 Welcome back, recall that in the last class we were discussing linearization process and linear stability analysis of non-linear systems. So, let me just recall what we are doing. So, this was our autonomous system and x bar isolated. So, that is standing assumption equilibrium point and then the linearized system is given by y dot equal to d f x bar y and this is the Jacobian of the vector f at x bar. So, it is an n by n matrix. So, when we analyze this one, we called it linear stability analysis of this non-linear problem. So, we are doing examples. So, let me again recall that Duffing's equation or it is also called Duffing oscillator. So, after some simplification, we reduce the equation with many parameters to only one parameter. So, and this is second order equation x double dot. So, second order equation, so x is a real valued function x dot plus or minus x plus x cube plus delta x dot is equal to 0 delta is positive. So, let me just concentrate on the negative sign and the positive sign is similar. In fact, there is only one equilibrium point in that case. So, let me write that negative sign as a system. So, this is x dot equal to y and y dot equal to if I take negative sign. So, this is x minus x cube minus delta y. So, the equilibrium points, so if you compute, so the first equation gives you y equal to 0 and so this is from first equation and second equation then implies x into 0 x equal to 0 plus or minus 1. So, therefore, in this case we have three equilibrium points 0 0 plus or minus 1 0 are the equilibrium points and now let us calculate the Jacobian at these three points. So, let me just calculate the general Jacobian at a general point x y. So, if you this is the now 2 by 2 matrix, we are just at two dimensional system. So, if you take the first function, so I differentiate with respect. So, first again go back. So, if you go back, so you see that the first equation contains only y and second equation is x minus x cube minus delta y. So, this will be 0 1 1 minus 3 x square minus delta. So, at 0 0 let us calculate the Jacobian. So, d f 0 0, so this is just 0 1 1 minus delta. So, the Eigen values you can compute them are half minus delta plus or minus square root of delta square plus 4. You see that for all delta non-negative, the Eigen values are real of opposite signs. So, 1 is positive and 1 is negative. So, we conclude that from the linear stability analysis. So, we invoke all the linear theory here. So, 0 0 is unstable. In this case, we also call it a saddle point. When the Eigen values are real and of opposite sign, the unstable equilibrium point is called a saddle point. And add to plus or minus 1 since the Jacobian has x square term. So, it is same. So, d f plus or minus 1 0 is just 0 1. Now, it is minus 2 and minus 1 minus 3 x square. So, that gives us minus 2. So, the Eigen values here easily computable 2 by 2 matrix. So, no problem. Eigen values are half minus delta plus or minus delta square minus 8. So, this in case of 0 0, you see that it is plus 4 and here it is minus 8 and that makes it difference. So, for all delta positive, the Eigen values negative real either they themselves are negative real numbers or certainly they have negative real parts. So, therefore, plus or minus 1 0 are asymptotically stable linearly asymptotically stable delta. So, what happens when delta is 0? When delta is 0, the Eigen values plus or minus. So, therefore, plus or minus in that case is both are same or linearly stable not asymptotically stable because here the real parts are 0. They are purely imaginary. So, this is again from the linear thing when delta is 0. So, delta equal to 0 also falls this case in this case the equation itself equation is in conservative form and we will study these things little later in detail. So, at that time again I will recall this. So, when delta equal to 0 the equation is referred to as undamped unforced Duffing equation. Duffing equation also referred that if you recall that non-linear term x cube is referred to as cubic stiffness term and that is used to describe the hardening spring effect observed in many mechanical systems. So, this is to describe hardening spring effect observed in many mechanical systems. So, with this we know to the next example. So, this is another important equation that is studied extensively and it has also given rise to new mathematics in the theory of non-linear dynamics. So, this is called Van der Pol equation or again oscillator. So, Van der Pol in the years around 1927 when he was working for the Philips company in Netherlands extensively studied this equation both theoretically as well as experimentally using electrical circuits. So, this is one of the again this is a second order equation. So, given by x double dot plus mu x square minus 1 x dot. So, this is unforced. So, there are there have been many studies you have to one with periodic forcing term I will give some reference. So, when mu 0 you see that this term vanishes and you get back our simple harmonic oscillator. So, when mu is equal to 0 this leads to linear harmonic oscillator simple harmonic oscillator. So, this non-linear term is added to that and see whether you still get periodic solutions. Even mu equal to 0 certainly we have periodic solutions and would like to see whether mu non equal to 0 also produces periodic solutions. So, you can imagine when mu is negative. So, that is interesting case. So, do the linear stability analysis for this also when mu is less than 0 and x is small x is small. So, this if you compare with the spring mass dash pot system. So, this is the damping term, but this is non-linear damping. So, this is non-linear damping. So, in spring mass dash pot system this was a constant, but here we have this non-linear this coefficient depends on the solution itself. So, in the engineering parlance this is called the oscillator is driven when x is small we will see what that means mathematically and damped or slowed down when x is large. So, when x is small means this x square minus 1 is negative and I have mu negative. So, this whole thing will become positive and that will produce exponential terms with positive terms. So, that is so there are large oscillations, but when x square is bigger than 1 then with mu negative this also becomes negative and that acts as damping and the oscillations will be slowed down. So, effectively what we expect is in this case. So, we expect this figure. So, we see that we will check that 0 0 is the only equilibrium point in this case whatever may be mu. So, this when you start the solution near 0. So, it just moves around and try to approach a periodic solution. So, let me just take. So, this is periodic solution and again when you start from far off that also try to come and approach this. And we will see later that this is the case all solutions starting at the different from different point then the equilibrium point all approach a periodic solution and that is indicated as a red circle here not circle, but some closed orbit we will see that later. So, let us do the linear stability analysis very simple here no I am not restricting mu to be negative. So, these two examples are important for us Duffing equation and van der Paul equation. So, again write as a system. So, x dot equal to y and y dot equal to minus mu x square minus 1 x dot is y and I have minus x. So, 0 if you solve again the right hand side equal to 0. So, first equation gives me y equal to 0 and if I put that in the second equation that gives me x equal to 0 whatever may be mu 0 0 is the only equilibrium point. So, hence isolated. So, even in the previous case they are all isolated equilibrium points. So, 0 0 is the only and if I calculate the Jacobian that is again simple here d of 0 0 and this is again 0 1 minus 1. So, the eigenvalues are given by 1 by 2 mu square minus 4. So, you see that when mu is positive. So, this always dominates and so you will have both the eigenvalues. So, this is the positive real parts. So, hence unstable. So, mu equal to 0 the equation itself is linear the equation is linear and it is 0 0 is stable, but not as simple to less than 0. So, the real parts real parts are negative and hence 0 0 is asymptotically stable. So, as I said in the beginning mu is positive that is interesting case and we will see that little later. So, let me now describe a situation when the linear stability analysis imply the non-linear stability. So, that means if linear system is stable then the non-linear system is also stable and if linear system is unstable then the non-linear system is also unstable and this goes by the name Hartmann-Gromman theorem. I will just explain this only in English. I will not even write down the precise terminology I just explain with only words. So, the situation is x bar is a hyperbolic equilibrium. So, last time we define this. So, let me again recall what that means is this d f x the Jacobian of f at x bar. So, this is a matrix has Eigen values all having non-zero real parts. So, this is important non-zero real parts. So, in this context when x is a hyperbolic x bar is a hyperbolic equilibrium point the Hartmann-Gromman theorem says that. So, you have this non-linear system and you have the linearized system. So, you just work here orbits around x bar the orbits and here also you this is 0 and you take orbits of this linear system here and then to compare these two. So, you just add x bar here and the orbits here. So, orbits around x bar and orbits around 0 of this linear linearized system add x bar to that. So, they are linked by a homeomorphism. So, it is an interesting and important theorem locally describes around a hyperbolic equilibrium point. So, the orbits of the non-linear system can be gotten from the linear system and vice versa. So, this is an important result and this happens only around a hyperbolic. A somewhat simpler one and older than this Hartmann-Gromman theorem referred to as Perron's theorem though it is not as precise as the Hartmann-Gromman theorem it gives us some sufficient conditions again for comparison of non-linear systems orbits of non-linear systems and linearized system. So, let me state this and give a proof though it is technical since we have developed all the technique technicalities in the linear algebra portion it is very easy to give a proof of this. So, let me just state this. So, we have the system a linear system perturbed by a non-linear. So, this f is different. So, this is a system. So, the hypothesis on a so a has eigenvalues all with negative real parts. So, Hartmann-Gromman theorem requires the eigenvalues all to have non-zero real parts, but here the Perron's theorem it only concerns about the eigenvalues when they all have negative real parts and f is a continuous function. Let me not bother about where it is defined, but it will be defined in a neighborhood of 0 and this is important hypothesis and this indicates small s. So, this is little o of as mod x tends to 0. So, uniformly n t I will clarify this little later. So, what does this mean is this limit mod f t x mod x mod x tends to 0 is 0 uniformly in t. So, that means this limit process does not depend on t that is uniformity. So, the usual epsilon deltas that appear in the definition of this limit they do not depend on t. So, that is what meant by this uniformly in t then 0 solution is asymptotically stable. So, again read the definition of asymptotic stability carefully stable for star put this star. So, that means if I start a solution of star near the origin, but not at the origin because origin is always a solution. I would like to show that the solution exists for all time and the solution tends to 0 as t goes to infinity. There are two steps. So, proof let me just indicate a proof of this and in our case when we want to apply to the linear stability analysis f t x does not depend on this is just a quadratic in x. So, it does not depend on t. So, automatically this condition is satisfied. So, there is no problem. So, this Perron theorem implies that 0 is asymptotically stable when the linearized problem has Eigen the matrix the Jacobian matrix in the linearized problem has Eigen values all with negative real part. So, let me just give a proof of this. So, you see some interplay between analysis and linear algebra and differential equations of course. So, proof of so the local existence theorem local in time theorem implies. So, whatever may be my x 0. So, given x 0 there exists a solution x t. So, let me indicate that x t of star with x 0 again that is for some time t in 0 t star. So, t star is possible. So, that is always guaranteed. What we would like to do now is we want to show that when this x 0 is small that the solution can be extended to all t positive that is the first step. And once we show that thing we are not interested in weakness here that is why there are no hypothesis on little f in star. So, this is by Pano existence theorem that there is always a solution. So, there are two steps. So, claim if x 0 is small. So, remember this is the standard Euclidean norm. So, this x 0 is a vector in R n. So, if this is sufficiently small then x t can be continued. So, this is also part of the existence theory we have done earlier when is it possible to continue a solution which exists for a short time to all times continued for all t and then. So, this is one and then limit x t at t tends to infinity equal to 0. So, both these things prove the results. So, 1 and 2 imply the theorem. And let me just stress as a remark this is not trivial. So, remark 1 is not. So, it does not follow automatically. So, we already seen an example where the solution does not exist for all time. So, for example, x dot equal to x square. So, no matter what x 0 is if x 0 is positive. However, small it is that does not matter the solution exists the solution in x t exists only in 0 1 by x 0. We have already seen that of course, this does not this equation does not fall under star because there is a linear term here. So, that linear term is 0 here. So, that is that a part the hypothesis on the Eigen values of a plays a crucial role in order to prove claim 1. For example, so if I want to compare this example with star. So, I take x dot equal to minus mu x plus x square and I take mu positive. So, this falls this is similar to star equation in 1 d. Then you can as an exercise check that x the solution x t exists for all t non-negative provided. So, this is the smallness as in what we say in the theorem provided 0 x 0 is less than. So, when once you exceed x 0 bigger than mu you can also check that the solution does not exist for all time. So, that is crucial. So, remark 1 this claim 1 is not trivial. So, this remark explains that. So, now you want to exploit the hypothesis on a namely the Eigen values of a all have negative real part in order to show that the solution exists for all time. The local solution the solution which I know exists for short time that 0 t star x t satisfies. So, this again you recall we have proved the existence by converting the given differential equation into an integral equation and I am going to write the same thing x t satisfies the relation equation x t is equal to a to the t a x 0. This comes from the homogeneous part x dot equal to a x and then for the inhomogeneous part you have this non-linear integral e t minus s a f of s x s. So, again remember this t is only for short. So, whenever that a t is a solution of the differential equation it can always be written in this. And one more remark here it is remarkable that the same technique works even for infinite dimensional systems where the matrix A will be replaced by a differential operator with good spectral properties. What are the good spectral properties like here the matrix A has all the Eigen values with negative real part. And similar to that if you assume the same process without any change even works for infinite dimensional systems. So, this is a powerful technique the methodology in Perron's theorem the proof of Perron's theorem is a powerful technique which is used even in other situations. So, now the hypothesis. So, let us by hypothesis. So, in the previous thing that e to the t a is the exponential matrix which you have defined in the linear algebra portion and I am going to use some more properties from that linear algebra. So, by the hypothesis on a. So, this will show in detail in linear algebra portion it follow that. So, e to the t a. So, this is matrix norm which we have introduced in the linear algebra part. So, this is less than or equal to some constant positive constant k e to the minus sigma for k positive and sigma this is important sigma positive. Let me just briefly explain where do these two constants come from. So, this essentially comes from the Jordan canonical form which we have already discussed in linear algebra portion. So, when we use the Jordan canonical form of a. So, this constant k is produced and sigma comes from the Eigen values. So, sigma is essentially half you can put any fraction there. So, minimum of minus of real part of lambda Eigen value and you see that is where the this. So, this is spectral property of a. So, Eigen values describe the spectrum. So, this comes from the knowledge of the Eigen values of a and by our hypothesis this minus real part of lambda are all positive and I am just taking there are only finitely many Eigen values and they are all this real part is now minus real part is positive. So, this sigma is positive. So, half you can replace by any fraction. So, that is no problem. So, this is a very crucial step in the proof of Perron's theorem. So, this estimate is very crucial and this we have done in linear algebra. Once you have this thing now I invoke the hypothesis on invoke the hypothesis I want to now write it in terms of epsilon and delta. So, given epsilon positive there exist delta positive such that whenever mod x less than equal to delta we have f of t x is less than or equal to epsilon by k. So, this is for a technical reason that k I wanted using it mod x uniformly in t. So, just this means again as I said earlier this delta does not depend on t. So, this does not depend on t. So, when f itself is independent of t. So, that uniformly in t does not arise, but when t is there. So, we want this delta to be independent of t. So, that we can just concentrate only on the x variable. So, now let me call this as 1 and this as 2. So, these 2 estimates play a crucial role. So, again go back to this. So, this solution satisfy this integral equation as t minus a s a and f of s x s. So, this implies. So, remember this value is only for a short t our aim is to extend this 2 for all t. So, taking norm and using estimate 1. So, this gives me minus sigma t x 0. So, this is Euclidean norm plus 0 to t e to the minus sigma t minus s and this f of. And now I want to use the hypothesis on f and for that I require only say if I want to use that estimate 2 on this thing I require that mod x t is less than or equal to delta and that I assume that for the time being and I will show that. So, this is less than or equal to epsilon by k x s provided norm x t is less than or equal to delta for 0 I will show that. So, if you know put all these things together in this inequality. So, what I get is e to the sigma t. So, I take that this side is less than or equal to. So, there is a k here. So, that is k x 0. So, this k cancels. So, I have that just. So, I take that e to the sigma t other side. So, what I am left is just e to the sigma s there is an epsilon here mod x s. So, provided this again I. So, this is valid provided that. But now look at this inequality. So, concentrate on this function the same function appears on the left hand side and this in integral and this is the situation where we can apply the Gronwall's inequality. So, by. So, this is also we have already seen the importance of Gronwall's inequality in proving uniqueness e to the sigma t x t is less than or equal to k x 0 e to the epsilon t. So, that epsilon comes there. So, if we choose now if 0 less than epsilon less than sigma, then we have x t less than or equal to minus sigma minus epsilon and now this with assumption on epsilon and this is just k x 0. So, this is the crucial inequality we have obtained provided see all these things you have to remember that all these things are derived provided x t is less than or equal to delta. So, we have to somehow now assure that that x t will be remain less than or equal to delta and now that is easy. So, now the hypothesis comes into picture. So, therefore, if mod x 0 is less than or equal to delta by k. So, this is the smallest assumption that is indicated in the theorem, then if you look at the previous inequality for all. So, indeed the solution has remained less than or equal to delta. So, all the steps we have derived are let me make and they are valid. So, hence the solution. So, this is the bound the solution x t can be continued. So, that means it exists continued and this is the smallest assumption. So, remember those this is just independent of. So, they are just come from essentially on the hypothesis of little f and on the matrix A and then we also have this k x 0 e to the minus sigma minus epsilon t. So, we have already chosen that epsilon is less than sigma. So, that goes to 0 as t goes to 0. So, thus we have proved both our claims 1 and 2 and that completes the proof of the Perron's theorem. So, what we have learnt? So, this the hypothesis in Perron's theorem is also a hypothesis on the equilibrium point x bar. So, A will be d f x bar. So, it all has negative real part the Eigen values have negative real part. So, in case of. So, this is a in case of hyperbolic equilibrium points. So, this is important the linear stability implies stability analysis implies non-linear stability. So, if the equilibrium point is hyperbolic then the linear stability analysis will be sufficient to conclude the non-linear stability analysis and linear stability analysis is much easier as we because we have explicit formulas and other things. So, the only case that will be left out is the case that is left out is that of non-hyperbolic. So, that means the Jacobian matrix now will have Eigen values with 0 real part and that we have seen through examples that the linear stability analysis may or may not imply the non-linear stability. We have seen example where in case of non-hyperbolic equilibrium points the equilibrium point may be stable in the linear linearization, but unstable or even asymptotically stable in non-linear case and this one is more effectively handled by the Lyapunov function and that will be the topic of our next discussion next class. So, the Hartmann-Grobbman theorem and the Perron's theorem take care of the case of hyperbolic equilibrium points and the non-hyperbolic equilibrium point case will be handled by Lyapunov functions. Thank you.