastX = xInitial;
>
> (c)
> ...,StepMonitor:>(If[x[t]*lastX<0,AppendTo[zeros,t]];lastX=x[t])
> This tests for zero crossings of x[t] between the current time step
> and the previous one.
>
> (d)
> When NDSolve is done, the list "zeros" contains a set of approximate t
> values with x[t]=0 that you can then use to refine using FindRoot.
> Let's say your solution (in the form of rules {x->..., y->...}) is
> stored in "sol", then say
> Map[FindRoot[Evaluate[x[t]/.sol],{t,#}]&,zeros]
>
> That should give you the values of t at which you would then evaluate
> y[t] and y'[t] to make the points for the Poincare section.
>
> Jens

See book by Stephen Lynch titled "Dynamical Systems with Applications
using Mathematica"

```

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