 Ok, quindi la seconda parte, diciamo, questa seconda parte è sulla stabilità... scusate. Sorry, sorry. The second part of the stability will concern the non-linear stability. So the question is, linear stability is very useful, but it's not enough sometimes. Ok, because it's just a part of the story I would say. And because non-linear temps can change the scenario. And so introduce non-linear effect that can bring to a failure of the linear stability. So if you make the linear stability you guess that your point is stable, but for some reason, or unstable for example, but for some reason the effect on the linear temps brings your motion near to the fixed point. So the fixed point that for the linear stability will be unstable becomes stable. And this is the case, no? Here I wrote the general form of the differential equation. Here this is the linear part, but of course you have also the non-linear part. And the effect of this part could be dramatic with respect to the stability. We study, for example, this simple system. Ok, this one. The linear stability matrix is, of course, you neglect a zero zero, of course it is the stable point, of course. Zero zero is the stable point. Ok, it's the fixed point. The fixed, sorry, non stable, you don't know. Neglet, in this part you soon have the linear stability matrix. If you diagonalize this, you see that you have two eigenvalues, lambda plus minus c. And according to the classification that I explained you before, this is an elliptic point. In fact, they stay on the imaginary axis. Ok, so the motion is some kind topological equivalent to a circle. Ok, but this problem, this model can be solved exactly in polar coordinate, for example. Or, so if you go in polar coordinate, you see, but it's quite boring in polar coordinate because you have the derivative of this quantity. Another trick that you can use is a clever multiplication term by term. For example, so you have your system, your system is, I should come back, come back. So your system is x dot one is equal to x two plus alpha x one, the sum of the square. Ok, and the second equation is x two is minus x one plus alpha x two, x one. So you see, basically it's an harmonic oscillator with this non linear friction. Ok, the system is artificial, but ok. Ok, it's a non linear oscillator plus a strange friction. Ok, if you multiply this by x one and this by x two, you obtain x one x two plus alpha x one squared, x one squared plus x two squared. And, again, you have x two, x two dot is equal to minus x one x two plus, correct me if I make mistake because the blackboard is very easy to, this is x two squared. Ok, now if you sum up these two equations, this cancel and this cancel, and you obtain x one x dot one plus x two x dot two is equal to, if you collect these, so you have alpha x one plus x two to the power, two to the power three. Is that correct? No, square squared. Is that correct? Oh, to the power two, to the power two. Ok. Sorry, why the power two? You are just summing, you know. Ah, no, no, ok. This way, this is ok, to the power two. Ok, yes, you are right. Ok, thank you. Ok, by what is this? This is the derivative of x one squared to squared. Ok, so if I am not mistaken, my equation of motion becomes alpha to the power four. Ok, if I am not mistaken. Ok, because this is the, in polar coordinate, this is the radius with respect to the origin and this is the derivative of this with a factor one half, I believe. Ok, let me check. Yes. Ok, so what happens now? If alpha is positive, if you look at this equation, now this is the distance of your trajectory from the fixed point that is zero, zero. You see this is raw with respect to this error and your trajectory moving on this way. What happens, no? This alpha is positive. The distance increases indefinitely, so it is unstable. Ok, but if alpha is negative, it means that this is negative. I come back to the origin. Ok, so the fixed point is stable. So the non linear terms can modify the scenario o your fixed point. Ok? So usually non linear terms can be relevant to the stability especially when the fixed point is marginal. Is marginal means is a center. In fact, this example, I show you that the fixed point is a center of the origin. Ok, while the fixed point is an attractor, a repeller, or saddle, the effect of non linear terms is an effective. So the stability remains unchanged. Ok? Question? It's clear, no? If I introduce, if I count also the effect of the non linear terms, I can have a different scenario of the stability. Ok, but anyway non linear terms may also give rise to other kind of motion, kind of motion, the limit cycles. We already introduced the limit cycle, if I'm not wrong. Ok. And now we see, for example, how this is possible, the non linear terms shift from the linear stability to limit cycle. Ok, definition. What is a limit cycle? Of course a limit cycle is an isolated closed orbit in the phase space. Oh, warning! The word isolated is important because in this case you have closed orbits. For example, if you have a center, you have a, if you have a situation like this you have lots of periodic orbits, but they are not isolated because if you have periodic orbit here you can find another isolated orbit, sorry, another periodic orbit very near. So this is not isolated. Ok. This is not the case for limit cycles. Ok, I have to exclude this possibility and then why I say limit cycle is an isolated closed orbit. Ok. And this is an example to show how the limit cycle arises in a dynamical system, in a dynamical system scenario. Another system like this you see, I use, no, but the trick is basically the same. Ok. The trick is basically the same. You have these two equations. Two equations, you see that zero is the fixed point. The linear stability matrix can be obtained simply neglecting the non-linear terms and this is this. You can diagonalize you can diagonalize your matrix and you see that you have these eigenvalues and these eigenvalues means if I put them into the complex plane I have one plus I omega and one minus I omega. What does it mean? What kind of what kind of what kind of what kind of fixed point is you have of course you remember if you write the the exponential of you have t because this is the real part minus plus t multiply minus I omega. So this is stable or stable unstable. You run away. But this gives you a spiral a circle motion. So this is a spiral no? From the fixed point you spiral away from the zero fixed point. But at this point enters in the game what? The non-linear terms is able to stop to stop this behavior and re-injecting your spiral into the zero around the zero. You have the competition between these two effect. The instability due to the spiral but also the fact that this term is basically pushing back to the origin. The competition gives rise to the fixed point in between these two effect. In fact again you can do the trick or in polar coordinate or by multiplication etc etc you can of course find the equation of motion of your you can solve this system exactly and you can have in polar coordinate this equation. Consider that this equation is easily obtained by even the multiplication but instead of summing if you sum multiply by homologous terms you have the R so the equation for R R squared. If you make the difference between these two terms you can have the equation so you multiply you cross multiply instead of multiplying this for the homologous you multiply by these two x dot one x one x dot two and you make the difference you can have the equation for the angle for the angle, the rotation angle and this one ok and in fact if you make this computation you see that the rotation angle is omega so and what is the solution of that what is this of course now the equation are the couplets ok so it's very simple to discuss to understand the behavior R equal to zero is a fixed point corresponding to the original fixed point in Cartesian coordinate R is equal one is another fixed point now that corresponds to this kind of motion ok R equal one in fact if you put R constant if you put R1 R equal to one you have zero so R is constant and this constant ok and this is the motion and this is the equation of motion of the trajectory that you obtained and this is a limit cycle to understand the stability of the limit cycle you can make this table analysis this analysis ok this is the function that you have here ok this is the function R square something like that no R 1 minus square ok, this is this function I look at this function I say if this function is in this way of course one is a stable fixed point and zero is a stable fixed point is a stable in both in both domains if I start inside or if I start outside ok on both directions so you see that the stability is here for R larger than one but also for R larger less than one so this is a stable in this case if I plot if the sign of course you have instability the stable point the fixed sorry the limit cycle isn't stable so if you start inside you spiral onto the stable point that's the origin while if you start outside you go to infinity ok and also there is a if you put absolute value of course you can have a mixed situation where it's stable from inside as unstable from outside and this is the mechanism how a fixed point can be generated ok sorry a limit cycle can be generated another example I don't know if I have time to discuss it no but just give you a sketch of the analysis this is called but the oscillation is very common in a non-linear dynamical system if you open a book of dynamical system you can find this example and is related to the radio technique circuit with a triode with a feedback consider here for example this is a voltage and the voltage oscillates basically like a non-harmonic oscillator but there is a friction of course which depends on the state of the system that can be considered like a feedback so if the system acquires energy the feedback lowers the energy instead if the system coin to a lower state the feedback inject energy into the system and make the system rising a higher energy state so it's a feedback of course you can write into the no I don't think so but the question is if there is a general general way to understand if there is a limit cycle in a system the question is generally not because there is a bifurcation theory that allows you for example hopf bifurcation is a way going from a stable fixed point which loses stability and the stability in favor in favor of a limit cycle this is called hopf bifurcation that is very important in fact I give you some information about this but as far as I know there is no general general possibility to understand if there is a fixed point or not you have to do the analysis direct the analysis look at the equations basically this is the call okay you are welcome, thank you this is the wonderful equation and of course you can do the stability analysis this is the two again vectors so the fixed point is here and is unstable since mu is considered positive the fixed point is unstable and we show that this kind of object can can develop a limit cycle and the way to do you can find in every book also in our book you can find this kind and the way is to to make this is to make linear transformation I don't consider this representation I would prefer to work with this one linear transformation means that you write this term of course in this form and you put at the other side at the other side this part you define this quantity that is called W and you reformulate your equation of motion with these two variables X dot and W dot you make another exchange of variables you introduce Y you introduce the quantity Y because you are assuming that Y is W divided by mu because you are assuming that mu is a very large parameter in this in this model and you arrive at these two these two equations if you study the Northline of the model you see the Northline is epsilon equal to f of X f of X is dysfunction is a cubic a cubic polynomial and the other Northline is of course the zero axis if you do the analysis around the Northline you see how the field goes in the Northline so you know that when you have the Northline the field is vertical or horizontal because you are putting to zero one of the component of the field Joshua show you the Northline analysis ok in this case for example for example I choose the Northline is X so the field is vertical in fact you have these lines the sign of course is defined by the sign of W W dot this way if X is positive this is the menu size means this goes sorry this is the Northline yes ok ok ok the shape ok the Northline goes like this ok maybe there is a wrong sign no if X is negative no sorry if X is negative this becomes positive so the vector points out upward the vector is negative and they point downward the same appears of course on the other side so if you follow the behavior of the Northline you see that the system develops a fixed limit cycle ok so the wonder pole has a limit cycle has a asymptotic solution question comments no ok ok ok yes ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok l'arizzazione dei cicoli limiti. Questo è un scenario in cui c'è un ciclo limito di bifurcazione. È molto importante. HOPF è la matematica che prima ha studiato questo tipo di bifurcazione. Quindi hai un punto fixato che perdono la stabilità e con l'apprensione del ciclo limito. E il radio, è importante dire che il radio di questo ciclo limito dipende della distanza dal punto di vista critico dal punto di vista critico della parametra in cui il stabilito punto perdono la stabilità. In generale è molto difficile determinare se un arbitro in un ciclo limito di bifurcazione non linear anche se l'existenza può essere provata. È molto difficile determinare l'espressione analitica e le proprietà di stabilità. Allo contrario, il progetto che un posto limitato di bifurcazione non ha un ciclo limito è molto semplice. È molto difficile provare che l'existenza e la natura del ciclo limito ma l'opposito è molto semplice. In generale, questo criterio usa il radio principale e l'absorbito. Assumiamo che c'è un ciclo limitato e puoi fare un po' di ipotesi ma dopo un po' di rispondimento puoi riuscire in una contraddizione, ok? E questo è per esempio il caso di... Ok, questo è un critico ben du Lac non credo che ci sia tempo e quattro, siamo finiti. Ok, basically, il critico dice che suppone che c'è un orbito fermo in un domenio D che è semplicemente connetto e ok, semplicemente connetto quindi non c'è un posto o non c'è... non può essere esplitato in pisci semplicemente connetto e questo è il tuo filo il tuo motore di equazione questo è in dimensione, il principale è in dimensione e si fa la divergenza se questa quantità ha un signo definito poi il sistema non ha una tradizionata in domenio quindi è la divergenza del filo ha un signo definito che non significa positivo o positivo o negativo non c'è nulla, non c'è nulla, ovviamente poi puoi escludere che, ovviamente, esiste e ok, l'applicazione... il profumo è molto semplice perché si applica ancora il fatto di fare la circolazione sul orbito fermo del tuo filo si applica... si applica il teoremo verde e si dice che per esempio, se applica il teoremo verde e si sa che questo è positivo si arriva in contraddizione perché si dice che può essere positivo ok, si applica il teoremo verde questo è nulla, se applica il teoremo verde questo è positivo per definizione ma puoi applicare il teoremo verde e si arriva da questa quantità per l'intera... l'intera... l'intera linea di questa scoperta e si scoperta di questo zero quindi si trasforma questa intera, l'intera dimensionale in un intera linea ma puoi vedere da la definizione che questo obiettivo canso e si arriva in absurdi perché si dice che questo è positivo ma dopo la computazione puoi vedere che questo è zero ovviamente non può essere in questo domenio e ho finito e lo stesso può essere fatto quando hai un sistema gradiente un sistema gradiente che puoi espressare quindi il vostro motore è espressivo ed è espressivo con un potenziale di energia quindi anche in questo caso se si assiume ok, puoi fare la stessa computazione si arriva in absurdi supponere che c'è un periodo t non? Si integra il vostro motore su un periodo t e quando si arriva in absurdi, la quantità è negativa ma la quantità non può essere negativa perché questo è una quantità positiva è un integrale di quantità positiva quindi è ok finalmente solo per concludere io credo solo l'ultima possibilità che hai per studiare la stabilità di un punto fiscale beyond beyond the linear approximation that somehow doesn't work is to construct the Lyapunov function if you are able to construct a function v of x such that this function is positive and zero at your fixed point and that the derivative of this quantity of this function on the trajectories is always negative then you can say this is a theorem due to Lyapunov that your fixed point of course is stable in fact you can do this is a steepest center behavior of your trajectories sooner or later you are arrived at this point and the problem here is to construct this function and there are some technique but of course there is no general there is no general strategy to construct the Lyapunov function and I think I finished ok what else I have just another ok this is tomorrow ok I finish I stop here for questions comments so just to summarize linear stability is very useful but linear stability has some some restriction the problem is that I would like to understand something that is due to the fact that there is also non linear stability but non linear stability is not so general the protocol of linear stability is very simple linearization, diagonalization studying of again vector and again values the program is very simple you can do on a computer if you want instead to study the stability when the linear stability fails you soon run in troubles you should be very clever to use one of this technique looking for example the existence of limit cycle using the Lyapunov stability using criteria some criteria that allows you to decide if a point is stable or not ok finished, thank you