0:06

So, okay, we've just wrapped up,

Â what does it mean to be positive definite, positive semi-definite?

Â Now we want to tie this back to stability, right?

Â We care about what if I perturb something slightly?

Â What if I perturb it a lot?

Â I don't want to have to linearize.

Â How do I now apply this?

Â This is going to fall into place very quickly.

Â This is the Lyapunov direct method basically.

Â It says now, [LAUGH] the next step is we're talking about, besides definiteness,

Â what is a Lyapunov function?

Â And once we've proven we have a Lyapunov function, actually we've proven stability.

Â You will see the last step is very anticlimactic.

Â So a Lyapunov function is always a scalar function like kinetic energy, right?

Â But it's in terms of whatever all your states are, which could be omegus.

Â If you use kinetic energy of a rigid body,

Â it's one over two omega transposed i omega, right?

Â And we just mentioned i is a symmetric

Â posit definite matrix which makes kinetic energy actually.

Â It's only going to be zero at zero speeds, and

Â everywhere else it's nonzero, that's something we'll be using a lot.

Â So a Lyapunov function is always a scalar function subject to this dynamical system.

Â And we're going to throw in our equations of motion, and attitude, and rotation and

Â everything.

Â And we're talking, if it is continuous this function and

Â there exists in neighborhood, such that for

Â any states, that we are arguing local stability here with Lyapunov.

Â There is a neighborhood that if you're staying within that one,

Â this v function is positive definite about x r.

Â So you know what that means.

Â It's kind of like x squared is the bowl shape, right?

Â There is a Lyapunov function, this Lyapunov function has continuous partial

Â derivatives, that's one of the requirements.

Â It takes two lectures to explain why, but that's important.

Â And then the last one is V dot is negative semi definite.

Â So once you define this V, you can take its time derivative.

Â Like we've done this with kinetic energy, and

Â then we took the time derivative kinetic energy which gave us the power equation.

Â And actually you had to derive that in your exam.

Â We're reusing the same math here.

Â It just doesn't have to be explicitly kinetic energy,

Â you can throw any mathematical function in here that satisfies these properties.

Â And you will see some things that we do there.

Â But if it has these properties, then we're good.

Â Then this function is a Lyapunov function.

Â 2:30

So this part in particular, V dot being negative semi-definite is important.

Â Because that means Vdot being negative everywhere, or zero,

Â if I have zero power loss.

Â That would be no external torque acting on it or it's acting in very particular ways.

Â Then the v function just stays constant and it's kind of like boundedness.

Â Okay you've got some tumble, you've got some air but

Â you can bound how big they are.

Â They're not going to get worse, they're not going to get bigger.

Â So an object tumbling without external torques may have a tumble.

Â But you can come up with bounds around that to say,

Â hey it's not going to spin up, it's not going to slow down and come to zero.

Â But it will be within this area, all right?

Â Being negative is good because that means we lose energy.

Â And our energies are,

Â these V functions are always defined relative to a reference.

Â So these things start to always decay, these Lyapunov functions getting smaller.

Â You're constraining yourself to be around that reference trajectory.

Â And now we start to get closer and closer, doesn't guarantee conversions yet.

Â So that's because they're semi-definite.

Â Now, there are sports where you might have saddle points and it doesn't go on.

Â But that's the key properties, the things we always check for is we come up with

Â the v, prove it's symmetric, sorry, prove that it's positive definite about xr.

Â And prove that v dot is negative definite.

Â When you evaluate this, v is a function of x.

Â So to get the time derivative you do basically chain rule.

Â You take the partial of v with respect to x.

Â And then the derivative of x with respect to time, which is x dot, and

Â x dot is nothing but f.

Â So, visually, that's a scalar function,

Â that's the gradient operator times your dynamics.

Â So they often draw these balls in this graph.

Â And what it guarantees is that projection is always, if this is less than here,

Â you're either getting closer to the origin.

Â Or you're staying the same distance if this v dot is equal to 0.

Â So, you're kind of orbiting the origin, and that's what it means graphically.

Â 4:50

And we can do this, you'll see examples in the homework.

Â We do this pendulum problem specifically in the homework where you come up with

Â the simple Lyapunov functions, do these derivatives, prove these properties.

Â And you can prove, for this one at least, it's stable.

Â Again, not asymptotically stable, Lyapunov stable just means you can pick any epsilon

Â finite, and then there's an initial set of conditions that will get you there.

Â 5:16

Any questions on this?

Â This is a lot of definitions today, Teebo.

Â >> Negative semi-definite, that means the derivative,

Â that means it actually goes like this?

Â >> Yes exactly.

Â >> Does that, okay.

Â >> No this is the bowl V, not the V dot part.

Â But the derivative, the gradients of that have to be always pointing inward or

Â not outward at least.

Â Your v function, my v function itself never gets bigger with this stuff.

Â I'm not driving in outward in a sense.

Â So let me draw a quick example, just to visually kind of highlight that.

Â If this is your v function, basically no matter where you are here.

Â And this is a multi-dimensional thing, if you keep losing energy levels

Â then these levels are going to, by the nature of this function are going to con

Â strain you closer and closer and closer to being through the origin.

Â But if we have a Lyapunov function that's not just an x squared,

Â in multi-dimensional space.

Â Let's say we have a Lyapunov function that does this, and then goes to zero.

Â You can see here we took the gradients protected by that.

Â You get regions where yeah, this works well here.

Â But all of a sudden on the outside with the gradients

Â you're actually going to be driving it further out.

Â This wouldn't be a globally stable system,

Â this would probably just be a locally stable system in this case.

Â >> My point is that the derivative V dot close to zero is positive.

Â 6:46

But it's negative on the left and positive on the right, I mean I guess I don't-

Â >> This is v not v dot.

Â >> Right.

Â >> I'm just trying v.

Â >> And v dot would look like?

Â >> And vdot would have to have a negative form, that if you're here,

Â if this is a stable region for example.

Â If you do the v dot times, that's the partial of v,

Â get it's gradient dotted with f, your equations of motion.

Â So now you have to put in the right equations of motion with this function.

Â I can't prove it without applying it to a specific dynamical system.

Â But basically, this is going to be a region where you are actually being

Â driven, the gradients are driving you towards there.

Â Whereas out here, the gradients are driving you away.

Â I mean, that's a little bit of a visual interpretation, not particularly rigorous,

Â but sometimes it helps people understand that.

Â 7:31

So we'll pick up here, this is called Lyapunov's Direct Method.

Â It's a way to prove the stability properties without

Â ever solving its differential equations.

Â So we get these energy like Lyapunov functions.

Â And so we've gone through definiteness.

Â This is the Lyapunov function that we have.

Â We just went through this three different conditions, and

Â it's all defined relative to a reference.

Â And this includes regulation and tracking and everything.

Â And basically v dot, these here,

Â this is not an explicitly time dependent function.

Â So to get the time derivative, you have to use chain rule.

Â Take the partial of v with respect to the states,

Â times the derivative of the states at the expected time, that's your x dot.

Â You plug in your differential equations, and it's basically the gradient

Â of your Lyapunov functions mapped on to your dynamical system has to be negative.

Â So it means that at best you're always converging to the origin,

Â at worst it's negative semi-definite.

Â If this v dot is zero, that means over time you're holding the same v value.

Â So if you think of a two dimensional ball it just means you're moving along a rim.

Â That's exactly what would happen with a spring mass system.

Â You could think of this as kinetic energy, and

Â in that case with a spring mass potential kinetic energy right.

Â Energy is constant and you have a v dot that is zero, that's what's happening.

Â So that's what makes this Lyapunov stability means you can find such

Â a function with all these properties around the states of interest.

Â Here I put it as the origin, but it also could be the [INAUDIBLE].

Â So let's look at spring mass system mathematics.

Â A spring mass is quite simple, we've go mx double + kx = 0.

Â No damping, I'd have to come up with the Lyapunov function, here, and

Â basically using total energy.

Â And you will find that anti, when we do the Lyapunov function,

Â they're often inspired by energy, but they don't have to be precisely energy.

Â Any function that satisfies these properties is useful.

Â But energy is a good source of inspiration to come up with mathematical structures

Â that are quite convenient, as we'll see.

Â So, this first part, mass over 2 velocity squared is kinetic energy.

Â Spring stiffness over 2, times deflection, squared is your.

Â 9:38

No, that's kinetic energy and then spring energy.

Â So that's potential, kinetic, that's a total energy.

Â If we use this, you take its time derivative.

Â So with chain rule you get mass over 2, times 2, times x dot, times x double dot.

Â 2s cancel out, and you're left with m, x double dot times x dot.

Â I factored out x dot, this one I'm going to do, again, same thing.

Â You're going to get k2 xx dot divided by 2.

Â The 2s all cancel, you end up with k times x times x dot.

Â Great, now the last step is here.

Â This is just taking this Lyapunov function, differentiating it.

Â You still have to plug in the equations of motion here.

Â So this bracketed term, you could solve for x double dot, plug it in.

Â Or just realize mx double dot + kx, by definition, is equal to 0, right?

Â So the v dot in this case is 0, which I'm labeling as negative semi-definite.

Â It satisfies the properties, it's 0 at the origin [COUGH] and it's negative or zero.

Â In this case, only zero, but that includes that case everywhere else, all right?

Â So this a Lyapunov function, and therefore,

Â the spring mass system is Lyapunov stable, within these things, which is good.

Â