In the last class I gave you this problem X.
= - Now we will just substitute the values
so that you can easily calculate. So this
is the state of equations that we started
with and I asked you to do two things. First,
as you have seen for any dynamical system,
if the system equations are given, it essentially
defines the vector field and we should first
try to get a grip on the character of the
vector field and for that we have understood
that the method is that we first locate the
equilibrium points. Then we locally linearize
around the equilibrium points and try to understand
the behavior around the equilibrium points
because these are pivots in a state space.
These are the places where you have good understanding.
So first understand those parts and then in
terms of that, if you can figure out how the
behavior is going to be, then it's fine.
If it is not possible to figure out then we
will look at the rest of the state space.
So go by that process because that will give
you some understanding. First where are the
equilibrium of this set of equations?
Obviously (0, 0, 0) is one but that is not
all because these guys are there. So it should
give rise to other equilibrium points. Can
you find out? So let us call them A B and
C three equilibrium points. The equilibrium
points are 
A (0,0,0,). You get the left hand side zero.
B is (root b(r-), root b(r -- 1), r -- 1).
The third equilibrium point, I can see that
it should be symmetrical. C is (-root b(r
-1), -b(r-), r-1). The moment we have it that
immediately tells you a few things. First
notice that if r is less than one, what happens?
This fellow gets imaginary and the position
of equilibrium point must be real because
it is in the real state space. eigenvalues
can be imaginary or complex whatever but the
position should be real and therefore these
equilibrium points B and C do not exist until
'r' the parameter reaches a value of one.
So that can be inferred immediately. But A
is always there. As you change the parameter
say starting from a value that is less than
one, suppose you are increasing the parameter,
then you find that this was there and this
was there. We will look at the stability later.
This fellow was there existing and at that
particular point, these two fellows come into
existence earlier it given not there. Only
that much can be inferred from these equations.
In order to infer more, we need to look at
the stability of the equilibrium points and
the stability of the equilibrium points are
obtained from the Jacobian matrices. So get
the Jacobian matrix.
We will obtain it in terms of a general thing
and then we will put the values. So this fellow
was our sigma and this fellow was our b. let
us do it generally so that you can use other
values of sigma and b also. So 
the first term will be - sigma + sigma 0.
Second equation with respect to x, it is - z
+ r. with respect to y,-1 and with respect
to z, -x. similarly here it is y, x and --b.
now we know these values. we have taken some
values here. We have kept r as the variable
parameter but for each equilibrium points
we can substitute these values x y and z.
Now if you substitute the value of (0,0,0),
then what you get? You have the matrix -10
+10 and 0. Here it is +r -1 and 0 and the
3rd row is 0 0 and -2/3. So can you find out
the eigenvalues of this 
in terms of r? There 
will be three eigenvalues. Lambda is --b,
-11+/- root (81+40 r) /2. Now let's see
what does this tell us. At r is equal to one,
you get 0 and -11. This one is square about
then 0 and -11. So there are two eigenvalues.
One is widely negative. So I can comfortably
assume that that fellow will remain negative
but the other fellow is zero. Zero means exactly
at the border line between stability and instability.
For r further negative then what happens?
There would be a for some values of r for
which this will be real. Some values of r
for which it will not be real. So find out
for which values it will be stable. For r
is equal to one we have found that it is minus
eleven and zero. So if you draw the the real
and imaginary parts, one is here and another
is here. As you change the parameter I can
see that either r values larger or smaller,
it should go this way and therefore the equilibrium
point will become unstable.
So here is something that is stable eigenvalue.
This part is also a stable eigenvalue and
the other part which is now having the value
zero is the suspect case which can make the
system unstable. For which value of the parameter
does it become unstable? It is one. There
are three equilibrium points A, B and C. this
fellow was stable at r =1. This fellow becomes
unstable and these two fellows come into existence.
So you are changing the parameter. One equilibrium
point was stable. So you are happy about its
behavior. Start from any initial condition
it will go into that and at that time all
the Eigenvalues were all real and therefore
it would nicely go into it. At this particular
value of r it becomes unstable. So what are
the Eigenvalues? Then it would remain real.
We simply substituted at r = 1 because we
had some hunch that something happens at r
= 1. We substitute r = 1 and realize that
that is a critical case. So move it this way
or that way it is going to be unstable. But
when it becomes unstable, is it still a real
pair of eigenvalues or is it a complex conjugate
pair of eigenvalues? Here it is still real
pair of Eigenvalue. They are still real pair
of eigenvalues.
So you have got the fellow unstable at that
point if you start from that that initial
condition or somewhere close to it will go
away. This is a stable eigenvalue. The other
one that was close to -11, that's also stable
eigenvalue. There would be Eigen directions
associated with them and any initial condition
along those eigendirection will not go away.
They will come close to the equilibrium point
even though the equilibrium point is unstable.
But a slight deviation along the unstable
eigendirection will grow exponentially and
it will go away along that direction. So it
is not just unstable anyway. It is unstable
in a specific way in a particular direction.
Now let us try to understand the question:
if it is unstable along that direction where
does it go? In order to answer that question
you have to find out the stability of these
two fellows who have come onto existence.
If they are stable they will ultimately land
of there and stay there. So find out their
stability for that. All you need to do is
to simply substitute these into here. At this
stage unless you put values you will find
it a little difficult to handle. Calculate
the stability when say, the value of r has
been pushed to an extend that is greater than
one say, two. So what will you do?
Now my Jacobian matrix is -10 +10 0. -1+2,
-1, root (8/3), root (8/3), root (8/3), -8/3.
So 
you can always find 
the eigenvalues and tell me whether this particular
equilibrium point is stable or unstable. Can
you figure out the type of eigenvalues that
you are going to get. One thing is to get
the values of the eigenvalue. You can always
plug it into Matlab and get it. So I am asking
you can you figure out whether it there going
to be real or complex or stable or unstable?
That's suffices our purpose? For our purpose
that will suffice the character of the equilibrium
point. I don't really need to know exactly
the eigenvalues. I think you will get one
real negative eigenvalue and 
two complex conjugate eigenvalues with negative
real. So at this parameter values where I
have assumed r = 2, that means I have pushed
it beyond the r =1 level then these two equilibria
are stable but with a spiral character. Now
try to understand what happens. I am just
writing the character of the eigenvalues.
First Eigenvalue is real negative. The other
two are complex conjugate. Can you infer then
in the 3D states space 
the kind of orbit that you are likely to see,
where are they? These two equilibrium points,
where is the position say b, where is it if
you substitute the values? It's root 8/3
positive.
So 
along the x coordinate, you have got something,
along the y coordinate you have got something
and along the z coordinate you have got something.
If these are the two equilibrium points, now
if you have the real eigenvalue, there should
be an eigenvector along which this fellow
will be stable. So you can see that? Suppose
this is the eigendirection along which it
is stable. Eigenvector associated with the
real eigenvalue. They should also be associated
an Eigen plane associated with the complex
conjugate Eigenvalues. How to calculate that?
These are concepts that possibly you need.
Here there was a complication that the the
equilibria are not at origin. So let us assume
that you have got an equilibrium at the origin.
Suppose this is one stable direction associated
with the real Eigenvalue which means that
if you have an initial condition here, it
will exponentially decay. Suppose there are
a pair of complex conjugate Eigenvalues. There
should be some kind of a plane associated
with it so that any perturbation along that
direction will die down like this. any deviation
from this plane will die down because of the
action of this stable Eigenvalue and any deviation
along this plane will die down because of
the action of the complex conjugate Eigenvalues
with negative real part so that if you start
from any initial condition away from here
it will behave like this and it will ultimately
go towards that. This is an Eigen plane in
the sense that if you start any initial condition
on this plane, it will forever remain on this
plane. In that sense it is an Eigen plane.
How to identify that Eigen plane?
Try to recall the way you solved the problem
with the pair of complex conjugate Eigenvalues.
What did you do you? First if you write down
the characteristic equation, in the case of
a two dimensional system, you got a quadratic
equation, in the case of the 3D system, a
cubic equation. Whatever it is but its solution
gave you the complex values. Now these are
Eigenvalues. How did you calculate the eigenvectors?
You plugged in these eigenvalues into the
equation (A-lambda) I X = 0. Thereby you obtain
the eigenvectors. Now these eigenvectors in
this case will also turn out to be complex
conjugate. So, Eigenvalues complex conjugate
if you obtain in the same way, blindly, I
have this eigenvalue and therefore I try to
obtain the eigenvectors. You will get complex
conjugate eigenvalues. Then what was the logic?
The logic was that you would say that this
complex conjugate Eigenvectors were consisting
of a real part and the imaginary part like
P + JQ and then you would say that P + JQ
is a linear combination of the P and Q and
therefore P and Q are both independent solutions
and thereby you would say that any solution
can then be constructed out of these two real
solutions. That was the essential logic. Let
us solve a problem simple problem.
x dot = sigma x - omega y and y dot = omega
x + sigma y. from this you will have x dot
= Ax where A is a matrix sigma -omega + omega
+ sigma. Then in order to calculate the eigenvalues,
you would write (A -- lambda I) = sigma -- lambda,
-omega, omega, sigma -- lambda. The determinant
of this would have to be 0 so that equation
then becomes lambda = sigma +/- j omega. The
equations have been written such that you
get it essentially. This is very familiar.
So I assume that you are feeding in the familiar
domain. You will have to find out the eigenvectors
associated with 
this eigenvalues.
Say an Eigenvector is v1 v 2. Then you would
write - j omega - omega omega -j omega times
v 1 v 2 =0. So that will be written as - j
omega v 1 - omega v 2. This is omega v 1 - j
omega v 2 = 0. That's what we write. Now
you see they both are the same equations as
happens for all Eigenvector equations. But
then from here it is possible to identify
the eigenvectors. So what is the eigenvectors?
From this line you can write v 1 = j v2. That
is the eigenvector equation. So it involves
j. in that sense, it is a complex conjugate
now. How would you identify this? We have
said that the eigenvector is this vector.
These two are related by this. Eigenvector
is any vector along that direction. So it
will be very convenient for us to simply say
that either v 1 or v 2 is 1 and it will be
convenient to say v 2 =1. So v 2 = 1 & v 1=
j. so what do we have? So here we started
with one eigenvector one eigenvalue. Its conjugate
take the negative one. It will turn up to
be complex conjugate. So we can safely work
with this Eigenvector associated with Eigenvalue.
Then what do you say? How do you proceed?
When you do that in the 3D, you have to do
exactly the same way. So only difference is
that this will be a three dimensional equation,
you will get a three dimensional thing but
nevertheless ultimately it will lead to conceptually
the same thing.
So we have here x(t) as the solution of differential
equation is the e to the power Eigenvalue
t times Eigenvector. That is one solution.
Eigenvalue is sigma + j omega t times the
Eigenvector which is j1. That is the solution.
So this is one solution. Now we can break
it into two parts: e to the power sigma t
and e to the power j omega t and we know the
e to the power of j omega can be written in
terms of sines and cosines. So just do that.
Then you will get e to the power sigma t [j
cos omega t - sin omega t cos omega t + j
sin omega t]. What have we done? We have separated
out into two parts: e to the power sigma t
times e to the power j omega t e to the power
j omega t. we have written as cos omega t
+ j sin omega t and then multiplied with this
vector you got this.
Now we can separate this out into the real
part and imaginary part. We will get e to
the power sigma t times (- sin omega t cos
omega t + j cos omega t and sin omega t).
So here we have separated out. Then we say
that let this part be called p and let this
part be called q. 
This complex number is nothing but a linear
combination on the p part and q part and so
we can say that ultimately x(t) therefore
would be possible to write it as c 1 e to
the power sigma t times. This minus sin omega
t cos omega t + c2 e to the power sigma t
cos omega t sin omega t. That was the logic.
notice that we had complex conjugate Eigenvectors
but we went by the same logic that in the
ultimate solution, one of the possible solution
is e to the power Eigenvalue t times the Eigenvector
but then that allowed us to separate out the
real part and imaginary part and ultimately
we get a real solution. That is solution of
this. Now this is a 2D system. In a three
dimensional case, all that will happen is
the matrix will be a three dimensional matrix
- 3 x 3 matrix and you will get a cubic equation.
Ultimately from there you will solve. A cubic
equation may lead to one real Eigenvalue and
two complex conjugate Eigenvalues or all real
Eigenvalues.
They cannot be in any other solution. Now
if you have one real Eigenvalue and two complex
conjugate Eigenvalues, if all are real Eigenvalues,
it is very trivial to obtain the eigenvectors.
They are all real eigenvectors but it is not
trivial. That is why this concept comes. What
happens if you have a complex conjugated pair
of Eigenvalues and one real? This situation
that I have just depicted here. Then how do
you obtain this plane? Notice the logic that
I was to following here. What was the logic
that the real part and the imaginary part
individually give two solutions I have notice
real part and imaginary part of the eigenvector
individually give solutions? Real part is
a vector in the real space imaginary part
that that real component of the imaginary
part is also real vector. So you will be able
to identify two vectors in this plane.
There can be only one plane passing through
these two vectors. The method of approach
is that if you have a complex conjugate pair
of eigenvalues, obtain the eigenvectors and
obtain the real and imaginary part separately.
The real part will give you one real vector
the imaginary part will give you another real
vector. The plane passing through that will
be the Eigen plane. Have you have you understood
the concept? Often this particular concept
is not given in the differential equation
classes but for us, in order to understand
what happens in the states space, it is definitely
necessary first to understand this Eigen directions
and Eigen planes. Next class I will show you
the eigenvalues. From the evolution of a system
you will be able to see here is an Eigenplane.
I can see that. But before seeing doing that
by simulation is blind. But what is really
necessary is to grasp the understanding that
if you have a three dimensional system and
if you have the one real eigenvalue and two
complex conjugate eigenvalues, there must
be one real eigendirection and one real Eigen
plane.
So in the problem that we were originally
attacking, that is the problem with this system,
here also there would be a real Eigen plane
and I don't know where it is but let me
schematically draw like this. This means that
any deviation from here will die down but
any deviation from here will also die down.
It will be an inward spiral. Similarly here
now because of this symmetry, we can say that
here the directions would be symmetrical to
that direction because all the values are
symmetrical. So we will say that here is also
another plane. So this fellow was an equilibrium
point that was unstable from here. It goes
away along that unstable eigendirection and
when it lands somewhere here it always goes
on to either this or that. Both are stable.
Will it go here or here? You really don't
know. It might go here it might go there.
There is the fun of non-linear system that
there can be two equilibria both stable and
you do not know which one it will go to. In
fact it can go to both the equilibrium points
depending on where the initial condition is.
If the initial condition is somewhere here,
common sense will tell you that will most
likely go to this one. if the initial condition
somewhere here, it is most likely it will
goes somewhere here but this also tells you
that the whole state space should be divide
into compartments. One compartment and those
initial conditions that will ultimately go
here, another compartment those initial conditions
that will ultimately go here. So even though
we have not calculated it today and I ask
you to actually calculate the Eigen directions,
first analytically work that out that and
then do the simulation. Don't do the simulation
first. Then you won't develop understanding.
but that has at least convinced that there
must be Eigen directions like this and there
must be Eigen planes like that and depending
on where I start from, this state space is
divided into compartments and that sort of
tells which equilibrium point ultimately I
will converge at. Now these have certain names.
A situation where there are two equilibria
and this state space is sort of divided into
compartments so that from this compartment
it goes here from this compartment it goes
there. These regions are called the "basins
of attraction". So from this basin all initial
conditions will ultimately hold on to this
equilibrium point. This basin ultimately all
hold on to that equilibrium point. Now what
is the attraction thing going on here? See,
in gravity you have heard that everybody is
sort of attracting everything. So sun is attracting
the earth. The earth is attracting the moon
and everything is attracting the others. So,
here in this state space. you can, in similarity
visualize these points are sort of gravitating
bodies and anything in this vicinity will
be attracted to it. In that sense, these equilibrium
points that are stable equilibrium points
are also called attractors. You might say
that the sink type of equilibrium point is
an attractor. Well, in linear systems, that
is the only type of attractor you have heard
of. These two are that type of attractors.
But in the non-linear system, you have heard
of another type of attractor called the limit
cycle. The limit cycle itself is an attractor.
We have discussed in the last class that a
limit cycle is itself an attractor. In the
sense that if you start outside, there is
a spiral that goes inwards. If you start inside
the limit cycle, there is a spiral that goes
outward. Ultimately it holds on to that. In
that sense limit cycle is also an attractor.
So linear systems theory will tell you that
there is only one type of attractor - "the
point attractor". Non-linear systems theory
will tell you that there is also another type
of attractor- "the periodical attractor"
that goes on and all oscillators as I told
you are attractors. Now from the stable equilibrium
point, how does the limit cycle kind of behavior
develop?
If 
the eigenvalue here where complex with negative
real part so that the behavior would be incoming
spiral kind and with the change of the parameter,
if they move like this so that at a certain
parameter value, they become outward spiral
because of the non-linear behavior of the
system, this behavior that I was talking about
- the outgoing spiral behavior that pertains
to only a very close neighborhood of the that
equilibrium point thereby the limit cycle
develops. So what has actually happening at
the equilibrium point if I calculate the Eigenvalues?
It is this. Eigenvalues that was complex with
negative real part were moving this way so
that the real part becomes positive. For our
three dimensional system that I have given,
does something like that happen? Try to find
out. This equilibrium points, you can find
the eigenvalues. So I will ask you to to workout.
Do these eigenvalues, for some parameter value,
do something like this? If that happens, then
I would expect, after sometime these behaviors
to develop. So that can be worked out from
looking at the local linear neighborhood.
beyond that these two equilibria are separate
and each of them have their own basins of
attraction and beyond that if you change the
parameter, you will not be able to infer what
happens from linear system. So the linear
systems theory will tell you what happens
near the equilibria point and you will be
able to infer the occurrence of limit cycles.
I don't want you to do this simulation.
That we will do later. Try to do things as
much as possible by analysis. That's all
for today.
