Question

Question: Consider the interaction of two species in a habitat. We are told that the change of the populations$\overrightarrow{\mathbf{x}}\left(t\right)\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\overrightarrow{\mathbf{y}}\left(t\right)$can be moderated by the equation

$\left|\begin{array}{c}\frac{dx}{dt}=5x-y\\ \frac{dy}{dt}=-2x+4y\end{array}\right|$

,

where timeis 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

a. The intersection of two species is competing.

b. The value $E{}_3=span\left[\begin{array}{c}1\\ 2\end{array}\right]andE{}_6=span\left[\begin{array}{c}1\\ -1\end{array}\right]$, the phase portrait of the system is

c. The species will die and will win if $\frac{x\left(0\right)}{y\left(0\right)}>2$, the species will die and will win if $\frac{x\left(0\right)}{y\left(0\right)}<2$and both species will prosper and$\underset{t\to \infty }{\lim }\frac{{\displaystyle x\left(t\right)}}{y{\displaystyle \left(t\right)}}=2$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.

$\left|\begin{array}{c}\frac{dx}{dt}=5x-y\\ \frac{dy}{dt}=-2x+4y\end{array}\right|$

From the term -2x in the 2^nd equation and -y in the 1^st equation, the species are hindered by the each other.

Hence, the intersection of two species is competing.

(b) Simplify the equation |dxdt=5x-ydydt=-2x+4y| .

Simplify the equation $\left|\begin{array}{c}\frac{dx}{dt}=5x-y\\ \frac{dy}{dt}=-2x+4y\end{array}\right|$as follows.

$$\begin{gathered} \left|\begin{array}{c}\frac{dx}{dt}=5x-y\\ \frac{dy}{dt}=-2x+4y\end{array}\right| \\ \left[\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\end{array}\right]=\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \end{gathered}$$

Determine the Eigen values

Compare the equations $\left[\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\end{array}\right]=\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$and $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$as follows.

$A=\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]$

Assume is an Eigen value of the matrix $A=\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]$ implies$\left|A-\lambda I\right|=0$ .

Substitute the values$A=\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]$ for A and$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation as follows.

$$\begin{gathered} \left|A-\lambda I\right|=0 \\ \left|\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0 \end{gathered}$$

Simplify the equation $\left|\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0$as follows.

$$\begin{gathered} \left|\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0 \\ \left|\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right|=0 \\ \left|\left[\begin{array}{cc}5-\lambda & -1\\ -2& 4-\lambda \end{array}\right]\right|=0 \\ \left(5-\lambda \right)\left(4-\lambda \right)-\left(-1\right)\left(-2\right)=0 \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} \left(5-\lambda \right)\left(4-\lambda \right)-\left(-1\right)\left(-2\right)=0 \\ \left(5-\lambda \right)\left(4-\lambda \right)-2=0 \\ {\lambda }^2-9\lambda +18=0 \\ {\lambda }^2-6\lambda -3\lambda +18=0 \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\lambda }^2-6\lambda -3\lambda +18=0 \\ \lambda \left(\lambda -6\right)-3\left(\lambda -6\right)=0 \\ \left(\lambda -3\right)\left(\lambda -6\right)=0 \end{gathered}$$

Therefore, the Eigen values of A are $\lambda =3,6$.

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$\lambda =3,6$corresponding to implies $\left|A-{\lambda }_{1}I\right|v_1=0and\left|A-{\lambda }_{2}I\right|v_2=0$.

Substitute the values for , for , for and for in the equation as $\left|\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0$follows.

$$\begin{gathered} \left|\left[\begin{array}{cc}5-\lambda & -1\\ -2& 4-\lambda \end{array}\right]\right|{v_{}}_1=0 \\ \left|\left[\begin{array}{cc}5-\left(-3\right)& -1\\ -2& 4-\left(-3\right)\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=0 \\ \left|\left[\begin{array}{cc}2& -1\\ -2& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \left[\begin{array}{cc}2x_1& -y_1\\ -2x_1& y_1\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \end{gathered}$$

Simplify the equations $2x_1-y_1=0and-2x_1+y_1=0$as follows.

$$\begin{gathered} 2x_1-y_1=0 \\ 2x_1=y_1 \end{gathered}$$

For x₁ =1 implies y₁=2.

Therefore, the Eigen vector corresponding to ${\lambda }_1=3$ is $\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right]$.

Substitute the valuesrole="math" localid="1660655693458" $\left[\begin{array}{cc}5& -1\\ -2& 4\end{array}\right]forA,6for{\lambda }_2,\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]for{v}_{2}and\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]forI$in the equation aslocalid="1660655330162" $\left|\left[\begin{array}{cc}5-\lambda & -1\\ -2& 4-\lambda \end{array}\right]\right|v_2=0$follows.

$$\begin{gathered} \left|\left[\begin{array}{cc}5-\lambda & -1\\ -2& 4-\lambda \end{array}\right]\right|v_2=0 \\ \left|\left[\begin{array}{cc}5-\left(6\right)& -1\\ -2& 4-\left(6\right)\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=0 \\ \left|\left[\begin{array}{cc}-1& -1\\ -2& -2\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \left[\begin{array}{cc}-x_2& -y_1\\ -2x_2& -2y_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} \left|\left[\begin{array}{cc}-1& -1\\ -2& -2\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \left[\begin{array}{cc}-x_2& -y_1\\ -2x_2& -2y_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ -x_2-y_2=0 \\ -2x_2-2y_2=0 \end{gathered}$$

Simplify the equations and as follows.

$$\begin{gathered} -x_2-y_2=0 \\ -x_2=y_2 \end{gathered}$$

For x₂=1 implies y₂ =-1.

Therefore, the Eigen vector corresponding to$\lambda =6$ is $\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{c}1\\ -1\end{array}\right]$.

The Eigen vectors are$v_1=\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]and{v}_2\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$ corresponding to the Eigen values role="math" localid="1660656303580" ${\lambda }_1=3and{\lambda }_2=6$respectively.

The value $E{}_3=span\left[\begin{array}{c}1\\ 2\end{array}\right]andE{}_6=span\left[\begin{array}{c}1\\ -1\end{array}\right]$, the phase portrait of the system is

 Step 5: Sketch the rough portraits for the system.

As${\lambda }_2>{\lambda }_1>0$ , 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.

From , in the long term the population will grow and as long as the population isn’t too small in the comparison to the , both the species have a steady growth as represented as follows.

If$\frac{x\left(0\right)}{y\left(0\right)}>2$ the species will die x and y will win.

If$\frac{x\left(0\right)}{y\left(0\right)}<2$ then the species will die x and ywill win.

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

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

The species will die and will win if$\frac{x\left(0\right)}{y\left(0\right)}>2$ , the species will die x and y will win if $\frac{x\left(0\right)}{y\left(0\right)}<2$and both species will prosper and$\underset{t\to \infty }{\lim }\frac{x\left(t\right)}{y\left(t\right)}=2$ if $\frac{x\left(0\right)}{y\left(0\right)}=2$.