Then we want to look at Asymptotic Stability. Here it's basically ditto, you do the same stuff you just did for stability, right? We kept with the definitions, began from boundedness to stable to asymptotic. There's extra layers that you add on arguments. So to go from stability to asymptotic stability you do everything you did before to prove that it's stable, again when I say stable it implies Lyapunov stable. But then there is an extra step and that extra step is you're talking about V dot be negative semi definite, that's what guarantees stability. Now you have to prove that neighbourhood V dot is negative definite. It basically means that V dot function, that space, there's no other flat spots where you can get stuck on. And that will only converge, you'll always have negative energy rate, essentially, until you reach the origin. There's no other saddle points or flat spots or equilibrias or whatever you want to describe them. So that's the property that we have to prove to guarantee asymptotic stability. Now we know it will get there. It doesn't guarantee you'll get there quick or slow, it just guarantees it'll get there if t goes to infinity all right? That's the distinctions. So lets look at our favorite system Spring Mass Damper it's a nice system. It's a linear system but all of the things we do should hold for linear systems as well. So with the spring mass damper system, I still need a Lyapunov function. Now here's a key thing when we design Lyapunov functions, you have to consider first you're looking at the stability about which point. If it's an equilibrium, then define your coordinates relative to that equilibrium. If it's a reference, you'll have delta x's, then we define the things we care about as a departure motion relative to that. Here, I'm just looking at the origin, right? Is the origin, which is an equilibrium point, is that one going to be, if I have x in here and x dot double dot, a zero, this whole thing is zero and it stays to x. The origin is an equilibrium. Now, when you write your v function, you have to do it in terms of all the states you care about. We talked earlier about the state vector, right? And for spring mass damper system, I need both x and x-dot to be 0 to remain at 0. If somebody gives you a system which is x = 0, You have no guarantee in a spring mass damper resist that you will remain at 0 because you might have a rate. You might just be crossing 0. So you need x and x dot to be 0 in this case. >> Daniel, your class notes the V, the V dot is missing the dot just FYI. >> Where? >> In the notes in the left side? >> This one? >> Yeah. >> I may have fixed that later, thanks. If you don't see that definitely put that in your notes. Thanks for pointing that out. [COUGH] So, this is why here we have, we care about both, what happens to x and x., this is the second order of differential system that we have. In one of the homework you're doing now, actually you only care about It's a first order differential equation, x dot equal to something. In that case, if I know x goes to zero, my x dots have to be zero and you're done. It's a first order system. In that case you only care about one of the variables. Here we're dealing with a second order system so I care about both position and rates. So we still need a Lyapunov Function. And what you'll find is rate this again is a very true. Once you have good Lyapunov functions don't throw it away, because you probably can use can use it for lots of different dynamical systems or different set ups. We will apply the same Lyapunov function and then we modify the control, and it's the same Lyapunov function by giving us slightly different v dots but we can still argue these properties. And then you modify it again with the same Lyapunov function. And hopefully if your life is good, you can still argue some other properties and it's conclusions and so forth right? So in that spirit, we're going to reuse total energy that we had. Kinetic plus potential energy, that's our Lyapunov function. It's 0 when x and x dot are 0 and everywhere else, it's non-zero. Which is great, and let me see where that goes. So, now if you plug it in, you get these equations of motion, this part, if you plug in these terms, this plus this has to be minus this. So I get minus c times x.squared. Now, is this v.negative semi-definite, or negative definite. What do you guys think? Debo? >> Definite, if c is nonzero and positive. >> Yep, c is a positive number, nonzero, yep. >> Actually, well, theta dot that's true, it's not x- >> So is that negative semi-definite, or negative definite? All right, this is a- >> Burn I mean, x dot is zero, only not zero when x is zero, first of all. If it's the spring system so that's not. >> I'm confused, what are you arguing, yes or which one? >> I'm not sure, I'm thinking about it. X-dot is not equal to 0 when x equals 0. >> Immediately, even a simple problem you go, wait a minute. And I'm glad, because this is often like, wait a minute, what's going on here? The beta definition means at the origin, when your states are 0, is this function 0? Well, If x and x dot are zero this function is zero so that's good. If for any non zero state, is this function guaranteed to be negative? If yes, it's negative definite. Is that true here? For any non zero state, is this function guaranteed to be negative? Yes? >> Yeah >> What if x is equal to 5 and x dot is equal to 0? What is my v dot going to be? 0? So what does that imply? All right? That's the crux of it, really. So if you look at this function. See that's why it's really important when you write out your v's, especially when you're just getting started with the Lyapunov functions, use this. Every time you write them, write out all your states. This V is a function of x and x dot. That means you care about both of them. We will talk today about untumbling. Just despinning objects in space and there we don't care about the attitude. I just want to stop tumbling. So then we only care about x dot and there'll be differentiation. Yes, sir Matt. >> The point where we're evaluating v dot to determine whether it is some sort of definite or semi-definite, we don't care at all what v was? >> No, v was actually posit-definite. In fact, it was globally posit-definite, it's good for any initial conditions. This v-dot though is only negative semi-definite, because x-dot has to be 0 for V dot to be zero. In fact, from this, you can argue with some electro math, that x dot must go to zero. The system will come to rest. It'll stop moving, but the question is where. Now we know from a spring mass damper system, it's going to come to rest at the origin, but this math doesn't prove it yet. Because v dot could be zero if x is five, x doesn't appear here. So we just don't know with these and this is the important stuff with this argument, it's an if statement. If you can prove this is negative definite then it is as particularly stable, if you can't prove it's negative definite. It doesn't mean it's not asymptotically stable. We just don't know. Maybe a smarter Lyapunov function would actually prove asymptotic stability. And so we have to use different arguments. So is everybody good why this is negative semi-definite? We care about x and x dot, but it's only negative definite in terms of x dot x doesn't appear, so it's not definite in terms of x. But the function is always zero, then, so it's, it's semi-definite. That means, we at least, we've still proven stability, which is nice, but we know the system asymptotically stable. So, how do we now prove asymptotic stability? If you've taken any classes on non-linear systems. Anybody here heard of Lasal's invariance principle by chance? Okay, a few of you, this a coronary I really like this extra theory that we're going to talk about here, this one's developed by Mukherjee and Cheng I'm giving you the original paper. So if you do a lot of work on those, I definitely recommend pulling it up and reading it. It's a great paper but what they're looking at is, well, I just don't know if it's asymptotically stable or not yet. And with this long dot so I need extra arguments to see this. And sadly for mechanical systems this is typically what we end up with. This is not an odd ball case, this is almost the dominant case you run into the mechanical systems that we can prove stability with the dot but we can't prove asymptotic stability. because we need functions that are built, anyway, just the way the mathematics work out. So what does Mukherjee/Chen do? They basically say, don't stop at the first derivative. We don't just look at V dot, we also look at V double dot, and V triple dot and V quad dot, and V five dot. Just, you keep taking derivatives. And you're looking for a very particular pattern that happens. So this is the process, the first thing you have to in identify here is the domain where the status, the set of states where v dot actually goes to zero. Because v dot here in this mechanical systems they always tend to go to zero when x dot is zero, that's great. That means it's negative definite and x dot. I know my rates are going to go to zero, I just don't know where I'm going to settle. So, in this here, in this state vector, would be basically the set of states where x dot is zero, that's my domain omega. So, I only need to look at that sub-space. And then I have to take higher order derivatives and you evaluate them. You take these derivatives of these functions and you evaluate them on the set of states here, when is x dot equal to zero? We plug it in and look what happens to the math. And what will happen is we ended up here with a first derivative. You take a second derivative. That's going to be zero typically and you keep on taking derivatives. For second order mechanical systems it'll always be the third derivative all of a sudden, that's non-zero. And if you can show the third derivative Is negative definite for all the states you care about here. From here we know x dot goes to zero. We don't know what happens to x. We will have a third derivative in terms of x that's actually negative definite. So this is like looking at higher order curvatures of these bowls. You've now proven you will actually converge in the system. If it's positive definite then you're in trouble. It's probably going to diverge. And there's whole instability theorems around this as well. As we're designing controls, we're not trying to be evil, we're trying to do good. So we're going to try to design controls that give us asymptotic conversions not asymptotic diversions out of the system. So that's the math. Let's go through that, we'll see when it get supplied, it's actually easy to follow. We got to this point, first thing we identify when is x.v., = 0 and that's only going to happen when x dot is 0, right? -c* d bowl c had to be a positive non 0 value then that's true. So I'm going to evaluate this on that set, here my V dot was simply minus cx., squared you take the derivative of that with chain rule you end up with -2cx.*x.. Now let's see, we know x dot will be zero. I don't know what happens to x. I'm not quiet sure what happens to x double dot. Different ways we can approach this subtlety here. One is I have x double dot always for my equations of motion, so you could plug that in. And that's what I'm doing here. So I multiple and divide by mass, so I can just plug this in directly. You end up with this system, and here I have to evaluate this function. When x dot is equal to zero, and this is zero, this is zero, this is something, and this whole thing is going to be zero. When is zero times something equal to zero? What does the something have to be? Finite, right? It just can't, it can be anything except for infinity. Otherwise, who knows what the answer is? Infinity and zero you have to look at the structure of that math that lead to that. So here, why do we know x is not going to be infinity? In the system. >> [INAUDIBLE] >> Are you saying I can't drive mechanical system unstable to infinity? I can prove you wrong. [LAUGH] Those sine errors, man, right. So just saying a mechanical system wouldn't necessarily fly. What's the key, why do we know x is not going to go to infinity? It is stable exactly. Don't ignore our early arguments. This already proved the system is stable, so you give it some finite response. There's going to be a finite region that the staff is going to be in, right? So we know we have local stability already. We just don't know if we have some asymptotic so x would never be infinity. And if x is stable actually you can also, this is another approach, you could keep it in this form and argue well, because the system is stable, the x double dots can't be infinity otherwise you would not have a stable system. It would have jumped infinitely faster than infinity with an acceleration like that. So finite times 0 would be 0 as well. So, as you do this in your homework and try this yourself, you can do either of those arguments. But you have to argue, hey, this is something stable, it's bounded times 0. So as we expected, the second derivative evaluated on the set where V dot vanishes. That second derivative also goes to zero. That's good, that's what we expect. Now let's take the next derivative. So we have this function, and taking another derivative, so a lot of chain rule from this function here, and then you plug in the equations of motion. Any time you get the derivative of x dot there will be an x double dot. You plug in the equations of motion. And I'm just pulling out the constants. This is what I get. Now I have to evaluate this triple derivative on the set where v dot vanished which means x dot is equal to zero. That means this is going to vanish, this is going to vanish, this is vanishing, this is vanishing. And again because everything is stable, we know x is finite and all that good stuff. What it leaves you is this one term, now that was vanished. Here's my term, it's going to leave you with this term times this stuff. So in the end my V triple dot is -2 which is a positive number, c, a positive number, k squared, a positive number, m squared, a positive number. So positive minus positive*x squred, this function is now negative definite in terms of the unknown state x. I know my rates go to 0, I just don't know where else I was going to settle all right. So this is the procedure that we use applied to a very simple spring mass damper system. And how we can now prove asymptotic stability. So it takes extra steps. And we'll do the same thing for attitude motion. Yes sir. >> Would you prove it's not asymptoticly statement? >> There, I'd have to go back to this one. Typically if you look in dynamical systems then thereâ€™s curvature results. The first none zero one is one of these that could be, thatâ€™s a big red flag that itâ€™s not going to work you're not going to have conversion. We know we have stability because of this. But we just don't know about convergence. So I think that's where it manifests but I want to just I have to go look at this paper again to see exactly where it manifests. So I do recommend if you work in this area, I'm showing you the one result when we're designing controls we have power of what we design. We will aim for something that makes this work and give us an asymptotically stable control. Yes Debo. >> So well first of all to rephrase this question then it's not an if and only if this theorem or another- >> Yes, that's correct. This is not an if and only if. This is, if this holds, then we have it. Not inverse. >> And I was going to ask, can you, I mean, it's more of a hands-on way. But can you just show that x dot, if x is zero, x dot will not be zero? More like within by integrating and looking at your function more closely. >> You could, but that's basically the following LaSalle's invariance. You're looking for the largest in variance set. That happens and that one is going to be is equal to 0. >> It's a local asymptotic stability if you showed that. >> Yeah, I have really discussed local and global but, in this case, this argument is true for any x which is nice. To argue global stability, that's on another slide right now that I'll show you. That's an extra thing is required of the v functions. Otherwise people came up with some weird degenerate cases. But this one would actually be globally asymptotically stable, which we know a linear spring mass damper should be in this case. V holds for everything. This argument, the definiteness holds for any x. But there's one extra statement that you really have to prove for global stability, okay? So if you think asymptotic on our mechanical systems, we typically have to look at second and third. Second should be 0, third should be negative def in the stuff that wasn't appearing in v-dot itself, that's the pattern.