(A Model of Species Cooperation-Symbiosis) Consider the symbiotic model governed by the system X1'(t) = -1/2 x1(t) + x2(t), x2'(t) = 1/4 x1(t) - 1/2 x2(t). In this model the population of each species increases proportionally to the population of the other and decreases proportionally to its own

Question

(A Model of Species Cooperation-Symbiosis) Consider the symbiotic model governed by the system

[latex]\begin{aligned}&x_{1}{ }^{\prime}(t)=-\frac{1}{2} x_{1}(t)+x_{2}(t) \&x_{2}{ }^{\prime}(t)=\frac{1}{4} x_{1}(t)-\frac{1}{2} x_{2}(t)\end{aligned}[/latex]

In this model the population of each species increases proportionally to the population of the other and decreases proportionally to its own population. Suppose that [latex]x_{1}(0)= 200[/latex] and [latex]x_{2}(0)=500[/latex]. Determine the population of each species for [latex]t>0[/latex].

Accepted Answer

Here [latex]A=\left(\begin{array}{rr}-\frac{1}{2} & 1 \ \frac{1}{4} & -\frac{1}{2}\end{array} ight)[/latex] with eigenvalues [latex]\lambda_{1}=0[/latex] and [latex]\lambda_{2}=-1[/latex] and corresponding eigenvectors [latex]\mathbf{v}_{1}=\left(\begin{array}{l}2 \ 1\end{array} ight)[/latex] and [latex]\mathbf{v}_{2}=\left(\begin{array}{r}2 \ -1\end{array} ight)[/latex]. Then

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

and [latex]\begin{aligned}e^{D t} &=\left(\begin{array}{cc}e^{0 t} & 0 \0 & e^{-t}\end{array} ight)=\left(\begin{array}{cc}1 & 0 \0 & e^{-t}\end{array} ight) ext {. Thus } \e^{A t} &=-\frac{1}{4}\left(\begin{array}{rr}2 & 2 \1 & -1\end{array} ight)\left(\begin{array}{cc}1 & 0 \0 & e^{-t}\end{array} ight)\left(\begin{array}{rr}-1 & -2 \-1 & 2\end{array} ight) \&=-\frac{1}{4}\left(\begin{array}{rr}2 & 2 \1 & -1\end{array} ight)\left(\begin{array}{ll}-1 & -2 \-e^{-t} & 2 e^{-t}\end{array} ight) \&=-\frac{1}{4}\left(\begin{array}{ll}-2 & -2 e^{-t} & -4+4 e^{-t} \-1+ & e^{-t} & -2-2 e^{-t}\end{array} ight)\end{aligned}[/latex]

and

[latex]\begin{aligned}\mathbf{x}(t)=e^{A t} \mathbf{x}(0) &=-\frac{1}{4}\left(\begin{array}{ll}-2-2 e^{-t} & -4+4 e^{-t} \-1+e^{-t} & -2-2 e^{-t}\end{array} ight)\left(\begin{array}{l}200 \500\end{array} ight) \&=-\frac{1}{4}\left(\begin{array}{l}-2400+1600 e^{-t} \-1200-800 e^{-t}\end{array} ight) \&=\left(\begin{array}{l}600-400 e^{-t} \300+200 e^{-t}\end{array} ight)\end{aligned}[/latex]

We have [latex]e^{-t} ightarrow 0[/latex] as [latex]t ightarrow \infty[/latex]. This means that as time goes on, the two cooperating species approach the equilibrium populations 600 and 300 , respectively. Neither population is eliminated.

Question 12.6.4: (A Model of Species Cooperation-Symbiosis) Consider the symb...

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