(A Competitive Model) Consider the system
$\begin{aligned}&x_{1}^{\prime}(t)=3 x_{1}(t)-x_{2}(t) \\&x_{2}^{\prime}(t)=-2 x_{1}(t)+2 x_{2}(t)\end{aligned}$
Here, an increase in the population of one species causes a decline in the grow rate of another. Suppose that the initial populations are $x_{1}(0)=90$ and $x_{2}(0)=150$ . Find the populations of both species for $t>0$

Question

(A Competitive Model) Consider the system

[latex]\begin{aligned}&x_{1}^{\prime}(t)=3 x_{1}(t)-x_{2}(t) \&x_{2}^{\prime}(t)=-2 x_{1}(t)+2 x_{2}(t)\end{aligned}[/latex]

Here, an increase in the population of one species causes a decline in the grow rate of another. Suppose that the initial populations are [latex]x_{1}(0)=90[/latex] and [latex]x_{2}(0)=150[/latex]. Find the populations of both species for [latex]t>0[/latex]

Accepted Answer

We have [latex]A=\left(\begin{array}{rr}3 & -1 \ -2 & 2\end{array} ight)[/latex]. The eigenvalues of [latex]A[/latex] are [latex]\lambda_{1}=1[/latex] and [latex]\lambda_{2}= 4[/latex] with corresponding eigenvectors [latex]\mathbf{v}_{1}=\left(\begin{array}{l}1 \ 2\end{array} ight)[/latex] and [latex]\mathbf{v}_{2}=\left(\begin{array}{r}1 \ -1\end{array} ight)[/latex]. Then

[latex]C=\left(\begin{array}{rr}1 & 1 \2 & -1\end{array} ight) \quad C^{-1}=-\frac{1}{3}\left(\begin{array}{rr}-1 & -1 \-2 & 1\end{array} ight) \quad D=\left(\begin{array}{ll}1 & 0 \0 & 4\end{array} ight) \quad e^{D t}=\left(\begin{array}{ll}e^{t} & 0 \0 & e^{4 t}\end{array} ight)[/latex]

[latex]\begin{aligned}e^{A t} &=C e^{D t} C^{-1}=-\frac{1}{3}\left(\begin{array}{rr}1 & 1 \2 & -1\end{array} ight)\left(\begin{array}{rr}e^{t} & 0 \0 & e^{4 t}\end{array} ight)\left(\begin{array}{rr}-1 & -1 \-2 & 1\end{array} ight) \&=-\frac{1}{3}\left(\begin{array}{rr}1 & 1 \2 & -1\end{array} ight)\left(\begin{array}{rr}-e^{t} & -e^{t} \-2 e^{4 t} & e^{4 t}\end{array} ight) \&=-\frac{1}{3}\left(\begin{array}{rr}-e^{t}-2 e^{4 t} & -e^{t}+e^{4 t} \-2 e^{t}+2 e^{4 t} & -2 e^{t}-e^{4 t}\end{array} ight)\end{aligned}[/latex]

Finally, the solution to the system is given by

[latex]\begin{aligned}\mathbf{x}(t) &=\left(\begin{array}{l}x_{1}(t) \x_{2}(t)\end{array} ight)=e^{A t} \mathbf{x}_{0}=-\frac{1}{3}\left(\begin{array}{cc}-e^{t}-2 e^{4 t} & -e^{t}+e^{4 t} \-2 e^{t}+2 e^{4 t} & -2 e^{t}-e^{4 t}\end{array} ight)\left(\begin{array}{r}90 \150\end{array} ight) \&=-\frac{1}{3}\left(\begin{array}{l}-240 e^{t}-30 e^{4 t} \-480 e^{t}+30 e^{4 t}\end{array} ight)=\left(\begin{array}{c}80 e^{t}+10 e^{4 t} \160 e^{t}-10 e^{4 t}\end{array} ight)\end{aligned}[/latex]

For example, after six months [latex]\left(t=\frac{1}{2} ight.[/latex] year [latex]), x_{1}(t)=80 e^{1 / 2}+10 e^{2} \approx 206[/latex] individuals, while [latex]x_{2}(t)=160 e^{1 / 2}-10 e^{2} \approx 190[/latex] individuals. More significantly, [latex]160 e^{t}-10 e^{4 t}=0[/latex] when [latex]16 e^{t}=e^{4t}[/latex] or [latex]16=e^{3 t}[/latex] or [latex]3 t=\ln 16[/latex] and [latex]t=(\ln 16) / 3 \approx 2.77 / 3 \approx 0.92[/latex] years [latex]\approx[/latex] 11 months. Thus the second species will be eliminated after only 11 months even though it started with a larger population. In Problems 10 and 11 you are asked to show that neither population will be eliminated if [latex]x_{2}(0)=2 x_{1}(0)[/latex] and that the first population will be eliminated if [latex]x_{2}(0)>2 x_{1}(0)[/latex]. Thus, as was well known to Darwin, survival in this very simple model depends on the relative sizes of the competing species when competition begins.

Question 12.6.2: (A Competitive Model) Consider the system x1'(t) = 3x1(t) - ...

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