A computational method for solution of the prey and predator problem

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Abstract

In this article a mathematical model of the problem of prey and predator being presented and Adomian decomposition method is employed to compute an approximation to the solution of the system of nonlinear Volterra differential equations governing on the problem. Some plots for the population of the prey and predator versus time are presented to illustrate the solution.

Introduction

There are some rabbits and foxes living together. Foxes eat the rabbits and rabbits eat clover. Suppose that there are enough clovers and the rabbits have enough food to eat. When there are a lot of rabbits, the foxes also grow and their population increase. When the number of foxes increase and they eat a lot of rabbits they enter into a short period of food and their number decrease. As the number of the foxes decrease, the rabbits will be safe and their population increase. When the number of rabbits increase the number of foxes would increase and by passing the time we can see an infinite repeatability of increase and decrease in the population of these two kinds of animals (Fig. 1, Fig. 2, Fig. 3, Fig. 4).

The governing equations to the problem would be as follows [1].dxdt=x(t)(a−by(t)),dydt=−y(t)(c−dx(t)),where x(t) and y(t) are respectively the populations of rabbits and the foxes at the time t.

Section snippets

Computational method for solving (1)

To solve the system of Eqs. (1) Adomian decomposition method, well addressed in [2], [3], is employed. The equivalent canonical form of this system is as follows:x(t)=x(t=0)+∫0tx(a−by)dt,y(t)=y(t=0)−∫0ty(c−dx)dt.As usual in Adomian decomposition method the solutions of Eqs. (2) are considered to be as the sum of the following series:x=∑n=0xn,y=∑n=0yn.And the integrand in Eq. (2), as the sum of the following series.f(x,y)=∑n=0An(x0,…,xn,y0,…,yn)g(x,y)=∑n=0Bn(x0,…,xn,y0,…,yn)where An, Bn are

Numerical results and discussion

For numerical study the following values are used.

Casex0y0abcd
11418110.11
214180.1111
316100.1111
41610110.11

According to the values introduced in the table the following solutions are derived.Case1:x(t)≈14−238t−271.6t2+20191.31333t3,y(t)≈18+250.2t−197.9t2−8794.533333t3,Case2:x(t)≈14−250.6t+604.87t2+19364.94233t3,y(t)≈18+234t−734.4t2−19546.40333t3,Case3:x(t)=16−158.4t−415.92t2+7516.536t3,y(t)=10+150t+333t2−7391.76t3,Case4:x(t)=16−144t−624t2+6602.4t3,y(t)=10+159t+544.05t2−6828.53t3.The following

References (5)

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