 Let us recall what we are trying to do in the previous lecture. So, we are considering this autonomous system f of x. So, let us call this equation 1. So, we are going to refer to this for the whole lecture. So, in expanded form, we have this x j dot is equal to f j x x n and these are n equations for e j. And we are defining orbit through orbit positive orbit through a given point x 0. The other terminology for orbit is a synonyms are also called trajectory. This terminology is also used. So, recall how did you define for example, the orbit. So, similar remarks hold for positive orbit. So, we started with a solution. So, let x be a solution of one passing through x 0 at t 0. What does this mean? That is x t 0 is x 0 and x satisfy the differential system x naught equal to f of x or in expanded form is component of x satisfies these equations. And then the orbit was defined. So, then the orbit of x 0 is just set of all x in R n such that x equal to x t. So, I am using all x x x, but it should be clear from the context. So, what I am just taking is all this I am varying t for every t I just pick up that x of t coming from the solution. So, a natural question arises, natural doubt arises. What if I have another solution of one passing through x naught at a different time and then I use that solution to define the orbit. Do I get a different set and obviously I should not get and let me state this as a lemma and so it does not depend on the particular solution used for defining the orbit. So, let me state it as a lemma 1 prime. So, let x t let me write that t y t be solutions of 1 satisfying x of t 0 t is of 0 x 0 and y of t 1 is equal to. So, both the solutions are passing through the given point x 0 may be at different times if they pass at pass through x naught at the same time then by uniqueness they must be same. So, they might pass at different times for some t 0 then the orbit x naught that is our purpose defined. So, this is important using the solutions x and y. So, I will not get a different set. So, that is important. So, that remove the ambiguity. So, I can just use any solution passing through x naught. So, that time is not important. So, the only that solution has to pass through that given point that is important. So, this will remove that ambiguity. So, no special attention is given to that time t naught only the solution. So, this proof is very quick. So, define z t is x of t t 0 minus. So, recall that for the autonomous system earlier we saw lemma 1 that if x t is a solution then x of t plus c for any constant is also a solution. So, that defines this that prove that this z t is also a solution. So, lemma 1 z is a solution. So, moreover now you compute z of t 1 also z of t 1. So, you substitute t equal to t 1 in this expression. So, you get x of t 0 and that is x 0, but x 0 is also equal to y of t. So, that is the hypothesis you see that is the hypothesis. So, now by uniqueness. So, this is important. So, all the time we use this thing. So, by uniqueness of the initial value problem uniqueness of I v p z t is equal to y t. So, the solutions z and y are the same because they have the same initial value at t equal to t 1. So, this condition implies that. So, uniqueness is a important thing. So, all the time we use that. Now, look at look at z look at y. So, z t is given by this expression. So, therefore, so this is an important thing for autonomous systems and it is not true in general for non-autonomous systems x of t plus t 0 sorry for that t 0 plus t 1 for all t. So, though x and y are two different solutions the only common thing they have they pass through the same given point x 0 at different timings they are related by this equation. So, they are not really different. So, similarly you can use y or you can just see here. So, similarly you get x of t y of t. Now, you have to just change these arguments carefully. So, you have t plus t 1. So, therefore, so if you now go back to the definition of the orbit. So, what is orbit? So, I use that orbit through x 0 I just use that solution passing through x 0 and collect all those points. So, therefore, if x is equal to x t or x tilde let me for some t tilde then x is also equal to just look at the relation between x tilde and x x of t tilde and y of t tilde you get that y of t tilde plus t 1 and this entire thing I might just call it. And similarly if x is equal to y of some t tilde I can get x is equal to x of t tilde. So, therefore, this completes therefore, O of x naught is unambiguously. So, let me stress what does that mean again. So, that means I just take any solution passing through x naught any solution of 1 system 1 passing through x naught and time is not important at all. I get the same set and similar you can also now work out and just look at the definition of the positive orbit. So, we will also get the same similarly. So, this positive orbit. So, what to all need is a solution passing through x naught. So, that simple clarification and now immediately we see that. So, this is a note remark. So, if x 1 belongs to an orbit passing through x 0. So, that means there is a solution I am using here in the definition of this orbit. So, there is a solution passing through x naught at some time that is not important and the same solution will also be now passing through x 1. So, I can use that same solution to define orbit passing through x 1. Since it is the same solution. So, I immediately get that x 0 is 0 of x 1. So, in lemma 1 prime we have seen that it is only a solution passing through a given point that is important to define the orbit through that point. So, since the same solution that passes through x naught also passes through x 1 when x 1 is in the orbit of x naught. So, I can use the same solution and that implies immediately that orbit passing through x naught and orbit passing through x 1 are the same. Whenever x 1 lies on the orbit of x naught and that also prove that x naught lies on the orbit of x naught x passing through x 1. So, this implies. So, it is a simpler proof last time I stated that. So, lemma 2 let me again recall that. So, if we take 2 points in R n x 0 x 1 in R n and then consider the orbits passing through x naught and x 1. So, then x 1 or e m t. So, this is m t. So, it says that any 2 orbits of the autonomous system 1 again. So, remember that everything is in reference to the autonomous system 1 x naught equal to f of x. So, that is important. So, everything. So, if you have 2 orbits either they are same or they are disjoint. So, there is nothing like no they meet at one point and they then they are not equal. So, if they intersect at one point then they must be equal. So, and this is now a simple proof. So, let me just. So, if x tilde belongs to o x 0. So, suppose they are intersection is non empty. So, if there is intersection empty then we do not have to prove anything. So, if suppose there is a common point between them then you look at this note. So, then since x tilde belongs to the orbit of orbit passing through x naught. So, I have this is equal to x tilde and x tilde also lies on the orbit passing through x 1 and that is equal to. So, if they have a common point then they must be equal. So, that is. So, now the question is the previous lemma says about 2 orbits. What about the same orbit can it intersect itself. So, this is the content of the next lemma. So, let me write it. So, if suppose x t is a solution of one satisfying. So, this is the condition I am. So, suppose that the orbit defined by the solution x intersects itself satisfying x t 0 equal to x t 0 plus t for some t 0 in R and t positive. So, that t 0 and t 0 plus t are two distinct points and suppose that intersects then. So, this is again a remarkable property. So, if it intersects at one point then it has to do that thing for all time not just for that one point. So, again let me stress that one is an autonomous system and these properties heavily use that fact. So, in general these properties of lemma 1 2 etcetera are not satisfied by non-autonomous system. So, this we have to bear it in mind. Again proof is very simple. More than proof the statement of the lemmas are important. So, again define. So, you can see the repeated use of the same set of ideas. So, x sub t of t is equal to x of t plus t and again by lemma 1. So, this is just a fixed translation t is some fixed capital t is some fixed number. So, this is just a translation. So, x t is a solution of one and look at now x t. So, look at x t at t naught this by definition is t 0 plus t and by hypothesis now look at the hypothesis. So, this is x of t. So, here we have two solutions of one x t and x and they have the same initial data at t equal to t 0 again by uniqueness by uniqueness of t is equal to x of t for all t and this is same as saying that x of t. So, that is and complete the proof. So, the solution remark the solution in lemma 3 is called a periodic solution a period. So, let me stress that a period this. So, if t is a period a period so is n t or n equal to 1 2. So, we will pick up the one which is least. So, the smallest t smallest t positive satisfying the hypothesis the period of that solution. So, these are some terminologies. So, we will see the importance of this periodic solution little later as we go along. So, another concept now we will introduce another concept. So, this is the equilibrium points or solutions. So, this is our next. So, definition. So, you pick a point now I generally use this x bar x bar is called equilibrium point or solution of 1. So, whatever the concept we are doing all refers to that 1 that is remember that if f of x bar is 0. So, let me explain. So, in case of when n is bigger than 1. So, this consists of so this let me explain this what that. So, let x bar be x 1 bar x n bar. So, this is in R n and let me expand the system of equations. So, that means I have f 1 of x 1 bar x n bar is 0 etcetera f n. So, that means the components of x bar this x 1 x 1 bar x 2 bar x n bar are the solutions of the simultaneous equations f 1 bar x 1 bar etcetera 0. So, in case of f 1 is linear. So, this is just a system of algebraic equations if f 1 are any of the f f as a non-linear. So, in many examples we consider them as polynomials then we have a set of non-linear equations and we are seeking a common 0 there may be 1 there may be many. So, you will see in examples. So, this you should remember this f x bar means this components x 1 bar x 2 bar x n bar they satisfy this system of equations that you should remember. So, another interpretation of this equilibrium points thus equilibrium points the constant solutions at stress time. So, constant solutions means it does not change with time. So, if x does not change with time then the derivative is 0 that means f of x is 0. So, just look at the condition here x naught equal to f x either 1 equation or more equation. So, when x is a constant solution that d x by d t is 0. So, that means f of x is 0 and conversely if it is an equilibrium point then f of x is 0 there and then we get that constant solution. So, equilibrium points are nothing, but the constant solutions of 1 they are solutions that is important. So, if so now immediately you see that if x bar is an equilibrium point and now I want to look at the orbit of that x bar since that itself is a solution. So, immediately obtain that x bar is positive orbit of course, y only positive orbit and also consider negative orbit, but we are interested in the future. So, that is why we consider the positive orbit. So, here is a situation. So, where we can immediately determine the entire orbit the positive orbit they are all equal to the single term set x bar. So, that is the beauty of this equilibrium points. So, in this case we can immediately determine its orbits. So, now another important thing that is connected with equilibrium point is this following lemmas. So, let me do this thing and then we go to examples. So, we also see the significance of this equilibrium points and what happens if the system does not have equilibrium points. See for example, this f may never be 0 then that is particular system will not have any equilibrium points. So, we will see through some examples. So, what is that significance and when there are equilibrium points what is their significance we will see all these things in detail. So, before that so if x bar is an equilibrium point then we know its orbit precisely. So, if I have a non-trivial orbit what does that mean? So, it is an orbit not passing through the equilibrium point then that cannot intersect this orbit. So, that is because we have seen in lemma 2 that either 2 orbits are the same or they are disjoint all the time. So, if I have a solution not it cannot pass through this thing because again uniqueness because that is the only solution passing through the given equilibrium point. So, if I start with another solution that orbit will never intersect this equilibrium point, but it can approach. So, that is what the lemma says. So, let x t be a solution of 1 such that limit x t as t tends to infinity exists. When we say exists so we always mean a finite number and equals x bar. So, this is in R n. So, let me again write it component wise. So, that is so if you take the j th component of the solution this is x bar. So, this is bar. So, that I am assuming this solution x t has this limit. So, then necessarily then x bar. So, if a solution has a finite limit as t goes to infinity. So, this also you can do it for t tending to minus infinity. So, necessarily the limit is an equilibrium point. So, now just for curiosity you can just argue that suppose the system does not have an equilibrium point. So, that means no solution can have this finite limit. So, in most of the cases in the absence of equilibrium points the all the solutions of the given system will be unbounded. We will see through some examples. So, that is one significance of the equilibrium points. So, let me just give a quick proof of this. So, let me write again component wise. So, fix h greater than 0 then x of t plus h is also a solution. So, if you are repeatedly using that lemma 1 you see that. So, that is an important thing is also a and as t tends to infinity t plus h also tends to infinity. So, therefore, I have limit x t plus h as t tends to infinity is also x bar. And since x bar is finite the difference if I take the difference. So, therefore, this is not of the infinity minus infinity form. So, we can separate the limit. So, this is calculus. So, since both are x bar and they are finite. So, this is 0. So, this is not of the. So, I can write this as limit of x t plus h minus limit x t because both are finite. And now look at this component wise. So, let me write this component wise for each j equal to 1 to n and u the mean value theorem each x j is a differentiable function. So, by mean value theorem. So, this is just x h x j dot at some point. So, this is theta j, but x is not an arbitrary function it is a solution of 1 x is a solution of 1. So, by using the equation. So, this is f of x of because x satisfies equation 1 system 1. And now you see look at this limit. So, I am using this limit star. So, the left hand side this one goes to 0 as t goes to infinity that is from star that we are seen it. And therefore, since h is positive. So, I conclude that therefore, limit f of f j for each j to infinity is 0. And now the continuity. So, this is again part of the preliminaries in analysis. So, this each f j is continuous function x is a continuous function. So, I can push this limit inside here. And so this limit let me just write that f j. So, that means so this implies f j x bar is 0 and this is true for all j. So, that implies. So, with these few preliminaries now we consider some examples. So, first simplest thing just consider this 1 d. So, most of the examples are 1 d examples. So, except perhaps the last one if I am able to come to that thing that will be a 2 d example. So, this is 1 d that means just a single equation. And so f of x is x square here. And so the 0's of the that function is just here in this case it is only 0. So, x bar equal to 0 is the only equivalent. So, given so we can also write the explicit solution. So, you should complete all these steps then this x t is a solution x of 0 is. So, for most calculation we just take t 0 equal to 0. But without this using this solution I would like to calculate the orbits positive orbits for arbitrary x naught in R. So, for that just draw this line and here is your equilibrium point. So, the orbit of that equilibrium point is obviously that singleton. So, any orbit starting here. So, automatically it has to stay there and any orbit starting in the left it will stay here because it cannot intersect with that orbit. So, orbit so this if x 0 is 0 then this orbit. So, any other orbit cannot intersect. So, if x 0 is positive let us see what happens. So, let me use a different color here. So, x 0 here and without the knowledge of the solution let us try to see how the solution moves. This shows some simple calculus would help us in most of the 1 d problems. So, look at the equation it says d x by d t is equal to x square and x square is always non negative. So, derivative of that solution is non negative that means the solution is always increasing with time. So, if I start here the solution has to only move to the right. Simple calculus I am not using anything here. So, using that fact so let me write here. So, you just see that x 0 is 0. The positive orbit and since it will stay here and it keep on increasing we can just conclude that this x t of course, it will not exist in this case particular case for all there is a t star. So, let me when this denominator become 0. So, it is infinity the solution is unbounded here. So, because it is the orbit is this entire 0 infinity the solutions in this case are unbounded. So, now look at a point here and the negative so x naught is again the solution is increasing. So, there is no because x naught is x square again it is increasing. So, it will move in this direction and there is a barrier here. So, it cannot cross that thing. So, at the most it can reach that in infinite time. So, in this case you verify that x 0 is minus infinity 0 0 and limit x t and in this case it exists for all time here. So, verify this facts. So, this just follows from this simple analysis. So, I do not really require this explicit solution. So, before moving to the next example. So, let me just in remaining 10 minutes. So, let me just write this calculus lemma. So, it is very easy to prove we will include in the we will discuss in the discussion on preliminaries on analysis. So, let chi be a real valued function defined on an interval a b and this could be entire r it could be finite infinite it does not matter semi infinite whatever any arbitrary interval of the real line and chi is a real valued function defined on that interval satisfy either. So, one of the two conditions. So, let me chi is bounded above is non-decrease chi is bounded below chi is non-increase then limit x t. So, again let me recall that when we say that limit exists it is a finite real number. It is very easy to prove in the first case this limit will be the supremum of chi t and in the second case it will be infimum of chi t. So, we will discuss more about this in when we discuss preliminaries. So, using this thing. So, let me now consider second example. So, this is logistic model of population growth. So, again 1 d. So, x dot equal to x into 1 minus x. So, usually there are some constants here. So, here I have just normalized and assume that the maximum plausible population is 1. Now, you can 1 can normalize that otherwise you can also put numbers that is no problem. So, in fact in the discussion on physical models you see actual numbers, but it does not matter. So, if. So, again here one can write down the explicit solution, but without that explicit solution I would like to draw the arbitrary through arbitrary points. So, let me just. So, since it is a population model we are interested only in positive solutions and see whether we get positive solutions if we start initially at positive initial data. So, here the equilibrium there are two equilibrium points. So, equilibrium points 0 and 1. So, look at the place where this right hand side is 0 that is x equal to 0 or x equal to 1. So, then by the discussion on lemma 1, lemma 2 etcetera. So, if I start a solution here its orbit cannot go beyond this, because that will intersect the orbit of this equilibrium point 0 and equilibrium point at 1. So, it cannot intersect. So, if I have an x 0 here the solution always remains in this interval 0 1 and similarly if I start a 0 here the orbit will always remain there. So, let me just do that thing. So, if 0 x 0 less than 1 and again look at the equation. Equation says x dot is equal to x into 1 minus x and in this case we have already seen that the solution lies between 0 and 1 again. So, then the corresponding solution lies in 0 1 and then you look at the equation and from the equation we see that the solution is increasing or at least not decreasing x into 1 minus x x always lies between 0 and 1. So, that product is positive. So, x dot is positive. So, x is increasing and in the other case. So, similarly if x 0 is bigger than 1 then the corresponding solution lies in 1 infinity and is decreasing. Now, if you look at x into 1 minus x, x is bigger than 1. So, the second factor is negative. So, it is decreasing. Simple calculus we are not using anything else here just simple calculus. So, if x 0 belongs to 0 1 now you can write down the orbits. So, the orbit 0 1 and the positive orbit is more important for us x 0 1 because it is increasing that is because it is increasing. So, if x 0 is bigger than 1 then the orbit in infinity and the positive orbit now it is decreasing. So, you see that 1. So, with this thing we will continue in the next lecture. So, what happens to the limit of this thing? So, go and do all this whatever gap is there. So, try to fill in all those gaps and we will see in the next lecture. Thank you.