LYDIA BOUROUIBA: Welcome
to this presentation
on the trace-determinant
diagram.
So here, you're asked to
label the regions and lines
of the trace-determinant diagram
for a 2 x 2 general system,
written in the form
x prime equals A*x,
and to indicate the
stability on your diagram.
So here, as a
reminder, this system
is simply a system of two
differential equations
in vector form.
The derivative of [x, y] equals
[a, b; c, d], a 2 x 2 matrix,
multiplying the vector [x, y].
Or in another form, it
would be x-dot equals
f of x, y, and y-dot equals j
of x, y, where the t wouldn't
appear in f and j
here, functions,
which means that the
system would be autonomous.
And we're dealing
with linear systems.
So why don't you
pause the video.
Take a few minutes
remind yourself
what the trace-determinant
diagram is and how to label it,
and I'll be right back.
Welcome back.
So let's remind ourselves where
this trace-determinant diagram
comes from.
So initially, what we saw
before was to solve this system,
we need to find the
eigenvalues of the matrix A.
The eigenvalues are solutions
of this equation, which
can be written in the form
lambda square minus trace
of lambda plus determinant
of A equals to 0.
Let's call this D, and this
T. And the trace of A is just
the sum of the
diagonals of the matrix,
and the determinant is just,
in this case for a 2 x 2,
a*d minus c*b.
So the solutions for-- I'm going
to call them plus and minus--
for this second-order
polynomial will simply
be T plus or minus the
root of the determinant
of second-order polynomial,
which is T square minus 4D.
So the sign of T
square minus 4D is
going to determine
whether we are dealing
with real eigenvalues
or complex eigenvalues,
or simply repeated
eigenvalues if we
have this determinant under
the root being equal to 0.
So just as a reminder here,
if we have T square over 4
equals to D, we are in the
case, well, this is equal to 0.
Lambda plus or
minus are the same,
and they're just equal
to the trace over 2.
So the sign of
the trace is going
to determine whether we have two
repeated negative eigenvalues
or two repeated
positive eigenvalues.
Now, if we have the
determinant of the matrix
A is larger than T square over
4, then we are above the curve
determined by this equation,
which is a parabola, which
I'll draw in a minute.
And in this case, we have
the number under the root
being negative, so we're dealing
with complex eigenvalues.
And basically, they're
just complex conjugate
of each other.
And you can notice here
that the trace would
be also the sum of
the two eigenvalues,
and basically we would just have
2 times the real part of lambda
plus or minus, just as a note.
And in the third case, where
we have that the determinant is
below T square over 4,
we're in the case where
this number under
the root is positive,
so we're dealing with
two real eigenvalues.
So here, I should
mention this is real.
Lambda plus, lambda minus real.
And we can have multiple cases.
We can have lambda plus
larger than lambda minus,
both positive.
We can have lambda plus,
less than lambda minus,
both negative.
Or we can have lambda plus
positive and lambda minus
negative.
And each case will give
us a different behavior
of the system.
So let's first summarize
this in this following
determinant-trace
diagram, and then we'll
start labeling this diagram.
I should probably keep a
little more space here.
So this is basically D
equal T square over 4
in our trace-determinant
diagram.
I'm going to also erase this
part and just keep it in dots.
So in the first case
that we looked at,
we are in the case where we
are right on this parabola.
So that's the case where we have
lambda plus equal lambda minus.
And you can see that if
the trace is positive,
so if we are on the right
hand side of the diagram,
that we're going to have two
positive eigenvalues that
are both real, and
they are repeated.
So we can have multiple cases.
We can have the case where we
have a defective matrix, where
basically here we only have
one eigenvector associated
with this repeated eigenvalue.
And we have a
defective case where
we need to come up with a
second eigenvector using
the generalized
eigenvector formula.
And I'm not going
to write this here,
but I'm going to just
to do the diagram.
So for example, we would
have one direction, v_1,
where we would have in this
case lambda_1 and lambda minus
positive.
So in the phase space, in
y-x, the little diagram
would show us that
the solution are
escaping from the critical
point, the equilibrium point.
And the second solution that we
build would have a dependence
in t*v_1, plus the
second eigenvector v_2,
also directed by the
positive eigenvalue.
And so, we would
have a solution,
for example, that would look
like this, with the solution
escaping from the critical
point because, again, we
are in the case where the
two eigenvalues are positive.
And so here, we could
have it in this form,
or we can also have the
diagram in the other direction.
And I'll to it in the other
direction on the other wing
of the diagram.
So this is the defective case.
Defective node.
The other possibility that we
could have on this parabola--
and I'm going to just to it
here-- will be the case where
we actually have
all the direction
could be eigenvectors
associated with this eigenvalue.
And this would be, for
example, for a diagonal matrix.
In which case,
all the directions
would be escaping-- all
the trajectories would
be escaping-- from
the critical point.
And this would be a star.
Obviously here, we are
in the unstable case
for the defective
node, and we are also
in the unstable
case for the star,
because all the solutions are
escaping and basically going
away.
So if I start at the equilibrium
and I perturb it a little bit,
the solutions would want to
escape from that equilibrium
point.
On the other side here, it would
be exactly the same structure,
except that I would
have stability
because the trace is negative.
So I would have
asymptotically stable star.
Why asymptotically?
Because basically,
it's when t goes
to infinity that the
trajectories reach
the critical point.
And this is again in the
phase space y-x diagram.
For the defective node,
we would have this time,
for example, direction
v_1 attracting
the solutions, the ray v_1.
And the new trajectory that
we will be constructing based
on v_1 again would be t*v_1 plus
v_2, generalized eigenvector.
Both of them would
give us solutions
that converge toward
the equilibrium,
because the two eigenvalues
here are negative.
And so, we have that for
large time, minus infinity,
the solution follows
v_1, for plus infinity
it's going toward the
equilibrium point,
also follows V1.
So the solutions would have to
look like this, for example.
And this would be asymptotically
stable defective node.
OK.
So we're done with the
points on the parabola.
So now let's look
at the other points,
and I'll go maybe a bit
less in the details.
So for the case where we have
the determinant larger then T
square over 4, so we are
just above this parabola now,
we have the case where we
have two complex eigenvalues.
They are two complex conjugates.
So let's assume that we can
just expand our solution
and write it in terms
of the exponential,
in terms of the real
part of the eigenvalue.
So it would be determined,
again, the trace
will give us the sign of the
real part of the eigenvalue,
multiplying a cosine and a sine.
So we have something that
is rotating in phase space,
because we have basically
the periodicity.
But the distance to the
critical point is changing,
and it's either
growing, if we have
a positive real part
for our eigenvalues,
so if we are on this side.
Or decaying if we are on the
left side of the diagram.
So that gives us
typically spirals.
So for example, that would
be a spiral, going toward 0,
toward the equilibrium
point in phase space.
And here, we would
be in the case
where we have instability
due to the fact
that the real part of the
eigenvalue determining
stability is positive.
And so the solution
of the trajectories
escape from the critical point.
And similarly here, we could
have the same structure,
but I'm just going to draw
it in another direction.
Where here, we would
have stability.
Oops.
They should not cross.
This is not the right
way to draw this.
And it would be going
toward the critical point.
So here, just a quick note.
You can have this trajectory
being drawn this way.
So here, we have basically
a clockwise motion.
But we could also
have it be drawn
in the other way for both cases,
giving us another direction
of rotation in phase space.
And the direction
that you choose
will be determined by the lowest
left entry of your matrix A,
and I will just explain
quickly how we do that.
OK.
So we have basically here
unstable spiral node,
and here it's again
asymptotically stable spiral.
So what happens now if
we are in this case?
We're still above
the parabola, so we
have still complex eigenvalues.
However, we have now
the fact that the trace
is equal to 0, which means
that the eigenvalues don't
have any real part.
So basically, we have
pure oscillation.
And in the phase
space, that corresponds
to closed trajectories that
could be circles or ellipses,
basically.
And so for example,
we would have
something in this form that
would be called the center.
And here again, you can
have either counterclockwise
or a clockwise rotation
in phase space depending
on the signs of the
entries of your matrix.
So one thing I want to note
is that here, the stability
for the center we say that
it's simply stable, and not
asymptotically stable,
because the solution
never actually reaches
the critical point,
but stays in the region
around the critical point.
So we're left with a
few other cases that
are now all below the parabola.
And for that, we just
have real eigenvalues.
So let's look at the case
where the eigenvalues have
two different signs.
So for that, we have to have
the determinant being negative.
So this whole lower
part of the diagram,
because the determinant is the
product of the two eigenvalues.
So if they have different
signs, we're in this region.
So in this case, we could have,
for example, one eigenvalue,
with associated eigenvector
v_1, being negative.
So the trajectories
along this ray
would be going toward
equilibrium point.
And then the other
eigenvalue will be positive.
So for example, lambda_2
here would correspond
to this other eigenvector v_2.
And so here, we
would have solutions
that would be close
to v_2 when we're
coming, for example, from
minus infinity approaching
the critical point.
And then going back,
approaching v_2
when we go at t plus
infinity, for example.
And so it gives us
pseudo-hyperbola form here,
of this form.
And this is said to be unstable,
because all those solutions do
not go to the critical point.
But we have some
stable manifolds.
For example, the v_1.
If we start on this
ray, we would be going
toward the critical point.
And if we start at the
critical point itself,
it would stay at
the critical point.
But it is unstable.
So now what happened in
these two different regions,
to finish.
In these two different regions--
and I'm going to have a little
bit more space here--
AUDIENCE: It's called a saddle.
LYDIA BOUROUIBA: Oh, yeah.
Thank you.
And this would be a saddle, and
it's just because of the shape.
I'm going to just add
a little bit of space
here, so that we can
complete the diagram.
I'll be basically looking at
the regions here in the wedge,
where we are below the parabola.
And I'm looking
for another color.
Maybe I'll just use white.
Now we are in the case where
we would have two eigenvalues.
Both real, so no oscillation.
But let's say both positive,
because we're basically
on the region where
the trace is positive.
Let's say that we
have one ray, v_1.
One ray, v_2.
The trajectories are going
away from the critical point,
because the two
eigenvalues are positive.
And now, what do the
other trajectories do,
where they follow--
so for example,
they would be following v_2.
And I'm going to explain how.
So here, we're in the case
where obviously we're unstable,
because again, the
solution, the trajectories
are going away from
the critical point.
And here, how do
you pick which ray
do you follow when you get
closer to the critical point?
Where in this case, we
would be in a situation
where we have lambda_2 smaller
than lambda_1, larger than 0.
So basically, the lambda 2
that is the closer to 0--
and that's also the
case for the positive
and the negative
eigenvalues-- is
the one that determines
the solution closer
to the critical point.
And the larger in absolute
value eigenvalue and its ray
then determines the behavior--
Oh, sorry.
There's a mistake here.
It should be lambda_1.
I said it, but I think
I wrote it reversely.
Yeah The eigenvalue closer
to 0 is the one that
determines the behavior at 0.
So here, when we're
going to infinity,
the larger eigenvalue lambda_2
will determine the behavior,
and the trajectories will become
more and more parallel to v_2.
So what happens
on this side would
be exactly the same
diagram, and I'm just
going to do it with the
same-- let's say, v_2 here.
Except that we would
have our trajectories
going toward the critical point.
And the trajectory
here again would
be closer to v_1, which means
that we would have a case where
we have lambda_1 less than
0 and larger than lambda_2.
So that finishes
roughly the diagram.
I didn't detail a few
borderline regions.
For example, the region where
the determinant equals to 0.
And we will discuss that
in another recitation.
And the case also at which
the determinant and the trace
are equal to 0.
So we'll detail that also.
But try to think about it,
to complete this diagram.
So the key points
here were to remember
what is the determinant-trace
diagram, how
to basically deduce the nature
of the eigenvalues based
on T and D, their sign.
And where to place the
different structures
on this determinant-trace
diagram.
And that ends this recitation.
