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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.
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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.
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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,
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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.
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And then the other
eigenvalue will be positive.
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So for example, lambda_2
here would correspond
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to this other eigenvector v_2.
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And so here, we
would have solutions
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that would be close
to v_2 when we're
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coming, for example, from
minus infinity approaching
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the critical point.
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00:14:33,080 --> 00:14:37,030
And then going back,
approaching v_2
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00:14:37,030 --> 00:14:39,350
when we go at t plus
infinity, for example.
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00:14:39,350 --> 00:14:44,750
And so it gives us
pseudo-hyperbola form here,
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00:14:44,750 --> 00:14:45,850
of this form.
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00:14:49,770 --> 00:14:57,330
And this is said to be unstable,
because all those solutions do
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00:14:57,330 --> 00:14:58,740
not go to the critical point.
253
00:14:58,740 --> 00:15:00,400
But we have some
stable manifolds.
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00:15:00,400 --> 00:15:01,980
For example, the v_1.
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00:15:01,980 --> 00:15:03,757
If we start on this
ray, we would be going
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00:15:03,757 --> 00:15:04,840
toward the critical point.
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00:15:04,840 --> 00:15:06,330
And if we start at the
critical point itself,
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00:15:06,330 --> 00:15:07,830
it would stay at
the critical point.
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00:15:07,830 --> 00:15:09,150
But it is unstable.
260
00:15:09,150 --> 00:15:11,350
So now what happened in
these two different regions,
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00:15:11,350 --> 00:15:12,506
to finish.
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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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00:15:15,130 --> 00:15:16,230
bit more space here--
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00:15:16,230 --> 00:15:18,037
AUDIENCE: It's called a saddle.
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00:15:18,037 --> 00:15:19,120
LYDIA BOUROUIBA: Oh, yeah.
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00:15:19,120 --> 00:15:19,715
Thank you.
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00:15:19,715 --> 00:15:22,340
And this would be a saddle, and
it's just because of the shape.
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00:15:27,350 --> 00:15:29,230
I'm going to just add
a little bit of space
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00:15:29,230 --> 00:15:34,750
here, so that we can
complete the diagram.
270
00:15:34,750 --> 00:15:38,820
I'll be basically looking at
the regions here in the wedge,
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00:15:38,820 --> 00:15:40,790
where we are below the parabola.
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00:15:40,790 --> 00:15:42,800
And I'm looking
for another color.
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00:15:42,800 --> 00:15:46,210
Maybe I'll just use white.
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00:15:46,210 --> 00:15:51,360
Now we are in the case where
we would have two eigenvalues.
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00:15:51,360 --> 00:15:53,480
Both real, so no oscillation.
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00:15:53,480 --> 00:15:57,480
But let's say both positive,
because we're basically
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00:15:57,480 --> 00:16:00,370
on the region where
the trace is positive.
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00:16:00,370 --> 00:16:02,810
Let's say that we
have one ray, v_1.
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00:16:02,810 --> 00:16:04,250
One ray, v_2.
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00:16:04,250 --> 00:16:07,100
The trajectories are going
away from the critical point,
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00:16:07,100 --> 00:16:10,840
because the two
eigenvalues are positive.
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00:16:10,840 --> 00:16:13,730
And now, what do the
other trajectories do,
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00:16:13,730 --> 00:16:17,845
where they follow--
so for example,
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00:16:17,845 --> 00:16:22,110
they would be following v_2.
285
00:16:22,110 --> 00:16:24,220
And I'm going to explain how.
286
00:16:24,220 --> 00:16:28,800
So here, we're in the case
where obviously we're unstable,
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00:16:28,800 --> 00:16:30,730
because again, the
solution, the trajectories
288
00:16:30,730 --> 00:16:32,530
are going away from
the critical point.
289
00:16:32,530 --> 00:16:36,530
And here, how do
you pick which ray
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00:16:36,530 --> 00:16:38,970
do you follow when you get
closer to the critical point?
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00:16:38,970 --> 00:16:41,630
Where in this case, we
would be in a situation
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00:16:41,630 --> 00:16:49,720
where we have lambda_2 smaller
than lambda_1, larger than 0.
293
00:16:49,720 --> 00:16:53,215
So basically, the lambda 2
that is the closer to 0--
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00:16:53,215 --> 00:16:55,090
and that's also the
case for the positive
295
00:16:55,090 --> 00:16:57,550
and the negative
eigenvalues-- is
296
00:16:57,550 --> 00:17:00,360
the one that determines
the solution closer
297
00:17:00,360 --> 00:17:01,390
to the critical point.
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00:17:01,390 --> 00:17:05,950
And the larger in absolute
value eigenvalue and its ray
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00:17:05,950 --> 00:17:07,635
then determines the behavior--
300
00:17:11,839 --> 00:17:12,339
Oh, sorry.
301
00:17:12,339 --> 00:17:13,297
There's a mistake here.
302
00:17:13,297 --> 00:17:14,530
It should be lambda_1.
303
00:17:17,540 --> 00:17:19,720
I said it, but I think
I wrote it reversely.
304
00:17:19,720 --> 00:17:23,079
Yeah The eigenvalue closer
to 0 is the one that
305
00:17:23,079 --> 00:17:24,329
determines the behavior at 0.
306
00:17:24,329 --> 00:17:26,280
So here, when we're
going to infinity,
307
00:17:26,280 --> 00:17:28,910
the larger eigenvalue lambda_2
will determine the behavior,
308
00:17:28,910 --> 00:17:32,530
and the trajectories will become
more and more parallel to v_2.
309
00:17:32,530 --> 00:17:34,510
So what happens
on this side would
310
00:17:34,510 --> 00:17:36,880
be exactly the same
diagram, and I'm just
311
00:17:36,880 --> 00:17:45,560
going to do it with the
same-- let's say, v_2 here.
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00:17:45,560 --> 00:17:54,790
Except that we would
have our trajectories
313
00:17:54,790 --> 00:17:56,260
going toward the critical point.
314
00:18:01,980 --> 00:18:03,950
And the trajectory
here again would
315
00:18:03,950 --> 00:18:07,240
be closer to v_1, which means
that we would have a case where
316
00:18:07,240 --> 00:18:12,600
we have lambda_1 less than
0 and larger than lambda_2.
317
00:18:12,600 --> 00:18:14,920
So that finishes
roughly the diagram.
318
00:18:14,920 --> 00:18:18,530
I didn't detail a few
borderline regions.
319
00:18:18,530 --> 00:18:22,360
For example, the region where
the determinant equals to 0.
320
00:18:22,360 --> 00:18:25,740
And we will discuss that
in another recitation.
321
00:18:25,740 --> 00:18:29,330
And the case also at which
the determinant and the trace
322
00:18:29,330 --> 00:18:30,057
are equal to 0.
323
00:18:30,057 --> 00:18:31,140
So we'll detail that also.
324
00:18:31,140 --> 00:18:35,890
But try to think about it,
to complete this diagram.
325
00:18:35,890 --> 00:18:38,280
So the key points
here were to remember
326
00:18:38,280 --> 00:18:42,960
what is the determinant-trace
diagram, how
327
00:18:42,960 --> 00:18:45,910
to basically deduce the nature
of the eigenvalues based
328
00:18:45,910 --> 00:18:47,710
on T and D, their sign.
329
00:18:47,710 --> 00:18:51,000
And where to place the
different structures
330
00:18:51,000 --> 00:18:53,540
on this determinant-trace
diagram.
331
00:18:53,540 --> 00:18:55,820
And that ends this recitation.