First of all, the way a nonlinear autonomous
system looks, you have had some practice with
it by now. This is nonlinear.
The right-hand side are no longer simple combinations ax
plus by. Nonlinear and autonomous,
these are function just of x and y.
There is no t on the right-hand side.
Now, most of today will be geometric.
The way to get a geometric picture of that is first by
constructing the velocity field whose components are the
functions f and g. This is a velocity field that
gives a picture of the system and has solutions.
The solutions to the system, from the point of view of
functions, they would look like pairs of functions,
x of t, y of t.
But, from the point of view of geometry, when you plot them as
parametric equations, they are called trajectories of
the field F, which simply means that they are curves everywhere
having the right velocity. So a typical curve would look
like -- There is a trajectory.
And we know it is a trajectory because at each point the vector
on it has, of course, the right direction,
the tangent direction, but more than that,
it has the right velocity. So here, for example,
the point is traveling more slowly.
Here it is traveling more rapidly because the velocity
vector is bigger, longer.
So this is a picture of a typical trajectory.
The only other things that I should mention are the critical
points. If you have worked the problems
for this week, the first couple of problems,
you have already seen the significance of the critical
points. Well, from Monday's lecture you
know from the point of view of solutions they are constant
solutions.
From the point of view of the field they are where the field
is zero. There is no velocity vector,
in other words. The velocity vector is zero.
And, therefore, a point being there has no
reason to go anywhere else. And, spelling it out,
it's where the partial derivatives, where the values of
the functions on the right-hand side, which give the two
components, the i and j components of the field,
where they are zero.
That is all I will need by way of a recall today.
I don't think I will need anything else.
The topic for today is another kind of behavior that you have
not yet observed at the computer screen, unless you have worked
ahead, and that is there are trajectories which go along to
infinity or end up at a critical point.
They are the critical points that just sit there all the
time. But there is a third type of
behavior that a trajectory can have where it neither sits for
all time nor goes off for all time.
Instead, it repeats itself. Such a thing is called a closed
trajectory. What does it look like?
Well, it is a closed curve in the plane that at every point,
it is a trajectory, i.e., the arrows at each point,
let's say it is traced in the clockwise direction.
And so the arrows of the field will go like this.
Here it is going slowly, here it is very slow and here
it picks up a little speed again and so on.
Now, for such a trajectory what is happening?
Well, it goes around in finite time and then repeats itself.
It just goes round and round forever if you land on that
trajectory. It represents a system that
returns to its original state periodically.
It represents periodic behavior of the system.
Now, we have seen one example of that, a simple example where
this simple system, x prime equals y,
y prime equals negative x.
We could write down the solutions to that directly,
but if you want to do eigenvalues and eigenvectors the
matrix will look like this. The equation will be lambda
squared plus zero lambda plus one equals zero,
so the eigenvalues will be plus
or minus i. In fact, from then on you could
work out in the usual ways the eigenvectors,
complex eigenvectors and separate them.
But, look, you can avoid all that just by writing down the
solution. The solutions are sines and
cosines. One basic solution will be x
equals cosine t, in which case what is y?
Well, y is the derivative of that.
That will be minus sine t.
Another basic solution, we will start with x equals
sine t. In which case y will be cosine
t, its derivative.
Now, if you do that, what do these things look like?
Well, either of these two basic solutions looks like a circle,
not traced in the usual way but in the opposite way.
For example, when t is equal to zero it
starts at the point one, zero.
Now, if the minus sign were not there this would be x equals
cosine t, y equals sine t,
which is the usual counterclockwise circle.
But if I change y from sine t to negative sine t
it is going around the other way.
So this circle is traced that way.
And this is a family of circles, according to the values
of c1 and c2, concentric, all of which go
around clockwise. So those are closed
trajectories. Those are the solutions.
They are trajectories of the vector field.
They are closed. They come around and they
repeat in finite time. Now, these are no good.
These are the kind I am not interested in.
These are commonplace, and we are interested in good
stuff today. And the good stuff we are
interested in is limit cycles.
A limit cycle is a closed trajectory with a couple of
extra hypotheses. It is a closed trajectory,
just like those guys, but it has something they don't
have, namely, it is king of the roost.
They have to be isolated, no other guys nearby.
And they also have to be stable.
See, the problem here is that none of these stands out from
any of the others. In other words,
there must be, isolated means,
no others nearby.
That is just what goes wrong here.
Arbitrarily close to each of these circles is yet another
circle doing exactly the same thing.
That means that there are some that are only of mild interest.
What is much more interesting is to find a cycle where there
is nothing nearby. Something, therefore,
that looks like this.
Here is our pink guy. Let's make this one go
counterclockwise. Here is a limit cycle,
it seems to be. And now what do nearby guys do?
Well, they should approach it. Somebody here like that does
this, spirals in and gets ever and every closer to that thing.
Now, it can never join it because, if it joined it at the
joining point, I would have two solutions
going through this point. And that is illegal.
All it can do is get arbitrarily close.
On the computer screen it will look as if it joins it but,
of course, it cannot. It is just the resolution,
the pixels. Not enough pixels.
The resolution isn't good enough.
And the ones that start further away will take longer to find
their way to the limit cycle and they will always stay outside of
the earlier guys, but they will get arbitrarily
close, too. How about inside?
Inside, well, it starts somewhere and does
the same thing. It starts here and will try to
join the limit cycle. That is what I mean by
stability. Stability means that nearby
guys, the guys that start somewhere else eventually
approach the limit cycle, regardless of whether they
start from the outside or start from the inside.
So that is stable. An unstable limit cycle --
But I am not calling it a limit cycle if it is unstable.
I am just calling it a closed trajectory, but let's draw one
which is unstable. Here is the way we will look if
it is unstable. Guys that start nearby will be
repelled, driven somewhere else. Or, if they start here,
they will go away from the thing instead of going toward
it. This is unstable.
And I don't call it a limit cycle.
It is just a closed trajectory.
Cycle because it cycles round and round.
Limit because it is the limit of the nearby curves.
The other case where it is unstable is not the limit.
Of course, you could have a case also where the curves
outside spiral in toward it but the ones inside are repelled and
do this. That would be called
semi-stable. And you can make up all sorts
of cases. And I think I,
at one point, drew them in the notes,
but I am not going to. The only interesting one,
of permanent importance that people study,
are the actual limit cycles. No, it was the stable closed
trajectories. Notice, by the way,
a closed trajectory is always a simple curve.
Remember what that means from 18.02?
Simple means it doesn't cross itself.
Why doesn't it cross itself? It cannot cross itself because,
if it tried to, what is wrong with that point?
At that point which way does the vector field go,
that way or that way? Why the interest of limit
cycles? Well, because there are systems
in nature in which just this type of behavior,
they have a certain periodic motion.
And, if you disturb it, gradually it returns to its
original periodic state. A simple example is breathing.
Now I have made you all self-conscious.
All of you are breathing. If you are here you are
breathing. At what rate are you breathing?
Well, you are unaware of it, of course, except now.
If you are sitting here listening.
There is a certain temperature and a certain air circulation in
the room. You are not thinking of
anything, certainly not of the lecture, and the lecture is not
unduly exciting, you will breathe at a certain
steady rate which is a little different for every person but
that is your rate. Now, you can artificially
change that. You could say now I am going to
breathe faster. And indeed you can.
But, as soon as you stop being aware of what you are doing,
the levels of various hormones and carbon dioxide in your
bloodstream and so on will return your breathing to its
natural state. In other words,
that system of your breathing, which is controlled by various
chemicals and hormones in the body, is exhibiting exactly this
type of behavior. It has a certain regular
periodic motion as a system. And, if disturbed,
if artificially you set it out somewhere else,
it will gradually return to its original state.
Now, of course, if I am running it will be
different. Sure.
If you are running you breathe faster, but that is because the
parameters in the system, the a's and the b's in the
equation, the f of (x, y) and g of (x,
y), the parameters in those
functions will be set at different levels.
You will have different hormones, a different of carbon
dioxide and so on. Now, I am not saying that
breathing is modeled by a limit cycle.
It is the sort of thing which one might look for a limit
cycle. That is, of course,
a question for biologists. And, in general,
any type of periodic behavior in nature, people try to see if
there is some system of differential equations which
governs it in which perhaps there is a limit cycle,
which contains a limit cycle. Well, what are the problems?
In a sense, limit cycles are easy to lecture about because so
little is known about them. At the end of the period,
if I have time, I will show you that the
simplest possible question you could ask, the answer to it is
totally known after 120 years of steady trying.
But let's first talk about what sorts of problems people address
with limit cycles. First of all is the existence
problem.
If I give you a system, you know, the right-hand side
is x squared plus 2y cubed minus 3xy,
and the g is something similar. I say does this have limit
cycles? Well, you know how to find its
critical points. But how do you find out if it
has limit cycles? The answer to that is nobody
has any idea. This problem,
in general, there are not much in the way of methods.
Not much.
Not much is known. There is one theorem that you
will find in the notes, a simple theorem called the
Poincare-Bendixson theorem which, for about 60 or 70 years
was about the only result known which enabled people to find
limit cycles. Nowadays the theorem is used
relatively little because people try to find limit cycles by
computer. Now, the difficulty is you have
to know where to look for them. In other words,
the computer screen shows that much and you set the axes and it
doesn't show any limit cycles. That doesn't mean there are not
any. That means they are over there,
or it means there is a big one like there.
And you are looking in the middle of it and don't see it.
So, in general, people don't look for limit
cycles unless the physical system that gave rise to the
pair of differential equations suggests that there is something
repetitive going on like breathing.
And, if it tells you that, then it often gives you
approximate values of the parameters and the variables so
you know where to look. Basically this is done by
computer search guided by the physical problem.
Therefore, I cannot say much more about it today.
Instead I am going to focus my attention on nonexistence.
When can you be sure that a system will not have any limit
cycles? And there are two theorems.
One, again, due to Bendixson who was a Swedish mathematician
who lived around 1900 or so. There is a criterion due to
Bendixson. And there is one involving
critical points. And I would like to describe
both of them for you today. First of all,
Bendixson's criterion.
It is very simply stated and has a marvelous proof,
which I am going to give you. We have D as a region of the
plane.
And what Bendixson's criterion tells you to do is take your
vector field and calculate its divergence.
We are set back in 1802, and this proof is going to be
straight 18.02. You will enjoy it.
Calculate the divergence. Now, I am talking about the
two-dimensional divergence. Remember that is fx,
the partial of f with respect to x, plus the partial of the g,
the j component with respect to y.
And assume that that is a continuous function.
It always will be with us. Practically all the examples I
will give you f and g will be simple polynomials.
They are smooth, continuous and nice and behave
as you want. And you calculate that and
assume -- Suppose, in other words,
that the divergence of f, I need more room.
The hypothesis is that the divergence of f is not zero in
that region D. It is never zero.
It is not zero at any point in that region.
The conclusion is that there are no limit cycles in the
region. If it is not zero in D,
there are no limit cycles. In fact, there are not even any
closed trajectories. You couldn't even have those
bunch of concentric circles, so there are no closed
trajectories of the original system whose divergence you
calculated. There are no closed
trajectories in D. For example,
let me give you a simple example to put a little flesh on
it. Let's see.
What do I have? I prepared an example.
x prime equals, here is a simple nonlinear
system, x cubed plus y cubed.
And y prime equals 3x plus y cubed plus 2y.
Does this system have limit cycles?
Well, even to calculate its critical points would be a
little task, but we can easily answer the question as to
whether it has limit cycles or not by Bendixson's criterion.
Let's calculate the divergence. The divergence of the vector
field whose components are these two functions is,
well, 3x squared, it's the partial of the first
guy with respect to x plus the partial of the second guy with
respect to y, which is 3y squared plus two.
Now, can that be zero anywhere
in the x,y-plane? No, because it is the sum of
these two squares. This much of it could be zero
only at the origin, but that plus two eliminates
even that. This is always positive in the
entire x,y-plane. Here my domain is the whole
x,y-plane and, therefore, the conclusion is
that there are no closed trajectories in the x,y-plane,
anywhere.
And we have done that with just a couple of lines of calculation
and nothing further required. No computer search.
In fact, no computer search could ever proof this.
It would be impossible because, no matter where you look,
there is always some other place to look.
This is an example where a couple lines of mathematics
dispose of the matter far more effectively than a million
dollars worth of calculation. Well, where does Bendixson's
theorem come from? Yes, Bendixson's theorem comes
from 18.02. And I am giving it to you both
to recall a little bit of 18.02 to you.
Because it is about the first example in the course that we
have had of an indirect argument.
And indirect arguments are something you have to slowly get
used to. I am going to give you an
indirect proof. Remember what that is?
You assume the contrary and you show it leads to a
contradiction. What would assuming the
contrary be? Contrary would be I will assume
the divergence is not zero, but I will suppose there is a
closed trajectory. Suppose there is a closed
trajectory that exists.
Let's draw a picture of it.
And let's say it goes around this way.
There is a closed trajectory for our system.
Let's call the curve C. And I am going to call the
inside of it R, the way one often does in
18.02. D is all this region out here,
in which everything is taking place.
This is to exist in D. Now, what I am going to do is
calculate a line integral around that curve.
A line integral of this vector field.
Now, there are two things you can calculate.
One of the line integrals, I will put in a few of the
vectors here. The vectors I know are pointing
this way because that is the direction in which the curve is
being traversed in order to make it a trajectory.
Those are a few of the typical vectors in the field.
I am going to calculate the line integral around that curve
in the positive sense. In other words,
not in the direction of the salmon-colored arrow,
but in the normal sense in which you calculate it using
Green's theorem, for example.
The positive sense means the one which keeps the region,
the inside on your left, as you walk around like that
the region stays on your left. That is the positive sense.
That is the sense in which I am integrating.
I am going to use Green's theorem, but the integral that I
am going to calculate is not the work integral.
I am going to calculate instead the flux integral,
the integral that represents the flux of F across C.
Now, what is that integral? Well, at each point,
you station a little ant and the ant reports the outward flow
rate across that point which is F dotted with the normal vector.
I will put in a few normal vectors just to remind you.
The normal vectors look like little unit vectors pointing
perpendicularly outwards everywhere.
These are the n's. F dotted with the unit normal
vector, and that is added up around the curve.
This quantity gives me the flux of the field across C.
Now, we are going to calculate that by Green's theorem.
But, before we calculate it by Green's theorem,
we are going to psych it out. What is it?
What is the value of that integral?
Well, since I am asking you to do it in your head there can
only be one possible answer. It is zero.
Why is that integral zero? Well, because at each point the
field vector, the velocity vector is
perpendicular to the normal vector.
Why? The normal vector points
perpendicularly to the curve but the field vector always is
tangent to the curve because this curve is a trajectory.
It is always supposed to be going in the direction given by
that white field vector. Do you follow?
A trajectory means that it is always tangent to the field
vector and, therefore, always perpendicular to the
normal vector. This is zero since F dot n is
always zero. Everywhere on the curve,
F dot n has to be zero. There is no flux of this field
across the curve because the field is always in the same
direction as the curve, never perpendicular to it.
It has no components perpendicular to it.
Good. Now let's do it the hard way.
Let's use Green's theorem. Green's theorem says that the
flux across C should be equal to the double integral over that
region of the divergence of F. It's like Gauss theorem in two
dimensions, this version of it. Divergence of F,
that is a function, I double integrate it over the
region, and then that is dx / dy, or let's say da because you
might want to do it in polar coordinates.
And, on the problem set, you certainly will want to do
it in polar coordinates, I think.
All right. How much is that?
Well, we haven't yet used the hypothesis.
All we have done is set up the problem.
Now, the hypothesis was that the divergence is never zero
anywhere in D. Therefore, the divergence is
never zero anywhere in R. What I say is the divergence is
either greater than zero everywhere in R.
Or less than zero everywhere in R.
But it cannot be sometimes positive and sometimes negative.
Why not? In other words,
I say it is not possible the divergence here is one and here
is minus two. That is not possible because,
if I drew a line from this point to that,
along that line the divergence would start positive and end up
negative. And, therefore,
have to be zero some time in between.
It's because it is a continuous function.
It is a continuous function. I am assuming that.
And, therefore, if it sometimes positive and
sometimes negative it has to be zero in between.
You cannot get continuously from plus one to minus two
without passing through zero. The reason for this is,
since the divergence is never zero in R it therefore must
always stay positive or always stay negative.
Now, if it always stays positive, the conclusion is then
this double integral must be positive.
Therefore, this double integral is either greater than zero.
That is if the divergence is always positive.
Or, it is less than zero if the divergence is always negative.
But the one thing it cannot be is not zero.
Well, the left-hand side, Green's theorem is supposed to
be true. Green's theorem is our bedrock.
18.02 would crumble without that so it must be true.
One way of calculating the left-hand side gives us zero.
If we calculate the right-hand side it is not zero.
That is called the contradiction.
Where did the contradiction arise from?
It arose from the fact that I supposed that there was a closed
trajectory in that region. The conclusion is there cannot
be any closed trajectory of that region because it leads to a
contradiction via Green's theorem.
Let me see if I can give you some of the argument for the
other, well, let's at least state the other criterion I
wanted to give you.
Suppose, for example, we use this system,
x prime equals --
Does this have limit cycles?
Does that have limit cycles?
Let's Bendixson it. We will calculate the
divergence of a vector field. It is 2x from the top function.
The partial with respect to x is 2x.
The second function with respect to y is negative 2y.
That certainly could be zero. In fact, this is zero along the
entire line x equals y. Its divergence is zero here
along that whole line. The best I could conclude was,
I could conclude that there is no limit cycle like this and
there is no limit cycle like this, but there is nothing so
far that says a limit cycle could not cross that because
that would not violate Bendixson's theorem.
In other words, any domain that contained part
of this line, the divergence would be zero
along that line. And, therefore,
I could conclude nothing. I could have limit cycles that
cross that line, as long as they included a
piece of that line in them. The answer is I cannot make a
conclusion. Well, that is because I am
using the wrong criterion. Let's instead use the critical
point criterion.
Now, I am going to say that it makes a nice positive statement
but nobody ever uses it this way.
Nonetheless, let's first state it
positively, even though that is not the way to use it.
The positive statement will be, once again, we have our region
D and we have a region of the xy plane and we have our C,
a closed trajectory in it. A closed trajectory of what?
Of our system. And that is supposed to be in
D. The critical point criterion
says something very simple. If you have that situation it
says that inside that closed trajectory there must be a
critical point somewhere.
It says that inside C is a critical point.
Now, this won't help us with the existence problem.
This won't help us find a closed trajectory.
We will take our system and say it has a critical point here and
a critical point there. Does it have a closed
trajectory? Well, all I know is the closed
trajectory, if it exists, will have to go around one or
more of those critical points. But I don't know where.
It is not going to go around it like this.
It might go around it like this.
And my computer search won't find it because it is looking at
too small a part of the screen. It doesn't work that way.
It works negatively by contraposition.
Do you know what the contrapositive is?
You will at least learn that. A implies B says the same thing
as not B implies not A.
They are different statements but they are equivalent to each
other. If you prove one you prove the
other. What would be the
contrapositive here? If you have a closed trajectory
inside is a critical point. The theorem is used this way.
If D has no critical points, it has no closed trajectories
and therefore has no limit cycle.
Because, if it did have a closed trajectory,
inside it would be a critical point.
But I said B had no critical point.
That enables us to dispose of this system that Bendixson could
not handle at all. We can dispose of this system
immediately. Namely, what is it?
Where are its critical points? Well, where is that zero?
x squared plus y squared is one,
plus one is never zero. This is positive.
Or, worse, zero. And then I add the one to it
and it is not zero anymore. This has no zeros and,
therefore, it does not matter that this one has a lot of
zeros. It makes no difference.
It has no critical points. It has none,
therefore, no limit cycles.
Now, I desperately wanted to give you the proof of this.
It is clearly impossible in the time remaining.
The proof requires a little time.
I haven't decided what to do about that.
It might leak over until Friday's lecture.
Instead, I will finish by telling you a story.
How is that?
And along side of it was little y prime.
I am not going to continue on with the letters of the
alphabet. I will prime the earlier one.
This has a total of 12 parameters in it.
But, in fact, if you change variables you can
get rid of all the linear terms. The important part of it is
only the quadratic terms in the beginning.
This sort of thing is called a quadratic system.
After you have departed from linear systems,
it is the simplest kind there is.
And the predictor-prey, the robin-earthworm example I
gave you is of a typical quadratic system and exhibits
typical nonlinear quadratic system behavior.
Now, the problem is the following.
A, b, c, d, e, f and so on,
those are just real numbers, parameters, so I am allowed to
give them any values I want. And the problem that has
bothered people since 1880 when it was first proposed is how
many limit cycles can a quadratic system have?
After 120 years this problem is totally unsolved,
and the mathematicians of the world who are interested in it
cannot even agree with each other on what the right
conjecture is. But let me tell you a little
bit of its history. There were attempts to solve it
in the 20 or 30 years after it was first proposed,
through the 1920 and `30s which all seemed to have gaps in them.
Until finally around 1950 two Russians mathematicians,
one of whom is extremely well-known, Petrovski,
a specialist in systems of ordinary differential equations
published a long and difficult, complicated 100 page paper in
which they proved that the maximum number is three.
I won't put down their names. Petrovski-Landis.
The maximum number was three. And then not many people were
able to read the paper, and those who did there seemed
to be gaps in the reasoning in various places until finally
Arnold who was the greatest Russian, in my opinion,
one of the greatest Russian mathematician,
certainly in this field of analysis and differential
equations, but in other fields, too, he still is great,
although he is somewhat older now, criticized it.
He said look, there is a really big gap in
this argument and it really cannot be considered to be
proven. People tried working very hard
to patch it up and without success.
Then about 1972 or so, '75 maybe, a Chinese
mathematician found a system with four.
Wrote down the numbers, the number a is so much,
b is so much, and they were absurd numbers
like 10^-6 and 40 billion and so on, nothing you could plot on a
computer screen, but found a system with four.
Nobody after this tried to fill in the gap in the
Petrovski-Landis paper. I was then chairman of the math
department, and one of my tasks was protocol and so on.
Anyway, we were trying very hard to attract a Chinese
mathematician to our department to become a full professor.
He was a really outstanding analyst and specialist in
various fields. Anyway, he came in for a
courtesy interview and we chatted.
At the time, I was very much interested in
limit cycles. And I had on my desk the Math
Society's translation of the Chinese book on limit cycles.
A collection of papers by Chinese mathematicians all on
limit cycles. After a certain point he said,
oh, I see you're interested in limit cycle problems.
I said yeah, in particular,
I was reading this paper of the mathematician who found four
limit cycles. And I opened to that system and
said the name is, and I read it out loud.
I said do you by any chance know him?
And he smiled and said yes, very well.
That is my mother. [LAUGHTER]
Well, bye-bye.