Moving on to new material now.

So, so far, globally stabilizing, that's what we had.

This Vdot is just negative semidefinite because the sigmas don't appear.

So we don't know if Vdot,

what sigma would be when Vdot goes to zero.

What we want to look at next is how do we prove stability,

and this is what we had with the spring-mass-damper system.

In fact, here this was -c * xdot squared.

And we ended up looking at higher order derivatives of V to figure out the stability.

This was the Mukai-Chen Theorem.

And that's what we're looking at here.

To apply Mukai -Chen,

you have to figure out for what set of states does Vdot vanish.

And with these systems, we always tend to write

Lyapunov rates like this – that's what we're enforcing.

So, it's simply when delta omega is zero my Vdot vanishes.

So now, I want to take higher-order derivatives and

evaluate it on the sets where Delta omega is zero.

Taking your first derivative,

if I do this from here,

with chain rule, you do this and it's a scalar, you can flip it.

You can quickly prove to yourself this becomes this.

So del omega times P,

posit-definite, symmetric, del omega dot.

And evaluate it when del omega is zero.

This whole thing is zero.

So here's my first argument to you then.

As you could plug in what is del omega dot,

put in the equations of motion,

and do a lot more complicated stuff.

This is kind of a faster route.

This is gonna be zero because,

you know, del omega is zero.

And what's the key argument about del omega dot?

Zero times something's gonna give me zero.

If that something is infinite,

finite.

Finite. Yes. Finite, exactly.

Only you don't really care what the true value is.

At this stage, zero times as long as you can say it's finite.

Why do we know it's finite?

Because we know that's stable. Exactly.

We've already proved the system is stable.

There's no way you'd have

infinite acceleration because that would move you all over the place.

Right? That can't happen.

So that's nice. This gives you a little bit of a shortcut.

I don't have to plug in all the equations,

which you can do, it's just this is kind of a,

but you have to argue in an exam, "Hey,

this is gonna be finite because of stability.

Therefore, this term is guaranteed zero."

Good. So Mukai -Chen,

we take higher-order derivatives.

The even derivatives evaluated on the set where Vdot vanished should be zero.

And for stability, it should be an odd derivative since we're dealing with

a second-order system and this kind of stuff we're not

controlling circuits and other kind of a thing.

It tends to always be the third derivative.

So if you differentiate this one again,

you're gonna get a del omega dot P del omega dot.

And the other one, if you differentiate this one,

it's del omega * P del omega double dot evaluated on the set where del omega's zero,

because of the stability,

we know that all the del omegas have to be bounded.

You can't have infinite jerks somewhere again.

This is gonna go to zero or you could plug in

the equations and prove that that would work too.

What we're left with, we have to figure out is what happened to del omega dot.

When we derived the equations and plugged in the conditions,

this was the linear closed-loop dynamics.

This was the condition we had to enforce to get our Vdots to

have this particular negative semidefinite form.

So, let me just show you where that was.

I see some frowns.

So when we did this,

we took the derivative, the body frame derivative of that.

All this had to be true, then this set, it equal to here.

This all comes over into one form.

Really, here.

So that's actually a body frame.

I think I have a dot in there,

but that should be a prime.

Yeah. Doesn't matter. It works.

So what we're gonna look at now is what happens when these things are zero,

[inaudible] are not zero.

Del omega dot, what is del omega dot when del omega is zero?

That's what we need from here.

And, so here, I'm just solving for del omega dot.

You invert the inertia tensor,

bring it to here, then set del omega equal to zero.

This is the expression. That's the angular acceleration you would have with

this control if there's

no rate error and there's still a state error. That's when it comes in.

And then the last step is you plug this part back up into here and simplify,

and you end up with this expression.

And this turns out to be negative definite in terms of sigmas.

And why is that? Well, minus,

that's good, K squared.

It's a positive number squared, definitely positive.

Then there's sigma transpose,

a big three-by-three matrix, times sigma.

So if this matrix is positive definite, we're good.

And the inertia tensor we know is posit-definite.

Therefore, the inverse of a posit-definite matrix is still posit-definite.

That's good. P is also a posit-definite matrix and symmetric.

So there's a theorem that says if you

multiply two symmetric posit-definite matrices with each other,

you end up with a symmetric posit-definite matrix again.

And that's what we're taking advantage of here.

So the product of I inverse P I is gonna be a posit-definite symmetric matrix,

and that's why now here the first nonzero higher-order derivative is

this negative definite quantity in terms of the sigma and this

essentially proves we would be converging.

This is not just stable, it's,

in fact, asymptoticly stable.

This result, as you can see,

is good for any sigmas.

This is also a globally asymptotically stabilizing result that you'd have here.

Any questions?

You kind of do these steps, you definitely should do this by hand.

I'm kind of showing the primary thing,

trying to plug in the right equations.

Go through the stuff. You will get there.

But that's Mukai-Chen that we can use to prove that.