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 there’s curvature results.
The first none zero one is one of these that could be, that’s a big red flag that
it’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.
