 Welcome back. So, we will continue the discussion in the previous class on stability analysis of non-linear autonomous systems. And last time we saw that at the hyperbolic equilibrium points the linear stability analysis implies the non-linear stability analysis. So, we may now concentrate on non-hyperbolic equilibrium points and this will be handled by the use of Lyapunov function. It is a very simple tool, but very powerful tool as we see it. So, before I go into the details. So, let me just mention a reference Takashi-Kanomoro. There is a nice write up by Takashi-Kanomoro on Duffing and Van der Poel oscillators. So, this is available in the web, it is scholarpedia.org. So, you can get much more information on this Duffing and Van der Poel oscillators. Last time we discussed little bit about that. So, I can refer to this write up by Takashi-Kanomoro. So, now we move on. So, we introduce the Lyapunov function. Definition is very simple. So, again consider the autonomous system x dot equal to f x. So, this is our standard system. So, system and let x bar and isolated. So, again we will be studying the stability of orbits of this system around this equilibrium point. So, we may assume. So, this is our here after we do this just for some simplification in writing we may assume x bar equal to 0. This we can always achieve by translating this system into a neighborhood of x bar. So, you can just check that. So, that is not this is no general is the loss of generality. So, just stick to this thing and let omega be a neighborhood n b d n b neighborhood of 0 which does not contain any other equilibrium point. So, this is possible because we are assuming 0 is isolated. So, a function. So, this is a definition a function. So, all our analysis now is restricted to that neighborhood omega. So, we define functions and other objects on that omega. So, v from omega to r. So, real valued function defined on that neighborhood of the origin omega is called positive definite respectively negative definite. So, any real valued function defined on omega respectively negative definite. So, this is definition v x is positive respectively v x negative for all x in omega and x naught 0 and v of 0 each. So, this is. So, I am not putting any other restrictions like continuity and other thing that those things will come as you go on. So, this is a real valued function which takes only positive values except at the 0 or negative value if it is negative definite. So, now definition of the Lyapunov function. So, this is very useful tool and it has been even adopted in other situations a function v from omega to r which is c 1. So, it is continuous differentiable and derivatives also continuous all partial derivatives is called a Lyapunov function for the system 1. So, that is where the system now enters otherwise just it is a real valued function right. So, what is the connection with one if so, two conditions right one v is positive definite in omega. So, that is the previous definition and second one that is connection with the system 1. So, del v dot f. So, del v is gradient of v. So, this is by definition. So, this is f is coming from the system. So, this is nothing but del v by del x 1 into f 1 plus etcetera. So, we are in n dimensional situation. So, remember v is a function from omega to r. So, it is a function of n variables. So, now this is a real valued function for all x in omega. So, this is less than or equal to 0. So, at all points in omega we require that this gradient v dot d f. So, this is scalar product and that is what I am defining. So, this is a vector and f is also vector grad v is a vector. So, this is the vector del v by del x 1 etcetera del v by del x n and f is the vector f 1 f 2 f n. So, I am taking the scalar product of those two vectors at every point in omega and we require that this is less than or equal to 0. So, that is where the given function is related to the system via the second condition. So, now theorem. So, these concepts were introduced by Lyapunov. This is stability theorem. So, suppose one the system one possesses a Lyapunov function 0 is stable. So, remember again by our notation 0 is an equilibrium point of 1 and so as and when the system one possesses a Lyapunov function 0 is stable if in addition. So, now we strengthen little bit the second condition grad v dot f. So, again look at the definition grad v dot f. So, that is again a function defined on omega to the real number. So, this is a real valued function defined on omega I want this is negative definite 0 is asymptotically steady. So, now you clearly see what role the Lyapunov function plays as far as stability and asymptotic stability of any equilibrium point of the given system place. So, that is the importance of Lyapunov function and proofs are very easy and they use simple calculus. It is not very deep analysis, but you can already see the interplay between analysis and the differential equation. So, I have stress this point again and again at many situations and here is another situation where you see that interplay very clearly. So, proof. So, again you go back to the definition of stability and asymptotic stability. I will not recall that here. So, consider the sphere. So, this is in n dimension sphere set of all x in omega is equal to epsilon. So, let me just put set of all x and since omega is a neighborhood of the origin. So, I can choose small epsilon such that this is in omega for some epsilon. So, to speak. So, here is a simple diagram I will put omega here and this is the origin. So, I am putting this one. So, we will be working in that particular neighborhood. So, we being continuous a positive minimum this is the because we is positive definite by definition we will have a positive minimum. So, let me call this S epsilon some notation on S epsilon call it M. So, M is positive and it is the minimum. So, this is S epsilon. So, again by continuity again since v 0 is 0 v is continuous. So, we can choose delta positive and delta less than or equal to epsilon such that v is strictly less than M on S delta. So, let me again stress this thing what does this mean. So, this means that is v of x is strictly less than x for all mod x equal to delta. So, let us go back to the previous page. So, this so delta will be some area. So, this is delta and I am choosing delta. So, that the v will be strictly less than and remember M is the minimum of v on the bigger sphere mod x equal to epsilon. Now, pick any x 0 such that mod x 0 is strictly less than delta and. So, you start your orbit and consider a solution x t of 1 our system with x 0. So, I am now starting the orbit in the ball of radius delta and we would like to show that this orbit through x 0 remains within x x 0 and that through the stability. So, wish to show wish to show. So, the r positive orbit is contained in mod x less than or equal to epsilon and this proves stability. So, implicitly I am assuming that the solution x t of 1 with x 0 equal to x 0 exists for all t positive. So, this is a standard assumption that I have been telling. So, then we would like to show that the positive orbit. So, this is remember positive orbit through x 0. So, that would like to show. So, once you show that thing that is through this stability for this thing consider v of x t. So, now, we are bringing in the. So, you compose the solution x with this real valued function v the Lyapunov function v. So, you get another a function from the real line into the real line. So, that is the composition of v and x and this by fundamental theorem of calculus one variable calculus. So, this is 0 to t d by d s. So, this you have already seen many times. I am just applying that. So, this is fundamental theorem of calculus. Remember this. So, just you see how simple things in calculus are utilized. Now, the second hypothesis come into picture. So, by chain rule. So, this is another thing from calculus. So, d by d t of v of x t is nothing but x 1 x 1 dot plus d v by d x 2 x 2 dot plus etcetera plus d v by d x n. But remember x is a solution of system one. So, this is nothing but d v by d x 1 f 1 plus d v by d x 2 f 2 and this is nothing but del v dot f that is defined. And by our assumption this is less than or equal to 0 in condition 2. So, now if you put back this. So, remember in the integrand we have this expression and now by chain rule I have computed this one and shown that is less than or equal to 0 and again now you go back. So, here you see the integrand I have that derivative. So, that will be. So, therefore, v of x t minus v of x 0 is less than or equal to 0 by that integral r. So, this v of x t is less than or equal to v of x 0 and remember x 0 is chosen within that ball of radius delta and by our choice this is less than m and this is minimum on the sphere mod x less than or equal to m and that proves that mod x t is less than or equal to m a epsilon for all t and that completes the proof. So, stability is one part. So, stability is easier. So, now let us go to the second part. So, we have the additional assumption. So, we have grad v dot f is negative definite in omega. So, again let me stress. So, that is grad v dot f at any point less than 0 in omega minus 0 and at 0 it is 0. So, this is important this is we are going to use now. So, again from the first part from the first part previous discussion we see that this function this composite function v of x t a function as a function of t is. So, again you can just use that fundamental theorem of calculus you integrate between any 2 times t 1 t 2 t 2 bigger than t 1 and you see that that is non increasing. So, that is v of x of t 2 is less than or equal to v of for all t 2 bigger than. So, we are just concentrating on that positive orbit. If this composite function is non increasing and it is bounded below the function is bounded below why bounded below because that is our assumption it is bigger than equal to 0. So, here we have a non increasing function which is bounded below and by calculus lemma this limit v of x t as t tends to infinity exists. So, here I have a non increasing function v of x t which is bounded below that is why this limit exists and since this is a non negative function call the limit call the limit equal to L. So, obviously L is bigger. So, we used to show that L equal to 0. So, if L equal to 0 you immediately see that since v 0 is 0 and v e vanishes only at 0. So, if we show L equal to 0 it follows that limit x t x t tends to infinity should be 0. Otherwise if this limit is different from 0 then you go back here. So, limit of as t tends to infinity of v of x t will be v of this limit whatever that limit and that we are a prove if you are going to show that that is going to be 0 then that will get a contradiction. So, this is important. So, once you show that the limit this limit L is 0 it follows automatically that this limit of x t is 0 otherwise we get a contradiction. So, assume to get a contradiction assume L is possible. So, again using continuity again using continuity of v and v 0 equal to 0. So, these are simple arguments you should check them using your knowledge of calculus. We can choose eta less than or equal to delta such that v of x will be strictly less than L bar mod x. So, here is the picture. So, we have this big let me again recall that. So, this is epsilon. So, this is origin and we have a delta here inside and now I am going to choose. So, maybe I will let me just and remember this excuse me this eta is chosen. So, that we have this condition this I can do because I am assuming L is positive and we will show how we get a contradiction on that. And now you consider this this x t consider the orbit positive orbit starting at x 0 and now I take mod x 0 less than or equal to eta which is automatically in in the delta. So, that is not and by assumption consider this thing by assumption and this construction you see that then O plus x 0 the entire orbit is continued in mod x less than or equal to epsilon. So, this comes from stability. So, this is part 1. So, no problem and now the second part this set of all x I want to show that it is bigger than or equal to. So, what I am claiming is the orbit lies in this green. So, it will not go to 0 and this part comes from follows from that V of x t. So, it cannot go beyond eta. So, the entire orbit lies in this annulus. So, therefore, O of x 0 lies in the shaded annulus. So, follow this geometry. So, now you the second hypothesis now. So, on the annulus so, that is a compact set on the annulus. What is the annulus? So, eta less than or equal to mod x less than or equal to epsilon is closed and bounded annulus. So, that is compact set the function. So, this is the second assumption grad V dot f is negative and attains its maximum. So, I want to put a notation for that. So, let minus k. So, since it is negative this is negative. So, I just put maximum of grad V dot f at x and x I am taking only on this. So, that is a continuous function and this is a compact set. So, it will have a maximum and since it is negative. So, this k is strictly positive and now back to the equation which is given by the fundamental theorem of calculus. Then V of x t minus V of x 0 is equal to 0 to t d by d s of V of x s d s and this we have already computed this is del V dot f x and now I know that this orbit remains only in this thing. So, this is just less than or equal to minus k. So, this is where I am using this definition of k and x the orbit is restricted to only this annulus. So, all these facts are used here and this I am replacing it by its maximum which is minus k and integral 0 to t d s is just t. So, therefore, so remember this is k is positive. So, as t becomes large. So, this is a negative number. So, I can always exceed this V x 0. So, if t is large V of x t will be negative and that is a contradiction because V is a positive definite function. So, this contradiction proves to be n equal to 0. Therefore, 0 is asymptotically stable. So, Lyapunov also has a theorem for the instability result. So, let me just state that also. So, theorem instability. So, suppose a C 1 function. So, this is Lyapunov function if you want to call it for the instability case a C 1 function V from omega 2 or satisfies the following satisfies 1 in every neighborhood every ball every neighborhood of 0 there exist a point. So, I am not requiring V to be positive in entire neighborhood, but in every neighborhood there is a point. So, just let me again. So, every neighborhood. So, there is some point call it x. So, where V x is I am not requiring that V to be positive everywhere in that neighborhood. And the second condition that again connecting the function with the system our system is positive definite. So, later on we will see that this condition we do not require no way we do not require this positive definiteness everywhere we require this positive definiteness only when in the region where V is positive then 0 is unstable. So, remark. So, I will just write that whatever I said now condition 2 is required in the regions. So, this was observed by Russian names. So, I may have might have mispelted, but this is and there are other as far as instability results are concerned there are more relaxed conditions given by others. So, this is sufficient for many purposes. So, let me sketch a proof of this thing and we will consider the examples next time. So, sketch. So, again the starting point is the fundamental theorem of calculus d by d s of V of x s d s and this we have already seen that this is del V dot evaluated at x s that is that is fine. And now you see this our assumption is that this is non negative. So, that means now we will be increasing. So, that is the idea. So, I am just sketching it and you start a solution where this is positive. So, that means if I start at a point where V is positive then I remain always bigger than that by this. So, this is an important for us. So, let me just sketch it. So, again you start with. So, fix one ball fix x as that mod x less than or equal to epsilon and this is in our omega there and this is x epsilon and now I take another one delta here. So, by hypothesis. So, choose x 0 such that mod x 0 is less than delta and V of x 0 is positive and this is possible by our assumption on V. So, in any neighborhood there is a point where V is positive. So, this is the condition one. Start the orbit at x 0. So, that is this same thing as saying. So, let x t be the solution of our system 1 of 1 with x 0 equal to x 0. So, then by hypothesis second hypothesis. So, therefore, by the integral fundamental theorem of integral calculus and the assumption we see that this is bigger than or equal to V of x 0. So, this is V of x 0 and call this c and this is positive. So, this is true for all t. So, that is important. So, if I start the orbit at such a point then V along that path remains bigger than this c. So, let me draw again a diagram. So, that is. So, now I have this. So, this is V of x equal to c and I want to remain. So, this is just V of x bigger than or equal to c. So, that is the reason and part of omega. Everything is in part of omega. So, the orbit is confined to this region and that is what that estimate shows. Now, would like to show that that orbit has to leave the epsilon ball at later stage. So, want to show that for some let me use t star, t star positive this x t star leaving the epsilon ball means this is bigger than or equal. Want to show that. Assume the contour again. So, this is just a sketch. You have to fill in the details using a simple analysis. Assume the contour V. Assume the contour V that the orbit the positive orbit starting at x 0 is contained in this x. Assume that. I already know that. So, we know that this orbit x 0 is already confined to this one. So, set of all x such that V x remember this is positive. So, in particular this is positive and now we are assuming that the orbit is also contained in this ball. So, it is in the intersection. So, let me put some name for that. So, put k is equal to set of all x such that mod x less than or equal to epsilon and at the same time I want this. So, by using continuity of V and this compactness of x epsilon. So, you see that this is a compact and important thing is 0 does not belong to k because V 0 is 0. So, if I have forgotten that to include that thing. So, include that V 0 is 0 that is always there V 0. So, since it is positive there. So, V 0 cannot be here. So, again by continuity I will come to that later. And now you recall that hypothesis and since now. So, you see that hypothesis the second hypothesis that grad V dot f be positive definite in entire omega is not required I am just applying only to this set. So, now grad V dot f is positive on k and k being compact set and this is 0 is not there. So, this. So, let let L be minimum of grad V and now you take x in k and this is positive. Now, we are more or less term. So, again the same argument. So, then V of x t minus V of x 0 again this is equal to 0 to t. So, I am writing it again and again. So, that you realize it is usefulness and now this is grad V dot f and we are assuming that our orbit is restricted to k. So, this is in k for all t bigger than 0. So, that is important assumption. So, that this whatever value you are taking now they are all in k and in k. So, this is always bigger than equal to L. So, L is positive remember L is the minimum of this grad V dot f. So, in particular. So, this proves in particular limit V of x t at t infinity is. So, that does not say anything about the x t as of now, but now you see on this set x less than or equal to epsilon. So, this is a compact set in the Euclidean space R n this is a compact set and V is continuous V is bounded. So, in particular V is bounded on k because k is a subset of this bounded on k and. So, if I call this contradicts a bounded function cannot have an infinite limit this contradicts. So, why this contradiction? The contradiction comes because we are assuming the orbit is restricted to. So, the epsilon bond in this bond we are assuming that thing and to get this contradiction. Therefore, the orbit therefore, orbit cannot remain within. So, now you again go back and see the definition of instability and this proves 0 is unstable. So, you see that the question of stability or instability in this method of Lyapunov can be analyzed by the construction of suitable Lyapunov functions. So, again where are the examples? This we will discuss the one in the next class. So, examples positive definite function or negative definite function because we are not seen any this part and again these come from linear algebra. We will see in the next class. So, they come from linear algebra quadratic forms. So, a quadratic form is of the form. So, this is V of. So, in n variables x 1, x 2, x n that is V of x is given by a i j x i x j. So, a i j are real and symmetric. So, these are termed as quadratic forms in n variables and there are well defined necessary and sufficient condition for this V to be positive definite. So, there are necessary and sufficient conditions. So, these are this is Sylvester's theorem in the theory of quadratic forms for V to be positive definite. So, we will discuss this. So, these are some important class of positive definite functions and for a suitable for a given system we have to choose a suitable quadratic form. Of course, not all the time this quadratic forms will work. So, sometimes we have to even take higher order polynomials, but in many cases just this quadratic forms will suffice and we will see some examples in the next class. So, we will discuss examples of positive definite functions, Lyapunov functions and other things in the next class. Thank you.