As a matter of fact, it plots them very accurately.
But it is something you also need to learn to do yourself,
as you will see when we study nonlinear equations.
It is a skill. And since a couple of important
mathematical ideas are involved in it, I think it is a very good
thing to spend just a little time on, one lecture in fact,
plus a little more on the problem set that I will give
out. The last problem set that I
will give out on Friday. I thought it might be a little
more fun to, again, have a simple-minded model.
No romance this time. We are going to have a little
model of war, but I have made it sort of
sublimated war. Let's take as the system,
I am going to let two of those be parameters,
you know, be variable, in other words.
And the other two I will keep fixed, so that you can
concentrate on them better. I will take a and d to be
negative 1 and negative 3. And the other ones we will
leave open, so let's call this one b times y,
and this other one will be c times x.
I am going to model this as a fight between two states,
both of which are trying to attract tourists.
Let's say this is Massachusetts and this will be New Hampshire,
its enemy to the North. Both are busy advertising these
days on television. People are making their summer
plans. Come to New Hampshire,
you know, New Hampshire has mountains and Massachusetts has
quaint little fishing villages and stuff like that.
So what are these numbers? Well, first of all,
what do x and y represent? x and y basically are the
advertising budgets for tourism, you know, the amount each state
plans to spend during the year. However, I do not want zero
value to mean they are not spending anything.
It represents departure from the normal equilibrium.
x and y represent departures --
-- from the normal amount of money they spend advertising for
tourists. The normal tourist advertising
budget.
If they are both zero, it means that both states are
spending what they normally spend in that year.
If x is positive, it means that Massachusetts has
decided to spend more in the hope of attracting more tourists
and if negative spending less. What is the significance of
these two coefficients? Those are the normal things
which return you to equilibrium. In other words,
if x gets bigger than normal, if Massachusetts spends more
there is a certain poll to spend less because we are wasting all
this money on the tourists that are not going to come when we
could be spending it on education or something like
that. If x gets to be negative,
the governor tries to spend less.
Then all the local city Chamber of Commerce rise up and start
screaming that our economy is going to go bankrupt because we
won't get enough tourists and that is because you are not
spending enough money. There is a push to always
return it to the normal, and that is what this negative
sign means. The same thing for New
Hampshire. What does it mean that this is
negative three and that is negative one?
It just means that the Chamber of Commerce yells three times as
loudly in New Hampshire. It is more sensitive,
in other words, to changes in the budget.
Now, how about the other? Well, these represent the
war-like features of the situation.
Normally these will be positive numbers.
Because when Massachusetts sees that New Hampshire has budgeted
this year more than its normal amount, the natural instinct is
we are fighting. This is war.
This is a positive number. We have to budget more,
too. And the same thing for New
Hampshire. The size of these coefficients
gives you the magnitude of the reaction.
If they are small Massachusetts say, well, they are spending
more but we don't have to follow them.
We will bucket a little bit. If it is a big number then oh,
my God, heads will roll. We have to triple them and put
them out of business. This is a model,
in fact, for all sorts of competition.
It was used for many years to model in simper times armaments
races between countries. It is certainly a simple-minded
model for any two companies in competition with each other if
certain conditions are met. Well, what I would like to do
now is try different values of those numbers.
And, in each case, show you how to sketch the
solutions at different cases. And then, for each different
case, we will try to interpret if it makes sense or not.
My first set of numbers is, the first case is --
-- x prime equals negative x plus 2y.
And y prime equals, this is going to be zero,
so it is simply minus 3 times y.
Now, what does this mean? Well, this means that
Massachusetts is behaving normally, but New Hampshire is a
very placid state, and the governor is busy doing
other things. And people say Massachusetts is
spending more this year, and the Governor says,
so what. The zero is the so what factor.
In other words, we are not going to respond to
them. We will do our own thing.
What is the result of this? Is Massachusetts going to win
out? What is going to be the
ultimate effect on the budget? Well, what we have to do is,
so the program is first let's quickly solve the equations
using a standard technique. I am just going to make marks
on the board and trust to the fact that you have done enough
of this yourself by now that you know what the marks mean.
I am not going to label what everything is.
I am just going to trust to luck.
The matrix A is negative 1, 2, 0, negative 3.
The characteristic equation,
the second coefficient is the trace, which is minus 4,
but you have to change its sign, so that makes it plus 4.
And the constant term is the determinant, which is 3 minus 0,
so that is plus 3 equals zero. This factors into lambda plus 3
times lambda plus one. And it means the roots
therefore are, one root is lambda equals
negative 3 and the other root is lambda equals negative 1.
These are the eigenvalues. With each eigenvalue goes an
eigenvector. The eigenvector is found by
solving an equation for the coefficients of the eigenvector,
the components of the eigenvector.
Here I used negative 1 minus negative 3, which makes 2.
The first equation is 2a1 plus 2a2 is equal to zero.
The second one will be, in fact, in this case simply
0a1 plus 0a2 so it won't give me any information at all.
That is not what usually happens, but it is what happens
in this case. What is the solution?
The solution is the vector alpha equals,
well, 1, negative 1 would be a good thing to
use. That is the eigenvector,
so this is the e-vector. How about lambda equals
negative 1? Let's give it a little more
room. If lambda is negative 1 then
here I put negative 1 minus negative 1.
That makes zero. I will write in the zero
because this is confusing. It is zero times a1.
And the next coefficient is 2 a2, is zero.
People sometimes go bananas over this, in spite of the fact
that this is the easiest possible case you can get.
I guess if they go bananas over it, it proves it is not all that
easy, but it is easy. What now is the eigenvector
that goes with this? Well, this term isn't there.
It is zero. The equation says that a2 has
to be zero. And it doesn't say anything
about a1, so let's make it 1.
Now, out of this data, the final step is to make the
general solution. What is it?
(x, y) equals, well, a constant times the
first normal mode. The solution constructed from
the eigenvalue and the eigenvector.
That is going to be 1, negative 1 e to the minus 3t.
And then the other normal mode
times an arbitrary constant will be (1, 0) times e to the
negative t.
The lambda is this factor which produces that,
of course. Now, one way of looking at it
is, first of all, get clearly in your head this
is a pair of parametric equations just like what you
studied in 18.02. Let's write them out explicitly
just this once. x equals c1 times e to the
negative 3t plus c2 times e to the negative t.
And what is y?
y is equal to minus c1 e to the minus 3t plus zero.
I can stop there.
In some sense, all I am asking you to do is
plot that curve. In the x,y-plane,
plot the curve given by this pair of parametric equations.
And you can choose your own values of c1,
c2. For different values of c1 and
c2 there will be different curves.
Give me a feeling for what they all look like.
Well, I think most of you will recognize you didn't have stuff
like this. These weren't the kind of
curves you plotted. When you did parametric
equations in 18.02, you did stuff like x equals
cosine t, y equals sine t.
Everybody knows how to do that. A few other curves which made
lines or nice things, but nothing that ever looked
like that. And so the computer will plot
it by actually calculating values but, of course,
we will not. That is the significance of the
word sketch. I am not asking you to plot
carefully, but to give me some general geometric picture of
what all these curves look like without doing any work.
Without doing any work. Well, that sounds promising.
Okay, let's try to do it without doing any work.
Where shall I begin? Hidden in this formula are four
solutions that are extremely easy to plot.
So begin with the four easy solutions, and then fill in the
rest. Now, which are the easy
solutions? The easy solutions are c1
equals plus or minus 1, c2 equals zero,
or c1 equals zero, or c1 = 0, c2 equals plus or
minus 1. By choosing those four values
of c1 and c2, I get simple solutions
corresponding to the normal mode.
If c1 is one and c2 is zero, I am talking about (1,
negative 1) e to the minus 3t,
and that is very easy plot. Let's start plotting them.
What I am going to do is color-code them so you will be
able to recognize what it is I am plotting.
Let's see. What colors should we use?
We will use pink and orange. This will be our pink solution
and our orange solution will be this one.
Let's plot the pink solution first.
The pink solution corresponds to c1 equals 1 and c2
equals zero. Now, that solution looks like--
Let's write it in pink.
No, let's not write it in pink. What is the solution?
It looks like x equals e to the negative 3t,
y equals minus e to the minus 3t.
Well, that's not a good way to look at it, actually.
The best way to look at it is to say at t equals zero,
where is it? It is at the point 1,
negative 1.
And what is it doing as t increases?
Well, it keeps the direction, but travels.
The amplitude, the distance from the origin
keeps shrinking. As t increases,
this factor, so it is the tip of this
vector, except the vector is shrinking.
It is still in the direction of 1, negative 1,
but it is shrinking in length because its amplitude is
shrinking according to the law e to the negative 3t.
In other words, this curve looks like this.
At t equals zero it is over here, and it goes along this
diagonal line until as t equals infinity, it gets to infinity,
it reaches the origin. Of course, it never gets there.
It goes slower and slower and slower in order that it may
never reach the origin. What was it doing for values of
t less than zero? The same thing,
except it was further away. It comes in from infinity along
that straight line. In other words,
the eigenvector determines the line on which it travels and the
eigenvalue determines which way it goes.
If the eigenvalue is negative, it is approaching the origin as
t increases. How about the other one?
Well, if c1 is negative 1, then everything is the
same except it is the mirror image of this one.
If c1 is negative 1, then at t equals zero it is at
this point. And, once again,
the same reasoning shows that it is coming into the origin as
t increases. I have now two solutions,
this one corresponding to c1 equals 1,
and the other one c2 equals zero.
This one corresponds to c1 equals negative 1.
How about the other guy, the orange guy?
Well, now c1 is zero, c2 is one, let's say.
It is the vector (1, 0), but otherwise everything is
the same. I start now at the point (1,
0) at time zero. And, as t increases,
I come into the origin always along that direction.
And before that I came in from infinity.
And, again, if c2 is 1 and if c2 is negative 1,
I do the same thing but on the other side.
That wasn't very hard. I plotted four solutions.
And now I roll up my sleeves and waive my hands to try to get
others. The general philosophy is the
following. The general philosophy is the
differential equation looks like this.
It is a system of differential equations.
These are continuous functions. That means when I draw the
velocity field corresponding to that system of differential
equations, because their functions are continuous,
as I move from one (x, y) point to another the
direction of the velocity vectors change continuously.
It never suddenly reverses without something like that.
Now, if that changes continuously then the
trajectories must change continuously,
too. In other words,
nearby trajectories should be doing approximately the same
thing. Well, that means all the other
trajectories are ones which come like that must be going also
toward the origin. If I start here,
probably I have to follow this one.
They are all coming to the origin, but that is a little too
vague. How do they come to the origin?
In other words, are they coming in straight
like that? Probably not.
Then what are they doing? Now we are coming to the only
point in the lecture which you might find a little difficult.
Try to follow what I am doing now.
If you don't follow, it is not well done in the
textbook, but it is very well done in the notes because I
wrote them myself. Please, it is done very
carefully in the notes, patiently follow through the
explanation. It takes about that much space.
It is one of the important ideas that your engineering
professors will expect you to understand.
Anyway, I know this only from the negative one because they
say to me at lunch, ruin my lunch by saying I said
it to my students and got nothing but blank looks.
What do you guys teach them over there?
Blah, blah, blah. Maybe we ought to start
teaching it ourselves. Sure.
Why don't they start cutting their own hair,
too?
Here is the idea. Let me recopy that solution.
The solution looks like (1, negative 1) e to the minus 3t
plus c2, (1, 0) e to the negative t.
What I ask is as t goes to infinity, I feel sure that the
trajectories must be coming into the origin because these guys
are doing that. And, in fact,
that is confirmed. As t goes to infinity,
this goes to zero and that goes to zero regardless of what the
c1 and c2 are. That makes it clear that this
goes to zero no matter what the c1 and c2 are as t goes to
infinity, but I would like to analyze it a little more
carefully. As t goes to infinity,
I have the sum of two terms. And what I ask is,
which term is dominant? Of these two terms,
are they of equal importance, or is one more important than
the other? When t is 10,
for example, that is not very far on the way
to infinity, but it is certainly far enough to illustrate.
Well, e to the minus 10 is an extremely
small number. The only thing smaller is e to
the minus 30. The term that dominates,
they are both small, but relatively-speaking this
one is much larger because this one only has the factor e to the
minus 10, whereas, this has the factor e
to the minus 30, which is vanishingly small.
In other words, as t goes to infinity --
Well, let's write it the other way.
This is the dominant term, as t goes to infinity.
Now, just the opposite is true as t goes to minus infinity.
t going to minus infinity means I am backing up along these
curves. As t goes to minus infinity,
let's say t gets to be negative 100, this is e to the 100,
but this is e to the 300,
which is much, much bigger.
So this is the dominant term as t goes to negative infinity.
Now what I have is the sum of two vectors.
Let's first look at what happens as t goes to infinity.
As t goes to infinity, I have the sum of two vectors.
This one is completely negligible compared with the one
on the right-hand side. In other words,
for a all intents and purposes, as t goes to infinity,
it is this thing that takes over.
Therefore, what does the solution look like as t goes to
infinity? The answer is it follows the
yellow line. Now, what does it look like as
it backs up? As it came in from negative
infinity, what does it look like?
Now, this one is a little harder to see.
This is big, but this is infinity bigger.
I mean very, very much bigger,
when t is a large negative number.
Therefore, what I have is the sum of a very big vector.
You're standing on the moon looking at the blackboard,
so this is really big. This is a very big vector.
This is one million meters long, and this is only 20
meters long. That is this guy,
and that is this guy. I want the sum of those two.
What does the sum look like? The answer is a sum is
approximately parallel to the long guy because this is
negligible. This does not mean they are
next to each other. They are slightly tilted over,
but not very much. In other words,
as t goes to negative infinity it doesn't coincide with this
vector. The solution doesn't,
but it is parallel to it. It has the same direction.
I am done. It means far away from the
origin, it should be parallel to the pink line.
Near the origin it should turn and become more or less
coincident with the orange line. And those were the solutions.
That's how they look.
How about down here? The same thing,
like that, but then after a while they turn and join.
Here, they have to turn around to join up, but they join.
And that is, in a simple way,
the sketches of those functions.
That is how they must look. What does this say about our
state? Well, it says that the fact
that the governor of New Hampshire is indifferent to what
Massachusetts is doing produces ultimately harmony.
Both states revert ultimately their normal advertising budgets
in spite of the fact that Massachusetts is keeping an eye
peeled out for the slightest misbehavior on the part of New
Hampshire. Peace reins,
in other words. Now you should know some names.
Let's see. I will write names in purple.
There are two words that are used to describe this situation.
First is the word that describes the general pattern of
the way these lines look. The word for that is a node.
And the fact that all the trajectories end up at the
origin for that one uses the word sink.
This could be modified to nodal sink.
That would be better. Nodal sink, let's say.
Nodal sink or, if you like to write them in
the opposite order, sink node.
In the same way there would be something called a source node
if I reversed all the arrows. I am not going to calculate an
example. Why don't I simply do it by
giving you -- For example,
if the matrix A produced a solution instead of that one.
Suppose it looked like 1, negative 1 e to the 3t.
The eigenvalues were reversed,
were now positive. And I will make the other one
positive, too. c2 1, 0 e to the t.
What would that change in the picture?
The answer is essentially nothing, except the direction of
the arrows. In other words,
the first thing would still be 1, negative 1.
The only difference is that now
as t increases we go the other way.
And here the same thing, we have still the same basic
vector, the same basic orange vector, orange line,
but it has now traversed the solution.
We traverse it in the opposite direction.
Now, let's do the same thing about dominance,
as we did before. Which term dominates as t goes
to infinity? This is the dominant term.
Because, as t goes to infinity, 3t is much bigger than t.
This one, on the other hand, dominates as t goes to negative
infinity.
How now will the solutions look like?
Well, as t goes to infinity, they follow the pink curve.
Whereas, as t starts out from negative infinity,
they follow the orange curve.
As t goes to infinity, they become parallel to the
pink curve, and as t goes to negative infinity,
they are very close to the origin and are following the
yellow curve. This is pink and this is
yellow. They look like this.
Notice the picture basically is the same.
It is the picture of a node. All that has happened is the
arrows are reversed. And, therefore,
this would be called a nodal source.
The word source and sink correspond to what you learned
in 18.02 and 8.02, I hope, also,
or you could call it a source node.
Both phrases are used, depending on how you want to
use it in a sentence. And another word for this,
this would be called unstable because all of the solutions
starting out from near the origin ultimately end up
infinitely far away from the origin.
This would be called stable. In fact, it would be called
asymptotically stable. I don't like the word
asymptotically, but it has become standard in
the literature. And, more important,
it is standard in your textbook.
And I don't like to fight with a textbook.
It just ends up confusing everybody, including me.
That is enough for nodes. I would like to talk now about
some of the other cases that can occur because they lead to
completely different pictures that you should understand.
Let's look at the case where our governors behave a little
more badly, a little more combatively.
It is x prime equals negative x as before,
but this time a firm response by Massachusetts to any sign of
increased activity by stockpiling of advertising
budgets. Here let's say New Hampshire
now is even worse. Five times, quintuple or
whatever increase Massachusetts makes, of course they don't have
an income tax, but they will manage.
Minus 3y as before. Let's again calculate quickly
what the characteristic equation is.
Our matrix is now negative 1, 3, 5 and negative 3.
The characteristic equation now
is lambda squared. What is that?
Again, plus 4 lambda. But now the determinant is 3
minus 15 is negative 12.
And this, because I prepared very carefully,
all eigenvalues are integers. And so this factors into lambda
plus 6 times lambda minus 2,
does it not? Yes.
6 lambda minus 2 is four lambda.
Good. What do we have?
Well, first of all we have our eigenvalue lambda,
negative 6. And the eigenvector that goes
with that is minus 1. This is negative 1 minus
negative 6 which makes, shut your eyes,
5. We have 5a1 plus 3a2 is zero.
And the other equation,
I hope it comes out to be something similar.
I didn't check. I am hoping this is right.
The eigenvector is, okay, you have been taught to
always make one of the 1, forget about that.
Just pick numbers that make it come out right.
I am going to make this one 3, and then I will make this one
negative 5. As I say, I have a policy of
integers only. I am a number theorist at
heart. That is how I started out life
anyway. There we have data from which
we can make one solution. How about the other one?
The other one will correspond to the eigenvalue lambda equals
2. This time the equation is
negative 1 minus 2 is negative 3.
It is minus 3a1 plus 3a2 is zero.
And now the eigenvector is (1, 1).
Now we are ready to draw pictures.
We are going to make this similar analysis,
but it will go faster now because you have already had the
experience of that. First of all,
what is our general solution? It is going to be c1 times 3,
negative 5 e to the minus 6t.
And then the other normal mode
times an arbitrary constant will be 1, 1 times e to the 2t.
I am going to use the same strategy.
We have our two normal modes here, eigenvalue,
eigenvector solutions from which, by adjusting these
constants, we can get our four basic solutions.
Those are going to look like, let's draw a picture here.
Again, I will color-code them. Let's use pink again.
The pink solution now starts at 3, negative 5.
That is where it is when t is
zero. And, because of the coefficient
minus 6 up there, it is coming into the origin
and looks like that. And its mirror image,
of course, does the same thing. That is when c1 is negative
one. How about the orange guy?
Well, when t is equal to zero, it is at 1, 1.
But what is it doing after
that? As t increases,
it is getting further away from the origin because the sign here
is positive. e to the 2t is
increasing, it is not decreasing anymore, so this guy is going
out. And its mirror image on the
other side is doing the same thing.
Now all we have to do is fill in the picture.
Well, you fill it in by continuity.
Your nearby trajectories must be doing what similar thing?
If I start out very near the pink guy, I should stay near the
pink guy. But as I get near the origin,
I am also approaching the orange guy.
Well, there is no other possibility other than that.
If you are further away you start turning a little sooner.
I am just using an argument from continuity to say the
picture must be roughly filled out this way.
Maybe not exactly. In fact, there are fine points.
And I am going to ask you to do one of them on Friday for the
new problem set, even before the exam,
God forbid. But I want you to get a little
more experience working with that linear phase portrait
visual because it is, I think, one of the best ones
this semester. You can learn a lot from it.
Anyway, you are not done with it, but I hope you have at least
looked at it by now. That is what the picture looks
like. First of all,
what are we going to name this? In other words,
forget about the arrows. If you just look at the general
way those lines go, where have you seen this
before? You saw this in 18.02.
What was the topic? You were plotting contour
curves of functions, were you not?
What did you call contours curves that formed that pattern?
A saddle point. You called this a saddle point
because it was like the center of a saddle.
It is like a mountain pass. Here you are going up the
mountain, say, and here you are going down,
the way the contour line is going down.
And this is sort of a min and max point.
A maximum if you go in that direction and a minimum if you
go in that direction, say.
Without the arrows on it, it is like a saddle point.
And so the same word is used here.
It is called the saddle. You don't say point in the same
way you don't say a nodal point. It is the whole picture,
as it were, that is the saddle. It is a saddle.
There is the saddle. This is where you sit.
Now, should I call it a source or a sink?
I cannot call it either because it is a sink along these lines,
it is a source along those lines and along the others,
it starts out looking like a sink and then turns around and
starts acting like a source. The word source and sink are
not used for saddle. The only word that is used is
unstable because definitely it is unstable.
If you start off exactly on the pink lines you do end up at the
origin, but if you start anywhere else ever so close to a
pink line you think you are going to the origin,
but then at the last minute you are zooming off out to infinity
again. This is a typical example of
instability. Only if you do the
mathematically possible, but physically impossible thing
of starting out exactly on the pink line, only then will you
get to the origin. If you start out anywhere else,
make the slightest error in measure and get off the pink
line, you end off at infinity. What is the effect with our
war-like governors fighting for the tourist trade willing to
spend any amounts of money to match and overmatch what their
competitor in the nearby state is spending?
The answer is, they all lose.
Since it is mostly this section of the diagram that makes sense,
what happens is they end up all spending an infinity of dollars
and nobody gets any more tourists than anybody else.
So this is a model of what not to do.
I have one more model to show you.
Maybe we better start over at this board here.
Massachusetts on top. New Hampshire on the bottom.
x prime is going to be, that is Massachusetts,
I guess as before. Let me get the numbers right.
Leave that out for a moment. y prime is 2x minus 3y.
New Hampshire behaves normally.
It is ready to respond to anything Massachusetts can put
out. But by itself,
it really wants to bring its budget to normal.
Now, Massachusetts, we have a Mormon governor now,
I guess. Imagine instead we have a
Buddhist governor. A Buddhist governor reacts as
follows, minus y. What does that mean?
It means that when he sees New Hampshire increasing the budget,
his reaction is, we will lower ours.
We will show them love. It looks suicidal,
but what actually happens? Well, our little program is
over. Our matrix a is negative 1,
negative 1, 2, negative 3.
The characteristic equations is
lambda squared plus 4 lambda.
And now what is the other term? 3 minus negative 2 makes 5.
This is not going to factor because I tried it out and I
know it is not going to factor. We are going to get lambda
equals, we will just use the quadratic formula,
negative 4 plus or minus the square root of 16 minus 4 times
5, that is 16 minus 20 or negative 4 all divided by 2,
which makes minus 2, pull out the 4,
that makes it a 2, cancels this 2,
minus 1 inside. It is minus 2 plus or minus i.
Complex solutions.
What are we doing to do about that?
Well, you should rejoice when you get this case and are asked
to sketch it because, even if you calculate the
complex eigenvector and from that take its real and imaginary
parts of the complex solution, in fact, you will not be able
easily to sketch the answer anyway.
But let me show you what sort of thing you can get and then I
am going to wave my hands and argue a little bit to try to
indicate what it is that the solution actually looks like.
You are going to get something that looks like --
A typical real solution is going to look like this.
This is going to produce e to the minus 2t times e
to the i t. e to the minus 2 plus i all
times t. This will be our exponential
factor which is shrinking in amplitude.
This is going to give me sines and cosines.
When I separate out the eigenvector into its real and
imaginary parts, it is going to look something
like this. a1, a2 times cosine t,
that is from the e to the it
part. Then there will be a sine term.
And all that is going to be multiplied by the exponential
factor e to the negative 2t.
That is just one normal mode. It is going to be c1 times this
plus c2 times something similar. It doesn't matter exactly what
it is because they are all going to look the same.
Namely, this is a shrinking amplitude.
I am not going to worry about that.
My real question is, what does this look like?
In other words, as a pair of parametric
equations, if x is equal to a1 cosine t plus b1 sine t
and y is a2 cosine plus b2 sine,
what does it look like?
Well, what are its characteristics?
In the first place, as a curve this part of it is
bounded. It stays within some large box
because cosine and sine never get bigger than one and never
get smaller than minus one. It is periodic.
As t increases to t plus 2pi,
it comes back to exactly the same point it was at before.
We have a curve that is repeating itself periodically,
it does not go off to infinity. And here is where I am waving
my hands. It satisfies an equation.
Those of you who like to fool around with mathematics a little
bit, it is not difficult to show this, but it satisfies an
equation of the form A x squared plus B y squared plus C xy
equals D.
All you have to do is figure out what the coefficients A,
B, C and D should be. And the way to do it is,
if you calculate the square of x you are going to get cosine
squared, sine squared and a cosine sine term.
You are going to get those same three terms here and the same
three terms here. You just use undetermined
coefficients, set up a system of simultaneous
equations and you will be able to find the A,
B, C and D that work. I am looking for a curve that
is bounded, keeps repeating its values and that satisfies a
quadratic equation which looks like this.
Well, an earlier generation would know from high school,
these curves are all conic sections.
The only curves that satisfy equations like that are
hyperbola, parabolas, the conic sections in other
words, and ellipses. Circles are a special kind of
ellipses. There is a degenerate case.
A pair of lines which can be considered a degenerate
hyperbola, if you want. It is as much a hyperbola as a
circle, as an ellipse say. Which of these is it?
Well, it must be those guys. Those are the only guys that
stay bounded and repeat themselves periodically.
The other guys don't do that. These are ellipses.
And, therefore, what do they look like?
Well, they must look like an ellipse that is trying to be an
ellipse, but each time it goes around the point is pulled a
little closer to the origin. It must be doing this,
in other words. And such a point is called a
spiral sink. Again sink because,
no matter where you start, you will get a curve that
spirals into the origin. Spiral is self-explanatory.
And the one thing I haven't told you that you must read is
how do you know that it goes around counterclockwise and not
clockwise? Read clockwise or
counterclockwise. I will give you the answer in
30 seconds, not for this particular curve.
That you will have to calculate.
All you have to do is put in somewhere.
Let's say at the point (1, 0), a single vector from the
velocity field. In other words,
at the point (1, 0), when x is 1 and y is 0 our
vector is minus 1, 2,
which is the vector minus 1, 2, it goes like this.
Therefore, the motion must be counterclockwise.
And, by the way, what is the effect of having a
Buddhist governor? Peace.
Everything spirals into the origin and everybody is left
with the same advertising budget they always had.
Thanks.