WEBVTT

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First of all,
the way a nonlinear autonomous

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system looks,
you have had some practice with

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it by now.
This is nonlinear.

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The right-hand side are no
longer simple combinations ax

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plus by.
Nonlinear and autonomous,

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these are function just of x
and y.

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There is no t on the right-hand
side.

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Now, most of today will be
geometric.

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The way to get a geometric
picture of that is first by

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constructing the velocity field
whose components are the

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functions f and g.
This is a velocity field that

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gives a picture of the system
and has solutions.

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The solutions to the system,
from the point of view of

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functions, they would look like
pairs of functions,

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x of t, y of t.

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But, from the point of view of
geometry, when you plot them as

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parametric equations,
they are called trajectories of

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the field F, which simply means
that they are curves everywhere

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having the right velocity.
So a typical curve would look

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like --
There is a trajectory.

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And we know it is a trajectory
because at each point the vector

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on it has, of course,
the right direction,

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the tangent direction,
but more than that,

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it has the right velocity.
So here, for example,

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the point is traveling more
slowly.

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Here it is traveling more
rapidly because the velocity

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vector is bigger,
longer.

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So this is a picture of a
typical trajectory.

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The only other things that I
should mention are the critical

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points.
If you have worked the problems

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for this week,
the first couple of problems,

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you have already seen the
significance of the critical

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points.
Well, from Monday's lecture you

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know from the point of view of
solutions they are constant

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solutions.

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From the point of view of the
field they are where the field

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is zero.
There is no velocity vector,

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in other words.
The velocity vector is zero.

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And, therefore,
a point being there has no

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reason to go anywhere else.
And, spelling it out,

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it's where the partial
derivatives, where the values of

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the functions on the right-hand
side, which give the two

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components, the i and j
components of the field,

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where they are zero.

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That is all I will need by way
of a recall today.

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I don't think I will need
anything else.

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The topic for today is another
kind of behavior that you have

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not yet observed at the computer
screen, unless you have worked

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ahead, and that is there are
trajectories which go along to

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infinity or end up at a critical
point.

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They are the critical points
that just sit there all the

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time.
But there is a third type of

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behavior that a trajectory can
have where it neither sits for

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all time nor goes off for all
time.

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Instead, it repeats itself.
Such a thing is called a closed

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trajectory.
What does it look like?

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Well, it is a closed curve in
the plane that at every point,

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it is a trajectory,
i.e., the arrows at each point,

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let's say it is traced in the
clockwise direction.

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And so the arrows of the field
will go like this.

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Here it is going slowly,
here it is very slow and here

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it picks up a little speed again
and so on.

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Now, for such a trajectory what
is happening?

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Well, it goes around in finite
time and then repeats itself.

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It just goes round and round
forever if you land on that

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trajectory.
It represents a system that

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returns to its original state
periodically.

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It represents periodic behavior
of the system.

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Now, we have seen one example
of that, a simple example where

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this simple system,
x prime equals y,

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y prime equals negative x.

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We could write down the
solutions to that directly,

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but if you want to do
eigenvalues and eigenvectors the

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matrix will look like this.
The equation will be lambda

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squared plus zero lambda plus
one equals zero,

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so the eigenvalues will be plus

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or minus i.
In fact, from then on you could

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work out in the usual ways the
eigenvectors,

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complex eigenvectors and
separate them.

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But, look, you can avoid all
that just by writing down the

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solution.
The solutions are sines and

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cosines.
One basic solution will be x

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equals cosine t,
in which case what is y?

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Well, y is the derivative of
that.

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That will be minus sine t.

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Another basic solution,
we will start with x equals

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sine t.
In which case y will be cosine

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t, its derivative.

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Now, if you do that,
what do these things look like?

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Well, either of these two basic
solutions looks like a circle,

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not traced in the usual way but
in the opposite way.

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For example,
when t is equal to zero it

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starts at the point one,
zero.

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Now, if the minus sign were not
there this would be x equals

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cosine t,
y equals sine t,

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which is the usual
counterclockwise circle.

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But if I change y from sine t
to negative sine t

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it is going around the
other way.

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So this circle is traced that
way.

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And this is a family of
circles, according to the values

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of c1 and c2,
concentric, all of which go

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around clockwise.
So those are closed

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trajectories.
Those are the solutions.

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They are trajectories of the
vector field.

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They are closed.
They come around and they

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repeat in finite time.
Now, these are no good.

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These are the kind I am not
interested in.

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These are commonplace,
and we are interested in good

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stuff today.
And the good stuff we are

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interested in is limit cycles.

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A limit cycle is a closed
trajectory with a couple of

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extra hypotheses.
It is a closed trajectory,

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just like those guys,
but it has something they don't

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have, namely,
it is king of the roost.

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They have to be isolated,
no other guys nearby.

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And they also have to be
stable.

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See, the problem here is that
none of these stands out from

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any of the others.
In other words,

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there must be,
isolated means,

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no others nearby.

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That is just what goes wrong
here.

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Arbitrarily close to each of
these circles is yet another

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circle doing exactly the same
thing.

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That means that there are some
that are only of mild interest.

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What is much more interesting
is to find a cycle where there

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is nothing nearby.
Something, therefore,

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that looks like this.

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Here is our pink guy.
Let's make this one go

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counterclockwise.
Here is a limit cycle,

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it seems to be.
And now what do nearby guys do?

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Well, they should approach it.
Somebody here like that does

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this, spirals in and gets ever
and every closer to that thing.

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Now, it can never join it
because, if it joined it at the

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joining point,
I would have two solutions

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going through this point.
And that is illegal.

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All it can do is get
arbitrarily close.

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On the computer screen it will
look as if it joins it but,

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of course, it cannot.
It is just the resolution,

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the pixels.
Not enough pixels.

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The resolution isn't good
enough.

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And the ones that start further
away will take longer to find

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their way to the limit cycle and
they will always stay outside of

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the earlier guys,
but they will get arbitrarily

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close, too.
How about inside?

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Inside, well,
it starts somewhere and does

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the same thing.
It starts here and will try to

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join the limit cycle.
That is what I mean by

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stability.
Stability means that nearby

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guys, the guys that start
somewhere else eventually

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approach the limit cycle,
regardless of whether they

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start from the outside or start
from the inside.

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So that is stable.
An unstable limit cycle --

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But I am not calling it a limit
cycle if it is unstable.

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I am just calling it a closed
trajectory, but let's draw one

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which is unstable.
Here is the way we will look if

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it is unstable.
Guys that start nearby will be

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repelled, driven somewhere else.
Or, if they start here,

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they will go away from the
thing instead of going toward

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it.
This is unstable.

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And I don't call it a limit
cycle.

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It is just a closed trajectory.

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Cycle because it cycles round
and round.

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Limit because it is the limit
of the nearby curves.

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The other case where it is
unstable is not the limit.

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Of course, you could have a
case also where the curves

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outside spiral in toward it but
the ones inside are repelled and

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do this.
That would be called

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semi-stable.
And you can make up all sorts

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of cases.
And I think I,

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at one point,
drew them in the notes,

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but I am not going to.
The only interesting one,

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of permanent importance that
people study,

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are the actual limit cycles.
No, it was the stable closed

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trajectories.
Notice, by the way,

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a closed trajectory is always a
simple curve.

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Remember what that means from
18.02?

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Simple means it doesn't cross
itself.

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Why doesn't it cross itself?
It cannot cross itself because,

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if it tried to,
what is wrong with that point?

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At that point which way does
the vector field go,

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that way or that way?
Why the interest of limit

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cycles?
Well, because there are systems

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in nature in which just this
type of behavior,

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they have a certain periodic
motion.

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And, if you disturb it,
gradually it returns to its

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original periodic state.
A simple example is breathing.

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Now I have made you all
self-conscious.

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All of you are breathing.
If you are here you are

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breathing.
At what rate are you breathing?

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Well, you are unaware of it,
of course, except now.

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If you are sitting here
listening.

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There is a certain temperature
and a certain air circulation in

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the room.
You are not thinking of

00:13:36.000 --> 00:13:42.000
anything, certainly not of the
lecture, and the lecture is not

00:13:40.000 --> 00:13:46.000
unduly exciting,
you will breathe at a certain

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steady rate which is a little
different for every person but

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that is your rate.
Now, you can artificially

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change that.
You could say now I am going to

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breathe faster.
And indeed you can.

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But, as soon as you stop being
aware of what you are doing,

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the levels of various hormones
and carbon dioxide in your

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bloodstream and so on will
return your breathing to its

00:14:11.000 --> 00:14:17.000
natural state.
In other words,

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that system of your breathing,
which is controlled by various

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chemicals and hormones in the
body, is exhibiting exactly this

00:14:23.000 --> 00:14:29.000
type of behavior.
It has a certain regular

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periodic motion as a system.
And, if disturbed,

00:14:32.000 --> 00:14:38.000
if artificially you set it out
somewhere else,

00:14:34.000 --> 00:14:40.000
it will gradually return to its
original state.

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Now, of course,
if I am running it will be

00:14:40.000 --> 00:14:46.000
different.
Sure.

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If you are running you breathe
faster, but that is because the

00:14:45.000 --> 00:14:51.000
parameters in the system,
the a's and the b's in the

00:14:48.000 --> 00:14:54.000
equation, the f of (x,
y) and g of (x,

00:14:50.000 --> 00:14:56.000
y), the parameters in those

00:14:53.000 --> 00:14:59.000
functions will be set at
different levels.

00:14:56.000 --> 00:15:02.000
You will have different
hormones, a different of carbon

00:14:59.000 --> 00:15:05.000
dioxide and so on.
Now, I am not saying that

00:15:04.000 --> 00:15:10.000
breathing is modeled by a limit
cycle.

00:15:07.000 --> 00:15:13.000
It is the sort of thing which
one might look for a limit

00:15:11.000 --> 00:15:17.000
cycle.
That is, of course,

00:15:13.000 --> 00:15:19.000
a question for biologists.
And, in general,

00:15:17.000 --> 00:15:23.000
any type of periodic behavior
in nature, people try to see if

00:15:22.000 --> 00:15:28.000
there is some system of
differential equations which

00:15:26.000 --> 00:15:32.000
governs it in which perhaps
there is a limit cycle,

00:15:30.000 --> 00:15:36.000
which contains a limit cycle.
Well, what are the problems?

00:15:36.000 --> 00:15:42.000
In a sense, limit cycles are
easy to lecture about because so

00:15:41.000 --> 00:15:47.000
little is known about them.
At the end of the period,

00:15:46.000 --> 00:15:52.000
if I have time,
I will show you that the

00:15:49.000 --> 00:15:55.000
simplest possible question you
could ask, the answer to it is

00:15:55.000 --> 00:16:01.000
totally known after 120 years of
steady trying.

00:16:00.000 --> 00:16:06.000
But let's first talk about what
sorts of problems people address

00:16:05.000 --> 00:16:11.000
with limit cycles.
First of all is the existence

00:16:09.000 --> 00:16:15.000
problem.

00:16:16.000 --> 00:16:22.000
If I give you a system,
you know, the right-hand side

00:16:19.000 --> 00:16:25.000
is x squared plus 2y cubed minus
3xy,

00:16:23.000 --> 00:16:29.000
and the g is something similar.
I say does this have limit

00:16:26.000 --> 00:16:32.000
cycles?
Well, you know how to find its

00:16:29.000 --> 00:16:35.000
critical points.
But how do you find out if it

00:16:34.000 --> 00:16:40.000
has limit cycles?
The answer to that is nobody

00:16:40.000 --> 00:16:46.000
has any idea.
This problem,

00:16:44.000 --> 00:16:50.000
in general, there are not much
in the way of methods.

00:16:50.000 --> 00:16:56.000
Not much.

00:17:00.000 --> 00:17:06.000
Not much is known.
There is one theorem that you

00:17:03.000 --> 00:17:09.000
will find in the notes,
a simple theorem called the

00:17:07.000 --> 00:17:13.000
Poincare-Bendixson theorem
which, for about 60 or 70 years

00:17:11.000 --> 00:17:17.000
was about the only result known
which enabled people to find

00:17:16.000 --> 00:17:22.000
limit cycles.
Nowadays the theorem is used

00:17:19.000 --> 00:17:25.000
relatively little because people
try to find limit cycles by

00:17:24.000 --> 00:17:30.000
computer.
Now, the difficulty is you have

00:17:27.000 --> 00:17:33.000
to know where to look for them.
In other words,

00:17:32.000 --> 00:17:38.000
the computer screen shows that
much and you set the axes and it

00:17:36.000 --> 00:17:42.000
doesn't show any limit cycles.
That doesn't mean there are not

00:17:40.000 --> 00:17:46.000
any.
That means they are over there,

00:17:43.000 --> 00:17:49.000
or it means there is a big one
like there.

00:17:46.000 --> 00:17:52.000
And you are looking in the
middle of it and don't see it.

00:17:50.000 --> 00:17:56.000
So, in general,
people don't look for limit

00:17:53.000 --> 00:17:59.000
cycles unless the physical
system that gave rise to the

00:17:56.000 --> 00:18:02.000
pair of differential equations
suggests that there is something

00:18:01.000 --> 00:18:07.000
repetitive going on like
breathing.

00:18:05.000 --> 00:18:11.000
And, if it tells you that,
then it often gives you

00:18:09.000 --> 00:18:15.000
approximate values of the
parameters and the variables so

00:18:14.000 --> 00:18:20.000
you know where to look.
Basically this is done by

00:18:18.000 --> 00:18:24.000
computer search guided by the
physical problem.

00:18:31.000 --> 00:18:37.000
Therefore, I cannot say much
more about it today.

00:18:35.000 --> 00:18:41.000
Instead I am going to focus my
attention on nonexistence.

00:18:40.000 --> 00:18:46.000
When can you be sure that a
system will not have any limit

00:18:46.000 --> 00:18:52.000
cycles?
And there are two theorems.

00:18:49.000 --> 00:18:55.000
One, again, due to Bendixson
who was a Swedish mathematician

00:18:54.000 --> 00:19:00.000
who lived around 1900 or so.
There is a criterion due to

00:18:59.000 --> 00:19:05.000
Bendixson.
And there is one involving

00:19:04.000 --> 00:19:10.000
critical points.
And I would like to describe

00:19:08.000 --> 00:19:14.000
both of them for you today.
First of all,

00:19:11.000 --> 00:19:17.000
Bendixson's criterion.

00:19:22.000 --> 00:19:28.000
It is very simply stated and
has a marvelous proof,

00:19:25.000 --> 00:19:31.000
which I am going to give you.
We have D as a region of the

00:19:29.000 --> 00:19:35.000
plane.

00:19:35.000 --> 00:19:41.000
And what Bendixson's criterion
tells you to do is take your

00:19:40.000 --> 00:19:46.000
vector field and calculate its
divergence.

00:19:43.000 --> 00:19:49.000
We are set back in 1802,
and this proof is going to be

00:19:48.000 --> 00:19:54.000
straight 18.02.
You will enjoy it.

00:19:51.000 --> 00:19:57.000
Calculate the divergence.
Now, I am talking about the

00:19:55.000 --> 00:20:01.000
two-dimensional divergence.
Remember that is fx,

00:20:00.000 --> 00:20:06.000
the partial of f with respect
to x, plus the partial of the g,

00:20:05.000 --> 00:20:11.000
the j component with respect to
y.

00:20:08.000 --> 00:20:14.000
And assume that that is a
continuous function.

00:20:12.000 --> 00:20:18.000
It always will be with us.
Practically all the examples I

00:20:16.000 --> 00:20:22.000
will give you f and g will be
simple polynomials.

00:20:20.000 --> 00:20:26.000
They are smooth,
continuous and nice and behave

00:20:24.000 --> 00:20:30.000
as you want.
And you calculate that and

00:20:27.000 --> 00:20:33.000
assume --
Suppose, in other words,

00:20:32.000 --> 00:20:38.000
that the divergence of f,
I need more room.

00:20:36.000 --> 00:20:42.000
The hypothesis is that the
divergence of f is not zero in

00:20:42.000 --> 00:20:48.000
that region D.
It is never zero.

00:20:45.000 --> 00:20:51.000
It is not zero at any point in
that region.

00:20:49.000 --> 00:20:55.000
The conclusion is that there
are no limit cycles in the

00:20:55.000 --> 00:21:01.000
region.
If it is not zero in D,

00:20:58.000 --> 00:21:04.000
there are no limit cycles.
In fact, there are not even any

00:21:04.000 --> 00:21:10.000
closed trajectories.
You couldn't even have those

00:21:09.000 --> 00:21:15.000
bunch of concentric circles,
so there are no closed

00:21:13.000 --> 00:21:19.000
trajectories of the original
system whose divergence you

00:21:18.000 --> 00:21:24.000
calculated.
There are no closed

00:21:21.000 --> 00:21:27.000
trajectories in D.
For example,

00:21:24.000 --> 00:21:30.000
let me give you a simple
example to put a little flesh on

00:21:29.000 --> 00:21:35.000
it.
Let's see.

00:21:32.000 --> 00:21:38.000
What do I have?
I prepared an example.

00:21:35.000 --> 00:21:41.000
x prime equals,
here is a simple nonlinear

00:21:40.000 --> 00:21:46.000
system, x cubed plus y cubed.

00:21:45.000 --> 00:21:51.000
And y prime equals 3x plus y
cubed plus 2y.

00:21:59.000 --> 00:22:05.000
Does this system have limit
cycles?

00:22:01.000 --> 00:22:07.000
Well, even to calculate its
critical points would be a

00:22:05.000 --> 00:22:11.000
little task, but we can easily
answer the question as to

00:22:09.000 --> 00:22:15.000
whether it has limit cycles or
not by Bendixson's criterion.

00:22:14.000 --> 00:22:20.000
Let's calculate the divergence.
The divergence of the vector

00:22:18.000 --> 00:22:24.000
field whose components are these
two functions is,

00:22:22.000 --> 00:22:28.000
well, 3x squared,
it's the partial of the first

00:22:26.000 --> 00:22:32.000
guy with respect to x plus the
partial of the second guy with

00:22:31.000 --> 00:22:37.000
respect to y,
which is 3y squared plus two.

00:22:34.000 --> 00:22:40.000
Now, can that be zero anywhere

00:22:39.000 --> 00:22:45.000
in the x,y-plane?
No, because it is the sum of

00:22:43.000 --> 00:22:49.000
these two squares.
This much of it could be zero

00:22:47.000 --> 00:22:53.000
only at the origin,
but that plus two eliminates

00:22:51.000 --> 00:22:57.000
even that.
This is always positive in the

00:22:55.000 --> 00:23:01.000
entire x,y-plane.
Here my domain is the whole

00:22:59.000 --> 00:23:05.000
x,y-plane and,
therefore, the conclusion is

00:23:03.000 --> 00:23:09.000
that there are no closed
trajectories in the x,y-plane,

00:23:07.000 --> 00:23:13.000
anywhere.

00:23:15.000 --> 00:23:21.000
And we have done that with just
a couple of lines of calculation

00:23:18.000 --> 00:23:24.000
and nothing further required.
No computer search.

00:23:21.000 --> 00:23:27.000
In fact, no computer search
could ever proof this.

00:23:24.000 --> 00:23:30.000
It would be impossible because,
no matter where you look,

00:23:27.000 --> 00:23:33.000
there is always some other
place to look.

00:23:30.000 --> 00:23:36.000
This is an example where a
couple lines of mathematics

00:23:35.000 --> 00:23:41.000
dispose of the matter far more
effectively than a million

00:23:41.000 --> 00:23:47.000
dollars worth of calculation.
Well, where does Bendixson's

00:23:46.000 --> 00:23:52.000
theorem come from?
Yes, Bendixson's theorem comes

00:23:51.000 --> 00:23:57.000
from 18.02.
And I am giving it to you both

00:23:56.000 --> 00:24:02.000
to recall a little bit of 18.02
to you.

00:24:01.000 --> 00:24:07.000
Because it is about the first
example in the course that we

00:24:06.000 --> 00:24:12.000
have had of an indirect
argument.

00:24:09.000 --> 00:24:15.000
And indirect arguments are
something you have to slowly get

00:24:14.000 --> 00:24:20.000
used to.
I am going to give you an

00:24:17.000 --> 00:24:23.000
indirect proof.
Remember what that is?

00:24:20.000 --> 00:24:26.000
You assume the contrary and you
show it leads to a

00:24:25.000 --> 00:24:31.000
contradiction.
What would assuming the

00:24:28.000 --> 00:24:34.000
contrary be?
Contrary would be I will assume

00:24:34.000 --> 00:24:40.000
the divergence is not zero,
but I will suppose there is a

00:24:40.000 --> 00:24:46.000
closed trajectory.
Suppose there is a closed

00:24:44.000 --> 00:24:50.000
trajectory that exists.

00:24:56.000 --> 00:25:02.000
Let's draw a picture of it.

00:25:08.000 --> 00:25:14.000
And let's say it goes around
this way.

00:25:11.000 --> 00:25:17.000
There is a closed trajectory
for our system.

00:25:15.000 --> 00:25:21.000
Let's call the curve C.
And I am going to call the

00:25:19.000 --> 00:25:25.000
inside of it R,
the way one often does in

00:25:22.000 --> 00:25:28.000
18.02.
D is all this region out here,

00:25:25.000 --> 00:25:31.000
in which everything is taking
place.

00:25:30.000 --> 00:25:36.000
This is to exist in D.
Now, what I am going to do is

00:25:35.000 --> 00:25:41.000
calculate a line integral around
that curve.

00:25:45.000 --> 00:25:51.000
A line integral of this vector
field.

00:25:47.000 --> 00:25:53.000
Now, there are two things you
can calculate.

00:25:51.000 --> 00:25:57.000
One of the line integrals,
I will put in a few of the

00:25:55.000 --> 00:26:01.000
vectors here.
The vectors I know are pointing

00:25:58.000 --> 00:26:04.000
this way because that is the
direction in which the curve is

00:26:03.000 --> 00:26:09.000
being traversed in order to make
it a trajectory.

00:26:08.000 --> 00:26:14.000
Those are a few of the typical
vectors in the field.

00:26:12.000 --> 00:26:18.000
I am going to calculate the
line integral around that curve

00:26:16.000 --> 00:26:22.000
in the positive sense.
In other words,

00:26:19.000 --> 00:26:25.000
not in the direction of the
salmon-colored arrow,

00:26:23.000 --> 00:26:29.000
but in the normal sense in
which you calculate it using

00:26:27.000 --> 00:26:33.000
Green's theorem,
for example.

00:26:31.000 --> 00:26:37.000
The positive sense means the
one which keeps the region,

00:26:34.000 --> 00:26:40.000
the inside on your left,
as you walk around like that

00:26:38.000 --> 00:26:44.000
the region stays on your left.
That is the positive sense.

00:26:42.000 --> 00:26:48.000
That is the sense in which I am
integrating.

00:26:45.000 --> 00:26:51.000
I am going to use Green's
theorem, but the integral that I

00:26:49.000 --> 00:26:55.000
am going to calculate is not the
work integral.

00:26:52.000 --> 00:26:58.000
I am going to calculate instead
the flux integral,

00:26:55.000 --> 00:27:01.000
the integral that represents
the flux of F across C.

00:27:00.000 --> 00:27:06.000
Now, what is that integral?
Well, at each point,

00:27:04.000 --> 00:27:10.000
you station a little ant and
the ant reports the outward flow

00:27:09.000 --> 00:27:15.000
rate across that point which is
F dotted with the normal vector.

00:27:15.000 --> 00:27:21.000
I will put in a few normal
vectors just to remind you.

00:27:20.000 --> 00:27:26.000
The normal vectors look like
little unit vectors pointing

00:27:25.000 --> 00:27:31.000
perpendicularly outwards
everywhere.

00:27:29.000 --> 00:27:35.000
These are the n's.
F dotted with the unit normal

00:27:34.000 --> 00:27:40.000
vector, and that is added up
around the curve.

00:27:38.000 --> 00:27:44.000
This quantity gives me the flux
of the field across C.

00:27:43.000 --> 00:27:49.000
Now, we are going to calculate
that by Green's theorem.

00:27:48.000 --> 00:27:54.000
But, before we calculate it by
Green's theorem,

00:27:52.000 --> 00:27:58.000
we are going to psych it out.
What is it?

00:27:55.000 --> 00:28:01.000
What is the value of that
integral?

00:28:00.000 --> 00:28:06.000
Well, since I am asking you to
do it in your head there can

00:28:04.000 --> 00:28:10.000
only be one possible answer.
It is zero.

00:28:06.000 --> 00:28:12.000
Why is that integral zero?
Well, because at each point the

00:28:10.000 --> 00:28:16.000
field vector,
the velocity vector is

00:28:13.000 --> 00:28:19.000
perpendicular to the normal
vector.

00:28:15.000 --> 00:28:21.000
Why?
The normal vector points

00:28:17.000 --> 00:28:23.000
perpendicularly to the curve but
the field vector always is

00:28:22.000 --> 00:28:28.000
tangent to the curve because
this curve is a trajectory.

00:28:27.000 --> 00:28:33.000
It is always supposed to be
going in the direction given by

00:28:34.000 --> 00:28:40.000
that white field vector.
Do you follow?

00:28:39.000 --> 00:28:45.000
A trajectory means that it is
always tangent to the field

00:28:47.000 --> 00:28:53.000
vector and, therefore,
always perpendicular to the

00:28:53.000 --> 00:28:59.000
normal vector.
This is zero since F dot n is

00:28:59.000 --> 00:29:05.000
always zero.
Everywhere on the curve,

00:29:03.000 --> 00:29:09.000
F dot n has to be zero.
There is no flux of this field

00:29:07.000 --> 00:29:13.000
across the curve because the
field is always in the same

00:29:12.000 --> 00:29:18.000
direction as the curve,
never perpendicular to it.

00:29:15.000 --> 00:29:21.000
It has no components
perpendicular to it.

00:29:19.000 --> 00:29:25.000
Good.
Now let's do it the hard way.

00:29:21.000 --> 00:29:27.000
Let's use Green's theorem.
Green's theorem says that the

00:29:25.000 --> 00:29:31.000
flux across C should be equal to
the double integral over that

00:29:30.000 --> 00:29:36.000
region of the divergence of F.
It's like Gauss theorem in two

00:29:35.000 --> 00:29:41.000
dimensions, this version of it.
Divergence of F,

00:29:39.000 --> 00:29:45.000
that is a function,
I double integrate it over the

00:29:42.000 --> 00:29:48.000
region, and then that is dx /
dy, or let's say da because you

00:29:46.000 --> 00:29:52.000
might want to do it in polar
coordinates.

00:29:48.000 --> 00:29:54.000
And, on the problem set,
you certainly will want to do

00:29:52.000 --> 00:29:58.000
it in polar coordinates,
I think.

00:30:00.000 --> 00:30:06.000
All right.
How much is that?

00:30:02.000 --> 00:30:08.000
Well, we haven't yet used the
hypothesis.

00:30:06.000 --> 00:30:12.000
All we have done is set up the
problem.

00:30:10.000 --> 00:30:16.000
Now, the hypothesis was that
the divergence is never zero

00:30:16.000 --> 00:30:22.000
anywhere in D.
Therefore, the divergence is

00:30:20.000 --> 00:30:26.000
never zero anywhere in R.
What I say is the divergence is

00:30:26.000 --> 00:30:32.000
either greater than zero
everywhere in R.

00:30:32.000 --> 00:30:38.000
Or less than zero everywhere in
R.

00:30:35.000 --> 00:30:41.000
But it cannot be sometimes
positive and sometimes negative.

00:30:40.000 --> 00:30:46.000
Why not?
In other words,

00:30:42.000 --> 00:30:48.000
I say it is not possible the
divergence here is one and here

00:30:47.000 --> 00:30:53.000
is minus two.
That is not possible because,

00:30:51.000 --> 00:30:57.000
if I drew a line from this
point to that,

00:30:55.000 --> 00:31:01.000
along that line the divergence
would start positive and end up

00:31:00.000 --> 00:31:06.000
negative.
And, therefore,

00:31:04.000 --> 00:31:10.000
have to be zero some time in
between.

00:31:06.000 --> 00:31:12.000
It's because it is a continuous
function.

00:31:09.000 --> 00:31:15.000
It is a continuous function.
I am assuming that.

00:31:12.000 --> 00:31:18.000
And, therefore,
if it sometimes positive and

00:31:15.000 --> 00:31:21.000
sometimes negative it has to be
zero in between.

00:31:18.000 --> 00:31:24.000
You cannot get continuously
from plus one to minus two

00:31:22.000 --> 00:31:28.000
without passing through zero.
The reason for this is,

00:31:27.000 --> 00:31:33.000
since the divergence is never
zero in R it therefore must

00:31:35.000 --> 00:31:41.000
always stay positive or always
stay negative.

00:31:40.000 --> 00:31:46.000
Now, if it always stays
positive, the conclusion is then

00:31:47.000 --> 00:31:53.000
this double integral must be
positive.

00:31:52.000 --> 00:31:58.000
Therefore, this double integral
is either greater than zero.

00:32:01.000 --> 00:32:07.000
That is if the divergence is
always positive.

00:32:04.000 --> 00:32:10.000
Or, it is less than zero if the
divergence is always negative.

00:32:10.000 --> 00:32:16.000
But the one thing it cannot be
is not zero.

00:32:14.000 --> 00:32:20.000
Well, the left-hand side,
Green's theorem is supposed to

00:32:18.000 --> 00:32:24.000
be true.
Green's theorem is our bedrock.

00:32:22.000 --> 00:32:28.000
18.02 would crumble without
that so it must be true.

00:32:26.000 --> 00:32:32.000
One way of calculating the
left-hand side gives us zero.

00:32:33.000 --> 00:32:39.000
If we calculate the right-hand
side it is not zero.

00:32:36.000 --> 00:32:42.000
That is called the
contradiction.

00:32:38.000 --> 00:32:44.000
Where did the contradiction
arise from?

00:32:41.000 --> 00:32:47.000
It arose from the fact that I
supposed that there was a closed

00:32:45.000 --> 00:32:51.000
trajectory in that region.
The conclusion is there cannot

00:32:48.000 --> 00:32:54.000
be any closed trajectory of that
region because it leads to a

00:32:52.000 --> 00:32:58.000
contradiction via Green's
theorem.

00:33:02.000 --> 00:33:08.000
Let me see if I can give you
some of the argument for the

00:33:06.000 --> 00:33:12.000
other, well, let's at least
state the other criterion I

00:33:10.000 --> 00:33:16.000
wanted to give you.

00:33:21.000 --> 00:33:27.000
Suppose, for example,
we use this system,

00:33:25.000 --> 00:33:31.000
x prime equals --

00:33:50.000 --> 00:33:56.000
Does this have limit cycles?

00:33:59.000 --> 00:34:05.000
Does that have limit cycles?

00:34:08.000 --> 00:34:14.000
Let's Bendixson it.
We will calculate the

00:34:12.000 --> 00:34:18.000
divergence of a vector field.
It is 2x from the top function.

00:34:18.000 --> 00:34:24.000
The partial with respect to x
is 2x.

00:34:22.000 --> 00:34:28.000
The second function with
respect to y is negative 2y.

00:34:29.000 --> 00:34:35.000
That certainly could be zero.
In fact, this is zero along the

00:34:34.000 --> 00:34:40.000
entire line x equals y.
Its divergence is zero here

00:34:39.000 --> 00:34:45.000
along that whole line.
The best I could conclude was,

00:34:44.000 --> 00:34:50.000
I could conclude that there is
no limit cycle like this and

00:34:50.000 --> 00:34:56.000
there is no limit cycle like
this, but there is nothing so

00:34:55.000 --> 00:35:01.000
far that says a limit cycle
could not cross that because

00:35:00.000 --> 00:35:06.000
that would not violate
Bendixson's theorem.

00:35:06.000 --> 00:35:12.000
In other words,
any domain that contained part

00:35:09.000 --> 00:35:15.000
of this line,
the divergence would be zero

00:35:13.000 --> 00:35:19.000
along that line.
And, therefore,

00:35:15.000 --> 00:35:21.000
I could conclude nothing.
I could have limit cycles that

00:35:20.000 --> 00:35:26.000
cross that line,
as long as they included a

00:35:23.000 --> 00:35:29.000
piece of that line in them.
The answer is I cannot make a

00:35:28.000 --> 00:35:34.000
conclusion.
Well, that is because I am

00:35:32.000 --> 00:35:38.000
using the wrong criterion.
Let's instead use the critical

00:35:36.000 --> 00:35:42.000
point criterion.

00:35:55.000 --> 00:36:01.000
Now, I am going to say that it
makes a nice positive statement

00:35:58.000 --> 00:36:04.000
but nobody ever uses it this
way.

00:36:01.000 --> 00:36:07.000
Nonetheless,
let's first state it

00:36:03.000 --> 00:36:09.000
positively, even though that is
not the way to use it.

00:36:07.000 --> 00:36:13.000
The positive statement will be,
once again, we have our region

00:36:13.000 --> 00:36:19.000
D and we have a region of the xy
plane and we have our C,

00:36:20.000 --> 00:36:26.000
a closed trajectory in it.
A closed trajectory of what?

00:36:26.000 --> 00:36:32.000
Of our system.
And that is supposed to be in

00:36:30.000 --> 00:36:36.000
D.
The critical point criterion

00:36:35.000 --> 00:36:41.000
says something very simple.
If you have that situation it

00:36:42.000 --> 00:36:48.000
says that inside that closed
trajectory there must be a

00:36:48.000 --> 00:36:54.000
critical point somewhere.

00:37:00.000 --> 00:37:06.000
It says that inside C is a
critical point.

00:37:15.000 --> 00:37:21.000
Now, this won't help us with
the existence problem.

00:37:18.000 --> 00:37:24.000
This won't help us find a
closed trajectory.

00:37:21.000 --> 00:37:27.000
We will take our system and say
it has a critical point here and

00:37:26.000 --> 00:37:32.000
a critical point there.
Does it have a closed

00:37:29.000 --> 00:37:35.000
trajectory?
Well, all I know is the closed

00:37:33.000 --> 00:37:39.000
trajectory, if it exists,
will have to go around one or

00:37:36.000 --> 00:37:42.000
more of those critical points.
But I don't know where.

00:37:40.000 --> 00:37:46.000
It is not going to go around it
like this.

00:37:43.000 --> 00:37:49.000
It might go around it like
this.

00:37:45.000 --> 00:37:51.000
And my computer search won't
find it because it is looking at

00:37:49.000 --> 00:37:55.000
too small a part of the screen.
It doesn't work that way.

00:37:53.000 --> 00:37:59.000
It works negatively by
contraposition.

00:37:56.000 --> 00:38:02.000
Do you know what the
contrapositive is?

00:38:00.000 --> 00:38:06.000
You will at least learn that.
A implies B says the same thing

00:38:07.000 --> 00:38:13.000
as not B implies not A.

00:38:20.000 --> 00:38:26.000
They are different statements
but they are equivalent to each

00:38:24.000 --> 00:38:30.000
other.
If you prove one you prove the

00:38:27.000 --> 00:38:33.000
other.
What would be the

00:38:29.000 --> 00:38:35.000
contrapositive here?
If you have a closed trajectory

00:38:35.000 --> 00:38:41.000
inside is a critical point.
The theorem is used this way.

00:38:44.000 --> 00:38:50.000
If D has no critical points,
it has no closed trajectories

00:38:53.000 --> 00:38:59.000
and therefore has no limit
cycle.

00:39:00.000 --> 00:39:06.000
Because, if it did have a
closed trajectory,

00:39:03.000 --> 00:39:09.000
inside it would be a critical
point.

00:39:07.000 --> 00:39:13.000
But I said B had no critical
point.

00:39:10.000 --> 00:39:16.000
That enables us to dispose of
this system that Bendixson could

00:39:15.000 --> 00:39:21.000
not handle at all.
We can dispose of this system

00:39:20.000 --> 00:39:26.000
immediately.
Namely, what is it?

00:39:22.000 --> 00:39:28.000
Where are its critical points?
Well, where is that zero?

00:39:29.000 --> 00:39:35.000
x squared plus y
squared is one,

00:39:32.000 --> 00:39:38.000
plus one is never zero.
This is positive.

00:39:35.000 --> 00:39:41.000
Or, worse, zero.
And then I add the one to it

00:39:38.000 --> 00:39:44.000
and it is not zero anymore.
This has no zeros and,

00:39:42.000 --> 00:39:48.000
therefore, it does not matter
that this one has a lot of

00:39:46.000 --> 00:39:52.000
zeros.
It makes no difference.

00:39:48.000 --> 00:39:54.000
It has no critical points.
It has none,

00:39:51.000 --> 00:39:57.000
therefore, no limit cycles.

00:40:01.000 --> 00:40:07.000
Now, I desperately wanted to
give you the proof of this.

00:40:04.000 --> 00:40:10.000
It is clearly impossible in the
time remaining.

00:40:07.000 --> 00:40:13.000
The proof requires a little
time.

00:40:09.000 --> 00:40:15.000
I haven't decided what to do
about that.

00:40:12.000 --> 00:40:18.000
It might leak over until
Friday's lecture.

00:40:15.000 --> 00:40:21.000
Instead, I will finish by
telling you a story.

00:40:18.000 --> 00:40:24.000
How is that?

00:40:37.000 --> 00:40:43.000
And along side of it was little
y prime.

00:40:39.000 --> 00:40:45.000
I am not going to continue on
with the letters of the

00:40:43.000 --> 00:40:49.000
alphabet.
I will prime the earlier one.

00:40:46.000 --> 00:40:52.000
This has a total of 12
parameters in it.

00:40:49.000 --> 00:40:55.000
But, in fact,
if you change variables you can

00:40:52.000 --> 00:40:58.000
get rid of all the linear terms.
The important part of it is

00:40:57.000 --> 00:41:03.000
only the quadratic terms in the
beginning.

00:41:01.000 --> 00:41:07.000
This sort of thing is called a
quadratic system.

00:41:05.000 --> 00:41:11.000
After you have departed from
linear systems,

00:41:09.000 --> 00:41:15.000
it is the simplest kind there
is.

00:41:12.000 --> 00:41:18.000
And the predictor-prey,
the robin-earthworm example I

00:41:16.000 --> 00:41:22.000
gave you is of a typical
quadratic system and exhibits

00:41:21.000 --> 00:41:27.000
typical nonlinear quadratic
system behavior.

00:41:25.000 --> 00:41:31.000
Now, the problem is the
following.

00:41:30.000 --> 00:41:36.000
A, b, c, d, e,
f and so on,

00:41:32.000 --> 00:41:38.000
those are just real numbers,
parameters, so I am allowed to

00:41:37.000 --> 00:41:43.000
give them any values I want.
And the problem that has

00:41:42.000 --> 00:41:48.000
bothered people since 1880 when
it was first proposed is how

00:41:47.000 --> 00:41:53.000
many limit cycles can a
quadratic system have?

00:42:03.000 --> 00:42:09.000
After 120 years this problem is
totally unsolved,

00:42:07.000 --> 00:42:13.000
and the mathematicians of the
world who are interested in it

00:42:12.000 --> 00:42:18.000
cannot even agree with each
other on what the right

00:42:16.000 --> 00:42:22.000
conjecture is.
But let me tell you a little

00:42:20.000 --> 00:42:26.000
bit of its history.
There were attempts to solve it

00:42:24.000 --> 00:42:30.000
in the 20 or 30 years after it
was first proposed,

00:42:28.000 --> 00:42:34.000
through the 1920 and `30s which
all seemed to have gaps in them.

00:42:35.000 --> 00:42:41.000
Until finally around 1950 two
Russians mathematicians,

00:42:39.000 --> 00:42:45.000
one of whom is extremely
well-known, Petrovski,

00:42:44.000 --> 00:42:50.000
a specialist in systems of
ordinary differential equations

00:42:49.000 --> 00:42:55.000
published a long and difficult,
complicated 100 page paper in

00:42:54.000 --> 00:43:00.000
which they proved that the
maximum number is three.

00:43:00.000 --> 00:43:06.000
I won't put down their names.
Petrovski-Landis.

00:43:03.000 --> 00:43:09.000
The maximum number was three.
And then not many people were

00:43:08.000 --> 00:43:14.000
able to read the paper,
and those who did there seemed

00:43:12.000 --> 00:43:18.000
to be gaps in the reasoning in
various places until finally

00:43:16.000 --> 00:43:22.000
Arnold who was the greatest
Russian, in my opinion,

00:43:20.000 --> 00:43:26.000
one of the greatest Russian
mathematician,

00:43:23.000 --> 00:43:29.000
certainly in this field of
analysis and differential

00:43:27.000 --> 00:43:33.000
equations, but in other fields,
too, he still is great,

00:43:31.000 --> 00:43:37.000
although he is somewhat older
now, criticized it.

00:43:37.000 --> 00:43:43.000
He said look,
there is a really big gap in

00:43:41.000 --> 00:43:47.000
this argument and it really
cannot be considered to be

00:43:46.000 --> 00:43:52.000
proven.
People tried working very hard

00:43:50.000 --> 00:43:56.000
to patch it up and without
success.

00:43:53.000 --> 00:43:59.000
Then about 1972 or so,
'75 maybe, a Chinese

00:43:58.000 --> 00:44:04.000
mathematician found a system
with four.

00:44:03.000 --> 00:44:09.000
Wrote down the numbers,
the number a is so much,

00:44:07.000 --> 00:44:13.000
b is so much,
and they were absurd numbers

00:44:11.000 --> 00:44:17.000
like 10^-6 and 40 billion and so
on, nothing you could plot on a

00:44:17.000 --> 00:44:23.000
computer screen,
but found a system with four.

00:44:22.000 --> 00:44:28.000
Nobody after this tried to fill
in the gap in the

00:44:26.000 --> 00:44:32.000
Petrovski-Landis paper.
I was then chairman of the math

00:44:32.000 --> 00:44:38.000
department, and one of my tasks
was protocol and so on.

00:44:36.000 --> 00:44:42.000
Anyway, we were trying very
hard to attract a Chinese

00:44:40.000 --> 00:44:46.000
mathematician to our department
to become a full professor.

00:44:44.000 --> 00:44:50.000
He was a really outstanding
analyst and specialist in

00:44:48.000 --> 00:44:54.000
various fields.
Anyway, he came in for a

00:44:50.000 --> 00:44:56.000
courtesy interview and we
chatted.

00:44:53.000 --> 00:44:59.000
At the time,
I was very much interested in

00:44:56.000 --> 00:45:02.000
limit cycles.
And I had on my desk the Math

00:44:59.000 --> 00:45:05.000
Society's translation of the
Chinese book on limit cycles.

00:45:05.000 --> 00:45:11.000
A collection of papers by
Chinese mathematicians all on

00:45:08.000 --> 00:45:14.000
limit cycles.
After a certain point he said,

00:45:11.000 --> 00:45:17.000
oh, I see you're interested in
limit cycle problems.

00:45:15.000 --> 00:45:21.000
I said yeah,
in particular,

00:45:16.000 --> 00:45:22.000
I was reading this paper of the
mathematician who found four

00:45:20.000 --> 00:45:26.000
limit cycles.
And I opened to that system and

00:45:23.000 --> 00:45:29.000
said the name is,
and I read it out loud.

00:45:26.000 --> 00:45:32.000
I said do you by any chance
know him?

00:45:30.000 --> 00:45:36.000
And he smiled and said yes,
very well.

00:45:32.000 --> 00:45:38.000
That is my mother.
[LAUGHTER]

00:45:43.000 --> 00:45:49.000
Well, bye-bye.