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Everything I say today is going
to be for n-by-n systems,
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but for your calculations and
the exams two-by-two will be
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good enough.
Our system looks like that.
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Notice I am talking today about
the homogeneous system,
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not the inhomogenous system.
So, homogenous.
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And we have so far two basic
methods of solving it.
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The first one,
on which we spent the most
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time, is the method of where you
calculate the eigenvalues of the
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matrix, the eigenvectors,
and put them together to make
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the general solution.
So eigenvalues,
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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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calculates the matrix e to the
At and says that the
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solution is e to the At times x
zero,
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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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At, although sometimes you can
do it by its definition as an
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infinite series,
most of the time the only way
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to calculate the matrix e to the
At is by using the fundamental
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matrix.
In other words,
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the normal way of doing it is
you have to calculate it as the
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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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in the recitations,
you have to find the
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fundamental matrix,
which, of course,
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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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inverse.
And that, by magic,
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turns out to be the same as the
exponential matrix.
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But, of course,
there has been no gain in
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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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you is because I have been
begged by various engineering
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departments to do so --
-- because that is the language
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they use.
In other words,
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each person who solves systems,
some like to use fundamental
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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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they talk exponential matrices.
But there are a certain number
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of engineers and scientists who
talk decoupling,
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express the problem and the
answer in terms of decoupling.
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And that is,
therefore, what I have to
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explain to you today.
So, the third method,
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today's method,
I stress is really no more than
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a change of language.
And I feel a little guilty
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about the whole business.
Instead of going more deeply
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into studying these equations,
what I am doing is like giving
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a language course and teaching
you how to say hello and
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good-bye in French,
German, Spanish,
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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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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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what you try to do is as
follows.
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You try to introduce new
variables.
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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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are interested in linear
systems.
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The problem is to find u and v
such that something wonderful
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happens, such that when you make
the change of variables to
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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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prime equals k1 times u
and v prime equals k2
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times v.
Such a system is called
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decoupled.
Why?
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Well, a normal system is called
coupled.
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Let's write out what it would
be.
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Well, let's not write that.
You know what it looks like.
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This is decoupled because it is
not really a system at all.
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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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This is two problems from the
first day of the term.
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It is not one problem from the
next to last day of the term,
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in other words.
To solve this all you say is u
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is equal to some constant times
e to the k1 t and v
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equals another constant times e
to the k2 t.
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Coupled means that the x and y
occur in both equations on the
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right-hand side.
And, therefore,
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you cannot solve separately for
x and y, you must solve together
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for both of them.
Here I can solve separately for
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u and v and, therefore,
the system has been decoupled.
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Now, obviously,
if you can do that it's an
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enormous advantage,
not just to the ease of
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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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There must be some insight.
There ought to be some physical
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reason for these new variables.
Now, that is where I plan to
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start with.
My plan for the lecture is
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first to work out,
in some detail,
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a specific example where
decoupling is done to show how
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that leads to the solution.
And then we will go back and
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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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that that particular method,
though it is suggested,
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will not work in general.
I would need a more general
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method.
But let's first go to the
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example.
It is a slight modification of
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one you should have done in
recitation.
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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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Sometimes it is called the two
compartment ice cube tray
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problem, a very old-fashion type
of ice cube tray.
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Not a modern one that is all
plastic where there is no
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leaking from one compartment to
another.
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The old kind of ice cube trays,
there were compartments and
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these were metal separated and
you leveled the liquid because
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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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talking about it makes no
difference.
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This is the side view.
This is meant to be twice as
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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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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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the flow rate through here.
That flow rate through the hole
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in units, let's say,
in liters per second.
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Just so you understand,
I am talking about the volume
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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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the hole.
And it is also times the
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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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Of course, you have to get the
sign right.
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I have just pointed out the
height difference is
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proportional to the pressure at
the hole.
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And it is that pressure at the
hole that determines the
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velocity with which the fluid
flows through.
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Where does this all produce our
equations?
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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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proportionality with the
pressure and the units and
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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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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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the rate of fluid flow is
determined.
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This time, if y prime is
positive, if this is rising,
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as it will be in this case,
it's because the fluid is
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flowing in that direction.
It is because x is higher than
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y.
So this should be the same
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constant x minus y.
But notice that right-hand side
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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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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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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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two.
And that is the reason for the
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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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which is only related to the
height, the rate at which the
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height is rising if you take
into account the cross-sectional
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area.
So there is the system.
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In order to use nothing but
integers here,
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I am going to take c equals to
two, so I don't have to put in
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halves.
The final system is x prime
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equals minus 2x,
you have to write them in the
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correct order,
and y prime equals,
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the twos cancel because c is
two, is x minus y.
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So there is our system.
Now the problem is I want to
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solve it by decoupling it.
I want, in other words,
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to find new variables,
u and v, which are more natural
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to the problem than the x and y
that are so natural to the
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problem that the new system will
just consistently be two
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side-by-side equations instead
of the single equation.
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The question is,
what should u and v be?
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Now, the difference between
what I am going to do now and
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what I am going to do later in
the period is later in the
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period I will give you a
systematic way of finding what u
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and v should be.
Now we are going to psyche out
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what they should be in the way
in which people who solve
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systems often do.
I am going to use the fact that
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this is not just an abstract
system of equations.
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It comes from some physical
problem.
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And I ask, is there some system
of variables,
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which somehow go more deeply
into the structure of what's
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going on here than the naīve
variables, which simply tell me
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how high the two tank levels
are?
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That is the obvious thing I can
see, but there are some
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variables that go more deeply.
Now, one of them is sort of
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obvious and suggested both the
form of the equation and by
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this.
Simply, the difference in
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heights is, in some ways,
a more natural variable because
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that is directly related to the
pressure difference,
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which is directly related to
the velocity of flow.
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They will differ by just
constant.
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I am going to call that the
second variable,
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or the difference in height
let's call it.
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That's x minus y.
That is a very natural variable
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for the problem.
The question is,
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what should the other one be?
Now you sort of stare at that
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for a while until it occurs to
you that something is constant.
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What is constant in this
problem?
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Well, the tank is sitting
there, that is constant.
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But what thing,
which might be a variable,
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clearly must be a constant?
It will be the total amount of
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water in the two tanks.
These things vary,
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but the total amount of water
stays the same because it is a
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homogenous problem.
No water is coming in from the
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outside, and none is leaving the
tanks through a little hole.
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Okay.
What is the expression for the
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total amount of water in the
tanks?
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x plus 2y.
Therefore, that is a natural
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variable also.
It is independent of this one.
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It is not a simple multiply of
it.
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It is a really different
variable.
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This variable represents the
total amount of liquid in the
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two tanks.
This represents the pressure up
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to a constant factor.
It is proportional to the
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pressure at the hole.
Okay.
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Now what I am going to do is
say this is my change of
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variable.
Now let's plug in and see what
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happens to the system when I
plug in these two variables.
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And how do I do that?
Well, I want to substitute and
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get the new system.
The new system,
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or rather the old system,
but what makes it new is in
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terms of u and v.
What will that be?
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Well, u prime is x prime plus
2y prime.
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But I know what x prime plus 2y
prime is because I
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can calculate it for this.
What will it be?
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x prime plus 2y prime is
negative 2x plus twice y prime,
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so it's plus 2x,
which is zero.
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And how about these two?
2y minus twice this,
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because I want this plus twice
that, so it 2y minus 2y,
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again, zero.
The right-hand side becomes
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zero after I calculate x prime
plus 2y.
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So that is zero.
That would just,
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of course, clear.
Now, that makes sense,
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of course.
Since the total amount is
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constant, that says that u prime
is zero.
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Okay.
What is v prime?
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v prime is x prime minus y
prime.
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What is that?
Well, once again we have to
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calculate.
x prime minus y prime is minus
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00:17:36 --> 00:17:42
2x minus x, which is minus 3x,
and 2y minus negative y,
266
00:17:43 --> 00:17:49
which makes plus 3y.
All right.
267
00:17:46 --> 00:17:52
What is the system?
The system is u prime equals
268
00:17:52 --> 00:17:58
zero and v prime
equals minus three times x minus
269
00:18:00 --> 00:18:06
y. But x minus y is v.
270
00:18:07 --> 00:18:13
In other words,
271
00:18:09 --> 00:18:15
these new two variables
decouple the system.
272
00:18:14 --> 00:18:20
And we got them,
as scientists often do,
273
00:18:18 --> 00:18:24
by physical considerations.
These variables go more deeply
274
00:18:24 --> 00:18:30
into what is going on in that
system of two tanks than simply
275
00:18:31 --> 00:18:37
the two heights,
which are too obvious as
276
00:18:35 --> 00:18:41
variables.
All right.
277
00:18:38 --> 00:18:44
What is the solution?
Well, the solution is,
278
00:18:42 --> 00:18:48
u equals a constant and v is
equal to?
279
00:18:45 --> 00:18:51
Well, the solution to this
equation is a different
280
00:18:50 --> 00:18:56
arbitrary constant from that
one.
281
00:18:53 --> 00:18:59
These are side-by-side
equations that have nothing
282
00:18:57 --> 00:19:03
whatever to do with each other,
remember?
283
00:19:02 --> 00:19:08
Times e to the minus 3t.
284
00:19:05 --> 00:19:11
Now, there are two options.
Either one leaves the solution
285
00:19:10 --> 00:19:16
in terms of those new variables,
saying they are more natural to
286
00:19:15 --> 00:19:21
the problem, but sometimes,
of course, one wants the answer
287
00:19:20 --> 00:19:26
in terms of the old one.
But, if you do that,
288
00:19:24 --> 00:19:30
then you have to solve that.
In order to save a little time,
289
00:19:29 --> 00:19:35
since this is purely linear
algebra, I am going to write --
290
00:19:36 --> 00:19:42
Instead of taking two minutes
to actually do the calculation
291
00:19:39 --> 00:19:45
in front of you,
I will just write down what the
292
00:19:42 --> 00:19:48
answer is --
293
00:19:43 --> 00:19:49
294
00:19:53 --> 00:19:59
-- in terms of u and x and y.
In other words,
295
00:19:56 --> 00:20:02
this is a perfectly good way to
leave the answer if you are
296
00:20:02 --> 00:20:08
allowed to do it.
But if somebody says they want
297
00:20:06 --> 00:20:12
the answer in terms of x and y,
well, you have to give them
298
00:20:10 --> 00:20:16
what they are paying for.
In terms of x and y,
299
00:20:13 --> 00:20:19
you have first to solve those
equations backwards for x and y
300
00:20:17 --> 00:20:23
in terms of u and v in which
case you will get x equals
301
00:20:21 --> 00:20:27
one-third of u plus 2v.
302
00:20:24 --> 00:20:30
Use the inverse matrix or just
do elimination,
303
00:20:27 --> 00:20:33
whatever you usually like to
do.
304
00:20:31 --> 00:20:37
And the other one will be
one-third of u minus v.
305
00:20:41 --> 00:20:47
And then, if you substitute in,
306
00:20:47 --> 00:20:53
you will see what you will get
is one-third of c1.
307
00:20:56 --> 00:21:02
Sorry.
u is c1.
308
00:21:00 --> 00:21:06
309
00:21:15 --> 00:21:21
c1 plus 2 c2 e to the negative
3t.
310
00:21:19 --> 00:21:25
And this is one-third of c1
minus c2 e to the minus 3t.
311
00:21:24 --> 00:21:30
And so, the final solution is,
312
00:21:28 --> 00:21:34
in terms of the way we usually
write out the answer,
313
00:21:33 --> 00:21:39
x will be what?
Well, it will be one-third c1
314
00:21:38 --> 00:21:44
times the eigenvector one,
one plus one-third times c2
315
00:21:43 --> 00:21:49
times the eigenvector two,
negative one times e to the
316
00:21:49 --> 00:21:55
minus 3t.
That is the solution written
317
00:21:54 --> 00:22:00
out in terms of x and y either
as a vector in the usual way or
318
00:22:00 --> 00:22:06
separately in terms of x and y.
But, notice,
319
00:22:05 --> 00:22:11
in order to do that you have to
have these backwards equations.
320
00:22:10 --> 00:22:16
In other words,
I need the equations in that
321
00:22:13 --> 00:22:19
form.
I need the equations because
322
00:22:16 --> 00:22:22
they tell me what the new
variables are.
323
00:22:19 --> 00:22:25
But I also have to have the
equations the other way in order
324
00:22:24 --> 00:22:30
to get the solution in terms of
x and y, finally.
325
00:22:27 --> 00:22:33
Okay.
That was all an example.
326
00:22:31 --> 00:22:37
For the rest of the period,
I would like to show you the
327
00:22:35 --> 00:22:41
general method of doing the same
thing which does not depend upon
328
00:22:40 --> 00:22:46
being clever about the choice of
the new variables.
329
00:22:43 --> 00:22:49
And then, at the very end of
the period, I will apply the
330
00:22:48 --> 00:22:54
general method to this problem
to see whether we get the same
331
00:22:52 --> 00:22:58
answer or not.
What is the general method?
332
00:22:55 --> 00:23:01
Our problem is the decouple.
Now, the first thing is you
333
00:23:00 --> 00:23:06
cannot always decouple.
To decouple the eigenvalues
334
00:23:05 --> 00:23:11
must all be real and
non-defective.
335
00:23:09 --> 00:23:15
In other words,
if they are repeated they must
336
00:23:13 --> 00:23:19
be complete.
You must have enough
337
00:23:16 --> 00:23:22
independent eigenvectors.
So they must be real and
338
00:23:21 --> 00:23:27
complete.
If repeated,
339
00:23:23 --> 00:23:29
they must be complete.
They must not be defective.
340
00:23:30 --> 00:23:36
As I told you at the time when
we studied complete and
341
00:23:34 --> 00:23:40
incomplete, the most common case
in which this occurs is when the
342
00:23:40 --> 00:23:46
matrix is symmetric.
If the matrix is real and
343
00:23:44 --> 00:23:50
symmetric then you can always
decouple the system.
344
00:23:49 --> 00:23:55
That is a very important
theorem, particularly since many
345
00:23:54 --> 00:24:00
of the equilibrium problems
normally lead to symmetric
346
00:23:59 --> 00:24:05
matrices and are solved by
decoupling.
347
00:24:04 --> 00:24:10
Okay.
So what are we looking for?
348
00:24:10 --> 00:24:16
349
00:24:17 --> 00:24:23
We are assuming this and we
need it.
350
00:24:19 --> 00:24:25
In general, otherwise,
you cannot decouple if you have
351
00:24:23 --> 00:24:29
complex eigenvalues and you
cannot decouple if you have
352
00:24:27 --> 00:24:33
defective eigenvalues.
353
00:24:29 --> 00:24:35
354
00:24:34 --> 00:24:40
Well, what are we looking for?
We are looking for new
355
00:24:39 --> 00:24:45
variables.
u, v equals a1,
356
00:24:42 --> 00:24:48
b1, a2, b2 times the x,
y.
357
00:24:47 --> 00:24:53
358
00:24:49 --> 00:24:55
And this matrix is called D,
the decoupling matrix and is
359
00:24:56 --> 00:25:02
what we are looking for.
How do I choose those new
360
00:25:01 --> 00:25:07
variables u and v when I don't
have any physical considerations
361
00:25:06 --> 00:25:12
to guide me as I did before?
Now, the key is to look instead
362
00:25:11 --> 00:25:17
at what you are going to need.
Remember, we are changing
363
00:25:15 --> 00:25:21
variables.
And, as I told you from the
364
00:25:18 --> 00:25:24
first days of the term,
when you change variables look
365
00:25:22 --> 00:25:28
at what you are going to need to
substitute in to make the change
366
00:25:27 --> 00:25:33
of variables.
Don't just start writing
367
00:25:32 --> 00:25:38
equations.
What we are going to need to
368
00:25:35 --> 00:25:41
plug into that system and change
it to the (u,
369
00:25:39 --> 00:25:45
v) coordinates is not u and v
in terms of x and y.
370
00:25:44 --> 00:25:50
What we need is x and y in
terms of u and v to do the
371
00:25:49 --> 00:25:55
substitution.
What we need is the inverse of
372
00:25:52 --> 00:25:58
this.
So, in order to do the
373
00:25:55 --> 00:26:01
substitution,
what we need is (x,
374
00:25:58 --> 00:26:04
y).
Oops.
375
00:26:00 --> 00:26:06
Let's call them prime.
Let's call these a1,
376
00:26:04 --> 00:26:10
b1, a2, b2 because these are
going to be much more important
377
00:26:09 --> 00:26:15
to the problem than the other
ones.
378
00:26:12 --> 00:26:18
Okay.
I am going to,
379
00:26:13 --> 00:26:19
I should call this matrix D
inverse, that would be a
380
00:26:18 --> 00:26:24
sensible thing to call it.
Since this is the important
381
00:26:22 --> 00:26:28
matrix, this is the one we are
going to need to do the
382
00:26:27 --> 00:26:33
substitution,
I am going to give it another
383
00:26:31 --> 00:26:37
letter instead.
And the letter that comes after
384
00:26:37 --> 00:26:43
D is E.
Now, E is an excellent choice
385
00:26:40 --> 00:26:46
because it is also the first
letter of the word eigenvector.
386
00:26:46 --> 00:26:52
And the point is the matrix E,
which is going to work,
387
00:26:51 --> 00:26:57
is the matrix whose columns are
the two eigenvectors.
388
00:26:58 --> 00:27:04
389
00:27:08 --> 00:27:14
The columns are the two
eigenvectors.
390
00:27:11 --> 00:27:17
Now, even if you didn't know
anything that would be
391
00:27:16 --> 00:27:22
practically the only reasonable
choice anybody could make.
392
00:27:21 --> 00:27:27
What are we looking for?
To make a linear change of
393
00:27:26 --> 00:27:32
variables like this really means
to pick new i and j vectors.
394
00:27:33 --> 00:27:39
You know, from the first days
of 18.02, what you want is a new
395
00:27:37 --> 00:27:43
coordinate system in the plane.
And the coordinate system in
396
00:27:41 --> 00:27:47
the plane is determined as soon
as you tell what the new i is
397
00:27:46 --> 00:27:52
and what the new j is in the new
system.
398
00:27:48 --> 00:27:54
To establish a linear change of
coordinates amounts to picking
399
00:27:53 --> 00:27:59
two new vectors that are going
to play the role of i and j
400
00:27:57 --> 00:28:03
instead of the old i and the old
j.
401
00:28:01 --> 00:28:07
Okay, so pick two vectors which
somehow are important to this
402
00:28:07 --> 00:28:13
matrix.
Well, there are only two,
403
00:28:11 --> 00:28:17
the eigenvectors.
What else could they possibly
404
00:28:17 --> 00:28:23
be? Now, what is the relation?
405
00:28:20 --> 00:28:26
I say with this,
what happens is I say that
406
00:28:25 --> 00:28:31
alpha one corresponds,
and alpha two,
407
00:28:29 --> 00:28:35
these are vectors in the
xy-system.
408
00:28:35 --> 00:28:41
Well, if I change the
coordinates to u and v,
409
00:28:38 --> 00:28:44
in the uv-system they will
correspond to the vectors one,
410
00:28:42 --> 00:28:48
zero. In other words,
411
00:28:45 --> 00:28:51
the vector that we would
normally call i in the u,
412
00:28:48 --> 00:28:54
v system.
And this one will correspond to
413
00:28:52 --> 00:28:58
the vector zero, one.
414
00:28:54 --> 00:29:00
Now, if you don't believe that
I will calculate it for you.
415
00:29:00 --> 00:29:06
The calculation is trivial.
Look.
416
00:29:04 --> 00:29:10
What have we got?
(x, y) equals a1,
417
00:29:09 --> 00:29:15
b1, a2, b2.
418
00:29:13 --> 00:29:19
This is the column vector alpha
one.
419
00:29:18 --> 00:29:24
This is the column vector alpha
two.
420
00:29:22 --> 00:29:28
Now, here is u and v.
Suppose I make u and v equal to
421
00:29:30 --> 00:29:36
one, zero, what happens to x and
y?
422
00:29:35 --> 00:29:41
Your matrix multiply.
One, zero.
423
00:29:38 --> 00:29:44
So a1 plus zero,
b1 plus zero.
424
00:29:45 --> 00:29:51
It corresponds to the column
vector (a1, b1).
425
00:29:50 --> 00:29:56
And in the same way zero,
one corresponds to
426
00:29:57 --> 00:30:03
(a2, b2).
427
00:30:00 --> 00:30:06
428
00:30:05 --> 00:30:11
Just by matrix multiplication.
And that shows that these
429
00:30:12 --> 00:30:18
correspond.
In the uv-system the two
430
00:30:16 --> 00:30:22
eigenvectors are now called i
and j.
431
00:30:21 --> 00:30:27
Well, that looks very
promising, but the program now
432
00:30:27 --> 00:30:33
is to do the substitution to
substitute into the system x
433
00:30:34 --> 00:30:40
prime equals Ax and
see if it is decoupled in the
434
00:30:42 --> 00:30:48
uv-coordinates.
Now, I don't dare let you do
435
00:30:47 --> 00:30:53
this by yourself because you
will run into trouble.
436
00:30:51 --> 00:30:57
Nothing is going to happen.
You will just get a mess and
437
00:30:54 --> 00:31:00
will say I must be missing
something.
438
00:30:57 --> 00:31:03
And that is because you are
missing something.
439
00:31:01 --> 00:31:07
What you are missing,
and this is a good occasion to
440
00:31:05 --> 00:31:11
tell you, is that,
in general, three-quarters of
441
00:31:09 --> 00:31:15
the civilized world does not
introduce eigenvalues and
442
00:31:14 --> 00:31:20
eigenvectors the way you learn
them in 18.03.
443
00:31:18 --> 00:31:24
They use a different definition
that is identical.
444
00:31:22 --> 00:31:28
I mean it is equivalent.
The concept is the same,
445
00:31:27 --> 00:31:33
but it looks a little
different.
446
00:31:31 --> 00:31:37
Our definition is what?
Well, what is an eigenvalue and
447
00:31:35 --> 00:31:41
eigenvector?
The basic thing is this
448
00:31:37 --> 00:31:43
equation.
449
00:31:39 --> 00:31:45
450
00:31:46 --> 00:31:52
This is a two-by-two matrix,
right?
451
00:31:48 --> 00:31:54
This is a column vector with
two entries.
452
00:31:51 --> 00:31:57
The product has to be a column
vector with two entries,
453
00:31:55 --> 00:32:01
but both entries are supposed
to be zero so I will write it
454
00:31:59 --> 00:32:05
this way.
This way first defines what an
455
00:32:03 --> 00:32:09
eigenvalue is.
It is something that makes the
456
00:32:06 --> 00:32:12
determinant zero.
And then it defines what an
457
00:32:09 --> 00:32:15
eigenvector is.
It is, then,
458
00:32:11 --> 00:32:17
a solution to the system that
you can get because the
459
00:32:15 --> 00:32:21
determinant is zero.
Now, that is not what most
460
00:32:19 --> 00:32:25
people do.
What most people do is the
461
00:32:21 --> 00:32:27
following.
They write this equation
462
00:32:24 --> 00:32:30
differently by having something
on both sides.
463
00:32:29 --> 00:32:35
Using the distributive law,
what goes on the left side is A
464
00:32:33 --> 00:32:39
alpha one.
What is that?
465
00:32:35 --> 00:32:41
That is a column vector with
two entries.
466
00:32:38 --> 00:32:44
What goes on the right?
Well, lambda one times the
467
00:32:42 --> 00:32:48
identity times alpha one.
Now, the identity matrix times
468
00:32:47 --> 00:32:53
anything just reproduces what
was there.
469
00:32:50 --> 00:32:56
There is no difference between
writing the identity times alpha
470
00:32:55 --> 00:33:01
one and just alpha one all by
itself.
471
00:33:00 --> 00:33:06
So that is what I am going to
do.
472
00:33:02 --> 00:33:08
This is the definition of
eigenvalue and eigenvector that
473
00:33:07 --> 00:33:13
all the other people use.
Most linear algebra books use
474
00:33:12 --> 00:33:18
this definition,
or most books use a different
475
00:33:16 --> 00:33:22
approach and say,
here is an eigenvalue and an
476
00:33:20 --> 00:33:26
eigenvector.
And it requires them to define
477
00:33:23 --> 00:33:29
them in the opposite order.
First what alpha one is and
478
00:33:28 --> 00:33:34
then what lambda one is.
See, I don't have any
479
00:33:33 --> 00:33:39
determinant now.
So what is the definition?
480
00:33:36 --> 00:33:42
And they like it because it has
a certain geometric flavor that
481
00:33:40 --> 00:33:46
this one lacks entirely.
This is good for solving
482
00:33:43 --> 00:33:49
differential equations,
which is why we are using it in
483
00:33:47 --> 00:33:53
18.03, but this has a certain
geometric content.
484
00:33:50 --> 00:33:56
This way thinks of A as a
linear transformation of the
485
00:33:54 --> 00:34:00
plane, a shearing of the plane.
You take the plane and do
486
00:33:57 --> 00:34:03
something to it.
Or, you squish it like that.
487
00:34:02 --> 00:34:08
Or, you rotate it.
That's okay,
488
00:34:04 --> 00:34:10
too.
And the matrix defines a linear
489
00:34:07 --> 00:34:13
transformation to the plane,
every vector goes to another
490
00:34:11 --> 00:34:17
vector.
The question it asks is,
491
00:34:14 --> 00:34:20
is there a vector which is
taken by this linear
492
00:34:18 --> 00:34:24
transformation and just left
alone or stretched,
493
00:34:21 --> 00:34:27
is kept in the same direction
but stretched?
494
00:34:25 --> 00:34:31
Or, maybe its direction is
reversed and it is stretched or
495
00:34:29 --> 00:34:35
it shrunk.
But, in general,
496
00:34:33 --> 00:34:39
if there are real eigenvalues
there will be such vectors that
497
00:34:38 --> 00:34:44
are just left in the same
direction but just stretched or
498
00:34:43 --> 00:34:49
shrunk.
And what is the lambda?
499
00:34:46 --> 00:34:52
The lambda then is the amount
by which they are stretched or
500
00:34:51 --> 00:34:57
shrunk, the factor.
This way, first we have to find
501
00:34:55 --> 00:35:01
the vector, which is left
essentially unchanged,
502
00:34:59 --> 00:35:05
and then the number here that
goes with it is the stretching
503
00:35:04 --> 00:35:10
factor or the shrinking factor.
But the end result is the pair
504
00:35:11 --> 00:35:17
alpha one and lambda one,
regardless of which order you
505
00:35:15 --> 00:35:21
find them, satisfied the same
equation.
506
00:35:18 --> 00:35:24
Now, a consequence of this
definition we are going to need
507
00:35:22 --> 00:35:28
in the calculation that I am
going to do in just a moment.
508
00:35:26 --> 00:35:32
Let me calculate that out.
What I want to do is calculate
509
00:35:31 --> 00:35:37
the matrix A times E.
I am going to need to calculate
510
00:35:36 --> 00:35:42
that.
Now, what is that?
511
00:35:38 --> 00:35:44
Remember, E is the matrix whose
columns are the eigenvectors.
512
00:35:43 --> 00:35:49
That is the matrix alpha one,
alpha two.
513
00:35:47 --> 00:35:53
Now, what is this?
Well, in both Friday's lecture
514
00:35:51 --> 00:35:57
and Monday's lecture,
I used the fact that if you do
515
00:35:55 --> 00:36:01
a multiplication like that it is
the same thing as doing the
516
00:36:01 --> 00:36:07
multiplication A alpha one and
putting it in the first column.
517
00:36:08 --> 00:36:14
And then A alpha two is the
column vector that goes in the
518
00:36:13 --> 00:36:19
second column.
But what is this?
519
00:36:16 --> 00:36:22
This is lambda one alpha one.
And this is lambda two alpha
520
00:36:22 --> 00:36:28
two by this other definition of
eigenvalue and eigenvector.
521
00:36:28 --> 00:36:34
And what is this?
Can I write this in terms of
522
00:36:34 --> 00:36:40
matrices?
Yes indeed I can.
523
00:36:36 --> 00:36:42
This is the matrix alpha one,
alpha two times this matrix
524
00:36:42 --> 00:36:48
lambda one, lambda two,
zero, zero.
525
00:36:46 --> 00:36:52
Check it out.
Lambda one plus zero,
526
00:36:50 --> 00:36:56
lambda one times this thing
plus zero, the first entry is
527
00:36:56 --> 00:37:02
exactly that.
And the same way the second
528
00:37:01 --> 00:37:07
column doing the same
calculation is exactly this.
529
00:37:06 --> 00:37:12
What is that?
That is e times this matrix
530
00:37:10 --> 00:37:16
lambda one, zero,
zero, lambda two.
531
00:37:14 --> 00:37:20
Okay.
We are almost finished now.
532
00:37:17 --> 00:37:23
Now we can carry out our work.
We are going to do the
533
00:37:22 --> 00:37:28
substitution.
I start with a system.
534
00:37:26 --> 00:37:32
Remember where we are.
I am starting with this system.
535
00:37:33 --> 00:37:39
I am going to make the
substitution x equal to this
536
00:37:39 --> 00:37:45
matrix E, whose columns are the
eigenvectors.
537
00:37:45 --> 00:37:51
I am in introducing,
in other words,
538
00:37:49 --> 00:37:55
new variables u and v according
to that thing.
539
00:37:55 --> 00:38:01
u is the column vector,
u and v.
540
00:38:00 --> 00:38:06
And x, as usual,
541
00:38:03 --> 00:38:09
is the column vector x and y.
So I am going to plug it in.
542
00:38:08 --> 00:38:14
Okay.
Let's plug it in.
543
00:38:10 --> 00:38:16
What do I get?
I take the derivative.
544
00:38:13 --> 00:38:19
E is a constant matrix so that
makes E times u prime,
545
00:38:17 --> 00:38:23
is equal to A times,
x is E times u again.
546
00:38:21 --> 00:38:27
Now, at this point,
547
00:38:24 --> 00:38:30
you would be stuck,
except I calculated for you A
548
00:38:28 --> 00:38:34
times E is E times that funny
diagonal matrix with the
549
00:38:32 --> 00:38:38
lambdas.
So this is E times that funny
550
00:38:38 --> 00:38:44
matrix of the lambdas,
the eigenvalues,
551
00:38:41 --> 00:38:47
and still the u at the end of
it.
552
00:38:45 --> 00:38:51
So where are we?
E times u prime equals E times
553
00:38:50 --> 00:38:56
this thing.
Well, multiply both sides by E
554
00:38:54 --> 00:39:00
inverse and you can cancel them
out.
555
00:38:59 --> 00:39:05
And so the end result is that
after you have made the
556
00:39:04 --> 00:39:10
substitution in terms of the new
variables u, what you get is u
557
00:39:10 --> 00:39:16
prime equals lambda one,
lambda two, zero,
558
00:39:14 --> 00:39:20
zero times u.
Let's write that out in terms
559
00:39:18 --> 00:39:24
of a system.
This is u prime is equal to,
560
00:39:22 --> 00:39:28
well, this is u,
v here.
561
00:39:24 --> 00:39:30
It is lambda one times u plus
zero times v.
562
00:39:30 --> 00:39:36
And the other one is v prime
equals zero times u plus lambda
563
00:39:40 --> 00:39:46
two times v.
We are decoupled.
564
00:39:45 --> 00:39:51
In just one sentence you would
say --
565
00:39:52 --> 00:39:58
In other words,
if you were reading a book that
566
00:39:55 --> 00:40:01
sort of assumed you knew what
was going on,
567
00:39:59 --> 00:40:05
all it would say is as usual.
That is to make you feel bad.
568
00:40:04 --> 00:40:10
Or, as is well-known to make
you feel even worse.
569
00:40:08 --> 00:40:14
Or, the system is decoupled by
choosing as the new basis for
570
00:40:13 --> 00:40:19
the system the eigenvectors of
the matrix and in terms of the
571
00:40:18 --> 00:40:24
resulting new coordinates,
the decoupled system will be
572
00:40:23 --> 00:40:29
the following where the
constants are the eigenvalues.
573
00:40:29 --> 00:40:35
And so the solution will be u
equals c1 times e to the lambda1
574
00:40:33 --> 00:40:39
t and v
is equal to c2 times e to the
575
00:40:38 --> 00:40:44
lambda2 t.
576
00:40:40 --> 00:40:46
Of course, if you want it back
in terms now of x and y,
577
00:40:44 --> 00:40:50
you will have to go back to
here, to these equations and
578
00:40:48 --> 00:40:54
then plug in for u and v what
they are.
579
00:40:51 --> 00:40:57
And then you will get the
answer in terms of x and y.
580
00:40:55 --> 00:41:01
Okay.
We have just enough time to
581
00:40:59 --> 00:41:05
actually carry out this little
program.
582
00:41:04 --> 00:41:10
It takes a lot longer to derive
than it does actually to do,
583
00:41:10 --> 00:41:16
so let's do it for this system
that we were talking about
584
00:41:16 --> 00:41:22
before.
Decouple the system,
585
00:41:19 --> 00:41:25
x, y prime equals the matrixes
negative two,
586
00:41:24 --> 00:41:30
two, one, negative one.
587
00:41:28 --> 00:41:34
588
00:41:31 --> 00:41:37
589
00:41:36 --> 00:41:42
Okay.
What do I do?
590
00:41:37 --> 00:41:43
Well, I first have to calculate
the eigenvalues in the
591
00:41:42 --> 00:41:48
eigenvectors,
so the Ev's and Ev's.
592
00:41:45 --> 00:41:51
The characteristic equation is
lambda squared.
593
00:41:49 --> 00:41:55
The trace is negative three,
but you have to change the
594
00:41:54 --> 00:42:00
sign.
The determinant is two minus
595
00:41:57 --> 00:42:03
two, so that is zero.
There is no constant term here.
596
00:42:02 --> 00:42:08
It is zero.
That is the characteristic
597
00:42:05 --> 00:42:11
equation.
The roots are obviously lambda
598
00:42:08 --> 00:42:14
equals zero, lambda equals
negative three.
599
00:42:12 --> 00:42:18
And what are the eigenvectors
that go with that?
600
00:42:15 --> 00:42:21
With lambda equals zero goes
the eigenvector,
601
00:42:19 --> 00:42:25
minus two.
Well, I subtract zero here,
602
00:42:22 --> 00:42:28
so the equation I have to solve
is minus 2 a1 plus --
603
00:42:28 --> 00:42:34
I am not going to write 2 a2,
which is what you have been
604
00:42:32 --> 00:42:38
writing up until now.
The reason is because I ran
605
00:42:36 --> 00:42:42
into trouble with the notation
and I had to use,
606
00:42:40 --> 00:42:46
as the eigenvector,
not a1, a2 but a1,
607
00:42:43 --> 00:42:49
b1.
So it should be a b1 here,
608
00:42:46 --> 00:42:52
not the a2 that you are used
to.
609
00:42:48 --> 00:42:54
The solution,
therefore, is alpha one equals
610
00:42:52 --> 00:42:58
one, one.
And for lambda equals negative
611
00:42:55 --> 00:43:01
three, the corresponding
eigenvector this time will be --
612
00:43:02 --> 00:43:08
Now I have to subtract negative
three from here,
613
00:43:06 --> 00:43:12
so negative two minus negative
three makes one.
614
00:43:10 --> 00:43:16
That is a1 plus 2 b1 equals
zero,
615
00:43:15 --> 00:43:21
a logical choice for the
eigenvector here.
616
00:43:19 --> 00:43:25
The second eigenvector would be
make b1 equal to one,
617
00:43:24 --> 00:43:30
let's say, and then a1 will be
negative two.
618
00:43:28 --> 00:43:34
Okay.
Now what do we have to do?
619
00:43:32 --> 00:43:38
Now, what we want is the matrix
E.
620
00:43:35 --> 00:43:41
The matrix E is the matrix of
eigenvectors,
621
00:43:38 --> 00:43:44
so it is the matrix one,
one, negative two,
622
00:43:42 --> 00:43:48
one.
623
00:43:44 --> 00:43:50
The next thing we want is what
the new variables u and v are.
624
00:43:50 --> 00:43:56
For that, we will need E
inverse.
625
00:43:52 --> 00:43:58
How do you calculate the
inverse of a two-by-two matrix?
626
00:43:59 --> 00:44:05
You switch the two diagonal
elements, there I have switched
627
00:44:04 --> 00:44:10
them, and you leave the other
two where they are but change
628
00:44:09 --> 00:44:15
their sign.
So it is two up here and
629
00:44:13 --> 00:44:19
negative one there.
Maybe I should make this one
630
00:44:17 --> 00:44:23
purple and then that one purple
to indicate that I have switched
631
00:44:23 --> 00:44:29
them.
I am not done yet.
632
00:44:25 --> 00:44:31
I have to divide by the
determinant.
633
00:44:30 --> 00:44:36
What is the determinant?
It is one minus negative two,
634
00:44:36 --> 00:44:42
which is three,
so I have to divide by three.
635
00:44:42 --> 00:44:48
I multiply everything here by
one-third.
636
00:44:47 --> 00:44:53
Okay.
And what is the decoupled
637
00:44:51 --> 00:44:57
system?
The new variables are u equals
638
00:44:56 --> 00:45:02
one-third.
639
00:44:59 --> 00:45:05
640
00:45:08 --> 00:45:14
In other words,
the new variables are given by
641
00:45:11 --> 00:45:17
D.
It is u, v equals one,
642
00:45:13 --> 00:45:19
two, negative one, one
times one-third times x,y.
643
00:45:16 --> 00:45:22
644
00:45:19 --> 00:45:25
That is the expression for u,
645
00:45:21 --> 00:45:27
v in terms of x and y.
It's this matrix D,
646
00:45:25 --> 00:45:31
the decoupling matrix which is
the one that is used.
647
00:45:30 --> 00:45:36
And that gives this system u
equals one-third of x plus 2y
648
00:45:37 --> 00:45:43
on top.
And what is the v entry?
649
00:45:44 --> 00:45:50
v is one-third of minus x plus
y.
650
00:45:51 --> 00:45:57
Now, are those the same
variables that I used before?
651
00:46:00 --> 00:46:06
652
00:46:08 --> 00:46:14
Yes.
This is my new and better you,
653
00:46:10 --> 00:46:16
the one I got by just blindly
following the method instead of
654
00:46:14 --> 00:46:20
looking for physical things with
physical meaning.
655
00:46:17 --> 00:46:23
It differs from the old one
just by a constant factor.
656
00:46:21 --> 00:46:27
Now, that doesn't have any
effect on the resulting equation
657
00:46:25 --> 00:46:31
because if the old one is u
prime equals zero the new one is
658
00:46:29 --> 00:46:35
one-third u prime equals zero.
It is still the same equation,
659
00:46:34 --> 00:46:40
in other words.
And how about this one?
660
00:46:36 --> 00:46:42
This one differs from the other
one by the factor minus
661
00:46:40 --> 00:46:46
one-third.
If I multiply that v through by
662
00:46:43 --> 00:46:49
minus one-third,
I get this v.
663
00:46:45 --> 00:46:51
And, therefore,
that too does not affect the
664
00:46:47 --> 00:46:53
second equation.
I simply multiply both sides by
665
00:46:51 --> 00:46:57
minus one-third.
The new v still satisfies the
666
00:46:54 --> 00:47:00
equation minus three times v.