 Hello folks Welcome to yet another session of our NPTEL on nonlinear and adaptive control. I am Srikanth Sukumar and from systems and control IIT Bombay We are as usual in front of this very nice motivating background image of this rover on Mars and We expect that we are moving forward towards Learning how to analyze and design algorithms that are going to drive autonomous systems such as these So without any further delay I'm going to sort of try to recap what we have done until last time so at the end of the last week's lectures, we started looking at stability in the sense of Lyapunov, right and in the beginning of it we sort of tried to look at what is the notion of existence of unique solutions, which is sort of a critical You know ingredient for us to be even able to talk about Existence of stability and so on and so forth and other properties of any differential equation for that matter. All right Once we understood that we require these Lipschitz notions in order to talk about Stability and talk about existence and uniqueness. We went on to define the notion of an equilibrium and especially the notion of an isolated equilibrium. We try to understand why this isolated equilibrium is such a key concept in studying stability Yeah, and the fact that a non-isolated equilibrium doesn't even allow us to talk about convergence and Comparison with one specific equilibrium point because there are so many of them arbitrarily close to each other All right. Now So this is what or this is why you know, this was such an important notion Okay Once we had under our belt the notion of An isolated equilibrium. We were ready to talk about the first definition Which was stability in the sense of Lyapunov Yeah, and this is in the format or very very standard format of an Epsilon Delta definition This is the standard format for any stability definition That we will be seeing now or in the future Okay, so what is it? It sort of comes as a challenge question. That is, you know Given every Epsilon positive or even any Epsilon positive. I Should be able to generate a Delta which depends on possibly the initial time and the Epsilon and There's also a positive number Such that if my initial conditions start within a Delta ball of the equilibrium then My state remains within the Delta ball of the equilibrium or remains within the Epsilon ball of the equilibrium for all time All right, and we do this very nice No illustrative picture in order to understand this notion. We also understood that Delta Has to be less than or equal to epsilon. What things to make sense Okay, so once we had the notion of Lyapunov stability, right Understood, right we want to Look at the notion of uniform stability. So what is it? So this is where we begin today. So What do we have in uniform stability the only difference between Lyapunov stability and Uniform stability or uniform stability in the sense of Lyapunov versus stability is the word uniform, of course, right? So we added a new word a new qualifier if you may so that's of course one new thing And what does it entail this uniformity? This uniformity is with respect to the initial time, right? So if you look at the deltas that you get In each of these definitions There is a difference that the delta in the case of uniform stability is independent of the initial time Okay, and so this is what is the difference everything else is exactly the same given an epsilon There has to exist a delta which now depends only on epsilon not on t naught Such that if my initial conditions are within the delta ball My solutions remain within the epsilon ball of the equilibrium for all time so this statement and This statement are absolutely the same if you notice there is no difference between the two the only difference lies in the Delta right there in Here it was dependent on t naught and here it was independent of t naught Okay, and we will of course see When such a situation arises we will look at examples subsequently of When we have one property and not the other Okay, now one of the important things to remember is that we say that the equilibrium is unstable Okay, if it's neither stable Nor uniformly stable if the equilibrium X e does not have either of these two properties Right, then we say that our equilibrium is Unstable, okay, then we say that our equilibrium is unstable Okay, so in order to sort of illustrate these properties. Let us look at an Example the very very standard textbook example from the book by Vidya Sagar This is from Vidya Sagar's book Yeah, you can find it in this Book of non-linear systems By Vidya Sagar. All right. This is in the chapter where you know where stability and other notions are in fact being defined Okay, the very very very standard example in that sense. So what is it? It's a very simple scalar system, right? You're given X As a scaler Quantity, so I will sort of explain I'll sort of explain X is in there are real numbers Okay, and you have some initial condition of course X t0 is just your initial condition I'm not denoting It's not denoted here is anything else but X T0 all right, so It's X dot is 60 sin t minus twice t times X Okay, so and then of course like I said, it's a real number with some initial condition at initial time now if I Once I've chosen an initial time and this initial state, of course I can integrate this and this is very easy to integrate because this is simply DX over X and this is 60 sin t minus twice t dt Right, and I can of course integrate both sides. Yeah, so from X t0 to X t on the left and t0 to t On the right So this is not very difficult to do I can actually carry out this integration And in fact the outcome of this integration is this expression Okay, so the outcome of this integration is exactly this expression. You see, okay, let's Not be very worried. So the left-hand side Actually gave me log X Right, so so basically from from this guy. I got log X Yeah, log X divided by X t0 Okay, therefore I got an exponential here because I had log X divided by X t0 And therefore I got an exponential here when I took it to the other side and the integral of this was simply 6 sin t minus 6 t cosine t minus t squared and this is corresponding to the initial time Like this is corresponding to the upper limit t and This is corresponding to the lower bound t0. You're very standard in a definite integral Okay, so now it's a rather long expression, but not too complicated You can actually take a derivative with respect to time of this guy And you can see that it satisfies your differential equation Okay, it's not difficult to see that it does satisfy your differential equation Okay, excellent, right, right. So I would advise you to do that diligently that take a derivative here and verify that It satisfies the differential equation. One thing is obvious if I plug in t0 instead of t here the exponential term is in fact 0 therefore X of 0 is 1 and So X t0 is X t0 that is it's verifying that initial condition And after taking a derivative I can also verify that it satisfies the differential equation Therefore, this is indeed the correct solution and the unique solution. All right. This is important to us This is a unique solution. Okay. So what is the equilibrium in this case? What is the equilibrium? We didn't discuss that. Let's see. We try to get rid of this. Yeah So what is the equilibrium? X sub e is Zero because zero is the equilibrium because you can see the right-hand side is zero Exactly at X equal to zero for all time Okay, so zero is the equilibrium. Great So I can succinctly write this quantity right as this Okay, now what have I done here? What I have done here is I have denoted I Defined a quantity gamma as X of minus 6 sine t0 plus 6 t0 cosine t0 plus t0 squared Okay, so notice. So this is what I have done. I have Defined a new quantity gamma Okay, which is just taking the initial time component of the exponential pulling it out Okay, and notice that this is a constant whenever t0 is fixed once I fix the initial time Gamma is a constant. So therefore I have written it as such Okay, so I have x t0 multiplying gamma and now only a Exponential containing only a function of the time tree Okay, so again, let's go back to our remember We talked about this abuse of notation, right? You see that the solution does depend on initial time and also on T and also on initial state Okay, however, when we sort of write the left-hand side, we just say that it's a function of time t Okay, so this is an abuse of notation between we've already Understood what this is and therefore we are okay with it But we do remember in our minds that the right-hand side does depend on the initial time as well as the initial state Okay, all right. Excellent. Excellent. So Once I have this sort of assignment this definition and I have this relatively succinct Expression for x of t solution I want to remind myself that There exists some finite time capital t beyond which this is negative right, why? Why is this? because t squared dominates 6 t cosine t and 6 Sine t Yeah, so this is just a property of the functions involved if you look at this minus t squared This is a quadratic term in t. So it's going to grow really fast Whereas this one is a linear term in t with a cosine, which is of course going to fluctuate only between plus minus one So it's just a linear increase in time And this is just a sinusoid term which is going to stay between plus minus six and This is of course going to stay between plus minus six t But this term is going to rapidly increase s t squared Okay, so beyond a certain time you can be sure that this function is going to become negative because this is there's a minus sign attached to the t squared Okay, so therefore there exists some capital t says that beyond that time this quantity is negative Okay, this is a rather critical thing for us Okay, so what do we do? We take this time and we find the supremum of this function Okay, what is it? so We say that m equal to zero and we denote Ms. This is actually a definition We denote Ms the supremum from initial time to cap t of Of this quantity six sin t minus 60 cosine t minus t squared Now, why do I find the supremum only over this time? because I know that this quantity is Possibly positive only in this window Only in this window is this quantity possibly positive beyond it. It is definitely negative Okay, so if I try to plot this function just to make things more clear if I try to plot this function I can expect it to look like So this is the time cap t Okay, so this is this function right here on the y-axis Okay, it should be evident that at t equal to zero what happens. Ah Let's see in fact, let me actually be careful I Should not have done that Okay, so let's say that this is Let's say this is t equal to t zero At t equal to t zero say it's some positive quantity. I don't know what it is Yeah, so this is as you can understand this term is exactly going to be six sine T zero minus six t zero cosine t zero Minus t zero square. That's what is this guy All right, and then beyond that say it's positive for some more time. Okay, and then I have this peak This peak is what is M? Then it drops and I know for sure that beyond this time cap t it is definitely Becoming negative therefore I need to consider the supremum of this function only in this window Okay, if I want to find the supremum maximum value of this function for all time Okay, if I want to find the maximum value of this function for all time I need only to look at this window t zero to capital T And that's what I do. That's what I do. I look at only this window t zero to capital T Okay, great. So I know that this function is definitely below M Now what I claim is that if I choose an if I'm given an epsilon positive I claim that a valid delta is this guy. I claim. This is a valid Okay, I claim this is a valid delta. Why why is that? this is true because if you Assume that your initial condition is Delta away from the origin Okay, what does that mean that means that my absolute value of x zero is less than Delta which is equal to Epsilon over gamma e to the power m Okay, then what can I say about the solutions? Let's look at the solutions right here. I'm going to write right here so Absolute value of x t. I'm taking absolute values on both sides and I'm going to freely use the Triangular inequality. So this is going to be equal to Gamma Gamma x t zero X of this whole thing, right? So Absolute value of this whole quantity But I know for sure that this is less than or equal to absolute value of gamma Absolute value of x t zero times the absolute value of this whole guy by my usual Triangle inequality Now I know for sure that this guy is upper bounded by m And because of what I just How m is defined This guy is up bounded Exactly by m Okay So what do I know? I know that this is Less than equal to also notice that gamma is basically You know gamma is basically an exponential of some quantity Right. So gamma is basically exponential of some quantity. Therefore, it is always positive So I can get rid of this then I have absolute value of x t naught and this is Capital M. Okay, this is capital M Now once I have this you see So this I mean, I have not been this is not very carefully done here. This is actually this should be X t zero because that is what we are using It is what? x t zero This is what we have been using. All right. So I have Once I reach this point that x of t is less than gamma m x t zero that is here Okay, I can substitute for absolute value of x t zero which is chosen to be less than delta right this guy So once I substitute this here, what do I get the gamma em cancels out? And I get that x t is less than epsilon Exactly as we required Exactly as we required. Okay, so great. So we have been able to show that Given an epsilon I can find a delta such that if my initial conditions start within the delta ball My solution remains within the epsilon ball. Okay, the important thing to remember is that The delta and hence gamma depends on T zero. Yeah, so in the definition of delta, I have a gamma and if you notice the gamma was defined as such and This is dependent on t zero and therefore delta is also dependent on t zero. Okay, so what have we shown until now? We have shown that Until this point if I see be careful here Until this point We have shown stability of Zero in the sense of re-apples Okay in the sense of Now what I want you to notice now. Okay, so we have already shown stability So the next question of course is is it uniformly stable? Yeah, because those are the only two definitions We've done so those are the only two things we can in fact test Okay, so the next question is is it uniformly stable, right? So what I would like you to notice is that the only dependence of delta to t zero comes from gamma Okay, that's the only way so we look at gamma itself So we know that gamma is this expression, right now. What can I say about gamma as? T zero goes to infinity as t zero becomes really large You can see that gamma also goes to infinity as t zero goes to infinity gamma goes to infinity This was sort of obvious and expected because the sign of t square is opposite in these two terms This is very standard when you do a definite integral Right, so what happens is t zero squared again the quadratic term dominates every other term So if I push my initial time larger and larger then My t zero squared dominates everything else in the exponential also goes to infinity in fact very fast Yeah, so if t zero goes to infinity gamma t zero also goes to infinity. So this is something you should remember now What we are claiming is that we cannot Choose a uniform delta independent of t zero All right So and and why we say that we say that because as t zero becomes large Gamma goes to infinity Okay, so let's see. Let's see why this is let's try to understand just by some crude example Okay, why this is Okay, what what am I saying? I'm saying that the delta we chose was Epsilon over gamma e to the power M Let's see. I'm sorry today. I think there was a slight error here. This is not gamma M Okay, I'm sorry here. It should not have been Gamma M, but this should have been gamma e to the power Okay, because the exponential was there so yeah, yeah, that's there was an error there. That's okay All right. Now the delta we chose was epsilon over gamma e to the power M Okay, now Now what are we saying that? We cannot choose a uniform delta for all initial time Okay, so why are we saying this we know one thing for sure that? as Beyond a certain time beyond say T zero greater than equal to some cap T Gamma of T zero increasing function Okay, gamma of T zero is an increasing function. This is not difficult to again claim right why because If you see the T zero square is dominating to beyond a certain time This entire quantity right this entire gamma quantity this entire quantity in the exponent of gamma Okay, is going to become positive and remain positive and this which will continue to increase Okay, so Is going to be an increasing function? Okay Another way of looking at it would be to sort of try to take a derivative of this exponent and confirm it All right, which is also not a difficult thing to do. So if I take the derivative also, I can confirm Right. So what is the derivative here very quickly? This will be of Of this exponent. So I'm trying to take a derivative of this guy DDT is Six Cos T zero minus six cos T zero Plus six Cos T zero Okay, minus Six T zero sin T zero plus twice T zero So here these two will cancel out and I'm left with just this much Okay, twice T zero minus Six T zero sin T zero all right. So let's try to See what happens here. Is this going to be always increasing let me be careful here before I claim this this is Going to be T zero sorry T zero So this is going to be T zero Times two minus six sin T zero I see this is not evidently increasing is it Let's see if I miss something here Six cos T zero Plus six cosine T zero minus six T zero sin T zero. Yeah, that seems about right Seems about right plus twice T zero okay, and This is going to sort of be subtracted by This quantity Okay No, this is not evidently Increasing unless I'm missing something All right. Okay. So this is something I need to verify so In any case we are sort of Almost at the end of the time for this lecture So what we will do is we will sort of continue our discussion on Whether this is uniformly stable or not In the next lecture All right. So anyway, what have we seen today? So what we have as Seen today is The definition of uniform stability Which is slightly different from stability in the sense that it is uniform with respect to the initial time The Delta doesn't just depend on initial time and we started to look at an example a very nice example Where we could actually solve the differential equation and verify stability at this point We are now trying to check whether the system is uniformly stable or not and this we will complete next time So that is it folks for this time we will meet again soon. Thank you