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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

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general theory.
And, in general,

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if you have a nonlinear system,

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how do you go about analyzing
it?

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The first step is to find the
simplest possible solutions,

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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

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single point?
Well, like the origin for a

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linear system,
those points which form

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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

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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.

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See the significance of that?
If you have such a point,

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let's say there is a critical
point, what is the velocity

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00:14:31,000 --> 00:14:37,000
field at that point?
Well, it is given by the

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vectors on the right-hand side,
but the components are zero.

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00:14:40,000 --> 00:14:46,000
That means, at this point the
velocity vector is zero.

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Well, that means if a solution
starts there,

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you put the mouse there and
tell it to move,

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where do you go?
It has no reason to go anywhere

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since the velocity vector is
zero there.

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So it sits there for all time.
And indeed it solves the

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system, doesn't it?
It makes the right-hand side

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zero and it makes the left-hand
side zero because x equals x

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zero, y equals y zero for

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all time.
If that is true for all time it

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sits there then the derivatives,
with respect to time are zero,

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so the left-hand sides are
zero, the right-hand sides are

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zero and everybody is happy.
Well, these are great points.

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How do I find them?
Well, by looking for points

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that make these two functions
zero.

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I find them by solving
simultaneously the equations f

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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

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00:16:01,000 --> 00:16:07,000
not linear equations.
Linear equations you know how

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to solve, but they are not
linear equations.

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They are nonlinear equations
that you don't know how to

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solve.
And, to some extent,

235
00:16:13,000 --> 00:16:19,000
nobody else does either.
There are very fat books in the

236
00:16:17,000 --> 00:16:23,000
library whose topic is how to
solve just a pair of equations,

237
00:16:22,000 --> 00:16:28,000
f of (x, y) equals zero,
g of (x, y) equals zero.

238
00:16:25,000 --> 00:16:31,000
And it is quite a hard problem.
It is even a hard problem by

239
00:16:30,000 --> 00:16:36,000
computer.
Because, if you know

240
00:16:33,000 --> 00:16:39,000
approximately where the solution
is going to be,

241
00:16:36,000 --> 00:16:42,000
you can make up the little
screen and then the computer

242
00:16:40,000 --> 00:16:46,000
will find it for you.
Or, even without a screen,

243
00:16:43,000 --> 00:16:49,000
it will calculate it by
Newton's method or something

244
00:16:47,000 --> 00:16:53,000
else, it will zero in.
The problem is,

245
00:16:50,000 --> 00:16:56,000
if you don't know in advance
roughly where the critical point

246
00:16:54,000 --> 00:17:00,000
is that you are looking for,
there are a lot of numbers.

247
00:16:58,000 --> 00:17:04,000
They go to infinity that way
and infinity that way.

248
00:17:03,000 --> 00:17:09,000
In general, it is almost an
impossible problem.

249
00:17:06,000 --> 00:17:12,000
The only thing that makes it
possible is that these problems

250
00:17:11,000 --> 00:17:17,000
always come from the real world
and one has some physical

251
00:17:15,000 --> 00:17:21,000
feeling, one hopes,
for where the critical point

252
00:17:19,000 --> 00:17:25,000
is.
I am going to assume breezily

253
00:17:21,000 --> 00:17:27,000
and cheerily we can solve those.
And I am only going to give you

254
00:17:26,000 --> 00:17:32,000
examples where it is possible to
solve them.

255
00:17:31,000 --> 00:17:37,000
But even there,
you have to watch out.

256
00:17:33,000 --> 00:17:39,000
There is a certain trickiness
that will be talked about in the

257
00:17:37,000 --> 00:17:43,000
recitations tomorrow.
Okay, so we found the critical

258
00:17:41,000 --> 00:17:47,000
points.
Let's do it for our example.

259
00:17:43,000 --> 00:17:49,000
Let's find the critical point.
What is the pair of equations

260
00:17:47,000 --> 00:17:53,000
we have to solve?
We have to solve the equations

261
00:17:50,000 --> 00:17:56,000
omega equals zero,
minus two sine theta minus

262
00:17:53,000 --> 00:17:59,000
omega equals zero.

263
00:17:56,000 --> 00:18:02,000
Now, it is not always this
easy.

264
00:18:00,000 --> 00:18:06,000
But the solution is omega is
zero.

265
00:18:03,000 --> 00:18:09,000
If omega is zero,
then sine theta is zero,

266
00:18:07,000 --> 00:18:13,000
and sine theta is zero at the

267
00:18:12,000 --> 00:18:18,000
integral multiples of pi.
The critical points are omega

268
00:18:18,000 --> 00:18:24,000
is always zero and theta is
zero, or it could be plus or

269
00:18:24,000 --> 00:18:30,000
minus pi, plus or minus two pi
and so on.

270
00:18:30,000 --> 00:18:36,000
In other words,
there are an infinity of

271
00:18:32,000 --> 00:18:38,000
critical points.
That seems a little

272
00:18:34,000 --> 00:18:40,000
discouraging.
On the other hand,

273
00:18:36,000 --> 00:18:42,000
there are really only two.
There are really physically

274
00:18:40,000 --> 00:18:46,000
only two because omega equals
zero means the mass is not

275
00:18:44,000 --> 00:18:50,000
moving.
The angular velocity is zero,

276
00:18:46,000 --> 00:18:52,000
so it is only the theta
position which is changing.

277
00:18:49,000 --> 00:18:55,000
Now, what are the possible
theta positions?

278
00:18:52,000 --> 00:18:58,000
Well, here is our nonlinear
pendulum.

279
00:18:56,000 --> 00:19:02,000
Here is the critical point,
theta equals zero,

280
00:18:59,000 --> 00:19:05,000
omega equals zero.
Theta equals zero means the rod

281
00:19:03,000 --> 00:19:09,000
is vertical.
Omega equals zero means that it

282
00:19:07,000 --> 00:19:13,000
is not moving,
despite the fact that it is

283
00:19:10,000 --> 00:19:16,000
moving.
Theoretically it is not moving.

284
00:19:14,000 --> 00:19:20,000
Now, what is the other one?
Well, there is theta equals pi.

285
00:19:18,000 --> 00:19:24,000
Theta is now,
starting from zero and

286
00:19:21,000 --> 00:19:27,000
increasing, I hope,
through positive theta.

287
00:19:25,000 --> 00:19:31,000
And when it gets to pi,
it is sticking straight up in

288
00:19:29,000 --> 00:19:35,000
the air.
And so the claim is that

289
00:19:34,000 --> 00:19:40,000
another critical point is theta
equals pi and omega equals zero.

290
00:19:40,000 --> 00:19:46,000
In other words,
if it gets to this position,

291
00:19:44,000 --> 00:19:50,000
it starts out in this position
and stays there for all time,

292
00:19:49,000 --> 00:19:55,000
as you see.
[LAUGHTER] But my point is what

293
00:19:53,000 --> 00:19:59,000
about negative pi?
That is the same as pi.

294
00:19:57,000 --> 00:20:03,000
Two pi is the same as zero.
So physically there really are

295
00:20:03,000 --> 00:20:09,000
only two critical points,
this one and that one.

296
00:20:08,000 --> 00:20:14,000
And they obviously have
something very different about

297
00:20:13,000 --> 00:20:19,000
them.
This critical point is stable.

298
00:20:16,000 --> 00:20:22,000
If I start near there,
I approach that critical point

299
00:20:21,000 --> 00:20:27,000
in infinite time.
This one, if I start near

300
00:20:25,000 --> 00:20:31,000
there, I do not stay near there.
I always leave it.

301
00:20:31,000 --> 00:20:37,000
Of the two critical points,
physically it is clear that the

302
00:20:38,000 --> 00:20:44,000
critical point is zero,
zero and the other guys that

303
00:20:44,000 --> 00:20:50,000
look like it,
two pi zero and so on is a

304
00:20:49,000 --> 00:20:55,000
stable critical point.
Whereas, pi zero,

305
00:20:53,000 --> 00:20:59,000
when theta is sticking straight
up in the air,

306
00:20:59,000 --> 00:21:05,000
is unstable.
Now, of course,

307
00:21:03,000 --> 00:21:09,000
we will want to see that
mathematically also,

308
00:21:06,000 --> 00:21:12,000
but basically there are just
physically two point.

309
00:21:10,000 --> 00:21:16,000
There are just two critical
points.

310
00:21:12,000 --> 00:21:18,000
Well, that raises the question
what about all the others?

311
00:21:16,000 --> 00:21:22,000
As you will see,
we have to have those.

312
00:21:19,000 --> 00:21:25,000
They are an essential part of
the problem.

313
00:21:22,000 --> 00:21:28,000
They are not just redundant
baggage that is trailing along.

314
00:21:26,000 --> 00:21:32,000
They are really important.
But you will see that when we

315
00:21:33,000 --> 00:21:39,000
talk about finally how the
trajectories look and how the

316
00:21:40,000 --> 00:21:46,000
solutions look.
Now what do we do?

317
00:21:43,000 --> 00:21:49,000
Well, we found the critical
points, and now the work begins.

318
00:21:50,000 --> 00:21:56,000
Virtually all the work is in
this next step.

319
00:21:56,000 --> 00:22:02,000
What do we do?
Step two.

320
00:22:00,000 --> 00:22:06,000
I can only describe it in
general terms,

321
00:22:05,000 --> 00:22:11,000
but here is what you do.
For each critical point x zero,

322
00:22:13,000 --> 00:22:19,000
y zero, a procedure that has to
be done at each one,

323
00:22:21,000 --> 00:22:27,000
you linearize the system near
that point.

324
00:22:28,000 --> 00:22:34,000
In other words,
you may find a linear system,

325
00:22:32,000 --> 00:22:38,000
the good kind,
the kind you know about,

326
00:22:37,000 --> 00:22:43,000
which is a good approximation
to the nonlinear system at that

327
00:22:43,000 --> 00:22:49,000
critical point.
Plot the trajectories of this

328
00:22:48,000 --> 00:22:54,000
linearized system.
And you do that near the

329
00:22:53,000 --> 00:22:59,000
critical point.

330
00:23:04,000 --> 00:23:10,000
How do you plot the
trajectories?

331
00:23:05,000 --> 00:23:11,000
Well, that you knew how to do
on Friday so I am assuming you

332
00:23:09,000 --> 00:23:15,000
still know how to do it on
Monday.

333
00:23:11,000 --> 00:23:17,000
In other words,
if the system is linear you

334
00:23:14,000 --> 00:23:20,000
know how to plot the
trajectories of it by

335
00:23:16,000 --> 00:23:22,000
calculating eigenvalues and
eigenvectors and maybe the

336
00:23:20,000 --> 00:23:26,000
direction of motion if it is a
spiral.

337
00:23:22,000 --> 00:23:28,000
I will give you a couple of
examples of that when we work

338
00:23:25,000 --> 00:23:31,000
out the pendulum.
But, on the other hand,

339
00:23:29,000 --> 00:23:35,000
how do I line arise a system?
Well, there are two methods.

340
00:23:33,000 --> 00:23:39,000
There is one method the book
gives you, which by and large I

341
00:23:37,000 --> 00:23:43,000
do not want you to use,
although I will give you an

342
00:23:41,000 --> 00:23:47,000
example of it now.
I want you to use another

343
00:23:44,000 --> 00:23:50,000
method because it is much
faster.

344
00:23:46,000 --> 00:23:52,000
Especially if you have to
handle several critical points

345
00:23:50,000 --> 00:23:56,000
it is much, much faster.
Let's first carry this out on

346
00:23:54,000 --> 00:24:00,000
an easy case,
and then I will show you how to

347
00:23:57,000 --> 00:24:03,000
do it in general.
Just this once I will use the

348
00:24:02,000 --> 00:24:08,000
book's method because I think it
is the method which would

349
00:24:08,000 --> 00:24:14,000
naturally occur to you.
Let's linearize this example at

350
00:24:13,000 --> 00:24:19,000
the point zero,
zero.

351
00:24:15,000 --> 00:24:21,000
What should be the linearized
system?

352
00:24:18,000 --> 00:24:24,000
In other words,
it's only the nonlinear terms I

353
00:24:23,000 --> 00:24:29,000
have to worry about.
Well, the minus omega is fine.

354
00:24:29,000 --> 00:24:35,000
It is that stupid sine theta
that I don't like.

355
00:24:33,000 --> 00:24:39,000
But if theta is small,
in other words,

356
00:24:37,000 --> 00:24:43,000
if I stay near zero,
I could replace sine theta by

357
00:24:42,000 --> 00:24:48,000
theta.
The linearized system is minus

358
00:24:46,000 --> 00:24:52,000
two.
You replace sine theta by

359
00:24:49,000 --> 00:24:55,000
theta, since sine theta is
approximately theta if theta is

360
00:24:55,000 --> 00:25:01,000
small, if theta is near zero.
It is the first term of its

361
00:25:02,000 --> 00:25:08,000
Taylor series you can think of,
or it's just the linear

362
00:25:06,000 --> 00:25:12,000
approximation starts out sine
theta equals theta.

363
00:25:10,000 --> 00:25:16,000
That is it.
Or, you draw a picture.

364
00:25:12,000 --> 00:25:18,000
I don't know.
There are a millions of ways to

365
00:25:16,000 --> 00:25:22,000
do it.
So we have it.

366
00:25:17,000 --> 00:25:23,000
Now what do I do?
Okay, now we will plot that.

367
00:25:21,000 --> 00:25:27,000
The matrix, let's do our little
routine, in other words.

368
00:25:25,000 --> 00:25:31,000
I am writing right to left for
no reason.

369
00:25:30,000 --> 00:25:36,000
The matrix is,
now I am just going to make

370
00:25:34,000 --> 00:25:40,000
marks on a board the way you did
on your exam,

371
00:25:39,000 --> 00:25:45,000
zero, one.
[LAUGHTER] I don't know what I

372
00:25:43,000 --> 00:25:49,000
am doing, but you know.
Negative two,

373
00:25:47,000 --> 00:25:53,000
negative one,
and then I write down the

374
00:25:51,000 --> 00:25:57,000
eigen-whatchamacallits.
Lambda squared,

375
00:25:55,000 --> 00:26:01,000
plus lambda,
minus the trace.

376
00:26:00,000 --> 00:26:06,000
The determinant is minus,
minus two, so it is plus two

377
00:26:05,000 --> 00:26:11,000
equals zero.
And then, since it doesn't

378
00:26:09,000 --> 00:26:15,000
occur to me how to factor this,
I will use the quadratic

379
00:26:14,000 --> 00:26:20,000
formula.
It is negative one plus or

380
00:26:18,000 --> 00:26:24,000
minus the square root of,
b squared minus 4ac,

381
00:26:22,000 --> 00:26:28,000
minus seven.
Complex.

382
00:26:24,000 --> 00:26:30,000
That means it is going to be a
spiral.

383
00:26:30,000 --> 00:26:36,000
I am going to get a spiral.
Will it be a source or a sink

384
00:26:34,000 --> 00:26:40,000
since they are complex roots?
Complex eigenvalues give a

385
00:26:39,000 --> 00:26:45,000
spiral.
A source or a sink?

386
00:26:41,000 --> 00:26:47,000
I tell that from the sine of
the real part.

387
00:26:44,000 --> 00:26:50,000
The real part is negative
one-half.

388
00:26:47,000 --> 00:26:53,000
Therefore, the amplitude is
shrinking like e to the minus t

389
00:26:52,000 --> 00:26:58,000
over two.
And, therefore,

390
00:26:55,000 --> 00:27:01,000
the spiral is coming into the
origin.

391
00:27:00,000 --> 00:27:06,000
It is a spiral sink since
lambda equals minus one-half

392
00:27:05,000 --> 00:27:11,000
plus some number times i.
Spiral sink.

393
00:27:10,000 --> 00:27:16,000
And the other thing to
determine is its direction of

394
00:27:15,000 --> 00:27:21,000
motion, which will be what?
I determine its direction of

395
00:27:21,000 --> 00:27:27,000
motion by putting in a single
vector from the velocity field.

396
00:27:28,000 --> 00:27:34,000
Here it is.
The vector at one,

397
00:27:32,000 --> 00:27:38,000
zero will be the
same as the first column of the

398
00:27:36,000 --> 00:27:42,000
matrix.
So that is zero,

399
00:27:38,000 --> 00:27:44,000
negative two.

400
00:27:39,000 --> 00:27:45,000
Here is a vector from the
velocity field,

401
00:27:42,000 --> 00:27:48,000
and that shows that the motion
is clockwise and is spiraling

402
00:27:46,000 --> 00:27:52,000
into the origin.
That is a picture,

403
00:27:49,000 --> 00:27:55,000
therefore, at the origin of how
that looks.

404
00:27:52,000 --> 00:27:58,000
Now, it is of the utmost
importance for your

405
00:27:55,000 --> 00:28:01,000
understanding of what comes now
that you understand in what

406
00:27:59,000 --> 00:28:05,000
sense this picture corresponds
to the physical behavior of the

407
00:28:03,000 --> 00:28:09,000
pendulum.
Let's start it over here.

408
00:28:08,000 --> 00:28:14,000
What is it doing?
That means that theta is some

409
00:28:13,000 --> 00:28:19,000
number like one,
for example.

410
00:28:16,000 --> 00:28:22,000
Let's make it smaller.
Let's say a little bit.

411
00:28:20,000 --> 00:28:26,000
And this is the omega access.
If it starts over here that

412
00:28:26,000 --> 00:28:32,000
means the angular velocity is
zero and theta is a small

413
00:28:32,000 --> 00:28:38,000
positive number.
Theta is a small positive

414
00:28:37,000 --> 00:28:43,000
number.
The angular velocity is zero.

415
00:28:40,000 --> 00:28:46,000
It's velocity zero,
theta small and positive.

416
00:28:44,000 --> 00:28:50,000
I release it and it does that.
What does this have to do with

417
00:28:50,000 --> 00:28:56,000
the spiral?
Well, the spiral is exactly a

418
00:28:54,000 --> 00:29:00,000
mathematical picture of this
motion.

419
00:28:57,000 --> 00:29:03,000
What happens?
Theta starts to decrease.

420
00:29:02,000 --> 00:29:08,000
And the angular velocity
increases but in the negative

421
00:29:06,000 --> 00:29:12,000
direction.
This is negative angular

422
00:29:08,000 --> 00:29:14,000
velocity.
Theta is decreasing.

423
00:29:11,000 --> 00:29:17,000
The angular velocity gets
bigger and bigger.

424
00:29:14,000 --> 00:29:20,000
And it is biggest,
most negative when theta is

425
00:29:18,000 --> 00:29:24,000
zero.
It has reached the vertical

426
00:29:20,000 --> 00:29:26,000
position is when the angular
velocity is biggest.

427
00:29:24,000 --> 00:29:30,000
It continues then.
Theta gets negative,

428
00:29:27,000 --> 00:29:33,000
but the angular velocity then
decreases to zero.

429
00:29:37,000 --> 00:29:43,000
Now the angular velocity is
zero and theta is at its most

430
00:29:41,000 --> 00:29:47,000
negative, and then it reverses.
Angular velocity gets positive

431
00:29:46,000 --> 00:29:52,000
as theta increases again,
and so on.

432
00:29:48,000 --> 00:29:54,000
These represent the successive
swings back and forth.

433
00:29:52,000 --> 00:29:58,000
Notice the fact that it is
damped is reflected in the fact

434
00:29:57,000 --> 00:30:03,000
that each successive swing,
the biggest that theta gets is

435
00:30:01,000 --> 00:30:07,000
a little less than it was
before.

436
00:30:05,000 --> 00:30:11,000
In other words,
this point is not quite as far

437
00:30:08,000 --> 00:30:14,000
out as that one.
And this one isn't as far out

438
00:30:11,000 --> 00:30:17,000
as that one.
In other words,

439
00:30:13,000 --> 00:30:19,000
it is spiraling in.

440
00:30:20,000 --> 00:30:26,000
Well, I hope that is clear
because we now have to go to the

441
00:30:26,000 --> 00:30:32,000
next critical point.
And now we have a little

442
00:30:31,000 --> 00:30:37,000
problem.
If I want to do the next

443
00:30:34,000 --> 00:30:40,000
critical point,
so what I want to do now is,

444
00:30:38,000 --> 00:30:44,000
in other words,
I want to linearize at the

445
00:30:42,000 --> 00:30:48,000
point pi zero where theta is pi
and the thing is sticking up in

446
00:30:48,000 --> 00:30:54,000
the air.
The question is,

447
00:30:50,000 --> 00:30:56,000
how am I going to do that?
This trick of replacing sine

448
00:30:55,000 --> 00:31:01,000
theta by theta,
that doesn't work at pi.

449
00:30:59,000 --> 00:31:05,000
That works at zero.

450
00:31:08,000 --> 00:31:14,000
Now we have to go to the next
step of the method.

451
00:31:11,000 --> 00:31:17,000
The way to do this,
in general, as you will read in

452
00:31:14,000 --> 00:31:20,000
the notes because,
as I say, this is not in the

453
00:31:17,000 --> 00:31:23,000
book, is to calculate the
Jacobian.

454
00:31:20,000 --> 00:31:26,000
I mean the Jacobian matrix.
The Jacobian is the

455
00:31:23,000 --> 00:31:29,000
determinant.
I mean, before you put the two

456
00:31:26,000 --> 00:31:32,000
bars down and made a
determinant, you called it just

457
00:31:29,000 --> 00:31:35,000
the Jacobian matrix.
And the formula for it is,

458
00:31:34,000 --> 00:31:40,000
it is calculated from f and g.
The top line is the partial of

459
00:31:39,000 --> 00:31:45,000
f with respect to x and y,
and the bottom line is the

460
00:31:44,000 --> 00:31:50,000
partial of g with respect to x
and y.

461
00:31:47,000 --> 00:31:53,000
So that is the Jacobian matrix.

462
00:31:57,000 --> 00:32:03,000
I hope I get a chance at the
end of the period to explain to

463
00:32:02,000 --> 00:32:08,000
you why, but I am most anxious
right now to at least get you

464
00:32:07,000 --> 00:32:13,000
familiar with the algorithm,
how to do it.

465
00:32:11,000 --> 00:32:17,000
The notes describe the y of it,
if we don't get a chance to get

466
00:32:17,000 --> 00:32:23,000
to it, but I hope we will.
What do you do?

467
00:32:20,000 --> 00:32:26,000
The Jacobian matrix.
You calculate it at the point x

468
00:32:25,000 --> 00:32:31,000
zero, y zero.

469
00:32:29,000 --> 00:32:35,000
I will indicate that by putting
a subscript zero on it.

470
00:32:32,000 --> 00:32:38,000
This means without the
subscript zero it is the

471
00:32:35,000 --> 00:32:41,000
Jacobian matrix calculated out
of those four partial

472
00:32:39,000 --> 00:32:45,000
derivatives.
When I put a subscript zero,

473
00:32:41,000 --> 00:32:47,000
I mean I evaluated it at the
critical point by plugging in

474
00:32:45,000 --> 00:32:51,000
each entry as a function of x
and y.

475
00:32:47,000 --> 00:32:53,000
You plug in for x equals
x zero, y equals y zero,

476
00:32:52,000 --> 00:32:58,000
and you get a numerical
matrix.

477
00:32:54,000 --> 00:33:00,000
That is the matrix which is the
matrix of the linearized system.

478
00:33:00,000 --> 00:33:06,000
This is the matrix of the
linearized system.

479
00:33:08,000 --> 00:33:14,000
Trust me, it is.
Now, since I don't expect you

480
00:33:11,000 --> 00:33:17,000
to trust me, let's calculate it.
Here, we got the matrix another

481
00:33:16,000 --> 00:33:22,000
way, by this procedure of saying
sine theta is theta-- What would

482
00:33:21,000 --> 00:33:27,000
we have gotten if we had done it
instead by the linearized

483
00:33:25,000 --> 00:33:31,000
system?
Let's do it that way.

484
00:33:27,000 --> 00:33:33,000
Let's do it via the Jacobian.

485
00:33:37,000 --> 00:33:43,000
I need to know the Jacobian of
the system, which I have

486
00:33:43,000 --> 00:33:49,000
conveniently covered up.
There is the system.

487
00:33:47,000 --> 00:33:53,000
The Jacobian matrix is what?
The top line,

488
00:33:52,000 --> 00:33:58,000
I take the partial derivative
of omega first with respect to

489
00:33:59,000 --> 00:34:05,000
theta and then with respect to
omega.

490
00:34:04,000 --> 00:34:10,000
I then take the second line on
the right-hand side.

491
00:34:08,000 --> 00:34:14,000
I take its partial with respect
to theta first.

492
00:34:12,000 --> 00:34:18,000
That is negative two cosine
theta.

493
00:34:16,000 --> 00:34:22,000
And, thinking ahead,
I erase the one and move it

494
00:34:20,000 --> 00:34:26,000
over a little bit.
And what is the partial of that

495
00:34:24,000 --> 00:34:30,000
thing with respect to omega?
It is negative one.

496
00:34:30,000 --> 00:34:36,000
Does everyone see how I
calculated that Jacobian matrix?

497
00:34:35,000 --> 00:34:41,000
And now I want to evaluate it.
Let's do our old case first.

498
00:34:40,000 --> 00:34:46,000
At zero, zero what
would this have

499
00:34:45,000 --> 00:34:51,000
amounted to?
This would have given me zero,

500
00:34:49,000 --> 00:34:55,000
one.
The cosine of theta at zero is

501
00:34:52,000 --> 00:34:58,000
one, so this is negative two,
negative one.

502
00:34:56,000 --> 00:35:02,000
I'm screwed.
[LAUGHTER]

503
00:35:00,000 --> 00:35:06,000
That is the same as that.
Now you see it,

504
00:35:04,000 --> 00:35:10,000
now you don't.
We got our old answer back.

505
00:35:08,000 --> 00:35:14,000
That should give us enough
confidence to use it in the new

506
00:35:14,000 --> 00:35:20,000
case, where I don't have an old
answer to compare it with.

507
00:35:20,000 --> 00:35:26,000
What is it going to be at pi,
zero?

508
00:35:23,000 --> 00:35:29,000
The answer is J zero is
now going to be --

509
00:35:30,000 --> 00:35:36,000
Well, everything is the same.
Zero, one, negative one.

510
00:35:34,000 --> 00:35:40,000
And here cosine of pi is
negative one,

511
00:35:38,000 --> 00:35:44,000
so negative two times negative
one is two.

512
00:35:41,000 --> 00:35:47,000
Everything is the same,
except there is a two there now

513
00:35:46,000 --> 00:35:52,000
instead of negative two.
Lambda squared plus lambda,

514
00:35:51,000 --> 00:35:57,000
the determinant,
is now negative two.

515
00:35:54,000 --> 00:36:00,000
And this factors into lambda
plus two times

516
00:36:00,000 --> 00:36:06,000
lambda minus one.

517
00:36:08,000 --> 00:36:14,000
So lambda equals one.
The corresponding eigenvector.

518
00:36:12,000 --> 00:36:18,000
I subtract one here,
so the equation is minus a1

519
00:36:17,000 --> 00:36:23,000
plus a2 is zero.

520
00:36:20,000 --> 00:36:26,000
The solution is one,
one, e to the t.

521
00:36:24,000 --> 00:36:30,000
And for the other one,
it's lambda is equal to

522
00:36:28,000 --> 00:36:34,000
negative two.
This is the sort of stuff you

523
00:36:33,000 --> 00:36:39,000
can do, so I am doing it fast.
Zero minus negative two is two.

524
00:36:38,000 --> 00:36:44,000
So the equation is 2a1 plus a2
equals zero.

525
00:36:43,000 --> 00:36:49,000
And the solution is now,
I give a1 the value one,

526
00:36:48,000 --> 00:36:54,000
a2 will be negative two,
and that is times e to the

527
00:36:52,000 --> 00:36:58,000
minus 2t.

528
00:37:00,000 --> 00:37:06,000
Well, what does the thing then
actually look like?

529
00:37:05,000 --> 00:37:11,000
What I am now going to do is,
I drew a picture before,

530
00:37:11,000 --> 00:37:17,000
that spiral picture we had
before of the way the thing

531
00:37:17,000 --> 00:37:23,000
looked at the point zero,
zero.

532
00:37:22,000 --> 00:37:28,000
So at the point pi,
zero how does it

533
00:37:27,000 --> 00:37:33,000
look, now?
Well, it looks like the origin,

534
00:37:32,000 --> 00:37:38,000
but I am thinking of it really
as the point pi,

535
00:37:37,000 --> 00:37:43,000
zero.
In other words,

536
00:37:39,000 --> 00:37:45,000
I am thinking of a linear
variable change sliding along

537
00:37:44,000 --> 00:37:50,000
the axis so that the point pi,
zero now looks like

538
00:37:50,000 --> 00:37:56,000
the origin.
If I do that then those two

539
00:37:53,000 --> 00:37:59,000
basic solutions,
there is the one,

540
00:37:56,000 --> 00:38:02,000
one solution which
is going out that way.

541
00:38:03,000 --> 00:38:09,000
And it is going out this way.
But the other guy is coming in

542
00:38:07,000 --> 00:38:13,000
along the vector one,
negative two.

543
00:38:09,000 --> 00:38:15,000
So one, negative two looks like
this.

544
00:38:12,000 --> 00:38:18,000
This guy is coming in at a
somewhat sharper angle,

545
00:38:15,000 --> 00:38:21,000
coming in because it is e to
the negative 2t.

546
00:38:19,000 --> 00:38:25,000
And we recognize this,
of course, as a saddle.

547
00:38:22,000 --> 00:38:28,000
And I would complete the
trajectories by putting in some

548
00:38:26,000 --> 00:38:32,000
of the typical saddle lines like
that.

549
00:38:30,000 --> 00:38:36,000
Now, I say that,
too, gives a picture of what is

550
00:38:36,000 --> 00:38:42,000
happening to the pendulum near
that point.

551
00:38:41,000 --> 00:38:47,000
Let's, for example,
look here.

552
00:38:44,000 --> 00:38:50,000
What is happening here?
This is theta,

553
00:38:49,000 --> 00:38:55,000
or really it is theta minus pi,
is this axis.

554
00:38:55,000 --> 00:39:01,000
In other words,
this will be zero.

555
00:39:01,000 --> 00:39:07,000
When theta is pi this will
correspond to the point zero.

556
00:39:08,000 --> 00:39:14,000
Here is omega.
What is theta doing?

557
00:39:12,000 --> 00:39:18,000
Starting up there this
represents a value of theta a

558
00:39:18,000 --> 00:39:24,000
little bit less than pi,
a little bit to the negative.

559
00:39:25,000 --> 00:39:31,000
A little bit less than pi.
Here is pi, so a little bit

560
00:39:33,000 --> 00:39:39,000
less than pi is over here.
A little bit less than pi.

561
00:39:40,000 --> 00:39:46,000
A little bit less.
And omega is zero.

562
00:39:44,000 --> 00:39:50,000
What happened?
Theta started decreasing ever

563
00:39:50,000 --> 00:39:56,000
more rapidly so that the omega
was zero here,

564
00:39:56,000 --> 00:40:02,000
now omega is negative and gets
much more negative.

565
00:40:04,000 --> 00:40:10,000
In other words,
both theta decreases from that

566
00:40:09,000 --> 00:40:15,000
point and omega decreases also.
What happened here?

567
00:40:14,000 --> 00:40:20,000
Here, theta is a little bit
more than pi.

568
00:40:19,000 --> 00:40:25,000
Now it is a little bit more
than pi.

569
00:40:23,000 --> 00:40:29,000
Theta now increases until it
gets to 2pi and then oscillates

570
00:40:29,000 --> 00:40:35,000
around 2pi.
So theta increases.

571
00:40:33,000 --> 00:40:39,000
And omega increases,
too, because the angular

572
00:40:37,000 --> 00:40:43,000
velocity started at zero but,
as theta gets more positive,

573
00:40:42,000 --> 00:40:48,000
omega gets positive,
too, because theta is

574
00:40:46,000 --> 00:40:52,000
increasing.
So omega increases and theta

575
00:40:49,000 --> 00:40:55,000
increases and it goes off like
that.

576
00:40:52,000 --> 00:40:58,000
Well, the final step is to put
them all together to be the big

577
00:40:58,000 --> 00:41:04,000
picture.
Here we are for three,

578
00:41:03,000 --> 00:41:09,000
let's say, the big picture.
Plot trajectories around each

579
00:41:12,000 --> 00:41:18,000
critical point and then add
some.

580
00:41:17,000 --> 00:41:23,000
It's that last step that can
cause you a little grief,

581
00:41:25,000 --> 00:41:31,000
but we will see how it works
out.

582
00:41:32,000 --> 00:41:38,000
Add some more,
according to your best

583
00:41:34,000 --> 00:41:40,000
judgment.
Let's make a big picture now of

584
00:41:37,000 --> 00:41:43,000
our pendulum the way it
apparently ought to look.

585
00:41:41,000 --> 00:41:47,000
Nice big axis since we are
going to accommodate a lot of

586
00:41:45,000 --> 00:41:51,000
critical points here.
Let's put in some critical

587
00:41:48,000 --> 00:41:54,000
points.
Here is the origin.

588
00:41:50,000 --> 00:41:56,000
And now here is the one at pi,
let's say, here is one at 2pi.

589
00:41:55,000 --> 00:42:01,000
I am going to add some of these
others, 3pi, 4pi.

590
00:42:00,000 --> 00:42:06,000
I won't in their values.
You can figure out what I mean.

591
00:42:05,000 --> 00:42:11,000
Zero here.
And then here it will be

592
00:42:08,000 --> 00:42:14,000
negative pi, and here negative
2pi.

593
00:42:11,000 --> 00:42:17,000
I guess I can stop there.
That is the theta axis,

594
00:42:15,000 --> 00:42:21,000
and here is the omega axis.
And now, at each one of these,

595
00:42:21,000 --> 00:42:27,000
I stay nearby and I draw the
linear trajectories,

596
00:42:25,000 --> 00:42:31,000
the trajectories of the
corresponding linearized system.

597
00:42:32,000 --> 00:42:38,000
We decided here that the spiral
went clockwise.

598
00:42:35,000 --> 00:42:41,000
Now, this point is physically,
of course, the same as that

599
00:42:40,000 --> 00:42:46,000
one.
But mathematically,

600
00:42:42,000 --> 00:42:48,000
the Jacobian matrix is the same
also because if theta is 2pi the

601
00:42:47,000 --> 00:42:53,000
cosine of 2pi is also one.
And this is the same matrix.

602
00:42:51,000 --> 00:42:57,000
So the analysis is identical.
And, therefore,

603
00:42:54,000 --> 00:43:00,000
this point will also correspond
to a counterclockwise spiral,

604
00:42:59,000 --> 00:43:05,000
as will this one.
I am sorry, a clockwise spiral.

605
00:43:04,000 --> 00:43:10,000
Here, too, all of these points
are the same.

606
00:43:08,000 --> 00:43:14,000
The behavior near them,
clockwise spirals everywhere.

607
00:43:12,000 --> 00:43:18,000
How about the other ones?
Well, the other ones correspond

608
00:43:17,000 --> 00:43:23,000
to these saddles,
so let's draw them efficiently

609
00:43:21,000 --> 00:43:27,000
by doing the same thing on every
one, mass production of saddles.

610
00:43:26,000 --> 00:43:32,000
There.
And these guys go out.

611
00:43:30,000 --> 00:43:36,000
And the other guys come in,
etc.

612
00:43:35,000 --> 00:43:41,000
And so here we have a little
bit there, a little here,

613
00:43:44,000 --> 00:43:50,000
here, there,
everywhere, etc.

614
00:43:58,000 --> 00:44:04,000
Now what?
Now you pray for inspiration.

615
00:44:01,000 --> 00:44:07,000
And what you have to do is add
trajectories that are compatible

616
00:44:08,000 --> 00:44:14,000
with the ones you have.
Let's start with this guy.

617
00:44:13,000 --> 00:44:19,000
Where is it going?
Well, a trajectory either goes

618
00:44:18,000 --> 00:44:24,000
off to infinity,
but generally they get trapped

619
00:44:22,000 --> 00:44:28,000
around critical points.
This guy must be surely doing

620
00:44:28,000 --> 00:44:34,000
this.
How about this one?

621
00:44:32,000 --> 00:44:38,000
Yeah, sure.
How about this one?

622
00:44:35,000 --> 00:44:41,000
Why not?
How about that one?

623
00:44:38,000 --> 00:44:44,000
Yeah.
But notice you are in trouble

624
00:44:42,000 --> 00:44:48,000
when two arrows you want to put
in are near each other and going

625
00:44:49,000 --> 00:44:55,000
in opposite directions.
That you cannot have.

626
00:44:54,000 --> 00:45:00,000
Continuity forbids it.
But notice if I,

627
00:44:59,000 --> 00:45:05,000
for example,
had gotten these eigenlines

628
00:45:01,000 --> 00:45:07,000
wrong, if I made this come in
and those go out because I made

629
00:45:05,000 --> 00:45:11,000
a simple error in drawing the
thing, I would have said but I

630
00:45:08,000 --> 00:45:14,000
cannot draw this picture because
this spiral wants to go that way

631
00:45:12,000 --> 00:45:18,000
but this, right next to it,
wants to go the other way.

632
00:45:16,000 --> 00:45:22,000
That is the way you would know
if you made a mistake.

633
00:45:19,000 --> 00:45:25,000
If you didn't make a mistake
you won't have any trouble

634
00:45:22,000 --> 00:45:28,000
filling these things out.
The directions of motions of

635
00:45:26,000 --> 00:45:32,000
the spirals, everything will be
compatible.

636
00:45:30,000 --> 00:45:36,000
Okay.
What is this guy doing?

637
00:45:33,000 --> 00:45:39,000
Oh, well, it must be joining up
with that.

638
00:45:37,000 --> 00:45:43,000
How about this one?
Well, it must be coming back

639
00:45:42,000 --> 00:45:48,000
there.
How about this one?

640
00:45:45,000 --> 00:45:51,000
Well, trajectories cannot
cross.

641
00:45:48,000 --> 00:45:54,000
This guy cannot cross so it
must be doing this.

642
00:45:53,000 --> 00:45:59,000
All right.
What is that trajectory?

643
00:45:57,000 --> 00:46:03,000
Starts zero.
Omega, on the other hand,

644
00:46:03,000 --> 00:46:09,000
is big and positive.
Omega big and positive.

645
00:46:10,000 --> 00:46:16,000
[LAUGHTER] I am scared.
Omega starts.

646
00:46:15,000 --> 00:46:21,000
Theta is zero.
Omega big and positive.

647
00:46:21,000 --> 00:46:27,000
It went around,
slowed up, but continued beyond

648
00:46:28,000 --> 00:46:34,000
pi.
And, in fact,

649
00:46:30,000 --> 00:46:36,000
went too far.
It continued to go here and

650
00:46:33,000 --> 00:46:39,000
then finally wound around that
one.

651
00:46:36,000 --> 00:46:42,000
Now do you see why we had to
have all the critical points?

652
00:46:40,000 --> 00:46:46,000
You have to have all the
critical points.

653
00:46:42,000 --> 00:46:48,000
And not just the two physicals
ones because the other critical

654
00:46:47,000 --> 00:46:53,000
points are necessary to describe
a complicated motion that goes

655
00:46:51,000 --> 00:46:57,000
round and round and round until
finally it crashes.

656
00:46:56,000 --> 00:47:02,000
You are going to practice
drawing these pictures and

657
00:46:59,000 --> 00:47:05,000
interpreting them in recitation
tomorrow.