PROFESSOR: Welcome to this recitation on linearization.
So here we consider a prey-predator system,
where we have x dot equals x, factorizing 3 minus x,
minus x*y, and y dot equals y factorizing 1 minus y,
plus x*y.
So here we can see that x is the prey, because y is basically
feeding on it.
And y is the predator, because feeding
on x gives a growth of the predator species, y.
So here, because it's a prey-predator system,
x and y are assumed to be positive.
So you're asked to interpret even further this system
and to find the critical points, linearize and sketch the phase
portrait and then discuss what your linearization tells you
about the behavior of this system.
So why don't you pause the video and take
a few minutes to do that.
I'll be right back.
Welcome back.
So I helped you already for the first question.
We interpreted that x was the prey and y was the predator.
The other thing that you could see
is that this term here is basically logistic growth.
So if we didn't have the predator species around,
this species x of prey would just
grow and eventually reach a value that corresponds
to the saturation level.
And if we started with a lot of prey, then
eventually they would die off and go back
to that same carrying capacity value
of prey in that environment.
Same thing here for the predator.
We have a logistic growth, with different growth coefficients.
And again, if we didn't have the prey around,
we would just have a logistic dynamic
for this species on its own.
But with the prey, we have a growth of this species.
So that basically ends the first part.
For the second part, we need to find the critical points.
So how do we find the critical points?
The critical points correspond to basically f and g
equals to 0.
So f equal to 0 equals to g.
You see from f equals to 0 that we can have either x equals 0,
if we factorize the x in both terms,
and if x is not equal to 0, then we end up
with y equals to 3 minus x.
For the second part, we can also here,
if g equals to 0, factorize y.
And we get either y equals to 0, or we
have that x is equal to y minus 1,
where I just bring that term to the other hand.
So the critical point would be either of these two entries.
It would be, for example, x equals to 0 and this entry,
which would give us y equals to 1;
or we would have y equals to 0, and then
this entry would give us x equals to 3.
And then the last combined case, where
we have these two entries, which corresponds to 1 and 2,
if you do it.
So let's look now at the stability
of the linearization of the system around each one
of these critical points.
So what do we do to linearize the system?
Very quickly, we need to compute the Jacobian
around each critical point-- and we're just
going to build a table after to do that-- where we basically
have f, a derivative of f around x, derivative of g
around x, derivative of f around y, derivative of g around y,
all this evaluated at the critical point.
And this corresponds to basically linearizing
our nonlinear system, because you see here
that there are a lot of nonlinear systems,
and studying the stability around the neighborhood
of the critical point, like if it was basically
linear around there.
This method has its limitations, and we'll discuss them later.
So first I'm going to just here give you
the results of computation that I did earlier,
where basically you can repeat this computation.
But I don't want to spend too much time with the algebra
here.
So we have four critical points.
(0,0), (0,1), (3,0), and (1,2).
So this is going to just replace the values for the Jacobian.
So here you compute basically the derivative
of f of x with respect to x.
And you evaluate this at then the value of (0,0).
And so that would give you [3, 0; 0, 1].
For this case, same thing.
We have the expression for the Jacobian.
We evaluate it at critical point (0,1).
And that gives us [2, 0; 1, -1].
For this one similarly, we evaluate the Jacobian at (3,0).
And we get this Jacobian value.
And for this last critical point,
we have this Jacobian value.
So now what's next?
So the Jacobian basically gives us
the expression around the critical point
to look at the system, like if it was linear around there.
So from this point, we're back to the linear methods
we learned before.
We need to compute the eigenvalues of the Jacobian
around each critical point, and then determine the structure
around each critical point of the structure of basically
the phase portrait.
So what are the eigenvalues?
The eigenvalues are 3, 1, and you can compute that
and verify for yourself.
Plus or minus root of 7i and the whole thing over 2.
So here basically, we have two real eigenvalues,
both positive.
So we're going to have, basically,
an instability, unstable node.
And it's just basically the local stability
around the critical point.
Here we would have one positive, one negative.
So this is a saddle, which would be unstable.
The -3, 4 would give us another saddle.
And these complex eigenvalues would basically
give us a spiral with the real part going
to 0, with the real part being negative.
So a spiral that's stable, asymptotically.
So let's just do the diagram here.
And I'll continue the discussion.
So let's consider only the case where x and y are positive,
because we're talking about populations.
And let's place our critical points.
So we have a first critical point here at (0,0).
We have a second critical point here at (0,1).
We have a third critical point at (3,0),
and the last critical point at basically (1,2),
or something around there.
So now, based on the information we have on this table,
we have the eigenvalues.
We could also compute the corresponding eigenvectors.
And you can compute that.
And I'm not going to get into the details here,
but basically the values of the eigenvectors
would be important to give you, for example,
the direction of your spiral.
But we will do that on the diagram as we go.
So here at the (0,0) point is an unstable node.
So basically, the solutions are going away
from this point in the x and y.
This point, (0,1) is a saddle.
So we basically are on the ray here.
You would compute that the eigenvector that corresponds
to this negative eigenvalue would actually
be in the direction [0,1] and would
converge toward this solution.
And you can compute that the other eigenvector,
corresponding to the eigenvalue 2,
would have a direction this form.
So here we neglect what's happening here,
but it would be in this direction.
And the solutions would be basically going away from here.
And we would have locally something like that.
For this point, which corresponds
to an unstable node, basically the solution
would be going away.
For this point, which corresponds to (3,0),
we have a saddle again, which is unstable.
And here you can compute that the eigenvector corresponding
to the negative eigenvalue would basically
be parallel to the x-axis.
So it would have coordinates (1,0).
And the eigenvector corresponding
to the eigenvalue 4 would be directed in this direction.
And basically, the solution would be fleeing from there.
So locally, we would have something like that,
like we did before.
So we'll complete the graph.
Now, let's focus on the last critical point, (1,2).
So this point corresponds to a spiral.
And it's a stable spiral -- asymptotically stable,
because this eigenvalue's negative.
So the solutions are going toward this point.
And you can look at the lower entry here of the matrix, which
is positive, which means that we would
be going counterclockwise.
So we would have something there would be looking like this,
going toward this point.
So that's roughly what we would have
for the three critical points.
Am I missing one?
For the four critical points.
And so now we can complete our diagram
by basically linking to different localized phase
portraits together.
And so what will we have here?
So for example, we would have a solution here
that would be escaping, if we start in this neighborhood,
would be escaping from this critical point.
But then eventually, it would get attracted
by this other critical point, which
when it enters its basin of attraction,
given that it's asymptotically stable would be attracted by it
and go here.
If we started from a very high y-value,
we would be going down and then eventually getting close
to this critical point.
That would then basically cause this trajectory
to, again, escape and go feed this spiral
by reaching the basin of attraction of this point.
If we look at the critical point (3,0),
then if we start with a population x that is very large
and we approach this point, then we would have a solution that
basically would eventually continue parallel
to this ray of the unstable part of the critical point (3,0),
follow this ray and eventually basically
just be far more parallel to this trajectory that
links this linear (3,0) critical point
to the (1,2) critical point.
And so we can complete the diagram
by having something like this.
Now for this point, what do we have?
So for this point, we also have the solutions
that would be basically fleeing from the point (0,0),
and eventually could be attracted to the spiral
as well.
So that we give us then something
like this, a trajectory that would be looking like that.
And basically these trajectories would be looking like that.
And I'm not going to complete the parts where y and x are not
positive.
So we can complete the phase diagram of this nonlinear
system in this way.
So now, how do we interpret this?
What does this mean?
Well, if we remind ourselves of what this is actually modeling,
and if we just look at, for example,
the different axes, what does the y equals 0,
x equals 0 point mean?
It means that basically we have 0 population
of prey, 0 of predator.
And so it makes sense that we have basically
an unstable point here, an unstable critical point,
because as soon as we add one prey or one predator,
we would have an increase of the prey population,
eventually the predator would grow,
and so we would have a solution that
basically escapes the area around the critical point
(0,0).
What would happen if we just looked at the axis y equals
to 0? y equals to 0 corresponds to dynamics
where we don't have the predator population.
So the prey is just living its life,
growing at logistic growth.
And so basically, it's attracted by the carrying capacity
that would be here set at 3.
And so if we start with a lot of prey,
eventually they die out until they reach population 3.
And if we start with not enough, they grow
and they reach population 3, without the predator around.
Same thing for the predator on its own.
Now if we put the two together then
we have a spiral, which means that we have oscillation.
If we have a lot of predators, we have very few prey.
And eventually the predators start dying off as well.
But then, because they start dying off,
the prey population starts increasing, which then gives
an increase of the predator.
So we get, eventually, an oscillation that
goes to this attractor, (1,2), where the system will stabilize
eventually, where we would have, basically, one
prey for two predators.
So that ends the interpretation of the system.
And the idea here was to basically
use what we learned for the linear systems,
in terms of the phase portrait, and to see
how we can apply that to the nonlinear cases,
after linearizing the system around each one
of the critical points.
What I should also mention here is
that in all these cases that we looked at,
we had cases that were structurally stable,
which means that in our determinant-trace diagram,
we weren't at any borderline case
where a little perturbation could
make the structure around the critical point change
radically.
So all these points are structurally stable,
and the linearization therefore is valid around them.
And that ends this recitation.