Everything I say today is going to be for n-by-n systems,
but for your calculations and the exams two-by-two will be
good enough. Our system looks like that.
Notice I am talking today about the homogeneous system,
not the inhomogenous system. So, homogenous.
And we have so far two basic methods of solving it.
The first one, on which we spent the most
time, is the method of where you calculate the eigenvalues of the
matrix, the eigenvectors, and put them together to make
the general solution. So eigenvalues,
e-vectors and so on. The second method,
which I gave you last time, I called "royal road," simply
calculates the matrix e to the At and says that the
solution is e to the At times x zero,
the initial condition. That is very elegant.
The only problem is that to calculate the matrix e to the
At, although sometimes you can do it by its definition as an
infinite series, most of the time the only way
to calculate the matrix e to the At is by using the fundamental
matrix. In other words,
the normal way of doing it is you have to calculate it as the
fundamental matrix time normalized at zero.
So, as I explained at the end of last time and you practiced
in the recitations, you have to find the
fundamental matrix, which, of course,
you have to do by eigenvalues and eigenvectors.
And then you multiply it by its value at zero,
inverse. And that, by magic,
turns out to be the same as the exponential matrix.
But, of course, there has been no gain in
simplicity or no gain in ease of calculation.
The only difference is that the language has been changed.
Now, today is going to be devoted to yet another method
which saves no work at all and only amounts to a change of
language. The only reason I give it to
you is because I have been begged by various engineering
departments to do so -- -- because that is the language
they use. In other words,
each person who solves systems, some like to use fundamental
matrices, some just calculate, some immediately convert the
system by elimination into a single higher order equation
because they are more comfortable with that.
Some, especially if they are writing papers,
they talk exponential matrices. But there are a certain number
of engineers and scientists who talk decoupling,
express the problem and the answer in terms of decoupling.
And that is, therefore, what I have to
explain to you today. So, the third method,
today's method, I stress is really no more than
a change of language. And I feel a little guilty
about the whole business. Instead of going more deeply
into studying these equations, what I am doing is like giving
a language course and teaching you how to say hello and
good-bye in French, German, Spanish,
and Italian. It is not going very deeply
into any of those languages, but you are going into the
outside world, where people will speak these
things. Here is an introduction to the
language of decoupling in which for some people is the exclusive
language in which they talk about systems.
Now, I think the best way to, well, in a general way,
what you try to do is as follows.
You try to introduce new variables.
You make a change of variables. I am going to do it two-by-two
just to save a lot of writing out.
And it's going to be a linear change of variables because we
are interested in linear systems.
The problem is to find u and v such that something wonderful
happens, such that when you make the change of variables to
express this system in terms of u and v it becomes decoupled.
And that means the system turns into a system which looks like u
prime equals k1 times u and v prime equals k2
times v. Such a system is called
decoupled. Why?
Well, a normal system is called coupled.
Let's write out what it would be.
Well, let's not write that. You know what it looks like.
This is decoupled because it is not really a system at all.
It is just two first-order equations sitting side by side
and having nothing whatever to do with each other.
This is two problems from the first day of the term.
It is not one problem from the next to last day of the term,
in other words. To solve this all you say is u
is equal to some constant times e to the k1 t and v
equals another constant times e to the k2 t.
Coupled means that the x and y occur in both equations on the
right-hand side. And, therefore,
you cannot solve separately for x and y, you must solve together
for both of them. Here I can solve separately for
u and v and, therefore, the system has been decoupled.
Now, obviously, if you can do that it's an
enormous advantage, not just to the ease of
solution, because you can write down the solution immediately,
but because something physical must be going on there.
There must be some insight. There ought to be some physical
reason for these new variables. Now, that is where I plan to
start with. My plan for the lecture is
first to work out, in some detail,
a specific example where decoupling is done to show how
that leads to the solution. And then we will go back and
see how to do it in general -- -- because you will see,
as I do the decoupling in this particular example,
that that particular method, though it is suggested,
will not work in general. I would need a more general
method. But let's first go to the
example. It is a slight modification of
one you should have done in recitation.
I don't think I worked one of these in the lecture,
but to describe it I have to draw two views of it to make
sure you know exactly what I am talking about.
Sometimes it is called the two compartment ice cube tray
problem, a very old-fashion type of ice cube tray.
Not a modern one that is all plastic where there is no
leaking from one compartment to another.
The old kind of ice cube trays, there were compartments and
these were metal separated and you leveled the liquid because
it could leak through the bottom that didn't go right to the
bottom. If you don't know what I am
talking about it makes no difference.
This is the side view. This is meant to be twice as
long. But, to make it quite clear,
I will draw the top view of this thing.
You have to imagine this is a rectangle, all the sides are
parallel and everything. This is one and this is two.
All I am trying to say is that the cross-sectional area of
these two chambers, this one has twice the
cross-sectional area of this one.
So I will write a two here and I will write a one there.
Of course, it is this hole here through which everything leaks.
I am going to let x be the height of this liquid,
the water here, and y the height of the water
in that chamber. Obviously, as time goes by,
they both reach the same height because of somebody's law.
Now, what is the system of differential equations that
controls this? Well, the essential thing is
the flow rate through here. That flow rate through the hole
in units, let's say, in liters per second.
Just so you understand, I am talking about the volume
of liquid. I am not talking about the
velocity. That is proportional to the
area of the hole. So the cross-sectional area of
the hole. And it is also times the
velocity of the flow, but the velocity of the flow
depends upon the pressure difference.
And that pressure difference depends upon the difference in
height. All those are various people's
laws. So times the height difference.
Of course, you have to get the sign right.
I have just pointed out the height difference is
proportional to the pressure at the hole.
And it is that pressure at the hole that determines the
velocity with which the fluid flows through.
Where does this all produce our equations?
Well, x prime is equal to, therefore, some constant,
depending on the area of the hole and this constant of
proportionality with the pressure and the units and
everything else times the pressure difference I am talking
about. Well, if fluid is going to flow
in this direction that must mean the y height is higher than the
x height. So, to make x prime positive,
it should be y minus x here.
Now, the y prime is different. Because, again,
the rate of fluid flow is determined.
This time, if y prime is positive, if this is rising,
as it will be in this case, it's because the fluid is
flowing in that direction. It is because x is higher than
y. So this should be the same
constant x minus y. But notice that right-hand side
is the rate at which fluid is flowing into this tank.
That is not the rate at which y is changing.
It is the rate at which 2y is changing.
Why isn't there a constant here?
There is. It's one.
That is the one, this one cross-section.
The area here is one and the cross-sectional area here is
two. And that is the reason for the
one here and the two here, because we are interested in
the rate at which fluid is being added to this,
which is only related to the height, the rate at which the
height is rising if you take into account the cross-sectional
area. So there is the system.
In order to use nothing but integers here,
I am going to take c equals to two, so I don't have to put in
halves. The final system is x prime
equals minus 2x, you have to write them in the
correct order, and y prime equals,
the twos cancel because c is two, is x minus y.
So there is our system. Now the problem is I want to
solve it by decoupling it. I want, in other words,
to find new variables, u and v, which are more natural
to the problem than the x and y that are so natural to the
problem that the new system will just consistently be two
side-by-side equations instead of the single equation.
The question is, what should u and v be?
Now, the difference between what I am going to do now and
what I am going to do later in the period is later in the
period I will give you a systematic way of finding what u
and v should be. Now we are going to psyche out
what they should be in the way in which people who solve
systems often do. I am going to use the fact that
this is not just an abstract system of equations.
It comes from some physical problem.
And I ask, is there some system of variables,
which somehow go more deeply into the structure of what's
going on here than the naīve variables, which simply tell me
how high the two tank levels are?
That is the obvious thing I can see, but there are some
variables that go more deeply. Now, one of them is sort of
obvious and suggested both the form of the equation and by
this. Simply, the difference in
heights is, in some ways, a more natural variable because
that is directly related to the pressure difference,
which is directly related to the velocity of flow.
They will differ by just constant.
I am going to call that the second variable,
or the difference in height let's call it.
That's x minus y. That is a very natural variable
for the problem. The question is,
what should the other one be? Now you sort of stare at that
for a while until it occurs to you that something is constant.
What is constant in this problem?
Well, the tank is sitting there, that is constant.
But what thing, which might be a variable,
clearly must be a constant? It will be the total amount of
water in the two tanks. These things vary,
but the total amount of water stays the same because it is a
homogenous problem. No water is coming in from the
outside, and none is leaving the tanks through a little hole.
Okay. What is the expression for the
total amount of water in the tanks?
x plus 2y. Therefore, that is a natural
variable also. It is independent of this one.
It is not a simple multiply of it.
It is a really different variable.
This variable represents the total amount of liquid in the
two tanks. This represents the pressure up
to a constant factor. It is proportional to the
pressure at the hole. Okay.
Now what I am going to do is say this is my change of
variable. Now let's plug in and see what
happens to the system when I plug in these two variables.
And how do I do that? Well, I want to substitute and
get the new system. The new system,
or rather the old system, but what makes it new is in
terms of u and v. What will that be?
Well, u prime is x prime plus 2y prime.
But I know what x prime plus 2y prime is because I
can calculate it for this. What will it be?
x prime plus 2y prime is negative 2x plus twice y prime,
so it's plus 2x, which is zero.
And how about these two? 2y minus twice this,
because I want this plus twice that, so it 2y minus 2y,
again, zero. The right-hand side becomes
zero after I calculate x prime plus 2y.
So that is zero. That would just,
of course, clear. Now, that makes sense,
of course. Since the total amount is
constant, that says that u prime is zero.
Okay. What is v prime?
v prime is x prime minus y prime.
What is that? Well, once again we have to
calculate. x prime minus y prime is minus
2x minus x, which is minus 3x, and 2y minus negative y,
which makes plus 3y. All right.
What is the system? The system is u prime equals
zero and v prime equals minus three times x minus
y. But x minus y is v.
In other words,
these new two variables decouple the system.
And we got them, as scientists often do,
by physical considerations. These variables go more deeply
into what is going on in that system of two tanks than simply
the two heights, which are too obvious as
variables. All right.
What is the solution? Well, the solution is,
u equals a constant and v is equal to?
Well, the solution to this equation is a different
arbitrary constant from that one.
These are side-by-side equations that have nothing
whatever to do with each other, remember?
Times e to the minus 3t.
Now, there are two options. Either one leaves the solution
in terms of those new variables, saying they are more natural to
the problem, but sometimes, of course, one wants the answer
in terms of the old one. But, if you do that,
then you have to solve that. In order to save a little time,
since this is purely linear algebra, I am going to write --
Instead of taking two minutes to actually do the calculation
in front of you, I will just write down what the
answer is --
-- in terms of u and x and y. In other words,
this is a perfectly good way to leave the answer if you are
allowed to do it. But if somebody says they want
the answer in terms of x and y, well, you have to give them
what they are paying for. In terms of x and y,
you have first to solve those equations backwards for x and y
in terms of u and v in which case you will get x equals
one-third of u plus 2v.
Use the inverse matrix or just do elimination,
whatever you usually like to do.
And the other one will be one-third of u minus v.
And then, if you substitute in,
you will see what you will get is one-third of c1.
Sorry. u is c1.
c1 plus 2 c2 e to the negative 3t.
And this is one-third of c1 minus c2 e to the minus 3t.
And so, the final solution is,
in terms of the way we usually write out the answer,
x will be what? Well, it will be one-third c1
times the eigenvector one, one plus one-third times c2
times the eigenvector two, negative one times e to the
minus 3t. That is the solution written
out in terms of x and y either as a vector in the usual way or
separately in terms of x and y. But, notice,
in order to do that you have to have these backwards equations.
In other words, I need the equations in that
form. I need the equations because
they tell me what the new variables are.
But I also have to have the equations the other way in order
to get the solution in terms of x and y, finally.
Okay. That was all an example.
For the rest of the period, I would like to show you the
general method of doing the same thing which does not depend upon
being clever about the choice of the new variables.
And then, at the very end of the period, I will apply the
general method to this problem to see whether we get the same
answer or not. What is the general method?
Our problem is the decouple. Now, the first thing is you
cannot always decouple. To decouple the eigenvalues
must all be real and non-defective.
In other words, if they are repeated they must
be complete. You must have enough
independent eigenvectors. So they must be real and
complete. If repeated,
they must be complete. They must not be defective.
As I told you at the time when we studied complete and
incomplete, the most common case in which this occurs is when the
matrix is symmetric. If the matrix is real and
symmetric then you can always decouple the system.
That is a very important theorem, particularly since many
of the equilibrium problems normally lead to symmetric
matrices and are solved by decoupling.
Okay. So what are we looking for?
We are assuming this and we need it.
In general, otherwise, you cannot decouple if you have
complex eigenvalues and you cannot decouple if you have
defective eigenvalues.
Well, what are we looking for? We are looking for new
variables. u, v equals a1,
b1, a2, b2 times the x, y.
And this matrix is called D, the decoupling matrix and is
what we are looking for. How do I choose those new
variables u and v when I don't have any physical considerations
to guide me as I did before? Now, the key is to look instead
at what you are going to need. Remember, we are changing
variables. And, as I told you from the
first days of the term, when you change variables look
at what you are going to need to substitute in to make the change
of variables. Don't just start writing
equations. What we are going to need to
plug into that system and change it to the (u,
v) coordinates is not u and v in terms of x and y.
What we need is x and y in terms of u and v to do the
substitution. What we need is the inverse of
this. So, in order to do the
substitution, what we need is (x,
y). Oops.
Let's call them prime. Let's call these a1,
b1, a2, b2 because these are going to be much more important
to the problem than the other ones.
Okay. I am going to,
I should call this matrix D inverse, that would be a
sensible thing to call it. Since this is the important
matrix, this is the one we are going to need to do the
substitution, I am going to give it another
letter instead. And the letter that comes after
D is E. Now, E is an excellent choice
because it is also the first letter of the word eigenvector.
And the point is the matrix E, which is going to work,
is the matrix whose columns are the two eigenvectors.
The columns are the two eigenvectors.
Now, even if you didn't know anything that would be
practically the only reasonable choice anybody could make.
What are we looking for? To make a linear change of
variables like this really means to pick new i and j vectors.
You know, from the first days of 18.02, what you want is a new
coordinate system in the plane. And the coordinate system in
the plane is determined as soon as you tell what the new i is
and what the new j is in the new system.
To establish a linear change of coordinates amounts to picking
two new vectors that are going to play the role of i and j
instead of the old i and the old j.
Okay, so pick two vectors which somehow are important to this
matrix. Well, there are only two,
the eigenvectors. What else could they possibly
be? Now, what is the relation?
I say with this, what happens is I say that
alpha one corresponds, and alpha two,
these are vectors in the xy-system.
Well, if I change the coordinates to u and v,
in the uv-system they will correspond to the vectors one,
zero. In other words,
the vector that we would normally call i in the u,
v system. And this one will correspond to
the vector zero, one.
Now, if you don't believe that I will calculate it for you.
The calculation is trivial. Look.
What have we got? (x, y) equals a1,
b1, a2, b2.
This is the column vector alpha one.
This is the column vector alpha two.
Now, here is u and v. Suppose I make u and v equal to
one, zero, what happens to x and y?
Your matrix multiply. One, zero.
So a1 plus zero, b1 plus zero.
It corresponds to the column vector (a1, b1).
And in the same way zero, one corresponds to
(a2, b2).
Just by matrix multiplication. And that shows that these
correspond. In the uv-system the two
eigenvectors are now called i and j.
Well, that looks very promising, but the program now
is to do the substitution to substitute into the system x
prime equals Ax and see if it is decoupled in the
uv-coordinates. Now, I don't dare let you do
this by yourself because you will run into trouble.
Nothing is going to happen. You will just get a mess and
will say I must be missing something.
And that is because you are missing something.
What you are missing, and this is a good occasion to
tell you, is that, in general, three-quarters of
the civilized world does not introduce eigenvalues and
eigenvectors the way you learn them in 18.03.
They use a different definition that is identical.
I mean it is equivalent. The concept is the same,
but it looks a little different.
Our definition is what? Well, what is an eigenvalue and
eigenvector? The basic thing is this
equation.
This is a two-by-two matrix, right?
This is a column vector with two entries.
The product has to be a column vector with two entries,
but both entries are supposed to be zero so I will write it
this way. This way first defines what an
eigenvalue is. It is something that makes the
determinant zero. And then it defines what an
eigenvector is. It is, then,
a solution to the system that you can get because the
determinant is zero. Now, that is not what most
people do. What most people do is the
following. They write this equation
differently by having something on both sides.
Using the distributive law, what goes on the left side is A
alpha one. What is that?
That is a column vector with two entries.
What goes on the right? Well, lambda one times the
identity times alpha one. Now, the identity matrix times
anything just reproduces what was there.
There is no difference between writing the identity times alpha
one and just alpha one all by itself.
So that is what I am going to do.
This is the definition of eigenvalue and eigenvector that
all the other people use. Most linear algebra books use
this definition, or most books use a different
approach and say, here is an eigenvalue and an
eigenvector. And it requires them to define
them in the opposite order. First what alpha one is and
then what lambda one is. See, I don't have any
determinant now. So what is the definition?
And they like it because it has a certain geometric flavor that
this one lacks entirely. This is good for solving
differential equations, which is why we are using it in
18.03, but this has a certain geometric content.
This way thinks of A as a linear transformation of the
plane, a shearing of the plane. You take the plane and do
something to it. Or, you squish it like that.
Or, you rotate it. That's okay,
too. And the matrix defines a linear
transformation to the plane, every vector goes to another
vector. The question it asks is,
is there a vector which is taken by this linear
transformation and just left alone or stretched,
is kept in the same direction but stretched?
Or, maybe its direction is reversed and it is stretched or
it shrunk. But, in general,
if there are real eigenvalues there will be such vectors that
are just left in the same direction but just stretched or
shrunk. And what is the lambda?
The lambda then is the amount by which they are stretched or
shrunk, the factor. This way, first we have to find
the vector, which is left essentially unchanged,
and then the number here that goes with it is the stretching
factor or the shrinking factor. But the end result is the pair
alpha one and lambda one, regardless of which order you
find them, satisfied the same equation.
Now, a consequence of this definition we are going to need
in the calculation that I am going to do in just a moment.
Let me calculate that out. What I want to do is calculate
the matrix A times E. I am going to need to calculate
that. Now, what is that?
Remember, E is the matrix whose columns are the eigenvectors.
That is the matrix alpha one, alpha two.
Now, what is this? Well, in both Friday's lecture
and Monday's lecture, I used the fact that if you do
a multiplication like that it is the same thing as doing the
multiplication A alpha one and putting it in the first column.
And then A alpha two is the column vector that goes in the
second column. But what is this?
This is lambda one alpha one. And this is lambda two alpha
two by this other definition of eigenvalue and eigenvector.
And what is this? Can I write this in terms of
matrices? Yes indeed I can.
This is the matrix alpha one, alpha two times this matrix
lambda one, lambda two, zero, zero.
Check it out. Lambda one plus zero,
lambda one times this thing plus zero, the first entry is
exactly that. And the same way the second
column doing the same calculation is exactly this.
What is that? That is e times this matrix
lambda one, zero, zero, lambda two.
Okay. We are almost finished now.
Now we can carry out our work. We are going to do the
substitution. I start with a system.
Remember where we are. I am starting with this system.
I am going to make the substitution x equal to this
matrix E, whose columns are the eigenvectors.
I am in introducing, in other words,
new variables u and v according to that thing.
u is the column vector, u and v.
And x, as usual,
is the column vector x and y. So I am going to plug it in.
Okay. Let's plug it in.
What do I get? I take the derivative.
E is a constant matrix so that makes E times u prime,
is equal to A times, x is E times u again.
Now, at this point,
you would be stuck, except I calculated for you A
times E is E times that funny diagonal matrix with the
lambdas. So this is E times that funny
matrix of the lambdas, the eigenvalues,
and still the u at the end of it.
So where are we? E times u prime equals E times
this thing. Well, multiply both sides by E
inverse and you can cancel them out.
And so the end result is that after you have made the
substitution in terms of the new variables u, what you get is u
prime equals lambda one, lambda two, zero,
zero times u. Let's write that out in terms
of a system. This is u prime is equal to,
well, this is u, v here.
It is lambda one times u plus zero times v.
And the other one is v prime equals zero times u plus lambda
two times v. We are decoupled.
In just one sentence you would say --
In other words, if you were reading a book that
sort of assumed you knew what was going on,
all it would say is as usual. That is to make you feel bad.
Or, as is well-known to make you feel even worse.
Or, the system is decoupled by choosing as the new basis for
the system the eigenvectors of the matrix and in terms of the
resulting new coordinates, the decoupled system will be
the following where the constants are the eigenvalues.
And so the solution will be u equals c1 times e to the lambda1
t and v is equal to c2 times e to the
lambda2 t.
Of course, if you want it back in terms now of x and y,
you will have to go back to here, to these equations and
then plug in for u and v what they are.
And then you will get the answer in terms of x and y.
Okay. We have just enough time to
actually carry out this little program.
It takes a lot longer to derive than it does actually to do,
so let's do it for this system that we were talking about
before. Decouple the system,
x, y prime equals the matrixes negative two,
two, one, negative one.
Okay. What do I do?
Well, I first have to calculate the eigenvalues in the
eigenvectors, so the Ev's and Ev's.
The characteristic equation is lambda squared.
The trace is negative three, but you have to change the
sign. The determinant is two minus
two, so that is zero. There is no constant term here.
It is zero. That is the characteristic
equation. The roots are obviously lambda
equals zero, lambda equals negative three.
And what are the eigenvectors that go with that?
With lambda equals zero goes the eigenvector,
minus two. Well, I subtract zero here,
so the equation I have to solve is minus 2 a1 plus --
I am not going to write 2 a2, which is what you have been
writing up until now. The reason is because I ran
into trouble with the notation and I had to use,
as the eigenvector, not a1, a2 but a1,
b1. So it should be a b1 here,
not the a2 that you are used to.
The solution, therefore, is alpha one equals
one, one. And for lambda equals negative
three, the corresponding eigenvector this time will be --
Now I have to subtract negative three from here,
so negative two minus negative three makes one.
That is a1 plus 2 b1 equals zero,
a logical choice for the eigenvector here.
The second eigenvector would be make b1 equal to one,
let's say, and then a1 will be negative two.
Okay. Now what do we have to do?
Now, what we want is the matrix E.
The matrix E is the matrix of eigenvectors,
so it is the matrix one, one, negative two,
one.
The next thing we want is what the new variables u and v are.
For that, we will need E inverse.
How do you calculate the inverse of a two-by-two matrix?
You switch the two diagonal elements, there I have switched
them, and you leave the other two where they are but change
their sign. So it is two up here and
negative one there. Maybe I should make this one
purple and then that one purple to indicate that I have switched
them. I am not done yet.
I have to divide by the determinant.
What is the determinant? It is one minus negative two,
which is three, so I have to divide by three.
I multiply everything here by one-third.
Okay. And what is the decoupled
system? The new variables are u equals
one-third.
In other words, the new variables are given by
D. It is u, v equals one,
two, negative one, one times one-third times x,y.
That is the expression for u,
v in terms of x and y. It's this matrix D,
the decoupling matrix which is the one that is used.
And that gives this system u equals one-third of x plus 2y
on top. And what is the v entry?
v is one-third of minus x plus y.
Now, are those the same variables that I used before?
Yes. This is my new and better you,
the one I got by just blindly following the method instead of
looking for physical things with physical meaning.
It differs from the old one just by a constant factor.
Now, that doesn't have any effect on the resulting equation
because if the old one is u prime equals zero the new one is
one-third u prime equals zero. It is still the same equation,
in other words. And how about this one?
This one differs from the other one by the factor minus
one-third. If I multiply that v through by
minus one-third, I get this v.
And, therefore, that too does not affect the
second equation. I simply multiply both sides by
minus one-third. The new v still satisfies the
equation minus three times v.