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PROFESSOR: Welcome
to this recitation

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on the phase portrait.

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So you're given matrix A, a
two-by-two matrix with entries

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minus 1, minus 1,
1 minus c, where

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c is a constant we will
be varying, and minus 1.

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As c increases from 0 to
other positive values,

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determine the path of the
system on the trace-determinant

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diagram that we saw in
previous recitations

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and draw the phase
portraits that

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corresponds to the critical
points and the nature

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of the critical points.

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So here, we're basically
talking of a system that

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would be a two-by-two system
where we have derivatives

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of entries x, y for vector,
equals to this matrix, A_c,

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multiplying the
unknown vector x_c.

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So take a few minutes and
work out through this problem

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and I'll be right back.

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Welcome back.

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So I already prepared the
trace-determinant diagram

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for you.

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So here as a reminder,
this is the parabola

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that determines whether
we're going to have

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repeated eigenvalues or not.

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Above this parabola, we have two
complex conjugate eigenvalues.

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Below this parabola, we have
real eigenvalues, either

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of the same signs,
positive or negative,

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or of different signs.

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And then, we have
the borderline cases.

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In our case here, we're
looking in a matrix A

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with the trace equals to minus
1 minus 1, so it's minus 2,

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and a determinant that
is equal to 2 minus c.

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So here, we are basically
along this dotted line

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where the trace
equals to minus 2.

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So either we're going
to have complex values

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with negative real parts
or negative eigenvalues.

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So we're in stable
configurations.

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So we're going to be
moving at c equal to 0

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from a case where-- I'm just
going to label this point 1.

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For example, we have the
matrix, minus 1, 1, 1, minus 1

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with the trace
equals to minus 2,

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and the determinant equals to 2.

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So clearly, we are this point.

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And we are just going to be
increasing the value of c

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and moving along these lines,
crossing this first boundary

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case where we will have
either defective or not case.

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And I'll discuss very
briefly what we have.

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And then, we have other
value where we are basically

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in this area where we have
real eigenvalues, both of them

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negative, so we have stability.

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And it would
basically be a sink.

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And then we cross this
other borderline case

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where the determinant
is equal to 0.

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And here, we have one eigenvalue
that is equal to 0 and one

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that is negative.

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And then, we get to do
the part of the diagram

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where the determinant
is negative,

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and the trace is negative.

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And we basically have
a saddle structure

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where we have one
eigenvalue as negative

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and another eigenvalue
that is positive.

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So let's get through
these five cases.

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And I'll just do more
detailed picture illustrating

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what I just said.

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So just write down a bit more
in detail the first case here.

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So we have the case
where the eigenvalues are

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both complex
conjugate, and we have

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their real part being negative.

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So basically, we have spirals,
asymptotically stable spirals.

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And the spirals could be either
clockwise or counterclockwise.

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And one way of determining
the direction of these spirals

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is to look at the
sign of the velocity

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vector of the trajectories
in the phase space.

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So the phase space
basically diagram here

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where we have x and y as
the axes of this space.

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So let's look, for example, at
a particular point here that

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would be velocity vector [1, 0].

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And this position vector [1, 0]
would give us a velocity vector

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of minus 1 and plus 1.

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So basically, this vector
would be directed upward,

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which means that we are in
the case where we would have

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a spiral coming this way
toward the critical point

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and basically with the velocity
vector here going this way.

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You don't have to
actually do all this

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to figure out which direction
of the spiral you should choose.

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You can just look-- for these
cases, two-by-two matrices--

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at the entry at the lower
left part of your matrix.

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And if the coefficient
here is positive,

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then it will determine the
direction of the velocity

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vector at this point.

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So if it's positive
in this case,

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you would have this
counterclockwise direction

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of rotation of the spiral.

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So now we can move on and do the
following cases a bit faster.

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So the second case, we are
in the case where we now

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have two repeated eigenvalues.

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Both of them are negative.

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So we can have either
a complete case.

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If we had a matrix
that was diagonal,

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we would have basically a star.

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But that's not the case,
because if we pick, for example,

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the value c equals to 1 that
puts us on the parabola,

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we can see that the matrix
A_c is not diagonal.

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So we are in a defective case.

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The defective case, if you
compute the eigenvalues

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and eigenvector would give you
something that looks like that.

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The first ray of the
eigenvector corresponding

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to your eigenvalues would be
in the direction of [1, 0].

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We have negative eigenvalues,
so it's going to 0.

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And basically, to
determine the direction,

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you have to see that we are
also following a transition here

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on the trace-determinant
diagram.

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And so basically, we
will have a spiral

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that would look like this.

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What am I doing?

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Am I doing the same thing?

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And it would basically
rotate, as you can see here,

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in the same direction as the
case that we had just before.

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Because we basically have the
transition of the structure

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of the critical point.

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And the arrows are pointing
toward the critical point

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again, because our two
eigenvalues are basically

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here negative.

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It is just a repeated
eigenvalue that is negative.

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So now we can move
on to a third case

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where here we move
into the wedge area

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where we have the two
eigenvalues now being real,

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and both of them are negative.

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So for example, we would
have two eigenvalues that

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would give us two eigenvectors
that are basically,

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they're rays.

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The direction of the
trajectories on these two rays

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would be toward
the critical point,

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because we have, again,
negative eigenvalues.

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And the trajectories then
would be following the lambda.

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So if I, for example, pick
lambda_1 less than lambda_2,

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then I would have something
like this going to 0.

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And we would have this
trajectory corresponding

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to the eigenvalue
that is closer to 0.

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And again, the
arrows would be going

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toward this critical point.

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So we can now move on.

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And if we keep increasing
c, we reach now the point

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4 where we are in
the special case,

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again, a boundary case
where the determinant is 0.

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But the trace is non-zero.

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If the determinant is 0, which
is the product of the two

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eigenvalues, it means that
we have one eigenvalue that

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is equal to 0 and another
eigenvalue that is actually,

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in this case, minus
2, so just negative.

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So what happens here?

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What happens is that we
have now the eigenvector

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that corresponds to the
eigenvalue equals to 0

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is basically just defining a
whole line of critical points.

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So all the points
on this line that

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would correspond, so
for example, to lambda 0

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are all critical points.

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So we don't need to
actually draw arrows here.

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All the points would
be critical points.

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And another direction
or ray would

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be determined by the second
eigenvalue, lambda_2.

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And we would have, for example,
directions in this way.

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It would correspond to
the direction of v_2.

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So all these would
be parallel to v_2.

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And we would have
the trajectories

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going toward all the
critical points on this line.

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So in this case, we're in
a case where we actually

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don't have one
localized critical point

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like in the previous
case that we saw.

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We've actually have a whole
line of critical points.

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So let's move one
more point down.

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So here, another
value of c, we now

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are in the lower
part of the diagram

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where we now have
two eigenvalues that

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are again real.

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But now, the
determinant is negative.

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Determinant is the
product of the two.

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So basically we know
that one is positive

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and the other one is negative.

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So for example here, we
would have one positive,

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the negative eigenvalue.

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So we would have the
positive eigenvalue

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and the negative
eigenvalue here.

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So here, again, we're just
moving in the structure

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of the critical point.

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So we should have a smooth
transition between the diagrams

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going 1 to 5.

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And so here, what we
see is that we actually

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have the ray that is basically
the stable space here

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that would give us the
trajectories going to 0.

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But it's the only region
where we have stability.

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And all the trajectories
will just then

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be tangent in minus
infinity to the ray v_1,

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corresponding to the
negative eigenvalue,

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and going towards the
direction of the ray

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with a positive eigenvalue
at plus infinity.

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So we would have
something like that, that

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would correspond to a saddle.

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This would be a comb.

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This is basically a sink
or basically a stable node.

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This would be
defective stable node.

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And here, we just have a
asymptotically stable spiral.

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So basically here
what you could see

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is that we have,
by just changing

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the value of the constant
c, had moved the structure

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of the critical
point of the system

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from the phase
diagrams one to five.

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And the transitions can
be seen to be smooth

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if you had a continuum
of values for c.

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So that ends this recitation.