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Q42E

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Found in: Page 428

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Question: Consider the interaction of two species in a habitat. We are told that the change of the populations $\stackrel{\mathbf{\to }}{\mathbf{x}}\left(t\right){\mathbf{}}{\mathbit{a}}{\mathbit{n}}{\mathbit{d}}{\mathbf{}}\stackrel{\mathbf{\to }}{\mathbf{y}}\left(t\right)$ can be moderated by the equation $|\begin{array}{c}\frac{dx}{dt}=1.4x-1.2y\\ \frac{dx}{dt}=1.8x-1.4y\end{array}|$ where time is a measured in years.What kind of intersection do we observe (symbiosis, competition, or predator-prey)?Sketch the phase portrait for the system. From the nature of the problem, we are interested only in the first quadrant.What will happen in the long term? Does the outcome depend on the initial populations? If so, how?

1. The intersection, pray on .
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 $\frac{x\left(0\right)}{y\left(0\right)}\ge 2$.

See the step by step solution

## Step 1: (a) Determine the kind of intersection.

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

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

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

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

## Step 2: (b) Simplify the equation |dxdt=1.4x-1.2ydxdt=0.8x-1.4y|.

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

role="math" localid="1660632114086" $\left|\begin{array}{c}\frac{dx}{dt}=1.4x-1.2y\\ \frac{dy}{dt}=0.8x-1.4y\end{array}\right|\phantom{\rule{0ex}{0ex}}\left[\begin{array}{c}\frac{dx}{dt}\\ \frac{dx}{dt}\end{array}\right]\left[\begin{array}{c}1.4x-1.2\\ 0.8x-1.4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$

## Step 3: Determine the Eigen values.

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

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

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

Substitute the values $\left[\begin{array}{c}1.4x-1.2\\ 0.8x-1.4\end{array}\right]$for A and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation $\left|A-\lambda I\right|=0$as follows.

role="math" localid="1660632725293" $\left|A-\lambda I\right|=0\phantom{\rule{0ex}{0ex}}\left|\left[\begin{array}{c}1.4x-1.2\\ 0.8x-1.4\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0\phantom{\rule{0ex}{0ex}}$

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

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

Further, simplify the equation as follows.

$\left(1.4-\lambda \right)\left(-1.4-\lambda \right)+\left(0.8\right)\left(1.2\right)=0\phantom{\rule{0ex}{0ex}}\left(1.4-\lambda \right)\left(-1.4-\lambda \right)+0.96=0\phantom{\rule{0ex}{0ex}}{\lambda }^{2}-1.96+0.96=0\phantom{\rule{0ex}{0ex}}{\lambda }^{2}-1=0\phantom{\rule{0ex}{0ex}}$

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

## Step 4: Find the Eigen vector corresponding to the Eigen values.

Assume ${v}_{1}=\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]and{v}_{2}=\left[\begin{array}{c}{x}_{2}\\ {y}_{2}\end{array}\right]$and are Eigen vector corresponding to $\lambda =-1,\text{}1$implies $\left|A-{\lambda }_{2}I\right|{v}_{2}=0and\left|A-{\lambda }_{2}I\right|{v}_{2}=0$

.

Substitute the values role="math" localid="1660634729871" $\left[\begin{array}{c}1.4x-1.2\\ 0.8x-1.4\end{array}\right]forA,-1for\lambda ,\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]for{v}_{1}and\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]forI\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$in the equation$\left[\begin{array}{c}1.4x-\lambda -1.2\\ 0.8x-1.4-\lambda \end{array}\right]{v}_{1}=0$ as follows.

$\left|\left[\begin{array}{c}1.4x-\lambda -1.2\\ 0.8x-1.4-\lambda \end{array}\right]\right|{v}_{1}=0\phantom{\rule{0ex}{0ex}}\left|\left[\begin{array}{c}1.4x-\left(-1\right)-1.2\\ 0.8x-1.4-\left(-1\right)\end{array}\right]\right|\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left|\left|\begin{array}{cc}2.4& -1.2\\ 0.8& -0.4\end{array}\right|\right|\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left|\left|\begin{array}{cc}2.4{x}_{1}& -1.2{y}_{1}\\ 0.8{x}_{1}& -0.4{y}_{1}\end{array}\right|\right|=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Simplify the equations $2.4{x}_{1}-1.2{y}_{1}=0and-0.2{x}_{1}-0.4{y}_{1}$ as follows.

$2.4{x}_{1}-1.2{y}_{1}=0\phantom{\rule{0ex}{0ex}}2.4{x}_{1}=1.2{y}_{1}\phantom{\rule{0ex}{0ex}}2{x}_{1}={y}_{1}\phantom{\rule{0ex}{0ex}}$

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 $\left[\begin{array}{c}1.4x-1.2\\ 0.8x-1.4\end{array}\right]forA,-1for{\lambda }_{2},\left[\begin{array}{c}{x}_{2}\\ {y}_{2}\end{array}\right]for{v}_{2}and\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]forI$ in the equation $\left|\left[\begin{array}{c}1.4x-\lambda -1.2\\ 0.8x-1.4-\lambda \end{array}\right]\right|{v}_{1}=0\phantom{\rule{0ex}{0ex}}$as follows.

role="math" localid="1660635035242" $\left|\left[\begin{array}{c}1.4x-\lambda -1.2\\ 0.8x-1.4-\lambda \end{array}\right]\right|{v}_{1}=0\phantom{\rule{0ex}{0ex}}\left|\left[\begin{array}{c}1.4x-\left(1\right)-1.2\\ 0.8x-1.4-\left(1\right)\end{array}\right]\right|\left[\begin{array}{c}{x}_{2}\\ {y}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left|\left|\begin{array}{cc}2.4& -1.2\\ 0.8& -2.4\end{array}\right|\right|\left[\begin{array}{c}{x}_{2}\\ {y}_{2}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left|\left|\begin{array}{cc}2.4{x}_{1}& -1.2{y}_{2}\\ 0.8{x}_{1}& -2.4{y}_{2}\end{array}\right|\right|=0$

Further, simplify the equation as follows.

$\left[\begin{array}{cc}2.4{x}_{1}& -1.2{y}_{2}\\ 0.8{x}_{1}& -2.4{y}_{2}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\begin{array}{cc}2.4{x}_{1}& -1.2{y}_{2}\end{array}=0\phantom{\rule{0ex}{0ex}}\begin{array}{cc}0.8{x}_{1}& -2.4{y}_{2}\end{array}=0$

Simplify the equations $\begin{array}{cc}2.4{x}_{1}& -1.2{y}_{2}\end{array}=0and\begin{array}{cc}0.8{x}_{1}& -2.4{y}_{2}\end{array}=0$ as follows.

$0.4{x}_{2}-1.2{y}_{2}=0\phantom{\rule{0ex}{0ex}}0.4{x}_{2}=1.2{y}_{2}\phantom{\rule{0ex}{0ex}}{x}_{2}=3{y}_{2}\phantom{\rule{0ex}{0ex}}$

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 and corresponding to the Eigen values ${\lambda }_{1}=-1and{\lambda }_{2}=1$ respectively.

## Step 5: Sketch the rough portraits for the system.

As $-1={\lambda }_{2}<0and1={\lambda }_{2}>0$ , 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.

## Step 6: (c) Determine the kind of intersection

The both species will prosper if$\frac{x\left(0\right)}{y\left(0\right)}<2$ implies , and the both species $\frac{x\left(0\right)}{y\left(0\right)}\ge 2$is died if .

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$\frac{x\left(0\right)}{y\left(0\right)}<2$ then both species will proper, and .

If $\frac{x\left(0\right)}{y\left(0\right)}\ge 2$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$\frac{x\left(0\right)}{y\left(0\right)}\ge 2$ .