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Consider a double pendulum:

Background

For the angles $varphi_i$ and the momenta $p_i$ we have (with equal lengths $l=1$, masses $m=1$ and gravitational constant $g=1$):

$dotvarphi_1 = 6frac2p_1 - 3p_2cos(varphi_1 - varphi_2)16 - 9cos^2(varphi_1 - varphi_2)$

$dotvarphi_2 = 6frac8p_2 - 3p_1cos(varphi_1 - varphi_2)16 - 9cos^2(varphi_1 - varphi_2)$

$dotp_1 = -frac12big( dotvarphi_1dotvarphi_2 sin(varphi_1 - varphi_2) +3sin(varphi_1) big)$

$dotp_2 = -frac12big( -dotvarphi_1dotvarphi_2 sin(varphi_1 - varphi_2) +sin(varphi_1) big)$

To see the relations more clearly:

$dotvarphi_1 = B(2p_1 + Ap_2)$

$dotvarphi_2 = B(8p_2 + Ap_1)$

$dotp_1 = -C + 3D$

$dotp_2 = +C + D $

with

$A = -3cos(varphi_1 - varphi_2)$

$B = 6/(16 - A^2)$

$C = dotvarphi_1dotvarphi_2sin(varphi_1 - varphi_2)/2$

$D = -sin(varphi_1)/2$

Observations

With initial angles $varphi_1^0 = varphi_2^0 = 0$ and different combinations of small values for $p_1^0$, $p_2^0$ a number of intriguiung patterns can be observed when plotting the tractory of the tip of the pendulum:

• 
  $p_1^0 = 1$, $p_2^0 = 1$
  

• $p_1^0 = 1, p_2^0 = -1$

• 
  $p_1^0 = 0$, $p_2^0 = 1$
  

• 
  $p_1^0 = 0$, $p_2^0 = 2$
  

• 
  $p_1^0 = 0$, $p_2^0 = 3$
  

• 
  $p_1^0 = 0$, $p_2^0 = 3.7$
  

• 
  $p_1^0 = 0$, $p_2^0 = 4$
  

What these patterns have in common: The tip of the pendulum draws a curve which more or less slowly "fills" an area enclosed by a specific envelope, intermediately exhibiting seemingly regular patterns which inevitably eventually vanish.

Questions

1. Can the envelope be given in closed form, depending only on the two parameters $p_1^0, p_2^0$?




2. Can the positions of the two inner cusps which can be seen clearly for $p_1=0$, $p_2=2,3$ be given in closed form, depending only on the two parameters $p_1^0, p_2^0$?

[The envelope for $p_1^0=0, p_2^0 = 1,2,dots$ looks like a canoe whose bow and stern bend to each other, eventually amalgamating. Can anyone guess what's the explicit formula for this shape?]

This is not a complete or rigorous answer, but it should point you in the right direction.

The double pendulum is a Hamiltonian system; energy is conserved. The most extreme points of the trajectory occur when both legs are at standstill ($p_1=p_2=0$) as all the energy is positional. The points where this is the case can be computed as parametrised curves as follows.

1. The parameter of your curve is the angle of excitation of the first pendulum ($φ_1$).


2. For each $φ_1$, you can calculate the two angles of the inner pendulum ($φ_2$) such that the potential energy of the system corresponds to your total energy.


3. From both angles ($φ_1$ and $φ_2$) obtain the position of the inner pendulums end as the point of your curve.

I would conjecture that the outermost of the two curves is the envelope in case of a chaotic dynamics such as your last example.
If the double pendulum is ergodic, the above follows, as all points on the proposed envelope have the same energy and thus have to be visited.