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PROFESSOR: Welcome to this
recitation on linearization.

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So here we consider a
prey-predator system,

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where we have x dot equals
x, factorizing 3 minus x,

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minus x*y, and y dot equals
y factorizing 1 minus y,

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plus x*y.

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So here we can see that x is
the prey, because y is basically

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feeding on it.

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And y is the predator,
because feeding

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on x gives a growth of
the predator species, y.

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So here, because it's
a prey-predator system,

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x and y are assumed
to be positive.

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So you're asked to interpret
even further this system

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and to find the critical points,
linearize and sketch the phase

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portrait and then discuss what
your linearization tells you

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about the behavior
of this system.

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So why don't you pause
the video and take

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a few minutes to do that.

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I'll be right back.

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Welcome back.

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So I helped you already
for the first question.

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We interpreted that x was the
prey and y was the predator.

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The other thing
that you could see

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is that this term here is
basically logistic growth.

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So if we didn't have the
predator species around,

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this species x of
prey would just

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grow and eventually reach
a value that corresponds

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to the saturation level.

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And if we started with
a lot of prey, then

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eventually they would
die off and go back

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to that same carrying
capacity value

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of prey in that environment.

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Same thing here
for the predator.

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We have a logistic growth, with
different growth coefficients.

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And again, if we didn't
have the prey around,

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we would just have
a logistic dynamic

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for this species on its own.

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But with the prey, we have
a growth of this species.

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So that basically
ends the first part.

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For the second part, we need
to find the critical points.

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So how do we find
the critical points?

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The critical points correspond
to basically f and g

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equals to 0.

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So f equal to 0 equals to g.

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You see from f equals to 0 that
we can have either x equals 0,

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if we factorize the
x in both terms,

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and if x is not equal
to 0, then we end up

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with y equals to 3 minus x.

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For the second part,
we can also here,

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if g equals to 0, factorize y.

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And we get either y
equals to 0, or we

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have that x is
equal to y minus 1,

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where I just bring that
term to the other hand.

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So the critical point would be
either of these two entries.

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It would be, for example, x
equals to 0 and this entry,

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which would give
us y equals to 1;

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or we would have y
equals to 0, and then

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this entry would give
us x equals to 3.

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And then the last
combined case, where

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we have these two entries,
which corresponds to 1 and 2,

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if you do it.

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So let's look now
at the stability

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of the linearization of
the system around each one

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of these critical points.

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So what do we do to
linearize the system?

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Very quickly, we need
to compute the Jacobian

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around each critical
point-- and we're just

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going to build a table after
to do that-- where we basically

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have f, a derivative of f
around x, derivative of g

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around x, derivative of f around
y, derivative of g around y,

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all this evaluated at
the critical point.

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And this corresponds to
basically linearizing

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our nonlinear system,
because you see here

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that there are a lot
of nonlinear systems,

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and studying the stability
around the neighborhood

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of the critical point,
like if it was basically

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linear around there.

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This method has its limitations,
and we'll discuss them later.

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So first I'm going
to just here give you

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the results of computation
that I did earlier,

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where basically you can
repeat this computation.

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But I don't want to spend too
much time with the algebra

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here.

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So we have four critical points.

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(0,0), (0,1), (3,0), and (1,2).

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So this is going to just replace
the values for the Jacobian.

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So here you compute
basically the derivative

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of f of x with respect to x.

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And you evaluate this at
then the value of (0,0).

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And so that would give
you [3,  0; 0, 1].

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For this case, same thing.

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We have the expression
for the Jacobian.

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We evaluate it at
critical point (0,1).

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And that gives us [2, 0; 1, -1].

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For this one similarly, we
evaluate the Jacobian at (3,0).

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And we get this Jacobian value.

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And for this last
critical point,

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we have this Jacobian value.

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So now what's next?

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So the Jacobian
basically gives us

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the expression around
the critical point

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to look at the system, like
if it was linear around there.

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So from this point, we're
back to the linear methods

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we learned before.

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We need to compute the
eigenvalues of the Jacobian

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around each critical point, and
then determine the structure

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around each critical point
of the structure of basically

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the phase portrait.

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So what are the eigenvalues?

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The eigenvalues are 3, 1,
and you can compute that

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and verify for yourself.

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Plus or minus root of 7i
and the whole thing over 2.

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So here basically, we
have two real eigenvalues,

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both positive.

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So we're going to
have, basically,

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an instability, unstable node.

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And it's just basically
the local stability

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around the critical point.

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Here we would have one
positive, one negative.

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So this is a saddle,
which would be unstable.

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The -3, 4 would give
us another saddle.

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And these complex
eigenvalues would basically

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give us a spiral with
the real part going

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to 0, with the real
part being negative.

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So a spiral that's
stable, asymptotically.

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So let's just do
the diagram here.

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And I'll continue
the discussion.

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So let's consider only the case
where x and y are positive,

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because we're talking
about populations.

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And let's place our
critical points.

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So we have a first critical
point here at (0,0).

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We have a second critical
point here at (0,1).

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We have a third
critical point at (3,0),

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and the last critical
point at basically (1,2),

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or something around there.

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So now, based on the information
we have on this table,

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we have the eigenvalues.

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We could also compute the
corresponding eigenvectors.

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And you can compute that.

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And I'm not going to get
into the details here,

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but basically the values
of the eigenvectors

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would be important to
give you, for example,

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the direction of your spiral.

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But we will do that on
the diagram as we go.

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So here at the (0,0)
point is an unstable node.

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So basically, the
solutions are going away

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from this point in the x and y.

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This point, (0,1) is a saddle.

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So we basically are
on the ray here.

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You would compute that the
eigenvector that corresponds

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to this negative
eigenvalue would actually

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be in the direction
[0,1] and would

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converge toward this solution.

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And you can compute that
the other eigenvector,

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corresponding to
the eigenvalue 2,

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would have a
direction this form.

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So here we neglect
what's happening here,

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but it would be
in this direction.

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And the solutions would be
basically going away from here.

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And we would have locally
something like that.

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For this point,
which corresponds

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to an unstable node,
basically the solution

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would be going away.

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For this point, which
corresponds to (3,0),

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we have a saddle again,
which is unstable.

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And here you can compute that
the eigenvector corresponding

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to the negative
eigenvalue would basically

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be parallel to the x-axis.

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So it would have
coordinates (1,0).

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And the eigenvector
corresponding

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to the eigenvalue 4 would be
directed in this direction.

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And basically, the solution
would be fleeing from there.

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So locally, we would
have something like that,

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like we did before.

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So we'll complete the graph.

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Now, let's focus on the
last critical point, (1,2).

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So this point
corresponds to a spiral.

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And it's a stable spiral
-- asymptotically stable,

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because this
eigenvalue's negative.

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So the solutions are
going toward this point.

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And you can look at the lower
entry here of the matrix, which

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is positive, which
means that we would

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be going counterclockwise.

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So we would have something there
would be looking like this,

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going toward this point.

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So that's roughly
what we would have

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for the three critical points.

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Am I missing one?

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For the four critical points.

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And so now we can
complete our diagram

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by basically linking to
different localized phase

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portraits together.

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And so what will we have here?

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So for example, we would
have a solution here

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that would be escaping, if we
start in this neighborhood,

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would be escaping from
this critical point.

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But then eventually,
it would get attracted

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by this other
critical point, which

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when it enters its
basin of attraction,

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given that it's asymptotically
stable would be attracted by it

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and go here.

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If we started from
a very high y-value,

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we would be going down and
then eventually getting close

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to this critical point.

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That would then basically
cause this trajectory

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to, again, escape and
go feed this spiral

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by reaching the basin of
attraction of this point.

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00:11:50,190 --> 00:11:55,470
If we look at the
critical point (3,0),

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then if we start with a
population x that is very large

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and we approach this point, then
we would have a solution that

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00:12:03,550 --> 00:12:08,290
basically would eventually
continue parallel

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00:12:08,290 --> 00:12:13,940
to this ray of the unstable part
of the critical point (3,0),

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00:12:13,940 --> 00:12:15,820
follow this ray and
eventually basically

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00:12:15,820 --> 00:12:21,500
just be far more parallel
to this trajectory that

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links this linear
(3,0) critical point

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00:12:24,886 --> 00:12:27,570
to the (1,2) critical point.

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00:12:27,570 --> 00:12:30,451
And so we can
complete the diagram

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00:12:30,451 --> 00:12:31,700
by having something like this.

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00:12:31,700 --> 00:12:34,230
Now for this point,
what do we have?

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So for this point, we
also have the solutions

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that would be basically
fleeing from the point (0,0),

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00:12:43,410 --> 00:12:46,930
and eventually could be
attracted to the spiral

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00:12:46,930 --> 00:12:48,810
as well.

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So that we give
us then something

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00:12:52,610 --> 00:12:57,681
like this, a trajectory that
would be looking like that.

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00:12:57,681 --> 00:13:02,220
And basically these trajectories
would be looking like that.

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00:13:02,220 --> 00:13:08,390
And I'm not going to complete
the parts where y and x are not

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positive.

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So we can complete the phase
diagram of this nonlinear

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00:13:13,310 --> 00:13:14,490
system in this way.

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So now, how do we
interpret this?

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What does this mean?

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Well, if we remind ourselves of
what this is actually modeling,

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and if we just look
at, for example,

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the different axes, what
does the y equals 0,

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00:13:27,825 --> 00:13:29,060
x equals 0 point mean?

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00:13:29,060 --> 00:13:31,260
It means that basically
we have 0 population

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of prey, 0 of predator.

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And so it makes sense
that we have basically

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an unstable point here, an
unstable critical point,

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00:13:40,770 --> 00:13:43,599
because as soon as we add
one prey or one predator,

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we would have an increase
of the prey population,

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eventually the
predator would grow,

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00:13:47,720 --> 00:13:49,220
and so we would
have a solution that

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00:13:49,220 --> 00:13:52,495
basically escapes the area
around the critical point

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00:13:52,495 --> 00:13:55,110
(0,0).

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00:13:55,110 --> 00:13:59,860
What would happen if we just
looked at the axis y equals

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00:13:59,860 --> 00:14:02,840
to 0? y equals to 0
corresponds to dynamics

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00:14:02,840 --> 00:14:08,130
where we don't have the
predator population.

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00:14:08,130 --> 00:14:09,820
So the prey is just
living its life,

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growing at logistic growth.

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And so basically, it's attracted
by the carrying capacity

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that would be here set at 3.

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And so if we start
with a lot of prey,

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eventually they die out until
they reach population 3.

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And if we start with
not enough, they grow

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and they reach population 3,
without the predator around.

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Same thing for the
predator on its own.

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Now if we put the
two together then

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we have a spiral, which means
that we have oscillation.

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If we have a lot of predators,
we have very few prey.

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And eventually the predators
start dying off as well.

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But then, because
they start dying off,

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the prey population starts
increasing, which then gives

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an increase of the predator.

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So we get, eventually,
an oscillation that

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goes to this attractor, (1,2),
where the system will stabilize

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eventually, where we
would have, basically, one

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prey for two predators.

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So that ends the
interpretation of the system.

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00:15:06,790 --> 00:15:10,060
And the idea here
was to basically

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00:15:10,060 --> 00:15:14,260
use what we learned
for the linear systems,

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in terms of the phase
portrait, and to see

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how we can apply that
to the nonlinear cases,

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after linearizing the
system around each one

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00:15:24,890 --> 00:15:26,100
of the critical points.

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00:15:26,100 --> 00:15:27,920
What I should also
mention here is

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that in all these cases
that we looked at,

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we had cases that were
structurally stable,

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00:15:32,800 --> 00:15:37,610
which means that in our
determinant-trace diagram,

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00:15:37,610 --> 00:15:39,900
we weren't at any
borderline case

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00:15:39,900 --> 00:15:41,730
where a little
perturbation could

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00:15:41,730 --> 00:15:44,940
make the structure around
the critical point change

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00:15:44,940 --> 00:15:45,780
radically.

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00:15:45,780 --> 00:15:48,080
So all these points are
structurally stable,

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00:15:48,080 --> 00:15:52,070
and the linearization
therefore is valid around them.

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00:15:52,070 --> 00:15:54,208
And that ends this recitation.