1
00:00:00 --> 00:00:06
2
00:00:08 --> 00:00:14
First of all,
the way a nonlinear autonomous
3
00:00:11 --> 00:00:17
system looks,
you have had some practice with
4
00:00:15 --> 00:00:21
it by now.
This is nonlinear.
5
00:00:17 --> 00:00:23
The right-hand side are no
longer simple combinations ax
6
00:00:22 --> 00:00:28
plus by.
Nonlinear and autonomous,
7
00:00:26 --> 00:00:32
these are function just of x
and y.
8
00:00:30 --> 00:00:36
There is no t on the right-hand
side.
9
00:00:33 --> 00:00:39
Now, most of today will be
geometric.
10
00:00:37 --> 00:00:43
The way to get a geometric
picture of that is first by
11
00:00:43 --> 00:00:49
constructing the velocity field
whose components are the
12
00:00:48 --> 00:00:54
functions f and g.
This is a velocity field that
13
00:00:53 --> 00:00:59
gives a picture of the system
and has solutions.
14
00:01:00 --> 00:01:06
The solutions to the system,
from the point of view of
15
00:01:05 --> 00:01:11
functions, they would look like
pairs of functions,
16
00:01:09 --> 00:01:15
x of t, y of t.
17
00:01:12 --> 00:01:18
But, from the point of view of
geometry, when you plot them as
18
00:01:18 --> 00:01:24
parametric equations,
they are called trajectories of
19
00:01:23 --> 00:01:29
the field F, which simply means
that they are curves everywhere
20
00:01:29 --> 00:01:35
having the right velocity.
So a typical curve would look
21
00:01:34 --> 00:01:40
like --
There is a trajectory.
22
00:01:38 --> 00:01:44
And we know it is a trajectory
because at each point the vector
23
00:01:42 --> 00:01:48
on it has, of course,
the right direction,
24
00:01:45 --> 00:01:51
the tangent direction,
but more than that,
25
00:01:48 --> 00:01:54
it has the right velocity.
So here, for example,
26
00:01:51 --> 00:01:57
the point is traveling more
slowly.
27
00:01:53 --> 00:01:59
Here it is traveling more
rapidly because the velocity
28
00:01:57 --> 00:02:03
vector is bigger,
longer.
29
00:02:00 --> 00:02:06
So this is a picture of a
typical trajectory.
30
00:02:03 --> 00:02:09
The only other things that I
should mention are the critical
31
00:02:09 --> 00:02:15
points.
If you have worked the problems
32
00:02:12 --> 00:02:18
for this week,
the first couple of problems,
33
00:02:16 --> 00:02:22
you have already seen the
significance of the critical
34
00:02:21 --> 00:02:27
points.
Well, from Monday's lecture you
35
00:02:24 --> 00:02:30
know from the point of view of
solutions they are constant
36
00:02:29 --> 00:02:35
solutions.
37
00:02:32 --> 00:02:38
38
00:02:37 --> 00:02:43
From the point of view of the
field they are where the field
39
00:02:41 --> 00:02:47
is zero.
There is no velocity vector,
40
00:02:43 --> 00:02:49
in other words.
The velocity vector is zero.
41
00:02:46 --> 00:02:52
And, therefore,
a point being there has no
42
00:02:49 --> 00:02:55
reason to go anywhere else.
And, spelling it out,
43
00:02:53 --> 00:02:59
it's where the partial
derivatives, where the values of
44
00:02:57 --> 00:03:03
the functions on the right-hand
side, which give the two
45
00:03:01 --> 00:03:07
components, the i and j
components of the field,
46
00:03:04 --> 00:03:10
where they are zero.
47
00:03:07 --> 00:03:13
48
00:03:11 --> 00:03:17
That is all I will need by way
of a recall today.
49
00:03:14 --> 00:03:20
I don't think I will need
anything else.
50
00:03:17 --> 00:03:23
The topic for today is another
kind of behavior that you have
51
00:03:22 --> 00:03:28
not yet observed at the computer
screen, unless you have worked
52
00:03:26 --> 00:03:32
ahead, and that is there are
trajectories which go along to
53
00:03:31 --> 00:03:37
infinity or end up at a critical
point.
54
00:03:35 --> 00:03:41
They are the critical points
that just sit there all the
55
00:03:39 --> 00:03:45
time.
But there is a third type of
56
00:03:42 --> 00:03:48
behavior that a trajectory can
have where it neither sits for
57
00:03:47 --> 00:03:53
all time nor goes off for all
time.
58
00:03:50 --> 00:03:56
Instead, it repeats itself.
Such a thing is called a closed
59
00:03:55 --> 00:04:01
trajectory.
What does it look like?
60
00:03:59 --> 00:04:05
Well, it is a closed curve in
the plane that at every point,
61
00:04:05 --> 00:04:11
it is a trajectory,
i.e., the arrows at each point,
62
00:04:10 --> 00:04:16
let's say it is traced in the
clockwise direction.
63
00:04:15 --> 00:04:21
And so the arrows of the field
will go like this.
64
00:04:20 --> 00:04:26
Here it is going slowly,
here it is very slow and here
65
00:04:25 --> 00:04:31
it picks up a little speed again
and so on.
66
00:04:31 --> 00:04:37
Now, for such a trajectory what
is happening?
67
00:04:35 --> 00:04:41
Well, it goes around in finite
time and then repeats itself.
68
00:04:42 --> 00:04:48
It just goes round and round
forever if you land on that
69
00:04:48 --> 00:04:54
trajectory.
It represents a system that
70
00:04:53 --> 00:04:59
returns to its original state
periodically.
71
00:04:57 --> 00:05:03
It represents periodic behavior
of the system.
72
00:05:04 --> 00:05:10
73
00:05:16 --> 00:05:22
Now, we have seen one example
of that, a simple example where
74
00:05:23 --> 00:05:29
this simple system,
x prime equals y,
75
00:05:29 --> 00:05:35
y prime equals negative x.
76
00:05:35 --> 00:05:41
77
00:05:40 --> 00:05:46
We could write down the
solutions to that directly,
78
00:05:43 --> 00:05:49
but if you want to do
eigenvalues and eigenvectors the
79
00:05:47 --> 00:05:53
matrix will look like this.
The equation will be lambda
80
00:05:51 --> 00:05:57
squared plus zero lambda plus
one equals zero,
81
00:05:55 --> 00:06:01
so the eigenvalues will be plus
82
00:05:58 --> 00:06:04
or minus i.
In fact, from then on you could
83
00:06:02 --> 00:06:08
work out in the usual ways the
eigenvectors,
84
00:06:04 --> 00:06:10
complex eigenvectors and
separate them.
85
00:06:07 --> 00:06:13
But, look, you can avoid all
that just by writing down the
86
00:06:11 --> 00:06:17
solution.
The solutions are sines and
87
00:06:13 --> 00:06:19
cosines.
One basic solution will be x
88
00:06:16 --> 00:06:22
equals cosine t,
in which case what is y?
89
00:06:19 --> 00:06:25
Well, y is the derivative of
that.
90
00:06:22 --> 00:06:28
That will be minus sine t.
91
00:06:24 --> 00:06:30
Another basic solution,
we will start with x equals
92
00:06:28 --> 00:06:34
sine t.
In which case y will be cosine
93
00:06:33 --> 00:06:39
t, its derivative.
94
00:06:36 --> 00:06:42
Now, if you do that,
what do these things look like?
95
00:06:41 --> 00:06:47
Well, either of these two basic
solutions looks like a circle,
96
00:06:47 --> 00:06:53
not traced in the usual way but
in the opposite way.
97
00:06:52 --> 00:06:58
For example,
when t is equal to zero it
98
00:06:56 --> 00:07:02
starts at the point one,
zero.
99
00:07:00 --> 00:07:06
Now, if the minus sign were not
there this would be x equals
100
00:07:04 --> 00:07:10
cosine t,
y equals sine t,
101
00:07:08 --> 00:07:14
which is the usual
counterclockwise circle.
102
00:07:11 --> 00:07:17
But if I change y from sine t
to negative sine t
103
00:07:15 --> 00:07:21
it is going around the
other way.
104
00:07:18 --> 00:07:24
So this circle is traced that
way.
105
00:07:20 --> 00:07:26
And this is a family of
circles, according to the values
106
00:07:24 --> 00:07:30
of c1 and c2,
concentric, all of which go
107
00:07:27 --> 00:07:33
around clockwise.
So those are closed
108
00:07:31 --> 00:07:37
trajectories.
Those are the solutions.
109
00:07:33 --> 00:07:39
They are trajectories of the
vector field.
110
00:07:36 --> 00:07:42
They are closed.
They come around and they
111
00:07:39 --> 00:07:45
repeat in finite time.
Now, these are no good.
112
00:07:42 --> 00:07:48
These are the kind I am not
interested in.
113
00:07:45 --> 00:07:51
These are commonplace,
and we are interested in good
114
00:07:49 --> 00:07:55
stuff today.
And the good stuff we are
115
00:07:51 --> 00:07:57
interested in is limit cycles.
116
00:07:55 --> 00:08:01
117
00:08:01 --> 00:08:07
A limit cycle is a closed
trajectory with a couple of
118
00:08:06 --> 00:08:12
extra hypotheses.
It is a closed trajectory,
119
00:08:10 --> 00:08:16
just like those guys,
but it has something they don't
120
00:08:15 --> 00:08:21
have, namely,
it is king of the roost.
121
00:08:19 --> 00:08:25
They have to be isolated,
no other guys nearby.
122
00:08:23 --> 00:08:29
And they also have to be
stable.
123
00:08:27 --> 00:08:33
See, the problem here is that
none of these stands out from
124
00:08:32 --> 00:08:38
any of the others.
In other words,
125
00:08:37 --> 00:08:43
there must be,
isolated means,
126
00:08:40 --> 00:08:46
no others nearby.
127
00:08:42 --> 00:08:48
128
00:08:49 --> 00:08:55
That is just what goes wrong
here.
129
00:08:51 --> 00:08:57
Arbitrarily close to each of
these circles is yet another
130
00:08:55 --> 00:09:01
circle doing exactly the same
thing.
131
00:08:58 --> 00:09:04
That means that there are some
that are only of mild interest.
132
00:09:03 --> 00:09:09
What is much more interesting
is to find a cycle where there
133
00:09:07 --> 00:09:13
is nothing nearby.
Something, therefore,
134
00:09:10 --> 00:09:16
that looks like this.
135
00:09:13 --> 00:09:19
136
00:09:22 --> 00:09:28
Here is our pink guy.
Let's make this one go
137
00:09:25 --> 00:09:31
counterclockwise.
Here is a limit cycle,
138
00:09:28 --> 00:09:34
it seems to be.
And now what do nearby guys do?
139
00:09:33 --> 00:09:39
Well, they should approach it.
Somebody here like that does
140
00:09:38 --> 00:09:44
this, spirals in and gets ever
and every closer to that thing.
141
00:09:44 --> 00:09:50
Now, it can never join it
because, if it joined it at the
142
00:09:50 --> 00:09:56
joining point,
I would have two solutions
143
00:09:54 --> 00:10:00
going through this point.
And that is illegal.
144
00:10:00 --> 00:10:06
All it can do is get
arbitrarily close.
145
00:10:02 --> 00:10:08
On the computer screen it will
look as if it joins it but,
146
00:10:06 --> 00:10:12
of course, it cannot.
It is just the resolution,
147
00:10:10 --> 00:10:16
the pixels.
Not enough pixels.
148
00:10:12 --> 00:10:18
The resolution isn't good
enough.
149
00:10:14 --> 00:10:20
And the ones that start further
away will take longer to find
150
00:10:18 --> 00:10:24
their way to the limit cycle and
they will always stay outside of
151
00:10:23 --> 00:10:29
the earlier guys,
but they will get arbitrarily
152
00:10:26 --> 00:10:32
close, too.
How about inside?
153
00:10:30 --> 00:10:36
Inside, well,
it starts somewhere and does
154
00:10:33 --> 00:10:39
the same thing.
It starts here and will try to
155
00:10:37 --> 00:10:43
join the limit cycle.
That is what I mean by
156
00:10:41 --> 00:10:47
stability.
Stability means that nearby
157
00:10:44 --> 00:10:50
guys, the guys that start
somewhere else eventually
158
00:10:48 --> 00:10:54
approach the limit cycle,
regardless of whether they
159
00:10:52 --> 00:10:58
start from the outside or start
from the inside.
160
00:10:56 --> 00:11:02
So that is stable.
An unstable limit cycle --
161
00:11:02 --> 00:11:08
But I am not calling it a limit
cycle if it is unstable.
162
00:11:06 --> 00:11:12
I am just calling it a closed
trajectory, but let's draw one
163
00:11:10 --> 00:11:16
which is unstable.
Here is the way we will look if
164
00:11:14 --> 00:11:20
it is unstable.
Guys that start nearby will be
165
00:11:17 --> 00:11:23
repelled, driven somewhere else.
Or, if they start here,
166
00:11:21 --> 00:11:27
they will go away from the
thing instead of going toward
167
00:11:25 --> 00:11:31
it.
This is unstable.
168
00:11:27 --> 00:11:33
And I don't call it a limit
cycle.
169
00:11:29 --> 00:11:35
It is just a closed trajectory.
170
00:11:33 --> 00:11:39
171
00:11:38 --> 00:11:44
Cycle because it cycles round
and round.
172
00:11:40 --> 00:11:46
Limit because it is the limit
of the nearby curves.
173
00:11:44 --> 00:11:50
The other case where it is
unstable is not the limit.
174
00:11:48 --> 00:11:54
Of course, you could have a
case also where the curves
175
00:11:52 --> 00:11:58
outside spiral in toward it but
the ones inside are repelled and
176
00:11:57 --> 00:12:03
do this.
That would be called
177
00:11:59 --> 00:12:05
semi-stable.
And you can make up all sorts
178
00:12:03 --> 00:12:09
of cases.
And I think I,
179
00:12:05 --> 00:12:11
at one point,
drew them in the notes,
180
00:12:07 --> 00:12:13
but I am not going to.
The only interesting one,
181
00:12:11 --> 00:12:17
of permanent importance that
people study,
182
00:12:14 --> 00:12:20
are the actual limit cycles.
No, it was the stable closed
183
00:12:19 --> 00:12:25
trajectories.
Notice, by the way,
184
00:12:21 --> 00:12:27
a closed trajectory is always a
simple curve.
185
00:12:25 --> 00:12:31
Remember what that means from
18.02?
186
00:12:29 --> 00:12:35
Simple means it doesn't cross
itself.
187
00:12:32 --> 00:12:38
Why doesn't it cross itself?
It cannot cross itself because,
188
00:12:37 --> 00:12:43
if it tried to,
what is wrong with that point?
189
00:12:41 --> 00:12:47
At that point which way does
the vector field go,
190
00:12:46 --> 00:12:52
that way or that way?
Why the interest of limit
191
00:12:50 --> 00:12:56
cycles?
Well, because there are systems
192
00:12:54 --> 00:13:00
in nature in which just this
type of behavior,
193
00:12:58 --> 00:13:04
they have a certain periodic
motion.
194
00:13:03 --> 00:13:09
And, if you disturb it,
gradually it returns to its
195
00:13:07 --> 00:13:13
original periodic state.
A simple example is breathing.
196
00:13:11 --> 00:13:17
Now I have made you all
self-conscious.
197
00:13:14 --> 00:13:20
All of you are breathing.
If you are here you are
198
00:13:18 --> 00:13:24
breathing.
At what rate are you breathing?
199
00:13:22 --> 00:13:28
Well, you are unaware of it,
of course, except now.
200
00:13:26 --> 00:13:32
If you are sitting here
listening.
201
00:13:30 --> 00:13:36
There is a certain temperature
and a certain air circulation in
202
00:13:34 --> 00:13:40
the room.
You are not thinking of
203
00:13:36 --> 00:13:42
anything, certainly not of the
lecture, and the lecture is not
204
00:13:40 --> 00:13:46
unduly exciting,
you will breathe at a certain
205
00:13:43 --> 00:13:49
steady rate which is a little
different for every person but
206
00:13:47 --> 00:13:53
that is your rate.
Now, you can artificially
207
00:13:50 --> 00:13:56
change that.
You could say now I am going to
208
00:13:53 --> 00:13:59
breathe faster.
And indeed you can.
209
00:13:57 --> 00:14:03
But, as soon as you stop being
aware of what you are doing,
210
00:14:01 --> 00:14:07
the levels of various hormones
and carbon dioxide in your
211
00:14:06 --> 00:14:12
bloodstream and so on will
return your breathing to its
212
00:14:11 --> 00:14:17
natural state.
In other words,
213
00:14:13 --> 00:14:19
that system of your breathing,
which is controlled by various
214
00:14:18 --> 00:14:24
chemicals and hormones in the
body, is exhibiting exactly this
215
00:14:23 --> 00:14:29
type of behavior.
It has a certain regular
216
00:14:27 --> 00:14:33
periodic motion as a system.
And, if disturbed,
217
00:14:32 --> 00:14:38
if artificially you set it out
somewhere else,
218
00:14:34 --> 00:14:40
it will gradually return to its
original state.
219
00:14:37 --> 00:14:43
Now, of course,
if I am running it will be
220
00:14:40 --> 00:14:46
different.
Sure.
221
00:14:41 --> 00:14:47
If you are running you breathe
faster, but that is because the
222
00:14:45 --> 00:14:51
parameters in the system,
the a's and the b's in the
223
00:14:48 --> 00:14:54
equation, the f of (x,
y) and g of (x,
224
00:14:50 --> 00:14:56
y), the parameters in those
225
00:14:53 --> 00:14:59
functions will be set at
different levels.
226
00:14:56 --> 00:15:02
You will have different
hormones, a different of carbon
227
00:14:59 --> 00:15:05
dioxide and so on.
Now, I am not saying that
228
00:15:04 --> 00:15:10
breathing is modeled by a limit
cycle.
229
00:15:07 --> 00:15:13
It is the sort of thing which
one might look for a limit
230
00:15:11 --> 00:15:17
cycle.
That is, of course,
231
00:15:13 --> 00:15:19
a question for biologists.
And, in general,
232
00:15:17 --> 00:15:23
any type of periodic behavior
in nature, people try to see if
233
00:15:22 --> 00:15:28
there is some system of
differential equations which
234
00:15:26 --> 00:15:32
governs it in which perhaps
there is a limit cycle,
235
00:15:30 --> 00:15:36
which contains a limit cycle.
Well, what are the problems?
236
00:15:36 --> 00:15:42
In a sense, limit cycles are
easy to lecture about because so
237
00:15:41 --> 00:15:47
little is known about them.
At the end of the period,
238
00:15:46 --> 00:15:52
if I have time,
I will show you that the
239
00:15:49 --> 00:15:55
simplest possible question you
could ask, the answer to it is
240
00:15:55 --> 00:16:01
totally known after 120 years of
steady trying.
241
00:16:00 --> 00:16:06
But let's first talk about what
sorts of problems people address
242
00:16:05 --> 00:16:11
with limit cycles.
First of all is the existence
243
00:16:09 --> 00:16:15
problem.
244
00:16:10 --> 00:16:16
245
00:16:16 --> 00:16:22
If I give you a system,
you know, the right-hand side
246
00:16:19 --> 00:16:25
is x squared plus 2y cubed minus
3xy,
247
00:16:23 --> 00:16:29
and the g is something similar.
I say does this have limit
248
00:16:26 --> 00:16:32
cycles?
Well, you know how to find its
249
00:16:29 --> 00:16:35
critical points.
But how do you find out if it
250
00:16:34 --> 00:16:40
has limit cycles?
The answer to that is nobody
251
00:16:40 --> 00:16:46
has any idea.
This problem,
252
00:16:44 --> 00:16:50
in general, there are not much
in the way of methods.
253
00:16:50 --> 00:16:56
Not much.
254
00:16:53 --> 00:16:59
255
00:17:00 --> 00:17:06
Not much is known.
There is one theorem that you
256
00:17:03 --> 00:17:09
will find in the notes,
a simple theorem called the
257
00:17:07 --> 00:17:13
Poincare-Bendixson theorem
which, for about 60 or 70 years
258
00:17:11 --> 00:17:17
was about the only result known
which enabled people to find
259
00:17:16 --> 00:17:22
limit cycles.
Nowadays the theorem is used
260
00:17:19 --> 00:17:25
relatively little because people
try to find limit cycles by
261
00:17:24 --> 00:17:30
computer.
Now, the difficulty is you have
262
00:17:27 --> 00:17:33
to know where to look for them.
In other words,
263
00:17:32 --> 00:17:38
the computer screen shows that
much and you set the axes and it
264
00:17:36 --> 00:17:42
doesn't show any limit cycles.
That doesn't mean there are not
265
00:17:40 --> 00:17:46
any.
That means they are over there,
266
00:17:43 --> 00:17:49
or it means there is a big one
like there.
267
00:17:46 --> 00:17:52
And you are looking in the
middle of it and don't see it.
268
00:17:50 --> 00:17:56
So, in general,
people don't look for limit
269
00:17:53 --> 00:17:59
cycles unless the physical
system that gave rise to the
270
00:17:56 --> 00:18:02
pair of differential equations
suggests that there is something
271
00:18:01 --> 00:18:07
repetitive going on like
breathing.
272
00:18:05 --> 00:18:11
And, if it tells you that,
then it often gives you
273
00:18:09 --> 00:18:15
approximate values of the
parameters and the variables so
274
00:18:14 --> 00:18:20
you know where to look.
Basically this is done by
275
00:18:18 --> 00:18:24
computer search guided by the
physical problem.
276
00:18:23 --> 00:18:29
277
00:18:31 --> 00:18:37
Therefore, I cannot say much
more about it today.
278
00:18:35 --> 00:18:41
Instead I am going to focus my
attention on nonexistence.
279
00:18:40 --> 00:18:46
When can you be sure that a
system will not have any limit
280
00:18:46 --> 00:18:52
cycles?
And there are two theorems.
281
00:18:49 --> 00:18:55
One, again, due to Bendixson
who was a Swedish mathematician
282
00:18:54 --> 00:19:00
who lived around 1900 or so.
There is a criterion due to
283
00:18:59 --> 00:19:05
Bendixson.
And there is one involving
284
00:19:04 --> 00:19:10
critical points.
And I would like to describe
285
00:19:08 --> 00:19:14
both of them for you today.
First of all,
286
00:19:11 --> 00:19:17
Bendixson's criterion.
287
00:19:14 --> 00:19:20
288
00:19:22 --> 00:19:28
It is very simply stated and
has a marvelous proof,
289
00:19:25 --> 00:19:31
which I am going to give you.
We have D as a region of the
290
00:19:29 --> 00:19:35
plane.
291
00:19:30 --> 00:19:36
292
00:19:35 --> 00:19:41
And what Bendixson's criterion
tells you to do is take your
293
00:19:40 --> 00:19:46
vector field and calculate its
divergence.
294
00:19:43 --> 00:19:49
We are set back in 1802,
and this proof is going to be
295
00:19:48 --> 00:19:54
straight 18.02.
You will enjoy it.
296
00:19:51 --> 00:19:57
Calculate the divergence.
Now, I am talking about the
297
00:19:55 --> 00:20:01
two-dimensional divergence.
Remember that is fx,
298
00:20:00 --> 00:20:06
the partial of f with respect
to x, plus the partial of the g,
299
00:20:05 --> 00:20:11
the j component with respect to
y.
300
00:20:08 --> 00:20:14
And assume that that is a
continuous function.
301
00:20:12 --> 00:20:18
It always will be with us.
Practically all the examples I
302
00:20:16 --> 00:20:22
will give you f and g will be
simple polynomials.
303
00:20:20 --> 00:20:26
They are smooth,
continuous and nice and behave
304
00:20:24 --> 00:20:30
as you want.
And you calculate that and
305
00:20:27 --> 00:20:33
assume --
Suppose, in other words,
306
00:20:32 --> 00:20:38
that the divergence of f,
I need more room.
307
00:20:36 --> 00:20:42
The hypothesis is that the
divergence of f is not zero in
308
00:20:42 --> 00:20:48
that region D.
It is never zero.
309
00:20:45 --> 00:20:51
It is not zero at any point in
that region.
310
00:20:49 --> 00:20:55
The conclusion is that there
are no limit cycles in the
311
00:20:55 --> 00:21:01
region.
If it is not zero in D,
312
00:20:58 --> 00:21:04
there are no limit cycles.
In fact, there are not even any
313
00:21:04 --> 00:21:10
closed trajectories.
You couldn't even have those
314
00:21:09 --> 00:21:15
bunch of concentric circles,
so there are no closed
315
00:21:13 --> 00:21:19
trajectories of the original
system whose divergence you
316
00:21:18 --> 00:21:24
calculated.
There are no closed
317
00:21:21 --> 00:21:27
trajectories in D.
For example,
318
00:21:24 --> 00:21:30
let me give you a simple
example to put a little flesh on
319
00:21:29 --> 00:21:35
it.
Let's see.
320
00:21:32 --> 00:21:38
What do I have?
I prepared an example.
321
00:21:35 --> 00:21:41
x prime equals,
here is a simple nonlinear
322
00:21:40 --> 00:21:46
system, x cubed plus y cubed.
323
00:21:45 --> 00:21:51
And y prime equals 3x plus y
cubed plus 2y.
324
00:21:51 --> 00:21:57
325
00:21:53 --> 00:21:59
326
00:21:59 --> 00:22:05
Does this system have limit
cycles?
327
00:22:01 --> 00:22:07
Well, even to calculate its
critical points would be a
328
00:22:05 --> 00:22:11
little task, but we can easily
answer the question as to
329
00:22:09 --> 00:22:15
whether it has limit cycles or
not by Bendixson's criterion.
330
00:22:14 --> 00:22:20
Let's calculate the divergence.
The divergence of the vector
331
00:22:18 --> 00:22:24
field whose components are these
two functions is,
332
00:22:22 --> 00:22:28
well, 3x squared,
it's the partial of the first
333
00:22:26 --> 00:22:32
guy with respect to x plus the
partial of the second guy with
334
00:22:31 --> 00:22:37
respect to y,
which is 3y squared plus two.
335
00:22:34 --> 00:22:40
Now, can that be zero anywhere
336
00:22:39 --> 00:22:45
in the x,y-plane?
No, because it is the sum of
337
00:22:43 --> 00:22:49
these two squares.
This much of it could be zero
338
00:22:47 --> 00:22:53
only at the origin,
but that plus two eliminates
339
00:22:51 --> 00:22:57
even that.
This is always positive in the
340
00:22:55 --> 00:23:01
entire x,y-plane.
Here my domain is the whole
341
00:22:59 --> 00:23:05
x,y-plane and,
therefore, the conclusion is
342
00:23:03 --> 00:23:09
that there are no closed
trajectories in the x,y-plane,
343
00:23:07 --> 00:23:13
anywhere.
344
00:23:10 --> 00:23:16
345
00:23:15 --> 00:23:21
And we have done that with just
a couple of lines of calculation
346
00:23:18 --> 00:23:24
and nothing further required.
No computer search.
347
00:23:21 --> 00:23:27
In fact, no computer search
could ever proof this.
348
00:23:24 --> 00:23:30
It would be impossible because,
no matter where you look,
349
00:23:27 --> 00:23:33
there is always some other
place to look.
350
00:23:30 --> 00:23:36
This is an example where a
couple lines of mathematics
351
00:23:35 --> 00:23:41
dispose of the matter far more
effectively than a million
352
00:23:41 --> 00:23:47
dollars worth of calculation.
Well, where does Bendixson's
353
00:23:46 --> 00:23:52
theorem come from?
Yes, Bendixson's theorem comes
354
00:23:51 --> 00:23:57
from 18.02.
And I am giving it to you both
355
00:23:56 --> 00:24:02
to recall a little bit of 18.02
to you.
356
00:24:01 --> 00:24:07
Because it is about the first
example in the course that we
357
00:24:06 --> 00:24:12
have had of an indirect
argument.
358
00:24:09 --> 00:24:15
And indirect arguments are
something you have to slowly get
359
00:24:14 --> 00:24:20
used to.
I am going to give you an
360
00:24:17 --> 00:24:23
indirect proof.
Remember what that is?
361
00:24:20 --> 00:24:26
You assume the contrary and you
show it leads to a
362
00:24:25 --> 00:24:31
contradiction.
What would assuming the
363
00:24:28 --> 00:24:34
contrary be?
Contrary would be I will assume
364
00:24:34 --> 00:24:40
the divergence is not zero,
but I will suppose there is a
365
00:24:40 --> 00:24:46
closed trajectory.
Suppose there is a closed
366
00:24:44 --> 00:24:50
trajectory that exists.
367
00:24:48 --> 00:24:54
368
00:24:56 --> 00:25:02
Let's draw a picture of it.
369
00:24:59 --> 00:25:05
370
00:25:08 --> 00:25:14
And let's say it goes around
this way.
371
00:25:11 --> 00:25:17
There is a closed trajectory
for our system.
372
00:25:15 --> 00:25:21
Let's call the curve C.
And I am going to call the
373
00:25:19 --> 00:25:25
inside of it R,
the way one often does in
374
00:25:22 --> 00:25:28
18.02.
D is all this region out here,
375
00:25:25 --> 00:25:31
in which everything is taking
place.
376
00:25:30 --> 00:25:36
This is to exist in D.
Now, what I am going to do is
377
00:25:35 --> 00:25:41
calculate a line integral around
that curve.
378
00:25:40 --> 00:25:46
379
00:25:45 --> 00:25:51
A line integral of this vector
field.
380
00:25:47 --> 00:25:53
Now, there are two things you
can calculate.
381
00:25:51 --> 00:25:57
One of the line integrals,
I will put in a few of the
382
00:25:55 --> 00:26:01
vectors here.
The vectors I know are pointing
383
00:25:58 --> 00:26:04
this way because that is the
direction in which the curve is
384
00:26:03 --> 00:26:09
being traversed in order to make
it a trajectory.
385
00:26:08 --> 00:26:14
Those are a few of the typical
vectors in the field.
386
00:26:12 --> 00:26:18
I am going to calculate the
line integral around that curve
387
00:26:16 --> 00:26:22
in the positive sense.
In other words,
388
00:26:19 --> 00:26:25
not in the direction of the
salmon-colored arrow,
389
00:26:23 --> 00:26:29
but in the normal sense in
which you calculate it using
390
00:26:27 --> 00:26:33
Green's theorem,
for example.
391
00:26:31 --> 00:26:37
The positive sense means the
one which keeps the region,
392
00:26:34 --> 00:26:40
the inside on your left,
as you walk around like that
393
00:26:38 --> 00:26:44
the region stays on your left.
That is the positive sense.
394
00:26:42 --> 00:26:48
That is the sense in which I am
integrating.
395
00:26:45 --> 00:26:51
I am going to use Green's
theorem, but the integral that I
396
00:26:49 --> 00:26:55
am going to calculate is not the
work integral.
397
00:26:52 --> 00:26:58
I am going to calculate instead
the flux integral,
398
00:26:55 --> 00:27:01
the integral that represents
the flux of F across C.
399
00:27:00 --> 00:27:06
Now, what is that integral?
Well, at each point,
400
00:27:04 --> 00:27:10
you station a little ant and
the ant reports the outward flow
401
00:27:09 --> 00:27:15
rate across that point which is
F dotted with the normal vector.
402
00:27:15 --> 00:27:21
I will put in a few normal
vectors just to remind you.
403
00:27:20 --> 00:27:26
The normal vectors look like
little unit vectors pointing
404
00:27:25 --> 00:27:31
perpendicularly outwards
everywhere.
405
00:27:29 --> 00:27:35
These are the n's.
F dotted with the unit normal
406
00:27:34 --> 00:27:40
vector, and that is added up
around the curve.
407
00:27:38 --> 00:27:44
This quantity gives me the flux
of the field across C.
408
00:27:43 --> 00:27:49
Now, we are going to calculate
that by Green's theorem.
409
00:27:48 --> 00:27:54
But, before we calculate it by
Green's theorem,
410
00:27:52 --> 00:27:58
we are going to psych it out.
What is it?
411
00:27:55 --> 00:28:01
What is the value of that
integral?
412
00:28:00 --> 00:28:06
Well, since I am asking you to
do it in your head there can
413
00:28:04 --> 00:28:10
only be one possible answer.
It is zero.
414
00:28:06 --> 00:28:12
Why is that integral zero?
Well, because at each point the
415
00:28:10 --> 00:28:16
field vector,
the velocity vector is
416
00:28:13 --> 00:28:19
perpendicular to the normal
vector.
417
00:28:15 --> 00:28:21
Why?
The normal vector points
418
00:28:17 --> 00:28:23
perpendicularly to the curve but
the field vector always is
419
00:28:22 --> 00:28:28
tangent to the curve because
this curve is a trajectory.
420
00:28:27 --> 00:28:33
It is always supposed to be
going in the direction given by
421
00:28:34 --> 00:28:40
that white field vector.
Do you follow?
422
00:28:39 --> 00:28:45
A trajectory means that it is
always tangent to the field
423
00:28:47 --> 00:28:53
vector and, therefore,
always perpendicular to the
424
00:28:53 --> 00:28:59
normal vector.
This is zero since F dot n is
425
00:28:59 --> 00:29:05
always zero.
Everywhere on the curve,
426
00:29:03 --> 00:29:09
F dot n has to be zero.
There is no flux of this field
427
00:29:07 --> 00:29:13
across the curve because the
field is always in the same
428
00:29:12 --> 00:29:18
direction as the curve,
never perpendicular to it.
429
00:29:15 --> 00:29:21
It has no components
perpendicular to it.
430
00:29:19 --> 00:29:25
Good.
Now let's do it the hard way.
431
00:29:21 --> 00:29:27
Let's use Green's theorem.
Green's theorem says that the
432
00:29:25 --> 00:29:31
flux across C should be equal to
the double integral over that
433
00:29:30 --> 00:29:36
region of the divergence of F.
It's like Gauss theorem in two
434
00:29:35 --> 00:29:41
dimensions, this version of it.
Divergence of F,
435
00:29:39 --> 00:29:45
that is a function,
I double integrate it over the
436
00:29:42 --> 00:29:48
region, and then that is dx /
dy, or let's say da because you
437
00:29:46 --> 00:29:52
might want to do it in polar
coordinates.
438
00:29:48 --> 00:29:54
And, on the problem set,
you certainly will want to do
439
00:29:52 --> 00:29:58
it in polar coordinates,
I think.
440
00:29:55 --> 00:30:01
441
00:30:00 --> 00:30:06
All right.
How much is that?
442
00:30:02 --> 00:30:08
Well, we haven't yet used the
hypothesis.
443
00:30:06 --> 00:30:12
All we have done is set up the
problem.
444
00:30:10 --> 00:30:16
Now, the hypothesis was that
the divergence is never zero
445
00:30:16 --> 00:30:22
anywhere in D.
Therefore, the divergence is
446
00:30:20 --> 00:30:26
never zero anywhere in R.
What I say is the divergence is
447
00:30:26 --> 00:30:32
either greater than zero
everywhere in R.
448
00:30:32 --> 00:30:38
Or less than zero everywhere in
R.
449
00:30:35 --> 00:30:41
But it cannot be sometimes
positive and sometimes negative.
450
00:30:40 --> 00:30:46
Why not?
In other words,
451
00:30:42 --> 00:30:48
I say it is not possible the
divergence here is one and here
452
00:30:47 --> 00:30:53
is minus two.
That is not possible because,
453
00:30:51 --> 00:30:57
if I drew a line from this
point to that,
454
00:30:55 --> 00:31:01
along that line the divergence
would start positive and end up
455
00:31:00 --> 00:31:06
negative.
And, therefore,
456
00:31:04 --> 00:31:10
have to be zero some time in
between.
457
00:31:06 --> 00:31:12
It's because it is a continuous
function.
458
00:31:09 --> 00:31:15
It is a continuous function.
I am assuming that.
459
00:31:12 --> 00:31:18
And, therefore,
if it sometimes positive and
460
00:31:15 --> 00:31:21
sometimes negative it has to be
zero in between.
461
00:31:18 --> 00:31:24
You cannot get continuously
from plus one to minus two
462
00:31:22 --> 00:31:28
without passing through zero.
The reason for this is,
463
00:31:27 --> 00:31:33
since the divergence is never
zero in R it therefore must
464
00:31:35 --> 00:31:41
always stay positive or always
stay negative.
465
00:31:40 --> 00:31:46
Now, if it always stays
positive, the conclusion is then
466
00:31:47 --> 00:31:53
this double integral must be
positive.
467
00:31:52 --> 00:31:58
Therefore, this double integral
is either greater than zero.
468
00:32:01 --> 00:32:07
That is if the divergence is
always positive.
469
00:32:04 --> 00:32:10
Or, it is less than zero if the
divergence is always negative.
470
00:32:10 --> 00:32:16
But the one thing it cannot be
is not zero.
471
00:32:14 --> 00:32:20
Well, the left-hand side,
Green's theorem is supposed to
472
00:32:18 --> 00:32:24
be true.
Green's theorem is our bedrock.
473
00:32:22 --> 00:32:28
18.02 would crumble without
that so it must be true.
474
00:32:26 --> 00:32:32
One way of calculating the
left-hand side gives us zero.
475
00:32:33 --> 00:32:39
If we calculate the right-hand
side it is not zero.
476
00:32:36 --> 00:32:42
That is called the
contradiction.
477
00:32:38 --> 00:32:44
Where did the contradiction
arise from?
478
00:32:41 --> 00:32:47
It arose from the fact that I
supposed that there was a closed
479
00:32:45 --> 00:32:51
trajectory in that region.
The conclusion is there cannot
480
00:32:48 --> 00:32:54
be any closed trajectory of that
region because it leads to a
481
00:32:52 --> 00:32:58
contradiction via Green's
theorem.
482
00:32:56 --> 00:33:02
483
00:33:02 --> 00:33:08
Let me see if I can give you
some of the argument for the
484
00:33:06 --> 00:33:12
other, well, let's at least
state the other criterion I
485
00:33:10 --> 00:33:16
wanted to give you.
486
00:33:12 --> 00:33:18
487
00:33:21 --> 00:33:27
Suppose, for example,
we use this system,
488
00:33:25 --> 00:33:31
x prime equals --
489
00:33:27 --> 00:33:33
490
00:33:50 --> 00:33:56
Does this have limit cycles?
491
00:33:53 --> 00:33:59
492
00:33:59 --> 00:34:05
Does that have limit cycles?
493
00:34:02 --> 00:34:08
494
00:34:08 --> 00:34:14
Let's Bendixson it.
We will calculate the
495
00:34:12 --> 00:34:18
divergence of a vector field.
It is 2x from the top function.
496
00:34:18 --> 00:34:24
The partial with respect to x
is 2x.
497
00:34:22 --> 00:34:28
The second function with
respect to y is negative 2y.
498
00:34:29 --> 00:34:35
That certainly could be zero.
In fact, this is zero along the
499
00:34:34 --> 00:34:40
entire line x equals y.
Its divergence is zero here
500
00:34:39 --> 00:34:45
along that whole line.
The best I could conclude was,
501
00:34:44 --> 00:34:50
I could conclude that there is
no limit cycle like this and
502
00:34:50 --> 00:34:56
there is no limit cycle like
this, but there is nothing so
503
00:34:55 --> 00:35:01
far that says a limit cycle
could not cross that because
504
00:35:00 --> 00:35:06
that would not violate
Bendixson's theorem.
505
00:35:06 --> 00:35:12
In other words,
any domain that contained part
506
00:35:09 --> 00:35:15
of this line,
the divergence would be zero
507
00:35:13 --> 00:35:19
along that line.
And, therefore,
508
00:35:15 --> 00:35:21
I could conclude nothing.
I could have limit cycles that
509
00:35:20 --> 00:35:26
cross that line,
as long as they included a
510
00:35:23 --> 00:35:29
piece of that line in them.
The answer is I cannot make a
511
00:35:28 --> 00:35:34
conclusion.
Well, that is because I am
512
00:35:32 --> 00:35:38
using the wrong criterion.
Let's instead use the critical
513
00:35:36 --> 00:35:42
point criterion.
514
00:35:38 --> 00:35:44
515
00:35:55 --> 00:36:01
Now, I am going to say that it
makes a nice positive statement
516
00:35:58 --> 00:36:04
but nobody ever uses it this
way.
517
00:36:01 --> 00:36:07
Nonetheless,
let's first state it
518
00:36:03 --> 00:36:09
positively, even though that is
not the way to use it.
519
00:36:07 --> 00:36:13
The positive statement will be,
once again, we have our region
520
00:36:13 --> 00:36:19
D and we have a region of the xy
plane and we have our C,
521
00:36:20 --> 00:36:26
a closed trajectory in it.
A closed trajectory of what?
522
00:36:26 --> 00:36:32
Of our system.
And that is supposed to be in
523
00:36:30 --> 00:36:36
D.
The critical point criterion
524
00:36:35 --> 00:36:41
says something very simple.
If you have that situation it
525
00:36:42 --> 00:36:48
says that inside that closed
trajectory there must be a
526
00:36:48 --> 00:36:54
critical point somewhere.
527
00:36:52 --> 00:36:58
528
00:37:00 --> 00:37:06
It says that inside C is a
critical point.
529
00:37:08 --> 00:37:14
530
00:37:15 --> 00:37:21
Now, this won't help us with
the existence problem.
531
00:37:18 --> 00:37:24
This won't help us find a
closed trajectory.
532
00:37:21 --> 00:37:27
We will take our system and say
it has a critical point here and
533
00:37:26 --> 00:37:32
a critical point there.
Does it have a closed
534
00:37:29 --> 00:37:35
trajectory?
Well, all I know is the closed
535
00:37:33 --> 00:37:39
trajectory, if it exists,
will have to go around one or
536
00:37:36 --> 00:37:42
more of those critical points.
But I don't know where.
537
00:37:40 --> 00:37:46
It is not going to go around it
like this.
538
00:37:43 --> 00:37:49
It might go around it like
this.
539
00:37:45 --> 00:37:51
And my computer search won't
find it because it is looking at
540
00:37:49 --> 00:37:55
too small a part of the screen.
It doesn't work that way.
541
00:37:53 --> 00:37:59
It works negatively by
contraposition.
542
00:37:56 --> 00:38:02
Do you know what the
contrapositive is?
543
00:38:00 --> 00:38:06
You will at least learn that.
A implies B says the same thing
544
00:38:07 --> 00:38:13
as not B implies not A.
545
00:38:11 --> 00:38:17
546
00:38:20 --> 00:38:26
They are different statements
but they are equivalent to each
547
00:38:24 --> 00:38:30
other.
If you prove one you prove the
548
00:38:27 --> 00:38:33
other.
What would be the
549
00:38:29 --> 00:38:35
contrapositive here?
If you have a closed trajectory
550
00:38:35 --> 00:38:41
inside is a critical point.
The theorem is used this way.
551
00:38:44 --> 00:38:50
If D has no critical points,
it has no closed trajectories
552
00:38:53 --> 00:38:59
and therefore has no limit
cycle.
553
00:39:00 --> 00:39:06
Because, if it did have a
closed trajectory,
554
00:39:03 --> 00:39:09
inside it would be a critical
point.
555
00:39:07 --> 00:39:13
But I said B had no critical
point.
556
00:39:10 --> 00:39:16
That enables us to dispose of
this system that Bendixson could
557
00:39:15 --> 00:39:21
not handle at all.
We can dispose of this system
558
00:39:20 --> 00:39:26
immediately.
Namely, what is it?
559
00:39:22 --> 00:39:28
Where are its critical points?
Well, where is that zero?
560
00:39:29 --> 00:39:35
x squared plus y
squared is one,
561
00:39:32 --> 00:39:38
plus one is never zero.
This is positive.
562
00:39:35 --> 00:39:41
Or, worse, zero.
And then I add the one to it
563
00:39:38 --> 00:39:44
and it is not zero anymore.
This has no zeros and,
564
00:39:42 --> 00:39:48
therefore, it does not matter
that this one has a lot of
565
00:39:46 --> 00:39:52
zeros.
It makes no difference.
566
00:39:48 --> 00:39:54
It has no critical points.
It has none,
567
00:39:51 --> 00:39:57
therefore, no limit cycles.
568
00:39:54 --> 00:40:00
569
00:40:01 --> 00:40:07
Now, I desperately wanted to
give you the proof of this.
570
00:40:04 --> 00:40:10
It is clearly impossible in the
time remaining.
571
00:40:07 --> 00:40:13
The proof requires a little
time.
572
00:40:09 --> 00:40:15
I haven't decided what to do
about that.
573
00:40:12 --> 00:40:18
It might leak over until
Friday's lecture.
574
00:40:15 --> 00:40:21
Instead, I will finish by
telling you a story.
575
00:40:18 --> 00:40:24
How is that?
576
00:40:20 --> 00:40:26
577
00:40:37 --> 00:40:43
And along side of it was little
y prime.
578
00:40:39 --> 00:40:45
I am not going to continue on
with the letters of the
579
00:40:43 --> 00:40:49
alphabet.
I will prime the earlier one.
580
00:40:46 --> 00:40:52
This has a total of 12
parameters in it.
581
00:40:49 --> 00:40:55
But, in fact,
if you change variables you can
582
00:40:52 --> 00:40:58
get rid of all the linear terms.
The important part of it is
583
00:40:57 --> 00:41:03
only the quadratic terms in the
beginning.
584
00:41:01 --> 00:41:07
This sort of thing is called a
quadratic system.
585
00:41:05 --> 00:41:11
After you have departed from
linear systems,
586
00:41:09 --> 00:41:15
it is the simplest kind there
is.
587
00:41:12 --> 00:41:18
And the predictor-prey,
the robin-earthworm example I
588
00:41:16 --> 00:41:22
gave you is of a typical
quadratic system and exhibits
589
00:41:21 --> 00:41:27
typical nonlinear quadratic
system behavior.
590
00:41:25 --> 00:41:31
Now, the problem is the
following.
591
00:41:30 --> 00:41:36
A, b, c, d, e,
f and so on,
592
00:41:32 --> 00:41:38
those are just real numbers,
parameters, so I am allowed to
593
00:41:37 --> 00:41:43
give them any values I want.
And the problem that has
594
00:41:42 --> 00:41:48
bothered people since 1880 when
it was first proposed is how
595
00:41:47 --> 00:41:53
many limit cycles can a
quadratic system have?
596
00:41:52 --> 00:41:58
597
00:42:03 --> 00:42:09
After 120 years this problem is
totally unsolved,
598
00:42:07 --> 00:42:13
and the mathematicians of the
world who are interested in it
599
00:42:12 --> 00:42:18
cannot even agree with each
other on what the right
600
00:42:16 --> 00:42:22
conjecture is.
But let me tell you a little
601
00:42:20 --> 00:42:26
bit of its history.
There were attempts to solve it
602
00:42:24 --> 00:42:30
in the 20 or 30 years after it
was first proposed,
603
00:42:28 --> 00:42:34
through the 1920 and `30s which
all seemed to have gaps in them.
604
00:42:35 --> 00:42:41
Until finally around 1950 two
Russians mathematicians,
605
00:42:39 --> 00:42:45
one of whom is extremely
well-known, Petrovski,
606
00:42:44 --> 00:42:50
a specialist in systems of
ordinary differential equations
607
00:42:49 --> 00:42:55
published a long and difficult,
complicated 100 page paper in
608
00:42:54 --> 00:43:00
which they proved that the
maximum number is three.
609
00:43:00 --> 00:43:06
I won't put down their names.
Petrovski-Landis.
610
00:43:03 --> 00:43:09
The maximum number was three.
And then not many people were
611
00:43:08 --> 00:43:14
able to read the paper,
and those who did there seemed
612
00:43:12 --> 00:43:18
to be gaps in the reasoning in
various places until finally
613
00:43:16 --> 00:43:22
Arnold who was the greatest
Russian, in my opinion,
614
00:43:20 --> 00:43:26
one of the greatest Russian
mathematician,
615
00:43:23 --> 00:43:29
certainly in this field of
analysis and differential
616
00:43:27 --> 00:43:33
equations, but in other fields,
too, he still is great,
617
00:43:31 --> 00:43:37
although he is somewhat older
now, criticized it.
618
00:43:37 --> 00:43:43
He said look,
there is a really big gap in
619
00:43:41 --> 00:43:47
this argument and it really
cannot be considered to be
620
00:43:46 --> 00:43:52
proven.
People tried working very hard
621
00:43:50 --> 00:43:56
to patch it up and without
success.
622
00:43:53 --> 00:43:59
Then about 1972 or so,
'75 maybe, a Chinese
623
00:43:58 --> 00:44:04
mathematician found a system
with four.
624
00:44:03 --> 00:44:09
Wrote down the numbers,
the number a is so much,
625
00:44:07 --> 00:44:13
b is so much,
and they were absurd numbers
626
00:44:11 --> 00:44:17
like 10^-6 and 40 billion and so
on, nothing you could plot on a
627
00:44:17 --> 00:44:23
computer screen,
but found a system with four.
628
00:44:22 --> 00:44:28
Nobody after this tried to fill
in the gap in the
629
00:44:26 --> 00:44:32
Petrovski-Landis paper.
I was then chairman of the math
630
00:44:32 --> 00:44:38
department, and one of my tasks
was protocol and so on.
631
00:44:36 --> 00:44:42
Anyway, we were trying very
hard to attract a Chinese
632
00:44:40 --> 00:44:46
mathematician to our department
to become a full professor.
633
00:44:44 --> 00:44:50
He was a really outstanding
analyst and specialist in
634
00:44:48 --> 00:44:54
various fields.
Anyway, he came in for a
635
00:44:50 --> 00:44:56
courtesy interview and we
chatted.
636
00:44:53 --> 00:44:59
At the time,
I was very much interested in
637
00:44:56 --> 00:45:02
limit cycles.
And I had on my desk the Math
638
00:44:59 --> 00:45:05
Society's translation of the
Chinese book on limit cycles.
639
00:45:05 --> 00:45:11
A collection of papers by
Chinese mathematicians all on
640
00:45:08 --> 00:45:14
limit cycles.
After a certain point he said,
641
00:45:11 --> 00:45:17
oh, I see you're interested in
limit cycle problems.
642
00:45:15 --> 00:45:21
I said yeah,
in particular,
643
00:45:16 --> 00:45:22
I was reading this paper of the
mathematician who found four
644
00:45:20 --> 00:45:26
limit cycles.
And I opened to that system and
645
00:45:23 --> 00:45:29
said the name is,
and I read it out loud.
646
00:45:26 --> 00:45:32
I said do you by any chance
know him?
647
00:45:30 --> 00:45:36
And he smiled and said yes,
very well.
648
00:45:32 --> 00:45:38
That is my mother.
[LAUGHTER]
649
00:45:35 --> 00:45:41
650
00:45:43 --> 00:45:49
Well, bye-bye.