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00:00:08,000 --> 00:00:14,000
Everything I say today is going
to be for n-by-n systems,

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00:00:12,000 --> 00:00:18,000
but for your calculations and
the exams two-by-two will be

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00:00:17,000 --> 00:00:23,000
good enough.
Our system looks like that.

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00:00:20,000 --> 00:00:26,000
Notice I am talking today about
the homogeneous system,

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00:00:25,000 --> 00:00:31,000
not the inhomogenous system.
So, homogenous.

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00:00:36,000 --> 00:00:42,000
And we have so far two basic
methods of solving it.

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00:00:40,000 --> 00:00:46,000
The first one,
on which we spent the most

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00:00:43,000 --> 00:00:49,000
time, is the method of where you
calculate the eigenvalues of the

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00:00:48,000 --> 00:00:54,000
matrix, the eigenvectors,
and put them together to make

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00:00:53,000 --> 00:00:59,000
the general solution.
So eigenvalues,

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00:00:56,000 --> 00:01:02,000
e-vectors and so on.
The second method,

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which I gave you last time,
I called "royal road," simply

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00:01:04,000 --> 00:01:10,000
calculates the matrix e to the
At and says that the

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00:01:08,000 --> 00:01:14,000
solution is e to the At times x
zero,

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00:01:11,000 --> 00:01:17,000
the initial condition.
That is very elegant.

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The only problem is that to
calculate the matrix e to the

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00:01:18,000 --> 00:01:24,000
At, although sometimes you can
do it by its definition as an

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00:01:22,000 --> 00:01:28,000
infinite series,
most of the time the only way

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00:01:26,000 --> 00:01:32,000
to calculate the matrix e to the
At is by using the fundamental

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00:01:30,000 --> 00:01:36,000
matrix.
In other words,

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00:01:33,000 --> 00:01:39,000
the normal way of doing it is
you have to calculate it as the

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00:01:37,000 --> 00:01:43,000
fundamental matrix time
normalized at zero.

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So, as I explained at the end
of last time and you practiced

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00:01:44,000 --> 00:01:50,000
in the recitations,
you have to find the

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00:01:46,000 --> 00:01:52,000
fundamental matrix,
which, of course,

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00:01:49,000 --> 00:01:55,000
you have to do by eigenvalues
and eigenvectors.

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And then you multiply it by its
value at zero,

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00:01:55,000 --> 00:02:01,000
inverse.
And that, by magic,

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00:01:57,000 --> 00:02:03,000
turns out to be the same as the
exponential matrix.

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00:02:02,000 --> 00:02:08,000
But, of course,
there has been no gain in

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00:02:05,000 --> 00:02:11,000
simplicity or no gain in ease of
calculation.

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The only difference is that the
language has been changed.

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Now, today is going to be
devoted to yet another method

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which saves no work at all and
only amounts to a change of

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language.
The only reason I give it to

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00:02:23,000 --> 00:02:29,000
you is because I have been
begged by various engineering

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00:02:28,000 --> 00:02:34,000
departments to do so --
-- because that is the language

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they use.
In other words,

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00:02:35,000 --> 00:02:41,000
each person who solves systems,
some like to use fundamental

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00:02:39,000 --> 00:02:45,000
matrices, some just calculate,
some immediately convert the

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system by elimination into a
single higher order equation

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because they are more
comfortable with that.

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Some, especially if they are
writing papers,

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00:02:55,000 --> 00:03:01,000
they talk exponential matrices.
But there are a certain number

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00:03:01,000 --> 00:03:07,000
of engineers and scientists who
talk decoupling,

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00:03:05,000 --> 00:03:11,000
express the problem and the
answer in terms of decoupling.

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00:03:09,000 --> 00:03:15,000
And that is,
therefore, what I have to

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00:03:12,000 --> 00:03:18,000
explain to you today.
So, the third method,

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00:03:16,000 --> 00:03:22,000
today's method,
I stress is really no more than

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00:03:20,000 --> 00:03:26,000
a change of language.
And I feel a little guilty

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00:03:23,000 --> 00:03:29,000
about the whole business.
Instead of going more deeply

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00:03:29,000 --> 00:03:35,000
into studying these equations,
what I am doing is like giving

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00:03:33,000 --> 00:03:39,000
a language course and teaching
you how to say hello and

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00:03:37,000 --> 00:03:43,000
good-bye in French,
German, Spanish,

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00:03:40,000 --> 00:03:46,000
and Italian.
It is not going very deeply

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into any of those languages,
but you are going into the

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outside world,
where people will speak these

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things.
Here is an introduction to the

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language of decoupling in which
for some people is the exclusive

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00:03:57,000 --> 00:04:03,000
language in which they talk
about systems.

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Now, I think the best way to,
well, in a general way,

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00:04:06,000 --> 00:04:12,000
what you try to do is as
follows.

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You try to introduce new
variables.

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00:04:13,000 --> 00:04:19,000
You make a change of variables.
I am going to do it two-by-two

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just to save a lot of writing
out.

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And it's going to be a linear
change of variables because we

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00:04:27,000 --> 00:04:33,000
are interested in linear
systems.

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00:04:32,000 --> 00:04:38,000
The problem is to find u and v
such that something wonderful

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00:04:37,000 --> 00:04:43,000
happens, such that when you make
the change of variables to

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00:04:43,000 --> 00:04:49,000
express this system in terms of
u and v it becomes decoupled.

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And that means the system turns
into a system which looks like u

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00:04:56,000 --> 00:05:02,000
prime equals k1 times u
and v prime equals k2

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00:05:01,000 --> 00:05:07,000
times v.
Such a system is called

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decoupled.
Why?

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00:05:08,000 --> 00:05:14,000
Well, a normal system is called
coupled.

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00:05:11,000 --> 00:05:17,000
Let's write out what it would
be.

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00:05:14,000 --> 00:05:20,000
Well, let's not write that.
You know what it looks like.

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00:05:18,000 --> 00:05:24,000
This is decoupled because it is
not really a system at all.

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00:05:23,000 --> 00:05:29,000
It is just two first-order
equations sitting side by side

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and having nothing whatever to
do with each other.

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00:05:34,000 --> 00:05:40,000
This is two problems from the
first day of the term.

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00:05:38,000 --> 00:05:44,000
It is not one problem from the
next to last day of the term,

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00:05:44,000 --> 00:05:50,000
in other words.
To solve this all you say is u

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00:05:48,000 --> 00:05:54,000
is equal to some constant times
e to the k1 t and v

85
00:05:54,000 --> 00:06:00,000
equals another constant times e
to the k2 t.

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00:06:00,000 --> 00:06:06,000
Coupled means that the x and y
occur in both equations on the

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00:06:04,000 --> 00:06:10,000
right-hand side.
And, therefore,

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00:06:06,000 --> 00:06:12,000
you cannot solve separately for
x and y, you must solve together

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00:06:10,000 --> 00:06:16,000
for both of them.
Here I can solve separately for

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00:06:14,000 --> 00:06:20,000
u and v and, therefore,
the system has been decoupled.

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00:06:17,000 --> 00:06:23,000
Now, obviously,
if you can do that it's an

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00:06:20,000 --> 00:06:26,000
enormous advantage,
not just to the ease of

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00:06:23,000 --> 00:06:29,000
solution, because you can write
down the solution immediately,

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but because something physical
must be going on there.

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00:06:33,000 --> 00:06:39,000
There must be some insight.
There ought to be some physical

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00:06:37,000 --> 00:06:43,000
reason for these new variables.
Now, that is where I plan to

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00:06:42,000 --> 00:06:48,000
start with.
My plan for the lecture is

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00:06:45,000 --> 00:06:51,000
first to work out,
in some detail,

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00:06:48,000 --> 00:06:54,000
a specific example where
decoupling is done to show how

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00:06:52,000 --> 00:06:58,000
that leads to the solution.
And then we will go back and

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00:06:57,000 --> 00:07:03,000
see how to do it in general --
-- because you will see,

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as I do the decoupling in this
particular example,

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00:07:05,000 --> 00:07:11,000
that that particular method,
though it is suggested,

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00:07:09,000 --> 00:07:15,000
will not work in general.
I would need a more general

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00:07:13,000 --> 00:07:19,000
method.
But let's first go to the

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example.
It is a slight modification of

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00:07:17,000 --> 00:07:23,000
one you should have done in
recitation.

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00:07:20,000 --> 00:07:26,000
I don't think I worked one of
these in the lecture,

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but to describe it I have to
draw two views of it to make

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sure you know exactly what I am
talking about.

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00:07:32,000 --> 00:07:38,000
Sometimes it is called the two
compartment ice cube tray

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00:07:35,000 --> 00:07:41,000
problem, a very old-fashion type
of ice cube tray.

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00:07:38,000 --> 00:07:44,000
Not a modern one that is all
plastic where there is no

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00:07:42,000 --> 00:07:48,000
leaking from one compartment to
another.

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00:07:44,000 --> 00:07:50,000
The old kind of ice cube trays,
there were compartments and

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00:07:48,000 --> 00:07:54,000
these were metal separated and
you leveled the liquid because

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00:07:52,000 --> 00:07:58,000
it could leak through the bottom
that didn't go right to the

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bottom.
If you don't know what I am

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00:07:58,000 --> 00:08:04,000
talking about it makes no
difference.

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00:08:02,000 --> 00:08:08,000
This is the side view.
This is meant to be twice as

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00:08:06,000 --> 00:08:12,000
long.
But, to make it quite clear,

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I will draw the top view of
this thing.

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You have to imagine this is a
rectangle, all the sides are

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parallel and everything.
This is one and this is two.

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All I am trying to say is that
the cross-sectional area of

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these two chambers,
this one has twice the

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00:08:29,000 --> 00:08:35,000
cross-sectional area of this
one.

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So I will write a two here and
I will write a one there.

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Of course, it is this hole here
through which everything leaks.

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I am going to let x be the
height of this liquid,

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the water here,
and y the height of the water

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in that chamber.
Obviously, as time goes by,

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they both reach the same height
because of somebody's law.

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Now, what is the system of
differential equations that

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controls this?
Well, the essential thing is

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00:09:04,000 --> 00:09:10,000
the flow rate through here.
That flow rate through the hole

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00:09:11,000 --> 00:09:17,000
in units, let's say,
in liters per second.

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00:09:16,000 --> 00:09:22,000
Just so you understand,
I am talking about the volume

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00:09:22,000 --> 00:09:28,000
of liquid.
I am not talking about the

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velocity.
That is proportional to the

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area of the hole.
So the cross-sectional area of

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00:09:37,000 --> 00:09:43,000
the hole.
And it is also times the

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00:09:40,000 --> 00:09:46,000
velocity of the flow,
but the velocity of the flow

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depends upon the pressure
difference.

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And that pressure difference
depends upon the difference in

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height.
All those are various people's

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laws.
So times the height difference.

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00:10:00,000 --> 00:10:06,000
Of course, you have to get the
sign right.

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00:10:03,000 --> 00:10:09,000
I have just pointed out the
height difference is

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00:10:06,000 --> 00:10:12,000
proportional to the pressure at
the hole.

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00:10:10,000 --> 00:10:16,000
And it is that pressure at the
hole that determines the

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00:10:14,000 --> 00:10:20,000
velocity with which the fluid
flows through.

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00:10:17,000 --> 00:10:23,000
Where does this all produce our
equations?

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00:10:21,000 --> 00:10:27,000
Well, x prime is equal to,
therefore, some constant,

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depending on the area of the
hole and this constant of

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00:10:29,000 --> 00:10:35,000
proportionality with the
pressure and the units and

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00:10:33,000 --> 00:10:39,000
everything else times the
pressure difference I am talking

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about.
Well, if fluid is going to flow

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00:10:43,000 --> 00:10:49,000
in this direction that must mean
the y height is higher than the

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x height.
So, to make x prime positive,

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it should be y minus x here.

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Now, the y prime is different.
Because, again,

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00:11:01,000 --> 00:11:07,000
the rate of fluid flow is
determined.

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00:11:03,000 --> 00:11:09,000
This time, if y prime is
positive, if this is rising,

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00:11:08,000 --> 00:11:14,000
as it will be in this case,
it's because the fluid is

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00:11:12,000 --> 00:11:18,000
flowing in that direction.
It is because x is higher than

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00:11:16,000 --> 00:11:22,000
y.
So this should be the same

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00:11:18,000 --> 00:11:24,000
constant x minus y.
But notice that right-hand side

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00:11:23,000 --> 00:11:29,000
is the rate at which fluid is
flowing into this tank.

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That is not the rate at which y
is changing.

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00:11:32,000 --> 00:11:38,000
It is the rate at which 2y is
changing.

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Why isn't there a constant
here?

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There is.
It's one.

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00:11:38,000 --> 00:11:44,000
That is the one,
this one cross-section.

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The area here is one and the
cross-sectional area here is

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00:11:45,000 --> 00:11:51,000
two.
And that is the reason for the

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00:11:47,000 --> 00:11:53,000
one here and the two here,
because we are interested in

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the rate at which fluid is being
added to this,

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00:11:54,000 --> 00:12:00,000
which is only related to the
height, the rate at which the

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00:11:58,000 --> 00:12:04,000
height is rising if you take
into account the cross-sectional

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00:12:03,000 --> 00:12:09,000
area.
So there is the system.

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00:12:07,000 --> 00:12:13,000
In order to use nothing but
integers here,

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00:12:11,000 --> 00:12:17,000
I am going to take c equals to
two, so I don't have to put in

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00:12:17,000 --> 00:12:23,000
halves.
The final system is x prime

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00:12:21,000 --> 00:12:27,000
equals minus 2x,
you have to write them in the

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00:12:25,000 --> 00:12:31,000
correct order,
and y prime equals,

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00:12:29,000 --> 00:12:35,000
the twos cancel because c is
two, is x minus y.

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00:12:35,000 --> 00:12:41,000
So there is our system.
Now the problem is I want to

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00:12:38,000 --> 00:12:44,000
solve it by decoupling it.
I want, in other words,

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00:12:42,000 --> 00:12:48,000
to find new variables,
u and v, which are more natural

191
00:12:46,000 --> 00:12:52,000
to the problem than the x and y
that are so natural to the

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00:12:50,000 --> 00:12:56,000
problem that the new system will
just consistently be two

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00:12:54,000 --> 00:13:00,000
side-by-side equations instead
of the single equation.

194
00:12:59,000 --> 00:13:05,000
The question is,
what should u and v be?

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00:13:01,000 --> 00:13:07,000
Now, the difference between
what I am going to do now and

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00:13:05,000 --> 00:13:11,000
what I am going to do later in
the period is later in the

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00:13:08,000 --> 00:13:14,000
period I will give you a
systematic way of finding what u

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00:13:12,000 --> 00:13:18,000
and v should be.
Now we are going to psyche out

199
00:13:15,000 --> 00:13:21,000
what they should be in the way
in which people who solve

200
00:13:18,000 --> 00:13:24,000
systems often do.
I am going to use the fact that

201
00:13:21,000 --> 00:13:27,000
this is not just an abstract
system of equations.

202
00:13:24,000 --> 00:13:30,000
It comes from some physical
problem.

203
00:13:28,000 --> 00:13:34,000
And I ask, is there some system
of variables,

204
00:13:31,000 --> 00:13:37,000
which somehow go more deeply
into the structure of what's

205
00:13:36,000 --> 00:13:42,000
going on here than the naīve
variables, which simply tell me

206
00:13:41,000 --> 00:13:47,000
how high the two tank levels
are?

207
00:13:44,000 --> 00:13:50,000
That is the obvious thing I can
see, but there are some

208
00:13:48,000 --> 00:13:54,000
variables that go more deeply.
Now, one of them is sort of

209
00:13:53,000 --> 00:13:59,000
obvious and suggested both the
form of the equation and by

210
00:13:58,000 --> 00:14:04,000
this.
Simply, the difference in

211
00:14:01,000 --> 00:14:07,000
heights is, in some ways,
a more natural variable because

212
00:14:05,000 --> 00:14:11,000
that is directly related to the
pressure difference,

213
00:14:09,000 --> 00:14:15,000
which is directly related to
the velocity of flow.

214
00:14:12,000 --> 00:14:18,000
They will differ by just
constant.

215
00:14:14,000 --> 00:14:20,000
I am going to call that the
second variable,

216
00:14:17,000 --> 00:14:23,000
or the difference in height
let's call it.

217
00:14:20,000 --> 00:14:26,000
That's x minus y.
That is a very natural variable

218
00:14:23,000 --> 00:14:29,000
for the problem.
The question is,

219
00:14:25,000 --> 00:14:31,000
what should the other one be?
Now you sort of stare at that

220
00:14:31,000 --> 00:14:37,000
for a while until it occurs to
you that something is constant.

221
00:14:35,000 --> 00:14:41,000
What is constant in this
problem?

222
00:14:38,000 --> 00:14:44,000
Well, the tank is sitting
there, that is constant.

223
00:14:42,000 --> 00:14:48,000
But what thing,
which might be a variable,

224
00:14:45,000 --> 00:14:51,000
clearly must be a constant?
It will be the total amount of

225
00:14:49,000 --> 00:14:55,000
water in the two tanks.
These things vary,

226
00:14:52,000 --> 00:14:58,000
but the total amount of water
stays the same because it is a

227
00:14:57,000 --> 00:15:03,000
homogenous problem.
No water is coming in from the

228
00:15:02,000 --> 00:15:08,000
outside, and none is leaving the
tanks through a little hole.

229
00:15:08,000 --> 00:15:14,000
Okay.
What is the expression for the

230
00:15:11,000 --> 00:15:17,000
total amount of water in the
tanks?

231
00:15:14,000 --> 00:15:20,000
x plus 2y.
Therefore, that is a natural

232
00:15:19,000 --> 00:15:25,000
variable also.
It is independent of this one.

233
00:15:23,000 --> 00:15:29,000
It is not a simple multiply of
it.

234
00:15:26,000 --> 00:15:32,000
It is a really different
variable.

235
00:15:31,000 --> 00:15:37,000
This variable represents the
total amount of liquid in the

236
00:15:36,000 --> 00:15:42,000
two tanks.
This represents the pressure up

237
00:15:40,000 --> 00:15:46,000
to a constant factor.
It is proportional to the

238
00:15:45,000 --> 00:15:51,000
pressure at the hole.
Okay.

239
00:15:48,000 --> 00:15:54,000
Now what I am going to do is
say this is my change of

240
00:15:53,000 --> 00:15:59,000
variable.
Now let's plug in and see what

241
00:15:57,000 --> 00:16:03,000
happens to the system when I
plug in these two variables.

242
00:16:04,000 --> 00:16:10,000
And how do I do that?
Well, I want to substitute and

243
00:16:09,000 --> 00:16:15,000
get the new system.
The new system,

244
00:16:13,000 --> 00:16:19,000
or rather the old system,
but what makes it new is in

245
00:16:19,000 --> 00:16:25,000
terms of u and v.
What will that be?

246
00:16:22,000 --> 00:16:28,000
Well, u prime is x prime plus
2y prime.

247
00:16:30,000 --> 00:16:36,000
But I know what x prime plus 2y
prime is because I

248
00:16:35,000 --> 00:16:41,000
can calculate it for this.
What will it be?

249
00:16:40,000 --> 00:16:46,000
x prime plus 2y prime is
negative 2x plus twice y prime,

250
00:16:45,000 --> 00:16:51,000
so it's plus 2x,
which is zero.

251
00:16:48,000 --> 00:16:54,000
And how about these two?
2y minus twice this,

252
00:16:52,000 --> 00:16:58,000
because I want this plus twice
that, so it 2y minus 2y,

253
00:16:57,000 --> 00:17:03,000
again, zero.
The right-hand side becomes

254
00:17:02,000 --> 00:17:08,000
zero after I calculate x prime
plus 2y.

255
00:17:06,000 --> 00:17:12,000
So that is zero.
That would just,

256
00:17:09,000 --> 00:17:15,000
of course, clear.
Now, that makes sense,

257
00:17:12,000 --> 00:17:18,000
of course.
Since the total amount is

258
00:17:15,000 --> 00:17:21,000
constant, that says that u prime
is zero.

259
00:17:19,000 --> 00:17:25,000
Okay.
What is v prime?

260
00:17:21,000 --> 00:17:27,000
v prime is x prime minus y
prime.

261
00:17:25,000 --> 00:17:31,000
What is that?
Well, once again we have to

262
00:17:31,000 --> 00:17:37,000
calculate.
x prime minus y prime is minus

263
00:17:36,000 --> 00:17:42,000
2x minus x, which is minus 3x,
and 2y minus negative y,

264
00:17:43,000 --> 00:17:49,000
which makes plus 3y.
All right.

265
00:17:46,000 --> 00:17:52,000
What is the system?
The system is u prime equals

266
00:17:52,000 --> 00:17:58,000
zero and v prime
equals minus three times x minus

267
00:18:00,000 --> 00:18:06,000
y. But x minus y is v.

268
00:18:07,000 --> 00:18:13,000
In other words,

269
00:18:09,000 --> 00:18:15,000
these new two variables
decouple the system.

270
00:18:14,000 --> 00:18:20,000
And we got them,
as scientists often do,

271
00:18:18,000 --> 00:18:24,000
by physical considerations.
These variables go more deeply

272
00:18:24,000 --> 00:18:30,000
into what is going on in that
system of two tanks than simply

273
00:18:31,000 --> 00:18:37,000
the two heights,
which are too obvious as

274
00:18:35,000 --> 00:18:41,000
variables.
All right.

275
00:18:38,000 --> 00:18:44,000
What is the solution?
Well, the solution is,

276
00:18:42,000 --> 00:18:48,000
u equals a constant and v is
equal to?

277
00:18:45,000 --> 00:18:51,000
Well, the solution to this
equation is a different

278
00:18:50,000 --> 00:18:56,000
arbitrary constant from that
one.

279
00:18:53,000 --> 00:18:59,000
These are side-by-side
equations that have nothing

280
00:18:57,000 --> 00:19:03,000
whatever to do with each other,
remember?

281
00:19:02,000 --> 00:19:08,000
Times e to the minus 3t.

282
00:19:05,000 --> 00:19:11,000
Now, there are two options.
Either one leaves the solution

283
00:19:10,000 --> 00:19:16,000
in terms of those new variables,
saying they are more natural to

284
00:19:15,000 --> 00:19:21,000
the problem, but sometimes,
of course, one wants the answer

285
00:19:20,000 --> 00:19:26,000
in terms of the old one.
But, if you do that,

286
00:19:24,000 --> 00:19:30,000
then you have to solve that.
In order to save a little time,

287
00:19:29,000 --> 00:19:35,000
since this is purely linear
algebra, I am going to write --

288
00:19:36,000 --> 00:19:42,000
Instead of taking two minutes
to actually do the calculation

289
00:19:39,000 --> 00:19:45,000
in front of you,
I will just write down what the

290
00:19:42,000 --> 00:19:48,000
answer is --

291
00:19:53,000 --> 00:19:59,000
-- in terms of u and x and y.
In other words,

292
00:19:56,000 --> 00:20:02,000
this is a perfectly good way to
leave the answer if you are

293
00:20:02,000 --> 00:20:08,000
allowed to do it.
But if somebody says they want

294
00:20:06,000 --> 00:20:12,000
the answer in terms of x and y,
well, you have to give them

295
00:20:10,000 --> 00:20:16,000
what they are paying for.
In terms of x and y,

296
00:20:13,000 --> 00:20:19,000
you have first to solve those
equations backwards for x and y

297
00:20:17,000 --> 00:20:23,000
in terms of u and v in which
case you will get x equals

298
00:20:21,000 --> 00:20:27,000
one-third of u plus 2v.

299
00:20:24,000 --> 00:20:30,000
Use the inverse matrix or just
do elimination,

300
00:20:27,000 --> 00:20:33,000
whatever you usually like to
do.

301
00:20:31,000 --> 00:20:37,000
And the other one will be
one-third of u minus v.

302
00:20:41,000 --> 00:20:47,000
And then, if you substitute in,

303
00:20:47,000 --> 00:20:53,000
you will see what you will get
is one-third of c1.

304
00:20:56,000 --> 00:21:02,000
Sorry.
u is c1.

305
00:21:15,000 --> 00:21:21,000
c1 plus 2 c2 e to the negative
3t.

306
00:21:19,000 --> 00:21:25,000
And this is one-third of c1
minus c2 e to the minus 3t.

307
00:21:24,000 --> 00:21:30,000
And so, the final solution is,

308
00:21:28,000 --> 00:21:34,000
in terms of the way we usually
write out the answer,

309
00:21:33,000 --> 00:21:39,000
x will be what?
Well, it will be one-third c1

310
00:21:38,000 --> 00:21:44,000
times the eigenvector one,
one plus one-third times c2

311
00:21:43,000 --> 00:21:49,000
times the eigenvector two,
negative one times e to the

312
00:21:49,000 --> 00:21:55,000
minus 3t.
That is the solution written

313
00:21:54,000 --> 00:22:00,000
out in terms of x and y either
as a vector in the usual way or

314
00:22:00,000 --> 00:22:06,000
separately in terms of x and y.
But, notice,

315
00:22:05,000 --> 00:22:11,000
in order to do that you have to
have these backwards equations.

316
00:22:10,000 --> 00:22:16,000
In other words,
I need the equations in that

317
00:22:13,000 --> 00:22:19,000
form.
I need the equations because

318
00:22:16,000 --> 00:22:22,000
they tell me what the new
variables are.

319
00:22:19,000 --> 00:22:25,000
But I also have to have the
equations the other way in order

320
00:22:24,000 --> 00:22:30,000
to get the solution in terms of
x and y, finally.

321
00:22:27,000 --> 00:22:33,000
Okay.
That was all an example.

322
00:22:31,000 --> 00:22:37,000
For the rest of the period,
I would like to show you the

323
00:22:35,000 --> 00:22:41,000
general method of doing the same
thing which does not depend upon

324
00:22:40,000 --> 00:22:46,000
being clever about the choice of
the new variables.

325
00:22:43,000 --> 00:22:49,000
And then, at the very end of
the period, I will apply the

326
00:22:48,000 --> 00:22:54,000
general method to this problem
to see whether we get the same

327
00:22:52,000 --> 00:22:58,000
answer or not.
What is the general method?

328
00:22:55,000 --> 00:23:01,000
Our problem is the decouple.
Now, the first thing is you

329
00:23:00,000 --> 00:23:06,000
cannot always decouple.
To decouple the eigenvalues

330
00:23:05,000 --> 00:23:11,000
must all be real and
non-defective.

331
00:23:09,000 --> 00:23:15,000
In other words,
if they are repeated they must

332
00:23:13,000 --> 00:23:19,000
be complete.
You must have enough

333
00:23:16,000 --> 00:23:22,000
independent eigenvectors.
So they must be real and

334
00:23:21,000 --> 00:23:27,000
complete.
If repeated,

335
00:23:23,000 --> 00:23:29,000
they must be complete.
They must not be defective.

336
00:23:30,000 --> 00:23:36,000
As I told you at the time when
we studied complete and

337
00:23:34,000 --> 00:23:40,000
incomplete, the most common case
in which this occurs is when the

338
00:23:40,000 --> 00:23:46,000
matrix is symmetric.
If the matrix is real and

339
00:23:44,000 --> 00:23:50,000
symmetric then you can always
decouple the system.

340
00:23:49,000 --> 00:23:55,000
That is a very important
theorem, particularly since many

341
00:23:54,000 --> 00:24:00,000
of the equilibrium problems
normally lead to symmetric

342
00:23:59,000 --> 00:24:05,000
matrices and are solved by
decoupling.

343
00:24:04,000 --> 00:24:10,000
Okay.
So what are we looking for?

344
00:24:17,000 --> 00:24:23,000
We are assuming this and we
need it.

345
00:24:19,000 --> 00:24:25,000
In general, otherwise,
you cannot decouple if you have

346
00:24:23,000 --> 00:24:29,000
complex eigenvalues and you
cannot decouple if you have

347
00:24:27,000 --> 00:24:33,000
defective eigenvalues.

348
00:24:34,000 --> 00:24:40,000
Well, what are we looking for?
We are looking for new

349
00:24:39,000 --> 00:24:45,000
variables.
u, v equals a1,

350
00:24:42,000 --> 00:24:48,000
b1, a2, b2 times the x,
y.

351
00:24:49,000 --> 00:24:55,000
And this matrix is called D,
the decoupling matrix and is

352
00:24:56,000 --> 00:25:02,000
what we are looking for.
How do I choose those new

353
00:25:01,000 --> 00:25:07,000
variables u and v when I don't
have any physical considerations

354
00:25:06,000 --> 00:25:12,000
to guide me as I did before?
Now, the key is to look instead

355
00:25:11,000 --> 00:25:17,000
at what you are going to need.
Remember, we are changing

356
00:25:15,000 --> 00:25:21,000
variables.
And, as I told you from the

357
00:25:18,000 --> 00:25:24,000
first days of the term,
when you change variables look

358
00:25:22,000 --> 00:25:28,000
at what you are going to need to
substitute in to make the change

359
00:25:27,000 --> 00:25:33,000
of variables.
Don't just start writing

360
00:25:32,000 --> 00:25:38,000
equations.
What we are going to need to

361
00:25:35,000 --> 00:25:41,000
plug into that system and change
it to the (u,

362
00:25:39,000 --> 00:25:45,000
v) coordinates is not u and v
in terms of x and y.

363
00:25:44,000 --> 00:25:50,000
What we need is x and y in
terms of u and v to do the

364
00:25:49,000 --> 00:25:55,000
substitution.
What we need is the inverse of

365
00:25:52,000 --> 00:25:58,000
this.
So, in order to do the

366
00:25:55,000 --> 00:26:01,000
substitution,
what we need is (x,

367
00:25:58,000 --> 00:26:04,000
y).
Oops.

368
00:26:00,000 --> 00:26:06,000
Let's call them prime.
Let's call these a1,

369
00:26:04,000 --> 00:26:10,000
b1, a2, b2 because these are
going to be much more important

370
00:26:09,000 --> 00:26:15,000
to the problem than the other
ones.

371
00:26:12,000 --> 00:26:18,000
Okay.
I am going to,

372
00:26:13,000 --> 00:26:19,000
I should call this matrix D
inverse, that would be a

373
00:26:18,000 --> 00:26:24,000
sensible thing to call it.
Since this is the important

374
00:26:22,000 --> 00:26:28,000
matrix, this is the one we are
going to need to do the

375
00:26:27,000 --> 00:26:33,000
substitution,
I am going to give it another

376
00:26:31,000 --> 00:26:37,000
letter instead.
And the letter that comes after

377
00:26:37,000 --> 00:26:43,000
D is E.
Now, E is an excellent choice

378
00:26:40,000 --> 00:26:46,000
because it is also the first
letter of the word eigenvector.

379
00:26:46,000 --> 00:26:52,000
And the point is the matrix E,
which is going to work,

380
00:26:51,000 --> 00:26:57,000
is the matrix whose columns are
the two eigenvectors.

381
00:27:08,000 --> 00:27:14,000
The columns are the two
eigenvectors.

382
00:27:11,000 --> 00:27:17,000
Now, even if you didn't know
anything that would be

383
00:27:16,000 --> 00:27:22,000
practically the only reasonable
choice anybody could make.

384
00:27:21,000 --> 00:27:27,000
What are we looking for?
To make a linear change of

385
00:27:26,000 --> 00:27:32,000
variables like this really means
to pick new i and j vectors.

386
00:27:33,000 --> 00:27:39,000
You know, from the first days
of 18.02, what you want is a new

387
00:27:37,000 --> 00:27:43,000
coordinate system in the plane.
And the coordinate system in

388
00:27:41,000 --> 00:27:47,000
the plane is determined as soon
as you tell what the new i is

389
00:27:46,000 --> 00:27:52,000
and what the new j is in the new
system.

390
00:27:48,000 --> 00:27:54,000
To establish a linear change of
coordinates amounts to picking

391
00:27:53,000 --> 00:27:59,000
two new vectors that are going
to play the role of i and j

392
00:27:57,000 --> 00:28:03,000
instead of the old i and the old
j.

393
00:28:01,000 --> 00:28:07,000
Okay, so pick two vectors which
somehow are important to this

394
00:28:07,000 --> 00:28:13,000
matrix.
Well, there are only two,

395
00:28:11,000 --> 00:28:17,000
the eigenvectors.
What else could they possibly

396
00:28:17,000 --> 00:28:23,000
be? Now, what is the relation?

397
00:28:20,000 --> 00:28:26,000
I say with this,
what happens is I say that

398
00:28:25,000 --> 00:28:31,000
alpha one corresponds,
and alpha two,

399
00:28:29,000 --> 00:28:35,000
these are vectors in the
xy-system.

400
00:28:35,000 --> 00:28:41,000
Well, if I change the
coordinates to u and v,

401
00:28:38,000 --> 00:28:44,000
in the uv-system they will
correspond to the vectors one,

402
00:28:42,000 --> 00:28:48,000
zero. In other words,

403
00:28:45,000 --> 00:28:51,000
the vector that we would
normally call i in the u,

404
00:28:48,000 --> 00:28:54,000
v system.
And this one will correspond to

405
00:28:52,000 --> 00:28:58,000
the vector zero, one.

406
00:28:54,000 --> 00:29:00,000
Now, if you don't believe that
I will calculate it for you.

407
00:29:00,000 --> 00:29:06,000
The calculation is trivial.
Look.

408
00:29:04,000 --> 00:29:10,000
What have we got?
(x, y) equals a1,

409
00:29:09,000 --> 00:29:15,000
b1, a2, b2.

410
00:29:13,000 --> 00:29:19,000
This is the column vector alpha
one.

411
00:29:18,000 --> 00:29:24,000
This is the column vector alpha
two.

412
00:29:22,000 --> 00:29:28,000
Now, here is u and v.
Suppose I make u and v equal to

413
00:29:30,000 --> 00:29:36,000
one, zero, what happens to x and
y?

414
00:29:35,000 --> 00:29:41,000
Your matrix multiply.
One, zero.

415
00:29:38,000 --> 00:29:44,000
So a1 plus zero,
b1 plus zero.

416
00:29:45,000 --> 00:29:51,000
It corresponds to the column
vector (a1, b1).

417
00:29:50,000 --> 00:29:56,000
And in the same way zero,
one corresponds to

418
00:29:57,000 --> 00:30:03,000
(a2, b2).

419
00:30:05,000 --> 00:30:11,000
Just by matrix multiplication.
And that shows that these

420
00:30:12,000 --> 00:30:18,000
correspond.
In the uv-system the two

421
00:30:16,000 --> 00:30:22,000
eigenvectors are now called i
and j.

422
00:30:21,000 --> 00:30:27,000
Well, that looks very
promising, but the program now

423
00:30:27,000 --> 00:30:33,000
is to do the substitution to
substitute into the system x

424
00:30:34,000 --> 00:30:40,000
prime equals Ax and
see if it is decoupled in the

425
00:30:42,000 --> 00:30:48,000
uv-coordinates.
Now, I don't dare let you do

426
00:30:47,000 --> 00:30:53,000
this by yourself because you
will run into trouble.

427
00:30:51,000 --> 00:30:57,000
Nothing is going to happen.
You will just get a mess and

428
00:30:54,000 --> 00:31:00,000
will say I must be missing
something.

429
00:30:57,000 --> 00:31:03,000
And that is because you are
missing something.

430
00:31:01,000 --> 00:31:07,000
What you are missing,
and this is a good occasion to

431
00:31:05,000 --> 00:31:11,000
tell you, is that,
in general, three-quarters of

432
00:31:09,000 --> 00:31:15,000
the civilized world does not
introduce eigenvalues and

433
00:31:14,000 --> 00:31:20,000
eigenvectors the way you learn
them in 18.03.

434
00:31:18,000 --> 00:31:24,000
They use a different definition
that is identical.

435
00:31:22,000 --> 00:31:28,000
I mean it is equivalent.
The concept is the same,

436
00:31:27,000 --> 00:31:33,000
but it looks a little
different.

437
00:31:31,000 --> 00:31:37,000
Our definition is what?
Well, what is an eigenvalue and

438
00:31:35,000 --> 00:31:41,000
eigenvector?
The basic thing is this

439
00:31:37,000 --> 00:31:43,000
equation.

440
00:31:46,000 --> 00:31:52,000
This is a two-by-two matrix,
right?

441
00:31:48,000 --> 00:31:54,000
This is a column vector with
two entries.

442
00:31:51,000 --> 00:31:57,000
The product has to be a column
vector with two entries,

443
00:31:55,000 --> 00:32:01,000
but both entries are supposed
to be zero so I will write it

444
00:31:59,000 --> 00:32:05,000
this way.
This way first defines what an

445
00:32:03,000 --> 00:32:09,000
eigenvalue is.
It is something that makes the

446
00:32:06,000 --> 00:32:12,000
determinant zero.
And then it defines what an

447
00:32:09,000 --> 00:32:15,000
eigenvector is.
It is, then,

448
00:32:11,000 --> 00:32:17,000
a solution to the system that
you can get because the

449
00:32:15,000 --> 00:32:21,000
determinant is zero.
Now, that is not what most

450
00:32:19,000 --> 00:32:25,000
people do.
What most people do is the

451
00:32:21,000 --> 00:32:27,000
following.
They write this equation

452
00:32:24,000 --> 00:32:30,000
differently by having something
on both sides.

453
00:32:29,000 --> 00:32:35,000
Using the distributive law,
what goes on the left side is A

454
00:32:33,000 --> 00:32:39,000
alpha one.
What is that?

455
00:32:35,000 --> 00:32:41,000
That is a column vector with
two entries.

456
00:32:38,000 --> 00:32:44,000
What goes on the right?
Well, lambda one times the

457
00:32:42,000 --> 00:32:48,000
identity times alpha one.
Now, the identity matrix times

458
00:32:47,000 --> 00:32:53,000
anything just reproduces what
was there.

459
00:32:50,000 --> 00:32:56,000
There is no difference between
writing the identity times alpha

460
00:32:55,000 --> 00:33:01,000
one and just alpha one all by
itself.

461
00:33:00,000 --> 00:33:06,000
So that is what I am going to
do.

462
00:33:02,000 --> 00:33:08,000
This is the definition of
eigenvalue and eigenvector that

463
00:33:07,000 --> 00:33:13,000
all the other people use.
Most linear algebra books use

464
00:33:12,000 --> 00:33:18,000
this definition,
or most books use a different

465
00:33:16,000 --> 00:33:22,000
approach and say,
here is an eigenvalue and an

466
00:33:20,000 --> 00:33:26,000
eigenvector.
And it requires them to define

467
00:33:23,000 --> 00:33:29,000
them in the opposite order.
First what alpha one is and

468
00:33:28,000 --> 00:33:34,000
then what lambda one is.
See, I don't have any

469
00:33:33,000 --> 00:33:39,000
determinant now.
So what is the definition?

470
00:33:36,000 --> 00:33:42,000
And they like it because it has
a certain geometric flavor that

471
00:33:40,000 --> 00:33:46,000
this one lacks entirely.
This is good for solving

472
00:33:43,000 --> 00:33:49,000
differential equations,
which is why we are using it in

473
00:33:47,000 --> 00:33:53,000
18.03, but this has a certain
geometric content.

474
00:33:50,000 --> 00:33:56,000
This way thinks of A as a
linear transformation of the

475
00:33:54,000 --> 00:34:00,000
plane, a shearing of the plane.
You take the plane and do

476
00:33:57,000 --> 00:34:03,000
something to it.
Or, you squish it like that.

477
00:34:02,000 --> 00:34:08,000
Or, you rotate it.
That's okay,

478
00:34:04,000 --> 00:34:10,000
too.
And the matrix defines a linear

479
00:34:07,000 --> 00:34:13,000
transformation to the plane,
every vector goes to another

480
00:34:11,000 --> 00:34:17,000
vector.
The question it asks is,

481
00:34:14,000 --> 00:34:20,000
is there a vector which is
taken by this linear

482
00:34:18,000 --> 00:34:24,000
transformation and just left
alone or stretched,

483
00:34:21,000 --> 00:34:27,000
is kept in the same direction
but stretched?

484
00:34:25,000 --> 00:34:31,000
Or, maybe its direction is
reversed and it is stretched or

485
00:34:29,000 --> 00:34:35,000
it shrunk.
But, in general,

486
00:34:33,000 --> 00:34:39,000
if there are real eigenvalues
there will be such vectors that

487
00:34:38,000 --> 00:34:44,000
are just left in the same
direction but just stretched or

488
00:34:43,000 --> 00:34:49,000
shrunk.
And what is the lambda?

489
00:34:46,000 --> 00:34:52,000
The lambda then is the amount
by which they are stretched or

490
00:34:51,000 --> 00:34:57,000
shrunk, the factor.
This way, first we have to find

491
00:34:55,000 --> 00:35:01,000
the vector, which is left
essentially unchanged,

492
00:34:59,000 --> 00:35:05,000
and then the number here that
goes with it is the stretching

493
00:35:04,000 --> 00:35:10,000
factor or the shrinking factor.
But the end result is the pair

494
00:35:11,000 --> 00:35:17,000
alpha one and lambda one,
regardless of which order you

495
00:35:15,000 --> 00:35:21,000
find them, satisfied the same
equation.

496
00:35:18,000 --> 00:35:24,000
Now, a consequence of this
definition we are going to need

497
00:35:22,000 --> 00:35:28,000
in the calculation that I am
going to do in just a moment.

498
00:35:26,000 --> 00:35:32,000
Let me calculate that out.
What I want to do is calculate

499
00:35:31,000 --> 00:35:37,000
the matrix A times E.
I am going to need to calculate

500
00:35:36,000 --> 00:35:42,000
that.
Now, what is that?

501
00:35:38,000 --> 00:35:44,000
Remember, E is the matrix whose
columns are the eigenvectors.

502
00:35:43,000 --> 00:35:49,000
That is the matrix alpha one,
alpha two.

503
00:35:47,000 --> 00:35:53,000
Now, what is this?
Well, in both Friday's lecture

504
00:35:51,000 --> 00:35:57,000
and Monday's lecture,
I used the fact that if you do

505
00:35:55,000 --> 00:36:01,000
a multiplication like that it is
the same thing as doing the

506
00:36:01,000 --> 00:36:07,000
multiplication A alpha one and
putting it in the first column.

507
00:36:08,000 --> 00:36:14,000
And then A alpha two is the
column vector that goes in the

508
00:36:13,000 --> 00:36:19,000
second column.
But what is this?

509
00:36:16,000 --> 00:36:22,000
This is lambda one alpha one.
And this is lambda two alpha

510
00:36:22,000 --> 00:36:28,000
two by this other definition of
eigenvalue and eigenvector.

511
00:36:28,000 --> 00:36:34,000
And what is this?
Can I write this in terms of

512
00:36:34,000 --> 00:36:40,000
matrices?
Yes indeed I can.

513
00:36:36,000 --> 00:36:42,000
This is the matrix alpha one,
alpha two times this matrix

514
00:36:42,000 --> 00:36:48,000
lambda one, lambda two,
zero, zero.

515
00:36:46,000 --> 00:36:52,000
Check it out.
Lambda one plus zero,

516
00:36:50,000 --> 00:36:56,000
lambda one times this thing
plus zero, the first entry is

517
00:36:56,000 --> 00:37:02,000
exactly that.
And the same way the second

518
00:37:01,000 --> 00:37:07,000
column doing the same
calculation is exactly this.

519
00:37:06,000 --> 00:37:12,000
What is that?
That is e times this matrix

520
00:37:10,000 --> 00:37:16,000
lambda one, zero,
zero, lambda two.

521
00:37:14,000 --> 00:37:20,000
Okay.
We are almost finished now.

522
00:37:17,000 --> 00:37:23,000
Now we can carry out our work.
We are going to do the

523
00:37:22,000 --> 00:37:28,000
substitution.
I start with a system.

524
00:37:26,000 --> 00:37:32,000
Remember where we are.
I am starting with this system.

525
00:37:33,000 --> 00:37:39,000
I am going to make the
substitution x equal to this

526
00:37:39,000 --> 00:37:45,000
matrix E, whose columns are the
eigenvectors.

527
00:37:45,000 --> 00:37:51,000
I am in introducing,
in other words,

528
00:37:49,000 --> 00:37:55,000
new variables u and v according
to that thing.

529
00:37:55,000 --> 00:38:01,000
u is the column vector,
u and v.

530
00:38:00,000 --> 00:38:06,000
And x, as usual,

531
00:38:03,000 --> 00:38:09,000
is the column vector x and y.
So I am going to plug it in.

532
00:38:08,000 --> 00:38:14,000
Okay.
Let's plug it in.

533
00:38:10,000 --> 00:38:16,000
What do I get?
I take the derivative.

534
00:38:13,000 --> 00:38:19,000
E is a constant matrix so that
makes E times u prime,

535
00:38:17,000 --> 00:38:23,000
is equal to A times,
x is E times u again.

536
00:38:21,000 --> 00:38:27,000
Now, at this point,

537
00:38:24,000 --> 00:38:30,000
you would be stuck,
except I calculated for you A

538
00:38:28,000 --> 00:38:34,000
times E is E times that funny
diagonal matrix with the

539
00:38:32,000 --> 00:38:38,000
lambdas.
So this is E times that funny

540
00:38:38,000 --> 00:38:44,000
matrix of the lambdas,
the eigenvalues,

541
00:38:41,000 --> 00:38:47,000
and still the u at the end of
it.

542
00:38:45,000 --> 00:38:51,000
So where are we?
E times u prime equals E times

543
00:38:50,000 --> 00:38:56,000
this thing.
Well, multiply both sides by E

544
00:38:54,000 --> 00:39:00,000
inverse and you can cancel them
out.

545
00:38:59,000 --> 00:39:05,000
And so the end result is that
after you have made the

546
00:39:04,000 --> 00:39:10,000
substitution in terms of the new
variables u, what you get is u

547
00:39:10,000 --> 00:39:16,000
prime equals lambda one,
lambda two, zero,

548
00:39:14,000 --> 00:39:20,000
zero times u.
Let's write that out in terms

549
00:39:18,000 --> 00:39:24,000
of a system.
This is u prime is equal to,

550
00:39:22,000 --> 00:39:28,000
well, this is u,
v here.

551
00:39:24,000 --> 00:39:30,000
It is lambda one times u plus
zero times v.

552
00:39:30,000 --> 00:39:36,000
And the other one is v prime
equals zero times u plus lambda

553
00:39:40,000 --> 00:39:46,000
two times v.
We are decoupled.

554
00:39:45,000 --> 00:39:51,000
In just one sentence you would
say --

555
00:39:52,000 --> 00:39:58,000
In other words,
if you were reading a book that

556
00:39:55,000 --> 00:40:01,000
sort of assumed you knew what
was going on,

557
00:39:59,000 --> 00:40:05,000
all it would say is as usual.
That is to make you feel bad.

558
00:40:04,000 --> 00:40:10,000
Or, as is well-known to make
you feel even worse.

559
00:40:08,000 --> 00:40:14,000
Or, the system is decoupled by
choosing as the new basis for

560
00:40:13,000 --> 00:40:19,000
the system the eigenvectors of
the matrix and in terms of the

561
00:40:18,000 --> 00:40:24,000
resulting new coordinates,
the decoupled system will be

562
00:40:23,000 --> 00:40:29,000
the following where the
constants are the eigenvalues.

563
00:40:29,000 --> 00:40:35,000
And so the solution will be u
equals c1 times e to the lambda1

564
00:40:33,000 --> 00:40:39,000
t and v
is equal to c2 times e to the

565
00:40:38,000 --> 00:40:44,000
lambda2 t.

566
00:40:40,000 --> 00:40:46,000
Of course, if you want it back
in terms now of x and y,

567
00:40:44,000 --> 00:40:50,000
you will have to go back to
here, to these equations and

568
00:40:48,000 --> 00:40:54,000
then plug in for u and v what
they are.

569
00:40:51,000 --> 00:40:57,000
And then you will get the
answer in terms of x and y.

570
00:40:55,000 --> 00:41:01,000
Okay.
We have just enough time to

571
00:40:59,000 --> 00:41:05,000
actually carry out this little
program.

572
00:41:04,000 --> 00:41:10,000
It takes a lot longer to derive
than it does actually to do,

573
00:41:10,000 --> 00:41:16,000
so let's do it for this system
that we were talking about

574
00:41:16,000 --> 00:41:22,000
before.
Decouple the system,

575
00:41:19,000 --> 00:41:25,000
x, y prime equals the matrixes
negative two,

576
00:41:24,000 --> 00:41:30,000
two, one, negative one.

577
00:41:36,000 --> 00:41:42,000
Okay.
What do I do?

578
00:41:37,000 --> 00:41:43,000
Well, I first have to calculate
the eigenvalues in the

579
00:41:42,000 --> 00:41:48,000
eigenvectors,
so the Ev's and Ev's.

580
00:41:45,000 --> 00:41:51,000
The characteristic equation is
lambda squared.

581
00:41:49,000 --> 00:41:55,000
The trace is negative three,
but you have to change the

582
00:41:54,000 --> 00:42:00,000
sign.
The determinant is two minus

583
00:41:57,000 --> 00:42:03,000
two, so that is zero.
There is no constant term here.

584
00:42:02,000 --> 00:42:08,000
It is zero.
That is the characteristic

585
00:42:05,000 --> 00:42:11,000
equation.
The roots are obviously lambda

586
00:42:08,000 --> 00:42:14,000
equals zero, lambda equals
negative three.

587
00:42:12,000 --> 00:42:18,000
And what are the eigenvectors
that go with that?

588
00:42:15,000 --> 00:42:21,000
With lambda equals zero goes
the eigenvector,

589
00:42:19,000 --> 00:42:25,000
minus two.
Well, I subtract zero here,

590
00:42:22,000 --> 00:42:28,000
so the equation I have to solve
is minus 2 a1 plus --

591
00:42:28,000 --> 00:42:34,000
I am not going to write 2 a2,
which is what you have been

592
00:42:32,000 --> 00:42:38,000
writing up until now.
The reason is because I ran

593
00:42:36,000 --> 00:42:42,000
into trouble with the notation
and I had to use,

594
00:42:40,000 --> 00:42:46,000
as the eigenvector,
not a1, a2 but a1,

595
00:42:43,000 --> 00:42:49,000
b1.
So it should be a b1 here,

596
00:42:46,000 --> 00:42:52,000
not the a2 that you are used
to.

597
00:42:48,000 --> 00:42:54,000
The solution,
therefore, is alpha one equals

598
00:42:52,000 --> 00:42:58,000
one, one.
And for lambda equals negative

599
00:42:55,000 --> 00:43:01,000
three, the corresponding
eigenvector this time will be --

600
00:43:02,000 --> 00:43:08,000
Now I have to subtract negative
three from here,

601
00:43:06,000 --> 00:43:12,000
so negative two minus negative
three makes one.

602
00:43:10,000 --> 00:43:16,000
That is a1 plus 2 b1 equals
zero,

603
00:43:15,000 --> 00:43:21,000
a logical choice for the
eigenvector here.

604
00:43:19,000 --> 00:43:25,000
The second eigenvector would be
make b1 equal to one,

605
00:43:24,000 --> 00:43:30,000
let's say, and then a1 will be
negative two.

606
00:43:28,000 --> 00:43:34,000
Okay.
Now what do we have to do?

607
00:43:32,000 --> 00:43:38,000
Now, what we want is the matrix
E.

608
00:43:35,000 --> 00:43:41,000
The matrix E is the matrix of
eigenvectors,

609
00:43:38,000 --> 00:43:44,000
so it is the matrix one,
one, negative two,

610
00:43:42,000 --> 00:43:48,000
one.

611
00:43:44,000 --> 00:43:50,000
The next thing we want is what
the new variables u and v are.

612
00:43:50,000 --> 00:43:56,000
For that, we will need E
inverse.

613
00:43:52,000 --> 00:43:58,000
How do you calculate the
inverse of a two-by-two matrix?

614
00:43:59,000 --> 00:44:05,000
You switch the two diagonal
elements, there I have switched

615
00:44:04,000 --> 00:44:10,000
them, and you leave the other
two where they are but change

616
00:44:09,000 --> 00:44:15,000
their sign.
So it is two up here and

617
00:44:13,000 --> 00:44:19,000
negative one there.
Maybe I should make this one

618
00:44:17,000 --> 00:44:23,000
purple and then that one purple
to indicate that I have switched

619
00:44:23,000 --> 00:44:29,000
them.
I am not done yet.

620
00:44:25,000 --> 00:44:31,000
I have to divide by the
determinant.

621
00:44:30,000 --> 00:44:36,000
What is the determinant?
It is one minus negative two,

622
00:44:36,000 --> 00:44:42,000
which is three,
so I have to divide by three.

623
00:44:42,000 --> 00:44:48,000
I multiply everything here by
one-third.

624
00:44:47,000 --> 00:44:53,000
Okay.
And what is the decoupled

625
00:44:51,000 --> 00:44:57,000
system?
The new variables are u equals

626
00:44:56,000 --> 00:45:02,000
one-third.

627
00:45:08,000 --> 00:45:14,000
In other words,
the new variables are given by

628
00:45:11,000 --> 00:45:17,000
D.
It is u, v equals one,

629
00:45:13,000 --> 00:45:19,000
two, negative one, one
times one-third times x,y.

630
00:45:19,000 --> 00:45:25,000
That is the expression for u,

631
00:45:21,000 --> 00:45:27,000
v in terms of x and y.
It's this matrix D,

632
00:45:25,000 --> 00:45:31,000
the decoupling matrix which is
the one that is used.

633
00:45:30,000 --> 00:45:36,000
And that gives this system u
equals one-third of x plus 2y

634
00:45:37,000 --> 00:45:43,000
on top.
And what is the v entry?

635
00:45:44,000 --> 00:45:50,000
v is one-third of minus x plus
y.

636
00:45:51,000 --> 00:45:57,000
Now, are those the same
variables that I used before?

637
00:46:08,000 --> 00:46:14,000
Yes.
This is my new and better you,

638
00:46:10,000 --> 00:46:16,000
the one I got by just blindly
following the method instead of

639
00:46:14,000 --> 00:46:20,000
looking for physical things with
physical meaning.

640
00:46:17,000 --> 00:46:23,000
It differs from the old one
just by a constant factor.

641
00:46:21,000 --> 00:46:27,000
Now, that doesn't have any
effect on the resulting equation

642
00:46:25,000 --> 00:46:31,000
because if the old one is u
prime equals zero the new one is

643
00:46:29,000 --> 00:46:35,000
one-third u prime equals zero.
It is still the same equation,

644
00:46:34,000 --> 00:46:40,000
in other words.
And how about this one?

645
00:46:36,000 --> 00:46:42,000
This one differs from the other
one by the factor minus

646
00:46:40,000 --> 00:46:46,000
one-third.
If I multiply that v through by

647
00:46:43,000 --> 00:46:49,000
minus one-third,
I get this v.

648
00:46:45,000 --> 00:46:51,000
And, therefore,
that too does not affect the

649
00:46:47,000 --> 00:46:53,000
second equation.
I simply multiply both sides by

650
00:46:51,000 --> 00:46:57,000
minus one-third.
The new v still satisfies the

651
00:46:54,000 --> 00:47:00,000
equation minus three times v.