Question

Consider the interaction of two species in a habitat. We are told that the change of the populationsand can be moderated by the equation

$\left[\begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}=1.4x-1.2y\\ \frac{\mathrm{dy}}{\mathrm{dt}}=0.8x-1.4y\end{array}\right]$

where time tis a measured in years.

1. What kind of intersection do we observe (symbiosis, competition, or predator-prey)?
2. Sketch the phase portrait for the system. From the nature of the problem, we are interested only in the first quadrant.
3. What will happen in the long term? Does the outcome depend on the initial populations? If so, how?

Short answer

1. The intersection, y pray on x.
2. The value $E_{-1}=span\left[\begin{array}{c}1\\ 2\end{array}\right]$and$E_{-1}=span\left[\begin{array}{c}3\\ 1\end{array}\right]$, the phase portrait of the system is
   
3. Both species will prosper if$\frac{x\left(0\right)}{y\left(0\right)}<2$ and die if $\frac{x\left(0\right)}{y\left(0\right)}⩾2$.

Step-by-step solution

(a) Determine the kind of intersection.

Consider the interaction of two species in a habitat such that the change of the populations $\overrightarrow{x}\left(t\right)$and $\overrightarrow{y}\left(t\right)$can be moderated by the equation.

role="math" localid="1668487818809" $\left[\begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}=1.4\mathrm{x}-1.2\mathrm{y}\\ \frac{\mathrm{dy}}{\mathrm{dt}}=0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]$

From the term 0.8x in the 2^nd equation, the species y is helped by x and the term - 1.2y in the 1^st equation, the species x is retarded by y.

Hence, the species, y pray on x during the intersection of two species in a habitat.

(b) Simplify the equation [dxdt=1.4x-1.2ydydt=0.8x-1.4y].

Simplify the equation$\left[\begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}=1.4\mathrm{x}-1.2\mathrm{y}\\ \frac{\mathrm{dy}}{\mathrm{dt}}=0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]$as follows.

$\left[\begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}=1.4\mathrm{x}-1.2\mathrm{y}\\ \frac{\mathrm{dy}}{\mathrm{dt}}=0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]$

role="math" localid="1668488272638" $\left[\begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}\\ \frac{\mathrm{dy}}{\mathrm{dt}}\end{array}\right]=\left[\begin{array}{c}1.4\mathrm{x}-1.2\mathrm{y}\\ 0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$

Determine the Eigen values.

Compare the equations$\left[\begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}\\ \frac{\mathrm{dy}}{\mathrm{dt}}\end{array}\right]=\left[\begin{array}{c}1.4\mathrm{x}-1.2\mathrm{y}\\ 0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$and$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$as follows.

$A=\left[\begin{array}{c}1.4\mathrm{x}-1.2\mathrm{y}\\ 0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]$

Assume is an Eigen value of the matrix role="math" localid="1668489372676" $\left[\begin{array}{c}1.4\mathrm{x}-1.2\mathrm{y}\\ 0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]$implies $\left|A-\lambda I\right|=0$.

Substitute the valuesrole="math" localid="1668488796998" $\left[\begin{array}{c}1.4\mathrm{x}-1.2\mathrm{y}\\ 0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]$for A androle="math" localid="1668489536484" $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation$\left|A-\lambda I\right|=0$as follows.

role="math" localid="1668488878799" $$\begin{gathered} \left|A-\lambda I\right|=0 \\ \left[\begin{array}{c}1.4\mathrm{x}-1.2\mathrm{y}\\ 0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0 \end{gathered}$$

Simplify the equation$\left[\begin{array}{c}1.4\mathrm{x}-1.2\mathrm{y}\\ 0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0$as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{c}1.4\mathrm{x}-1.2\mathrm{y}\\ 0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|& =& 0\\ \left|\left[\begin{array}{c}1.4\mathrm{x}-1.2\mathrm{y}\\ 0.8\mathrm{x}-1.4\mathrm{y}\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right|& =& 0\\ \left|\left[\begin{array}{c}1.4\mathrm{x}-\lambda \\ 0.8\mathrm{x}\end{array}\begin{array}{c}1.2\mathrm{y}\\ -1.4\mathrm{y}-\mathrm{\lambda }\end{array}\right]\right|& =& 0\\ \left(1.4-\lambda \right)\left(-1.4-\lambda \right)+\left(0.8\right)\left(1.2\right)& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left(1.4-\lambda \right)\left(-1.4-\lambda \right)+\left(0.8\right)\left(1.2\right)& =& 0\\ \left(1.4-\lambda \right)\left(-1.4-\lambda \right)+0.96& =& 0\\ {\lambda }^2-1.96+0.96& =& 0\\ {\lambda }^2-1& =& 0\end{array}$

Therefore, the Eigen values of A are $\lambda =\pm 1$, as eigenvalues are opposite sign means equilibrium point is a saddle point.

Find the Eigen vector corresponding to the Eigen values.

Assume $v_1=\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]$and $v_2=\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$are Eigen vector corresponding to $\lambda =-1,1$implies $\left|A-{\lambda }_{1}I\right|v_1=0$and $\left|A-{\lambda }_{2}I\right|v_2=0$.

Substitute the values role="math" localid="1668490765074" width="90" height="48">1.40.8-1.2-1.4for A, -1 for role="math" localid="1668490066114" $\lambda ,\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]$, for v₁ and role="math" localid="1668490329090" $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation role="math" localid="1668490442265" $\begin{array}{rcl}& & \left|\left[\begin{array}{c}1.4\mathrm{x}-\lambda \\ 0.8\end{array}\begin{array}{c}1.2\\ -1.4\mathrm{y}-\mathrm{\lambda }\end{array}\right]\right|\end{array}{v}_1\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 0\end{array}$as follows.

role="math" localid="1668490720038" $\begin{array}{rcl}\left|\left[\begin{array}{c}1.4\mathrm{x}-\lambda \\ 0.8\end{array}\begin{array}{c}1.2\\ -1.4\mathrm{y}-\mathrm{\lambda }\end{array}\right]\right|v_1& =& 0\\ \left|\left[\begin{array}{c}1.4\mathrm{x}-\left(-1\right)\\ 0.8\end{array}\begin{array}{c}1.2\\ -1.4\mathrm{y}-\left(-1\right)\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left|\left[\begin{array}{c}2.4\\ 0.8\end{array}\begin{array}{c}-1.2\\ -0.4\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left|\left[\begin{array}{c}2.4x_1\\ -0.2x_1\end{array}\begin{array}{c}-1.2y_1\\ -0.4y_1\end{array}\right]\right|& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

Simplify the equations $2.4x_1-1.2y_1=0and-0.2x_1-0.4y_1$and as follows.

$\begin{array}{rcl}2.4x_1-1.2y_1& =& 0\\ 2.4x_1& =& 1.2y_1\\ 2x_1& =& y_1\end{array}$

For x₁ = 1 implies y₁ = 2.

Therefore, the Eigen vector corresponding to ${\lambda }_1=-1$is $\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right]$.

Substitute the values role="math" localid="1668490789384" width="90" height="48">1.40.8-1.2-1.4for A, 1 for role="math" localid="1668491120669" ${\lambda }_2,\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$for v₂ and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation role="math" localid="1668490467858" $\begin{array}{rcl}\left|\left[\begin{array}{c}1.4\mathrm{x}-\lambda \\ 0.8\end{array}\begin{array}{c}1.2\\ -1.4\mathrm{y}-\mathrm{\lambda }\end{array}\right]\right|v_1& =& 0\end{array}$as follows.

role="math" localid="1668490924708" $\begin{array}{rcl}\left|\left[\begin{array}{c}1.4\mathrm{x}-\lambda \\ 0.8\end{array}\begin{array}{c}1.2\\ -1.4\mathrm{y}-\mathrm{\lambda }\end{array}\right]\right|v_1& =& 0\\ \left|\left[\begin{array}{c}1.4-\left(-1\right)\\ 0.8\end{array}\begin{array}{c}1.2\\ -1.4-\left(-1\right)\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left|\left[\begin{array}{c}0.4\\ 0.8\end{array}\begin{array}{c}-1.2\\ -2.4\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left[\begin{array}{c}0.4x_2\\ 0.8x_2\end{array}\begin{array}{c}-1.2y_2\\ -2.4y_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left[\begin{array}{c}0.4x_2\\ 0.8x_2\end{array}\begin{array}{c}-1.2y_2\\ -2.4y_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ 0.4x_2-1.2y_2& =& 0\\ 0.8x_2-2.4y_2& =& 0\end{array}$

Simplify the equations $0.4x_2-1.2y_2=0$and $0.8x_2-2.4y_2=0$as follows.

$\begin{array}{rcl}0.4x_2-1.2y_2& =& 0\\ 0.4x_2& =& 1.2y_2\\ x_2& =& 3y_2\end{array}$

For $x_2=3$implies $y_2=1$.

Therefore, the Eigen vector corresponding to ${\lambda }_2=1$is $\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{c}3\\ 1\end{array}\right]$.

The Eigen vectors are $v_1=\left[\begin{array}{c}1\\ 2\end{array}\right]$and $v_2=\left[\begin{array}{c}3\\ 1\end{array}\right]$corresponding to the Eigen values ${\lambda }_1=-1$and${\lambda }_2=1$respectively.

Sketch the rough portraits for the system.

As $-1={\lambda }_1<0$and $1={\lambda }_2>0$, Sketch the rough phase portraits for the systems in the first quadrant as follows.

Hence, the rough phase portraits for the systems in the first quadrant is sketched.

(c) Determine the kind of intersection.

The both species will prosper if $\frac{x\left(0\right)}{y\left(0\right)}<2$implies role="math" localid="1668491536198" $\underset{t\to \infty }{lim}\frac{x\left(t\right)}{y\left(t\right)}=\frac{1}{3}$, and the both species is died if $\frac{x\left(0\right)}{y\left(0\right)}⩾2$.

From $\text{Figure}-1$, in the long term the population will grow and as long as the x population isn’t too small in the comparison to the y, both the species have a steady growth as represented as follows.

If$\frac{x\left(0\right)}{y\left(0\right)}<2$then both species will proper, and $\underset{t\to \infty }{lim}\frac{x\left(t\right)}{y\left(t\right)}=\frac{1}{3}$.

If$\frac{x\left(0\right)}{y\left(0\right)}⩾2$then both species will die.

Hence, the both species will prosper if$\frac{x\left(0\right)}{y\left(0\right)}<2$and die if $\frac{x\left(0\right)}{y\left(0\right)}⩾2$.