WEBVTT

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LYDIA BOUROUIBA: Welcome
to this presentation

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on the trace-determinant
diagram.

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So here, you're asked to
label the regions and lines

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of the trace-determinant diagram
for a 2 x 2 general system,

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written in the form
x prime equals A*x,

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and to indicate the
stability on your diagram.

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So here, as a
reminder, this system

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is simply a system of two
differential equations

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in vector form.

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The derivative of [x, y] equals
[a, b; c, d], a 2 x 2 matrix,

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multiplying the vector [x, y].

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Or in another form, it
would be x-dot equals

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f of x, y, and y-dot equals j
of x, y, where the t wouldn't

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appear in f and j
here, functions,

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which means that the
system would be autonomous.

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And we're dealing
with linear systems.

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So why don't you
pause the video.

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Take a few minutes
remind yourself

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what the trace-determinant
diagram is and how to label it,

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and I'll be right back.

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Welcome back.

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So let's remind ourselves where
this trace-determinant diagram

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comes from.

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So initially, what we saw
before was to solve this system,

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we need to find the
eigenvalues of the matrix A.

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The eigenvalues are solutions
of this equation, which

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can be written in the form
lambda square minus trace

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of lambda plus determinant
of A equals to 0.

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Let's call this D, and this
T. And the trace of A is just

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the sum of the
diagonals of the matrix,

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and the determinant is just,
in this case for a 2 x 2,

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a*d minus c*b.

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So the solutions for-- I'm going
to call them plus and minus--

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for this second-order
polynomial will simply

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be T plus or minus the
root of the determinant

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of second-order polynomial,
which is T square minus 4D.

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So the sign of T
square minus 4D is

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going to determine
whether we are dealing

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with real eigenvalues
or complex eigenvalues,

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or simply repeated
eigenvalues if we

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have this determinant under
the root being equal to 0.

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So just as a reminder here,
if we have T square over 4

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equals to D, we are in the
case, well, this is equal to 0.

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Lambda plus or
minus are the same,

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and they're just equal
to the trace over 2.

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So the sign of
the trace is going

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to determine whether we have two
repeated negative eigenvalues

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or two repeated
positive eigenvalues.

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Now, if we have the
determinant of the matrix

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A is larger than T square over
4, then we are above the curve

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determined by this equation,
which is a parabola, which

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I'll draw in a minute.

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And in this case, we have
the number under the root

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being negative, so we're dealing
with complex eigenvalues.

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And basically, they're
just complex conjugate

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of each other.

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And you can notice here
that the trace would

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be also the sum of
the two eigenvalues,

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and basically we would just have
2 times the real part of lambda

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plus or minus, just as a note.

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And in the third case, where
we have that the determinant is

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below T square over 4,
we're in the case where

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this number under
the root is positive,

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so we're dealing with
two real eigenvalues.

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So here, I should
mention this is real.

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Lambda plus, lambda minus real.

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And we can have multiple cases.

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We can have lambda plus
larger than lambda minus,

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both positive.

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We can have lambda plus,
less than lambda minus,

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both negative.

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Or we can have lambda plus
positive and lambda minus

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negative.

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And each case will give
us a different behavior

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of the system.

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So let's first summarize
this in this following

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determinant-trace
diagram, and then we'll

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start labeling this diagram.

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I should probably keep a
little more space here.

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So this is basically D
equal T square over 4

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in our trace-determinant
diagram.

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I'm going to also erase this
part and just keep it in dots.

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So in the first case
that we looked at,

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we are in the case where we
are right on this parabola.

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So that's the case where we have
lambda plus equal lambda minus.

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And you can see that if
the trace is positive,

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so if we are on the right
hand side of the diagram,

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that we're going to have two
positive eigenvalues that

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are both real, and
they are repeated.

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So we can have multiple cases.

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We can have the case where we
have a defective matrix, where

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basically here we only have
one eigenvector associated

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with this repeated eigenvalue.

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And we have a
defective case where

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we need to come up with a
second eigenvector using

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the generalized
eigenvector formula.

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And I'm not going
to write this here,

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but I'm going to just
to do the diagram.

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So for example, we would
have one direction, v_1,

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where we would have in this
case lambda_1 and lambda minus

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positive.

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So in the phase space, in
y-x, the little diagram

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would show us that
the solution are

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escaping from the critical
point, the equilibrium point.

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And the second solution that we
build would have a dependence

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in t*v_1, plus the
second eigenvector v_2,

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also directed by the
positive eigenvalue.

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And so, we would
have a solution,

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for example, that would look
like this, with the solution

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escaping from the critical
point because, again, we

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are in the case where the
two eigenvalues are positive.

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And so here, we could
have it in this form,

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or we can also have the
diagram in the other direction.

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And I'll to it in the other
direction on the other wing

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of the diagram.

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So this is the defective case.

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Defective node.

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The other possibility that we
could have on this parabola--

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and I'm going to just to it
here-- will be the case where

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we actually have
all the direction

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could be eigenvectors
associated with this eigenvalue.

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And this would be, for
example, for a diagonal matrix.

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In which case,
all the directions

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would be escaping-- all
the trajectories would

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be escaping-- from
the critical point.

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And this would be a star.

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Obviously here, we are
in the unstable case

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for the defective
node, and we are also

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in the unstable
case for the star,

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because all the solutions are
escaping and basically going

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away.

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So if I start at the equilibrium
and I perturb it a little bit,

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the solutions would want to
escape from that equilibrium

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point.

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On the other side here, it would
be exactly the same structure,

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except that I would
have stability

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because the trace is negative.

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So I would have
asymptotically stable star.

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Why asymptotically?

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Because basically,
it's when t goes

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to infinity that the
trajectories reach

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the critical point.

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And this is again in the
phase space y-x diagram.

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For the defective node,
we would have this time,

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for example, direction
v_1 attracting

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the solutions, the ray v_1.

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And the new trajectory that
we will be constructing based

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on v_1 again would be t*v_1 plus
v_2, generalized eigenvector.

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Both of them would
give us solutions

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that converge toward
the equilibrium,

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because the two eigenvalues
here are negative.

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And so, we have that for
large time, minus infinity,

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the solution follows
v_1, for plus infinity

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it's going toward the
equilibrium point,

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also follows V1.

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So the solutions would have to
look like this, for example.

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And this would be asymptotically
stable defective node.

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OK.

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So we're done with the
points on the parabola.

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So now let's look
at the other points,

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and I'll go maybe a bit
less in the details.

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So for the case where we have
the determinant larger then T

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square over 4, so we are
just above this parabola now,

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we have the case where we
have two complex eigenvalues.

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They are two complex conjugates.

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So let's assume that we can
just expand our solution

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and write it in terms
of the exponential,

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in terms of the real
part of the eigenvalue.

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So it would be determined,
again, the trace

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will give us the sign of the
real part of the eigenvalue,

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multiplying a cosine and a sine.

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So we have something that
is rotating in phase space,

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because we have basically
the periodicity.

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But the distance to the
critical point is changing,

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and it's either
growing, if we have

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a positive real part
for our eigenvalues,

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so if we are on this side.

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Or decaying if we are on the
left side of the diagram.

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So that gives us
typically spirals.

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So for example, that would
be a spiral, going toward 0,

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toward the equilibrium
point in phase space.

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And here, we would
be in the case

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where we have instability
due to the fact

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that the real part of the
eigenvalue determining

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stability is positive.

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And so the solution
of the trajectories

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escape from the critical point.

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And similarly here, we could
have the same structure,

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but I'm just going to draw
it in another direction.

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Where here, we would
have stability.

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Oops.

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They should not cross.

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This is not the right
way to draw this.

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And it would be going
toward the critical point.

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So here, just a quick note.

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You can have this trajectory
being drawn this way.

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So here, we have basically
a clockwise motion.

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But we could also
have it be drawn

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in the other way for both cases,
giving us another direction

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of rotation in phase space.

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And the direction
that you choose

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will be determined by the lowest
left entry of your matrix A,

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and I will just explain
quickly how we do that.

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OK.

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So we have basically here
unstable spiral node,

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and here it's again
asymptotically stable spiral.

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So what happens now if
we are in this case?

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We're still above
the parabola, so we

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have still complex eigenvalues.

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However, we have now
the fact that the trace

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is equal to 0, which means
that the eigenvalues don't

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have any real part.

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So basically, we have
pure oscillation.

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And in the phase
space, that corresponds

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to closed trajectories that
could be circles or ellipses,

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basically.

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And so for example,
we would have

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something in this form that
would be called the center.

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And here again, you can
have either counterclockwise

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or a clockwise rotation
in phase space depending

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on the signs of the
entries of your matrix.

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So one thing I want to note
is that here, the stability

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for the center we say that
it's simply stable, and not

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asymptotically stable,
because the solution

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never actually reaches
the critical point,

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but stays in the region
around the critical point.

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So we're left with a
few other cases that

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are now all below the parabola.

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And for that, we just
have real eigenvalues.

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So let's look at the case
where the eigenvalues have

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two different signs.

00:13:40.500 --> 00:13:43.430
So for that, we have to have
the determinant being negative.

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So this whole lower
part of the diagram,

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because the determinant is the
product of the two eigenvalues.

00:13:48.685 --> 00:13:51.080
So if they have different
signs, we're in this region.

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So in this case, we could have,
for example, one eigenvalue,

00:14:00.690 --> 00:14:04.420
with associated eigenvector
v_1, being negative.

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So the trajectories
along this ray

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would be going toward
equilibrium point.

00:14:11.800 --> 00:14:15.915
And then the other
eigenvalue will be positive.

00:14:15.915 --> 00:14:19.230
So for example, lambda_2
here would correspond

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to this other eigenvector v_2.

00:14:23.340 --> 00:14:25.610
And so here, we
would have solutions

00:14:25.610 --> 00:14:28.910
that would be close
to v_2 when we're

00:14:28.910 --> 00:14:32.289
coming, for example, from
minus infinity approaching

00:14:32.289 --> 00:14:33.080
the critical point.

00:14:33.080 --> 00:14:37.030
And then going back,
approaching v_2

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when we go at t plus
infinity, for example.

00:14:39.350 --> 00:14:44.750
And so it gives us
pseudo-hyperbola form here,

00:14:44.750 --> 00:14:45.850
of this form.

00:14:49.770 --> 00:14:57.330
And this is said to be unstable,
because all those solutions do

00:14:57.330 --> 00:14:58.740
not go to the critical point.

00:14:58.740 --> 00:15:00.400
But we have some
stable manifolds.

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For example, the v_1.

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If we start on this
ray, we would be going

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toward the critical point.

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And if we start at the
critical point itself,

00:15:06.330 --> 00:15:07.830
it would stay at
the critical point.

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But it is unstable.

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So now what happened in
these two different regions,

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to finish.

00:15:12.506 --> 00:15:15.130
In these two different regions--
and I'm going to have a little

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bit more space here--

00:15:16.230 --> 00:15:18.037
AUDIENCE: It's called a saddle.

00:15:18.037 --> 00:15:19.120
LYDIA BOUROUIBA: Oh, yeah.

00:15:19.120 --> 00:15:19.715
Thank you.

00:15:19.715 --> 00:15:22.340
And this would be a saddle, and
it's just because of the shape.

00:15:27.350 --> 00:15:29.230
I'm going to just add
a little bit of space

00:15:29.230 --> 00:15:34.750
here, so that we can
complete the diagram.

00:15:34.750 --> 00:15:38.820
I'll be basically looking at
the regions here in the wedge,

00:15:38.820 --> 00:15:40.790
where we are below the parabola.

00:15:40.790 --> 00:15:42.800
And I'm looking
for another color.

00:15:42.800 --> 00:15:46.210
Maybe I'll just use white.

00:15:46.210 --> 00:15:51.360
Now we are in the case where
we would have two eigenvalues.

00:15:51.360 --> 00:15:53.480
Both real, so no oscillation.

00:15:53.480 --> 00:15:57.480
But let's say both positive,
because we're basically

00:15:57.480 --> 00:16:00.370
on the region where
the trace is positive.

00:16:00.370 --> 00:16:02.810
Let's say that we
have one ray, v_1.

00:16:02.810 --> 00:16:04.250
One ray, v_2.

00:16:04.250 --> 00:16:07.100
The trajectories are going
away from the critical point,

00:16:07.100 --> 00:16:10.840
because the two
eigenvalues are positive.

00:16:10.840 --> 00:16:13.730
And now, what do the
other trajectories do,

00:16:13.730 --> 00:16:17.845
where they follow--
so for example,

00:16:17.845 --> 00:16:22.110
they would be following v_2.

00:16:22.110 --> 00:16:24.220
And I'm going to explain how.

00:16:24.220 --> 00:16:28.800
So here, we're in the case
where obviously we're unstable,

00:16:28.800 --> 00:16:30.730
because again, the
solution, the trajectories

00:16:30.730 --> 00:16:32.530
are going away from
the critical point.

00:16:32.530 --> 00:16:36.530
And here, how do
you pick which ray

00:16:36.530 --> 00:16:38.970
do you follow when you get
closer to the critical point?

00:16:38.970 --> 00:16:41.630
Where in this case, we
would be in a situation

00:16:41.630 --> 00:16:49.720
where we have lambda_2 smaller
than lambda_1, larger than 0.

00:16:49.720 --> 00:16:53.215
So basically, the lambda 2
that is the closer to 0--

00:16:53.215 --> 00:16:55.090
and that's also the
case for the positive

00:16:55.090 --> 00:16:57.550
and the negative
eigenvalues-- is

00:16:57.550 --> 00:17:00.360
the one that determines
the solution closer

00:17:00.360 --> 00:17:01.390
to the critical point.

00:17:01.390 --> 00:17:05.950
And the larger in absolute
value eigenvalue and its ray

00:17:05.950 --> 00:17:07.635
then determines the behavior--

00:17:11.839 --> 00:17:12.339
Oh, sorry.

00:17:12.339 --> 00:17:13.297
There's a mistake here.

00:17:13.297 --> 00:17:14.530
It should be lambda_1.

00:17:17.540 --> 00:17:19.720
I said it, but I think
I wrote it reversely.

00:17:19.720 --> 00:17:23.079
Yeah The eigenvalue closer
to 0 is the one that

00:17:23.079 --> 00:17:24.329
determines the behavior at 0.

00:17:24.329 --> 00:17:26.280
So here, when we're
going to infinity,

00:17:26.280 --> 00:17:28.910
the larger eigenvalue lambda_2
will determine the behavior,

00:17:28.910 --> 00:17:32.530
and the trajectories will become
more and more parallel to v_2.

00:17:32.530 --> 00:17:34.510
So what happens
on this side would

00:17:34.510 --> 00:17:36.880
be exactly the same
diagram, and I'm just

00:17:36.880 --> 00:17:45.560
going to do it with the
same-- let's say, v_2 here.

00:17:45.560 --> 00:17:54.790
Except that we would
have our trajectories

00:17:54.790 --> 00:17:56.260
going toward the critical point.

00:18:01.980 --> 00:18:03.950
And the trajectory
here again would

00:18:03.950 --> 00:18:07.240
be closer to v_1, which means
that we would have a case where

00:18:07.240 --> 00:18:12.600
we have lambda_1 less than
0 and larger than lambda_2.

00:18:12.600 --> 00:18:14.920
So that finishes
roughly the diagram.

00:18:14.920 --> 00:18:18.530
I didn't detail a few
borderline regions.

00:18:18.530 --> 00:18:22.360
For example, the region where
the determinant equals to 0.

00:18:22.360 --> 00:18:25.740
And we will discuss that
in another recitation.

00:18:25.740 --> 00:18:29.330
And the case also at which
the determinant and the trace

00:18:29.330 --> 00:18:30.057
are equal to 0.

00:18:30.057 --> 00:18:31.140
So we'll detail that also.

00:18:31.140 --> 00:18:35.890
But try to think about it,
to complete this diagram.

00:18:35.890 --> 00:18:38.280
So the key points
here were to remember

00:18:38.280 --> 00:18:42.960
what is the determinant-trace
diagram, how

00:18:42.960 --> 00:18:45.910
to basically deduce the nature
of the eigenvalues based

00:18:45.910 --> 00:18:47.710
on T and D, their sign.

00:18:47.710 --> 00:18:51.000
And where to place the
different structures

00:18:51.000 --> 00:18:53.540
on this determinant-trace
diagram.

00:18:53.540 --> 00:18:55.820
And that ends this recitation.