Numerical computational solution of the Volterra integral equations system of the second kind by using an expansion method

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Abstract

An expansion method is used for treatment of second kind Volterra integral equations system. This method gives an analytic solution for the system. The method reduces the system of integral equations to a linear system of ordinary differential equations. After constructing boundary conditions, this system reduces to a system of equations that can be solved easily with any of the usual methods. Finally, for showing the efficiency of the method we use some numerical examples.

Introduction

In this paper, we use a modified Taylor-series expansion method for solving Volterra integral equations system of the second kind. This method was first presented in 2 for solving Fredholm integral equations of second kind and then in [3], [4] for solving Volterra integral equations system and Fredholm integral equations system of second kind.

Consider the second kind Volterra integral equations system of the form:F(s)=G(s)+0sK(s,t)F(t)dt,0s1,whereF(s)=[f1(s),f2(s),,fn(s)]T,G(s)=[g1(s),g2(s),,gn(s)]T,K(s,t)=[kij(s,t)],i,j=1,2,,n.

In Eq. (1) the functions K and G are given, and F is the solution to be determined [1].

We assume that (1) has a unique solution.

Consider the ith equation of (1):fi(s)=gi(s)+0sj=1nkij(s,t)fj(t)dt,i=1,2,,n.

A Taylor-series expansion can be made for the solution fj(t) in the integral Eq. (2):fj(t)=fj(s)+fj(s)(t-s)++1m!fj(m)(s)(t-s)m+E(t),where E(t) denotes the error between fj(t) and its Taylor-series expansion (3).E(t)=1(m+1)!fj(m+1)(s)(t-s)(m+1)+

If we use the first m terms of the Taylor-series expansion (3) and neglige the term 0sj=1nkij(s,t)E(t)dt in Eq. (2), then by substituting (3) for fj(t) in the integral in Eq. (2), we have:fi(s)gi(s)+0sj=1nkij(s,t)r=0m1r!(t-s)rfj(r)(s)dt,i=1,2,,n,fi(s)gi(s)+j=1nr=0m1r!fj(r)(s)0skij(s,t)(t-s)rdt,i=1,2,,n,fi(s)-j=1nr=0m1r!fj(r)(s)0skij(s,t)(t-s)rdtgi(s),i=1,2,,n.If the integrals in Eq. (6) can be solved analytically, then the bracketed quantities are functions of s alone. So Eq. (6) becomes a linear system of ordinary differential equations that can be solved. However, this requires the construction of an appropriate number of boundary conditions.

Section snippets

Boundary conditions

In order to construct boundary conditions, we first differentiate both sides of Eq. (2) to get the interval 0 < s < 1 and i = 1, 2,  , n:fi(s)=gi(s)+j=1nkij(s,s)fj(s)+0sj=1nkijs(s,t)fj(t)dt,fi(m)(s)=gi(m)(s)+j=1nr=1mmrkijs(m-r)(s,s)fj(r-1)(s)+0sj=1nkijs(m)(s,t)fj(t)dt,where kijs(m)(s,t)=mkij(s,t)/sm. Substitute fj(s) for fj(t) in the integrals in Eqs. (7), (8) to obtain for 0 < s < 1 and i = 1, 2,  , n:fi(s)-j=1nkij(s,s)+0sj=1nkijs(s,t)dtfj(s)gi(s),fi(m)(s)-j=1nr=1mmkkijs(m-r)(s,s)fj(r-1)(s

Numerical examples

Example 1

For the first example, consider the following Volterra system of integral equations:f1(s)=g1(s)+0s(s-t)3f1(t)dt+0s(s-t)2f2(t)dt,f2(s)=g2(s)+0s(s-t)4f1(t)dt+0s(s-t)3f2(t)dt.g1(s) and g2(s) are chosen such that the exact solution is f1(s) = s2 + 1, f2(s) = 1  s3 + s. Numerical results are given in Table 1.

Example 2

For the second example, consider the following Volterra system of integral equations:f1(s)=g1(s)+0s(sin(s-t)-1)f1(t)dt+0s(1-tcoss)f2(t)dt,f2(s)=g2(s)+0sf1(t)dt+0s(s-t)f2(t)dt.g1(s) and g2(s) are

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