 Hello, welcome to yet another session of our NPTEL on nonlinear internet with control. This is Shrekan Supermart from Systems and Control IIT Bombay. As always, we are in front of this very nice background, this Mars rover and what we learned in this course are the sort of algorithms that drive systems like this autonomously without communication to any earth station on Mars and the moon. So let's go forward with what we have been doing in this course until now. And so we had started off with talking about the Babalat's Lemma and its corollary last time. So we had sort of discussed the corollary to the Babalat's Lemma and in order to motivate how to use this Babalat's Lemma, we had started by discussing a standard Sprigman's damper system. We, of course, wrote the dynamics of the system, created error states corresponding to reference trajectories, wrote everything in, wrote the error dynamics in state space form. So we wrote the error dynamics in the state space form. Then we created a target system that we wanted to achieve and we did that by choosing a very specific feedback controller. It was, of course, evident to us by standard linear systems methods that this is a exponentially stable system. However, we want to prove the same using energy functionals like these, because this is what will eventually help us when the target systems turn out to be nonlinear, and of course, also in adaptive control cases. So what we did with Babalat's Lemma was only to prove asymptotic convergence. So we cannot say anything about transient performance. Back things could happen in the trans genes, but we are only using the Babalat's Lemma to claim that for large time, the errors converge to zero. So we did a few steps. So we took the derivative of the candidate potential function and we found that the derivative was less than equal to zero, while the potential function was greater than equal to zero. And using this, we started our signal chasing analysis. And we did a few steps. The first step was to show that V infinity, that is the limit as T grows to infinity, V exists, and it's finite. Then the second step was to show that all the closed loop signals, that is, E1 and E2 in this case, were bounded. And the third and a rather critical step was to show that the signal E2 is in fact an L2 signal. Yeah, very quickly recapping this last point, because this is a very important point. How we did this was we integrated both sides of this V dot. And it was possible to integrate the left hand side only because we know that limit as T goes to infinity, VT is finite, that is, V infinity is a finite quantity. And since we can do that, we have this right hand side, which is essentially the integral of E2 squared. And we also want to remind ourselves that the two norm of E2 looks something like an equation 313. Yeah, and we see that this is very similar to what we have here. And therefore, what we have is V infinity minus V zero is just the square of the two norm V2 multiplied by a gate gate. And so from this, we can compute what is the two norm exactly. And this turns out to be finite and real because of the fact that V infinity is finite and V is non-increasing. And this is essentially what is the definition of a signal to be an N2 that it is L2 norm V finite. Okay, so excellent. So this is where we were until last time. And so today we start off with our lecture number four. So we are on lecture four of the week two. Okay, we are on lecture four of week two. All right, so what do we have here? So the next step is to prove that the derivative of E2 is L infinity. Okay, that is just bounded. Yeah, so you want to prove that the derivative of E2 is bounded. Why do we want to do that? Well, I mean, if you all remember the statement of the Babelatz Lemma in the corollary, you would remember that if a derivative of the function is E is bounded or L infinity, then we can claim that the function itself is uniformly continuous. And we know that uniform continuity plays a rather critical role in the Babelatz Lemma. Okay, so that's essentially why we care about the derivative of the signal being L infinity. All right, excellent. Right, so how do we claim that? We look at the derivative itself. So the derivative is rather straightforward in this case. It's given by this expression. E2 log is minus k1 even and minus k2 E2. You should remember that k1 and k2 are just some positive constants. And we've already proven in step number two that E1 and E2 are bounded. Okay, and because even in E2 are bounded, we have that k1 E1 and k2 E2 are also bounded. And therefore, this quantity is also bounded. So, excellent. Right, so now if we use the corollary to the Babelatz Lemma, not the Babelatz Lemma itself, but the corollary. Yeah, this is corollary 2.2. What did it say? It said that if a signal is L infinity and some Lp and further its derivative is L infinity, then the signal itself goes to 0 as t goes to infinity. Okay. All right. So, and that's exactly what we have here. That's exactly what we have here. We have that E2 is L infinity by the fact that E2 is bounded in step two. Right. So, this I will in fact write it for this particular case. So, this is step two. E2 is L2 was in fact step three. Right. We prove E2 is L2 in step three and the fact that the derivative is L infinity was proven in step four. Okay. And with all these ingredients, we can invoke this corollary 2.2. Yeah, let's go back again and look at the corollary just to remind ourselves. If a function is in any L infinity and any Lp and further the derivative is in L infinity, then the function itself goes to 0 as t goes to infinity. Okay. Great. Right. So, therefore, we have proved that limit as t goes to infinity, E2 of t is exactly zero. Right. So, we have actually completed the proof of one part. Yeah. Or we have completed one part of the proof. I'm sorry. Yeah. Because we started by trying to prove that both signals E1 and E2 are going to go to zero as t goes to infinity. And so, we have been successfully improving one part of it. Okay. What does this part say in the context of the spring mass damper? It says that your velocities, yeah, converge to the desired velocity trajectories. Okay. We've already proved in 318 that the velocities converge to the desired velocity trajectories. Okay. All right. So, moving on. So, this is only one part of the proof. So, one half of the proof. Therefore, we would like to move on and try to complete the proof. Okay. And let's see how it looks. All right. So, in the next step, we want to prove that the signal E2 dot is an integrable signal. Okay. Notice where we started. Okay. Notice where we started. This is very, very critical again because these steps are rather standard. So, you need to understand these steps very well. So, notice where we started in step six. Until step five, we've been able to prove that E2 goes to zero. And what exactly is E2? Okay. E2 is the only quantity that shows up in V dot. Okay. So, the first half of the proof essentially lets you conclude that terms appearing in V dot converge to zero. Okay. So, this is what we will be able to prove in the first half of the proof of, you know, first half of the sequence of steps. Yeah. That all the terms that appear in V dot, not all the terms. Of course, I mean, we are doing an inequality that so we essentially have all quadratic terms. Well, I don't want to say quadratic. Okay. I'm sorry. I don't want to say quadratic per se. But all the terms that, you know, you have sort of in the end in V dot. And you will see more examples of this. It will become clear. This is merely a guideline. It's not a law or a theorem. Okay. So, if you see V dot in this case, it contained only E2. Okay. And so in the first half of the proof, we are able to show that E2 goes to zero. Then where do we start in the second half of the proof? Now, this is an important question. Okay. So, in step six, we start exactly at the derivative of E2. Okay. Until now, we have proven that E2 goes to zero. And now let's all recall all this myths that we spoke about. Yeah. We know that E2 converging to zero does not necessarily mean that the derivative of E2 is going to go to a constant. Okay. It may not even have a limit. We looked at several examples. I mean, really, you know, it's rather easy to construct such examples. Right. So, I mean, something like for example, sine t squared, let's see, divided by t. Okay. This quantity has a limit as t goes to infinity. Right. So, this is, this was the example. Right. But if I take the derivative, right, if I take the derivative minus sine t squared over t squared plus 2 cosine t squared has no limit as t goes to infinity. Okay. So, this is, this was sort of, yeah, this funny looking box is sort of this, you know, what we proved. Right. So, sine, so just the fact that a signal goes to zero does not mean that its derivative is going to converge. Okay. And so, this is exactly what we start to do in this next step. Yeah. We start looking at the derivative of the signals we proved are going to zero. Okay. In this first half, we proved that all the terms appearing in v dot go to zero. And now we start with the derivative of those signals. Okay. So, we prove that E2 dot E2 goes to zero. So, then we start with E2 dot in the next stage. And this is important to remember. Okay. So, I'm going to write this. Start with derivative converging terms from above. We start here with all the converging terms from above. So, the derivative of all the converging terms from above. So, in this case, this is only E2 dot. Okay. So, what do we show? We show that E2 dot is integrable. Why? Just integrate it. How do we conclude this? Simply by integrating it. If I take this integral right here, you see that at infinity, so this is again same idea as before. So, you have this is DE2 by DT and DT. So, this is just integral of DE2, which is essentially E2 at infinity minus E2 at zero. And again, why was this possible? This is possible because E2 infinity is nothing but zero. E2 infinity is the limit of E2 as T goes to infinity. And that's zero by the previous step. Okay. Otherwise, this would not have been possible. If we had not proven that the E2 had a limit as T goes to infinity, then this step would have been impossible. Okay. Excellent. Excellent. So, I hope we understand that quite well. Great. So, then because we have this step, this gives me that this integral of E2 dot DT as T goes zero to infinity is just minus E2 at zero. And this is a finite integral because your error started at a finite value. I mean, we wouldn't make any sense otherwise. Okay. So, this is a finite integral. Yeah. And therefore, this essentially helps us conclude that E2 dot is an integrable signal. Okay. E2 dot is an integrable signal. Again, you should remind yourself of the Bapalach Lemma and its corollary because now we are moving towards the original Bapalach Lemma statement. Yeah. Where we need integrability and uniform continuity. Excellent. So, what do we do for uniform continuity? Because we already have integrability, right? So, what do we do for uniform continuity? Evaluate the derivative as simple as that. So, we now claim that E2 double dot is an infinity, right? Because so, this will imply that E2 dot is uniformly continuous. Yeah. Because the derivative of E2 dot is bounded. Therefore, the signal itself has to be uniformly continuous by the lemma we have seen before. Excellent. So, what do I do? How do I compute E2 double dot? Just take the E2 dot and take derivatives again, right? So, just take E2 dot here and I take the derivative again of both sides. So, this gives me minus k1 E1 dot minus k2 E2 dot here and then I substitute for E1 dot and E2 dot using the dynamics again. So, this is minus k1 E2 and minus k2 times minus k1 even minus k2 E2, right? And you will notice that all these terms, again, k1 and k2 are constant. So, they don't play any role in signals becoming unbounded. No question of that happening, right? But what remains are just states here, E1 and E2. And we've already proven in step two, in fact, long, long time ago, step two, that even E2 are bounded, right? Therefore, E2 double dot also has to be bounded, right? Even E2 is bounded. So, the right-hand side is just bounded quantities. So, the left-hand side is obviously bounded. Yeah, some of bounded quantities or difference of bounded quantities is necessarily bounded, right? So, we've essentially now shown that E2 double dot is L infinity, okay? And this immediately tells me that E2 dot is uniformly continuous, okay? Excellent. Now, let's go back and look at the statement of the original Barbara Lutz lemma. What does it say? It says that if a signal is integrable and it's uniformly continuous, then the signal is going to 0 st equals to infinity, okay? So, let's look at what we have. We have just shown that E2 dot is integrable. And because E2 double dot is L infinity, E2 dot is uniformly continuous. So, by the original Barbara Lutz lemma statement, I can immediately conclude that E2 dot, in fact, goes to 0, okay? All right. So, I hope this sinks in well. Yeah, we started after the end of the previous half of steps with the derivative of the terms that we had proven to be going to 0, which is only E2 in this case. In a more general setup, this could be multiple signals, yeah? E2, E3, E4, and so on and so forth, yeah? And in the next stage, we start with the derivative of all of those. So, E3 dot E4 dot E2 dot E3 dot E4 dot E5 dot so on and so forth. And we prove their integrability. We know, why do we start with this step? Because we know this is immediately possible, yeah? Because I prove that I have proven that all of these signals go to 0 as T goes to infinity, right? Therefore, if I integrate it, it's simply going to depend on the endpoints. Yeah, and these endpoints, at these endpoints, that is at infinity and 0, signal is finite. And so we are done. And this is true for all signals, yeah? Or even if you have E2, E3, E4, E5, E100, it doesn't matter. It will come out to be like this. As long as you have proven that these signals converge to 0. Excellent. And in the next stage, we prove that E2 dot is uniformly continuous by proving, by taking its derivative and showing that it's bounded. It is also very easy. I can take many, many derivatives further. This will always be bounded, yeah? Because you'll always get terms in E1 and E2, which are already known to be bounded, okay? Excellent. So, with this, with the Bob Lutz lemma, we have proven that the derivative of E2 also goes to 0, right? So, we've actually taken a nice formal procedure to prove that. After we've proven that E2 goes to 0, we've made a nice, you know, nice formal process to prove that E2 dot also goes to 0. We didn't just conclude it, you know, just like that, all right? Okay, great. So, once I have E2 dot goes to 0, the rest of the proof is not difficult at all. Why is that? I immediately claim that E1 also goes to 0. So, look at the expression for E2 dot. It is minus k1 E1 minus k2 E2. And here I know, right, that the limit as t goes to infinity E2 dot is 0. And I know that limit as t goes to infinity E2 is 0. So, what do I do? I can simply take limit, apologize. I simply take the limit as t goes to infinity of both sides, right? And they have to match because I have an equivalence here, yeah? If I say a is equal to b, a function a of t is equal to function v of t, then the limit as t goes to infinity cannot be different, right? Therefore, if I take limits on both sides, what happens? I know that this guy is going to 0. I know that this guy is going to 0. So, the only way that the left hand side can go to 0 is if this quantity is also going to 0. And k1 is just a positive constant. So, this is possible only if E1 itself also goes to 0 as t goes to infinity, okay? All right? So, this is very, very important, right? So, what have we shown? We have shown with a bunch of, well, nine steps to be precise, yeah? Looks like a lot of steps. But when your problem gets more complicated and you cannot resort to linear analysis, or you have an adaptive system which inherently makes your closed loop system nonlinear, these nine steps will look like a blessing to you, yeah? As of now, it looks like, oh my god, I could just have computed the, you know, eigenvalues and be done in a moment. But we took nine steps. But like I said, when problems become nonlinear and more complex, these nine steps will be super easy, simple, significantly simple, or in fact, just possible, okay? The other step wouldn't even be possible, okay? Excellent. So, we have, as we planned, we've been able to show that E1 and E2 both go to 0 as t goes to infinity, all right? Of course, we could have used this Lyapunov analysis with Lassal invariance. We've not done this still, yeah? In this particular case, we could also have used Lyapunov analysis and Lassal invariance. However, Bablats lemma can be used in a larger scheme of things. So, it's well known that this will work only for time invariant systems, okay? Only for time invariant systems. Of course, there are modern versions which work for some, some sort of time varying behavior, but they're not all encompassing, okay? So, in general, Lassal invariance principle, which again, we are yet to cover, works only for time invariant systems. So, if I have something like this where my gains, yeah, instead of being constant gains, they're time varying gains, again, something that's not uncommon in adaptive control, then only Bablats lemma can be applied. These cannot be applied anymore. Yeah, only Bablats lemma can then be applied to prove convergence, okay? So, of course, this is an exercise, yeah? You are to prove convergence of u and a2 to 0 using these similar steps. Any additional assumptions, you're free to make. And you're, but you have to state all these assumptions. You have to state all these. Now, I want to remind you that we only proved convergence. So, we have only been, we only completed the proof of convergence of these signals to origin, okay? So, we are saying something about the steady state behavior only. We're not saying anything about the transient behavior, yeah? So, remember this, yeah, convergence only has been proven. Stability or the fact that the trajectories remain bounded. See, one thing should be obvious to you already is that by step two here, let's see, by step two here, I mean, it's not like we have got nothing beyond convergence. We did get something more than convergence. We're also saying that even an e2 remain bounded. So, all the closed loop signals. So, notice that our trajectories we chose were themselves bounded with bounded derivatives, all right? So, what does it mean? It means that if even an e2 are bounded, then x1 and x2 are also bounded, okay? So, we do have convergence from this, but we also prove that all our closed loop signals remain bounded, okay? So, this sort of thing that, you know, I gave this funny sort of example, this cannot happen. This cannot happen, yeah? This cannot go to infinity and come back. It can become very large and come back, but everything is still bounded, okay? So, all the closed loop trajectories are going to be bounded and on top of it is proven convergence, okay? So, that's sort of the good thing, okay? So, remember that's what we have. So, we have completed the proof of how to use the Baab-Latz lemma to conclude asymptotic convergence of signals, okay? Stability is a separate question which we address slightly later, all right? So, what did we do today? We continued our proof of convergence of signals using the Baab-Latz lemma, right? And we had finished, you know, proving that a signal was L2, then we proved that whatever terms that appear in V dot, they converge to zero. That's the first thing we proved. Then we start off our next set of steps with the derivative of those signals that we proved already go to zero. And then from that, we proved that these derivatives of those signals are integrable by virtue of their convergence and then that the derivatives are bounded. And from this, we can prove that the derivatives of signals that we proved were going to zero, themselves also go to zero. So, if E2 goes to zero, we prove that E2 dot also goes to zero, okay? And using the dynamics that is how the equations of E2 dot look, we show that even itself also goes to zero. Of course, later on we will also see where these things fail, yeah? And we get into detectability obstacles and adaptive control, okay? So, we will discuss that soon, right? Okay. So, this is where we conclude today. Thank you for your attention.