 ...in progrez. Čekaj. Zdaj mi se skonnešljaj, vsega vsega in fraktalna dimensja, dimensja in atrakta. Čekaj? To je konjuč. Kaplj in jork. Zdaj nekaj je vsega vsega in več, več, več vsega situacija. Zato v 2D vsega uniformi, vsega uniformi in perbolik map. In zelo je je to vsočnje. Zato je vsočnje, ker da počkajte počkajte vsočnje vsočnje odavilu odvega. Nelj ne zelo videli, ko je vsočnja na kori. Nelj ne zelo videli, da počkajte počkajte počkajte Lebo je všeččo počite v 1 over n2. Skolje nekaj tih? Ko bom se počakajte? To je tem, ko bi pačljadi taki taj taki, tati taki taki taki, tati taki taki taki, te vse ni počakajte. Nelakajte vse ni počaki taka, kar je začasno dopoznil počakael. This effort is to in order to have these fit, so you need, they say, 2 decades. Just to say, 1 decandalf, just to be generous. They say 2 decandalf. If you, so between. If you have 100, this means that how many points you need in order to have something like that. po materijalnih početih, oče, tako, nekaj, ki bo vših, in nekaj nekaj, ki je je, kot bo našli početih, početih, kot je, kot bo našli početih, ko se ti zrbi našli početih, kot da tega za 100, začnej da je zrbi našli početih. Kaj je da se bo kaj, kaj je hrv, nekaj 10, kaj je 10-20, kaj je to početih, kaj je nekaj posibili. kaj je, da je zakončiti, zelo si, da jih je zelo zelo, da jih je zelo, da je odrečilo, da je zelo to. In zelo, da jih je zelo, da jih je zelo, da jih je zelo, da jih je zelo, da jih je zelo, da jih je zelo. Mi je izelo, da je izelo, da jih je zelo. So, to je pravda limitacija. To je nemojmo, da jih je vsega počjeva, da se zelo naredimo, da je tebe, da je. Na različno, to je problem. To je pravdjev, to je počkot. In pri vse predstavne, ni je zelo naredimo, da je ne vse način, da je prečo, da bi je vse način, da je prečo vse način. Pohledajte stvari, da je vse način, da je vse način, da je prečo, da je vse način način, da je prečo vse način. Tako, ideja Kapplán-Ejorka je začala. Vse je lambda 1, lambda 2 in so on. Vse je vse. Vse je vse. Vse je vse, vse je vse. Vse je začala. Tako, vse je vse. Tukaj je 1, 2, 3, vse. Tukaj jel se vse tačil, je vse je zvršen. Vse je 1-j. Vse je zvršen. Vse je tukaj. Vse je tukaj. Vse je tukaj. Vse je tukaj. Vse je tukaj. Vse je tukaj. Zelo se načinaj, če je jasnja. Jasnja je zelo, ki je zelo pošelj, da je vse pozitiv. Zelo, da je jasnja plus 1, da je vse pozitiv. Zelo, ki je tudi način, je, da je začelj, če je začelj, če je začelj. Če je začelj, če je začelj, če je začelj, če je začelj, če je začelj. D is equal J star, so, the, the, the, the, the, the, the, this point, okay, the, J star plus something, so, this is linear interpolation, okay, this is the congeature. Now, before to discuss the origin of this congeature, this congeature has been proved only, only in, only in very few particular system. But the, the, this is very important. The, the important is that in this way you, you can compute the, the dimension. You can compute the dimension because there is no difficulty in the computation of Lyapun's point. Okay, and for example, you can check this, you can check, so far in the, in the, in the 2D maps, like Enona. This formula, say that D is equal 1 plus lambda 1 over lambda 2. So, in, in Lorentz, in Lorentz, you have to remind that in Lorentz the second Lyapun's point is 0. And so, this formula give 2 plus lambda 1 over lambda 3. So, and then you can, you can check, for example, you can check if there is a good agreement. And actually this is, the agreement is good, just empirical, empirical remark, okay. But how this gentleman had this idea? What is the, the reasoning? The reasoning is the following. The reason is the following. So, you have this attractor. So, the attractor is something which is, by definition, is invariant under evolution. And then you take, when you look at the K Lyapun's point, Kz, you know, we discussed yesterday. And I look at the volume, at the increasing or decreasing of the volume given by this, this tangent vector. Z1, 90, Z2, and so on. Okay, this, this volume, let me call this volume, volume K, okay. So, what happened? So, this volume go like the sum of the Lyapun's point up to K. So, if, if K is smaller than this quantity, if K is smaller than, is smaller equal J star, you have that, the volume diverging time, when you go to infinite, okay. If K is greater or equal than J star plus one, this go to zero. So, you, you expect that the dimension must be between this and this, because by definition, the dimension, the, the vol, the, the, the volume compute with the, the, the, the, because this volume increases. So, this, this is not enough. This decreases, so it's too much, okay. And so, it must be between this, okay. Between, there is no particular argument to say how in between, just to, to be simple. You take a linear interpolation, okay. This is the argument, very end-wave argument. I admit, it's very end-wave argument. But in, in, in 2D, in 2D map, you can produce another argument, which is more convincing, but you are at the same result. Let me explain the argument in 2D, in 2D with, and you can, you can repeat the argument even for Lawrence, because it's the same. The argument is the following. So, for, for the 2D map, 2D map. You have lambda 1 positive, and lambda 2 negative. In such a way, lambda 1 plus lambda 2 are negative, okay. So, this means, if I start, we have an attractor. So, this means, if I start for something like that, start with a square, okay. This in time evolve, evolve, and yesterday we understood that it must be something like that. This is not the proper picture, okay. And, so, this type T was zero, this T large. So, let me call this L. So, you have that at zero. So, the, the, the length of this curve will increase like the, the, the, the firstly upon the spawn. The length, time T is like L zero, lambda 1, time T. The transversal, this, this thickness, let me call epsilon, increase in time, decrease in time, according to the second, the upon the spawn. It's like L zero, L, L two, where this is like, this is negative. This L zero, L minus L two, okay. You have this, okay. This is, this is here. Now, now you try to cut, try to, now after a certain time, you have an attractor asymptotically. So, now you can decide, you have to count how many, how many square of size epsilon T I need to cover this stuff. And so on. The number of epsilon T is this, this length divided by this. This is the argument. Okay. This is correct. This L of T divided epsilon of T. And so, I have, what I have, I have L one plus modulus of lambda two. Okay. Now, I wanted to write this in term of epsilon. So, if I write this in term of epsilon, this epsilon, I have to write L minus L two times T, two something, two something, okay. What is the something? The something is one. You see, you can consider, you can see that this is correct. If I take D equal one plus lambda one over lambda two, no, sorry, minus. You see that this is true. This is the argument. Okay. Okay. I understand. This is an underway argument. It's not. It's not. But it's a bit convincing. It's a bit convincing. So, if you repeat it with Lawrence, it's the same. But you have to remind that there is a, here there is a factor two because there is a neutral, neutral direction. So, you see that magically this coincide with the previous argument. And so, you can check numerically, and the agreement is rather good. And, okay, some mathematicians. What is the Chinese lady in New York? Lai Sang Yang, Lai Sang Yang, this Chinese lady, who is professor in New York, proved that this is true, where this is the one, what he has said that he called the one. For two dimensional maps uniformly, whatever it means, don't worry. Okay. This is the connection between these two problems. Now, let me spend the last time discussing chaos in Hamiltonian system, because up to now. Sorry, Angelo. What is the fact that lambda one plus lambda two must be negative? Must be negative because it must be dissipative. Okay. This is dissipation. This is the contraction of volume. So, let me now discuss the Hamiltonian system. So, chaos in Hamiltonian system is extremely important because, okay, Hamiltonian system is an important part of the mathematics applied to physics. And the Hamiltonian system are very specific situation, which somehow are not generic. So, usually the system is not Hamiltonian. So, if you have a Hamiltonian system, you perturb slightly, you destroy the Hamiltonian character. Okay. But it's important. I guess it's not necessary, I convinced you that Hamiltonian system is important. So, what means Hamiltonian system? Hamiltonian system means that x is equal to given by a set p of q, and you have a certain quantity, which is called Hamiltonian. Maybe it can also depend from time, but now, let me... So, and you have the Hamiltonian equation. You have q dot n equal this, and p dot n equal minus. Okay. Typically, we have something like that, but it's not necessary. Yeah, it's not necessary that we have something like that. Typically, we have something like interacting stuff. Typically, we have something like this, but it's not necessary. And so, q is taking r to something. Okay. So, the case, the simple case, the simple case is when n equal to 1. n equal to 1 is trivial. So, you know, n equal to 1 is trivial. It's trivial in the sense that n equal to 1, you have the Hamiltonian. It's something like p squared over 2m plus this. Okay. And this is absolutely trivial. It's absolutely elementary. Why is elementary? Because you have... Imagine that v diverge to infinity. So, you have something like that, for example. This is v. If you do the plot here, here q, here p. So, if you fix a certain energy here. So, you have a... You know that in your undergraduate course, that you have periodic motion here. Maybe it's not exactly... Maybe it's not exactly sine cosine, because this... No, but whoever is, for sure, is periodic. So, if you are here, you have something like... Something like that. Okay. I am not very smart. So, you have always periodic motion. Okay. So, the case n equal to 1 and h, autonomous, the system is three in the sense that it's periodic. And you can find the solution just solving an equation, solving an integral. It's difficult, but don't worry. So, apart the difficulty of performing the explicit, the integral, the situation is trivial. The... What is not trivial is when n is in larger dimension. In larger dimension... In larger dimension, the situation is not trivial. And you can wonder if it's possible to solve explicitly the problem in analogy what you are able to do in 1D. So, what is the approach? So, you study in your course in analytical mechanics that... So, you have q, p. You perform a canonical transformation, which preserves the Hamiltonian structure. And you try to obtain a new Hamiltonian, which depends only from the action, and not the angle. If you are able to do this, if you are able, then you are very happy. You are very happy, because now, if you write, again, this equation, what do you have? You have that theta point n is equal derivative of this n. And so, you see, this is just a function of i. Let me call it omega n i. And you have i dot is equal to zero. So, you see that this is constant, and this is trivial, this is just a rotation. And that's it. So, if you are able to put the system in this form, you solve the problem. And the question is that, is it possible to do this, or not, because if it's possible generically, means that we are able to solve all the Hamiltonian problems. So, the problem is the following. So, in the very general question, is it possible to find the canonical transformation of such, that, blah, blah, blah? If yes, that's it. No, for example, in some cases we know. But how many cases you know that this is possible to do? There are only two cases. Harmonic oscillator and two-body problem, two-body gradation problem. These are in classical mechanics. In quantum mechanics, the same. Hydrogen atoms, quick corresponding to Keper problem, and harmonic potential. So, the unique two situations where it's possible to do this explicitly, apart of some particular, the harmonic oscillator, or set harmonic oscillator, the Kepelerian problem. For the rest, you can wonder, why the people didn't try other, no particular pathological situation, such that there is some deep reason. Okay, this problem has been investigated from the very beginning of, from the very beginning means starting from Newton. So, and in particular the people try to understand the problem in astronomy. So, because in astronomy we have the two-body problem. Two-body problem means sun plus planet. One planet, one. Plus one planet, we know, we solve the problem, we have the Kepeler law and so on. This is okay, this is easy. Now, the next step is three-body. Sun plus planet plus asteroid. For example, sun, Jupiter and an asteroid. In first approximation, the solar system is Jupiter. Jupiter, the mass of Jupiter is much larger compared with the mass of the rest of the solar system, apart of the sun. So, now the next problem is to consider the sun, the planet and an asteroid. So, this problem is integrable. So, it is in this form. The problem of the three-body problem, you have something like that. So, you have an integrable part, which corresponds to the situation where you neglect the asteroid and then you have an interaction up here. So, typically the problem in mechanics is that you have an evident integral part plus something disturbing. So, this is very typical, in quantum mechanics you have a problem you can solve and then you have a set of perturbation. And so, you want to understand the problem in terms of the starting point. So, the idea is to find, is to repeat this game, to look for a canonical transformation such that So, the problem is that look for a canonical transformation such that this canonical transformation go to i prime theta prime in such a way this go like if I am able to do this it is the end of the game because now for this I have a conservation law. If this is true, then it means that I have a prime n equal constant. And that's it. This is the end of the story. So, you know that you can formulate the problem in this way or alternative but it's the same. So, you see that in the case where epsilon equals zero you have n integral of the motion all the i are constant. Then, you can wonder if epsilon is smaller question there exist non-trivial integrable because if there exist and then I repeat the game this is consistent. These are equivalent. You can formulate the problem in term of canonical transformation or look for the existence of a non-trivial conservation law. Of course, the energy is conserved. Then, remove the trivial conservation law by the center and so on. So, this is the problem. This problem has been considered. So, the story of the three-body problem is very long. It starts from Newton and all the giant mathematics try to work in these are Euler, Laplace, Lagrange, all these eminent all the people who are in the books of mathematics and with the very weak, very poor result so the existence of some particular situation are very, very poor. The person who solved the problem no, solved the problem in the sense that he was able to find the solution who explained the reason, the difficulty was Poincare. Poincare, so now let me present this very important result of Poincare which is at the basis of Re Poincare around 1200 roughly maybe something less improve the following result. Theorem. So, you have H0 plus epsilon. If epsilon is different from 0 they are not integrable motion, non-trivial. Of course there are trivial, but you remove the trivial, you write the system in such a way you already remove the trivial. Okay, and this is very bad news. Because it means that the idea to find the change of variable such that it is not possible. So the reason why this giant in the past didn't succeed to find the solution was not because it was not smart enough but because it is really impossible. They didn't realize this. How is possible to prove this? This is very simple computation, very nice. Poincare wrote an enormous paper. The paper is 274 pages. It's true that at the time people wrote all the detainer. There are many things in this work. What I presented is just a few pages of this but it is very important from in general point of view. The paper is important for the problem of the body. The computation is important. It is interesting not only of the conclusion but also why you understand where is the difficulty of the problem. The area is the following. For epsilon equals zero, so you have a certain function which is constant. For example, let me kf. Just a certain action, for example. And then you can wonder. Now you can wonder where epsilon is different from zero, smaller. I look for an f which now must depend on i and theta. Of course, you start with this plus epsilon f1i theta plus epsilon square f2i theta and so on, constant. You look for this. You use essentially the idea in elementary quantum mechanics, in turbo teori. Of course, you have to be careful. You have to compute the term and so on. And apparently the problem seems simple. So you look for a set of equations for f1, f2 in such a way this is constant. How is possible with this? So what do you mean that a certain quantity is constant that is Poisson bracket is zero? This constant means that ffi theta h must be zero. Then I write this explicitly. So here I have f0 plus epsilon f1 plus epsilon square f2 and so on. Here h0 plus epsilon h1. This is equal to zero. I write this explicitly. So if I write this explicitly, I start f0 h0 plus epsilon plus epsilon f1 h0 plus epsilon plus epsilon square f2 h0 plus epsilon hi and so on. All this must be zero. Now this is f0 h0 plus epsilon f0 hi plus epsilon f1 h0 plus epsilon square f1 hi plus epsilon square f2 h0 plus epsilon cube f2 and so on. All this stuff must be zero. Must be zero for any epsilon. This is zero for sure because this is the input. Now you look at the term order epsilon. This is the term order epsilon and this is the term order epsilon square. Then you have the other term. The first term is already okay but look for the first order. So the order epsilon. So the order epsilon you have to kill this term. So this must be equal to this. So you have f1 f1 h0 equal to minus f0 order epsilon square. You have f2 h0 equal to minus f1. If you go to the other order you take something. If you go to the other order you have f3 h0 equal to minus f2 and so on. So apparently this is usual chain of equation. Maybe boring but you see. From this you expect this I know I solve f1. So from this I solve f1. So I found f1. I take f1 and put here. So I have an equation for f2. And I found f2. I put here. And f3. And so on. It's boring but so. At least I can stop when I will be exhausted. I consider this is enough. So apparently this is a usual perturbatory computation. It's not particularly nobody love too much these things but the life is tough. So apparently the problem seems a usual problem in perturbatory system. Then you say if you are a mathematician you have to do this then you have to check the information before to worry about the convergence you have to compute this stuff. And apparently there is a precise protocol. The problem is that now let it looks carefully and look at the f1. Look at f1 and we realize that f1 doesn't exist. One career realize that f1 doesn't exist. And then in the most optimistic situations the most optimistic situation means that h i is analytical very smooth with all the infinite derivative maybe I stop here. So the result that I obtain I stop here that this quantity doesn't exist. It's not possible to find this and this is the end of the story. This is the end of the story. The problem is not to check the convergence because the first term of the series doesn't exist. And so this means that this procedure is impossible. So you can rephrase this in term of canonical transformation. You arrive exactly the same result. There appear a trouble and the trouble is called small denominator. There is a certain quantity in the denominator of the expression which you are not able to control in the sense that this can be arbitrary small since it's in the denominator means that the quantity must be arbitrarily large. So this is the end of the story. It was the end of the story at that time. And it was necessary to arrive at 1945 which is an important year for the dynamical system for tourism because I was born there. I was only personal because there is an important theorem not from me, from Kulmogorov. This is very important here and what is necessary to spend some years from the work of Kulmogorov and the work of Kulmogorov was able to put some order in a situation which apparently the idea was hopeless. Actually it's not so hopeless but at least it's difficult. This is Kulmogorov. What I am discussing and this is Kulmogorov. Kulmogorov I just mentioned because entering the work of Kulmogorov is 20h roughly Kulmogorov and then the other name Kulmogorov Arnold what is called KEM Theorem. Ok, now I stop after the break I show you the origin of the trouble which is not a mathematical problem which is not a mathematical difficulty. This is the real origin. It's not possible to do this because we are not smart enough. No, this is the trouble. The trouble is real. It's a physical trouble. So we take 5 minutes break. Recording stopped.