Question

Competing Species. Let pi(t) denote, respectively, the populations of three competing species ${\mathbf{S}}_{\mathbf{i}}\mathbf{,}\mathbf{i}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{.}$Suppose these species have the same growth rates, and the maximum population that the habitat can support is the same for each species. (We assume it to be one unit.) Also, suppose the competitive advantage that${\mathbf{S}}_{\mathbf{1}}$ has over${\mathbf{S}}_{\mathbf{2}}$ is the same as that of${\mathbf{S}}_{\mathbf{2}}$ over${\mathbf{S}}_{\mathbf{3}}$ and over. This situation is modeled by the system

$$\begin{gathered} \mathbf{p}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{=}{\mathbf{p}}_{\mathbf{1}}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{p}}_{\mathbf{1}}\mathbf{-}{\mathbf{ap}}_{\mathbf{2}}\mathbf{-}{\mathbf{bp}}_{\mathbf{3}}\mathbf{)} \\ \mathbf{p}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{=}{\mathbf{p}}_{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{b}{\mathbf{p}}_{\mathbf{1}}\mathbf{-}{\mathbf{p}}_{\mathbf{2}}\mathbf{-}{\mathbf{ap}}_{\mathbf{3}}\mathbf{)} \\ \mathbf{p}{\mathbf{\text{'}}}_{\mathbf{3}}\mathbf{=}{\mathbf{p}}_{\mathbf{3}}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{a}{\mathbf{p}}_{\mathbf{1}}\mathbf{-}{\mathbf{bp}}_{\mathbf{2}}\mathbf{-}{\mathbf{p}}_{\mathbf{3}}\mathbf{)} \end{gathered}$$

where a and b are positive constants. To demonstrate the population dynamics of this system when a = b = 0.5, use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the populations over the time interval [0, 10] under each of the following initial conditions:

$$\begin{gathered} \mathbf{(}\mathbf{a}\mathbf{)}\mathbf{}{\mathbf{p}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}{\mathbf{p}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1} \\ \mathbf{(}\mathbf{b}\mathbf{)}\mathbf{}{\mathbf{p}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}{\mathbf{p}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1} \\ \mathbf{(}\mathbf{c}\mathbf{)}\mathbf{}{\mathbf{p}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}{\mathbf{p}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0} \end{gathered}$$

Short answer

In all cases, the population approaches to 0.5.

Step-by-step solution

Given conditions

Given that the system is:

$$\begin{gathered} \mathrm{p}{\text{'}}_1={\mathrm{p}}_1(1-{\mathrm{p}}_1-{\mathrm{ap}}_2-{\mathrm{bp}}_3) \\ \mathrm{p}{\text{'}}_2={\mathrm{p}}_2(1-\mathrm{b}{\mathrm{p}}_1-{\mathrm{p}}_2-{\mathrm{ap}}_3) \\ \mathrm{p}{\text{'}}_3={\mathrm{p}}_3(1-\mathrm{a}{\mathrm{p}}_1-{\mathrm{bp}}_2-{\mathrm{p}}_3) \end{gathered}$$

And

The initial conditions are:

${\mathrm{p}}_1\left(0\right)=1.0,{\mathrm{p}}_2=0.1,{\mathrm{p}}_3=0.1$

T	${\mathrm{p}}_1$	${\mathrm{p}}_2$	${\mathrm{p}}_3$
0	1	0.1	0.1
0.1	0.99035	0.103	0.1035
0.5	0.9574	0.1189	0.1189
1	0.9245	0.1406	0.140
1.5	0.8960	0.1647	0.164
3	0.817	0.245	0.245
4	0.766	0.298	0.298
5	0.7187	0.349	0.349
9	0.583	0.452	0.452
10.1	0.565	0.463	0.4638

Solve for part (b)

The initial conditions are${\mathrm{p}}_1\left(0\right)=0.1,{\mathrm{p}}_2=1.0,{\mathrm{p}}_3=0.1.$

T	$p_1$	$p_2$	$p_3$
0	0.1	1	0.1
0.1	0.103	0.990	0.1035
0.5	0.118	0.957	0.1189
1	0.1406	0.9245	0.1406
1.5	0.1647	0.8960	0.164
3	0.245	0.8177	0.245
4	0.298	0.7668	0.298
5	0.344	0.7187	0.3449
9	0.452	0.583	0.452
10.1	0.463	0.565	0.4638

Find the result of part (c)

The initials conditions are ${\mathrm{p}}_1\left(0\right)=0.1,{\mathrm{p}}_2=0.1,{\mathrm{p}}_3=1.0.$

T	$p_1$	$p_2$	$p_3$
0	0.1	0.1	1
0.1	0.103	0.103	0.990
0.5	0.118	0.118	0.957
1	0.1406	0.1406	0.9245
1.5	0.1647	0.1647	0.8960
3	0.2452	0.2453	0.8117
4	0.2982	0.2982	0.7618
5	0.344	0.3449	0.7187
9	0.4521	0.4521	0.5834
10.1	0.463	0.4638	0.565

In all cases, the population approaches 0.5.

This is the required result.