(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
We have A=\left(\begin{array}{rr}3 & -1 \\ -2 & 2\end{array}\right). The eigenvalues of A are \lambda_{1}=1 and \lambda_{2}= 4 with corresponding eigenvectors \mathbf{v}_{1}=\left(\begin{array}{l}1 \\ 2\end{array}\right) and \mathbf{v}_{2}=\left(\begin{array}{r}1 \\ -1\end{array}\right). Then
C=\left(\begin{array}{rr}1 & 1 \\2 & -1\end{array}\right) \quad C^{-1}=-\frac{1}{3}\left(\begin{array}{rr}-1 & -1 \\-2 & 1\end{array}\right) \quad D=\left(\begin{array}{ll}1 & 0 \\0 & 4\end{array}\right) \quad e^{D t}=\left(\begin{array}{ll}e^{t} & 0 \\0 & e^{4 t}\end{array}\right) \begin{aligned}e^{A t} &=C e^{D t} C^{-1}=-\frac{1}{3}\left(\begin{array}{rr}1 & 1 \\2 & -1\end{array}\right)\left(\begin{array}{rr}e^{t} & 0 \\0 & e^{4 t}\end{array}\right)\left(\begin{array}{rr}-1 & -1 \\-2 & 1\end{array}\right) \\&=-\frac{1}{3}\left(\begin{array}{rr}1 & 1 \\2 & -1\end{array}\right)\left(\begin{array}{rr}-e^{t} & -e^{t} \\-2 e^{4 t} & e^{4 t}\end{array}\right) \\&=-\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}\right)\end{aligned}Finally, the solution to the system is given by
\begin{aligned}\mathbf{x}(t) &=\left(\begin{array}{l}x_{1}(t) \\x_{2}(t)\end{array}\right)=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}\right)\left(\begin{array}{r}90 \\150\end{array}\right) \\&=-\frac{1}{3}\left(\begin{array}{l}-240 e^{t}-30 e^{4 t} \\-480 e^{t}+30 e^{4 t}\end{array}\right)=\left(\begin{array}{c}80 e^{t}+10 e^{4 t} \\160 e^{t}-10 e^{4 t}\end{array}\right)\end{aligned}For example, after six months \left(t=\frac{1}{2}\right. year ), x_{1}(t)=80 e^{1 / 2}+10 e^{2} \approx 206 individuals, while x_{2}(t)=160 e^{1 / 2}-10 e^{2} \approx 190 individuals. More significantly, 160 e^{t}-10 e^{4 t}=0 when 16 e^{t}=e^{4t} or 16=e^{3 t} or 3 t=\ln 16 and t=(\ln 16) / 3 \approx 2.77 / 3 \approx 0.92 years \approx 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 x_{2}(0)=2 x_{1}(0) and that the first population will be eliminated if x_{2}(0)>2 x_{1}(0). 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.