Chapter 7.6, Problem 3E

(a)

To determine

To Find: Whether the model describe the cooperation, or competition or the predator-prey relationship.

Answer to Problem 3E

The value $c_1,c_2$ is negative this shows that the other species is hurting both of them so the relationship is competitive relationship.

Explanation of Solution

Given:

The given equation are,

  $\begin{array}{l}\frac{dx}{dt}=0.5x-0.004x^2-0.001xy\\ \frac{dy}{dt}=0.4y-0.001y^2-0.002xy\end{array}$

Calculation:

Consider the given system is,

  $\begin{array}{l}\frac{dx}{dt}=a_{1}x-b_1{x}^2+c_{1}xy\\ \frac{dy}{dt}=a_{2}x-b_2{y}^2+c_{2}xy\end{array}$

The system is said to be cooperative if $c_1>0$ and $c_2>0$ . The equation is cooperative for $c_1>0$ and $c_2>0$ , it is competitive if $c_1<0$ and $c_2<0$ . It is predator if-prey if $c_1{c}_2<0$ .

Consider the given equation is,

  $\begin{array}{l}\frac{dx}{dt}=0.5x-0.004x^2-0.001xy\\ \frac{dy}{dt}=0.4y-0.001y^2-0.002xy\end{array}$

Since, the value $c_1,c_2$ is negative this shows that the other species is hurting both of them so the relationship is competitive relationship.

(b)

To determine

To Find: The equilibrium solutions and the explanation of its significance.

Answer to Problem 3E

The population of both the species satisfy the equilibrium the population stops changing son the population of both the species remains constant.

Explanation of Solution

Given:

The given equation are,

  $\begin{array}{l}\frac{dx}{dt}=0.5x-0.004x^2-0.001xy\\ \frac{dy}{dt}=0.4y-0.001y^2-0.002xy\end{array}$

Calculation:

Consider the condition for the equilibrium is,

  $\begin{array}{l}\frac{dx}{dt}=0\\ 0.5x-0.004x^2-0.001xy=0\end{array}$

For above $x=0$ is one of the solution. Then, divide both the sides 0.001x as,

  $\begin{array}{l}500-4x-y=0\\ 4x+y=500\end{array}$

Consider for $\frac{dy}{dt}=0$ as,

  $\begin{array}{l}\frac{dy}{dt}=0\\ 0.4y-0.001y^2-0.002xy=0\end{array}$

For above $y=0$ is one of the solution. Then, divide both the sides $0.001y$ as,

  $\begin{array}{l}400-y-2x=0\\ 2x+y=400\end{array}$

From both the equations.

  $\begin{array}{l}x=50\\ y=300\end{array}$

Consider the solution for $x=0$ as,

  $\begin{array}{l}x=0\\ y=400\end{array}$

Consider the solution for $y=0$ as,

  $\begin{array}{l}x=125\\ y=0\end{array}$

Thus, as the population of both the species satisfy the equilibrium the population stops changing son the population of both the species remains constant.