We continue our discussion on interacting species. This is part two of three. We are
in the process of computing an interior steady state or an equilibrium
point that lies in the interior of the first quadrant
in the phase plane. So we had to use a calculation to the equation
of a line. In fact, we had two lines, both of which have a negative slope and both of
these lines have a positive intercept with the vertical axis. Let's take a look
at the... this is a graphical representation of a line
that has this equation. Clearly the intercept with the vertical axis is at Lambda over b
and is also another thing we point in this sketch is that the
intercept with the horizontal axis happens at lambda over a.
On the right-hand side we have the other line and it's easy to see that the intercept with
the vertical axis happens at mu over d and it crosses the x-axis
at mu over b. So what we would really like to do is to put both of these lines in the
same figure,
okay? And this is what we do next. There are four possibilities as to how this figure may
look like. The information we have at hand is that both
lines have negative slope, right? Minus a over b; minus beta over d,
and that they both have positive intercepts with the vertical axis, so this
is the figure for the first case. Here we describe
the cases with these inequalities. We note that the first inequality refers to
these points on the X axis. So we have mu over beta is smaller than lambda over a.
The second inequality refers to conditions as they happen on the vertical
axis, mu over d is less than lambda over b. If this is the case you can see that there
is no intersection in the first quadrant. In fact these two lines
would be intersecting at somewhere on the fourth quadrant. Let's take a look at the
second case. The second case is when lambda over a is less than mu over beta,
together with lambda over b, here this intercept, is below mu over d. In this case the intersection
between these two lines doesn't happen on the first quadrant, as suggested
here with the sketch that these two lines will intersect
somewhere on the second quadrant. Case number three: In this case, lambda over
a is smaller than mu over beta and mu over d is less than
the lambda over b. In this case, these lines intersect on the first
quadrant and this intersection, this point of intersection, will define or give rise
to steady state or an equilibrium point of our system. There is a fourth
case, when mu over beta is smaller than the lambda over a and when lambda over b is less
than mu over d. This also gives rise to another point on the first quadrant.
So now that we have identified possible scenarios
we're gonna do some, we're gonna solve the system numerically and we're gonna illustrate
some of these cases. In summary, we had previously we had previously summarized, we had points
lying on the boundary of the first quadrant. Here is one of those points,lambda
over a, zero. Then we have 00. This first point is what we would
that we would label as extinction of species number two. Here the second species goes to
zero, basically, the species number one out runs species number two. The trivial
steady state is when both species go extinct and then this other point is
the extinction of species number one meaning that, basically, species number two drives
species number one to extinction. Okay. Now,
the new point is this point that lies that may exist
under some conditions and lies in the interior. When that happens that point
has these coordinates. The first coordinate and the second coordinate are expressed here.
These are the four equilibrium points or steady states of the system. Now we're gonna
revisit some illustrations as to to see when and how some of these points are
stable. We'll see that this total extinction equillibrium point is
never stable. Here is an illustration for case number one
and this is the scenario in which species number two goes extinct.
Here is that point. What we did was to set some values for the parameters and this is
a phase plot so we had, we have y(t) versus x(t). And we have three curves.
Each of them with a different set of initial conditions. All the simulations are calculated
using the package, r package Lsoda, so here we have a curve that starts at x0=1,y0=1
and this curve converges at this point
then we set initial conditions 10 and 40 and this curve quickly converges here and, similarly,
for the numerical solution obtained using initial condition 16 20.
And you can see that this point lamda over a, 0, given the numerical
parameter values used here, that point is basically (20,0). Now for case number two which is the
scenario in which species one goes extinct and this is the equilibrium point, 0,
mu over d. We have, wer'e testing numerical solutions with three different
initial conditions. Initialize at (1,1). Initialize at (35,1). This would be X at zero is 35,
Y at zero is one and you can see how these curves are quickly converging
to the points 0, mu over d. This is the third initial condition. In this
case, mu over d. We chose parameters so that mu over d is 50, so this is converging to
point (0,50).
The next case, case number three, is what we call coexistence. In the previous of cases,
there was an a exclusion. Basically one species drove
the other species to extinction. In this case, the
both species would maintain levels. They would stabilize at equilibrium levels that are positive.
We depict that a equilibrium point of phase plane with the
circle here. Here is the expression we had presented before. In this case, there are
three solutions. We chose parameter values and tested
different values for initial conditions. Here is one curve. It starts at (1,1) and
quickly converges to this equilibrium point. Then we had (15,5) and (10,50) and the three
solutions converge to that equilibrium point that is called coexsistence.
In the last case, is a case that allows what stability of either
extinction of species 2 or extinction of species 1. So both of these steady states are stable
and therefore the name bistability meaning that there is stability
of both and here is the equilibrium point in the interior. In
this case, this equilibrium point is unstable and the sense that is unstable can be explained
that, in the sense that stable points are limits. The solutions converge
to the point that the equilibrium point we call stable. When a point
is unstable and so the stable point is attracting solutions and the stability has that sense
of converging to it. Right? This like a magnet. There is an attraction
to it. On the other hand, when a point is unstable
then there is a repulsion. The point is propelling trajectories and so
you can see that there are four curves here. There are four initial conditions tested.
This solution starting at (1,10) converged to zero, mu over d and this at (12,40)
also converged to that point, but when we look at the solution with initial
condition (1,1) then that converges to the other point. And with this initial conditions
(45,60) does converges to the other points.So in this case, the interior equilibrium
point is repelling, is unstable and solutions are moving away from
this point. They're moving away towards either extinction
of species one or extinction of species two. Okay. So this allows
for what is called bistability. This was an area in which parameter values were set and
for the calculation of these four curves, the parameter values are the same. The only
thing that changed from one scenario to the next was the values used for the initial
conditions. And this concludes our discussion of interacting
species part two.