WEBVTT
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So the topic for today is we
have a system like the kind we
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have been studying,
but there is now a difference.
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A system of first order
differential equations,
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just two of them.
It is an autonomous system
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meaning, of course,
that there is no t explicitly
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on the right-hand side.
But what makes this different,
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now, is that it is nonlinear.
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In other words,
the functions on the right-hand
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side are no longer simple things
like ax plus by,
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cx plus dy.
Those are the kind we have been
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studying.
But we are going to allow them
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to have quadratic terms,
sines, cosines,
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different stuff there that are
not linear functions anymore.
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And the problem is,
if it's a linear system you
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know how to get a sketch of its
trajectories without using the
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computer by using eigenlines.
You were very good at that on
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the exam on Friday.
Most of you could do that very
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well.
But what do you do if you have
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a nonlinear system?
The problem is to sketch its
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trajectories.
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In general, there are not
analytic formulas for the
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solutions to nonlinear systems
like that.
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There are only computer-drawn
things.
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But sometimes you have to get
qualitative information,
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a quick idea of how the
trajectories look.
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And, especially on Friday,
I will give you examples of
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stuff that you can do that the
computer cannot do very well at
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all.
Okay, so the problem is to
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sketch those trajectories.
Now, what I am going to do is
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--
The way I will give the lecture
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is, this is the general problem.
We have to do two things sort
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of simultaneously.
I will give a general
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explanation using x and y,
but then, as we do each step of
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the process and talk about it in
general, I would like to carry
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it out on a specific example.
And so we will do it with a
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specific example.
The example I am going to carry
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out is that of the nonlinear
pendulum.
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I am using this because it
illustrates virtually
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everything.
And, in addition,
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it has the great advantage
that, since we know how a
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pendulum swings,
we will be able to,
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when we get the answer,
verify it and,
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at various stages of the
procedure, verify that the
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mathematics is,
in fact, in agreement with our
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physical intuition.
It is going to be a lightly
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damped pendulum because I am
going to have to put in numbers
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in order to do the calculations.
And that seems like a good case
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which illustrates several types
of behavior.
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Let's first of all,
before we talk in general,
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remind you of the pendulum.
The pendulum I am talking about
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has the vertex from which it
swings.
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This is a rigid rod.
It is not one of these
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string-type pendulums.
There is a mass here.
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The rigid rod is of length l.
And so it swings in a circular
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orbit like that back and forth
in a circle.
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And let's put in the vertical
distance, the vertical position
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rather.
And now, as variables,
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of course normally we use
neutral variables like x and y.
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But here x and y are not
relevant variables to describing
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the way the mass moves.
The obviously relevant variable
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is theta, this angle.
Now, I am taking it in the
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positive direction.
Here theta is zero.
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As it swings,
theta becomes positive.
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Over here, when it is
horizontal, theta has the value
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pi over two and then so on it
goes.
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Values here correspond to
negative values of theta.
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That is how it swings.
Now, just to remind you of the
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equation that this satisfies,
it satisfies F equals ma,
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rather ma equals F.
Now, the acceleration is along
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the circular path.
And that is different from the
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angular acceleration.
I have to put in the factor of
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the length.
You had that a lot in 8.01 so I
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am simply going to write it
down.
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It is the mass.
Therefore, the linear
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acceleration along the circular
path is equal to the angular
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acceleration times l.
It is l times theta prime
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prime, or double dot if you
prefer.
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And so this much of it is the
acceleration vector.
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Now, once the force is acting
on it, well, there is a force of
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gravity which is pulling it
straight down.
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But, of course,
that is not the relevant force.
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I am interested in the force
that acts along that circular
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line.
And that will not be all of the
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pink line but only its component
in that direction.
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And the component,
I fill out the little right
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triangle.
And then the way to get the
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component in that direction,
the vertical pink part has the
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magnitude mg.
But, since I only want this
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part, I have to multiply by the
sign of this angle.
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Now, sometimes I have given on
a diagnostic test to students
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when they enter to what angle is
that?
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But, of course,
anybody can guess it must be
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theta.
Otherwise, why would he be
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asking it?
So this is still the angle
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theta.
You can prove these two
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triangles are similar.
One of them I haven't even
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written in, but it would be the
right triangle whose leg is
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perpendicular to it.
So the right angle is here.
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If that is theta then the
length of this small pink line
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is mg times the sine of theta.
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That is the force due gravity.
It is mg sine theta.
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Except in what direction is it?
It is acting in the negative
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direction.
This is theta increasing.
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This is the opposite direction,
so I should put a negative sign
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in front of it.
But that is not the only force.
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There is also a damping force
that goes with a velocity.
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And that also occurs if the
angular velocity is positive,
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the angle theta is increasing,
in other words,
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the damping force resists that.
It is opposite to the velocity.
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So the velocity is going to be
l times theta prime.
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There is my velocity v.
Linear velocity,
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not angular velocity.
And so this is going to be a
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negative times some constant
times that c1.
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Now, that is the equation.
But let's make it look a little
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better by getting rid of some of
these constants.
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If I write it out this way and
put everything on the left-hand
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side, the way it is usually done
in writing a second order
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differential equation.
Theta I am going to divide
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through by ml and put everything
on the left-hand side and in the
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right order.
Next should come the theta
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prime term.
And so that is going to be c1l
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divided by ml,
so that is c1 over m.
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The l's cancel out.
And, finally,
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the last term on the right,
I will move this over to the
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left, but remember everything is
being divided by ml,
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so the m's cancel out,
and it is plus g over l times
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the sine of theta.
That is our differential
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equation.
But let's make it look still a
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little bit better by lumping
these constants and giving them
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new names.
It is going to be finally theta
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double prime.
I will simply call this thing
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the damping constant.
I will lump those two together
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into single damping constant.
And then g over l,
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I will lump those together,
too.
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And we usually call that k.
It is k sine theta.
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Now, this is a second-order
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different equation,
but it is not linear.
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If this were a month and a half
ago and I said solve that,
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you would stare at me.
But, anyway,
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you couldn't solve that.
And nobody can,
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in some sense.
It is a nonlinear equation.
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It doesn't have any exact
solution.
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The only thing you could do is
look for a solution in infinite
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series or something like that.
Well, what do you do?
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You throw it on the computer,
that is the easy answer,
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but what does the computer do?
Well, the first thing the
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computer does is turns it into a
system because the computer is
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going to use numerical methods
to solve it.
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But only those methods,
the formulas it uses,
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Euler or modified or improved
Euler or the Runge-Cutta method,
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they are always expressed not
for single higher-order
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equations, but instead they
always assume that the equation
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has been converted to a first
order system.
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Let's do that for the computer,
even though it will do it
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itself if nobody tells it not
to.
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Theta prime is equal to,
now I have to figure out what
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new variable to introduce.
Normally we use x prime and
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call that y.
That really doesn't seem to be
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very suitable here.
But what do the physicists call
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it?
This is the angular velocity.
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And the standard designation
for that is omega.
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Two Greek letters.
I told you this was going to be
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hard.
Omega prime equals what?
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Well, omega prime equals,
now you do it in the standard
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way, you convert the system,
but remember you have to put
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the theta first.
So it is minus k sine theta.
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The theta first and then the
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omega term first.
So minus c times omega,
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minus c theta prime,
but theta prime is,
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in real life,
omega.
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And now we have our acceptable
system.
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The only problem is I have not
put in any numbers yet.
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The numbers I am going to put
in will make it lightly damped.
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I am going to give c,
think of it here,
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this is the damping,
and this is the stuff
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representing,
well, if I want to make it
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lightly damped all I am saying
is that c should be small
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compared with k,
but it doesn't have to be very
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small.
I am going to take c equal one
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and k equal two,
and that will make it
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lightly enough damped.
This is the lightly damped
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value, values which give
underdamping.
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In other words,
they are going to allow the
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pendulum to swing back and forth
instead of strictly going ug and
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ending up there.
Finally, therefore,
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the system that we are going to
calculate is where theta prime
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equals omega and omega prime
equals negative two sine theta
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minus omega.
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And now what do we do with
that?
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There is our example.
That is our system that
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represents a pendulum swinging
back and forth,
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damped away.
And now let's go back to the
00:13:08.000 --> 00:13:14.000
general theory.
And, in general,
00:13:10.000 --> 00:13:16.000
if you have a nonlinear system,
00:13:15.000 --> 00:13:21.000
how do you go about analyzing
it?
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The first step is to find the
simplest possible solutions,
00:13:27.000 --> 00:13:33.000
solutions that you hope can be
found by inspection.
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Now, what would they be?
They are the solutions that
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consist of a single point.
How could a solution be a
00:13:40.000 --> 00:13:46.000
single point?
Well, like the origin for a
00:13:44.000 --> 00:13:50.000
linear system,
those points which form
00:13:47.000 --> 00:13:53.000
solutions all by themselves are
called the critical points.
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I am looking for the critical
points of the system.
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That is the first step.
The definition is a critical
00:14:03.000 --> 00:14:09.000
point x zero,
y zero.
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For that to be a critical point
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means that it makes the
right-hand side zero.
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The f is zero there,
and the g is zero there.
00:14:19.000 --> 00:14:25.000
See the significance of that?
If you have such a point,
00:14:25.000 --> 00:14:31.000
let's say there is a critical
point, what is the velocity
00:14:31.000 --> 00:14:37.000
field at that point?
Well, it is given by the
00:14:36.000 --> 00:14:42.000
vectors on the right-hand side,
but the components are zero.
00:14:40.000 --> 00:14:46.000
That means, at this point the
velocity vector is zero.
00:14:43.000 --> 00:14:49.000
Well, that means if a solution
starts there,
00:14:46.000 --> 00:14:52.000
you put the mouse there and
tell it to move,
00:14:49.000 --> 00:14:55.000
where do you go?
It has no reason to go anywhere
00:14:52.000 --> 00:14:58.000
since the velocity vector is
zero there.
00:14:55.000 --> 00:15:01.000
So it sits there for all time.
And indeed it solves the
00:14:58.000 --> 00:15:04.000
system, doesn't it?
It makes the right-hand side
00:15:03.000 --> 00:15:09.000
zero and it makes the left-hand
side zero because x equals x
00:15:08.000 --> 00:15:14.000
zero, y equals y zero for
00:15:12.000 --> 00:15:18.000
all time.
If that is true for all time it
00:15:15.000 --> 00:15:21.000
sits there then the derivatives,
with respect to time are zero,
00:15:20.000 --> 00:15:26.000
so the left-hand sides are
zero, the right-hand sides are
00:15:25.000 --> 00:15:31.000
zero and everybody is happy.
Well, these are great points.
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How do I find them?
Well, by looking for points
00:15:37.000 --> 00:15:43.000
that make these two functions
zero.
00:15:41.000 --> 00:15:47.000
I find them by solving
simultaneously the equations f
00:15:46.000 --> 00:15:52.000
of (x, y) equals zero and g of
(x, y) equals zero.
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A pair of equations.
But the trouble is those are
00:16:01.000 --> 00:16:07.000
not linear equations.
Linear equations you know how
00:16:04.000 --> 00:16:10.000
to solve, but they are not
linear equations.
00:16:07.000 --> 00:16:13.000
They are nonlinear equations
that you don't know how to
00:16:11.000 --> 00:16:17.000
solve.
And, to some extent,
00:16:13.000 --> 00:16:19.000
nobody else does either.
There are very fat books in the
00:16:17.000 --> 00:16:23.000
library whose topic is how to
solve just a pair of equations,
00:16:22.000 --> 00:16:28.000
f of (x, y) equals zero,
g of (x, y) equals zero.
00:16:25.000 --> 00:16:31.000
And it is quite a hard problem.
It is even a hard problem by
00:16:30.000 --> 00:16:36.000
computer.
Because, if you know
00:16:33.000 --> 00:16:39.000
approximately where the solution
is going to be,
00:16:36.000 --> 00:16:42.000
you can make up the little
screen and then the computer
00:16:40.000 --> 00:16:46.000
will find it for you.
Or, even without a screen,
00:16:43.000 --> 00:16:49.000
it will calculate it by
Newton's method or something
00:16:47.000 --> 00:16:53.000
else, it will zero in.
The problem is,
00:16:50.000 --> 00:16:56.000
if you don't know in advance
roughly where the critical point
00:16:54.000 --> 00:17:00.000
is that you are looking for,
there are a lot of numbers.
00:16:58.000 --> 00:17:04.000
They go to infinity that way
and infinity that way.
00:17:03.000 --> 00:17:09.000
In general, it is almost an
impossible problem.
00:17:06.000 --> 00:17:12.000
The only thing that makes it
possible is that these problems
00:17:11.000 --> 00:17:17.000
always come from the real world
and one has some physical
00:17:15.000 --> 00:17:21.000
feeling, one hopes,
for where the critical point
00:17:19.000 --> 00:17:25.000
is.
I am going to assume breezily
00:17:21.000 --> 00:17:27.000
and cheerily we can solve those.
And I am only going to give you
00:17:26.000 --> 00:17:32.000
examples where it is possible to
solve them.
00:17:31.000 --> 00:17:37.000
But even there,
you have to watch out.
00:17:33.000 --> 00:17:39.000
There is a certain trickiness
that will be talked about in the
00:17:37.000 --> 00:17:43.000
recitations tomorrow.
Okay, so we found the critical
00:17:41.000 --> 00:17:47.000
points.
Let's do it for our example.
00:17:43.000 --> 00:17:49.000
Let's find the critical point.
What is the pair of equations
00:17:47.000 --> 00:17:53.000
we have to solve?
We have to solve the equations
00:17:50.000 --> 00:17:56.000
omega equals zero,
minus two sine theta minus
00:17:53.000 --> 00:17:59.000
omega equals zero.
00:17:56.000 --> 00:18:02.000
Now, it is not always this
easy.
00:18:00.000 --> 00:18:06.000
But the solution is omega is
zero.
00:18:03.000 --> 00:18:09.000
If omega is zero,
then sine theta is zero,
00:18:07.000 --> 00:18:13.000
and sine theta is zero at the
00:18:12.000 --> 00:18:18.000
integral multiples of pi.
The critical points are omega
00:18:18.000 --> 00:18:24.000
is always zero and theta is
zero, or it could be plus or
00:18:24.000 --> 00:18:30.000
minus pi, plus or minus two pi
and so on.
00:18:30.000 --> 00:18:36.000
In other words,
there are an infinity of
00:18:32.000 --> 00:18:38.000
critical points.
That seems a little
00:18:34.000 --> 00:18:40.000
discouraging.
On the other hand,
00:18:36.000 --> 00:18:42.000
there are really only two.
There are really physically
00:18:40.000 --> 00:18:46.000
only two because omega equals
zero means the mass is not
00:18:44.000 --> 00:18:50.000
moving.
The angular velocity is zero,
00:18:46.000 --> 00:18:52.000
so it is only the theta
position which is changing.
00:18:49.000 --> 00:18:55.000
Now, what are the possible
theta positions?
00:18:52.000 --> 00:18:58.000
Well, here is our nonlinear
pendulum.
00:18:56.000 --> 00:19:02.000
Here is the critical point,
theta equals zero,
00:18:59.000 --> 00:19:05.000
omega equals zero.
Theta equals zero means the rod
00:19:03.000 --> 00:19:09.000
is vertical.
Omega equals zero means that it
00:19:07.000 --> 00:19:13.000
is not moving,
despite the fact that it is
00:19:10.000 --> 00:19:16.000
moving.
Theoretically it is not moving.
00:19:14.000 --> 00:19:20.000
Now, what is the other one?
Well, there is theta equals pi.
00:19:18.000 --> 00:19:24.000
Theta is now,
starting from zero and
00:19:21.000 --> 00:19:27.000
increasing, I hope,
through positive theta.
00:19:25.000 --> 00:19:31.000
And when it gets to pi,
it is sticking straight up in
00:19:29.000 --> 00:19:35.000
the air.
And so the claim is that
00:19:34.000 --> 00:19:40.000
another critical point is theta
equals pi and omega equals zero.
00:19:40.000 --> 00:19:46.000
In other words,
if it gets to this position,
00:19:44.000 --> 00:19:50.000
it starts out in this position
and stays there for all time,
00:19:49.000 --> 00:19:55.000
as you see.
[LAUGHTER] But my point is what
00:19:53.000 --> 00:19:59.000
about negative pi?
That is the same as pi.
00:19:57.000 --> 00:20:03.000
Two pi is the same as zero.
So physically there really are
00:20:03.000 --> 00:20:09.000
only two critical points,
this one and that one.
00:20:08.000 --> 00:20:14.000
And they obviously have
something very different about
00:20:13.000 --> 00:20:19.000
them.
This critical point is stable.
00:20:16.000 --> 00:20:22.000
If I start near there,
I approach that critical point
00:20:21.000 --> 00:20:27.000
in infinite time.
This one, if I start near
00:20:25.000 --> 00:20:31.000
there, I do not stay near there.
I always leave it.
00:20:31.000 --> 00:20:37.000
Of the two critical points,
physically it is clear that the
00:20:38.000 --> 00:20:44.000
critical point is zero,
zero and the other guys that
00:20:44.000 --> 00:20:50.000
look like it,
two pi zero and so on is a
00:20:49.000 --> 00:20:55.000
stable critical point.
Whereas, pi zero,
00:20:53.000 --> 00:20:59.000
when theta is sticking straight
up in the air,
00:20:59.000 --> 00:21:05.000
is unstable.
Now, of course,
00:21:03.000 --> 00:21:09.000
we will want to see that
mathematically also,
00:21:06.000 --> 00:21:12.000
but basically there are just
physically two point.
00:21:10.000 --> 00:21:16.000
There are just two critical
points.
00:21:12.000 --> 00:21:18.000
Well, that raises the question
what about all the others?
00:21:16.000 --> 00:21:22.000
As you will see,
we have to have those.
00:21:19.000 --> 00:21:25.000
They are an essential part of
the problem.
00:21:22.000 --> 00:21:28.000
They are not just redundant
baggage that is trailing along.
00:21:26.000 --> 00:21:32.000
They are really important.
But you will see that when we
00:21:33.000 --> 00:21:39.000
talk about finally how the
trajectories look and how the
00:21:40.000 --> 00:21:46.000
solutions look.
Now what do we do?
00:21:43.000 --> 00:21:49.000
Well, we found the critical
points, and now the work begins.
00:21:50.000 --> 00:21:56.000
Virtually all the work is in
this next step.
00:21:56.000 --> 00:22:02.000
What do we do?
Step two.
00:22:00.000 --> 00:22:06.000
I can only describe it in
general terms,
00:22:05.000 --> 00:22:11.000
but here is what you do.
For each critical point x zero,
00:22:13.000 --> 00:22:19.000
y zero, a procedure that has to
be done at each one,
00:22:21.000 --> 00:22:27.000
you linearize the system near
that point.
00:22:28.000 --> 00:22:34.000
In other words,
you may find a linear system,
00:22:32.000 --> 00:22:38.000
the good kind,
the kind you know about,
00:22:37.000 --> 00:22:43.000
which is a good approximation
to the nonlinear system at that
00:22:43.000 --> 00:22:49.000
critical point.
Plot the trajectories of this
00:22:48.000 --> 00:22:54.000
linearized system.
And you do that near the
00:22:53.000 --> 00:22:59.000
critical point.
00:23:04.000 --> 00:23:10.000
How do you plot the
trajectories?
00:23:05.000 --> 00:23:11.000
Well, that you knew how to do
on Friday so I am assuming you
00:23:09.000 --> 00:23:15.000
still know how to do it on
Monday.
00:23:11.000 --> 00:23:17.000
In other words,
if the system is linear you
00:23:14.000 --> 00:23:20.000
know how to plot the
trajectories of it by
00:23:16.000 --> 00:23:22.000
calculating eigenvalues and
eigenvectors and maybe the
00:23:20.000 --> 00:23:26.000
direction of motion if it is a
spiral.
00:23:22.000 --> 00:23:28.000
I will give you a couple of
examples of that when we work
00:23:25.000 --> 00:23:31.000
out the pendulum.
But, on the other hand,
00:23:29.000 --> 00:23:35.000
how do I line arise a system?
Well, there are two methods.
00:23:33.000 --> 00:23:39.000
There is one method the book
gives you, which by and large I
00:23:37.000 --> 00:23:43.000
do not want you to use,
although I will give you an
00:23:41.000 --> 00:23:47.000
example of it now.
I want you to use another
00:23:44.000 --> 00:23:50.000
method because it is much
faster.
00:23:46.000 --> 00:23:52.000
Especially if you have to
handle several critical points
00:23:50.000 --> 00:23:56.000
it is much, much faster.
Let's first carry this out on
00:23:54.000 --> 00:24:00.000
an easy case,
and then I will show you how to
00:23:57.000 --> 00:24:03.000
do it in general.
Just this once I will use the
00:24:02.000 --> 00:24:08.000
book's method because I think it
is the method which would
00:24:08.000 --> 00:24:14.000
naturally occur to you.
Let's linearize this example at
00:24:13.000 --> 00:24:19.000
the point zero,
zero.
00:24:15.000 --> 00:24:21.000
What should be the linearized
system?
00:24:18.000 --> 00:24:24.000
In other words,
it's only the nonlinear terms I
00:24:23.000 --> 00:24:29.000
have to worry about.
Well, the minus omega is fine.
00:24:29.000 --> 00:24:35.000
It is that stupid sine theta
that I don't like.
00:24:33.000 --> 00:24:39.000
But if theta is small,
in other words,
00:24:37.000 --> 00:24:43.000
if I stay near zero,
I could replace sine theta by
00:24:42.000 --> 00:24:48.000
theta.
The linearized system is minus
00:24:46.000 --> 00:24:52.000
two.
You replace sine theta by
00:24:49.000 --> 00:24:55.000
theta, since sine theta is
approximately theta if theta is
00:24:55.000 --> 00:25:01.000
small, if theta is near zero.
It is the first term of its
00:25:02.000 --> 00:25:08.000
Taylor series you can think of,
or it's just the linear
00:25:06.000 --> 00:25:12.000
approximation starts out sine
theta equals theta.
00:25:10.000 --> 00:25:16.000
That is it.
Or, you draw a picture.
00:25:12.000 --> 00:25:18.000
I don't know.
There are a millions of ways to
00:25:16.000 --> 00:25:22.000
do it.
So we have it.
00:25:17.000 --> 00:25:23.000
Now what do I do?
Okay, now we will plot that.
00:25:21.000 --> 00:25:27.000
The matrix, let's do our little
routine, in other words.
00:25:25.000 --> 00:25:31.000
I am writing right to left for
no reason.
00:25:30.000 --> 00:25:36.000
The matrix is,
now I am just going to make
00:25:34.000 --> 00:25:40.000
marks on a board the way you did
on your exam,
00:25:39.000 --> 00:25:45.000
zero, one.
[LAUGHTER] I don't know what I
00:25:43.000 --> 00:25:49.000
am doing, but you know.
Negative two,
00:25:47.000 --> 00:25:53.000
negative one,
and then I write down the
00:25:51.000 --> 00:25:57.000
eigen-whatchamacallits.
Lambda squared,
00:25:55.000 --> 00:26:01.000
plus lambda,
minus the trace.
00:26:00.000 --> 00:26:06.000
The determinant is minus,
minus two, so it is plus two
00:26:05.000 --> 00:26:11.000
equals zero.
And then, since it doesn't
00:26:09.000 --> 00:26:15.000
occur to me how to factor this,
I will use the quadratic
00:26:14.000 --> 00:26:20.000
formula.
It is negative one plus or
00:26:18.000 --> 00:26:24.000
minus the square root of,
b squared minus 4ac,
00:26:22.000 --> 00:26:28.000
minus seven.
Complex.
00:26:24.000 --> 00:26:30.000
That means it is going to be a
spiral.
00:26:30.000 --> 00:26:36.000
I am going to get a spiral.
Will it be a source or a sink
00:26:34.000 --> 00:26:40.000
since they are complex roots?
Complex eigenvalues give a
00:26:39.000 --> 00:26:45.000
spiral.
A source or a sink?
00:26:41.000 --> 00:26:47.000
I tell that from the sine of
the real part.
00:26:44.000 --> 00:26:50.000
The real part is negative
one-half.
00:26:47.000 --> 00:26:53.000
Therefore, the amplitude is
shrinking like e to the minus t
00:26:52.000 --> 00:26:58.000
over two.
And, therefore,
00:26:55.000 --> 00:27:01.000
the spiral is coming into the
origin.
00:27:00.000 --> 00:27:06.000
It is a spiral sink since
lambda equals minus one-half
00:27:05.000 --> 00:27:11.000
plus some number times i.
Spiral sink.
00:27:10.000 --> 00:27:16.000
And the other thing to
determine is its direction of
00:27:15.000 --> 00:27:21.000
motion, which will be what?
I determine its direction of
00:27:21.000 --> 00:27:27.000
motion by putting in a single
vector from the velocity field.
00:27:28.000 --> 00:27:34.000
Here it is.
The vector at one,
00:27:32.000 --> 00:27:38.000
zero will be the
same as the first column of the
00:27:36.000 --> 00:27:42.000
matrix.
So that is zero,
00:27:38.000 --> 00:27:44.000
negative two.
00:27:39.000 --> 00:27:45.000
Here is a vector from the
velocity field,
00:27:42.000 --> 00:27:48.000
and that shows that the motion
is clockwise and is spiraling
00:27:46.000 --> 00:27:52.000
into the origin.
That is a picture,
00:27:49.000 --> 00:27:55.000
therefore, at the origin of how
that looks.
00:27:52.000 --> 00:27:58.000
Now, it is of the utmost
importance for your
00:27:55.000 --> 00:28:01.000
understanding of what comes now
that you understand in what
00:27:59.000 --> 00:28:05.000
sense this picture corresponds
to the physical behavior of the
00:28:03.000 --> 00:28:09.000
pendulum.
Let's start it over here.
00:28:08.000 --> 00:28:14.000
What is it doing?
That means that theta is some
00:28:13.000 --> 00:28:19.000
number like one,
for example.
00:28:16.000 --> 00:28:22.000
Let's make it smaller.
Let's say a little bit.
00:28:20.000 --> 00:28:26.000
And this is the omega access.
If it starts over here that
00:28:26.000 --> 00:28:32.000
means the angular velocity is
zero and theta is a small
00:28:32.000 --> 00:28:38.000
positive number.
Theta is a small positive
00:28:37.000 --> 00:28:43.000
number.
The angular velocity is zero.
00:28:40.000 --> 00:28:46.000
It's velocity zero,
theta small and positive.
00:28:44.000 --> 00:28:50.000
I release it and it does that.
What does this have to do with
00:28:50.000 --> 00:28:56.000
the spiral?
Well, the spiral is exactly a
00:28:54.000 --> 00:29:00.000
mathematical picture of this
motion.
00:28:57.000 --> 00:29:03.000
What happens?
Theta starts to decrease.
00:29:02.000 --> 00:29:08.000
And the angular velocity
increases but in the negative
00:29:06.000 --> 00:29:12.000
direction.
This is negative angular
00:29:08.000 --> 00:29:14.000
velocity.
Theta is decreasing.
00:29:11.000 --> 00:29:17.000
The angular velocity gets
bigger and bigger.
00:29:14.000 --> 00:29:20.000
And it is biggest,
most negative when theta is
00:29:18.000 --> 00:29:24.000
zero.
It has reached the vertical
00:29:20.000 --> 00:29:26.000
position is when the angular
velocity is biggest.
00:29:24.000 --> 00:29:30.000
It continues then.
Theta gets negative,
00:29:27.000 --> 00:29:33.000
but the angular velocity then
decreases to zero.
00:29:37.000 --> 00:29:43.000
Now the angular velocity is
zero and theta is at its most
00:29:41.000 --> 00:29:47.000
negative, and then it reverses.
Angular velocity gets positive
00:29:46.000 --> 00:29:52.000
as theta increases again,
and so on.
00:29:48.000 --> 00:29:54.000
These represent the successive
swings back and forth.
00:29:52.000 --> 00:29:58.000
Notice the fact that it is
damped is reflected in the fact
00:29:57.000 --> 00:30:03.000
that each successive swing,
the biggest that theta gets is
00:30:01.000 --> 00:30:07.000
a little less than it was
before.
00:30:05.000 --> 00:30:11.000
In other words,
this point is not quite as far
00:30:08.000 --> 00:30:14.000
out as that one.
And this one isn't as far out
00:30:11.000 --> 00:30:17.000
as that one.
In other words,
00:30:13.000 --> 00:30:19.000
it is spiraling in.
00:30:20.000 --> 00:30:26.000
Well, I hope that is clear
because we now have to go to the
00:30:26.000 --> 00:30:32.000
next critical point.
And now we have a little
00:30:31.000 --> 00:30:37.000
problem.
If I want to do the next
00:30:34.000 --> 00:30:40.000
critical point,
so what I want to do now is,
00:30:38.000 --> 00:30:44.000
in other words,
I want to linearize at the
00:30:42.000 --> 00:30:48.000
point pi zero where theta is pi
and the thing is sticking up in
00:30:48.000 --> 00:30:54.000
the air.
The question is,
00:30:50.000 --> 00:30:56.000
how am I going to do that?
This trick of replacing sine
00:30:55.000 --> 00:31:01.000
theta by theta,
that doesn't work at pi.
00:30:59.000 --> 00:31:05.000
That works at zero.
00:31:08.000 --> 00:31:14.000
Now we have to go to the next
step of the method.
00:31:11.000 --> 00:31:17.000
The way to do this,
in general, as you will read in
00:31:14.000 --> 00:31:20.000
the notes because,
as I say, this is not in the
00:31:17.000 --> 00:31:23.000
book, is to calculate the
Jacobian.
00:31:20.000 --> 00:31:26.000
I mean the Jacobian matrix.
The Jacobian is the
00:31:23.000 --> 00:31:29.000
determinant.
I mean, before you put the two
00:31:26.000 --> 00:31:32.000
bars down and made a
determinant, you called it just
00:31:29.000 --> 00:31:35.000
the Jacobian matrix.
And the formula for it is,
00:31:34.000 --> 00:31:40.000
it is calculated from f and g.
The top line is the partial of
00:31:39.000 --> 00:31:45.000
f with respect to x and y,
and the bottom line is the
00:31:44.000 --> 00:31:50.000
partial of g with respect to x
and y.
00:31:47.000 --> 00:31:53.000
So that is the Jacobian matrix.
00:31:57.000 --> 00:32:03.000
I hope I get a chance at the
end of the period to explain to
00:32:02.000 --> 00:32:08.000
you why, but I am most anxious
right now to at least get you
00:32:07.000 --> 00:32:13.000
familiar with the algorithm,
how to do it.
00:32:11.000 --> 00:32:17.000
The notes describe the y of it,
if we don't get a chance to get
00:32:17.000 --> 00:32:23.000
to it, but I hope we will.
What do you do?
00:32:20.000 --> 00:32:26.000
The Jacobian matrix.
You calculate it at the point x
00:32:25.000 --> 00:32:31.000
zero, y zero.
00:32:29.000 --> 00:32:35.000
I will indicate that by putting
a subscript zero on it.
00:32:32.000 --> 00:32:38.000
This means without the
subscript zero it is the
00:32:35.000 --> 00:32:41.000
Jacobian matrix calculated out
of those four partial
00:32:39.000 --> 00:32:45.000
derivatives.
When I put a subscript zero,
00:32:41.000 --> 00:32:47.000
I mean I evaluated it at the
critical point by plugging in
00:32:45.000 --> 00:32:51.000
each entry as a function of x
and y.
00:32:47.000 --> 00:32:53.000
You plug in for x equals
x zero, y equals y zero,
00:32:52.000 --> 00:32:58.000
and you get a numerical
matrix.
00:32:54.000 --> 00:33:00.000
That is the matrix which is the
matrix of the linearized system.
00:33:00.000 --> 00:33:06.000
This is the matrix of the
linearized system.
00:33:08.000 --> 00:33:14.000
Trust me, it is.
Now, since I don't expect you
00:33:11.000 --> 00:33:17.000
to trust me, let's calculate it.
Here, we got the matrix another
00:33:16.000 --> 00:33:22.000
way, by this procedure of saying
sine theta is theta-- What would
00:33:21.000 --> 00:33:27.000
we have gotten if we had done it
instead by the linearized
00:33:25.000 --> 00:33:31.000
system?
Let's do it that way.
00:33:27.000 --> 00:33:33.000
Let's do it via the Jacobian.
00:33:37.000 --> 00:33:43.000
I need to know the Jacobian of
the system, which I have
00:33:43.000 --> 00:33:49.000
conveniently covered up.
There is the system.
00:33:47.000 --> 00:33:53.000
The Jacobian matrix is what?
The top line,
00:33:52.000 --> 00:33:58.000
I take the partial derivative
of omega first with respect to
00:33:59.000 --> 00:34:05.000
theta and then with respect to
omega.
00:34:04.000 --> 00:34:10.000
I then take the second line on
the right-hand side.
00:34:08.000 --> 00:34:14.000
I take its partial with respect
to theta first.
00:34:12.000 --> 00:34:18.000
That is negative two cosine
theta.
00:34:16.000 --> 00:34:22.000
And, thinking ahead,
I erase the one and move it
00:34:20.000 --> 00:34:26.000
over a little bit.
And what is the partial of that
00:34:24.000 --> 00:34:30.000
thing with respect to omega?
It is negative one.
00:34:30.000 --> 00:34:36.000
Does everyone see how I
calculated that Jacobian matrix?
00:34:35.000 --> 00:34:41.000
And now I want to evaluate it.
Let's do our old case first.
00:34:40.000 --> 00:34:46.000
At zero, zero what
would this have
00:34:45.000 --> 00:34:51.000
amounted to?
This would have given me zero,
00:34:49.000 --> 00:34:55.000
one.
The cosine of theta at zero is
00:34:52.000 --> 00:34:58.000
one, so this is negative two,
negative one.
00:34:56.000 --> 00:35:02.000
I'm screwed.
[LAUGHTER]
00:35:00.000 --> 00:35:06.000
That is the same as that.
Now you see it,
00:35:04.000 --> 00:35:10.000
now you don't.
We got our old answer back.
00:35:08.000 --> 00:35:14.000
That should give us enough
confidence to use it in the new
00:35:14.000 --> 00:35:20.000
case, where I don't have an old
answer to compare it with.
00:35:20.000 --> 00:35:26.000
What is it going to be at pi,
zero?
00:35:23.000 --> 00:35:29.000
The answer is J zero is
now going to be --
00:35:30.000 --> 00:35:36.000
Well, everything is the same.
Zero, one, negative one.
00:35:34.000 --> 00:35:40.000
And here cosine of pi is
negative one,
00:35:38.000 --> 00:35:44.000
so negative two times negative
one is two.
00:35:41.000 --> 00:35:47.000
Everything is the same,
except there is a two there now
00:35:46.000 --> 00:35:52.000
instead of negative two.
Lambda squared plus lambda,
00:35:51.000 --> 00:35:57.000
the determinant,
is now negative two.
00:35:54.000 --> 00:36:00.000
And this factors into lambda
plus two times
00:36:00.000 --> 00:36:06.000
lambda minus one.
00:36:08.000 --> 00:36:14.000
So lambda equals one.
The corresponding eigenvector.
00:36:12.000 --> 00:36:18.000
I subtract one here,
so the equation is minus a1
00:36:17.000 --> 00:36:23.000
plus a2 is zero.
00:36:20.000 --> 00:36:26.000
The solution is one,
one, e to the t.
00:36:24.000 --> 00:36:30.000
And for the other one,
it's lambda is equal to
00:36:28.000 --> 00:36:34.000
negative two.
This is the sort of stuff you
00:36:33.000 --> 00:36:39.000
can do, so I am doing it fast.
Zero minus negative two is two.
00:36:38.000 --> 00:36:44.000
So the equation is 2a1 plus a2
equals zero.
00:36:43.000 --> 00:36:49.000
And the solution is now,
I give a1 the value one,
00:36:48.000 --> 00:36:54.000
a2 will be negative two,
and that is times e to the
00:36:52.000 --> 00:36:58.000
minus 2t.
00:37:00.000 --> 00:37:06.000
Well, what does the thing then
actually look like?
00:37:05.000 --> 00:37:11.000
What I am now going to do is,
I drew a picture before,
00:37:11.000 --> 00:37:17.000
that spiral picture we had
before of the way the thing
00:37:17.000 --> 00:37:23.000
looked at the point zero,
zero.
00:37:22.000 --> 00:37:28.000
So at the point pi,
zero how does it
00:37:27.000 --> 00:37:33.000
look, now?
Well, it looks like the origin,
00:37:32.000 --> 00:37:38.000
but I am thinking of it really
as the point pi,
00:37:37.000 --> 00:37:43.000
zero.
In other words,
00:37:39.000 --> 00:37:45.000
I am thinking of a linear
variable change sliding along
00:37:44.000 --> 00:37:50.000
the axis so that the point pi,
zero now looks like
00:37:50.000 --> 00:37:56.000
the origin.
If I do that then those two
00:37:53.000 --> 00:37:59.000
basic solutions,
there is the one,
00:37:56.000 --> 00:38:02.000
one solution which
is going out that way.
00:38:03.000 --> 00:38:09.000
And it is going out this way.
But the other guy is coming in
00:38:07.000 --> 00:38:13.000
along the vector one,
negative two.
00:38:09.000 --> 00:38:15.000
So one, negative two looks like
this.
00:38:12.000 --> 00:38:18.000
This guy is coming in at a
somewhat sharper angle,
00:38:15.000 --> 00:38:21.000
coming in because it is e to
the negative 2t.
00:38:19.000 --> 00:38:25.000
And we recognize this,
of course, as a saddle.
00:38:22.000 --> 00:38:28.000
And I would complete the
trajectories by putting in some
00:38:26.000 --> 00:38:32.000
of the typical saddle lines like
that.
00:38:30.000 --> 00:38:36.000
Now, I say that,
too, gives a picture of what is
00:38:36.000 --> 00:38:42.000
happening to the pendulum near
that point.
00:38:41.000 --> 00:38:47.000
Let's, for example,
look here.
00:38:44.000 --> 00:38:50.000
What is happening here?
This is theta,
00:38:49.000 --> 00:38:55.000
or really it is theta minus pi,
is this axis.
00:38:55.000 --> 00:39:01.000
In other words,
this will be zero.
00:39:01.000 --> 00:39:07.000
When theta is pi this will
correspond to the point zero.
00:39:08.000 --> 00:39:14.000
Here is omega.
What is theta doing?
00:39:12.000 --> 00:39:18.000
Starting up there this
represents a value of theta a
00:39:18.000 --> 00:39:24.000
little bit less than pi,
a little bit to the negative.
00:39:25.000 --> 00:39:31.000
A little bit less than pi.
Here is pi, so a little bit
00:39:33.000 --> 00:39:39.000
less than pi is over here.
A little bit less than pi.
00:39:40.000 --> 00:39:46.000
A little bit less.
And omega is zero.
00:39:44.000 --> 00:39:50.000
What happened?
Theta started decreasing ever
00:39:50.000 --> 00:39:56.000
more rapidly so that the omega
was zero here,
00:39:56.000 --> 00:40:02.000
now omega is negative and gets
much more negative.
00:40:04.000 --> 00:40:10.000
In other words,
both theta decreases from that
00:40:09.000 --> 00:40:15.000
point and omega decreases also.
What happened here?
00:40:14.000 --> 00:40:20.000
Here, theta is a little bit
more than pi.
00:40:19.000 --> 00:40:25.000
Now it is a little bit more
than pi.
00:40:23.000 --> 00:40:29.000
Theta now increases until it
gets to 2pi and then oscillates
00:40:29.000 --> 00:40:35.000
around 2pi.
So theta increases.
00:40:33.000 --> 00:40:39.000
And omega increases,
too, because the angular
00:40:37.000 --> 00:40:43.000
velocity started at zero but,
as theta gets more positive,
00:40:42.000 --> 00:40:48.000
omega gets positive,
too, because theta is
00:40:46.000 --> 00:40:52.000
increasing.
So omega increases and theta
00:40:49.000 --> 00:40:55.000
increases and it goes off like
that.
00:40:52.000 --> 00:40:58.000
Well, the final step is to put
them all together to be the big
00:40:58.000 --> 00:41:04.000
picture.
Here we are for three,
00:41:03.000 --> 00:41:09.000
let's say, the big picture.
Plot trajectories around each
00:41:12.000 --> 00:41:18.000
critical point and then add
some.
00:41:17.000 --> 00:41:23.000
It's that last step that can
cause you a little grief,
00:41:25.000 --> 00:41:31.000
but we will see how it works
out.
00:41:32.000 --> 00:41:38.000
Add some more,
according to your best
00:41:34.000 --> 00:41:40.000
judgment.
Let's make a big picture now of
00:41:37.000 --> 00:41:43.000
our pendulum the way it
apparently ought to look.
00:41:41.000 --> 00:41:47.000
Nice big axis since we are
going to accommodate a lot of
00:41:45.000 --> 00:41:51.000
critical points here.
Let's put in some critical
00:41:48.000 --> 00:41:54.000
points.
Here is the origin.
00:41:50.000 --> 00:41:56.000
And now here is the one at pi,
let's say, here is one at 2pi.
00:41:55.000 --> 00:42:01.000
I am going to add some of these
others, 3pi, 4pi.
00:42:00.000 --> 00:42:06.000
I won't in their values.
You can figure out what I mean.
00:42:05.000 --> 00:42:11.000
Zero here.
And then here it will be
00:42:08.000 --> 00:42:14.000
negative pi, and here negative
2pi.
00:42:11.000 --> 00:42:17.000
I guess I can stop there.
That is the theta axis,
00:42:15.000 --> 00:42:21.000
and here is the omega axis.
And now, at each one of these,
00:42:21.000 --> 00:42:27.000
I stay nearby and I draw the
linear trajectories,
00:42:25.000 --> 00:42:31.000
the trajectories of the
corresponding linearized system.
00:42:32.000 --> 00:42:38.000
We decided here that the spiral
went clockwise.
00:42:35.000 --> 00:42:41.000
Now, this point is physically,
of course, the same as that
00:42:40.000 --> 00:42:46.000
one.
But mathematically,
00:42:42.000 --> 00:42:48.000
the Jacobian matrix is the same
also because if theta is 2pi the
00:42:47.000 --> 00:42:53.000
cosine of 2pi is also one.
And this is the same matrix.
00:42:51.000 --> 00:42:57.000
So the analysis is identical.
And, therefore,
00:42:54.000 --> 00:43:00.000
this point will also correspond
to a counterclockwise spiral,
00:42:59.000 --> 00:43:05.000
as will this one.
I am sorry, a clockwise spiral.
00:43:04.000 --> 00:43:10.000
Here, too, all of these points
are the same.
00:43:08.000 --> 00:43:14.000
The behavior near them,
clockwise spirals everywhere.
00:43:12.000 --> 00:43:18.000
How about the other ones?
Well, the other ones correspond
00:43:17.000 --> 00:43:23.000
to these saddles,
so let's draw them efficiently
00:43:21.000 --> 00:43:27.000
by doing the same thing on every
one, mass production of saddles.
00:43:26.000 --> 00:43:32.000
There.
And these guys go out.
00:43:30.000 --> 00:43:36.000
And the other guys come in,
etc.
00:43:35.000 --> 00:43:41.000
And so here we have a little
bit there, a little here,
00:43:44.000 --> 00:43:50.000
here, there,
everywhere, etc.
00:43:58.000 --> 00:44:04.000
Now what?
Now you pray for inspiration.
00:44:01.000 --> 00:44:07.000
And what you have to do is add
trajectories that are compatible
00:44:08.000 --> 00:44:14.000
with the ones you have.
Let's start with this guy.
00:44:13.000 --> 00:44:19.000
Where is it going?
Well, a trajectory either goes
00:44:18.000 --> 00:44:24.000
off to infinity,
but generally they get trapped
00:44:22.000 --> 00:44:28.000
around critical points.
This guy must be surely doing
00:44:28.000 --> 00:44:34.000
this.
How about this one?
00:44:32.000 --> 00:44:38.000
Yeah, sure.
How about this one?
00:44:35.000 --> 00:44:41.000
Why not?
How about that one?
00:44:38.000 --> 00:44:44.000
Yeah.
But notice you are in trouble
00:44:42.000 --> 00:44:48.000
when two arrows you want to put
in are near each other and going
00:44:49.000 --> 00:44:55.000
in opposite directions.
That you cannot have.
00:44:54.000 --> 00:45:00.000
Continuity forbids it.
But notice if I,
00:44:59.000 --> 00:45:05.000
for example,
had gotten these eigenlines
00:45:01.000 --> 00:45:07.000
wrong, if I made this come in
and those go out because I made
00:45:05.000 --> 00:45:11.000
a simple error in drawing the
thing, I would have said but I
00:45:08.000 --> 00:45:14.000
cannot draw this picture because
this spiral wants to go that way
00:45:12.000 --> 00:45:18.000
but this, right next to it,
wants to go the other way.
00:45:16.000 --> 00:45:22.000
That is the way you would know
if you made a mistake.
00:45:19.000 --> 00:45:25.000
If you didn't make a mistake
you won't have any trouble
00:45:22.000 --> 00:45:28.000
filling these things out.
The directions of motions of
00:45:26.000 --> 00:45:32.000
the spirals, everything will be
compatible.
00:45:30.000 --> 00:45:36.000
Okay.
What is this guy doing?
00:45:33.000 --> 00:45:39.000
Oh, well, it must be joining up
with that.
00:45:37.000 --> 00:45:43.000
How about this one?
Well, it must be coming back
00:45:42.000 --> 00:45:48.000
there.
How about this one?
00:45:45.000 --> 00:45:51.000
Well, trajectories cannot
cross.
00:45:48.000 --> 00:45:54.000
This guy cannot cross so it
must be doing this.
00:45:53.000 --> 00:45:59.000
All right.
What is that trajectory?
00:45:57.000 --> 00:46:03.000
Starts zero.
Omega, on the other hand,
00:46:03.000 --> 00:46:09.000
is big and positive.
Omega big and positive.
00:46:10.000 --> 00:46:16.000
[LAUGHTER] I am scared.
Omega starts.
00:46:15.000 --> 00:46:21.000
Theta is zero.
Omega big and positive.
00:46:21.000 --> 00:46:27.000
It went around,
slowed up, but continued beyond
00:46:28.000 --> 00:46:34.000
pi.
And, in fact,
00:46:30.000 --> 00:46:36.000
went too far.
It continued to go here and
00:46:33.000 --> 00:46:39.000
then finally wound around that
one.
00:46:36.000 --> 00:46:42.000
Now do you see why we had to
have all the critical points?
00:46:40.000 --> 00:46:46.000
You have to have all the
critical points.
00:46:42.000 --> 00:46:48.000
And not just the two physicals
ones because the other critical
00:46:47.000 --> 00:46:53.000
points are necessary to describe
a complicated motion that goes
00:46:51.000 --> 00:46:57.000
round and round and round until
finally it crashes.
00:46:56.000 --> 00:47:02.000
You are going to practice
drawing these pictures and
00:46:59.000 --> 00:47:05.000
interpreting them in recitation
tomorrow.