 Well, we stopped yesterday when I was discussing the Huygens experiment and trying to at least see why a certain kind of solution was more stable than the other. So I'd like to resume from that point. But before I go much further, I was yesterday's pace and I'm sort of comprehensible. Yeah? Okay. And if it was a too slow yesterday or it was okay, means I have to downward revise what I'm going to cover in this course because I thought it might have been a little too slow. But today's lecture is going to be really about trying to understand this experiment as quickly as possible and then going on to generalizations of it in particular very important model called the Kuramoto model. So the experiment itself, as we saw, was just these two pendulums on a common beam. And this Lagrangian can be written down essentially by inspection and there is one unfamiliar time. You see, this is just the kinetic energy of the entire system. This is the kinetic energy of the two pendulums themselves. This is the potential energy of the pendula, the potential energy, so to speak of this system which is, you can think of it as being bound. And this is an unfamiliar term because this actually captures the interaction of the base with the pendulums. And it involves, it's a torque of some kind. But I mean, you'll see this in the derived in the paper, so I'm not going to actually spend too much time on it. From this equation, if you wrote down the equations from this Lagrangian, if you wrote down the equations of motion for the phi's and the x's and the x, then you get the following two, this is actually three equations because k goes from 1 to 2 and x is over here. And there are two extra terms, two types of terms that have been added. One is the damping term, just linear damping and a forcing term over here where the pendulums will stop if they are not given some energy. And in a actual pendulum that works like a clock, which is the one which Huygens invented, this extra forcing term comes by weights that have to, this whole thing is driven by gravity. In the sense that there are weights, has anyone seen what used to be called grandfather clocks in my time and probably great, great grandfather clocks in your times? But if you've seen them, you know that they are actually housed in very big containers, cupboards maybe. And there is a weight which gradually keeps falling and that gives energy to the system. So this is just a phenomenological forcing term. You could put it down as cosine omega t if you like, whatever, something to give energy to the system. And this is the damping of the base for want of a better world, right? Now if you were to extend this to the case of n pendulums on this thing, it's just a very straightforward way in which you can do that. I introduce it here only because partly it's not been studied that well. The case of n pendulums hanging on a common support, although I will present some results of some numerical simulations in a minute or two. Now there is some tedious algebra to be done, but once you do that, you find that these are the equations of motion that will come out. The tedious algebra is just, oops, so the tedious algebra is going from there to essentially this without these extra terms. And then there is a little more stuff to do. You change variables and eventually come to a dimensionless set of equations, one for y, which is the same as x, rescaled, so it becomes, you know, l is the length of the pendulum and x is this other distance, so y is dimensionless variable, tau is also now dimensionless time and instead of the double derivative there in time, you go to the double derivative in tau. And having done that you get these two coupled equations, which one would then like to analyze. So I just thought I will do this little part of it on the board so that you can see what exactly happens. So you have these two equations for, what is it, phi i double prime plus 2 gamma phi prime plus sin phi i is equal to y double prime. Now these are actually two equations, one for phi 1 and the other for phi 2. So let me just go to a new set of variables, delta is phi 1 minus phi 2 and sigma is phi 1 plus phi 2. So if I take these two new variables, I should get two new equations from there and for the moment I am just ignoring the equation for y, although because that does not change much. So now I will just write phi 1 plus, which one, phi 1 minus, phi 1 double prime minus phi 2 double prime will just give me delta double prime and then I will get 2 gamma delta plus delta is equal to 0 because everything cancels otherwise. And sigma double prime plus 2 gamma sigma prime plus sigma is equal to minus 2 and finally y double prime plus 2 gamma y prime. So I am sorry, I have just recopied the first two equations, the last two equations over there, then taken the sums and the differences, yeah, okay. I mean there is a lot more analysis one can do, but let me just tell you what is the main thing I want you to notice. The equation for delta is now independent of the motion of y and y is the motion of the entire frame, yeah. So what one notices over here is that this is just a damped oscillator because this is the, oh, I am sorry, there should be a prime over there, okay. So you have got a damped oscillator, an equation for a damped oscillation for the difference variable, right. And therefore the difference between, you see, what is the solutions that we are looking for? We've seen in the experiment, in the metronome experiment we saw that the surviving variable was, when everything was moving together, right. So when everything is moving together, phi 1 minus phi 2 goes to 0 and this survives. In the pendulum equation, observations tell you that it is the anti-phase motion that survives. The anti-phase motion says that phi 1 is equal to minus of phi 2, so this survives and this goes away. So here we notice that the equation for delta doesn't involve y, so we can solve it. In particular, if you just solved it for small gamma, gamma equals 0, we can see that delta double prime plus delta is equal to 0 or delta goes as cosine top, plus some phase or whatever, right. Now if, okay, now the equation for, so delta never dies, it just keeps oscillating. And similarly, if I now put gamma equals to 0 over here and put in the solutions everywhere and so on and so forth, then I'll get some, I'll get the following equation for, so this now is just an equation for sigma prime and it involves y. So because it involves y, I can actually solve this and show that sigma, this will go to 0, this gets damped as a function of time, all right. What I'm trying to tell you is that the algebra is pretty straightforward. You can see that this will not, this will oscillate about a constant value, this will go to 0. And if this goes to 0, what means basically is phi 1 is equal to minus phi 2, yeah. I'm skipping over a lot of algebra over here only because it's, I don't know, took me several pages, yeah. Gamma is a phenomenological damping for, small gamma is the phenomenological damping for the pendulums, I mean, that's just the, right. You see, small gamma I'm saying is equal to 0, big gamma is not, yeah. See, it's because I don't actually have to, all right. In the sense that if I have damping, everything goes to 0 and so on and so forth, okay. So it's just that I don't have, I'm doing this without a model for energy input into the system. So I'm just appealing to the fact that the base is heavy, all right. And the, and its interaction with whatever on what it's resting, that is going to be something that I cannot tune out. But I'm saying, consider small gamma is 0, effectively, all right. If it is effectively 0, then, then you can see automatically that that doesn't die, okay, all right, yeah. Everything is small angle regime, then the 0 damping for the, or small damping for the pendulums, and some finite damping for the base, okay, yeah. This one, yeah, I don't have it in my notes, but it might well be there, okay. Find the missing factor to, it's okay. See, the point is that I try to be super careful about the algebra, but I may not have been, leave that out. The argument is really independent of any of these particular details. All that I have is that this quantity will, it can be shown, let's say, will go to 0, and I get this solution. The actual point I want to get to is the following. If I now made a plot of phi 1 minus phi 2, what is it that I expect to see if there are two pendulums which are not coupled? We know that one pendulum is phi 1 versus phi 1 dot is a circle, right? It's an oscillator, it's just oscillating, it's a circle, yeah? The other is also another circle, right? And in some, it could be some, well, not in the periodic case, of course. When they are uncoupled, it's going to be a circle of this kind, all right? Finally, however, we see that this solution begins to predominate as a function of time, because the damping will just push it onto this solution. So effectively, what damping does and coupling in all these various things, it just sort of moves all this around as energy flows. But finally, it comes on to this anti-diagonal, yeah? So at initial time, let's say they are uncoupled, and I gradually switch on coupling, all right? Supposing I do that. Then what I will see initially is that in phi 1 versus phi 2 space, I have a circle. Then eventually, this will transition at long times to, this captures one important idea in synchronization. This motion, of course, it's one dimensional because it's on a circle, but it's occupying a certain part of the plane, phi 1, phi 2 plane, yeah? When these objects synchronize, then we find that it goes on to this line. And this line is what we will learn to call the synchronization manifold. So this is an important part of the whole business of synchronization, that when two systems are in synchrony, the motion eventually occurs on a lower dimensional subspace of the full configuration space. See, the configuration space over here is whatever it is, right? I mean, in this particular case, it is just the, it's a torus. Phi 1 is a circle, phi 2 is a circle, and so on, right? When it synchronizes, it goes on to this manifold. In the case of the metronomes, again, the initial condition was that. But eventually, they go on to that synchronization manifold, because that in this case, phi 1 is equal to phi 2. And so if you do the analysis over there, you will see that delta goes to zero, and sigma stays non-zero. Yeah, no, it's a Lisa-Juice figure. In general, it's a Lisa-Juice figure. I was just trying to avoid a little excess. You see, the thing is that it depends on the frequency, the rate of rotation of each of the circles. I'm just taking the case when both of them are the same, then you get one circle, all right? If they are not quite the same, then you find a curve which is, you all are familiar with Lisa-Juice figures? Must have seen that at some undergraduate thing, right? So, no? Maybe my probably pronounced it wrong, Lisa-Ju. Mateo, do we have access to textbooks electronically? Okay, it's probably worth just, I mean, if you all have access to a good plotting routine, what I'm saying in one word is the following, that A1 cosine phi 1t plus B1, and A2 cosine phi 2t plus B2 can take A1 and A2, I mean, just take, you know, pendulums with the same amplitude, and take any values you like for these quantities, and then you will, if you just plot one versus the other, you will find that they fill out some space, but not exactly like a circle, all right? And these are curves that go around and around and around, and they fill the, all right? In the case where the frequencies are essentially the same, then it'll be a curve that looks like a circle. My main motivation in sort of at least introducing you to this analysis, and I'm going to give you the references where you can see the rest of it, is basically to have this important idea that when two systems go in anti-phase, in, sorry, in phase synchrony, then your synchronization manifold is a straight line, the diagonal, whereas if they are in anti-phase, then it goes like that. There can be more complicated relations and we will study those in one of these lectures during this week, okay? Now, the case of metronome, yeah? In the case of the pendulums, almost all initial conditions, if the base is sufficiently heavy, will give rise to anti-phase, all right? But I mean, in general, you're right, both solutions are possible. I'm going to show you an example just now, all right? Okay, I've chosen not to sit down and write the Lagrangian for this case. And all is exactly the same kind of methodology one uses, because these are mechanical systems. So you start with a Lagrangian, you then go ahead and try to get some equations of motion for phi one and phi two, where phi one and phi two are now the angles are, see for the metronome, the base is so. And there's a point of pivot and the mass is there. I mean, it's a little messier because the center of mass is not obvious, and all that stuff. It's discussed at great length in this paper by Panta Leone, all right? And the angles are measured in with reflection, okay? So I mean, meaning that it's done carefully, but the analysis essentially is the same. Again, you look for difference and some variables, and there you show that the other one, the difference variable goes to zero. So one important thing that you observed in the demonstration that we saw yesterday was that for a little while it's looked as if it was going to the solution of this, sorry, of this one going to zero, right? Because it looked as if they were all going in phase with each other and that was a little unstable. And then it eventually went out to be all of them in phase, right? So we'll see it again if absolutely necessary. But I'd just like you to remember yesterday that they started off randomly. It looked for a while as if they were going anti-phase. Then that solution just again sort of started and moved around and then it came into the in phase, right? So the synchronized, so the whole process of synchronization requires two things. One is that you go on to a sub-manifold. See, this itself is this configuration space is a manifold of some kind. Your motion goes on to a sub-manifold of this manifold, okay? Which has got, at least in this particular case, one lower dimension where it can be much less. So when you go on to this sub-manifold, that sub-manifold has got to be stable. What that means is that any small perturbations from this sub-manifold should bring the motion back because there is damping. It should bring the motion back on to the sub-manifold, okay? So in addition to confining the motion on to a lower dimensional space, we need that space to be stable, all right? And then tomorrow and day after when we discuss this whole business of stability and chaos and so on and so forth, this is usually put in the language that the transverse Lyapunov exponents, that is the Lyapunov exponents in the direction which are orthogonal to the subspace should always be negative, okay? So that the whole system is stable, okay? But I'll come back to that, all right? Now the question of, okay, so we understand the following, namely that when you have two pendulums of almost the same length, right? Where the natural frequencies are omega one is roughly omega two. This goes to the case of slightly altered frequencies which are exactly identical. Now, increasingly, we have to consider systems where not just two pendulums or two oscillators are interacting, but maybe there are many of them. And where this number is actually quite large, right? One complex system which is interesting to study and all of us do that is you look at dynamics, let's say, within a cell, right? As you know, and I know that you have many lectures on biological systems. Every cell has got a very large number of oscillators in it. Namely, stuff that is periodically expressed. And there are many, many regulatory elements in a cell, all the biology depends so much on timing that each cell can be thought of consisting of a very large number like 10,000 or so oscillators. So one very simple example of, and these oscillators have all got intrinsically different frequencies, right? There are things that have to finish in a certain amount of time, certain biological processes or biochemical processes that finish in a certain amount of time, show you examples again. Or various things have to happen and the timing has to be absolutely perfect. Because the fundamental part of cellular dynamics, namely replication, right, is mitosis. It has to happen in a way which is completely concerted. One cell has to make enough material for two identical copies to then split. And it can't work if timing is off, right? So you've got a large number of interacting oscillators in a system like that. So a complete cartoon of that process is to consider this system that we did. Namely, you take a ring and you just put n pendulums on that ring. But let these pendulums have different lengths. So the intrinsic timescales of each of the pendulums are very different, right? So you've got, and this entire ring on which they are suspended. This is suspended to something by springs, okay? I mean, what I'm trying to do over here is to think of a large number of oscillators that are coupled to each other and are externally driven so that they don't, this ring is oscillating at some level. And all these have random lengths. So the Lagrangian looks like this. This is the kinetic energy of the pendula themselves. This is the potential energy of the pendula. Why represents the motion of the ring? It's kinetic energy and it's potential energy. And this term over here, you've noticed that the coupling of the pendula to the base in the Huygens case over here is just now represented by a phi dot y dot, okay? Similar kind of interaction. Now this, after you go ahead and adding damping forces and damping terms, forcing terms, etc., we just get a set of couple second order differential equations. There are many of them. This is not going to be easy to solve because I'm considering intrinsically that the pendulums have all got different lengths and there is some distribution. So before I show you what actually happens, any guess of what you think might happen? Everything goes into one grand oscillation? No, I mean, you see the complexity and we're trying to build up some intuition. Yeah, no, just so that we don't have to worry about them. They're only swinging outward radially. Yeah, yeah, yeah, normal to the circumference, if you like. Yeah, yeah, the circular structure goes only up and down, all right? There is some simplifying up. No, I just want to see, when you've got two of them, it's straightforward. Either they synchronize either for omega one equals omega two or if it's close to some rational number, it's going to synchronize to that, depending on the strength of the coupling. What do you think might happen over here? Well, it turns out that the answer is, I mean, I wouldn't be presenting it over here if there was no. But what I want you to sort of try to get out of this is that when a large number of dissimilar objects interact with one another, all right, is very different from when a large number of similar objects interact with one another. So having a homogeneous ensemble where every element is identical to every other element is quite different from having a heterogeneous ensemble where every element is a little different and there's a range of differences, okay? So this is what happens if I take all the lengths to be identical, start them off at random initial positions, right? And one notices over here that you're asking whether some initial conditions will push you to in phase or some will push you to anti-phase. What happens in this system is that the entire set of interventula separate into two groups, okay? There are some which are anti-phase with, okay? So there are some which are all in phase with each other. The other group is also all in phase with each other in that group. But these two groups are anti-phase with respect to one another. Is it clear what's happening over here? When you've got a large number of these oscillators, here I've taken them all to be almost identical, adjusted parameters, just a moment. And so you find some of them go into, basically you just separate into two clusters. One cluster is oscillating in phase, all the elements are in phase there, all the other elements are in phase there. And you've just got a lower dimensional. See, now there are only two kinds of oscillators. There are either oscillators that are in one group or the other, and they are in phase. Yeah, in the sense what one, three, five, seven, no. Because it see all this depends a lot on randomness. What I'm trying to do over here at least is to systematically build up what's likely to happen, right? So random initial conditions, random lengths, all the masses are taken to be the same just because, yeah, okay. So the intuition I would like, yeah, no. There are different numbers in this and in that, okay? Because it depends on initial conditions, all right? Yeah, I'm pretty sure they could, this is sort of this is the result of one simulation for a certain length of time, then you'll find, okay, what happens in the case when all of them are random is a little more interesting. I mean, this is interesting itself, but actually you can see it on this curve that they are changing from one group to another because it takes a little while for them to settle in over here and you can see that the oscillations are not exactly in phase over there. You know, it's a little sort of ragged at the top. So that's probably because things are moving from one group to the other, okay? What happens when you've got just some, you take the initial frequencies, the lengths which gives rise to the frequencies, you take them at random from some distribution, all right? In this particular case, I think we just did some uniform distribution and just picked out some, okay? What happens in this case is all the oscillators whose intrinsic frequencies are very close to one another or the intrinsic lengths are very close to one another, they all synchronize. Just one second. And so a very natural thing that happens over here is what's called dynamical clustering, that is the entire system separates into different groups that are all in sync with each other. Not necessarily exactly in sync, as I show you because clustering is a complicated methodology. But basically what you find is that if I start with arbitrary lengths and then asymptotically I try to find out what is the amplitude and what is the time period, then you get this kind of a distribution and you can make clusters that look like so. So let me show you what a clustering looks like. What you find is that the dynamics can be, when I say clustering, are people familiar with the process of identifying cluster, classification, clusters and classification and so on? Yeah, all right. So there is a method called hierarchical clustering, we just go ahead and apply that. And you find what are the pendulah that are in one group and what are the pendulah in another group? So what you see below most clearly, right? Is that there is, so these are all the short pendulah which have got, which are all oscillating together. These are all the long pendulah that are roughly oscillating together, right? And this is not just all of them haven't split into two, okay? There are many, many clusters, many different groups of pendulah. I'm just showing you the ones with the largest components, all right? So you can see what it means for this kind of clustering to happen, right? See, the blue dynamics doesn't necessarily look as if they're all in sync, but they are in some level of approximation, yeah? Absolutely, so this is the minimum and the maximum. Cluster on the frequency or on the period or what? I mean, you can go ahead and do all the, see, in this particular case over here, I think we clustered on the period, right? Which is why you can see that they're all, I mean, the amplitudes are a little different, but it's, big part? Yeah, I'm sorry, I didn't, yeah, my silhouette, because I mean, clustering is, as if you've studied it, you know that it's a highly subjective. It's not even mathematics, it's an art form. Yeah, no, no, no, no, I don't mean to be facetious. See, a lot of, the number of clusters is your choice. So what's called k-means or k-medoids or some sort, I think we use k-medoids. So the number of clusters depends on the scale at which you want to differentiate things, okay? The only point I want to make over here is that if you take a large number of interacting oscillators with intrinsically different frequencies as coming from the length, then it seems that very naturally, it will separate out, depending on the coupling, all right? Into a smaller number of groups, and this smaller number of groups essentially is the analogy of synchronization, okay? That is, within each group, there is synchrony. The two groups may not be in synchrony with each other, okay? When, yeah, not really, because this depends on, these are all simulations of random ensembles, right? What we noticed was that if you start with n pendula, right? The number of groups that you get, let's say, is some m. But m is always much, much smaller than n, okay? Part of the motivation for this came from, all right, so one has done a number of studies of this. If you increase the mass of the beam, and you take, you know, so you can ask questions like what happens if the beam is very heavy? What happens if the lengths are very small or in a narrow distribution and so on and so forth? Okay, so one can do all that. The point is that, let me just go and show you the initial motivation for this. This is a nice study from Said Tavazoi and George Church at Harvard, where they looked at the time evolution of all the genes in yeast. So over a cell cycle, they were sampling the mRNA and trying to see how many, what is the population of different mRNAs in yeast? And this technology that can do that, so let me not worry about it. And what they saw was that the data actually separated out into, so the number of genes in yeast is about 6,000, okay? So, and these have all got to be expressed in some fashion over a cell cycle. So this cell cycle time is the period of overall replication. And so there are many, many different, oops, yeah, okay? So here is one, here is one pattern that emerges over a large number of different genes, okay? And they all, I mean, so this is the average is plotted out over here. And this is your variance in the spread. Again, the same process as I described for the pendulum was done. Namely, you looked at the time sequence, you look at the clustering of it. And you find one group has something like this. There's another group which has this kind of a time pattern and so on and so forth, okay? Now the important part over here is the biologically relevant part is that out of the 6,000 plus genes that are there, they only separate into some 30 odd, okay? So in our model, this is 6,000 and that would be 30, 30 different groups, all right? And again, they're not all identical. They don't lie on top of one another, but there is a spread and that is pretty much, or at least I like to think that it's pretty much like this. There are some groups that are very tight together. There are some groups that are somewhat looser together, all right? And because it's a biological system, nothing is exactly periodic the way it should be. So you find, there's more, those are more chaotic in some sense, yeah? I'm giving you an experiment where there's a large number of naturally oscillatory systems, right? Now I'm not saying that an oscillator is an mRNA thing. I'm just saying it's something that, the oscillator gives me an example of what I can play with numerically, analytically. In this case, I have no hope. At least right now, I have no hope of being able to do a simulation of this kind, all right? But I'm just suggesting that when you've got a large number of interacting oscillators, they separate out into groups that will have the ones that are together separate out into a naturally correlated set of oscillators. These are also oscillating, these genes are also oscillating in time, right? So the same, I'm just just a suggestion that the pendulum system is like a model for understanding what happens when a large number of these dynamical systems are interacting, okay? And this kind of oscillations in biology are actually very, very common, all right? And when you have to calculate them, you use things like K-means clustering or whatever and you see how good the clustering is by what's called a silhouette and so on, but this is not, okay? So I want to stop on pendulums of any kind for the moment and go into a very important model of large number of oscillators that are interacting with one another. It's the same story, meaning that our interest in when we look at complex systems is really in the large number game, all right? Then the Kuramoto model was one of the first models to be sort of solved analytically in the 70s. And I found a slightly nicer picture of the Firefly's nature. So I don't have a movie of this one. The picture was just so nice that I thought that you come back and show it to you, all right? Okay, so this phenomenon actually inspired Winfrey to come up with a model which then eventually led over the years to Kuramoto simplifying everything and solving it in a way which we'd like to discuss now, yeah, all right? So we're trying to understand the collective behavior of populations of oscillators, and the model is actually a very simple one. The, consider an oscillator now just to, there is no variation in amplitude. So these are just oscillators that are moving around on a circle. That is the phase, you know, so, all right. So you replace an oscillator now by just its phase, okay? And this phase goes from 0 to 2 pi at some rate, and that rate is omega, okay? So the angular frequency, if you like, is omega. And so an i goes from 1 to n tells you you've got n oscillators that you are considering, okay? Now, Kuramoto took this coupling, which is just sinusoidal, right? And the reason for that is that the omega, this group of oscillators that he is looking at, the intrinsic frequency is drawn from some distribution g of omega, okay? And in the model that Kuramoto solved exactly, g of omega was taken to be a Lorentzian function. So you just take some frequency, distribution, and it's sort of basically something that looks, every possible variation of this has been studied, so, you know, the phenomena pretty robust. With the Lorentzian function, its solutions come out nice, all right? Then you have the coupling, strength is k, and n is like a normalization, so, all right? And the coupling itself is just a non-linear diffusive coupling term. And because it's trigonometric, many things get solved nicely, all right? Like I said, you can take any two pi periodic function for the coupling, yeah? I said you can take any two pi periodic function for the coupling, and the formalism and methodology will go through. So, this isn't sort of an experiment of what happens when you have zero, not zero, but very, very small coupling. And what you find is that the oscillators are essentially independent. All right? You increase the coupling, I'm not sure that the notation is the same, but I'm just saying this is something which is one by n, this is six by n, and that's 12 by n. So this is increasing the coupling. Now once you increase the coupling, the phase, the phi's don't go, or the thetas, whatever, don't become identical, but they all now are moving at the same common velocity, all right? So, and in between you have a situation where some of them are coupled to one another. Or rather, some of them are locked, phase locked with one another, and the others are moving around randomly. So, basically what you have is a situation where Omega is now both positive and negative, so you've got Omega positive meaning that they are moving in this direction, Omega negative meaning that they're moving in that direction. As you switch on the coupling, some of the ones that are moving in one direction, depending on whichever one is going to take over, sort of it drags the same phenomena that we had of things dragging each other to a common angle of velocity, all right? Now, what Coromoto did was to introduce the following order parameter. And I mean, this was actually a crucially brilliant move that he made. Was to just look at the, look at the sum of the exponentials of these phases, all right? And he defined this to be R e i psi. And one can see that if all the phases are identical, then R is just equal to one. And if all the phases are random, that sum goes to zero, and R is zero. So, this order parameter is able to detect the overall amount of order in the system, whether all these oscillators are in phase or in outer phase. For the situation where you are neither zero nor one, what R will measure is the degree of coherence, right? And what psi will tell you is what is the average phase, okay? So, R measures the phase coherence and psi is the average phase. And this is now used in so many systems that any expression that looks like this is called a Coromoto order parameter, all right? Not quite, not quite. Although it's an interesting question, I'm not sure. I think that when the distribution of omega is unimodal, it probably doesn't. But if it is bimodal or, see, the Lorentzian for which things were solved is basically something that looks like that, right? But you can actually solve this, and I was originally thinking of doing it, but it's a little more complicated if you take a bimodal distribution, all right? And as we will see later on today, things are very different if you take a uniform distribution, right? Because what figures in the solution, if this is called, see, g of omega is this distribution from which things are drawn. The solution will actually involve g double prime at zero, okay? So when, it requires differentiability and so on and so forth. But for convenience, okay? Because you can solve it well in that particular system. But any kind of coupling between phase variables has to be two pi periodic. And a simple one is sine. It works, I mean this entire formalism works for different kinds of coupling also if you want, yeah? If the argument is small, it is just linear and it's theta one minus theta two. See, there are situations, some initial condition. The r and psi make sense only in the infinite time limit. And there, they are assumed to be essentially constant. Initially, they will be changing and fluctuating to groups that are moving out, but that is not a long term solution, right? You, as if you, yes, yes, you're right. This is, that is the partial ordering is very much like one phase that is moving together and another phase that is running around completely uncorrelated. And we'll see the implication of that also, yeah? Yeah, yeah. It's actually a second order phase transition, all right? But that's on the next slide. You had a little bet whether it was first or continuous. Well, we see that. So that's the, it was coming on a slide and we are sharing the, how much did you bet? Okay, all right. So this is a model which is known to exhibit a second order phase transition and you, this is from a numerical simulation. So you find that as, you know, you start out as you increase the coupling strength, you are essentially in an incoherent state. There is no coherence. And then at large coupling, you go almost to complete coherence, which is one. And at a critical value of K, this seconds, you know, there are two solutions. There's always the possibility of having an incoherent state. But you will then find that you go at a value Kc to the coherent state. So KuroMoto's solution goes like this. Is there a place where you hide the chalk material? Because, you know, every piece has now gone to the epsilon limit. So this is our KuroMoto model. And you've got Re to the i of psi is equal to 1 over n. Just the sum of, so that's our, let me just remind you. Using standard trigonometry, you can show that this is just equal to omega i plus Kr into sin. You can see roughly how that comes. Write this in exponential form. Use the summation. And I mean, even the colored ones are gone over here, so okay. So you can see roughly what's happening over here. Now, theta i will try to keep aligning itself to the average psi. Because, when theta i is equal to psi i, then exactly this term will go away. And you just have that theta i is equal to omega i, right? I mean, I'm just trying to sort of re-explain what the psi being the average is. And effectively, this is the same KuroMoto equation. And the coupling now is it's coupled to the average of the entire set of oscillators with the coupling is mean field. Every oscillator is connected to every other oscillator. So, this is a globally coupled system. Now, I can ask the following that basically, r e to the i psi in the limit of large numbers of oscillators. In the large n limit, omega is taken from some distribution g of omega. So, I can rewrite this equation by changing the sum to an integral going from minus pi to pi of d theta, because the theta is r from minus pi to pi. And j equals 1 to n delta of theta minus theta j. So, I am going to use this delta function to represent that, changing the sum to an integral. And this I can now rewrite as the integral from minus pi to pi over d theta. And minus infinity to infinity d omega g of omega times some distribution function theta omega. So, I am going to consider where all these oscillators are drawn from some distribution and they have a density function rho. And this function rho is such that, when I integrate over minus infinity to infinity of rho of theta omega t d omega that is equal to 1. I mean it is just a normalized distribution function. And I am saying that this is equal to r e to the i psi. So, what we are going to try to do is to figure out what this distribution is and solve it in the way in which we can understand why there is a phase transition at all. Now, as k goes to 0 supposing the coupling is non-existent, each of the theta i's is just equal to omega i times t. So, since the theta i's are just equal to the omega i times t, then what I can do is to replace this quantity by e to the i omega t. And I have already gone through this other part here minus pi to pi minus infinity to infinity g of omega times e to the i omega t rho of theta omega e theta d. Basically, I am using the following feature that when this is equal to 0, then theta i is just omega i and theta itself is just going to be omega times t. And then I am going to use the Riemann-Lebesgue lemma to basically say that this is going to go to 0, k is equal to 0 and I can make this approximation where this is just e to the i omega t. This is oscillating so fast that all the integral just goes to 0. In other words, as k goes to 0, r goes to 0. First what I want to do is to say that when k is equal to 0, this is just the simple linear equation. So, I can replace theta by omega t plus some initial phase which is unimportant. Then put it into this equation and regardless of this function, just e to the i omega t integral for all ranges of omega, that should vanish. e to the i omega t, you see I am sorry I rub that out but I will just remind you that this I wrote as this double integral d theta d omega e to the i theta rho of theta omega t g of omega. Which one? Oh, thank you. Yes. No, no, no, this is oscillating. As a function of omega, this will oscillate. And omega is taken from the distribution that goes from plus to minus. It's a Fourier, I mean this is basically a Fourier transform and in the theory of Fourier transforms, if you go and look up Riemann-Lebesgue, it will tell you that this integral goes to 0. No, thanks for alerting me to that. That's how you just replace it by e to the i omega. Now as k goes to infinity, so in the large k limit, what's going to happen is that every oscillator is going to align with itself. So as k goes to infinity, this will go to 1. So we know the two ends of this story as you like. It's 1 at very large coupling, 0 at very small coupling. But you can do a lot more, especially because now you can write an equation for rho. Now going to the theory of the Fokker-Planck equation, what we will do is to write a continuity equation for rho. And the continuity equation, so let me just use the colored chalk to remind you that theta i dot is equal to, so this is the drift velocity if you like and this drift velocity is just equal to omega i plus k r psi minus. So let me, I'm also going to make an approximation of this argument being small when they are aligning together psi minus theta i for all of them is actually small. So I may leave the sign out of this occasionally, so let me just keep that aside. If I have this density rho of theta omega and t, this satisfies the continuity equation partial of rho with respect to time plus partial with respect to theta of nu times rho less equal to 0. This is just the first two terms of the Fokker-Planck equation, there is no noise in the system. So I don't have the second derivative with respect to theta, I presume that it's been covered at some point. You are familiar with the Fokker-Planck equation? So omitting all the arguments of rho partial with respect to time plus partial with respect to theta of omega plus k r to psi minus theta. So this is the equation that we are going to try to solve and see when we can expect to have a stationary solution. If you have a stationary solution rho doesn't change and let me keep my sign over there because this is not, I mean I can consider the case of small variations later on but for the moment at least let me just keep that over there. Now given the fact that you have this normalization over here, there is one solution which will just always solve, I mean there is this normalization, there is also this other normalization which will be with respect to theta. So rho is equal to just uniform 1 by 2 pi is the solution and if rho is just 1 by 2 pi that tells you that r is equal to 0. So this uniform solution where rho is just distributed uniformly around the circle is an incoherent solution and that tells you that r is equal to 0 and that exists for this particular case. Now our basic idea is going to be the following. You want this argument, you want a stationary solution so in long time so d rho by dt has to be 0. And if d rho by dt is 0, one way to make that happen is to have this argument over here equal to 0, namely 0 is equal to omega plus k r sin of theta minus psi. So for some of these, you see this now is a very simple equation and solutions will exist only if omega by k r modulus is less than 1 because it is the sin. Remember omega is both positive and negative. So so long as the modulus of that is less than 1 then sin theta minus psi will have a solution. You can find some value when that is equal to 0. So this sort of gives you an idea that as you keep increasing k, more and more solutions can be found because for any omega this ratio has to be less than 1. So when k is this you cannot of course go to k equals 0 but for small k you are going to find very few omegas are synchronizable and that means that very few of the oscillators can come together. See what I am trying to get an intuition of is that as you keep increasing k more and more there is a larger range of theta that can synchronize. Some of it depends on the theta minus psi my notes are whimsical, so I changed the argument. So what we want is theta minus psi is between pi over 2 and minus pi over 2 and given a value of omega by k r there are going to be two solutions. The arc sign will have two solutions in this particular interval and that means that if this is equal to 0, I mean if the sign of this quantity is equal to 0 this basically means that the equation theta dot i is equal to omega plus k r psi minus theta that is equal to 0. So there is this is what stationality implies right stationary point is a fixed point derivative is equal to 0. So that will exist only if this settles. So basically what I am trying to say is that rho is going to split into two sets. One set for which the omegas are much bigger than you see you want omega over k r right. So r can go between 0 and 1 we do not know what it is but this is of some quantity. Depending on k there is going to be a range of omegas that is just excluded from consideration because you are never going to find a solution which is part of the stationary set. So this describes the other oscillators that are moving around the circle almost uniformly. They are not entrained by this common phase I am sorry I am not I just want is you want a stationary distribution. In order to get that stationary distribution we realize that this condition has to be met. If k is very large everything is going to meet it right but for intermediate k what happens there are some omegas that are moving much faster. So omega is large compared to k and that means there is no solution right what is what is the behavior of this on an average it is just stuff moving around in a circle that is equivalent to this kind of case where it is a uniform distribution for the very fast movers. The ones that are smaller sorry the ones for which this condition is met smaller basically they will have the following distribution the theta's are just going to be given from that. So it is a delta function that theta minus psi and now I need to get minus sin inverse of omega by kr okay and I told you that there were two solutions that will come for this little algebra will show you that there is only one solution which is acceptable and that is the one for which cosine of theta is positive. So it is traditional to add a heavy side step function so rho for this particular case where a solution exists is just given as a sum of delta functions over here as you sweep through theta you just pick up each of these ones. The ones that are moving around right they have some constant distribution divided by the velocity so it is omega plus kr sin minus sorry it is a plus okay. So this describes the group of phase oscillators that have omega less than kr and modulus of omega and this is for modulus of omega greater than kr okay. So the invariant distribution solution that we find over here has these two natural parts what are we going to do with these? Yeah this one this distribution see you want rho dot to be equal to zero I mean once you have written down your continuity equation you do not want this distribution to change so you would like rho dot to be zero rho dot is equal to zero has only one possibility and that is the argument over here is equal to zero and this argument equal to zero gives us that we are going to find that okay let me just stop over here for today and I am going to come back to this at little before this point tomorrow but let me tell you what we are trying to do okay the first things we see that without doing any algebra one limit gives us r is equal to zero the other limit gives us r is equal to one yeah correct what we would like to know is how does r depend on k right this so we eventually want to get r as a function of k that is going to be our objective now to find r is a function of k I need to be able to at least go to some continuum limit so what I did was to say that instead of looking at each oscillator one to n n goes to infinity and I look at the distribution of these oscillators and this distribution rho is the distribution in theta omega and it is time dependent but if I want to look at some stable behavior I have to look at the situation where this distribution doesn't change in order to get an equation for a distribution that doesn't change I go to the theory of Fokker Planck equation right which basically is in some notation it is just large my equation and so on and so forth these are all related to one another just one second please let me right so what do I do I get this continuity equation there is no noise so the second term which would be d squared by d theta squared that doesn't exist the diffusion term doesn't exist now I want to know under what conditions am I going to get a stationary solution so that has to be zero which means this is zero which means that argument is zero there is one possibility there is another possibility of course where the derivative itself is equal to zero but I am not considering that okay so I just want it now if this has got to be equal to zero then this rewriting that gives me the condition on the sign yeah and this additional assertion that you have to only consider the case of cosine of theta positive is just very trivial stability analysis that you can do alright so I will come back so tomorrow what I will try to sort of wrap up the coromotor problem is as follows will start with this distribution right and then show under what circumstances you can evaluate it and this gives us the phase transition essentially it gives us a second order phase transition because the dependence of see as I said we are looking at the dependence of r as a function of k I am looking for r as a function of k I know that in zero over here and one over here but what we will see is that it will that the solution will be a square root solution that comes up at some value which is k sub c and this distribution goes as square root of k yeah okay so that I will try to be a little brief and a little better organized about the derivations tomorrow