 Okay, so now we are starting to continuing the actions by Professor Diaz Ramona. Okay, thanks. We're going on, but now turn to the opposite task. Where were we yesterday? Do you remember? We were talking about the linen cycles. Yesterday, the linen cycles, the topic was how to rule out that some dynamical signal has a good cycle. Vote tasks are not easy for the general system at all. So I just showed you yesterday some most frequently used methods to try to rule out some cycle from some dynamical system. So now we will see how we do the opposite task. So, to establish that there is an encyclical system. One of the most important results concerning the existence of the linen cycle, but as you will see at the end of this chapter, this is very important. More general sense, the result called Poincaré-Bendixson theorem. This theoretically very important result in the nonlinear dynamics, because due to lack of Poincaré-Bendixson theorem, there is no dimension. Then in plain, we are linked to the health theories that generally are the points. So in contrast to the process of linearization and apartment theorem, which I said, we have quite an empty dimension. So no matter what's the dimension of your learning system, you're always talking about meditation and art, you are always talking about the application of Poincaré-Bendixson theorem is already in the play. You don't have even in the space. So what is generally the Poincaré-Bendixson theorem it says. If we have a region which is closely bound and subset of the play, our vector field is, as always, continuously different. And Fort is the most important assumptions. It says that this region does not contain any fixed point. Okay. So inside the region, there is no fixed point and port. If there exists some trajectory that starts in that region. Then it will remain in that region for all the future time. Then either this is itself a closed orbit, or its spiral approach to some closed orbit. So either way, surely have the closed orbit. Okay. First, how we, in practice, applied in this theorem. So the first three assumptions are quite easy. So the first two just constructs a closed bounded region in the plane. The vector field as we discussed earlier is usually continuously fresh or we don't have problems in that direction. You exclude the fixed point from that region. But the most tricky one is how to track your trajectory inside some region. How to be sure that trajectory will not leave the region. So if it starts there, it will remain in that region for all the future time. That is basically what should be done. So in practice, we are constructing so-called the tracking region. And how to check that something is the trapping region. Yes, or try to construct it. But how to conclude that trapping region. So by plotting the vector fields on the boundary. Okay, that is one more reason why we at the beginning. I show you how to plot the vector field. Of course, by yourself, we use a computer software. And you also saw yesterday, according to some examples, we also use the vector field directions in order to rule out the close orbit. But also here, in order to prove the existence of the close orbit also the vector field directions are important. You check on the boundary if the direction field point inwards. Okay. On the both boundaries. So the trajectories cannot point that region. It's quite clear. Okay. So let's see. The examples. So you remember our first two example from the yesterday. When I plot you do the limit cycle. It was the basic examples. In all the books. Yesterday we concluded about the existence of the limit cycle by transforming the system to the power coordinate and then discuss. So on the left. The system. Our limit cycle here. And the vector field. This is the table that we used yesterday to establish the stability of the limit cycle but it also helps in constructing the trapping region. Yeah, because now you know that between zero and one the vector field is pointing to the right. So it's enough to take some cycle here example with radius one half. That is what I plotted there. You have this green green field and inside this edge to take one radius bigger than one. That is what I wanted here. As you see the vector field. The ones that the trajectory enters in this region from this side. Always remain there and also once the trajectory enters region from the other boundary also remains trapped in this region. So this green ring is now the trapping region. And by application of current band exon theorem. We can conclude that there is surely limit cycle. The other picture is for. That second example when I just constructed so let's go on some example. My next example which is not so obvious. So consider this non linear system and prove that prove that it has a cycle. Let's start with the fixed point. The origin is the only fixed point and by linearization. We now conclude that it is unstable folks. I construct the rectangle. The axis between minus one and one and absolutely between minus two and two. So, on the left, just plotted to the face portrait. Oh, it is quite obvious that we have the lemon cycle cycle here, but let's prove it. Consider the boundaries of this. This alliance. And here is the calculus. So let's take, for example, absolute to and see what's the science is usually what's the sign of X prime and absolute prime. So the sign of X prime is going to change. What absolute prime is less than zero. Okay, what's that mean that absolute prime is less than zero. We are going down. Yeah, surely, but maybe X will be there, but anyway, this direction. So anyway, the vector field points like this. Okay, we are absolutely equal to so this boundary. So here we either go there or there. The vector field point in one. So the trajectory will never leave this region crossing and pretty much the similar. Okay, we are pretty much the single similar calculation on each on the on the boundary. So on the left as you see we also calculated points here. We also come to the conclusion by calculating the the first derivative is going here and from here, definitely this rectangle is the tracking region. Yeah. The trajectory start in this region. What my question is, can we apply the point curve band extend here? Not yet. Why? Because now we deal with the fourth assumption. Nicely, we construct the trapping region, but the third assumption remember, third and fourth is what's important. Not supposed to have a fixed point inside, because if we have a fixed point inside, then we cannot come to the conclusion that there is some attractor inside maybe the fixed point will attract the trajectories. If there is no fixed point inside the region, then surely something will attract in that region. The trajectories that has to be non-linear, the nine experiments. In the play. Okay. So what do we do? Luckily, in this situation, our fixed point is unstable. What that means? That means that we can construct a neighborhood like a small circle around the fixed point. And since it is unstable by the formal mathematical definition by the intuition, what means the unstable fixed point? That the trajectory which starts close to the fixed point will leave the neighborhood after some time because it is unstable. This small region, like the circle around the fixed point, we are sure by the definition of the stability that the trajectory will leave that neighborhood enter into the now boundary region, which this is yellow one, yeah, without that circle, which is now the trapping region on which we can look by the Poincaré-Bandit's theorem and conclude that there is a fixed limit. Of course, in the application, usually have the parameters and usually the stability of the fixed point depends on that parameter. If you want to prove that for some parameters, there is a limit cycle, then first you have to assure that your fixed point inside is unstable. So, to repeal the trajectory, then to construct the trapping region because if the, remember yesterday at the end, I told you, let's keep this part to the lack of time, that by the index theory, come to a very important conclusion that the close arbitrary must recycle at least one fixed point. Okay, so the trapping region and the inside cycle always have a fixed point inside, but that fixed point has to be unstable. Now, to repeal the region, you construct the small circle, then you construct the trapping region on which boundary trajectories, the vector field is too large. Yeah. Why should it be unstable? Unstable. Why should it be unstable? The earlier example, it's not because you have an institutional answer. Yes, but I will come to that point. Yeah, we can just, if you go back for time, then you can make another. So, okay, first to answer to this question, why the fixed point has to be unstable? That's what you said. Yeah. And why so? The fixed point is because if not, then you cannot make a boundary. Of course, you can make another, but in this case. Yes. Yeah. Yeah. Because it's actually unstable. Yeah, because it's unstable. Yeah. But the other question and other, if we just reverse the time, we can apply it in a reverse time. What means that on the both boundary, the vector will be closed outwards. Yeah. But then we will have the unstable cycle in that conclusion. I think, yeah, I just remember so, yeah, so this is generally the situation in the previous example you see. That's a larger limit cycle is unstable. Yeah. See on the green ring. The vector field points out. And the tracking region for unstable. Okay. So, in caribandis and theorem as a measure is very, very important. The nonlinear dynamics. I didn't say that the nonlinear dynamics is a play and say the compare with a higher dimension. It's quite limited. So if the trajectory strapped the close bounded region. And without any fixed point inside. In fact, we must approach orbit cycle. We will say but as I mentioned in the higher dimension the point caribandis no more applies. And we come to much more complicated discussion and dynamics so the transaction is as seem a wonder around. Or, without into some either to fix point or those orbit has to be a tractor, but as you probably heard, there can be some so called strange attractors, right. So I don't. Just get to the idea. Introduction to house. Mine. Sorry. Somewhere. Share some. You should hear this one. Yeah. I don't know. Maybe it's when he stopped sharing. I'm going to make a comment. That's the basic idea about what my vacation is, because in the application, of course, you have the photo. I think that really I don't know much about the energy. For example, in the, in the system. The vector field is constant. So when you are in that factor, you also are busy. You fall into that factor and use always. So once you start the cycle, you're consuming the energy. Energy while getting close to the inside. Less and less energy. As you say, So you can. No, but you have to be patient. In every place of space. When you are in that factor, maybe you are compensating. Always look at the divergences. So you are, you are losing energy at a constant base. Always like that. I'm not sure each dynamic systems. We don't, we don't see it because we are training or like all this in my part system. Now that the normal system for the beginner, it's one dimensional parameter, but also we have the theory of vitrification and two parametric vitrification and so on and so on. And of course, for the beginning, I decided just to show you the global idea what to do by for a very basic each tax book. The examples of the bar from page. So what is the verification. Asia is simply if we change the parameter. And come to some what you call the critical value. Then the dynamical system will be radical change what that means that mean that two systems will no longer be what we call the topological equivalent. Okay, so we will have the change either in the number of the fixed points. Okay, maybe the fixed point will disappear, or we will have more fixed points. One, then there might be the equal number of the fixed point that they will change this ability. We have one to stable fixed point but one will become unstable point or we have one stable one stable this that is what you say so the change in the number or either the stability of the fixed point. So ironically change the dynamic of the system. If it is happening from some values of the parameter, we call it verification. Also, there is a change in the existence of the cycle. Yeah, so it can be that the system has a limit cycle then the cycle disappear or the cycle appear after some while of the parameter that we call the called the bifurcation critical critical line. So I will represent you three. First three basic bifurcation concerning the number of the fixed points and either their stability. Several nodes bifurcation is the most typical which equilibrium are destroyed or created. Also known as the fault bifurcation and the typical what we call the canonical normal form of the saddle noted bifurcation is the nonlinear system of this form. What is what are the fixed points of the system. Obviously, absolute zero, but depending on me. Yeah, x square x square is me so if me is negative, then we will obviously don't have any fixed point. If M is zero, we have only one fixed point. Yeah, and if M is Um, um, great and zero. Yeah, we will have two solutions of this equation. So definitely the number of the fixed point depending on the parameter. So when me is negative, no fixed point, and the flow is from the right to the left. Okay, here you have the face portrait in all three. In all three cases on the left me is minus two or negative in the middle me is zero and on the right me is great and zero so we have a it is not seen very well because of this color, but there is one fixed point. Okay, the saddle, obviously, and there is another fixed point now is always symmetric fixed points. This is obviously the node. Yeah, this is the note now we have to explain but what is interesting in the case when me is zero, we have only one fixed point at the origin. And it is non hyperbolic. So you remember I told you that I will talk about that case. We said that if by the linearization, we have a hyperbolic point, then we can conclude about the dynamics in the system by the dynamics of the. But if the fixed point is non hyperbolic, then we can come to the conclusion. And yesterday we see the situation about the center, which is one example of the non hyperbolic fixed point, which has complex 18 values with the real part zero. We haven't still considered the cases where there is one agent while you reach zero. And this comes to the situation when we have here we have the fixed point, which is no hyperbolic, you see that one. The reason why there is zero, another reason why there is minus one. So, by the linearization we cannot say anything about the dynamics inside of this, but throughout this big rotation. So, in the case me is greater than zero we have to symmetric fixed point, and by the linearization and usual technique. We either plot agent wires, of course here, you already have a diagonalize so canonical form. Yeah. So, you know, the 18 miles. This is the numbers on the, the main diagonal so we don't have to calculate. So one fixed point is stable nodes and another fixed point is a saddle point. And the saddle node of this location, the number of fixed points is n is negative. No, if m is zero to one, one non hyperbolic fixed point and when the perimeter is positive, we have two fixed points, one stable and one unstable fixed point. The usual way to show the bifurcation is throughout the bifurcation diagram. What is bifurcation diagram so it is plot in me, you are explained. Okay. Just plot this normal form, so the first equation graph, and just plot the vector field, and it is a usual way to denote the stable part with this and unstable part with the dashed line. So here we have when means no fixed points. Okay. As we increase, now we have one fixed point. This is the vector field of this and here, obviously approach to this part is stable fixed point. Because we have one fixed point which is stable and another fixed point which is unstable. So this is the best way to show another type is a transcritical bifurcation. Now first equation is also the second order polynomial but with x next to me. What is now the situation. This equation. Yeah. Always have roots. Yeah. x one and the other is x me. Yeah. So now we will always have two fixed points. Okay. Absolutely zero and x is either zero or me. So in transcritical bifurcation there is no change in the number of the fixed points. Change in their stability. Okay. When means less than zero, we have to fix point one is origin and another is and by the linearization. We show that 18 values are both negative. This is stable notes. Yeah. Both ideas are real and negative. We have the stable note and the other fixed point. Now we have one. A value which is positive. Another is negative. So several. Yeah. One positive one. We have to fix point one is stable. One is unstable. When me is zero obviously we have one. Again. At the bifurcation value, which is critical, we have no hyperbolic fixed points. The bifurcation at critical value always happen in non hyperbolic fixed points. When me is great than zero. Now the origin became the saddle. Okay. The origin is always the fixed point. First it was stable. After the bifurcation the origin becomes unstable. So the saddle point, but another fixed point appear on the right. And it is now a stable note by the linearization. There are the base portrait. This way. Yeah. We have saddle. We have the origin, which is the stable note. As me enlarge these two points are close, close, close, close, close. These two points can become one. Yeah, they approach to each other as we increase the parameter. This is obvious because this coordinated depend on the parameter. Yeah, so as we increase parameter, these two fixed points are closer and closer and closer become one. That point, the origin change the stability. It is no more stable fixed point trajectory. It became the saddle, but another attractor another fixed point appear, which will now attract the directories and we have another. Another stable stable stable fix. So this, this bifurcation is only considered not in a change of the number of the fixed point. The stability of the fixed point. Okay, this is the conclusion about the trust critical bifurcation. The origin is stable. We have another fixed point, which is a saddle. At me, let's do fixed points. Of course, so joined in and became non hyperbolic fix fixed point and after further increase the parameter origin became unstable so settle point and another fixed point appear. So, okay, this is the bifurcation diagram. Here we plot. This is the origin. So you equal to zero. Yeah. And this is the line you equal to these two lines. Okay. So here left so when me is negative. The origin is stable. This is the stable part. But another fixed point, which lies on this. This line, it is unstable. Okay, clear. Yeah, when we increase what's going on the origin become unstable so we now have the dashed lines. Yeah. And another fixed point, which is on this line so me equal to you become the stable. Of course, the critical bifurcation is in each of these four cases. Similarly, the normal formula like this. Okay, we also can minus here or plus here or plus here, minus here so all the combination. This much gives the same dynamics. But maybe before the critical value was here, for example, the only difference is that now for negative parameter the origin is unstable, then becomes stable on the discussion. The dynamics is much the same. Yeah, you have here. Next, the next is the fork, I think that I pronounce it. Correctly. Okay, the normal form is this. Now we have minus me X minus X. What will go on here. How many solution we have here how many roots real roots we have here. X one always, but the other two also depend on the sign of me. Yeah. Zero is always the solution but the other two depends on the sign of me so we will either have one or two. So when we is less than zero, there's only one fixed point at the origin by the linearization we come to the conclusion that is the stable note. We have two negative agent values. In the case means zero again, the situation is the same so we have non hyperbolic x point origin, but with increasing further the parameter, and it becomes positive. Now, two more fixed points against symmetric according to the, to the origin and by the linearization now for the origin. Positive one negative. So saddle. And for other to fix point the agent values are both negative. So, we have now to stable notes. Okay. So, super critical fork be vocation for me negative only one stable origin fixed point. We still fix point for the critical value but non hyperbolic. And when me become positive origin becomes unstable, but to other stable fix points up here, this is stable notes. This is bifurcation diagram on the left. Okay. So we have fixed points here. This is me. Zero, and the parabola. When me is positive, the origin is unstable dashed line, and this is. But now we change in the normal form here this time. We have another type of bifurcation. Why it is so radically different here. The discussion is pretty much the same as before. Okay, but now when me is negative. The origin will be stable. And we will have to unstable fix me is zero. The origin is again non hyperbolic fixed point. But if we increase me after bifurcation value to unstable fixed point disappear. And only stay one unstable fixed point. And you tell me why do you think that this is radically different than the previous case. There is there is always always one, always one. Here we have always one. We have three after the bifurcation value only one that sense doesn't make any difference just before the barcation. What is the situation when we have only one on the left that only one is stable. But on the right when we have only one attractor only one fixed point unstable. It's the main difference. We no longer have anything stable in the system. Now after the bifurcation value, the system, which was for some values of the parameter stable has at least one fixed point which will attract the trajectories. Now after the bifurcation value is increased, we have no longer anything stable. We don't have any stable fixed point. There is much radical change in the behavior of the trajectories. And you can see here also the face portrait. No cases. Obviously here you have three fixed points. No saddle saddle note. The stable. Well, you have also for me zero the saddle point and great. Now the origin is the only fixed point. But as you see it is the saddle point. So unstable. Wherever you the only fixed point is unstable. So whatever the initial value for this system is go the solution to infinity. They will not approach any. They will not approach any finite value throughout the time. Okay. This was so called the normal form of bifurcation. As you see, according to the techniques that we learned before it is very, very simple to analyze this form. The normal form is no problem. Then some examples in some examples before. But of course, the question is when you got some system, which is not in a normal form with arbitrary function. Is there a way to see this occurs some bifurcation and which one. Not not very pleasant calculation. But with course using computers. You have some software which do that. Of course, you can do that for arbitrary nonlinear system. So for some function at one after according you. You have to calculate the partial derivatives of this function. Okay. So the Jacobian matrix is, let's say, first capital D derivative of this vector field, but we also define this expression. So this is the vector field of the second partial derivatives. This is with the third order. What we have to do in order to apply the calculation, we have to use a vector, but the one corresponding to the lambda zero. So to the eigenvalue. Okay. The first assumption is, of course, this is the fixed point. Yeah, and to this fixed point we have the verification. The second assumption is. As you see in all examples, when me is zero. Yeah, at the bifurcation value, as I told, the origin was non hyperbolic fixed point with one eigenvalue to this is necessary. So the first condition that has to be checked in order to have the bifurcation because if the fixed point is at the void, we will not have a problem. After that you calculate the eigenvector v corresponding to zero eigenvalue. And with the w w, the eigenvector corresponding to the transform matrix, which of course again has a value zero. And after that with the calculation, so you multiply w with the vector field partial derivatives according to the parameter. Yeah, this is different from zero. And if the so this is the vector field. If this is different from zero a saddle note bifurcation occurs this fixed point for the bifurcation value and zero. If this first value is zero, then you check further. This is different from zero. This is different from zero. This is the Jacobian now off the vector field. According to the parameter now. So if these two values are different from zero, we have transcritical bifurcation. And in the third case, if the first one is zero, but the third one is also zero. You have to check this to have to be different from zero. And in this case, we have the work bifurcation. So here you have a really one. For example, just maybe the most important steps. Just arbitrary systems. Okay. What you first check, because the origin is the fixed point. Yeah. You calculate the curve in matrix. And how you determine the critical value. Yeah, but yes, but that means that. Determinant is. Yeah, this is a rule. Yeah. Yeah, yeah. Determinant of this. Zero now you have two solutions and I considered here, for example, the bifurcation but I think that also for the other bifurcation value also have what was their transcritical bifurcation if I remember yeah, transcritical bifurcation. So, next, you fixed that bifurcation value. And now you calculate the eigenvector. Yeah, this matrix and transpose matrix. And that just calculate the partial derivatives of this function course for for all the time. The bifurcation values from all the further calculation you took me in the system and calculate the partial derivatives and the fixed point. Also, I said, every point of view is not pleasant with the computer. It's not no problem. And here is, for example, the face portraits. We have the transcritical bifurcation. We have to fix point. I consider the bifurcation for me equal to one. This is for less than bifurcation value. We have to fix point one, the origin is stable, but here the focus. Okay. In the normal form it is a node. We have now the arbitrary system. The other is unstable. It's a saddle point. It will close close close close. Not to the bifurcation value. We have here and this is obviously the note unstable unstable note. But anyway, we have a two fixed point just different difference. And the bifurcation that we're considering existence appearance or disappearance of David cycle is the hopf bifurcation. Here we also have supercritical subcritical. It will be different. Let's look at the normal form which will be more easier. So I put the system in a usual form, but this is also called the normal form of the hopf bifurcation. But in order to analyze the system as you own the previous example and we wanted to do that and once discussion about the limit cycle transform to the polar coordinate. When we do the transformation again we obtain a very, very simple system of differential equation uncoupled the first equation. So now it is like the pitchfork bifurcation for fixed points. So there is zero corresponding to the origin. Yeah, so the origin will always exist. Okay, but depending on me. Yeah, our quadrat error square is me. So if me is positive, you will have two solutions. Of course, one with the positive is a limit cycle. But if me is negative, you will not have any limit cycle here and there is system. So when me is negative. Biggest point is a stable table for zero still stable focus. But as we increase, I thought it to me too, because for as me increase the radius cycle also. Because the equation for the limit cycle is areas equal the square root of me so increasing me the limit cycle radius increase so I thought it for to be obvious. To have a good pitch. So for me, for me too. Yeah, but the before bifurcation for the system in the polar coordinate. Yeah, it's the same. Yeah, so but now that fixed points are actually the limits. Except that, of course, the limit cycle cannot be your equal minus square root, because there is a distance, it has to be positive. Yeah, so only one. That's true, like in pitch for bifurcation, obviously, because every distance. Yeah, that is why only only one actually. Yes. And you get radius is the other side of the cycle. Then we have this stable point. Yes. Yeah, yeah, the stable. Yeah, you can also plot the stability of the length cycle. Yes. Yeah. As you see, now the door is going to become unstable, unstable focus, replying the trajectories, but there exists a limit cycle and obviously limit cycle is stable. So the trajectories, both from outside and from the inside will approach the cycle so come to the conclusion that this table in the cycle up here. The calculation is, as we did it before. Determine the sign of first derivative, positive, positive here so we have the stable in the cycle here. So in the other case, so this is the bifurcation theorem on the left. So when means less than zero on the origin. Okay. Now the parabola but we plot only parabola for the positive values of error. Now the origin become unstable dashed line. And we have the limit cycle, which has the parameter increase the radius of the cycle. So this is super critical of vacation and very similar. We have, we have a subcritical of the vocation, but as in the case of the pitch for bifurcation also quite radical. Now, we have the origin which is the stable, but the limit cycle around that origin is unstable. And as me increasing, so what's going on here, the radius is getting smaller and smaller and smaller and smaller, approaching the fixed point, then on one fixed point and the limit cycle is. That's me increasing, but the origin became stable after the bifurcation value as in the subcritical pitch for now, we also don't have any chapters in the system. We have only one unstable x point in the system. Yeah. Yes, yes, there is another type of vacation also but the vocation of course can occur in just here but represented you and the first equation is one dimensional. Yeah, but you can have the same of course the calculation in the theorem that I will do is a little bit different to establish the vocation and the higher dimension. What is actually the situation. How to explain this. So, any system that occurs to any of the previous bifurcation or the half bifurcation can be reduced to the normal after some transformation that is the point that is why we first to do this normal form and we call it normal. If you have arbitrary system, and if you see that it satisfies condition for some bifurcation, they're the transformation of the function or the time dependent value. So let's put the system in the normal form and after that you discuss that is how the things are done. And in the higher dimension. There is a hop bifurcation but for example if you have the four dimensional course because the hop bifurcation occurs only when the origin is the center. The non hyperbolic fixed points. But with the real power is zero. This bifurcation was also in a non hyperbolic fixed point, but one real. If you have the four dimensional system, the whole bifurcation always occur in the manifolds, which is called center manifolds and which corresponds to the few imaginary, imaginary. So the other two dimension, you can just ignore. You can put in the system in the normal form, and you just discuss the system in the normal form, which the first two equations then will be uncoupled in compare with the rest of the system. You can just ignore the equation third and fourth. Pretty much the calculation will be of course a little different. Okay, yeah, we have now unstable limit cycle. The origin is stable. Bifurcation value origin becomes unstable fixed point. And up to the refrotation of the limit cycle disappear and we have only unstable unstable fixed point. Remember from yesterday we had the problem in the center, but we show that for some systems and hope the bifurcation occur. When the center of the linearism system, the bifurcation value will become. The center of the non-linear system, you don't have bifurcation, because you have already proposed trajectories around, doesn't disappear. Am I clear? So the bifurcation value by the linearization, you will come to the conclusion that the fixed point is the center, it is a linear center. Actually, as you see on the vector field plot is a non-linear box is not the center. This is everything example from the yesterday. When I said that the center can remain the center as in the polynomial system, but the linear center can actually be the non-linear box. That's the point which is critical. And when usually the bifurcation occurs in the system, it's of course changing the appropriate parameter. And this is also sometimes a good way to show the existence. When you notice something like this, just put the parameter inside and show that for this value of parameters. Sometimes if you cannot construct the trapping region, which we see it is not so easy, you can come to the conclusion that there is a limit cycle throughout hope bifurcation in the system. Yeah, again, I put the calculation is, calculation is its technique. The calculation is not pleasant at all, but using the computer, everything can be done. So again, the question considering arbitrary systems, not in a normal form, two dimensional systems. There is a well-famous hope bifurcation theorem, giving the conditions that some system occurs the hope bifurcation. First condition, the origin is the fixed point, but for the bifurcation value, it is a center of the linear system to have pure imaginary eigenvalues. Okay, that is the first condition that has to be satisfied. The real part of the bifurcation value is positive. It means, and the real part is zero. This means that we have a pair of pure imaginary eigenvalues. Now the eigenvalues crossing the pool line with this target, so the derivative of the real part of the bifurcation value is different from zero. And the third task, which is more, most difficult, at least for calculation, is to calculate the so-called, first, the pool of number. We have for two parametric bifurcation, so on, we'll then calculate it also. The second, the pool of number and so on. Is there a table of exponents? No. Just another type. There is too much in mathematics called the opponent. The opponent's stability of the function is from the exponent, the opponent's coefficient. Mixed up, but no. It has to be different from zero. Why? Just because the main reason is, I talked a couple minutes ago, so if you want to change the coordinate and put the initial system in the normal form for the hopf bifurcation. In the calculation, all of this condition pretty much appear and with you in the way to prove the hopf bifurcation here and it is obvious that it has to be satisfied. And then according, because these two numbers has to be zero, according to the sign of these two numbers, is it lesser than zero, everything is described here. So it just depends if we're going to have the subcritical or subcritical value and is it going to happen when we increase the me or we decrease the department. So the conclusion is here. So if this L delta is greater, is less than zero, the unique cycle appears as me increase. This is positive. It decreases value. The stability of the fixed point is determined with delta. So if delta is less than zero, it is stable for great then bifurcation unstable. This is the contrary and always the limit cycle and the fixed point has a difference to build. So if the fixed point is unstable, the limit cycle around is stable. If the limit cycle around is stable, the limit cycle around is pretty much the conclusion. Okay, conclusion again here. So you see, if you put the system in the, okay, what's the steps, just two questions and just to present this more. We translated the fixed point to the origin. That's the first step. The other step to put it in a canonical force that is always done. And after that, your system becomes like this. Okay, the matrix of the system is, you see, for this system, the matrix of the system is zero, zero minus beta, beta, which means that we have a center. Real eigenvalues, real part is zero. So you will always put the system in this one. And after that, you just plot the partial derivative of this one. Mostly up to the third order. So this formula, of course, all the values are in the origin, as the fixed point. So X and I is equal to zero. So this is the number or if you already have the system, which is the functions are expressed using the Taylor expansion in the polynomial form. Then you can calculate it in this way. So, okay, this is the system. This is the inner part. This is from the second. Yeah. Second degree polynomial third degree polynomial and in order to calculate the coefficient, you use the second and coefficient. And you have here the formula. So, as I said, using the computers, it's not the problem to calculate the first coefficient. Of course, you can do software or design for money. So, okay, I think I will stop it there. Any questions? Yeah, yeah, it's quite obvious as I said, so when you start to transform your system in order to put it to the normal form. Then it's obvious. Some points that this coefficient appears coefficient and it has to be, it has to be different from zero. Otherwise, you cannot put the original system that normal form that we discussed at the beginning. So, the system will be in another form, another type of verification. Is it stable or not? This is then the bifurcation, this is the, all this bifurcation is local and it is the global bifurcation. You can have two limit cycles. How is it related to the bifurcation? I mean, not. You cannot decide just for up to, you need the further restructuring. You just now know that there is one limit cycle using the first bifurcation or if all the assumptions on the bifurcation are verified doesn't mean that there will be, maybe there is another way. You'll have to continue with your system in order to see if there is a multiple limit cycle. There is examples of the half bifurcation. There is only one limit cycle appears after some time. Because this will happen because of the rest of the dynamics. Because, for example, I just showed you the good example. Let's imagine that you have a limit cycle around this bifurcus. Let's imagine that you have the saddle point also. That's why we call it the global bifurcation. It's not only concerning this fixed point, but on that kind of bifurcation, another fixed point. Imagine you have the saddle, then what will happen in some dynamical system? That limit cycle goes large, large, large, yeah, closing to the saddle, then at what moment the saddle will attract a homoclinic. The tractor will appeal, you understand me? After some time you increase the parameter, it will rock the homoclinic and nothing else. So no, no homoclinic, no limit cycle, no, no. You can have the situation, it's two limit cycles, but it's another fixed point, which appears in the dynamics, you know, influence. Of course you have also the saddle, for example, saddle nodes bifurcation of the limit cycle. The same as the saddle node bifurcation of the fixed point, so two cycles, which will closely approach, approach, approach each other, become one, and then disappear, but that is what is called the saddle node bifurcation of the limit cycle. Of the limit cycle, yeah, two limit cycle, yeah, yeah. These parameters, they are close close close, one, and then disappear. Just like saddle node bifurcation of the fixed point, you'll have two, they approach each other, there is one, and then everything disappears. This is everything that will want to disappear. How is it possible, if it is really possible to calculate the number of cycles? Of course the calculation is usually not so pleasant, especially if you have a lot of parameters, like the coefficient in dynamics, which I find sometimes you can make it sometimes not. Yes, there is everything pretty much can be calculated. Theoretically, we know what conditions has to be to put the system in the form so that you have. But the question is, are you going, of course with the computer or plot the face portraits and you will see this. Another point is how to obtain the bifurcation value, which is also possible, but to formally prove not only by seeing the face portrait, but sometimes. The model is actually for the lamprope of the system and it's starting for the button on the side, I can identify. My query is, what a new cycle, it is mandatory to have the fixed point in the cycle. What would be the guarantee fixed point in the schedule? What the new cycle, this new cycle you will not get. If it is centered around that, how isolated goes the office? Only for the lamprope of the system, the idea is not in the cycle, the center will be the fixed point. What is going to be the center part? For other systems, what would be the guarantee to send that as a fixed point? You only can have a fixed point. That sends the closed order. The lamprope and my experience with you will happen. You have a center, you have orbit, you circle in a round, you have a lot of limits. That is also one way that I missed, you have an index theory, index of all the fixed points of the saddle is plus one. And the addition of all the fixed points inside the cycle must be one. Also the limit cycle must run several times. I didn't do any system. As a summary, I think that the summary event was that in a two-dimensional system, you can only have fixed points, limit cycles, orbiters, from one point to another or to the same. That's okay. One turning to a limit cycle. Thank you very much for your questions. That is special. I hope you enjoy your stay in the niche. For example, I'm concerned. If you have any point in your wish to all assess the basic knowledge of you, the mathematical point of view is in my best. Thank you very much. Thank you.