 Hello everyone. Welcome to yet another session of our NPTEL on Non-linear and Adaptive Control. I am Srikanth Sukumar from Systems and Control, IIT Bombay. So we are almost at the end of our lectures for week three. And we have sort of learned over the course of this week some nice tools to analyze algorithms that will potentially drive systems such as that we see in our background. So what we were doing until last time was basically look at a few more examples of these particular stability properties that we had defined. So we saw a few more illustrative examples. So I really hope that all of you got a pretty fair idea of what kind of systems are going to have what kind of properties and what do these properties sort of mean for that particular system. So we defined stability, attractivity and several notions of asymptotic and exponential stability. We also saw at the end of the last lecture the particular notion of linear system stability. So basically how does the stability definitions in terms of epsilons and deltas that we defined how does it become more specific for linear systems. So we sort of had a brief overview of that. Of course we didn't really prove these results as such. So I just gave you a little bit of a sketch of what you have in linear system stability. So today we jump on to a new set of lectures. Again like I said don't worry about when we say week four and so on although we are not into week four lectures. This is more for the purpose of the homework assignments. So we will sort of go forward and delve into the week four lectures depending on how our schedule is because as the lectures become more involved we will need more and more. We will need more and more time. So therefore I don't want to delay getting into week four lectures any further. So as the outline states we discuss function classes and Lyapunov theorems in this particular set of lectures with a designated week four lectures. So you already start. So what is the motivation? What is the motivation? So we already saw several stability definitions. We saw a few examples where we did conclude what kind of stability properties a system might have. However what we found was that in all these cases we were required to find the solution or in some cases of course we made a call on stability based on phase plane portraits. But to be honest that cannot be considered a final sort of final technical answer on whether a system is stable or asymptotically stable or not. Because a lot of times in phase plane portrait again a phase plane portrait is based on a numerical method. So you sort of take a software and draw these phase plane portraits. It is based on a numerical method. So it is very, very, very possible in numerical methods to ignore or forget to consider certain problematic initial conditions. And you might actually end up claiming something about the stability of a system which actually does not hold true. So the biggest issues with these epsilon delta definitions of stability is the usability and it is very difficult to actually use these definitions. So that is the motivation for studying these Lyapunov theorems that we are going to look at now. So Lyapunov theorems really make our life easy in terms of concluding stability of non-linear systems. And we will very quickly get to that. But before we even start on to Lyapunov stability theorems or what is also called the Lyapunov direct method. We will of course need to define a few preliminary class of functions. So these are rather powerful class of functions. So there are three. So first is called a class k function. So what is a class k function? A function, again, so these are all scalar valued functions. So they take values in non-negative reals and map to non-negative reals. So this is required for all these functions. So they take values in non-negative reals and they map it to also non-negative reals. So a function phi of this kind is said to be of class k if it satisfies three properties. First, it is continuous. Second, it is strictly increasing. And third, it is 0 at 0. So this is what is a class k function. So we have of course given examples of what is a class k function. So fx equal to x, so let us be a little bit more precise. So examples, I will sort of write it again. fx or let us use the notation that we are in fact using here for x in non-negative reals is a class k function. Or if you want to make it much more easy to understand, so phi x is x squared with x in r plus is also class k function. So notice that I keep saying x is in r plus because I can either way up to find it, this map has to be from non-negative reals to non-negative reals. The argument also has to be a non-negative real number. So this is important. The next example we consider is, I do not want to number these because I have sort of given a few different examples. The next example that I give that is stated here is 1e minus x for again x in r plus. So how do I know that this is 1e minus x not 1e e to the 1 minus e to the power minus x? How do I know this is a class k function? First of all continuity is obvious. How do I know it's a class k function? Just compute phi prime of x. It is simply going to be e to the power minus x which is always greater than 0 for all x less than infinity. So which is basically saying that x is in r plus. Non-negative reals does not include infinity. Infinity is not included in non-negative reals or in reals for that matter. So that's how real numbers are defined. So this is of course, you can see that this is a positive derivative and therefore it's obvious that it is a strictly increasing part. So you can have many different such examples. So this is what is a class k function. So what is a class l function? It's sort of the opposite. A function phi is class l again with the same domain and range. If it is again continuous strictly decreasing and the initial condition has to be finite. Some finite initial condition. So one example again of such a function is 1 over x plus 1. So this is a class l function. Why this is continuous? Because again this is assuming x is in r plus. We are already assuming that x is in r plus. So therefore non-negative reals, therefore the least value x can take is 0. The denominator is non-zero there. So it's in fact a continuous function. So it's continuous. It's strictly decreasing is obvious again because as I increase x, the denominator is becoming larger and therefore the fraction is becoming smaller. And the initial value is of course finite. In fact the initial value, if you plug x equals 0, it's just 1. So this is a class l function. So we have class k and then class l. It sort of seems like it's opposite but I would in fact put a nice note here. If phi is in class k, this is the notation for saying that a function is in a particular class. This does not imply, I hope you understand, that minus phi in class l. This does not imply this. I hope that's clear because why? Because just look at this, minus phi becomes a map from r plus to r minus. So this is not even allowed. Not allowed. The only candidates that are allowed are in fact your functions which are mapping from non-negative reals to non-negative reals. So negative of phi is not a class l function if phi is a class k function. Excellent. So we know that. Let's look at the next class of functions. This is again another very important class of functions. And that is the class kr function. This is the class kr function. So function phi again, non-negative reals to non-negative reals. Let's always remember that. So we don't make a mistake. So this is of class kr, if again it's at least class k. First of all, we want it to be class k. And further, phi of r goes to infinity as r goes to infinity. So it's first of all, it's class k and it should go to infinity as r goes to infinity. So just to distinguish, one of the examples we considered as a class k function phi x equals 1 minus e minus x with x in non-negative reals is not class kr. Why? If you sort of try to make a picture of this function, if you make a plot of this, what this function looks like. What happens on the x-axis, I have of course x itself. And on the y-axis, I have phi of x. Phi of 0, as you can see, is exactly 1. So phi of 0 is exactly 0. And what I will do is I will draw the line corresponding to 1 because as x goes to infinity, minus x goes to minus infinity, this goes to 0. So phi tends to 1 as x goes to infinity. So the thing is the function is in fact increasing, if I may. Of course, I cannot do justice to this very much. The phi continues to increase, but it hits at most 1, never crosses 1. So this is not a class kr function. So this is a rather critical distinction. So this is not a class kr function. On the other hand, the other examples that we considered like phi x equals x squared, x is in r plus. So this phi is of course class kr. So it should be obvious to you that any function which is... So again, I will put this as a note. Just a note. Phi in class kr implies phi in class k. So if a function is in class kr, then it belongs to class k for sure. But it doesn't hold otherwise. The other way around doesn't hold because we just saw this example, which is a class k function. So this is a class k function. Let me reiterate that. So phi is in class k and not class kr. So hold this way, not the other way. So class kr function is a stronger requirement. So there are less number of functions in class kr than class k. So class kr is actually a subset of the class k functions because class k is a requirement of being a class kr function. So these are of course rather important function classes. So anyway, this we already mentioned last time too. We did talk about this in the earlier lectures also. We do require that our equilibrium is at the origin. If not, as always, we do a change of coordinates. We already spoke about this again last time and the time before. So whenever we have an equilibrium which is non-zero, we just do a simple change of coordinates. It's always possible. This is not a very stringent requirement as such. Excellent. So using these notions of class k and class kr and so on, we now talk about definiteness of functions. So before we go into definiteness of functions, if you remember, just a reminder definiteness of matrices. Yeah. I hope all of you remember. We sort of had a short discussion or whatever. We stated a few things on symmetric matrices. And then we talked about conditions under which a symmetric matrix is positive definite. There were several equivalent conditions. So a equals a transpose, of course, a sum n by n, a is positive definite. So this is the notation. If any of these conditions hold, any of these conditions hold, first is that it is, you know, first is that it is X transpose AX, strictly positive for all X in Rn, X not equal to the zero vector. The next one is that all. So I'll write it in short time because you already done this. Next was that Eigen values of A are strictly positive. All principal minors have positive determinant. So these are sort of the conditions under which a matrix is considered to be positive definite. So positive determinants, like I said, I will just expand this. All principal minors have positive determinant. So I want you to keep this in mind when we talk about positive definiteness or definiteness of functions. Now we are talking about definiteness of functions. Excellent. So what is the, you know, sort of the first such definition? All right. We want to take a look at that. Okay. Right. So the first definition is that of a positive definite function. Right. So if you have a scalar valued function from R cross Br to R. So what is this R cross Br and all that. Right. So I think we already know the notation. So Br is basically the set of all X in Rn such that norm X is less than R. So this is the ball of radius R around the origin. Right. Notice that origin has been assumed to be our equilibrium. Right. And therefore we sort of measure distances from the origin. So the ball of radius R is around the origin. So we don't specify the center. So what is this Br cross Br goes to R and all that before I even, you know, sort of go forward with the rest of the discussion. So here R represents time. This represents the sort of ball around the origin of the states. Okay. So this V function takes time and states and gives me a real number. Right. So this is the very specific kind of functions for which we are discussing definiteness. Right. We are not discussing definiteness for some arbitrary functions. Right. We are discussing definiteness for very, very specific kind of functions. Right. Which map time and which map states to some real value. Okay. Right. So what do we require? We of course require that this function is continuous. It's already scalar value is obvious because I map to reals. Right. We require that the function takes zero value for all time as long as the state is zero. So if I plug in the state value to be zero, that is origin, zero means the origin of course. So if the state value is zero, that is if I am at the origin, then this scalar value function definitely gives me zero irrespective of what time I choose. Okay. Irrespective of what time I choose. Right. The second important point is that there should exist a class K function phi such that this function VTX dominates this class K function of phi x, sorry, of norm x. Right. Notice this. Notice the argument. Norm is always non negative. Yeah. Remember the argument of a class K function also has to be non negative. It definitely maps into non negative values, but the argument itself has to be non negative. So this class K function takes as its argument the norm of the states, which we know to be non negative always. Okay. So phi in there has to exist such a function phi, such a function phi such that the VTX dominates this function phi of the norm of the x for all t and for all x in Br. So notice. So remember this function phi. So is a class K function in the norm of the states. Okay. So the important thing is if these two conditions are satisfied along with continuity, then we say that the function V is positive definite and it is denoted by V greater than zero. I mean, just like we denote for matrices A greater than, just like we denote for matrices A greater than zero, we have a similar notation for functions also, we say V is greater than zero. Okay. The other important point that I really want to highlight is that notice that the class K function, the way we have defined, right, is it's continuous, it's strictly increasing for all values of the argument. And it is zero at zero. Okay. So it is continuous, strictly increasing for all values of the argument and it is zero at zero. This is important. Right. But we need the function V to only dominate this function. Okay. We do not particularly require V to be also strictly increasing. Okay. This condition does not, first of all, does not imply V is strictly increasing. Okay. We just need to dominate a strictly increasing function. Also, we need the domination to happen only for a small ball around the origin. And we don't need the function V to dominate this class K function for all time. Okay. So let's look at a sort of picture. And we will see what is the significance of this BR in, I mean, the BR is just some local domain around the origin. Okay. If you remember our stability definitions, right, also have, you know, the attractivity property, if you may. Also, as this epsilon delta, right, where this delta could depend on initial time or not. But if you have initial conditions starting in this delta ball, then you converse to the origin. It doesn't say that you can start at any initial condition and go to the origin. Okay. So that was also a local notion. So that local notion is being sort of going to be portrayed using this BR, ball BR. Okay. So we don't need this function to dominate a class K function for all states. Okay. Definitely we want it to dominate for all time. Yeah, the time thing is not flexible. We need this domination to happen for all time because notice that the right hand side doesn't contain any time argument at all. Okay. The left hand side contains time, but the right hand side does not contain any time argument. Right. So that's to hold for all time, no doubt about it. But it doesn't need to hold for all values of the state. Okay. So if I, again, if I try to make this sort of a picture, right, here I have X. So this is sort of, right. So suppose I have some, you know, some kind of a class K function, then let me make it very simple. Right. Suppose I have like a linear class K function, something very simple. Okay. So this is my phi of norm X. In this case, X is a scalar. So norm X is, you know, absolute value of X. Well, let's not worry about it. I will stick to the notation to be consistent here. Right. Now I want my VX to dominate this function. Of course, notice that V at zero is zero. So obviously I have to still start at zero. Right. Then I can do things like this. Okay. I can do things like this. Right. So until, if I may, again, let me try to pick another color. Right. So until here, right, which is sort of say R. Right. So until here, until R. I'm just calling it V of X because I have not shown the time argument. I mean, I can even do that. But, you know, let's say this is V T of X, but the, but it becomes complicated. Okay. So I'm just going to call VX. I'm going to say that there is no time argument involved in this particular case. Right. Why am I saying that is because if I put the time argument, then I'll have to draw another axis and show how the function also evolves with time. So, instead, I'll just say it's a function of just the state X. Okay. So even if V is just a function of X and time argument doesn't appear at all. And that is also allowed, by the way. That's also allowed. Right. So if it happens so, even in that case, I don't really need my V to be strictly increasing. Okay. I need it to dominate this class K function only. And that too, not necessarily for all values of the state, but until a particular, you know, until the norm is less than R. Right. I need the domination only until a particular point. Okay. And if this happens, this function VX, so they, in this case, V is said to be positive definite. In this case, V is said to be positive definite. Right. As we have mentioned, BR is an open ball around origin and positive definiteness is related to the notion of asymptotic stability. Okay. In the Lyapunov theorems, positive definiteness is going to be directly connected to asymptotic stability. Okay. So, what would be simple examples? So, simple example would be, say, you know, I know that, I mean, I'll start to construct a simple example from the class K function itself. So, I know that 1 over norm X plus 1 is a class K function. Right. So, if say X is in R2. Right. Then I can use the, you know, 2 norm. This is X1 square plus X2 squared. Right. And in fact, I will use a squared here. That's fine. So, this I know is a class K function. Right. This I know is a class K function. Right. So, if I sort of, you know, sort of work backwards and if I say my VTX is something like, I'm just introducing some time argument here. I can say something like e to the power T divided by norm X squared plus 1. Yeah. This, it should be obvious to you that this is greater than equal to, let's see. Yeah. So, in fact, I can just simply take it as T. Right. I make it more. Yeah. So, this is greater than equal to phi norm X. Right. This is greater than equal to phi norm X. Okay. All right. And so, this is obviously a V is a positive definite function. Okay. Because it dominates a class K function. So, this is a valid example of a class K function. Sorry. This is a valid example of a positive definite V function. All right. So, anyway. So, I think this is, we figured out some important things here. So, what we have sort of seen today is a few different function classes, right, which are critical in talking about definiteness. And from there, we talked about the first notion of definiteness, which is positive definite. And we saw an example also of what is a positive definite function. So, we will continue again in this vein in the upcoming sessions. And we will progress on slowly towards stating the Lyapunov direct method or the Lyapunov stability theorems. All right. That's it, folks. Thank you.