The question is, is global stability only possible if there's one equilibrium?
I would lean towards yes, but live right now, teaching, I wouldn't bet my life on it
that some smarter mathematician hasn't come up with a weird degenerate case
where something could asymptotically go crazy and never actually reach that equilibrium.
But it's probably a good sign.
With reference trajectory tracking though,
it changes a lot of that because you're creating artificial equilibria, right?
I wanna follow this path and things get more complicated.
So, there'll be some rigorous tools that we will have to argue it.
But we said, ok, so, we know this pen cannot be asymptotically globally stable
'cause I found a set of states that don't make it converge.
But, is it still stable, right?
For stable it means, can you pick an Epsilon and a Delta,
such that once it enters that condition, it stays within that set of conditions.
What would your-- Daniel, David.
Basically, yeah.
So, if you just make your bound 180, which you can, you're gonna stay in that bound.
Of course, that attitude's almost cheating because we can't get worse than 180,
but some people go to 200 and higher.
But that's right, right?
With 180 plus with zero rates.
That's gonna be the kind of, there's a full state spacing.
That's gonna be the bound that you're going to have.
So, you could say that this system is bounded.
You're never gonna have-- With friction,
you're never gonna have a set of initial conditions where the attitude angle keeps--
Even if you went beyond 180.
So, let's forget that it's, you know, the same thing.
360 is same thing as zero.
There is no way the system would, you know,
asymptotically spin out of control because with damping you would always be losing energy,
and it eventually it would settle there and there would be a bound that you can have.
If you do the clipping of multi revolutions,
it's 180, otherwise you can come up with something slightly different.
But that's an example quickly that hopefully starts to illustrate what we're doing, right?
Our controls.
We will design them to be asymptotic.
But then once we throw in other issues, you will see where it comes up.
So, asymptotic stability, the Epsilon now can go to zero.
We converge.
Andrew.
You know...
Tough demo.
I've had two hours of sleep last night, sorry.
So, just the pendulum problem.
It's that what you're talking about.
It's both.
You said stable as both Lagrange and Lyaponuv stable.
Not asymptotically but.
Yes.
Even though it's for all Epsilon, but that Epsilon would-- You wouldn't say Lagrange
and Lyapunov stable typically, cause if it's Lyapunov stable, it's a stronger argument.
Yeah, yeah.
That already overrides Lagrange stability.
But it's not--the key is, it's not just locally stable,
it's actually globally as Lyapunov stable.
But it's not globally asymptotically stable
because I could give it initial conditions where if you get that error,
you're stuck, right?
And this happens a lot then in stuff.
And what if you happen to be just exactly upside down?
You're gonna stay there.
And if you're off a little bit, you might recover, but very often what happens then,
kind of like with the separate tricks,
it might take a really long time to get there as well.
