Identify the known information and the dynamical system.

\(\displaystyle \\A=\text{number of species A monkey} \\B=\text{number of species B monkey} \\g_A=\text{growth rate of species A per year} \\g_B=\text{growth rate of species B per year} \\c_A=\text{competition on species A} \\c_B=\text{competition on species B} \\\\g_A=0.04A(1-A/125,000) \\g_B=0.07B(1-B/450,000) \\c_A=c_B=\alpha AB \\A\geq 0, b\geq 0 \\\alpha \epsilon \mathbb{R}>0\)

\(\displaystyle \\\frac{dx_1}{dt}=f_1(x_1,...,x_n) \\\vdots \\\\\frac{dx_n}{dt}=f_n(x_1,...,x_n)\)

The dynamical system equation is:

\(\displaystyle \frac{dx}{dt}=F(x)\)

From here solve by using the Euler Method for a discrete time dynamical system.

\(\displaystyle \frac{\Delta x}{\Delta t}=F(x)\)

Now formulate the model

\(\displaystyle \\\frac{dx_1}{dt}=f_1(x_1,x_2)=0.04x_1\left(1-\frac{x_1}{125,000} \right )-\alpha x_1x_2 \\\\\frac{dx_2}{dt}=f_1(x_1,x_2)=0.07x_1\left(1-\frac{x_1}{450,000} \right )-\alpha x_1x_2\)

\(\displaystyle \\x_1=\text{species A monkey} \\x_2=\text{species B monkey}\)

Now transform these equations into a set of difference equations.

\(\displaystyle \\\Delta x_1=f_1(x_1,x_2)\Delta t \\\Delta x_2=f_2(x_1,x_2)\Delta t\)

\(\displaystyle \\\Delta x_1= \text{change in the species A monkey population} \\\Delta x_2= \text{change in the species B monkey population}\)

Assume 

\(\displaystyle \\\alpha =10^{-8} \\x_1(0)=4,000 \\x_2(0)=80,000\\N=50\)

where \(\displaystyle N\) represents the number of iterations the computer will use when implementing the Euler Method.

The time step \(\displaystyle h=\Delta t\) will need to be changed and the model reformulated to calculate how large the time step can be before it doesn't resemble a continuous time model.

Using computer implementation of the Euler Method on the population of species B monkeys the following graph is produced. 

It is seen that  \(\displaystyle h=47\) is the largest time step before the function begins to have chaos and not represent the behavior of a continuous system.

Since there is competition between the two differing species of flowers, the growth rate function is,

\(\displaystyle g(P)=rP-aP^2\)

where \(\displaystyle r\) is the intrinsic growth rate and \(\displaystyle a< < r\) measures the strength of the limitations of resources.

Now identify the question as a dynamic model in steady state.

In other words,

\(\displaystyle \\f_1(x_1,...,x_n) \\f_2(x_1,...,x_n)\)

From here, identify what is known about the problem in mathematical terms.

\(\displaystyle \\D=\text{Yellow Daisy Population} \\C=\text{Green Floor Crawler Population} \\g_D=\text{growth rate of yellow daisies} \\g_C=\text{growth rate of green floor crawler population} \\c_D=\text{loss due to competition for yellow daisies} \\c_C=\text{loss due to competition for green floor crawler}\)

\(\displaystyle \\g_D=r_1D-a_1D^2 \\g_C=r_2C-a_2C^2 \\c_D=b_1CD \\c_C=b_2CD \\D\geq0, C\geq0 \\r_1,r_2,a_1,a_2,b_1, b_2 \text{are positive reals}\)

To answer the question determine whether \(\displaystyle C\rightarrow0\) or \(\displaystyle D\rightarrow0\).

To formulate the model let 

\(\displaystyle \\x_1=D \\x_2=C \\\begin{Bmatrix} (x_1,x_2):x_1\geq0, x_2\geq0 \end{Bmatrix}\)

where the steady state equations are as follows.

\(\displaystyle \\\\r_1x_1-a_1x_1^2-b_1x_1x_2=0 \\r_2x_2-a_2x_2^2-b_2x_1x_2=0\)

For the particular problem, the points on interests are at those where the two functions intersect.

\(\displaystyle \\(0,0) \\\left( 0,\frac{r_2}{a_2}\right) \\\left(\frac{r_1}{a_1},0\right)\)

and the intersection of 

\(\displaystyle \\a_1x_1+b_1x_2=r_1 \\b_2x_1+a_2x_2=r_2\)

Using Cramer's rule to solve results in the following.

\(\displaystyle \\x_1=\frac{r_1a_2-r_2b_1}{a_1a_2-b_1b_2} \\\\\\x_2=\frac{a1_r2-b_2r_1}{a_1a_2-b_1b_2}\)

and when the question of coexistence is ask it is found that,

\(\displaystyle \frac{r_2}{a_2}< \frac{r_1}{b_1}\) and \(\displaystyle \frac{r_1}{a_1}< \frac{r_2}{b_2}\)

Which after further examination learns to the conclusion that the steady state analysis cannot answer the question.