 Okay, good morning and sometime today you should get a bibliography set of books and papers that would be useful for this week and I'd similarly do one for next week and hopefully you all will have it. Big one? It will depend on the powers that be, if you make some sacrifice to the gods maybe that will happen. I don't have any objection, right? See, but as you... Yeah, I mean I will make them available. I don't know if they have the... They're all on Keynote. I'll make PDFs and give them. Yeah. No, because Keynote, Powerpoint, et cetera, et cetera. Okay, alright. So today, I mean that's the only slide on display today. There's nothing else. I was talking with some people about volcanoes last night and there's some Indonesian volcano. What I actually want to do today is what I promised yesterday that we will try to just go through the algebra of the Kuramoto because A, I think it's one sort of very useful model that was invented. It's highly applicable in a variety of situations and there are some features of it which generalize easily to other systems which I would... I noticed that that was not... Okay, so one could start by writing a model quite generally like that where you have N oscillators. N of these phase oscillators and just to tell you how some of these generalizations can happen, one can think of having an adjacency matrix. So if oscillator I is influenced by oscillator J, then that quantity, you know, we can take K, I, J is equal to coupling between I and J. The model that Kuramoto solved was all K, I, J equals K. That is, they're all the same and they're all equal to one another. And this is an example of what today is called global coupling. When every oscillator is connected with every other oscillator, it's just all of them are equal to K. But clearly you can think of having a variety of different... You can put, for example, the Kuramoto model on networks and I know that several of you are working with networks. So I mean, if you put them on a network which is just linear where I and J are only... Where K, I, J is non-zero when I and J differ by one and zero otherwise, right? So that's just nearest neighbor coupling which is local coupling. You can also think in terms of having K, I, J depend on this difference I minus J. So that the coupling goes down or goes up depending on your whimsy as a difference in the indices. So I mean, it's up to you. Put it on a Bethe lattice, put it on a hypergraph, put it on whatever. And the behavior is different and interesting. And since Kuramoto described his model in 1975 or maybe even a little earlier, it's probably been studied. So these models have been explored in great detail. So we are going to consider the case, as I said, of all of them the same. And we have this order parameter R e to the r f psi is just summation 1 over n of e. And we can write this equation in the form of theta sub i is just omega i plus K r sine. So we'll deal with all forms of this. I hope that, I'm just recapping yesterday is going to be very little. Okay, so just to tell you that this is where we stopped yesterday. And what I'm going to try to do is to give you a better, I mean, I hope hopefully a better intuition of what's going on and why this problem is solvable also. Okay, so given this, we'd like to take the limit of n going to infinity. And then instead of having individual phase oscillators described with this index i, I will think in terms of having a density, which is a function of theta, omega and t. And theta goes from minus pi to pi. Omega goes from plus to minus infinity and is drawn from some distribution g omega. And all that I'm going to describe today is only for the following kind of distribution unimodal with one maximum. Now, clearly the integral of this from minus pi to pi d theta, that's just a normalization condition. And we've, okay, again as I pointed out yesterday, one simple solution for this is that rho is just equal to 1 by 2 pi, independent of omega and t and independent of theta as well. So that just tells you that with the uniform density, everything is just swirling around that circle. It's incoherent because when it's all swirling around the circle, r immediately goes to zero because it is just a large sum of random numbers over there. Why does the, I mean, you don't have to also integrate over omega because the omegas are given to you. They are drawn from some distribution. I mean, there will be a, there will be a normalization, which is rho of theta, omega t times g of omega, that is equal to 1. Yeah, absolutely. But, you know, the whole point about this exercise is that in the long, I mean, we're going to look for stationary solutions. So we want rho to be invariant in time. All right? So, I mean, there will be stuff moving around, but the whole density itself will be invariant, or we're going to try to find solutions like that. So, the r e to the i of psi, which is that sum, gets replaced by the integral pi e theta e to the i theta delta of theta minus the same as the integral from minus infinity to infinity d omega minus pi to pi d theta e to the i theta g of omega. At the end of the day, okay, the objective of this algebraic exercise is the following. We want to find r for arbitrary k, and in particular, we want to know what happens when k increases such that r, which initially is 0, for k is equal to 0, when there is no coupling whatsoever, theta is just equal to omega t. If theta is equal to omega t, then you can put down i omega t over here, and that integral, because of the fact that we also will be assuming, I should say that this is equal to, okay. So, g is symmetric about, the distribution is symmetric, positive and negative values. Take one? Sorry. Okay, so, if it's symmetric around that, then that's, I mean, we'll take all those into consideration. So, for k equals 0, r is 0, and the point is that if you start with some distribution, initially the r may not be 0, right? But if they are, as a function of time asymptotically, you will find that r will go to 0, okay? So, now the continuity equation that rho satisfies is partial rho with time plus partial with respect to theta. Okay, the number of oscillators is conserved. You've got this density that's moving around, and this is the equation that it, that rho has to satisfy. Now, for the stationary solution, we want this to be 0, so we would like this quantity to be 0, and that basically says that if the inner bracket over here, if this is equal to, sorry, if this is equal to 0, then this implies the following, that omega by kr modulus is equal to sin of psi minus theta, and clearly there are only real solutions for this theta and psi if omega by kr is less than 1, yeah? Unimodal, see, one maximum, no, on the width, no. I mean, eventually everything, you know, if you increase case large enough, you will be able to get everything down to some common psi, right? But, you know, for today at least, other than the fact that we want g to have a single maximum, right? I don't want to, there is no other restriction on g. Things like a flat distribution you'll see are difficult, or at least this doesn't naturally extend to the flat distribution, all right? But there is, I have given a paper there in the bibliography today by Martens where they look at a bimodal distribution, and then you will see what happens when your distributions are not as simple as just, okay? Today we will just look at the Lorentzian, where g is just equal to gamma over gamma squared plus omega squared, we'll just look at that, all right? So, I mean, I had said all this, yes, yes? No, no, no. And see, we want this entire thing to be zero. I mean, we want a, this is your continuity equation. Given that I want to have a stationary distribution in rho, one way of doing it is to have this quantity equal to zero. It's not the only solution, I'm just, I'm looking for the easy way through. Other solutions can be found in the bibliography that I've given you. No, I mean, see, you have to do a certain amount of work for it, and I want to largely focus on the essential ideas over here, all right? Okay, one can just imagine that there's some complicated rho, which when you take the derivative is all equal to zero, but it's not easy to find, okay? All right, so this condition tells us that there's a range of thetas that can be a solution. That solution will then give us a value for psi, and that is the common frequency, the common phase that these oscillators have. What, for example, what we saw yesterday was this, that in the coupling is, sorry, when the coupling is very weak, you've got stuff moving left and right pretty much without being influenced by each other in the extreme picture. For intermediate coupling, you find that many of them align onto a common phase, and that is just this condition that only when this quantity is less than one, those are those oscillators, the ones for which this is not true are going around in their own merry way, yeah? You increase the coupling, they also fall in line, okay? So, I mean, this is exactly the effect of increasing k, because as you keep increasing k, more and more of the omegas will come down. See, this is the, okay, this is basically this definition. All right? When I go to a large number, I have to choose, okay, the theta has come with it, you know, there is a density of oscillators. This density is rho, because the variables of this problem are theta, omega, you could, if you want, you know, you can get rid of all the distribution of thetas and go for something, you know, you can have maybe a uniform distribution in omega, and then you will not need, okay? Meaning, look, for the moment now, I'm just allowing myself to have random, initial random velocities, phase velocities, omega, and then I want to see how this entire population evolves, and obviously, you know, we need, I mean, for two, okay, maybe I will give this as a homework, just take two phase oscillators and see what happens, okay, because that you can do analytically. So, what is this? Given the fact that this solves, you know, the particular equation, let me call these as the locked solutions. So, this quantity is then, okay? So, we'd like one, this would be smaller than one, and then you've got that going. For the others, which are moving around the, they are moving around this particular circle, right, because they are not able to solve this equation, they set up a current which is just equal to this velocity, I'll just put down as a modulus, times rho of the unlocked, I'll just call them the drift solution. So, the point is that for arbitrary k, the oscillators separate into two groups, with different numbers of oscillators in each group. In one group, you've got all the ones that have somehow locked onto upside. In the other group, they are moving around, okay? So, instead of having a single rho that describes the entire lot, I will basically have two components, one which are moving around independently, sorry, in phase with each other, and another which are just moving around at random, okay? So, this helps me now to write down the overall density as a sum of two components. And this, the sum of two components is the following. Let me just write down rho is equal to either, okay? So, for one group, they all satisfy this particular condition, right? And for the other group, it is just c upon that quantity, okay? The drift part is just here. The reason I'm looking, I seem to, you know, always there's difference between psi minus phi and phi minus psi. I'm not being super careful over here, but hopefully it'll all come out in the end. Notice that the sine inverse has got two roots, and of the two roots, only the stable one has to be taken, all right? And that comes out as a heavyside function on cosine of theta. So, of the two roots that will come for any given value of omega and k, all right? You choose the one which has got the positive root. It's a simple stability analysis, which unfortunately I will only cover tomorrow, but I will, you'll see why. Anyway, so that's the point. So, here is, for, you know, for any given value of k and r, this is the solution that you get, yeah? Okay, what I'm pointing out are that there are two roots for the sine inverse. One of them is stable, one of them is unstable. It turns out that the one that is stable has cosine, which is positive. And I'm just asserting it, all right? So, given the fact that we now have an expression for rho, the expression for r is, as we have already seen, d omega d theta, e to the i, now it is theta minus psi g of omega rho of lot plus rho of drift, okay? What I've done is to take the e to the i psi over here, over onto this side. The rest of the expression is the same, and I'm just pointing out that these two, that these two are now, it's broken into the parts which are the drifting and the locked part. I wrote it down a little differently yesterday if I wrote it down at all, I'm sorry, a little change, okay? Now, the reason we consider these two separately is that this integral, which is d omega d theta e to the i theta minus psi g of omega rho drift, this rho drift is just a constant divided by omega plus k r psi and theta under modulus, right? So this is just the part of r that comes from the drifting ones. And you can see that supposing all the oscillators were just drifting positive, negative, r would be 0. That was the first thing that we started by considering k was equal to 0 and the actual integral r, r turned out to be a good order parameter because it sensed that, right? When r is equal to 0, all the oscillators are uncorrelated and moving around by themselves. The same is true over here, that this part just goes to 0. And here you can see it by the following fact as well, that g of omega is equal to, okay, so there are these symmetries, g of omega is equal to g of minus omega and rho of theta plus pi minus omega is equal to rho of theta omega. One is the distribution coming from the locked ones, the purple ones. Yeah, so the delta. The other is the part coming from the drifting ones, the green circles. If you have essentially independent oscillators moving around in a circle, their contribution to the real part of the order parameter is 0, okay, because they are incoherent, all right? And this symmetry when you apply it onto this will just tell you that it cancels out, okay? So this integral just goes to 0 and we are left with the second integral and the second integral is just this delta function, so let me put that down. So this is our contribution from the per k symmetry, symmetry. You know, yeah, r is 0 because they are uncorrelated, they are moving around by themselves, yeah? I am doing it for that case, all right? I mean, remember this is, I am describing Kuramoto's work 50 years later, all right? So, yeah. Okay, now this part is actually much simpler because we know what to do with delta functions and this entire integral, therefore, just becomes an integral of G of omega, d omega, okay? So what I am going to do over here is to do the integral over theta, all right? And replace theta minus psi over here by sine inverse of omega by kr, all right? So e to the i sine inverse omega by kr. So this, the delta function evaluated just gives us this and now I won't, okay? And remember that r is real, r that is coming from the locked part is real. So let me just take the real part of this integral and rewrite that as cosine of omega by kr. Oh, cosine of, that's the real part of that exponential, okay? And now it's just standard doing some algebra, asset omega by kr is equal to theta and this, therefore, becomes d omega becomes kr times d theta, et cetera, et cetera. So this will just give me the integral from minus over omega less than kr of kr d theta, that's the d omega, g of omega but omega is just kr sine theta. Just one second, let me, omega by kr I am putting down is equal to theta. So of course, my mistake, I am sorry, I have that that I want, I want this to be equal to sine theta, right? I want the argument of the cosine, time out, okay? So this is my integral, we've got that right, okay? Now I want to change variables to sine inverse omega by kr is equal to theta or omega by kr is sine theta or omega is that quantity, right? And d omega, therefore, is just cosine theta times kr times d theta. They do, but I mean, that's right now not the most important part, okay? Okay, I'm almost there to get you the critical case, so that's part of the, okay? So now I've got, yeah, it's not that theta which we, it's a new variable, I can call it del, no, not delta, call it something else. Z, no, not Z, Z is complex. Language is so restrictive. This is an integration, integration variable that we're going to be taking, okay? So this integral now just becomes from minus pi by 2 to pi by 2 of kr g of now omega and omega is this quantity. So g of kr sine theta, right? Times cosine squared. See, there's one cosine theta coming from here by definition and one coming from the derivative. So cosine squared of theta d theta, okay? So this is now our expression for the order parameter, and I'm just going to rub this part out because this is r is equal to that. This is one solution that you can already see which is that r is equal to 0 that comes out of this because you've got r on this side and r on this side and r over here. So let me be selective about just dividing out this r that one is equal to that quantity. This is a k constant, so I can bring that out. And now I have that expression to evaluate, all right? Now the point is that the, as we've seen in this picture, as k keeps increasing, the order parameter is actually 0 for a while. And it's only above this critical k, k sub c that you get a solution that comes out of the r is equal to 0 solution. So at that particular point we must have this, okay? At that point we will have this solution that one is equal to k c integral over d theta g of when r is 0 at that particular point as it's coming out. So g of 0 cosine squared of theta. This is just an argument over here that, I mean, we know this, let's just say this is a fact that we know experimentally that r is 0 when you start out. We know that r goes to 1 as k goes to infinity. We're looking for the transition. I mean, in a sense, I'm just, I'm trying to, in a sense what is happening is that there are solutions. We've seen r is equal to 0 is always a solution, okay? From that. Since r is equal to 0 is a solution, we would like to see our new solution rising out of it. It's essentially bifurcating. The new solution is coming out of the r is equal to, out of the incoherent one. So one way to estimate when this will happen is to just ask, when does this integral give us 1 if I start out as r is equal to 0. So this is the integral that I now need to evaluate. And see, so 1 is equal to k sub c g of 0 integral of d theta cosine squared of theta. And this quantity is an easy solution which then gives us that k sub c or the critical k is just 2 over pi of y not g of 1. See, if I have r is equal to 1, that is that line over there. This solution doesn't, I mean, you can't get the onset of this particular curve from this part. Okay, so let me just draw your attention to what are the important points over here. The first is that the details of this distribution as far as determining the critical k, they are important only so far as the value at 0, which we have taken to be the position of the maximum. So the value of the distribution at its maximum is the only thing that really matters. Okay, and the final expression is quite simple. Okay, and now my problem is that I don't actually remember, as you've seen this is a very sort of schematic. I wish I remembered what the model was taken and so on, but it is basically Lorentzian. I forgot the parameters of the Lorentzian. So here if you look at what is g of 0, g of 0 is just 1 over pi gamma. So kc for this Lorentzian model is just going to be 2 times gamma. Yeah, see the coherent ones don't necessarily have k equals 1. It's only when r is equal to 1. It's only when everything is in coherence that you will have 1. But what is true is that r is equal to 0 is always going to be a solution. Yeah, and just formally this is the way of doing it. Namely, I'm asking for solutions of this equation, which in a sense, my only handle over here is r. r is equal to 0 is the solution. So I'm asking, I mean obviously I can't put it in the earlier equations because we had r is equal to, I don't know what was over here, maybe there was an r as well. Now if I put r is equal to 0 over here, there's nothing. So I'm just choosing to put that out over here. I should have a better mathematical argument for doing it and I will try to do that at a point, but I'm just trying to give you an intuition. At the transition, r is equal to 0. At the transition, that's when it's going to be moving up. So I'm just trying to figure out the point where it moves up in terms of k by solving this equation. The circle is just trying to give an idea of what does r mean. Here when you've got k much below kc, all the phases are distributed around the circle and the value of r is some small number. Yeah, yeah, this is this r with this funny arrow. The arrow is just the length of r. Over here, when all of them are together, the same arrow is just grown. That theta is not this theta. I'm going to be wrapping myself into knots over here, but I'm looking for a solution of theta. I'm not looking for theta equals 0 over there. I'm looking for kr times sine of this is equal to 0. Let's discuss the mathematical nicety of it. I do understand that there is a need to give you a better feeling of why we are looking for this bifurcation out of this. r is equal to 0 solution and we need to come out of there. For that, we just determine the validity of this equation. This equation gives us the integral is just that, kc times g0 times whatever and then that's just pi by 2. Finally, the result that I would like you to look at is that the critical coupling for a non-zero value of r is just whatever, 2 over pi times the maximum of the distribution. This is just for the Lorentzian I'm giving as an example. If you had a Gaussian, whatever. I just did it as a Lorentzian to tell you that because this simulation, if I remember correctly, was with a Lorentzian, you can do it with whatever unimodal distribution you like and that will be the answer. Coromoto did it with the Lorentzian 1970 whatever and he did no simulations at all. It was just the theory. It's going to be a critical point and the new solution comes only after that point. r is 0. See, it's like this. There are many solutions to this particular equation, many solutions. One we can immediately identify is r is equal to 0 and r is equal to 0 will turn out to be a solution, a coherent solution. Overall, there is no discussion of the stability of any of these states. So the coherent state, is it stable or not? We don't know. Is the incoherent state that is stable, presumably, but at some point it has to become unstable otherwise the stable solution will never come out of it. You'll never see a phase transition. If the incoherent state was always stable, right? Meaning there are those issues which I'm really not getting into because they are quite complicated as to is this state stable, is that state stable? For the moment, I just want to take one feature. Below kc, this nonzero r solution doesn't exist. So where can I find this transition point? It also turns out that this solution that we get over here is in fact the stable one for k above kc. And for k below kc, the incoherent solution is the stable one. Put it this way. If I started out with a coherent distribution of thetas below kc, they will just spread apart. So it is just the determining of the transition point that I need to set. r is equal to 0 in this equation and solve it. This is just, by the way, it's for this example because I showed this phase transition curve. Yes, Tomasso? No, I wrote down the order parameter as a real time phase. No, r e to the i psi. By definition it was. Now, what I want to now do is to consider a somewhat different and very powerful way of solving the same problem. This is a method that was introduced in 2007, 2008. It's called the Ott-Antonsson method. Some of you might have seen it before, but this is now the method of choice for how to address some of these issues. Can we discuss some of this later in person with gloves or without gloves? I'd like you to remember this. Using this solution and a little more algebra and work, you can actually figure out how r depends on k above kc. I will not do the derivation, but I'll just point out that r, in the case of the Lorentzian actually, it turns out that r goes as 1 minus kc by k. So this curve over here is given by 1 minus kc by k, and in the limit of k going to infinity, you get that r goes to 1. I'm just giving a result without deriving it, but it's straightforward. I have skipped a fair amount of discussion of this transition, so I'll probably just scan this page and put it up, because I do want to consider the Ott-Antonsson ansatz, because it's important, it's powerful, it's necessary to learn. So let's go back to this continuity equation. d rho by dt of theta omega t is equal to minus d rho. Let's rewrite, I'm now going to go back to the complex definition of r and call it r twiddle. And I want to rewrite this as kr by 2i e to the i psi minus theta minus e to the minus i psi minus theta. Let me just tell you what I'm doing right now. I've expanded this sign over here. I want to go to this variable r twiddle, which is just the complex order parameter. So r times e to the i psi is just going to give me r twiddle and r times e to the minus i psi will give me the complex conjugate of that. So given that partial of rho with time is just plus k over 2i times r e to the minus i theta minus r complex conjugate e to the i theta times rho. This derivative is on this side. Now rho is periodic in theta. So it seems like a good thing to try to do a Fourier expansion of rho. In other words, let me say that rho of theta omega t is 1 over 2 pi summation n going from minus infinity to infinity rho twiddle of n omega t e to the i n theta. That allows us now because I have a derivative in time on one side. I have a derivative on theta on the other side. And what I'm going to do is just to tell you what happens when I differentiate and collect all terms with a common n, everything multiplying the same phase e to the i n theta. That will give me the following equation partial of rho sub n with time plus i n omega rho n. I will discuss the origin of these various terms in a moment. k n by 2 r twiddle rho twiddle n plus 1 minus r twiddle star rho twiddle n minus 1 is equal to 0. And n goes from minus infinity to infinity. It is just the Fourier components. It's a standard discrete Fourier transform that I'm doing. It's just a number. I'm letting n go from plus to minus infinity. Rho twiddle of n is the Fourier components. It's a standard thing. Just one second. Go ahead. It's like when in doubt, take the Fourier transform. Or the Laplace transform is more appropriate. Meaning, no. No, no, no, no. Everything, we are looking at phases. Stuff is all 2 pi periodic. And it's the kind of thing that you need to do. Meaning that of all the mathematical tools that you might like to apply, whenever you see a periodic function, let's just see what happens. I mean, this is not exactly how it's done. But what I'm saying is that there is no a priori reason. See, we've already solved the problem in one way. There's no a priori reason that you have to do it. But I'm just pointing out, so far you have conserved difficulty. You had an integral over all of space and all of frequency, et cetera. It's converted into n equations. And they're all coupled. I'm just pointing out now just the question of taking derivatives. If you take the derivative in time over here, you'll just get rho n twiddle times e to the i n theta. Omega is just multiplying rho. So again, there is a derivative in theta. So it just gives you i n omega times rho n. In these two terms, there is one factor of i n that comes. The appropriate term to take the derivative of will be rho n plus 1 because you're losing one power of e to the i theta over here, gaining one over here. I mean, I'm just helping you to keep count. So you get this equation. And now comes the brilliant part. Ott and Antonin made the assumption that let us consider a class of functions such that rho n twiddle is just equal to some complex number alpha to the power n. So they took the subscript and made it a power. But that was done with a lot of intuition and a lot of motivation from other studies looking at various values. And this is a restrictive class of functions. See, remember what this rho tells you one second. Let me point out that this is a function of, these are time dependent functions. And we are going to try to find a solution for rho n of t, which asymptotically goes to stationary distribution. So if I start with, I mean, there's a lot of mathematics in this paper and I've given you a reference to the paper. But you are equipped to understand most of it. I mean, there's some stuff about, there are always things that Eddard does which two people understand, him and God. But jokes apart, this is the definition which is restrictive. You're asking for all these Fourier coefficients to go down as a power law. And all that is being asserted is that this is not an unreasonable thing to do. Also post facto, when having seen that it works and it works very well, then you say that, aha, this must be the way in which it is. So I mean, there's a little bit of both. But just see what happens. I've got n over there. This is a step which will appear in your tutorial, so you will have to do it. But finally what happens is that once you take the derivatives and you throw out all the common parts, you get an equation, partial of alpha with respect to time, plus i omega alpha plus k over 2 r alpha squared minus r. And magically, n has disappeared. This is an ansatz, well justified, but it's just ab ovo out of the blue. But the use of this ansatz, I mean, is that it is good and valuable and all that. But what it helps us to do is to take an infinite dimensional system because this is after all an infinite set of coupled equations. They are coupled because the rho n and rho n plus and minus 1 and n are all mixed in over here. And now you finally have one equation. And this one equation, if you can solve it, which you can, it should give you the correct solution for the problem in hand. So the fact that n drops out is extremely important. And you can do a lot of things with it. In particular, big one? Alpha is complex. Yeah, alpha. Alpha is complex, r and in, everything is complex. Now, let me remind you that r twiddle was just the integral of d omega, d theta, g of omega, e to the i theta rho of theta omega t. And I don't want to do too much algebra over here. So I'm now going to replace this by the summation of rho n omega t. And out of all these, if I just look over here and do this integral, then the only term that's going to survive when I do an integral over d theta is just going to be the term rho n1, because that's the only thing that will be constant. So this is an integral over d theta of e to the i theta, summation of rho twiddle n e to the i n theta. The only term that will survive will be the term which does not depend on theta, because everything else will integrate out to zero. And the one that does not depend on theta, here there's one. And so the only term that will be important is n minus 1, that entire sum will vanish when I take the integral over theta. Yes, I mean I've taken the integral. 1 over 2 pi will also go because of that. So finally our twiddle is integral of d omega g of omega rho twiddle of n minus 1. Of n will go as to zero, so this is minus n star 1. I'm using the fact that the complex conjugate of rho is the same as the negative of v. We have the conditions which I'm going to let you read in Kuramoto's paper. But I want all the alphas to be less than 1. I want this entire series to be convergent. And then n has to be positive, et cetera, et cetera. See, eventually we're going to try to solve many of these problems by summing geometric series, because these are powers. So all the conditions that require convergence will be there. I'm just leaving them as mathematical niceties for the moment, because I want to just get ahead and try to show you that our main aim is to go for r. And r basically depends on the first component of the first term in this Fourier series. I just did the Fourier integral. See, this is, again, because this is e to the i theta summation of rho n e to the i n theta. This is going to be the operator part of the integral. So let me bring that in. Now if I take the integral from theta going from minus pi to pi, I integrate over the circle. The only term that will contribute is when this exponent is 0, because there are as many pluses and minuses on everything else. So the only term which will actually be that n plus 1 is equal to 0. And so n is equal to minus 1. But then a little thought over there on the Fourier expansion will tell you that rho n, rho twiddle of minus 1 is rho twiddle star of 1. And that's what I'm just using over here. 1 is symmetrical in theta. And the advantage is that this will help you to find for any given value of r what is alpha or vice versa. Because I can change that now to alpha star of omega. I should sort of just alert you to the fact that the Kuramoto model is one that tends to equilibrium in the long term. I mean there is both dynamics and ideas of statistical mechanics over here. Because at any finite time as you have seen in all these calculations, you don't get clear behavior until you go to asymptotics. So we are going to start with some arbitrary alpha. That arbitrary alpha corresponds to arbitrary initial values of rho. And then ask, how does this evolve? And where does it evolve? Does it evolve towards r is equal to 1? Or does it evolve towards r is equal to 0? And that looking for the change in behavior is what is going to give us the transition point. So finally, we just got one expression for r. And this is going from minus infinity to infinity. And what Ott and Antonson did, I mean that was the first level of brilliance to say that all the coefficients are just power laws related by a power law. The second part is to say that we can now extend this into the complex omega plane. By analytic continuation. And try to evaluate this integral by the residue theorem. So that's a second assumption that you can do it. And they showed that if g is sufficiently well behaved, the number of singularities that exist in the complex plane is finite. And in this particular case, it just turns out that when you evaluate it by the residue theorem, you can just ask what this whole thing comes out to be. If I take the Lorentzian, the two poles are just as gamma plus or minus i omega. Or omega is equal to plus or minus i gamma. Those are the two poles for the Lorentzian. So this is now specific to the Lorentzian, not to arbitrary cases. In arbitrary cases, I'll have to take some g, find out where its poles are. And then all you have is that r twiddle is just equal to alpha star of this quantity. So i gamma t. And r twiddle star is just equal to alpha of i gamma t. We are almost home in the sense that what needs to be done now is to substitute the alphas over here with r. Once I've done that, then all I have to do is to separate out real and imaginary parts. And I will get a single equation for the real order parameter. So partial of alpha with respect to time just becomes partial. You see, alpha is just equal to r twiddle. I'm just writing it down over here. See, i omega is just gamma. Alpha is just r. Let me write it. And then plus k by 2 alpha squared r twiddle minus r twiddle star is equal to 0. So I've just replaced various things over here. And this squared, I can go into r twiddle squared and r, okay. Yeah. So this is now specific to the Lorentzian case. Here there is a single root, a single pole in the complex plane which is at minus i omega, minus i gamma. And for that particular solution, I find that r is just alpha and r star is just complex conjugate. r is alpha star and so on. Anyway, putting these two solutions for alpha wherever they appear gives me an equation that just involves r. And now remember that r star, this is just equal to r e to the minus i psi. And this will give me, and this is just equal to r e to the i psi and so on, okay. So putting that in, taking the derivatives and proceeding, you get the following equation dr by dt, okay. There is a derivative of psi and there is a derivative of r. You know, you've got to take the derivatives of both the parts. I'm just looking at all the real parts now. Okay, dr by dt plus gamma r plus k over 2 into r cubed minus r is equal to 0. And the same algebra will tell you that d psi by dt is equal to 0, okay. So this is, as far as we are concerned now, the solution for the Kuramoto model with the Lorentzian distribution of frequencies. Using the Ortt-Antons and Ansatz, you get the following equation, which, because there is only a single pole, I mean the residue theorem, that's not more complicated. If there are more poles, you have to add over all of them. And so only you know that well enough, right. And then finally, you come down to a single first order differential equation for the real part of the order parameter. And this will give us, in principle now, r as a function of time, which is what we needed. I'm not going to give you the solution of this because this is the first order differential equation which can all solve, right. The point is that this works amazingly well, not just for the Kuramoto model, but for all sorts of variations of the Kuramoto model or for all sorts of other oscillator problems. So this is a technique of great generality. And it applies in such an amazing variety of places, as you will discover, I hope. Yeah, just one, start at the back. Yeah, physically it just means that this is a non-equilibrium, you know, any finite n, this is a non-equilibrium system. And so what you will find is that this is, you know, r will take some time to go to its asymptotic value, that's all. The coupling has to be 2 pi periodic. I mean, all sorts of things have got to hold well for this, right. I mean, so many natural systems are 2 pi periodic in their coupling. The sannanas, for example, in some appropriate units of time, 2 pi. Meaning it is not as unusual as all that, okay. The real restriction is that you are stuck with a certain class of functions that you can play with. But what Orton-Antonsson and then later others have shown is that this is a very mathematically sound methodology. You can actually use it for an amazing range of functions. I'm sure there will be situations where it doesn't apply. But the point is more to know when it applies and why it applies. Because it could turn out that the problem that you choose to study, you know, I will not have the time to do all the applications of such things. But you can just look around in the literature and you'll see that it's actually, it applies to places where you don't imagine such a thing. Supposing you have two communities of Kuramoto's. One Kuramoto with one strength of coupling, another Kuramoto with another strength of coupling, two of them are coupled with some intermediate, you know, the study of cliques or communities, for example. And I know that many of you are interested in this kind of a problem that the intra-clic coupling is much stronger than the coupling between cliques. You find that Orton-Antonsson can be used in that situation also. Split row into two, you know, two parts, one for each clique, and then you can apply Orton-Antonsson there and so on and so forth. So it's an amazingly powerful kind of technique. And it's just something that you should know. I mean, today's lecture was mainly to show you, I mean, to sort of bring it into the class and for discussion. There are a lot of mathematical issues. But most of them are solvable. That is, they don't necessarily apply to the range of problems that you might be interested in. There will always be a singular perturbation somewhere. So let's stop here for today and I'll try to get a homework out to you all. Bye-bye.