Now
we turn
our attention on mathematical models for interacting species. This is part one of
three, most of what we will discuss
is, the main references here are a couple of chapters, one chapter
from each of these references
and we're very interested in considering
competition between two species and we'll write some mathematical equations
for this matter but
this is a context in which two species may be competing for common
resources
or common territory or common habitat
and specifically
we will assume that the presence of
each species implies a reduction in the growth rate of the competitor.
Now we are going to use the notation
x of t and y of t to denote population size
or density for species one and two
respectively at time t. And we write
the equations in this
form. You can perhaps notice that this
resembles, this resembles
the... this is kind of a combination with the original Lotka
Lotka-Volterra equations and
we also discuss taking into account self-regulation.
So this equation is for species one.
We have here
a portion that takes into account that self-regulation
for species one; similar term
taking into account self-regulation for species two
and then we have interaction terms,
the terms that accompany X times
Y. So, the term b
XY and beta XY denote the interactions between species one and two,
and they are subtracted
because of what we point out here,
the presence of each species implies a reduction
on the growth rate, so that
that explains the subtraction and
as we mentioned the terms square,
X square and Y square are
those terms that are denoting self-regulation
for external resources on each species.
So now that we have a set of equations what we do next is to
go after steady states. We'd like to compute equilibrium points for which
all we have to do is to set the equations equal to zero and solve.
For X and Y, this would be a
set of two equations, two unknowns, and
these are very non-linear equations. Luckily
there are trivial solutions. For example,
X equals zero and Y equals zero when we evaluate
these expressions at X equals zero and Y equals zero,
then we can clearly see the first equation becomes zero on the left
which coincides with zero on the right and for the second
equation, the same is true. There are
other solutions that are not exactly trivial
but in a similar spirt, if
X is zero and Y is different from zero,
if X is zero, then we assume X is zero,
then that takes care of the first equation in which case we would get
a zero equals zero for the first equation, the left hand side of the
first equation vanishes and
that reduces the second equation
to
mu minus dY equals 0. We are left there to do is to solve for Y
and that gives us
the steady-state has form zero comma
mu over d, OK? This is the equilibrium point
and this is what we consider the extinction of species one.
Now, we consider the extinction
the extinction of species two, what we are doing is considering
Y equals 0. That will take care of
the second equation. If Y is 0,
then, the second equation
becomes an identity, zero equals zero,
and the first equation becomes a little simpler. So it reduces
the first equation to solving for X
in the expression lambda minus ax equals zero, solving for X and calling that solution X bar,
gives us lambda over a. In this case, the steady state or equilibrium point
has the form lambda over a comma zero, and this
is referring to the extinction
of species two. So, in summary we have
3 steady states,
one that is mathematically called the trivial
steady-state and this refers to the extinction
of both species one and species two, and then we have a
steady-state that refers to extinction of species one;
this would be a steady state zero comma mu over d
and then another steady state for extinction of species two,
lambda over a comma zero; these three equilibrium points, these three steady states
lie on the boundary of the first quadrant
of the phase plane. What we would like to do
next is to compute and know yet another steady state that
lies in the interior, not in the boundary but in the interior
of the first quadrant,
and to do that, what we are left with,
to work with is this set of equations,
right? We have eliminated the possibility that the both X
and Y are zero and we're now working with
this set of equations. We're solving were solving
for X and Y of two equations, two unknowns,
so what we can do here
is to solve for, take the first equation
and solve for Y, OK? And by
doing that we obtain the equation of a line. We are assuming
all parameters on this model, all paramaters
of this model are positive. That means
that this line, this, seen
as a function of X, is the equation of a line. This line has
a positive intercept with the vertical axis
and it has a negative slope. Now, let's take the second equation,
solve for the Y on the second equation and obtain another expression
that again has a positive intercept
with the vertical axis and it has a negative slope, so, what we'll do
on the next video
is to use graphical means,
to use a graphical approach
to continue our analysis. We're
going after the calculation of
the steady state line of the interior
of the first quadrant.