 Hello, welcome to yet another session of our NPTEL on on linear and adaptive control. I am Srikanth Sukumar from Systems and Control IIT Bombay. We are as always in front of our nice motivational image of this autonomously operating rover on Mars and we sort of are advancing towards analyzing and developing algorithms that we hope will drive systems such as these. So last time we were looking at the notion of radial unboundedness. We had started with these special functions which we are going to use subsequently in Lyapunov theorems. And the first property that we looked at was the property of definiteness. So we looked at positive definite functions. And then last time we were looking at radial unbounded functions in some detail. So as always we sort of gave some definition for radial unboundedness, which is sort of an extension of positive definite functions. And just like in the case of positive definite functions, we also proposed some easier tests for verifying radial unboundedness. So just like the case of positive definite functions, the definition itself may not be amenable for use in a lot of cases. And so we sort of came up with some easy tests because the definitions themselves require us to come up with some class K and some class KR, which may not always be easy to find. So of course, also for the case of radial unboundedness, we came up with these easier foundations. And we saw some pretty, hopefully some pretty good representative examples. Of course, the exercises that we are providing have more of these examples and more of these examples that you have to consider and discuss definiteness and unboundedness, etc. So this was the notion of radial unboundedness. So we also pointed out that positive definiteness was connected to local stability notions and radial unboundedness is connected to global stability notions. This is again something that we will see subsequently. Now, there's also, you know, if you go back and sort of remember the definitions that we had looked at, we also looked at the notion of uniform stability, right? And this uniform stability notions are what are connected to decreases. So that is what we are going to look at today. So this is where we start. We say lecture 4.3, this third lecture of the fourth week, right? And we want to define what is a decrease in function, okay? So if we, as always, we have a scalar valued function, which is going to take as its arguments time. And again, just like we mentioned last time, there isn't really no harm. In fact, it might even be better to assume that this function takes non-negative real time because that's what we are going to consider almost everywhere in this course. In fact, no scores are in non-linear control, right? So this is a function which takes in the first argument time, r plus, and the second argument as state, which is now in some bounded ball around the origin. So this is Br, Br. We have already introduced this notation. And as usual, it maps to reals. So all these functions have a similar range and domain. And of course we require that vt0 be equal to 0 for all t in r plus. This is again the same condition as before. Remember that we really care about this function taking zero values at the origin, right? Because origin is where we want to convert and this has the notion of some kind of an energy function. And we want this energy to be to take its minimum value, right? When you are at this equilibrium, right? Which is the origin. So for decreasing, the second condition that we require is the existence of a class k function such that absolute value of vtx is less than equal to phi norm x for all t in r plus and for all x in Br, right? So this is, you know, sort of completely opposite to what we saw for positive definiteness in some sense, right? So at least it looks like it's completely opposite to what we saw for positive definiteness, right? So what's the idea? It has to be dominated by a class k function, okay? So if you look at the picture, this should help us identify the difference. Here we have x and here, of course, we have all the different functions of x so that we will sort of mark later on. So this is say my class k function. I mean, I mean, again, this is only until some r beyond that we don't care. I mean, so this is just some class k function, right? And what we want is that, in fact, yeah, let me, yeah, let me sort of draw it in a different way. So to indicate that, again, in fact, go below it. Oh no, let's leave it at me. So we want this v, absolute value of v to be less, right? So of course, what I will do is I will draw something on the other end too, something like this. And we want, so this blue thing is, of course, the phi norm x and this is minus phi norm x. And we want the v to be bounded inside this, so we want, so v is allowed to go in the negative direction, but it has to remain within this sort of envelope if you may, right? So this is what is a decrescent function. I mean, this is an example of a decrescent function. So this is an example of a decrescent function, okay? So as I mentioned before, decrescence is related to the notion of uniform stability. And so basically it's somehow, if you look at this, well, I mean, it's not evident here, but anyway, I mean, essentially there is some kind of an independence with respect to time that you achieve using the notion of decrescence, okay? So what are the examples? So let's look at examples and counter examples just like we have been doing. So let's see, let me try to expand this and say, look at some example one, right? What is it? Suppose I take something like v tx is equal to, what we do for the positive definite case, let's try to see that. So this was t plus 1 over 2 times norm of x squared, right? I mean, you can also take norm of x squared divided by 1 plus norm of x squared. So that's fine. So we know that this is a positive definite. So v positive definite, we have already sort of discussed this a couple of times before. Because this is of course greater than equal to this guy for all non-negative time, right? So it's positive definite, but is it decrescent? No, it's not decrescent, right? Because there's no way I can, you know, so this is sort of not very easy to argue again because we are not giving any easy test in this case, right? So it's not very easy to argue. So we want to claim is, ask if v is less than equal to, I mean, this is already non-negative, so I'm not using absolute value of u. So is v less than equal to some phi norm x, right? Let's think of a simple situation. Let's say, is this true? Is this true for some positive gamma? What do you folks think? So I have essentially taken this guy out and I'm saying does this inequality hold true for some gamma? Because if it does, obviously this right-hand side is a class k function, right? And so we are done. I mean, well, fine. The right-hand side is not exactly a class k function, but the right-hand side is a positive definite function, right? So we are fine. We are sort of, you know, it's not a big deal, right? Because I can always say that this is, let's see, less than, okay, I apologize. I can always sort of say that this is less than equal to gamma norm x squared and then I'm done, okay? Because gamma norm x squared is, of course, a class k function, right? In fact, it is a class k r function, right? So that's okay. Class k r function is also a class k function, okay? But does this even hold true? Does this hold true? Yeah? So the point is this guy, I'm going to put a nice square around it. So this is not, let's say, I'm sorry, not true. Why? Why is this not true? If you give me any gamma, for example, say gamma is 100, okay? If you give me a really large gamma, say gamma equal to 100, I choose some arbitrary large gamma. What do I know? I know that at t greater than 99, right? v tx becomes greater than gamma norm x squared divided by 1 plus norm x squared, okay? And I immediately get a contradiction, right? And it's so easy. I got a contradiction. See it very, very easy, right? And so if you give me any choice of gamma, arbitrary large, it doesn't matter. I can always find a t. It says that it will start to dominate this function rather than being less than this function. It in fact starts to dominate this function. And that is a problem, right? And so you can see that this is not decrescent, right? So this is, this v is not decrescent, right? So this is positive definite, but it is not decrescent, okay? So let's look at another example and say the other way around. Suppose I take v tx and I am going to make it bigger. Suppose I take my second example with v tx as equal to, I sort of divide, right? So suppose I take 1 over 2 t plus 1, norm x squared divided by 1 plus norm x squared. In fact, I'll keep my life simple and I'll just get rid of this guy, okay? And just say it's norm x squared because anyway, it doesn't matter whether it's positive definite or radially unbounded, okay? Because we are more concerned with decrescent here, right? What can I say? What can I say? I know that this is for sure less than norm x squared over 2 for all t greater than, for all t in R plus, right? Because as t becomes larger than 0, this is definitely smaller than norm x squared by 2, right? That's obvious. So implies decrescent, right? Implies decrescent, sorry, okay? But, but v not positive definite, right? Why? Let's look at the easier condition. Let's look at the easier condition. What do we require? We require v to be strictly positive for all norm x not equal to 0, okay? On the face of it, it looks like it, right? But if I take any non-zero x, right? The problem is I have division by t plus 1, this guy, yeah? So what happens? Again, I'll say no. So let limit as t goes to infinity 1 over 2 t plus 1 non-x when is 0, right? It doesn't matter what norm x is. It is irrelevant. It is irrelevant what norm x is, all right? As I push t larger and larger, this is going to go to 0. Therefore, v is not positive definite, right? So we have some sort of contradictory examples, right? We found that for when, you know, sort of I give you an example where if it was a function of state and time, where v was a function of state and time. And it was decrescent like this case here. It turned out that it was not positive definite, right? Because if it was upper bounded by a class k function, I could not lower bounded by a class k function. I mean, I did not really see it in terms of lower bound in class k, but I did the equivalent easier test, right? On the other hand, if whenever I could lower bounded by a class k function, like this, this is, you know, whenever I could prove that it was positive definite, right? Which we did already in the previous lectures, we could show that it was not decrescent, right? Of course, in this case, we only showed with a particular example of a class k function. We could have said that it is dominated, it's upper bound by some very large class k function. But the point is whatever function of x I choose here, remember that the right hand side has to be independent of time. So whatever function of states I choose, right, this has to be true for all x, notice. So if I choose really small x, then this is right hand side is potentially not very big. But as I keep pushing my time up and up and up, VTX is always going to dominate something like this. Any function of only the state on the right hand side. Okay, so although I chose this particular example to verify that it's not decrescent, it doesn't matter. I could have chosen any function of state here and this would always dominate this guy as T gets pushed up. As T, sorry, as T goes to infinity, right? So this was not decrescent. So the example where I had positive definiteness, I did not get decrescence. The example where I got decrescence, I did not get positive definiteness. So it begs the question, is there an example of a, does there even exist an example of a V, which is function of both T and X, which is both decrescent and positive definiteness? So the answer is yes. The answer is yes, it's not that difficult. And what is it? You'll of course first choose the nice positive definiteness sort of structure. Let's see. I choose something like this. I will keep a long x squared over to here. What I will do is I will take something like a bounded function of time, if you may. I'll take a bounded function of time. So this is something like 1 plus sine squared T. Right? So what do I know about this VTX? I know that VTX is definitely greater than equal to non x squared by 2. Why is V positive definite? Because I could find lower bounding class K function. Why this is true is because the lowest value this can take is 0. Largest value is of course 1, but the lowest value is 0. Right? And so it dominates this class K function. Okay? Simple. So this becomes 1. So now I also know that this has the largest value. The largest value this can take is 2, because sine squared T is upper bounded at 1. Right? So this whole thing can at most become 2. Right? So I also know that VTX is upper bounded by non x squared. It implies V is also decrescent. It's upper bounded by x squared. Why? Because the largest value this can take is 2. So the 2 and 2 here cancel, and that's the largest possible value of this function. Right? So I have bounds on both sides. So there do exist, of course, functions of both state and time, which are in fact both decrescent and positive definiteness. Positive definite. Right? So it's not like, you know, one excludes the other or something like that. Okay? Let's not. I hope we are not, you know, sort of getting into this preconceived notion that one does not mean the other. Because if that was the case, then it would never be possible to get uniform stability and, you know, asymptotic stability together. Right? As you can see further. So now the other thing to remember, of course, is that if you have a function, yeah, which is actually just a function of the state, right? So this is obviously positive definite. So in such cases, only, so, and in fact, make a remark also, remark, if Z only dependent on X, then decrescent is free. It's trivially decrescent. Because of course it's upper bounded by norm X squared. It's upper bounded by norm X squared. And I'm done. It's trivially decrescent because there's no time appearing, which will not allow this domination. Okay? I hope that's clear. Then the final property, which is the, actually rather weak property is that of positive semi-definiteness. Okay? So the positive definite, semi-definite is a rather simple property. It just says that, you know, anything, as usual, if you have a function going from time to the state, and of course we want vt0 to be zero, further, if the function is just non-negative, just takes non-negative values. Okay? That's all we need for semi-definite. If the function v takes non-negative values for all values of time and state in a ball, then v is said to be positive definite, semi-definite. And it is denoted by greater than equal to zero. Okay? So positive definiteness was denoted greater than zero. So positive semi-definite is denoted greater than equal to zero. Thank you. So these are rather simple functions. Basically any function taking a non-negative value, right? So it means so many examples. So v tx equals tx1 square plus x2 square is fine. It is greater than equal to zero. v tx equal to x1 square by 2 plus x1 square by 4, which was not positive definite, remember, is in fact positive semi-definite. Because whatever happens, this is never going to take a negative value, right? Similarly, if I choose v, all the examples that did not satisfy other properties, yeah, this is again positive semi-definite. Because whatever happens, this function is never going to take a negative value. And remember that positive semi-definite, semi-definiteness also plays an important role in Lyapunov stability theorems. So it's not a class that is to be taken lightly, yeah? But of course it is a rather weak requirement. It just requires that the function itself never take a non-negative value. Yeah, that's it. That's all we require in positive semi-definiteness. All right. All right. I hope that all made sense. So beyond this, we are of course ready to look at Lyapunov theorems. I look at the Lyapunov theorem, but we will stop here and sort of try to conclude a discussion for this session. So what is it that we saw? We went on to sort of look at the final property, if you may, and this is the property of declassence. So we have seen three properties of functions, which is positive definiteness connected to asymptotic stability, radial and boundedness, which we said is connected to global asymptotic stability, and declassence, which we have mentioned is connected to uniform stability. So you've seen all the three properties in this week's lectures. We've also seen a good number of examples to hopefully help us distinguish between these properties. So I really hope that it helps clarify what kinds of functions are positive definite and which are end positive definite. So it's more important to sort of remember the ones that are not positive definite. Yeah, so we don't make any errors in our Lyapunov theorem applications. And of course, these are rather nice potential functions and we will see subsequently the setup for the Lyapunov theorems. And we will also see how these three properties of functions that we have defined, we have looked at easier tests to actually talk about these, to actually verify these definitions. So we will actually see how these three kinds of classes of functions play a pivotal role in helping us state the Lyapunov theorems. And so this will be rather critical in what is going to come in next. Right. So, yeah, so I mean, what's upcoming in the lectures now is what is the most most critical aspect of nonlinear control, which is the Lyapunov stability theorems and which is what helps us to conclude that a system or an autonomous system such as what we see in the background it performs satisfactory in any given environment. That it follows a given trajectory or reaches a particular point or has a particular orientation. So all of these are posed as stability questions by control engineers and nonlinear control theorists verify these using the Lyapunov stability theorem. So what we will see subsequently are probably one of the most seminal results in nonlinear systems analysis. All right. Excellent. So that's where we will stop today. And I hope to see you again soon next time. Thank you.