To start things off I'd say that noting the $L_z$ component is conserved seems to mean pretty much nothing, since you're considering the motion as restricted to the $\mathcal{X}\mathcal{Y}$ plane. If you had assumed the motion along the $\mathcal{Z}$ axis to be possible, then we'd be talking about the spherical double pendulum instead of the planar one (which is the case, since the lagrangian has two degrees of freedom).

The energy will be conserved because the system is autonomous (time-independent). Notice also that an integrable autonomous system with $n$ freedoms has $n$ conserved quantities, one of them being the energy. So, since our system has two degrees of freedom, there is one constant of motion missing for it to be integrable. Differentiating $L$ with respect to $\dot{\theta}_1$ and $\dot{\theta}_2$ will give us the canonical momenta of the system, but notice that the derivatives of $L$ with respect to $\theta_1$ and $\theta_2$ are not zero. Thus, the canonical momenta are not conserved quantities. Also, the total energy of the system doesn't factor as a sum of individual energies, since the Lagrangian has mixed terms in it. You cannot extract any other quantity from the Lagrangian which should be conserved, since mechanics talks about the conservation of momenta and energy (and sometimes their projections). Since there are not as much conserved quantities as there are degrees of freedom, the system is non-integrable.

P.S.: In one dimension we can clearly see that (using simple examples as the harmonic oscillator or the simple pendulum) some systems do not conserve their momenta. Even though, if they're autonomous then they're integrable, because they have one degree of freedom and one conserved quantity: energy.

EDIT.: Since the question is directly oriented to "is it possible to obtain, from the Lagrangian, an answer about integrability?", then (as suggested in the comments) I'll sketch something on Noether's theorem. The theorem relates Lie groups to invariant quantities, so it's just one more way to find conserved quantities. It basically says that if you find a transformation that leaves the Lagrangian invariant, then there's a conserved quantity associated to that transformation. As an example, when the transformation reduces to a translation, invariance of the Lagrangian implies momentum conservation; much the same way, if the transformation is a rotation, then invariance implies the conservation of angular momentum along the axis of rotation. So this is basically a way of using symmetries of the Lagrangian to obtain conservation laws (knowing Noether's theorem content well is very important to a clear understanding of many concepts of Quantum Mechanics and Quantum Field Theory). I'm not being quantitative because proving the double pendulum's Lagrangian is not an invariant is tedious, even thought it should be somehow obvious by just looking at it.

**Short answer: No. General trajectories of double pendulum are not periodic**.

You need to distinguish between two aspects: the trajectory in the spatial coordinate system and the trajectory in phase space.

Your claim about $\gamma$ is about the first aspect and is thus false. It is perfectly okay for trajectories to intersect in the real space, and this doesn't mean the solution is periodic.

However, in the phase space it is forbidden for different trajectories to intersect (because of the uniqueness of the solution of ODEs given initial conditions). And if they do, you are correct that the dynamic is periodic. Indeed, notice that it may be that the mass travels through the same spatial point twice, but it can be with different velocities.

As @agemO suggested in a comment below, it is important to stress that although the solution is not periodic, it seem like it is getting close to there (which is probably what confuses you). Suppose for example that the mass starts from a point $(x,y)$ in the XY plane with velocity vector $(v_{x},v_{y})$. Then according to **Poincare Recurrence Theorem**, after some time the mass will travel as close as you want to that point with a very very similar velocity - but they are not guaranteed to be the same. In other words, the motion is as close as you want to be periodic, but it misses, and the resulting behavior is chaotic.

There is another very interesting theorem that should also be worth stating in this case. It is called **Poincare Bendixson Theorem**, and it states that a traped trajectory in 2D phase space must eventually repeat itself (given that the trapping region doesn't contain fixed points). But in this case the phase space is 4D and the theorem doesn't apply.

## Best Answer

I assume that you are talking about an undamped undriven double pendulum.

In this case the motion exhibited by the double pendulum may be chaotic (depending on the intial conditions, lengths of the pendulum arms and masses), but it exhibits no attractors in the sense that trajectories converge to a certain invariant set (the attractor) in phase space. Instead all initial conditions are on some invariant set. For a reason, see this question of mine. So, it’s

strange, but no strange attractor.I understand the dynamics you are describing as one where both pendulum arms are always aligned and the pendulum behaves like a single pendulum. This corresponds to a regular, non-chaotic attractor. However, as already said, the double pendulum (as assumed above) does not have attractors at all. For a suited small excitation of both pendulums, you get a behaviour close to what you are describing, but it immediately starts to exhibit that motion – not just after a certain time.

To make such a dynamics happen, you would, e.g., have to have some friction within the double pendulum’s joint and nowhere else. (If you have friction elsewhere, the pendulum has a single fixed point as an attractor, namely its resting position.)