% 2009-03-4
% Population Dynamics : Interspecific Competition
% Project 8 for Math 481

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\title[Project 8: Population dynamics: Interspecific Competition]{
  Math 481, Spring 2009\medskip \\
  Project 8 \\
  Population dynamics: Interspecific Competition
}
\author{Katherine Miller}
\address{Department of Mathematics and Statistics\\UMBC\\Baltimore, MD, 21250}
\email{kmiller2@umbc.edu}
\date{May 14, 2009}

\begin{abstract}
We first present a model of a two populations, each competing for the same 
resources. We will show the conditions under which this system is unstable,
and then introduce a predator under those same conditions. Then we will
be able to find a set of coefficients which will result in stability. In
this way, we will show that the introduction of a predator into an unstable
system of two competing populations can produce a stable system. Lastly
we will examine the system of two competing species and a predator without
any assumptions about the preference of the predator or how much it benefits
from preying upon each species. The material
explored in this project is heavily influenced by the work of Parish and 
Saila~\cite{parish-saila-1970} and Cramer and May~\cite{cramer-may-1972}.
\end{abstract}
\maketitle
\section{Introduction}\label{sec-intro}
Along with single species dynamics and predator-prey situations, another interesting
facet of population modeling is interspecific competition. This occurs when two 
species both compete for the same set of resources, such as grass and weeds in a plot
of land. The grass and the weeds are both looking to occupy the land, and when one
takes up an amount of space on that land the other cannot have access to it. This also 
occurs when two species feed on the same thing, or wish to occupy the same habitat.
When we examine this two species system in the next section, we will notice that there
are instances in which the two cannot both survive; one species may take over and cause
the other to die out. In this instance of an unstable system, it is interesting to 
think about the possibility of a predator who preys on both species, an idea examined 
by Cramer and May~\cite{cramer-may-1972} which we will analyze in this project. While it
may seem contradictory, the introduction of a predator may cause the unstable two species
system to stabilize at a nonzero equilibrium. This is possibly because the predator will
prey on a percentage of each of the competing species, so that no one species will
become so large that they take over all the resources. In this paper we will examine
the stability of the two species system, and how this changes with the introduction of a predator.

\section{The two species model}\label{sec-2model}
The first system we will look at is that of two populations
competing with each other for resources. Because they are influenced both
by overcrowding of their own species, and decreased resources due to 
the other species, we can write our mathematical model as such:
\begin{align*}
    \frac{dx}{dt} &= (a_1 - b_1x - c_1y)x, \\
    \frac{dy}{dt} &= (a_2 - b_2x - c_2y)y,
\end{align*}
with all positive constants. Solving $\frac{dx}{dt}=0$ and $\frac{dy}{dt}=0$ 
gives four equilibrium points. Because we are looking to find the stability of
the system, we are interested only in the behavior of the nonzero equilibrium 
point, which is
\begin{equation}\label{equilibrium}
    \bar{x} = \frac{-a_2 c_1+c_2 a_1}{-b_2 c_1+b_1 c_2}, \quad 
    \bar{y} = \frac{-b_2a_1-b_1a_2}{-b_2c_1+b_1c_2}.
\end{equation}
If this equilibrium point is unstable, then we know that both populations
cannot converge to a nonzero equilibrium, and therefore the system is
unstable. If this point is stable, however, then obviously the system
is also stable. It is presented in the paper by Cramer and May~\cite{cramer-may-1972} 
that the two species system is unstable if $b_1c_2-b_2c_1 \leq 0$, and stable
otherwise. However, this is not necessarily true. For example, consider the coefficients
$a_1=6, b_1=3, c_1=2$ and $a_2=b_2=c_2=1$. In this case $b_1c_2-b_2c_1=1$ which is
obviously positive, and so according to the previous claim should cause the
system to be stable. However calculating the Jacobian of the system evaluated
at the nonzero equilibrium point with the aforementioned coefficients gives
a negative determinant, indicating that the system is in fact unstable.
In order to understand the true conditions that imply stability or instability
we must use linearization techniques to examine the signs of the eigenvalues. 
The Jacobian of the system is
\[ 
\begin{bmatrix}
-2b_1x+a_1-c_1y & -c_1x\\
-b_2x & -2c_2y+a_2-b_2x \end{bmatrix}.
\] 
If we evaluate it at the nonzero equilibrium point and simplify, we get
\[ 
\begin{bmatrix}
-b_1\Big(\frac{c_2a_1-a_2c_1}{b_1c_2-b_2c_1}\Big) & -c_1\Big(\frac{c_2a_1-a_2c_1}{b_1c_2-b_2c_1}\Big)\\
-b_2\Big(\frac{b_1a_2-b_2a_1}{b_1c_2-b_2c_1}\Big) & -c_2\Big(\frac{b_1a_2-b_2a_1}{b_1c_2-b_2c_1}\Big) \end{bmatrix} =
\begin{bmatrix}
-b_1\bar{x} & -c_1\bar{x}\\
-b_2\bar{y} & -c_2\bar{y}\end{bmatrix}.
\] 
Therefore, the determinant is $(b_1c_2 - b_2c_1)\bar{x}\bar{y}$ and the trace is $-(b_1\bar{x} + c_2\bar{y})$.
We know that if the determinant of the Jacobian of a system evaluated at an equilibrium point is negative,
that this implies that equilibrium point is unstable. So, in order for the system to be unstable we must
have $(b_1c_2 - b_2c_1)\bar{x}\bar{y} \leq 0$. Cramer and May were partly correct in saying that 
$b_1c_2 - b_2c_1$ must be negative in order for the system to be unstable, but they did not mention another
important condition: the equilibrium populations $\bar{x}$ and $\bar{y}$ must be positive. Some sets of
coefficients, like the set $a_1=6, b_1=3, c_1=2$ and $a_2=b_2=c_2=1$ mentioned earlier produce
a nonzero equilibrium with negative values (in this case $\bar{x}=4, \bar{y}=-3$). If we impose
the additional condition that $\bar{x}$ and $\bar{y}$ must be positive, it follows that the trace of
the Jacobian is always negative. Because the trace is equal to the sum of the eigenvalues, this means
that there exists at least one negative eigenvalue. Therefore, when $(b_1c_2 - b_2c_1)$ is positive,
this implies that there are two negative eigenvalues and the equilibrium point in question is a stable
node. This shows that adding the additional condition that $\bar{x}$ and $\bar{y}$ must be positive
fixes the original claim.

\section{Introduction of a predator}\label{sec-3model}
Now, we will consider a three species system that consists of the two species from 
section~\ref{sec-2model} along with a predator that preys on both of the other 
species. This new predator will obviously affect the other two species negatively,
and preying on the other two species will affect the predator positively. Additionally,
we will assume that overcrowding is not a problem for the predator population. Therefore,
the new model looks like
\begin{align*}
    \frac{dx}{dt} &= (a_1 - b_1x - c_1y - d_1z)x, \\
    \frac{dy}{dt} &= (a_2 - b_2x - c_2y - d_2z)y, \\
    \frac{dz}{dt} &= (-a_3 + b_3x + c_3y)z,
\end{align*}
with all positive constants. In the Cramer and May article, they have taken $d_1 = d_2$
and $b_3 = c_3$. This essentially means that the predator has no preference between
population $x$ and population $y$; they are both preyed upon equally. Additionally,
the predator population benefits equally from the consumption of each of the prey 
populations. We will assign new constants $d_1 = d_2 = u$ and $b_3 = c_3 = v$
in order to remember that these coefficients are the same, giving the new system
\begin{align*}
    \frac{dx}{dt} &= (a_1 - b_1x - c_1y - uz)x, \\
    \frac{dy}{dt} &= (a_2 - b_2x - c_2y - uz)y, \\
    \frac{dz}{dt} &= (-a_3 + v(x + y))z.
\end{align*}
We are looking to show that adding this new predator to a previously unstable system
of two competing species can result in a stable system. In order for this to be true we 
must recall our conditions that imply an unstable system in the case of two 
species: the nonzero equilibrium populations must be positive, and $b_1c_2-b_2c_1$ must
be less than or equal to zero. From here, we can use linearization techniques on the
three species system and find the conditions which cause the nonzero equilibrium point
to be stable. The Jacobian of this new system of three equations evaluated at the new nonzero 
equilibrium point $(\bar{x},\bar{y},\bar{z})$  is 
\[ 
J = \begin{bmatrix}
-b_1\bar{x} & -c_1\bar{x} & -u\bar{x} \\
-b_2\bar{y} & -c_2\bar{y} & -u\bar{y} \\
v\bar{z} & v\bar{z} & 0 \end{bmatrix},
\] 
which gives a trace of $-(b_1\bar{x} + c_2\bar{y})$ and the following determinant:
\begin{align*}
\det(J) &= -b_1\bar{x}u\bar{y}v\bar{z} + b_2\bar{y}v\bar{z}u\bar{x} + v\bar{z}c_1\bar{x}u\bar{y} - v\bar{z}c_2\bar{y}u\bar{x} \\
&= (-b_1 + b_2 + c_1 - c_2)(\bar{x}\bar{y}\bar{z}uv).
\end{align*}
Again, we must explicitly assume the equilibrium populations $\bar{x}$, $\bar{y}$, and $\bar{z}$ to be positive, 
which will imply that the trace will always be negative. Because the trace is the sum of the three eigenvalues, this indicates
that at least one of the three eigenvalues is negative. Additionally, we know that the determinant is the product 
of the three eigenvalues, and so if $\det(J)$ is positive, this indicates not all eigenvalues are negative and
so the nonzero equilibrium is not stable. If the determinant is positive, this implies
\[
  (b_1 - b_2 - c_1 + c_2)(\bar{x}\bar{y}\bar{z}uv) \leq 0.
\]
This condition holds when the equilibrium populations $\bar{x}$, $\bar{y}$, and $\bar{z}$ are all positive and
$b_1 - b_2 - c_1 + c_2$ is negative. If $b_1 - b_2 - c_1 + c_2$ is positive, then the determinant is negative, and while
it is possible for the nonzero equilibrium to be stable, this is not guaranteed. Recall that the determinant is the product 
of the three eigenvalues. If the determinant is negative, this could mean that all of the eigenvalues are negative,
or one of the eigenvalues is negative and the other two are positive. From here, we can only prove that there is a stable
population by using our conditions to find coefficients that cause the nonzero equilibrium to be a stable node. 

We are most
interested in small integer coefficients, and so we can use a programming tool of some sort to loop every possible coefficient
from one to some maximum, and test all of the conditions which make the two species system unstable, and make the three
species system possibly stable. We can leave the constant $u$ out of these tests, as its value does not affect the value
of our conditions. If we set the maximum at six, we will notice that there are twelve possible sets of coefficients which 
satisfy our conditions and produce positive nonzero equilibrium populations. If we take the maximum to be seven, this number
increases dramatically to 448 different sets of coefficients. Phase portraits describing the unstable two species system
made stable by the introduction of a predator are displayed in figure~\ref{fig-phaseportraits}, using one of the twelve sets
of constants that are possible when taking coefficients to be integers between one and six.
\begin{figure}
\centering
\includegraphics[width=2.3in]{twospecies.pdf} 
\includegraphics[width=2.3in]{threespecies.pdf}
\caption{On the left is the phase portrait of the unstable system of interspecific competition between two species. On the 
right is the phase portrait of the stable system of the same two species in competition for resources along with a predator
who preys on the both of them. The coefficients used are $a_1=5, a_2=3, a_3=5, b_1=6, b_2=4, c_1=2, c_2=1, u=2, v=3$. These values
produce one negative real eigenvalue and two complex eigenvalues with negative real parts in the three species system. }
\label{fig-phaseportraits}
\end{figure}

\section{A more general three species model}\label{sec-general}
In section~\ref{sec-3model}, we examined a three species model but made the assumptions
that the predator both benefits equally from preying upon both species, and has no
preference between the two. While these assumptions certainly made it easier to prove
the claim introduced in the article by Cramer and May~\cite{cramer-may-1972}, it is
possible to derive a more general condition for stability in a three species system
where these assumptions cannot be made. Recall that the three species model without
assumptions is
\begin{align*}
    \frac{dx}{dt} &= (a_1 - b_1x - c_1y - d_1z)x, \\
    \frac{dy}{dt} &= (a_2 - b_2x - c_2y - d_2z)y, \\
    \frac{dz}{dt} &= (-a_3 + b_3x + c_3y)z,
\end{align*}
where $x$ and $y$ are competing for resources and $z$ is preying upon the both of
them. The Jacobian for this system is
\[ 
J = \begin{bmatrix}
-b_1\bar{x} & -c_1\bar{x} & -d_1\bar{x} \\
-b_2\bar{y} & -c_2\bar{y} & -d_2\bar{y} \\
b_3\bar{z} & c_3\bar{z} & 0 \end{bmatrix},
\] 
which gives the following determinant:
\begin{align*}
\det(J) &= -b_1\bar{x}d_2\bar{y}c_3\bar{z} + b_2\bar{y}c_3\bar{z}d_1\bar{x} + b_3\bar{z}c_1\bar{x}d_2\bar{y} - b_3\bar{z}c_2\bar{y}d_1\bar{x} \\
&= (d_1(b_2c_3-b_3c_2) + d_2(b_3c_1-b_1c_3))(\bar{x}\bar{y}\bar{z}).
\end{align*}
Similarly to the determinant in section~\ref{sec-3model}, in order for the system
to be unstable the determinant, which is the product of the three eigenvalues must be positive.
Again, this requires the equilibrium populations $\bar{x}$, $\bar{y}$, and $\bar{z}$ to
be positive as well as for $d_1(b_2c_3-b_3c_2) + d_2(b_3c_1-b_1c_3)$ to be greater than zero.
If it is less than or equal to zero, this could mean that the population will stabilize to
the nonzero equilibrium point, but one cannot be sure without checking that each of the
eigenvalues of the Jacobian evaluated at that point are negative or have negative real parts.

\begin{bibdiv}
\begin{biblist}

\bib{Math481Textbook}{book}{
  author   = {Barnes, Belinda},
  author   = {Fulford, Glenn R.}, 
  title    = {Mathematical modelling with case studies},
  subtitle = {A differential equation approach using Maple},
  publisher= {Taylor \& Francis},
  place    = {London},
  date     = {2002},
}

\bib{parish-saila-1970}{article}{
  author = {Parrish, J. D.},
  author = {Saila, S. B.},
  title  = {Interspecific Competition, Predation and Species Diversity},
  journal= {Journal of Theoretical Biology},
  volume = {27},
  number = {2},
  date   = {1970},
  pages  = {207--220},
}

\bib{cramer-may-1972}{article}{
  author = {Cramer, N. F.},
  author = {May, R. M.},
  title  = {Interspecific Competition, Predation and Species Diversity:
                A~Comment},
  journal= {Journal of Theoretical Biology},
  volume = {34},
  number = {2},
  date   = {1972},
  pages  = {289--293},
}

\end{biblist}
\end{bibdiv}

\end{document}