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The stadium billiard is known to be a chaotic system. This means that the only integral of motion (quantity which is conserved along any trajectory of motion) is the energy $E=(p_x^2+p_y^2)/2m$.

Why are we sure that no other, independent on $E$, integrals of motion exist in this system? One can assume existence of some, perhaps infinitely complicated, function $I(x,y,p_x,p_y)$, which is conserved and independent on $p_x^2+p_y^2$. Why is this assumption wrong?

In other words: the simplest prototypical examples of integrable billiards (rectangular, circular, elliptical) have some obvious symmetries allowing us to find two independent integrals of motion. What if, in some other billiard, such integrals do also exist, but are not so obvious and have no simple analytical form? How can we distinguish two situations:

  1. there exist two independent integrals of motion, so the system is integrable, but their form is very complicated,

  2. the only integral of motion is $E$ and other independent integrals are absent?

I'm not a specialist in dynamical systems and related complicated mathematics, so any simple explanations will be appreciated. I've found the related questions Idea of integrable systems and Non-integrability of the 2D double pendulum but didn't get any simple answer.

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    $\begingroup$ Integrability requires "smooth" continuity within a given interval. Chaotic motion is abrupt and even discontinuous, therefore, not integrable - unless you are interested in only a part of the total motion where the requirements for integrability are met. $\endgroup$
    – Guill
    Commented Jun 27, 2017 at 21:43
  • $\begingroup$ @AlexeySokolik did you even reach an answer by yourself to your question? $\endgroup$
    – Lo Scrondo
    Commented Mar 20, 2021 at 14:50
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    $\begingroup$ @LoScrondo No, I have not yet obtained an answer which would be clear for me. For myself, I have made preliminary conclusion that there could be no sharp distinction between existence and non-existence of (perhaps, arbitrarily complicated) integrals of motion, in the same way as integrable and chaotic systems cannot always be clearly resolved as the KAM theory says. $\endgroup$ Commented Mar 22, 2021 at 11:02
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    $\begingroup$ Ok, so I dare to add my two cents...1) Instead of speculating on the admissibility of any symmetry that could take to an integral of motion (IoM) of any form, what is commonly done is more empiric, e.g. demonstrate the existence of a polynomial/algebraic/transcendental IoM in a specific system...2) Even if the argument cited by Vittorio Moretti (and echoed in some other authors) seems conclusive, in the original Poincaré works (and other authors) highly irregular (e.g. discontinuous) IoMs are taken into consideration - and are not covered by the former accounts. So AFAIK the question is open $\endgroup$
    – Lo Scrondo
    Commented Mar 22, 2021 at 22:33
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    $\begingroup$ related: useful terminology here physics.stackexchange.com/q/55861/226902 . Wiki for Dynamical Billard: en.wikipedia.org/wiki/Dynamical_billiards "billiards with positive Lyapunov exponents were thought to need convex scatters, such as the disk in the Sinai billiard, to produce the exponential divergence of orbits. Bunimovich showed that by considering the orbits beyond the focusing point of a concave region it was possible to obtain exponential divergence." $\endgroup$
    – Quillo
    Commented Jun 18, 2022 at 10:34

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I think one must distinguish between chaotic "Hamiltonian" systems and chaotic dissipative systems. In the latter, the phase space volume is not conserved, so it is much more difficult to find "integrals of motion" because Liouville's theorem is broken. Remember, a quantity "A" is an integral of motion if

$\frac{dA}{dt} = \frac{\partial A}{\partial t} + \{A,H\} = 0$, where $H$ is the Hamiltonian. For dissipative chaotic systems, you can't even write down $H$, so it is difficult to see how one could generally find integrals/constants of motion of the system.

However, there is an important class of systems that show up in cosmology for example where you have Hamiltonian "chaos", where essentially the trajectories of the system exhibit all of the properties of chaos: sensitive dependence on initial conditions, diverging trajectories over time, but, the system still has attractors: a famous example is the dynamics of a closed anisotropic universe / Bianchi IX, in shameless self-promotion here: https://arxiv.org/pdf/1311.0389.pdf (in particular, see Page 27) This has of course led to wide debates for years in the cosmology community of whether this is "Really" chaos, since, in principle, the trajectories are predictable, but, I hope this answers your question.

Further, with respect to your Billiards problem / the famous Hadamard billiards, as you can see it is the same as the diagram on Page 27. Therefore, the billiard problem is also an example of Hamiltonian / deterministic / non-dissipative chaos. The phase space has an asymptotic attractor. This hopefully demonstrates that integrals of motion such as the one you found above ($E$ is the total energy of the system, and in this case, is the Hamiltonian, $H$) are only really possible if one can write down a Hamiltonian by Liouville's theorem.

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  • $\begingroup$ Sp there is some connection between attractors in phase space and existence of integrals of motion? And how the difference between a) integrable system with some very complicated integral of motion and b) chaotic system without it looks like from this point of view? $\endgroup$ Commented Jun 23, 2017 at 12:59
  • $\begingroup$ Yes, you can say that. You can only have a Hamiltonian system if you have no attractors / asymptotic stability. $\endgroup$ Commented Jun 23, 2017 at 13:02
  • $\begingroup$ I'm afraid I have no enough knowledge to fully understand it. How it works in the case of the stadium billiard? How its attractor looks like and why, e.g., rectangular billiard doesn't have the same property? $\endgroup$ Commented Jun 26, 2017 at 10:51
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I am not an expert on these issues, but if a further integral existed the orbit would be confined in a codimesion-1 embedded submanifold (for almost all the possible values of that function due to Sard's theorem). An embedded submanifold is a very regular subset, it cannot have self-intersections and cannot be dense in the space for example. Instead the orbit of a chaotic system does not seem to belong to such a regular set...However without a precise definition, all that remains just a suggestion...

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  • $\begingroup$ Asking for details after 4 years does probably not have much sense...anyway even if your observation is suggestive, it is still not conclusive to me - so could I ask you if the conditions you have exemplificated (an embedded submanifold cannot be dense) are absolute? $\endgroup$
    – Lo Scrondo
    Commented Mar 20, 2021 at 14:48
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    $\begingroup$ A first integral is a map $I :M\to R$ where $M$ is the space of phases. Sard'stheorem says that for almost all values of $I$ it holds $dI\neq0$. The preimage of each of such values is therefore a codimension 1 embedded submanifold as a consequence of the theorem of regular values. A codimension 1 embedded submanifold cannot be dense: around each point it appears to be as a n-1 plane in $R^n$. However I am not an expert on these issues... $\endgroup$ Commented Mar 21, 2021 at 8:24
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The Bunimovich stadium is well known to be ergodic. Here is a nice description by Terry Tao. This notion extends naturally to quantum (or wave) chaos where instead of the trajectories being asymptotically uniformly distributed, it is the nodal domains of the eigenfunctions that are asymptotically uniform. The new book by Steve Zelditch "Eigenfunctions of the Laplacian on a Riemannian manifold" is a thorough technical look at these issues for the quantum/wave case. Free pdf here

The existence of scars (asymptotic concentrations in phase space) is a reflection of unstable periodic trajectories. If there are no concentrations then the system is called uniquely ergodic.

Integrability requires not just N constants of motion, but also that the constants be in involution to one another. This means that the Poisson Brackets of any pair is zero. By the way, there is a difference between constants of motion and integrals of motion. Integrals of motion comprise a subset of the constants of motion. Here is a reference to ergodicity vs integrability.

The term chaos is used to mean different things in different contexts. In addition to the spreading of trajectories you need mixing to get classical chaos. Thus you'll find extensive discussion of these issues in books on Riemannian Geometry since the mixing usually comes from boundary curvature. My favorite is Marcel Berger's A Panoramic View of Riemannian Geometry. Berger has an extensive discussion of stadium billiards.

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  1. On one hand, a positive maximal Lyapunov exponent (MLE) is often taken as a de-facto definition of (deterministic) chaos. (Note that chaos also requires topological mixing.)

  2. On the other hand, Poincare showed that an autonomous Liouville-integrable Hamiltonian system has only zero Lyapunov exponents along periodic orbits, cf. e.g. my Phys.SE answer here. (If the level-sets are compact, then every orbit is periodic, cf. the Liouville-Arnold theorem. For details, see e.g. my Phys.SE answer here.)

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This depends in ways on the system. Yet integrable domains can exist. The logistics map $x_{n+1}~=~rx_n(1~-~x_n)$ has for the parameter $r$ zones of stability. The image

enter image description here

illustrates bifurcation regions of regular dynamics. These are islands embedded in this region of "scarring" with chaotic dynamics.

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    $\begingroup$ Yes, I've heard about regions of stability, but how can it be described in terms of integrals of motion? Does it mean some integral of motion exists in the stadium billiard, which is conserved only on a subset of trajectories? $\endgroup$ Commented Jun 22, 2017 at 13:59
  • $\begingroup$ @AlexeySokolik for example a back-and-forth trajectory, resulting from reflection from walls perpendicular to velocity, has a conserved zero component of momentum. $\endgroup$
    – Ruslan
    Commented Jun 22, 2017 at 19:39
  • $\begingroup$ zero component? A fourvector or are you introducing some strange C programming indexing into physics? $\endgroup$ Commented Jun 22, 2017 at 20:34
  • $\begingroup$ @VladimirF no, he meant literally that one of the components of the momentum vector is always zero. $\endgroup$
    – LLlAMnYP
    Commented Jun 23, 2017 at 8:09

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