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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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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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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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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,

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nobody else does either.
There are very fat books in the

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library whose topic is how to
solve just a pair of equations,

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f of (x, y) equals zero,
g of (x, y) equals zero.

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And it is quite a hard problem.
It is even a hard problem by

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computer.
Because, if you know

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approximately where the solution
is going to be,

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you can make up the little
screen and then the computer

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will find it for you.
Or, even without a screen,

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it will calculate it by
Newton's method or something

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else, it will zero in.
The problem is,

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if you don't know in advance
roughly where the critical point

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is that you are looking for,
there are a lot of numbers.

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They go to infinity that way
and infinity that way.

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In general, it is almost an
impossible problem.

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The only thing that makes it
possible is that these problems

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always come from the real world
and one has some physical

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feeling, one hopes,
for where the critical point

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is.
I am going to assume breezily

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and cheerily we can solve those.
And I am only going to give you

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examples where it is possible to
solve them.

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But even there,
you have to watch out.

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There is a certain trickiness
that will be talked about in the

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recitations tomorrow.
Okay, so we found the critical

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points.
Let's do it for our example.

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Let's find the critical point.
What is the pair of equations

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we have to solve?
We have to solve the equations

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omega equals zero,
minus two sine theta minus

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00:17:53 --> 00:17:59
omega equals zero.

268
00:17:56 --> 00:18:02
Now, it is not always this
easy.

269
00:18:00 --> 00:18:06
But the solution is omega is
zero.

270
00:18:03 --> 00:18:09
If omega is zero,
then sine theta is zero,

271
00:18:07 --> 00:18:13
and sine theta is zero at the

272
00:18:12 --> 00:18:18
integral multiples of pi.
The critical points are omega

273
00:18:18 --> 00:18:24
is always zero and theta is
zero, or it could be plus or

274
00:18:24 --> 00:18:30
minus pi, plus or minus two pi
and so on.

275
00:18:30 --> 00:18:36
In other words,
there are an infinity of

276
00:18:32 --> 00:18:38
critical points.
That seems a little

277
00:18:34 --> 00:18:40
discouraging.
On the other hand,

278
00:18:36 --> 00:18:42
there are really only two.
There are really physically

279
00:18:40 --> 00:18:46
only two because omega equals
zero means the mass is not

280
00:18:44 --> 00:18:50
moving.
The angular velocity is zero,

281
00:18:46 --> 00:18:52
so it is only the theta
position which is changing.

282
00:18:49 --> 00:18:55
Now, what are the possible
theta positions?

283
00:18:52 --> 00:18:58
Well, here is our nonlinear
pendulum.

284
00:18:56 --> 00:19:02
Here is the critical point,
theta equals zero,

285
00:18:59 --> 00:19:05
omega equals zero.
Theta equals zero means the rod

286
00:19:03 --> 00:19:09
is vertical.
Omega equals zero means that it

287
00:19:07 --> 00:19:13
is not moving,
despite the fact that it is

288
00:19:10 --> 00:19:16
moving.
Theoretically it is not moving.

289
00:19:14 --> 00:19:20
Now, what is the other one?
Well, there is theta equals pi.

290
00:19:18 --> 00:19:24
Theta is now,
starting from zero and

291
00:19:21 --> 00:19:27
increasing, I hope,
through positive theta.

292
00:19:25 --> 00:19:31
And when it gets to pi,
it is sticking straight up in

293
00:19:29 --> 00:19:35
the air.
And so the claim is that

294
00:19:34 --> 00:19:40
another critical point is theta
equals pi and omega equals zero.

295
00:19:40 --> 00:19:46
In other words,
if it gets to this position,

296
00:19:44 --> 00:19:50
it starts out in this position
and stays there for all time,

297
00:19:49 --> 00:19:55
as you see.
[LAUGHTER] But my point is what

298
00:19:53 --> 00:19:59
about negative pi?
That is the same as pi.

299
00:19:57 --> 00:20:03
Two pi is the same as zero.
So physically there really are

300
00:20:03 --> 00:20:09
only two critical points,
this one and that one.

301
00:20:08 --> 00:20:14
And they obviously have
something very different about

302
00:20:13 --> 00:20:19
them.
This critical point is stable.

303
00:20:16 --> 00:20:22
If I start near there,
I approach that critical point

304
00:20:21 --> 00:20:27
in infinite time.
This one, if I start near

305
00:20:25 --> 00:20:31
there, I do not stay near there.
I always leave it.

306
00:20:31 --> 00:20:37
Of the two critical points,
physically it is clear that the

307
00:20:38 --> 00:20:44
critical point is zero,
zero and the other guys that

308
00:20:44 --> 00:20:50
look like it,
two pi zero and so on is a

309
00:20:49 --> 00:20:55
stable critical point.
Whereas, pi zero,

310
00:20:53 --> 00:20:59
when theta is sticking straight
up in the air,

311
00:20:59 --> 00:21:05
is unstable.
Now, of course,

312
00:21:03 --> 00:21:09
we will want to see that
mathematically also,

313
00:21:06 --> 00:21:12
but basically there are just
physically two point.

314
00:21:10 --> 00:21:16
There are just two critical
points.

315
00:21:12 --> 00:21:18
Well, that raises the question
what about all the others?

316
00:21:16 --> 00:21:22
As you will see,
we have to have those.

317
00:21:19 --> 00:21:25
They are an essential part of
the problem.

318
00:21:22 --> 00:21:28
They are not just redundant
baggage that is trailing along.

319
00:21:26 --> 00:21:32
They are really important.
But you will see that when we

320
00:21:33 --> 00:21:39
talk about finally how the
trajectories look and how the

321
00:21:40 --> 00:21:46
solutions look.
Now what do we do?

322
00:21:43 --> 00:21:49
Well, we found the critical
points, and now the work begins.

323
00:21:50 --> 00:21:56
Virtually all the work is in
this next step.

324
00:21:56 --> 00:22:02
What do we do?
Step two.

325
00:22:00 --> 00:22:06
I can only describe it in
general terms,

326
00:22:05 --> 00:22:11
but here is what you do.
For each critical point x zero,

327
00:22:13 --> 00:22:19
y zero, a procedure that has to
be done at each one,

328
00:22:21 --> 00:22:27
you linearize the system near
that point.

329
00:22:28 --> 00:22:34
In other words,
you may find a linear system,

330
00:22:32 --> 00:22:38
the good kind,
the kind you know about,

331
00:22:37 --> 00:22:43
which is a good approximation
to the nonlinear system at that

332
00:22:43 --> 00:22:49
critical point.
Plot the trajectories of this

333
00:22:48 --> 00:22:54
linearized system.
And you do that near the

334
00:22:53 --> 00:22:59
critical point.

335
00:22:56 --> 00:23:02

336
00:23:04 --> 00:23:10
How do you plot the
trajectories?

337
00:23:05 --> 00:23:11
Well, that you knew how to do
on Friday so I am assuming you

338
00:23:09 --> 00:23:15
still know how to do it on
Monday.

339
00:23:11 --> 00:23:17
In other words,
if the system is linear you

340
00:23:14 --> 00:23:20
know how to plot the
trajectories of it by

341
00:23:16 --> 00:23:22
calculating eigenvalues and
eigenvectors and maybe the

342
00:23:20 --> 00:23:26
direction of motion if it is a
spiral.

343
00:23:22 --> 00:23:28
I will give you a couple of
examples of that when we work

344
00:23:25 --> 00:23:31
out the pendulum.
But, on the other hand,

345
00:23:29 --> 00:23:35
how do I line arise a system?
Well, there are two methods.

346
00:23:33 --> 00:23:39
There is one method the book
gives you, which by and large I

347
00:23:37 --> 00:23:43
do not want you to use,
although I will give you an

348
00:23:41 --> 00:23:47
example of it now.
I want you to use another

349
00:23:44 --> 00:23:50
method because it is much
faster.

350
00:23:46 --> 00:23:52
Especially if you have to
handle several critical points

351
00:23:50 --> 00:23:56
it is much, much faster.
Let's first carry this out on

352
00:23:54 --> 00:24:00
an easy case,
and then I will show you how to

353
00:23:57 --> 00:24:03
do it in general.
Just this once I will use the

354
00:24:02 --> 00:24:08
book's method because I think it
is the method which would

355
00:24:08 --> 00:24:14
naturally occur to you.
Let's linearize this example at

356
00:24:13 --> 00:24:19
the point zero,
zero.

357
00:24:15 --> 00:24:21
What should be the linearized
system?

358
00:24:18 --> 00:24:24
In other words,
it's only the nonlinear terms I

359
00:24:23 --> 00:24:29
have to worry about.
Well, the minus omega is fine.

360
00:24:29 --> 00:24:35
It is that stupid sine theta
that I don't like.

361
00:24:33 --> 00:24:39
But if theta is small,
in other words,

362
00:24:37 --> 00:24:43
if I stay near zero,
I could replace sine theta by

363
00:24:42 --> 00:24:48
theta.
The linearized system is minus

364
00:24:46 --> 00:24:52
two.
You replace sine theta by

365
00:24:49 --> 00:24:55
theta, since sine theta is
approximately theta if theta is

366
00:24:55 --> 00:25:01
small, if theta is near zero.
It is the first term of its

367
00:25:02 --> 00:25:08
Taylor series you can think of,
or it's just the linear

368
00:25:06 --> 00:25:12
approximation starts out sine
theta equals theta.

369
00:25:10 --> 00:25:16
That is it.
Or, you draw a picture.

370
00:25:12 --> 00:25:18
I don't know.
There are a millions of ways to

371
00:25:16 --> 00:25:22
do it.
So we have it.

372
00:25:17 --> 00:25:23
Now what do I do?
Okay, now we will plot that.

373
00:25:21 --> 00:25:27
The matrix, let's do our little
routine, in other words.

374
00:25:25 --> 00:25:31
I am writing right to left for
no reason.

375
00:25:30 --> 00:25:36
The matrix is,
now I am just going to make

376
00:25:34 --> 00:25:40
marks on a board the way you did
on your exam,

377
00:25:39 --> 00:25:45
zero, one.
[LAUGHTER] I don't know what I

378
00:25:43 --> 00:25:49
am doing, but you know.
Negative two,

379
00:25:47 --> 00:25:53
negative one,
and then I write down the

380
00:25:51 --> 00:25:57
eigen-whatchamacallits.
Lambda squared,

381
00:25:55 --> 00:26:01
plus lambda,
minus the trace.

382
00:26:00 --> 00:26:06
The determinant is minus,
minus two, so it is plus two

383
00:26:05 --> 00:26:11
equals zero.
And then, since it doesn't

384
00:26:09 --> 00:26:15
occur to me how to factor this,
I will use the quadratic

385
00:26:14 --> 00:26:20
formula.
It is negative one plus or

386
00:26:18 --> 00:26:24
minus the square root of,
b squared minus 4ac,

387
00:26:22 --> 00:26:28
minus seven.
Complex.

388
00:26:24 --> 00:26:30
That means it is going to be a
spiral.

389
00:26:30 --> 00:26:36
I am going to get a spiral.
Will it be a source or a sink

390
00:26:34 --> 00:26:40
since they are complex roots?
Complex eigenvalues give a

391
00:26:39 --> 00:26:45
spiral.
A source or a sink?

392
00:26:41 --> 00:26:47
I tell that from the sine of
the real part.

393
00:26:44 --> 00:26:50
The real part is negative
one-half.

394
00:26:47 --> 00:26:53
Therefore, the amplitude is
shrinking like e to the minus t

395
00:26:52 --> 00:26:58
over two.
And, therefore,

396
00:26:55 --> 00:27:01
the spiral is coming into the
origin.

397
00:27:00 --> 00:27:06
It is a spiral sink since
lambda equals minus one-half

398
00:27:05 --> 00:27:11
plus some number times i.
Spiral sink.

399
00:27:10 --> 00:27:16
And the other thing to
determine is its direction of

400
00:27:15 --> 00:27:21
motion, which will be what?
I determine its direction of

401
00:27:21 --> 00:27:27
motion by putting in a single
vector from the velocity field.

402
00:27:28 --> 00:27:34
Here it is.
The vector at one,

403
00:27:32 --> 00:27:38
zero will be the
same as the first column of the

404
00:27:36 --> 00:27:42
matrix.
So that is zero,

405
00:27:38 --> 00:27:44
negative two.

406
00:27:39 --> 00:27:45
Here is a vector from the
velocity field,

407
00:27:42 --> 00:27:48
and that shows that the motion
is clockwise and is spiraling

408
00:27:46 --> 00:27:52
into the origin.
That is a picture,

409
00:27:49 --> 00:27:55
therefore, at the origin of how
that looks.

410
00:27:52 --> 00:27:58
Now, it is of the utmost
importance for your

411
00:27:55 --> 00:28:01
understanding of what comes now
that you understand in what

412
00:27:59 --> 00:28:05
sense this picture corresponds
to the physical behavior of the

413
00:28:03 --> 00:28:09
pendulum.
Let's start it over here.

414
00:28:08 --> 00:28:14
What is it doing?
That means that theta is some

415
00:28:13 --> 00:28:19
number like one,
for example.

416
00:28:16 --> 00:28:22
Let's make it smaller.
Let's say a little bit.

417
00:28:20 --> 00:28:26
And this is the omega access.
If it starts over here that

418
00:28:26 --> 00:28:32
means the angular velocity is
zero and theta is a small

419
00:28:32 --> 00:28:38
positive number.
Theta is a small positive

420
00:28:37 --> 00:28:43
number.
The angular velocity is zero.

421
00:28:40 --> 00:28:46
It's velocity zero,
theta small and positive.

422
00:28:44 --> 00:28:50
I release it and it does that.
What does this have to do with

423
00:28:50 --> 00:28:56
the spiral?
Well, the spiral is exactly a

424
00:28:54 --> 00:29:00
mathematical picture of this
motion.

425
00:28:57 --> 00:29:03
What happens?
Theta starts to decrease.

426
00:29:02 --> 00:29:08
And the angular velocity
increases but in the negative

427
00:29:06 --> 00:29:12
direction.
This is negative angular

428
00:29:08 --> 00:29:14
velocity.
Theta is decreasing.

429
00:29:11 --> 00:29:17
The angular velocity gets
bigger and bigger.

430
00:29:14 --> 00:29:20
And it is biggest,
most negative when theta is

431
00:29:18 --> 00:29:24
zero.
It has reached the vertical

432
00:29:20 --> 00:29:26
position is when the angular
velocity is biggest.

433
00:29:24 --> 00:29:30
It continues then.
Theta gets negative,

434
00:29:27 --> 00:29:33
but the angular velocity then
decreases to zero.

435
00:29:33 --> 00:29:39

436
00:29:37 --> 00:29:43
Now the angular velocity is
zero and theta is at its most

437
00:29:41 --> 00:29:47
negative, and then it reverses.
Angular velocity gets positive

438
00:29:46 --> 00:29:52
as theta increases again,
and so on.

439
00:29:48 --> 00:29:54
These represent the successive
swings back and forth.

440
00:29:52 --> 00:29:58
Notice the fact that it is
damped is reflected in the fact

441
00:29:57 --> 00:30:03
that each successive swing,
the biggest that theta gets is

442
00:30:01 --> 00:30:07
a little less than it was
before.

443
00:30:05 --> 00:30:11
In other words,
this point is not quite as far

444
00:30:08 --> 00:30:14
out as that one.
And this one isn't as far out

445
00:30:11 --> 00:30:17
as that one.
In other words,

446
00:30:13 --> 00:30:19
it is spiraling in.

447
00:30:15 --> 00:30:21

448
00:30:20 --> 00:30:26
Well, I hope that is clear
because we now have to go to the

449
00:30:26 --> 00:30:32
next critical point.
And now we have a little

450
00:30:31 --> 00:30:37
problem.
If I want to do the next

451
00:30:34 --> 00:30:40
critical point,
so what I want to do now is,

452
00:30:38 --> 00:30:44
in other words,
I want to linearize at the

453
00:30:42 --> 00:30:48
point pi zero where theta is pi
and the thing is sticking up in

454
00:30:48 --> 00:30:54
the air.
The question is,

455
00:30:50 --> 00:30:56
how am I going to do that?
This trick of replacing sine

456
00:30:55 --> 00:31:01
theta by theta,
that doesn't work at pi.

457
00:30:59 --> 00:31:05
That works at zero.

458
00:31:03 --> 00:31:09

459
00:31:08 --> 00:31:14
Now we have to go to the next
step of the method.

460
00:31:11 --> 00:31:17
The way to do this,
in general, as you will read in

461
00:31:14 --> 00:31:20
the notes because,
as I say, this is not in the

462
00:31:17 --> 00:31:23
book, is to calculate the
Jacobian.

463
00:31:20 --> 00:31:26
I mean the Jacobian matrix.
The Jacobian is the

464
00:31:23 --> 00:31:29
determinant.
I mean, before you put the two

465
00:31:26 --> 00:31:32
bars down and made a
determinant, you called it just

466
00:31:29 --> 00:31:35
the Jacobian matrix.
And the formula for it is,

467
00:31:34 --> 00:31:40
it is calculated from f and g.
The top line is the partial of

468
00:31:39 --> 00:31:45
f with respect to x and y,
and the bottom line is the

469
00:31:44 --> 00:31:50
partial of g with respect to x
and y.

470
00:31:47 --> 00:31:53
So that is the Jacobian matrix.

471
00:31:51 --> 00:31:57

472
00:31:57 --> 00:32:03
I hope I get a chance at the
end of the period to explain to

473
00:32:02 --> 00:32:08
you why, but I am most anxious
right now to at least get you

474
00:32:07 --> 00:32:13
familiar with the algorithm,
how to do it.

475
00:32:11 --> 00:32:17
The notes describe the y of it,
if we don't get a chance to get

476
00:32:17 --> 00:32:23
to it, but I hope we will.
What do you do?

477
00:32:20 --> 00:32:26
The Jacobian matrix.
You calculate it at the point x

478
00:32:25 --> 00:32:31
zero, y zero.

479
00:32:29 --> 00:32:35
I will indicate that by putting
a subscript zero on it.

480
00:32:32 --> 00:32:38
This means without the
subscript zero it is the

481
00:32:35 --> 00:32:41
Jacobian matrix calculated out
of those four partial

482
00:32:39 --> 00:32:45
derivatives.
When I put a subscript zero,

483
00:32:41 --> 00:32:47
I mean I evaluated it at the
critical point by plugging in

484
00:32:45 --> 00:32:51
each entry as a function of x
and y.

485
00:32:47 --> 00:32:53
You plug in for x equals 
x zero, y equals y zero,

486
00:32:52 --> 00:32:58
and you get a numerical
matrix.

487
00:32:54 --> 00:33:00
That is the matrix which is the
matrix of the linearized system.

488
00:33:00 --> 00:33:06
This is the matrix of the
linearized system.

489
00:33:03 --> 00:33:09

490
00:33:08 --> 00:33:14
Trust me, it is.
Now, since I don't expect you

491
00:33:11 --> 00:33:17
to trust me, let's calculate it.
Here, we got the matrix another

492
00:33:16 --> 00:33:22
way, by this procedure of saying
sine theta is theta-- What would

493
00:33:21 --> 00:33:27
we have gotten if we had done it
instead by the linearized

494
00:33:25 --> 00:33:31
system?
Let's do it that way.

495
00:33:27 --> 00:33:33
Let's do it via the Jacobian.

496
00:33:31 --> 00:33:37

497
00:33:37 --> 00:33:43
I need to know the Jacobian of
the system, which I have

498
00:33:43 --> 00:33:49
conveniently covered up.
There is the system.

499
00:33:47 --> 00:33:53
The Jacobian matrix is what?
The top line,

500
00:33:52 --> 00:33:58
I take the partial derivative
of omega first with respect to

501
00:33:59 --> 00:34:05
theta and then with respect to
omega.

502
00:34:04 --> 00:34:10
I then take the second line on
the right-hand side.

503
00:34:08 --> 00:34:14
I take its partial with respect
to theta first.

504
00:34:12 --> 00:34:18
That is negative two cosine
theta.

505
00:34:16 --> 00:34:22
And, thinking ahead,
I erase the one and move it

506
00:34:20 --> 00:34:26
over a little bit.
And what is the partial of that

507
00:34:24 --> 00:34:30
thing with respect to omega?
It is negative one.

508
00:34:30 --> 00:34:36
Does everyone see how I
calculated that Jacobian matrix?

509
00:34:35 --> 00:34:41
And now I want to evaluate it.
Let's do our old case first.

510
00:34:40 --> 00:34:46
At zero, zero what 
would this have

511
00:34:45 --> 00:34:51
amounted to?
This would have given me zero,

512
00:34:49 --> 00:34:55
one.
The cosine of theta at zero is

513
00:34:52 --> 00:34:58
one, so this is negative two,
negative one.

514
00:34:56 --> 00:35:02
I'm screwed.
[LAUGHTER]

515
00:35:00 --> 00:35:06
That is the same as that.
Now you see it,

516
00:35:04 --> 00:35:10
now you don't.
We got our old answer back.

517
00:35:08 --> 00:35:14
That should give us enough
confidence to use it in the new

518
00:35:14 --> 00:35:20
case, where I don't have an old
answer to compare it with.

519
00:35:20 --> 00:35:26
What is it going to be at pi,
zero?

520
00:35:23 --> 00:35:29
The answer is J zero is
now going to be --

521
00:35:30 --> 00:35:36
Well, everything is the same.
Zero, one, negative one.

522
00:35:34 --> 00:35:40
And here cosine of pi is
negative one,

523
00:35:38 --> 00:35:44
so negative two times negative
one is two.

524
00:35:41 --> 00:35:47
Everything is the same,
except there is a two there now

525
00:35:46 --> 00:35:52
instead of negative two.
Lambda squared plus lambda,

526
00:35:51 --> 00:35:57
the determinant,
is now negative two.

527
00:35:54 --> 00:36:00
And this factors into lambda
plus two times 

528
00:36:00 --> 00:36:06
lambda minus one.

529
00:36:04 --> 00:36:10

530
00:36:08 --> 00:36:14
So lambda equals one.
The corresponding eigenvector.

531
00:36:12 --> 00:36:18
I subtract one here,
so the equation is minus a1

532
00:36:17 --> 00:36:23
plus a2 is zero.

533
00:36:20 --> 00:36:26
The solution is one,
one, e to the t.

534
00:36:24 --> 00:36:30
And for the other one,
it's lambda is equal to

535
00:36:28 --> 00:36:34
negative two.
This is the sort of stuff you

536
00:36:33 --> 00:36:39
can do, so I am doing it fast.
Zero minus negative two is two.

537
00:36:38 --> 00:36:44
So the equation is 2a1 plus a2
equals zero.

538
00:36:43 --> 00:36:49
And the solution is now,
I give a1 the value one,

539
00:36:48 --> 00:36:54
a2 will be negative two,
and that is times e to the

540
00:36:52 --> 00:36:58
minus 2t.

541
00:36:55 --> 00:37:01

542
00:37:00 --> 00:37:06
Well, what does the thing then
actually look like?

543
00:37:05 --> 00:37:11
What I am now going to do is,
I drew a picture before,

544
00:37:11 --> 00:37:17
that spiral picture we had
before of the way the thing

545
00:37:17 --> 00:37:23
looked at the point zero,
zero.

546
00:37:22 --> 00:37:28
So at the point pi,
zero how does it

547
00:37:27 --> 00:37:33
look, now?
Well, it looks like the origin,

548
00:37:32 --> 00:37:38
but I am thinking of it really
as the point pi,

549
00:37:37 --> 00:37:43
zero.
In other words,

550
00:37:39 --> 00:37:45
I am thinking of a linear
variable change sliding along

551
00:37:44 --> 00:37:50
the axis so that the point pi,
zero now looks like

552
00:37:50 --> 00:37:56
the origin.
If I do that then those two

553
00:37:53 --> 00:37:59
basic solutions,
there is the one,

554
00:37:56 --> 00:38:02
one solution which
is going out that way.

555
00:38:03 --> 00:38:09
And it is going out this way.
But the other guy is coming in

556
00:38:07 --> 00:38:13
along the vector one,
negative two.

557
00:38:09 --> 00:38:15
So one, negative two looks like
this.

558
00:38:12 --> 00:38:18
This guy is coming in at a
somewhat sharper angle,

559
00:38:15 --> 00:38:21
coming in because it is e to
the negative 2t.

560
00:38:19 --> 00:38:25
And we recognize this,
of course, as a saddle.

561
00:38:22 --> 00:38:28
And I would complete the
trajectories by putting in some

562
00:38:26 --> 00:38:32
of the typical saddle lines like
that.

563
00:38:30 --> 00:38:36
Now, I say that,
too, gives a picture of what is

564
00:38:36 --> 00:38:42
happening to the pendulum near
that point.

565
00:38:41 --> 00:38:47
Let's, for example,
look here.

566
00:38:44 --> 00:38:50
What is happening here?
This is theta,

567
00:38:49 --> 00:38:55
or really it is theta minus pi,
is this axis.

568
00:38:55 --> 00:39:01
In other words,
this will be zero.

569
00:39:01 --> 00:39:07
When theta is pi this will
correspond to the point zero.

570
00:39:08 --> 00:39:14
Here is omega.
What is theta doing?

571
00:39:12 --> 00:39:18
Starting up there this
represents a value of theta a

572
00:39:18 --> 00:39:24
little bit less than pi,
a little bit to the negative.

573
00:39:25 --> 00:39:31
A little bit less than pi.
Here is pi, so a little bit

574
00:39:33 --> 00:39:39
less than pi is over here.
A little bit less than pi.

575
00:39:40 --> 00:39:46
A little bit less.
And omega is zero.

576
00:39:44 --> 00:39:50
What happened?
Theta started decreasing ever

577
00:39:50 --> 00:39:56
more rapidly so that the omega
was zero here,

578
00:39:56 --> 00:40:02
now omega is negative and gets
much more negative.

579
00:40:04 --> 00:40:10
In other words,
both theta decreases from that

580
00:40:09 --> 00:40:15
point and omega decreases also.
What happened here?

581
00:40:14 --> 00:40:20
Here, theta is a little bit
more than pi.

582
00:40:19 --> 00:40:25
Now it is a little bit more
than pi.

583
00:40:23 --> 00:40:29
Theta now increases until it
gets to 2pi and then oscillates

584
00:40:29 --> 00:40:35
around 2pi.
So theta increases.

585
00:40:33 --> 00:40:39
And omega increases,
too, because the angular

586
00:40:37 --> 00:40:43
velocity started at zero but,
as theta gets more positive,

587
00:40:42 --> 00:40:48
omega gets positive,
too, because theta is

588
00:40:46 --> 00:40:52
increasing.
So omega increases and theta

589
00:40:49 --> 00:40:55
increases and it goes off like
that.

590
00:40:52 --> 00:40:58
Well, the final step is to put
them all together to be the big

591
00:40:58 --> 00:41:04
picture.
Here we are for three,

592
00:41:03 --> 00:41:09
let's say, the big picture.
Plot trajectories around each

593
00:41:12 --> 00:41:18
critical point and then add
some.

594
00:41:17 --> 00:41:23
It's that last step that can
cause you a little grief,

595
00:41:25 --> 00:41:31
but we will see how it works
out.

596
00:41:32 --> 00:41:38
Add some more,
according to your best

597
00:41:34 --> 00:41:40
judgment.
Let's make a big picture now of

598
00:41:37 --> 00:41:43
our pendulum the way it
apparently ought to look.

599
00:41:41 --> 00:41:47
Nice big axis since we are
going to accommodate a lot of

600
00:41:45 --> 00:41:51
critical points here.
Let's put in some critical

601
00:41:48 --> 00:41:54
points.
Here is the origin.

602
00:41:50 --> 00:41:56
And now here is the one at pi,
let's say, here is one at 2pi.

603
00:41:55 --> 00:42:01
I am going to add some of these
others, 3pi, 4pi.

604
00:42:00 --> 00:42:06
I won't in their values.
You can figure out what I mean.

605
00:42:05 --> 00:42:11
Zero here.
And then here it will be

606
00:42:08 --> 00:42:14
negative pi, and here negative
2pi.

607
00:42:11 --> 00:42:17
I guess I can stop there.
That is the theta axis,

608
00:42:15 --> 00:42:21
and here is the omega axis.
And now, at each one of these,

609
00:42:21 --> 00:42:27
I stay nearby and I draw the
linear trajectories,

610
00:42:25 --> 00:42:31
the trajectories of the
corresponding linearized system.

611
00:42:32 --> 00:42:38
We decided here that the spiral
went clockwise.

612
00:42:35 --> 00:42:41
Now, this point is physically,
of course, the same as that

613
00:42:40 --> 00:42:46
one.
But mathematically,

614
00:42:42 --> 00:42:48
the Jacobian matrix is the same
also because if theta is 2pi the

615
00:42:47 --> 00:42:53
cosine of 2pi is also one.
And this is the same matrix.

616
00:42:51 --> 00:42:57
So the analysis is identical.
And, therefore,

617
00:42:54 --> 00:43:00
this point will also correspond
to a counterclockwise spiral,

618
00:42:59 --> 00:43:05
as will this one.
I am sorry, a clockwise spiral.

619
00:43:04 --> 00:43:10
Here, too, all of these points
are the same.

620
00:43:08 --> 00:43:14
The behavior near them,
clockwise spirals everywhere.

621
00:43:12 --> 00:43:18
How about the other ones?
Well, the other ones correspond

622
00:43:17 --> 00:43:23
to these saddles,
so let's draw them efficiently

623
00:43:21 --> 00:43:27
by doing the same thing on every
one, mass production of saddles.

624
00:43:26 --> 00:43:32
There.
And these guys go out.

625
00:43:30 --> 00:43:36
And the other guys come in,
etc.

626
00:43:35 --> 00:43:41
And so here we have a little
bit there, a little here,

627
00:43:44 --> 00:43:50
here, there,
everywhere, etc.

628
00:43:50 --> 00:43:56

629
00:43:58 --> 00:44:04
Now what?
Now you pray for inspiration.

630
00:44:01 --> 00:44:07
And what you have to do is add
trajectories that are compatible

631
00:44:08 --> 00:44:14
with the ones you have.
Let's start with this guy.

632
00:44:13 --> 00:44:19
Where is it going?
Well, a trajectory either goes

633
00:44:18 --> 00:44:24
off to infinity,
but generally they get trapped

634
00:44:22 --> 00:44:28
around critical points.
This guy must be surely doing

635
00:44:28 --> 00:44:34
this.
How about this one?

636
00:44:32 --> 00:44:38
Yeah, sure.
How about this one?

637
00:44:35 --> 00:44:41
Why not?
How about that one?

638
00:44:38 --> 00:44:44
Yeah.
But notice you are in trouble

639
00:44:42 --> 00:44:48
when two arrows you want to put
in are near each other and going

640
00:44:49 --> 00:44:55
in opposite directions.
That you cannot have.

641
00:44:54 --> 00:45:00
Continuity forbids it.
But notice if I,

642
00:44:59 --> 00:45:05
for example,
had gotten these eigenlines

643
00:45:01 --> 00:45:07
wrong, if I made this come in
and those go out because I made

644
00:45:05 --> 00:45:11
a simple error in drawing the
thing, I would have said but I

645
00:45:08 --> 00:45:14
cannot draw this picture because
this spiral wants to go that way

646
00:45:12 --> 00:45:18
but this, right next to it,
wants to go the other way.

647
00:45:16 --> 00:45:22
That is the way you would know
if you made a mistake.

648
00:45:19 --> 00:45:25
If you didn't make a mistake
you won't have any trouble

649
00:45:22 --> 00:45:28
filling these things out.
The directions of motions of

650
00:45:26 --> 00:45:32
the spirals, everything will be
compatible.

651
00:45:30 --> 00:45:36
Okay.
What is this guy doing?

652
00:45:33 --> 00:45:39
Oh, well, it must be joining up
with that.

653
00:45:37 --> 00:45:43
How about this one?
Well, it must be coming back

654
00:45:42 --> 00:45:48
there.
How about this one?

655
00:45:45 --> 00:45:51
Well, trajectories cannot
cross.

656
00:45:48 --> 00:45:54
This guy cannot cross so it
must be doing this.

657
00:45:53 --> 00:45:59
All right.
What is that trajectory?

658
00:45:57 --> 00:46:03
Starts zero.
Omega, on the other hand,

659
00:46:03 --> 00:46:09
is big and positive.
Omega big and positive.

660
00:46:10 --> 00:46:16
[LAUGHTER] I am scared.
Omega starts.

661
00:46:15 --> 00:46:21
Theta is zero.
Omega big and positive.

662
00:46:21 --> 00:46:27
It went around,
slowed up, but continued beyond

663
00:46:28 --> 00:46:34
pi.
And, in fact,

664
00:46:30 --> 00:46:36
went too far.
It continued to go here and

665
00:46:33 --> 00:46:39
then finally wound around that
one.

666
00:46:36 --> 00:46:42
Now do you see why we had to
have all the critical points?

667
00:46:40 --> 00:46:46
You have to have all the
critical points.

668
00:46:42 --> 00:46:48
And not just the two physicals
ones because the other critical

669
00:46:47 --> 00:46:53
points are necessary to describe
a complicated motion that goes

670
00:46:51 --> 00:46:57
round and round and round until
finally it crashes.

671
00:46:56 --> 00:47:02
You are going to practice
drawing these pictures and

672
00:46:59 --> 00:47:05
interpreting them in recitation
tomorrow.