Research ArticleOn Solution of Integrodifferential Equation with DelayParameter by Sinc Basis Functions

I Najafi Khalilsaraye and K Maleknejad

Department of Mathematics College of Basic Sciences Karaj Branch Islamic Azad University Alborz Iran

Correspondence should be addressed to K Maleknejad maleknejadiustacir

Received 10 March 2014 Accepted 18 April 2014 Published 8 May 2014

Academic Editor Reza Ezzati

Copyright copy 2014 I Najafi Khalilsaraye and K MaleknejadThis is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We want to find a numerical solution for an integrodifferential equation with an integral boundary condition and delay parameterThis type of problems arises in mathematical physics mechanics population growth and other fields of physics and mathematicalchemistry So convergence of this approach is discussed by presenting a theorem which gives exponential type convergence rateand guarantees the accuracy of that Finally by some numerical examples we show the efficiency and accuracy of this numericalmethod

1 Introduction

Discussing integrodifferential equationswith integral bound-ary condition consisting of delay parameter is a worthyand significant branch of nonlinear applied mathematicsIt is important that integrodifferential equation with delayparameter is generated often in investigations connectedwithchemical engineering mathematical physics undergroundwater flow engineering and so on (see [1 2]) Note thatthe problems with integral boundary conditions have variousapplications in applied fields such as population growthproblems and blood flow problems For a detailed descriptionof the integrodifferential equations with delay parameter andproblems with integral boundary conditions the reader canrefer to references of [3ndash7]

In this paper we discuss the following problem whichshows a first-order integrodifferential equation with integralboundary condition consisting of delay parameter The exis-tence and uniqueness of solution for this problem are provedin [8] But analytic solving and reaching an exact solutionare impossible Then in this paper we approximate the exactsolution by numerical method

119889119910

119889119909= 119892 (119909 119910 (119909) 119910 (120574

1(119909)) 119880119910 (119909) 119881119910 (119909)) equiv 119866119910 (119909)

119910 (0) = 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

(1)

where 119909 isin 119868 = [0 119887] (119887 gt 0) 119892 isin 119862(119868 times R4R) where 119862 isfamily of all continuous functions and 120574

1isin 119862(119868 119868) 120581 isin (0 119887]

120596 isin 119862(119868 times RR) 1205781 1205782 119886 isin R

And

(119880119910) (119909) = int

1205742(119909)

0

119896 (119909 119903) 119910 (119861 (119903)) 119889119903

(119881119910) (119909) = int

119887

0

ℎ (119909 119903) 119910 (119863 (119903)) 119889119903

(2)

Here 1205742 119861 119863 isin 119862(119868 119868) 119896(119909 119903) isin 119862[119860R+] and ℎ(119909 119903) isin

119862[1198600R+] that

119860 = (119909 119903) isin R2| 0 le 119903 le 120574

2(119909) 119909 isin 119868

1198600= (119909 119903) isin R

2| 0 le 119903 le 119909 119909 isin 119868

(3)

Here if 1205781= 1 120578

2= 119886 = 0 and 120581 = 119887 then we have a problem

with boundary condition of kind periodic and if 1205781= 0 then

we have a problem with integral boundary condition and if1205781

= 1205782

= 0 we have a problem with an initial condition Soproblem (1) is general type of these cases

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 564632 8 pageshttpdxdoiorg1011552014564632

2 Abstract and Applied Analysis

Now the following functional integral equation can beeasily concluded from (1)

119910 (119909) = 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892(119905 119910 (119905) 119910 (1205741(119905)) int

1205742(119905)

0

119896 (119905 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119905 119903) 119910 (119863 (119903)) 119889119903) 119889119905

(4)

and in this paper we want to discuss solution of (1) in point ofequivalent integral equation (4) But we know the fact that wecannot solve this integral equation to give an exact solutionso numerical approaches are used to reach an approximatedsolution

Numerical approaches for estimated solution of integrod-ifferential and integral equation and search for existenceand uniqueness of solution for some problems have beenresearched bymany authors and reader can see thesemethodsin [9ndash15] In these references authors use methods on anestimate by basis function such as wavelets polynomialsand so forth or use some quadrature formulas But thesetechnics usually have convergence rate of polynomial orderwith respect to 119872 where 119872 represents the cardinal of termsof sum in the expansion or the cardinal of points of thequadrature formula In [16] author showed that if we employthe Sinc approach the convergence rate is exponential ordersuch as 119874(exp(minus119888119872

12)) with some 119888 gt 0 We know the

exponential rate is much faster than that of polynomial rateSo in this paper we employ the Sinc function instead ofbase function and an iterative technique to estimate exactsolution of (4) in marked points Our approach dose notcontain of changing the solution of (4) to a systemof algebraicequations by expanding 119910(119909) as basis functionwith unknowncoefficients so this technique has computations less thanother methods and exponential rate in accuracy Also inthis present paper we prove a theorem to guarantee theconvergence of numerical technique

2 Main Results

In this section we introduce basic requirements and theoremto prove existence and uniqueness of solution for first-order integrodifferential equation with integral boundarycondition consisting of delay parameter (1) For detaileddescriptions we refer the reader to [17 18]

Definition 1 A function as 119910 isin 1198621(119868 119868) is called a lower

solution of (1) if

119889119910

119889119909le 119866119910 (119909) 119909 isin 119868

119910 (0) le 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

(5)

and it is an upper solution of (1) if

119889119910

119889119909ge 119866119910 (119909) 119909 isin 119868

119910 (0) ge 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

(6)

Theorem2 Let 1199060 V0isin 1198621(119868R) be lower and upper solutions

of (1) respectively and 1199060(119909) le V

0(119909) 119909 isin 119868

In addition consider the following

(1198711) 119892 isin 119862(119868 times R4R) 120574

1 1205742 119861 119863 isin 119862(119868 119868) 120574

1(119909) 1205742(119909)

119861(119909) 119863(119909) le 119909 forall119909 isin 119868 120581 isin (0 119887) 120596 isin 119862(119868 times RR)and 1205781 1205782ge 0

(1198712) there are nonnegative bounded integrable functions

1198721(119909) 119872

2(119909) 119872

3(119909) 119872

4(119909) on 119868 that

int

119887

0

[1198721(119905) + 119872

2(119909) + 119872

3(119909) int

1205742(119909)

0

119896 (119909 119903) 119889119903

+1198724(119909) int

119887

0

ℎ (119909 119903) 119889119903] 119889119909 le 1

(7)

such that

119892 (119909 1206011 1206012 1198801206011 1198811206011) minus 119892 (119909 120595

1 1205952 1198801205951 1198811205951)

ge minus1198721(119909) (120601

1minus 1205951) minus 119872

2(119909) (120601

2minus 1205952)

minus 1198723(119909)119880 (120601

1minus 1205951) minus 119872

4(119909) 119881 (120601

1minus 1205951)

(8)

if1199060le 1205951le 1206011le V01199060(1205741(119909)) le 120595

2le 1206012le V0(1205741(119909))

(1198713) there is 120579(119909) isin 119862(119868R+) such that120596(119909 120595) minus 120596(119909 120595

minus) ge

120579(119909)(120595 minus 120595minus) and if 119906

0(119905) le 120595

minusle 120595 le V

0(119909) then

problem (1) has extremal solutions 119906 V isin [1199060 V0] In

addition there are monotone sequences 119906119899(119909) V119899(119909) sub

[1199060 V0] such that 119906

119899rarr 119906 V

119899rarr V for 119899 rarr +infin and

these are convergent uniformly on 119909 isin 119868 where 119906119899(119909)

V119899(119909) are defined as

119906119899(119909) = int

119909

0

119890minusint119909

1199031198721(119904)119889119904

times [119892 (119903 119906119899minus1

(119903) 119906119899minus1

(1205741(119903))

119880119906119899minus1

(119903) 119881119906119899minus1

(119903) ) + 1198721(119903) 119906119899minus1

(119903)

minus 1198722(119903) (119906119899minus 119906119899minus1

) (1205741(119903))

minus 1198723(119903) 119880 (119906

119899minus 119906119899minus1

) (119903)

minus1198724(119903) 119881 (119906

119899minus 119906119899minus1

) (119903)] 119889119903

+ 119890minusint119909

01198721(119904)119889119904

[1205781119906119899minus1

(120581)

+ 1205782int

119887

0

120596 (119903 119906119899minus1

(119903)) 119889119903 + 119886]

forall119909 isin 119868 119899 = 1 2 3

Abstract and Applied Analysis 3

V119899(119909) = int

119909

0

119890minusint119909

1199031198721(119904)119889119904

times [119892 (119903 V119899minus1

(119903) V119899minus1

(1205741(119903))

119880V119899minus1

(119903) 119881V119899minus1

(119903)) + 1198721(119903) 119911119899minus1

(119903)

minus 1198722(119903) (V119899minus V119899minus1

) (1205742(119903))

minus 1198723(119903) 119880 (V

119899minus V119899minus1

) (119903)

minus1198724(119903) 119881 (V

119899minus V119899minus1

) (119903)] 119889119903

+ 119890minusint119909

01198721(119904)119889119904

[1205781V119899minus1

(120581)

+ 1205782int

119887

0

120596 (119903 V119899minus1

(119903)) 119889119903 + 119886]

forall119909 isin 119868 119899 = 1 2 3

1199060le 1199061le sdot sdot sdot le 119906

119899le sdot sdot sdot le 119906 le V le sdot sdot sdot le V

119899

le sdot sdot sdot le V1le V0

(9)

Proof See [18]

Theorem 3 Consider that assumptions of Theorem 2 holdMoreover consider the following

(1198714) there are nonnegative bounded functions 120572

1(119909) 120572

2(119909)

1205723(119909) 120572

4(119909) on 119868 such that

119892 (119909 1206011 1206012 1198801206011 1198811206011) minus 119892 (119909 120595

1 1205952 1198801205951 1198811205951)

le 1205721(119909) (120601

1minus 1205951) + 1205722(119909) (120601

2minus 1205952)

+ 1205723(119909)119880 (120601

1minus 1205951) + 1205724(119909)119881 (120601

1minus 1205951)

(10)

if 1199060(119909) le 120595

1le 1206011

le V0(119909) 119906

0(1205741(119909)) le 120595

2le 1206012

le

V0(1205741(119909))

(1198715) there is120573(119909) isin C(119868R+) such that120596(119909 120595)minus120596(119909 120595

minus) le

120573(119909)(120595 minus 120595minus) if 1199060(119905) le 120595

minusle 120595 le V

0(119909)

Then problem (1) has a unique solution 119906minus

isin [1199060 V0]

In addition there are sequences 119906119899(119909) V119899(119909) sub [119906

0 V0]

that these are monotone and 119906119899

rarr 119906minus V119899

rarr 119906minus

for 119899 rarr +infin This convergence is uniformly on 119909 isin

119868 where 119906119899(119909) V

119899(119909) are defined as (9) such that

119906119899minus 119906minus119888le 119871119899V0minus 1199060119888 119899 isin N where

119871 = 1205781+ int

119887

0

[1205782120573 (119909) + 120572

1(119909) + 119872

1(119909) + 120572

2(119909) + 119872

2(119909)

+ (1205723(119909) + 119872

3(119909)) int

1205742(119909)

0

119870 (119909 119903) 119889119903

+ (1205724(119909) + 119872

4(119909)) int

119887

0

ℎ (119909 119903) 119889119903] 119889119909 lt 1

(11)

Proof See [18]

3 Sinc Function

In this section we recall the basis function and some ofits applicabilities In here definition of sinc(119909) function isfollowed by

Sinc (119909) =

sin (120587119909)

120587119909 119909 = 0

1 119909 = 0

(12)

Now for ℎ gt 0 and integer 119895 we define 119895th Sinc functionwith step size ℎ by

119878 (119895 ℎ) (119909) =sin (120587 (119909 minus 119895ℎ) ℎ)

120587 (119909 minus 119895ℎ) ℎ (13)

31 Sinc Estimation on [119886 119887] Let 119909 = 120593(119908) be a transforma-tion that denotes a conformal transformation which transfersthe simply connected domain 119860 onto a strip region 119860

119889such

that

120593 ((119886 119887)) = (minusinfininfin) lim119909rarr119886

120593 (119909) = minusinfin

lim119909rarr119887

120593 (119909) = infin

(14)

In here 120597119860 is boundary of 119860 and in order to have the Sincestimation on (119886 119887) conformal transformation is applied asfollows

120593 (119905) = ln(119905 minus 119886

119887 minus 119905) (15)

This function transfers the eye-shaped complex region

119908 = 119909 + 119894119910

1003816100381610038161003816100381610038161003816arg(

119908 minus 119886

119887 minus 119908)

1003816100381610038161003816100381610038161003816lt 119889 le

120587

2 (16)

onto 119860119889that it is a strip region

119860119889

= 120590 = 120572 + 120573119894 1003816100381610038161003816120573

1003816100381610038161003816 lt 119889 lt120587

2 (17)

The basis functions on finite interval (119886 119887) are given by

119878 (119895 ℎ) ∘ 120593 (119909) =sin (120587 (120593 (119909) minus 119895ℎ) ℎ)

120587 (120593 (119909) minus 119895ℎ) ℎ (18)

and also Sinc function for interpolation points 119909119895

= 119895ℎ isgiven by

119878 (119895 ℎ) (119896ℎ) = 120575(0)

119895119896=

1 119895 = 119896

0 119895 = 119896(19)

Then 119878(119895 ℎ) ∘ 120593(119909) shows behavior of Kronecker deltafunction on the network points

119909119895= 120593minus1

(119895ℎ) =119886 + 119887119890

119895ℎ

1 + 119890119895ℎ (20)

4 Abstract and Applied Analysis

The approximation of 119892(119909) by interpolation and quadratureformulas for int

119887

119886119892(119909)119889119909 is

119892 (119909) asymp

119872

sum

119895=minus119872

119892 (119909119895) 119878 (119895 ℎ) ∘ 120593 (119909)

int

119887

119886

119892 (119909) 119889119909 asymp ℎ

119872

sum

119895=minus119872

119892 (119909119895)

1205931015840 (119909119895)

(21)

Theorem 4 Consider that for a map 119908 = 120593minus1

(120585) the map119892(120593minus1

(120585)) satisfies

(1) 119892 isin 1198671(119863119889) for 119889 gt 0

(2) 119892 decays exponentially on the real line such that

1003816100381610038161003816119892 (119909)1003816100381610038161003816 le 120572 exp (minus120573 |119909|) forall119909 isin R 120572 120573 gt 0 (22)

with some 120572 120573 and 119889 Then one has

sup119886lt119909lt119887

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119909) minus

119872

sum

119895=minus119872

119892 (120593minus1

(119895ℎ)) 119878 (119895 ℎ) ∘ 120593 (119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

le 119862radic119872 exp(minusradic120587119889120573119872)

(23)

That there is some 119862 and ℎ is ℎ = radic120587119889120573119872

Proof See [16]

Definition 5 Let 119871120572(119860) be the set of all analytic functions 119892

for which there exists a constant 119862 such that

1003816100381610038161003816119892 (119911)1003816100381610038161003816 le 119862

|119890120593(119911)

|120572

(1 + |119890120593(119911)|)2120572

119911 isin 119860 0 lt 120572 le 1 (24)

Theorem6 Let1198921205931015840isin 119871120572(119860) with 0 lt 120572 le 1 and 0 lt 119889 le 120587

also let ℎ = radic120587119889120572119872 Then there exists a constant 1198621 which

is independent of 119872 such that

10038161003816100381610038161003816100381610038161003816100381610038161003816

int

119909119895

119886

119892 (119905) 119889119905 minus ℎ

119872

sum

119896=minus119872

120575(minus1)

119895119896

119892 (119909119895)

1205931015840 (119909119896)

10038161003816100381610038161003816100381610038161003816100381610038161003816

le 1198621119890minusradic120587119889120572119872

(25)

where

120575(minus1)

119895119896=

1

2+ int

119895minus119896

0

sin (120587119905)

120587119905119889119905 (26)

and 120593(119909) 119909119896are defined as above

Proof See [16]

32 Sinc-Qudrature Method In this section for solvingequation

119910 (119909) = 1205821119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892(119904 119910 (119904) 119910 (1205741(119904))

int

1205742(119904)

0

119896 (119904 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910 (119863 (119904)) 119889119903) 119889119904

(27)

we try to discrete integral equation by quadrature formula as

int

119887

119886

119892 (119904) 119889119904 = ℎ

119872

sum

119895=minus119872

119892 (119904119895)

1205931015840 (119904119895)

+ 119874(exp(minus2120587119889119872

log (2120587119889119872120573)))

int

119904

119886

119891 (119909) 119889119909 = ℎ

119872

sum

119895=minus119872

119892 (119909119895)

1205931015840 (119909119895)120578ℎ119895

(119904)

+ 119874(log119872

119872exp(minus

120587119889119872

log (120587119889119872120573)))

(28)

where

120578ℎ119895

(119904) =1

2+

1

120587119903119894(120587

119903 minus 119895ℎ

ℎ) 119903

119894= int

119909

0

sin (119905)

119905119889119905

(29)

with 119909119895

= 119904119895

= (119886 + 119887119890119895ℎ

)(1 + 119895ℎ) 119895 = minus119872 119872 and ℎ =

(1119872) log(120587119889119872120573) (see [16])Now by substituting quadrature formulas in the integral

equation (4) we have

119910119872

(119909) = 1205781119910 (120581) + 120578

2ℎ

119872

sum

119895=minus119872

120596 (119904119895 119910 (119904119895))

1205931015840 (119904119895)

+ 119886

+ ℎ

119872

sum

119895=minus119872

119892(119904119895 119910 (119904119895) 119910 (120572 (119904

119895))

ℎ

119872

sum

119894=minus119872

119870(119904119895 119903119894) 119910 (119861 (119903

119894))

1205931015840 (119903119894)

120578ℎ119894

(119905)

ℎ

119872

sum

119894=minus119872

ℎ (119904119895 119903119894) 119910 (119863 (119903

119894))

1205931015840 (119903119894)

)

times (1205931015840(119904119895))minus1

120578ℎ119895

(119909)

(30)

Abstract and Applied Analysis 5

Now for 119899 = 1 2 3 let

1199101119872

(119909) = 1205781119910 (120581)

119910119899+1119872

(119909)

= 1205781119910119899119872

(120581) + 1205782ℎ

119872

sum

119895=minus119872

120596 (119904119895 119910119899119872

(119904119895))

1205931015840 (119904119895)

+ 119886

+ ℎ

119872

sum

119895=minus119872

119892(119904119895 119910119899119872

(119904119895) 119910119899119872

(120572 (119904119895))

ℎ

119872

sum

119894=minus119872

119886 (119904119895 119903119894) 119910119899119872

(119861 (119903119894))

1205931015840 (119903119894)

120578ℎ119894

(119909)

ℎ

119872

sum

119894=minus119872

ℎ (119904119895 119903119894) 119910119899119872

(120575 (119903119894))

1205931015840 (119903119894)

)

times (1205931015840(119904119895))minus1

120578ℎ119895

(119909)

(31)

4 Convergence of Method

In this section we present a theorem that shows a boundfor 119910(119909) minus 119910

119899(119909) with the real norm where 119910(119909) is the exact

solution of problem (4) and 119910119899(119909) is an estimation for 119910(119909) by

using Sinc function in interpolaion and quadrature formulaThe result is shown as follows

Theorem 7 Under the assumptions (1198711)ndash(1198715) iterative

approximation approach (31) is convergent to exact solution if1199101(119909) is closed enough to the it and

1198732le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

(32)

where 1198731

= sup120573(119909) 119909 isin 119868 1198732

= max120572119894(119909) 119894 = 1 2 3 4

119909 isin 119868 1198733

= sup119870(119909 119903) 119909 119903 isin 119868 and 1198734

= supℎ(119909 119903)

119909 119903 isin 119868

Proof For a fixed 119872 let

119910119899+1119872

(119909) = 1205781119909119899119872

(120591) + 1205782int

119887

0

120596 (119903 119910119899119872

(119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119910119899119872

(119904) 119910119899119872

(1205741(119904)) 119880119910

119899119872(119904)

119881119910119899119872

(119904)) 119889119904

119910 (119909) = 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119909 (119904) 119910 (1205741(119904)) 119880119910 (119904) 119881119910 (119904)) 119889119904

(33)

Now1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816

+ 1205782

100381610038161003816100381610038161003816100381610038161003816

int

119887

0

(120596 (119903 119910119899(119903)) minus 120596 (119903 119910 (119903))) 119889119903

100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

int

119909

0

[119892(119904 119910119899(119904) 119910119899(1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910119899(119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910119899(119863 (119903)) 119889119903)

minus 119891(119904 119910 (119904) 119910 (1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910 (119863 (119903)) 119889119903)]119889119904

100381610038161003816100381610038161003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816 + 12057821198871198731

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816

+ 1198871198732

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816 + 119887119873

2

1003816100381610038161003816119910119899 (1205741 (119909)) minus 119910 (1205741(119909))

1003816100381610038161003816

+ 11988711987321198733

1003816100381610038161003816119909119899 (119861 (119903)) minus 119910 (119861 (119903))1003816100381610038161003816

+ 11988711987321198724

1003816100381610038161003816119910119899 (119863 (119903)) minus 119910 (119863 (119903))1003816100381610038161003816

le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))

1003816100381610038161003816119910119899119872 (119909) minus 119910 (119909)1003816100381610038161003816

(34)

Then1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816 le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))119899

times10038161003816100381610038161199101119872 (119909) minus 119910 (119909)

1003816100381610038161003816

(35)

Now by assumption in theorem we have 1205781+12057821198871198731+1198871198732(2+

1198733+ 1198734) lt 1

Because 1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734) lt 1205781+ 12057821198871198731+

119887((1 minus (1205781+ 12057821198871198731))119887(2 + 119873

3+ 1198734))(2 + 119873

3+ 1198734) = 1 and

then lim119899rarr+infin

119910119899+1119872

(119909) = 119910(119909) 0 le 119909 le 119887

5 Illustrative Examples

In this section for showing efficiency of the iterative schemeand in order to show the facts of the exact solution wegive some examples below All routines have been written inMathematica 7 and a Dual-Core CPU 200GHz is used torun the programs Also about efficiency and accuracy of theproposed numerical method we present absolute errors fordifferent examples Therefore numerical results are shown infigures to illustrate the efficiency of this scheme

6 Abstract and Applied Analysis

5 10 15 20 25 30

0001

01

n

En60

10minus5

10minus7

10minus9

Figure 1 Absolute errors of Example 1 For different values ofiteration 119899

Example 1 Consider the following problem

119889119910

119889119909=

1

151199093[119909 minus 119910 (119909)] minus

1

5119909119910 (1205741(119909)) +

1

10011990921199102(1205741(119909))

+1

500119909[1199093minus int

(12)119909

0

8119909119903119910 (1199032) 119889119903]

5

minus1

6001199092[1199093minus int

(12)119909

0

8119909119903119910(1199032)119889119903]

6

+1

7001199092[1199092minus int

1

0

21199092119903119910 (1199032) 119889119903]

7

minus1

8001199093[1199092minus int

1

0

21199092119903119910(1199032)119889119903]

8

equiv 119866119910 (119909)

119909 isin 119868 = [0 1]

119910 (0) =1

4119910 (120581) +

1

4int

1

0

(2119903 + 119910 (119903)) 119889119903 +1

4

(36)

where 120581 isin (0 1] 1205741isin 119862(119868 119868) and 120574

1(119909) le 119909 on 119868

Now we want to show that assumptions (1198711)ndash(1198715) and

convergence criterion which was prepared in Theorem 7 aresatisfied for integral equation (4)

Noting that 1205742(119909) = (12)119909 119861(119909) = 119863(119909) = 119909

2 forall119909 isin 119868then (119871

1)ndash(1198715) is true For details see [6]

In addition we have 120573(119909) = 1 1205721(119909) = 0 120572

2(119909) =

(150)1199092 1205723(119909) = 120572

4(119909) = (1100)119909

17 119870(119909 119903) = 8119909119903ℎ(119909 119903) = 2119909

2119903 and 120578

1= 1205782

= 14 then 1198731

= 1 1198732

= 1501198733= 8 and 119873

4= 2 and then

1198732=

1

50le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

=1

24 (37)

02 04 06 08 10

090

092

094

096

t

x(t)

Figure 2 Approximate solution for Example 1

Therefore the iterative method (31) converges to the exactsolution of this equation Now based on Sinc quadraturescheme we can have some successive approximations forsolution of this example for 120581 = 12 Absolute error for 119899thapproximation and 119873 points quadrature method is definedby

119890119899119872

= max119909isin[01]

1003816100381610038161003816119910119899+1119872 (119909) minus 119910

119899119872(119909)

1003816100381610038161003816 119899 = 1 2 3

(38)

For different values of 119899 absolute errors are depicted inFigure 1 and approximate solution for 119899 = 30 is depicted inFigure 2

Example 2 In Example 1 of [19] authors considered theintegrodifferential equation

119889119910

119889119909= 1 minus

1

31199093+ int

1

0

11990931199102(119911) 119889119911

119910 (0) = 0

(39)

The exact solution is 119910(119909) = 119909 Maximum absolute error foreach iteration and different values of quadrature points aredepicted in Figures 3 and 4 By comparing these results withthe numerical results given in [19] in Table 2 efficiency andaccuracy of current approach are guaranteed

6 Conclusion

In this paper we apply a numerical approach by Sinc functionfor reaching the estimated solution of integrodifferentialequation with enteral boundary condition and with delayparameter To reach this aim we change this problem to afunctional enteral equation The Sinc estimation has expo-nential convergence rate such as119874(minus119888119890

11987212

) that this propertyis an advantage so we applied it to solve our problem byusing collocation method Finally some examples are solvedby this numericalmethod to show the efficiency and accuracyof Sinc estimation It is worthy to note that this method can

Abstract and Applied Analysis 7

2 4 6 8 10 12 14

01

n

En300

10minus4

10minus7

10minus10

Figure 3 Maximum absolute error related to each iteration inExample 2

100 150 200 250 300

0001

N

E15N

10minus5

10minus7

10minus9

10minus11

Figure 4 Maximum error related to different quadrature points inExample 2

be used for solving integrodifferential equations with integralboundary conditions with deviating arguments arising in allsciences such as chemistry physics and other fields of appliedmathematics

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authors thank the referee for hisher careful reading ofthe paper and useful suggestion

References

[1] R P Agarwal D OrsquoRegan and P J Y Wong Positive Solutionsof Differential Difference and Integral Equations Kluwer Aca-demic Publishers Dordrecht The Netherlands 1999

[2] T A Burton ldquoDifferential inequalities for integral and delaydifferential equationsrdquo in Comparison Methods and StabilityTheory vol 162 of Lecture Notes in Pure and Applied Mathemat-ics pp 35ndash56 Dekker New York NY USA 1994

[3] B Ahmad and B S Alghamdi ldquoApproximation of solutionsof the nonlinear Duffing equation involving both integral andnon-integral forcing terms with separated boundary condi-tionsrdquo Computer Physics Communications vol 179 no 6 pp409ndash416 2008

[4] B Ahmad ldquoOn the existence of p-periodic solutions forDuffing type integro-differential equations with p-LaplacianrdquoLobachevskii Journal ofMathematics vol 29 no 1 pp 1ndash4 2008

[5] Y K Chang and J J Nieto ldquoExistence of solutions for impulsiveneutral integro-differential inclusions with nonlocal initial con-ditions via fractional operatorsrdquo Numerical Functional Analysisand Optimization vol 30 no 3-4 pp 227ndash244 2009

[6] Z Luo and J J Nieto ldquoNew results for the periodic boundaryvalue problem for impulsive integro-differential equationsrdquoNonlinear Analysis Theory Methods amp Applications vol 70 no6 pp 2248ndash2260 2009

[7] T Jankowski ldquoDifferential equations with integral boundaryconditionsrdquo Journal of Computational andAppliedMathematicsvol 147 no 1 pp 1ndash8 2002

[8] G Wang G Song and L Zhang ldquoIntegral boundary valueproblems for first order integro-differential equations withdeviating argumentsrdquo Journal of Computational and AppliedMathematics vol 225 no 2 pp 602ndash611 2009

[9] K Maleknejad P Torabi and R Mollapourasl ldquoFixed pointmethod for solving nonlinear quadratic Volterra integral equa-tionsrdquo Computers amp Mathematics with Applications vol 62 no6 pp 2555ndash2566 2011

[10] K Maleknejad and M N Sahlan ldquoThe method of momentsfor solution of second kind Fredholm integral equations basedon B-spline waveletsrdquo International Journal of Computer Math-ematics vol 87 no 7 pp 1602ndash1616 2010

[11] K Maleknejad K Nouri and R Mollapourasl ldquoExistence ofsolutions for some nonlinear integral equationsrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no6 pp 2559ndash2564 2009

[12] W Volk ldquoThe iterated Galerkin method for linear integro-differential equationsrdquo Journal of Computational and AppliedMathematics vol 21 no 1 pp 63ndash74 1988

[13] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces vol 60 Marcel Dekker New York NY USA 1980

[14] A Mohsen and M El-Gamel ldquoA sinc-collocation method forthe linear Fredholm integro-differential equationsrdquo Journal ofApplied Mathematics and Physics vol 58 no 3 pp 380ndash3902007

[15] A Avudainayagam and C Vani ldquoWavelet-Galerkin method forintegro-differential equationsrdquoAppliedNumericalMathematicsvol 32 no 3 pp 247ndash254 2000

[16] F StengerNumerical Methods Based on Sinc and Analytic Func-tions vol 20 Springer New York NY USA 1993

[17] L Zhang ldquoBoundary value problem for first order impulsivefunctional integro-differential equationsrdquo Journal of Computa-tional and Applied Mathematics vol 235 no 8 pp 2442ndash24502011

[18] G Wang L Zhang and G Song ldquoExtremal solutions for thefirst order impulsive functional differential equations with

8 Abstract and Applied Analysis

upper and lower solutions in reversed orderrdquo Journal of Com-putational and AppliedMathematics vol 235 no 1 pp 325ndash3332010

[19] M Dehghan and R Salehi ldquoThe numerical solution of thenon-linear integro-differential equations based on the meshlessmethodrdquo Journal of Computational and Applied Mathematicsvol 236 no 9 pp 2367ndash2377 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

Now the following functional integral equation can beeasily concluded from (1)

119910 (119909) = 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892(119905 119910 (119905) 119910 (1205741(119905)) int

1205742(119905)

0

119896 (119905 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119905 119903) 119910 (119863 (119903)) 119889119903) 119889119905

(4)

and in this paper we want to discuss solution of (1) in point ofequivalent integral equation (4) But we know the fact that wecannot solve this integral equation to give an exact solutionso numerical approaches are used to reach an approximatedsolution

Numerical approaches for estimated solution of integrod-ifferential and integral equation and search for existenceand uniqueness of solution for some problems have beenresearched bymany authors and reader can see thesemethodsin [9ndash15] In these references authors use methods on anestimate by basis function such as wavelets polynomialsand so forth or use some quadrature formulas But thesetechnics usually have convergence rate of polynomial orderwith respect to 119872 where 119872 represents the cardinal of termsof sum in the expansion or the cardinal of points of thequadrature formula In [16] author showed that if we employthe Sinc approach the convergence rate is exponential ordersuch as 119874(exp(minus119888119872

12)) with some 119888 gt 0 We know the

exponential rate is much faster than that of polynomial rateSo in this paper we employ the Sinc function instead ofbase function and an iterative technique to estimate exactsolution of (4) in marked points Our approach dose notcontain of changing the solution of (4) to a systemof algebraicequations by expanding 119910(119909) as basis functionwith unknowncoefficients so this technique has computations less thanother methods and exponential rate in accuracy Also inthis present paper we prove a theorem to guarantee theconvergence of numerical technique

2 Main Results

In this section we introduce basic requirements and theoremto prove existence and uniqueness of solution for first-order integrodifferential equation with integral boundarycondition consisting of delay parameter (1) For detaileddescriptions we refer the reader to [17 18]

Definition 1 A function as 119910 isin 1198621(119868 119868) is called a lower

solution of (1) if

119889119910

119889119909le 119866119910 (119909) 119909 isin 119868

119910 (0) le 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

(5)

and it is an upper solution of (1) if

119889119910

119889119909ge 119866119910 (119909) 119909 isin 119868

119910 (0) ge 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

(6)

Theorem2 Let 1199060 V0isin 1198621(119868R) be lower and upper solutions

of (1) respectively and 1199060(119909) le V

0(119909) 119909 isin 119868

In addition consider the following

(1198711) 119892 isin 119862(119868 times R4R) 120574

1 1205742 119861 119863 isin 119862(119868 119868) 120574

1(119909) 1205742(119909)

119861(119909) 119863(119909) le 119909 forall119909 isin 119868 120581 isin (0 119887) 120596 isin 119862(119868 times RR)and 1205781 1205782ge 0

(1198712) there are nonnegative bounded integrable functions

1198721(119909) 119872

2(119909) 119872

3(119909) 119872

4(119909) on 119868 that

int

119887

0

[1198721(119905) + 119872

2(119909) + 119872

3(119909) int

1205742(119909)

0

119896 (119909 119903) 119889119903

+1198724(119909) int

119887

0

ℎ (119909 119903) 119889119903] 119889119909 le 1

(7)

such that

119892 (119909 1206011 1206012 1198801206011 1198811206011) minus 119892 (119909 120595

1 1205952 1198801205951 1198811205951)

ge minus1198721(119909) (120601

1minus 1205951) minus 119872

2(119909) (120601

2minus 1205952)

minus 1198723(119909)119880 (120601

1minus 1205951) minus 119872

4(119909) 119881 (120601

1minus 1205951)

(8)

if1199060le 1205951le 1206011le V01199060(1205741(119909)) le 120595

2le 1206012le V0(1205741(119909))

(1198713) there is 120579(119909) isin 119862(119868R+) such that120596(119909 120595) minus 120596(119909 120595

minus) ge

120579(119909)(120595 minus 120595minus) and if 119906

0(119905) le 120595

minusle 120595 le V

0(119909) then

problem (1) has extremal solutions 119906 V isin [1199060 V0] In

addition there are monotone sequences 119906119899(119909) V119899(119909) sub

[1199060 V0] such that 119906

119899rarr 119906 V

119899rarr V for 119899 rarr +infin and

these are convergent uniformly on 119909 isin 119868 where 119906119899(119909)

V119899(119909) are defined as

119906119899(119909) = int

119909

0

119890minusint119909

1199031198721(119904)119889119904

times [119892 (119903 119906119899minus1

(119903) 119906119899minus1

(1205741(119903))

119880119906119899minus1

(119903) 119881119906119899minus1

(119903) ) + 1198721(119903) 119906119899minus1

(119903)

minus 1198722(119903) (119906119899minus 119906119899minus1

) (1205741(119903))

minus 1198723(119903) 119880 (119906

119899minus 119906119899minus1

) (119903)

minus1198724(119903) 119881 (119906

119899minus 119906119899minus1

) (119903)] 119889119903

+ 119890minusint119909

01198721(119904)119889119904

[1205781119906119899minus1

(120581)

+ 1205782int

119887

0

120596 (119903 119906119899minus1

(119903)) 119889119903 + 119886]

forall119909 isin 119868 119899 = 1 2 3

Abstract and Applied Analysis 3

V119899(119909) = int

119909

0

119890minusint119909

1199031198721(119904)119889119904

times [119892 (119903 V119899minus1

(119903) V119899minus1

(1205741(119903))

119880V119899minus1

(119903) 119881V119899minus1

(119903)) + 1198721(119903) 119911119899minus1

(119903)

minus 1198722(119903) (V119899minus V119899minus1

) (1205742(119903))

minus 1198723(119903) 119880 (V

119899minus V119899minus1

) (119903)

minus1198724(119903) 119881 (V

119899minus V119899minus1

) (119903)] 119889119903

+ 119890minusint119909

01198721(119904)119889119904

[1205781V119899minus1

(120581)

+ 1205782int

119887

0

120596 (119903 V119899minus1

(119903)) 119889119903 + 119886]

forall119909 isin 119868 119899 = 1 2 3

1199060le 1199061le sdot sdot sdot le 119906

119899le sdot sdot sdot le 119906 le V le sdot sdot sdot le V

119899

le sdot sdot sdot le V1le V0

(9)

Proof See [18]

Theorem 3 Consider that assumptions of Theorem 2 holdMoreover consider the following

(1198714) there are nonnegative bounded functions 120572

1(119909) 120572

2(119909)

1205723(119909) 120572

4(119909) on 119868 such that

119892 (119909 1206011 1206012 1198801206011 1198811206011) minus 119892 (119909 120595

1 1205952 1198801205951 1198811205951)

le 1205721(119909) (120601

1minus 1205951) + 1205722(119909) (120601

2minus 1205952)

+ 1205723(119909)119880 (120601

1minus 1205951) + 1205724(119909)119881 (120601

1minus 1205951)

(10)

if 1199060(119909) le 120595

1le 1206011

le V0(119909) 119906

0(1205741(119909)) le 120595

2le 1206012

le

V0(1205741(119909))

(1198715) there is120573(119909) isin C(119868R+) such that120596(119909 120595)minus120596(119909 120595

minus) le

120573(119909)(120595 minus 120595minus) if 1199060(119905) le 120595

minusle 120595 le V

0(119909)

Then problem (1) has a unique solution 119906minus

isin [1199060 V0]

In addition there are sequences 119906119899(119909) V119899(119909) sub [119906

0 V0]

that these are monotone and 119906119899

rarr 119906minus V119899

rarr 119906minus

for 119899 rarr +infin This convergence is uniformly on 119909 isin

119868 where 119906119899(119909) V

119899(119909) are defined as (9) such that

119906119899minus 119906minus119888le 119871119899V0minus 1199060119888 119899 isin N where

119871 = 1205781+ int

119887

0

[1205782120573 (119909) + 120572

1(119909) + 119872

1(119909) + 120572

2(119909) + 119872

2(119909)

+ (1205723(119909) + 119872

3(119909)) int

1205742(119909)

0

119870 (119909 119903) 119889119903

+ (1205724(119909) + 119872

4(119909)) int

119887

0

ℎ (119909 119903) 119889119903] 119889119909 lt 1

(11)

Proof See [18]

3 Sinc Function

In this section we recall the basis function and some ofits applicabilities In here definition of sinc(119909) function isfollowed by

Sinc (119909) =

sin (120587119909)

120587119909 119909 = 0

1 119909 = 0

(12)

Now for ℎ gt 0 and integer 119895 we define 119895th Sinc functionwith step size ℎ by

119878 (119895 ℎ) (119909) =sin (120587 (119909 minus 119895ℎ) ℎ)

120587 (119909 minus 119895ℎ) ℎ (13)

31 Sinc Estimation on [119886 119887] Let 119909 = 120593(119908) be a transforma-tion that denotes a conformal transformation which transfersthe simply connected domain 119860 onto a strip region 119860

119889such

that

120593 ((119886 119887)) = (minusinfininfin) lim119909rarr119886

120593 (119909) = minusinfin

lim119909rarr119887

120593 (119909) = infin

(14)

In here 120597119860 is boundary of 119860 and in order to have the Sincestimation on (119886 119887) conformal transformation is applied asfollows

120593 (119905) = ln(119905 minus 119886

119887 minus 119905) (15)

This function transfers the eye-shaped complex region

119908 = 119909 + 119894119910

1003816100381610038161003816100381610038161003816arg(

119908 minus 119886

119887 minus 119908)

1003816100381610038161003816100381610038161003816lt 119889 le

120587

2 (16)

onto 119860119889that it is a strip region

119860119889

= 120590 = 120572 + 120573119894 1003816100381610038161003816120573

1003816100381610038161003816 lt 119889 lt120587

2 (17)

The basis functions on finite interval (119886 119887) are given by

119878 (119895 ℎ) ∘ 120593 (119909) =sin (120587 (120593 (119909) minus 119895ℎ) ℎ)

120587 (120593 (119909) minus 119895ℎ) ℎ (18)

and also Sinc function for interpolation points 119909119895

= 119895ℎ isgiven by

119878 (119895 ℎ) (119896ℎ) = 120575(0)

119895119896=

1 119895 = 119896

0 119895 = 119896(19)

Then 119878(119895 ℎ) ∘ 120593(119909) shows behavior of Kronecker deltafunction on the network points

119909119895= 120593minus1

(119895ℎ) =119886 + 119887119890

119895ℎ

1 + 119890119895ℎ (20)

4 Abstract and Applied Analysis

The approximation of 119892(119909) by interpolation and quadratureformulas for int

119887

119886119892(119909)119889119909 is

119892 (119909) asymp

119872

sum

119895=minus119872

119892 (119909119895) 119878 (119895 ℎ) ∘ 120593 (119909)

int

119887

119886

119892 (119909) 119889119909 asymp ℎ

119872

sum

119895=minus119872

119892 (119909119895)

1205931015840 (119909119895)

(21)

Theorem 4 Consider that for a map 119908 = 120593minus1

(120585) the map119892(120593minus1

(120585)) satisfies

(1) 119892 isin 1198671(119863119889) for 119889 gt 0

(2) 119892 decays exponentially on the real line such that

1003816100381610038161003816119892 (119909)1003816100381610038161003816 le 120572 exp (minus120573 |119909|) forall119909 isin R 120572 120573 gt 0 (22)

with some 120572 120573 and 119889 Then one has

sup119886lt119909lt119887

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119909) minus

119872

sum

119895=minus119872

119892 (120593minus1

(119895ℎ)) 119878 (119895 ℎ) ∘ 120593 (119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

le 119862radic119872 exp(minusradic120587119889120573119872)

(23)

That there is some 119862 and ℎ is ℎ = radic120587119889120573119872

Proof See [16]

Definition 5 Let 119871120572(119860) be the set of all analytic functions 119892

for which there exists a constant 119862 such that

1003816100381610038161003816119892 (119911)1003816100381610038161003816 le 119862

|119890120593(119911)

|120572

(1 + |119890120593(119911)|)2120572

119911 isin 119860 0 lt 120572 le 1 (24)

Theorem6 Let1198921205931015840isin 119871120572(119860) with 0 lt 120572 le 1 and 0 lt 119889 le 120587

also let ℎ = radic120587119889120572119872 Then there exists a constant 1198621 which

is independent of 119872 such that

10038161003816100381610038161003816100381610038161003816100381610038161003816

int

119909119895

119886

119892 (119905) 119889119905 minus ℎ

119872

sum

119896=minus119872

120575(minus1)

119895119896

119892 (119909119895)

1205931015840 (119909119896)

10038161003816100381610038161003816100381610038161003816100381610038161003816

le 1198621119890minusradic120587119889120572119872

(25)

where

120575(minus1)

119895119896=

1

2+ int

119895minus119896

0

sin (120587119905)

120587119905119889119905 (26)

and 120593(119909) 119909119896are defined as above

Proof See [16]

32 Sinc-Qudrature Method In this section for solvingequation

119910 (119909) = 1205821119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892(119904 119910 (119904) 119910 (1205741(119904))

int

1205742(119904)

0

119896 (119904 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910 (119863 (119904)) 119889119903) 119889119904

(27)

we try to discrete integral equation by quadrature formula as

int

119887

119886

119892 (119904) 119889119904 = ℎ

119872

sum

119895=minus119872

119892 (119904119895)

1205931015840 (119904119895)

+ 119874(exp(minus2120587119889119872

log (2120587119889119872120573)))

int

119904

119886

119891 (119909) 119889119909 = ℎ

119872

sum

119895=minus119872

119892 (119909119895)

1205931015840 (119909119895)120578ℎ119895

(119904)

+ 119874(log119872

119872exp(minus

120587119889119872

log (120587119889119872120573)))

(28)

where

120578ℎ119895

(119904) =1

2+

1

120587119903119894(120587

119903 minus 119895ℎ

ℎ) 119903

119894= int

119909

0

sin (119905)

119905119889119905

(29)

with 119909119895

= 119904119895

= (119886 + 119887119890119895ℎ

)(1 + 119895ℎ) 119895 = minus119872 119872 and ℎ =

(1119872) log(120587119889119872120573) (see [16])Now by substituting quadrature formulas in the integral

equation (4) we have

119910119872

(119909) = 1205781119910 (120581) + 120578

2ℎ

119872

sum

119895=minus119872

120596 (119904119895 119910 (119904119895))

1205931015840 (119904119895)

+ 119886

+ ℎ

119872

sum

119895=minus119872

119892(119904119895 119910 (119904119895) 119910 (120572 (119904

119895))

ℎ

119872

sum

119894=minus119872

119870(119904119895 119903119894) 119910 (119861 (119903

119894))

1205931015840 (119903119894)

120578ℎ119894

(119905)

ℎ

119872

sum

119894=minus119872

ℎ (119904119895 119903119894) 119910 (119863 (119903

119894))

1205931015840 (119903119894)

)

times (1205931015840(119904119895))minus1

120578ℎ119895

(119909)

(30)

Abstract and Applied Analysis 5

Now for 119899 = 1 2 3 let

1199101119872

(119909) = 1205781119910 (120581)

119910119899+1119872

(119909)

= 1205781119910119899119872

(120581) + 1205782ℎ

119872

sum

119895=minus119872

120596 (119904119895 119910119899119872

(119904119895))

1205931015840 (119904119895)

+ 119886

+ ℎ

119872

sum

119895=minus119872

119892(119904119895 119910119899119872

(119904119895) 119910119899119872

(120572 (119904119895))

ℎ

119872

sum

119894=minus119872

119886 (119904119895 119903119894) 119910119899119872

(119861 (119903119894))

1205931015840 (119903119894)

120578ℎ119894

(119909)

ℎ

119872

sum

119894=minus119872

ℎ (119904119895 119903119894) 119910119899119872

(120575 (119903119894))

1205931015840 (119903119894)

)

times (1205931015840(119904119895))minus1

120578ℎ119895

(119909)

(31)

4 Convergence of Method

In this section we present a theorem that shows a boundfor 119910(119909) minus 119910

119899(119909) with the real norm where 119910(119909) is the exact

solution of problem (4) and 119910119899(119909) is an estimation for 119910(119909) by

using Sinc function in interpolaion and quadrature formulaThe result is shown as follows

Theorem 7 Under the assumptions (1198711)ndash(1198715) iterative

approximation approach (31) is convergent to exact solution if1199101(119909) is closed enough to the it and

1198732le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

(32)

where 1198731

= sup120573(119909) 119909 isin 119868 1198732

= max120572119894(119909) 119894 = 1 2 3 4

119909 isin 119868 1198733

= sup119870(119909 119903) 119909 119903 isin 119868 and 1198734

= supℎ(119909 119903)

119909 119903 isin 119868

Proof For a fixed 119872 let

119910119899+1119872

(119909) = 1205781119909119899119872

(120591) + 1205782int

119887

0

120596 (119903 119910119899119872

(119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119910119899119872

(119904) 119910119899119872

(1205741(119904)) 119880119910

119899119872(119904)

119881119910119899119872

(119904)) 119889119904

119910 (119909) = 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119909 (119904) 119910 (1205741(119904)) 119880119910 (119904) 119881119910 (119904)) 119889119904

(33)

Now1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816

+ 1205782

100381610038161003816100381610038161003816100381610038161003816

int

119887

0

(120596 (119903 119910119899(119903)) minus 120596 (119903 119910 (119903))) 119889119903

100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

int

119909

0

[119892(119904 119910119899(119904) 119910119899(1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910119899(119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910119899(119863 (119903)) 119889119903)

minus 119891(119904 119910 (119904) 119910 (1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910 (119863 (119903)) 119889119903)]119889119904

100381610038161003816100381610038161003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816 + 12057821198871198731

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816

+ 1198871198732

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816 + 119887119873

2

1003816100381610038161003816119910119899 (1205741 (119909)) minus 119910 (1205741(119909))

1003816100381610038161003816

+ 11988711987321198733

1003816100381610038161003816119909119899 (119861 (119903)) minus 119910 (119861 (119903))1003816100381610038161003816

+ 11988711987321198724

1003816100381610038161003816119910119899 (119863 (119903)) minus 119910 (119863 (119903))1003816100381610038161003816

le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))

1003816100381610038161003816119910119899119872 (119909) minus 119910 (119909)1003816100381610038161003816

(34)

Then1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816 le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))119899

times10038161003816100381610038161199101119872 (119909) minus 119910 (119909)

1003816100381610038161003816

(35)

Now by assumption in theorem we have 1205781+12057821198871198731+1198871198732(2+

1198733+ 1198734) lt 1

Because 1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734) lt 1205781+ 12057821198871198731+

119887((1 minus (1205781+ 12057821198871198731))119887(2 + 119873

3+ 1198734))(2 + 119873

3+ 1198734) = 1 and

then lim119899rarr+infin

119910119899+1119872

(119909) = 119910(119909) 0 le 119909 le 119887

5 Illustrative Examples

In this section for showing efficiency of the iterative schemeand in order to show the facts of the exact solution wegive some examples below All routines have been written inMathematica 7 and a Dual-Core CPU 200GHz is used torun the programs Also about efficiency and accuracy of theproposed numerical method we present absolute errors fordifferent examples Therefore numerical results are shown infigures to illustrate the efficiency of this scheme

6 Abstract and Applied Analysis

5 10 15 20 25 30

0001

01

n

En60

10minus5

10minus7

10minus9

Figure 1 Absolute errors of Example 1 For different values ofiteration 119899

Example 1 Consider the following problem

119889119910

119889119909=

1

151199093[119909 minus 119910 (119909)] minus

1

5119909119910 (1205741(119909)) +

1

10011990921199102(1205741(119909))

+1

500119909[1199093minus int

(12)119909

0

8119909119903119910 (1199032) 119889119903]

5

minus1

6001199092[1199093minus int

(12)119909

0

8119909119903119910(1199032)119889119903]

6

+1

7001199092[1199092minus int

1

0

21199092119903119910 (1199032) 119889119903]

7

minus1

8001199093[1199092minus int

1

0

21199092119903119910(1199032)119889119903]

8

equiv 119866119910 (119909)

119909 isin 119868 = [0 1]

119910 (0) =1

4119910 (120581) +

1

4int

1

0

(2119903 + 119910 (119903)) 119889119903 +1

4

(36)

where 120581 isin (0 1] 1205741isin 119862(119868 119868) and 120574

1(119909) le 119909 on 119868

Now we want to show that assumptions (1198711)ndash(1198715) and

convergence criterion which was prepared in Theorem 7 aresatisfied for integral equation (4)

Noting that 1205742(119909) = (12)119909 119861(119909) = 119863(119909) = 119909

2 forall119909 isin 119868then (119871

1)ndash(1198715) is true For details see [6]

In addition we have 120573(119909) = 1 1205721(119909) = 0 120572

2(119909) =

(150)1199092 1205723(119909) = 120572

4(119909) = (1100)119909

17 119870(119909 119903) = 8119909119903ℎ(119909 119903) = 2119909

2119903 and 120578

1= 1205782

= 14 then 1198731

= 1 1198732

= 1501198733= 8 and 119873

4= 2 and then

1198732=

1

50le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

=1

24 (37)

02 04 06 08 10

090

092

094

096

t

x(t)

Figure 2 Approximate solution for Example 1

Therefore the iterative method (31) converges to the exactsolution of this equation Now based on Sinc quadraturescheme we can have some successive approximations forsolution of this example for 120581 = 12 Absolute error for 119899thapproximation and 119873 points quadrature method is definedby

119890119899119872

= max119909isin[01]

1003816100381610038161003816119910119899+1119872 (119909) minus 119910

119899119872(119909)

1003816100381610038161003816 119899 = 1 2 3

(38)

For different values of 119899 absolute errors are depicted inFigure 1 and approximate solution for 119899 = 30 is depicted inFigure 2

Example 2 In Example 1 of [19] authors considered theintegrodifferential equation

119889119910

119889119909= 1 minus

1

31199093+ int

1

0

11990931199102(119911) 119889119911

119910 (0) = 0

(39)

The exact solution is 119910(119909) = 119909 Maximum absolute error foreach iteration and different values of quadrature points aredepicted in Figures 3 and 4 By comparing these results withthe numerical results given in [19] in Table 2 efficiency andaccuracy of current approach are guaranteed

6 Conclusion

In this paper we apply a numerical approach by Sinc functionfor reaching the estimated solution of integrodifferentialequation with enteral boundary condition and with delayparameter To reach this aim we change this problem to afunctional enteral equation The Sinc estimation has expo-nential convergence rate such as119874(minus119888119890

11987212

) that this propertyis an advantage so we applied it to solve our problem byusing collocation method Finally some examples are solvedby this numericalmethod to show the efficiency and accuracyof Sinc estimation It is worthy to note that this method can

Abstract and Applied Analysis 7

2 4 6 8 10 12 14

01

n

En300

10minus4

10minus7

10minus10

Figure 3 Maximum absolute error related to each iteration inExample 2

100 150 200 250 300

0001

N

E15N

10minus5

10minus7

10minus9

10minus11

Figure 4 Maximum error related to different quadrature points inExample 2

be used for solving integrodifferential equations with integralboundary conditions with deviating arguments arising in allsciences such as chemistry physics and other fields of appliedmathematics

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authors thank the referee for hisher careful reading ofthe paper and useful suggestion

References

[1] R P Agarwal D OrsquoRegan and P J Y Wong Positive Solutionsof Differential Difference and Integral Equations Kluwer Aca-demic Publishers Dordrecht The Netherlands 1999

[2] T A Burton ldquoDifferential inequalities for integral and delaydifferential equationsrdquo in Comparison Methods and StabilityTheory vol 162 of Lecture Notes in Pure and Applied Mathemat-ics pp 35ndash56 Dekker New York NY USA 1994

[3] B Ahmad and B S Alghamdi ldquoApproximation of solutionsof the nonlinear Duffing equation involving both integral andnon-integral forcing terms with separated boundary condi-tionsrdquo Computer Physics Communications vol 179 no 6 pp409ndash416 2008

[4] B Ahmad ldquoOn the existence of p-periodic solutions forDuffing type integro-differential equations with p-LaplacianrdquoLobachevskii Journal ofMathematics vol 29 no 1 pp 1ndash4 2008

[5] Y K Chang and J J Nieto ldquoExistence of solutions for impulsiveneutral integro-differential inclusions with nonlocal initial con-ditions via fractional operatorsrdquo Numerical Functional Analysisand Optimization vol 30 no 3-4 pp 227ndash244 2009

[6] Z Luo and J J Nieto ldquoNew results for the periodic boundaryvalue problem for impulsive integro-differential equationsrdquoNonlinear Analysis Theory Methods amp Applications vol 70 no6 pp 2248ndash2260 2009

[7] T Jankowski ldquoDifferential equations with integral boundaryconditionsrdquo Journal of Computational andAppliedMathematicsvol 147 no 1 pp 1ndash8 2002

[8] G Wang G Song and L Zhang ldquoIntegral boundary valueproblems for first order integro-differential equations withdeviating argumentsrdquo Journal of Computational and AppliedMathematics vol 225 no 2 pp 602ndash611 2009

[9] K Maleknejad P Torabi and R Mollapourasl ldquoFixed pointmethod for solving nonlinear quadratic Volterra integral equa-tionsrdquo Computers amp Mathematics with Applications vol 62 no6 pp 2555ndash2566 2011

[10] K Maleknejad and M N Sahlan ldquoThe method of momentsfor solution of second kind Fredholm integral equations basedon B-spline waveletsrdquo International Journal of Computer Math-ematics vol 87 no 7 pp 1602ndash1616 2010

[11] K Maleknejad K Nouri and R Mollapourasl ldquoExistence ofsolutions for some nonlinear integral equationsrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no6 pp 2559ndash2564 2009

[12] W Volk ldquoThe iterated Galerkin method for linear integro-differential equationsrdquo Journal of Computational and AppliedMathematics vol 21 no 1 pp 63ndash74 1988

[13] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces vol 60 Marcel Dekker New York NY USA 1980

[14] A Mohsen and M El-Gamel ldquoA sinc-collocation method forthe linear Fredholm integro-differential equationsrdquo Journal ofApplied Mathematics and Physics vol 58 no 3 pp 380ndash3902007

[15] A Avudainayagam and C Vani ldquoWavelet-Galerkin method forintegro-differential equationsrdquoAppliedNumericalMathematicsvol 32 no 3 pp 247ndash254 2000

[16] F StengerNumerical Methods Based on Sinc and Analytic Func-tions vol 20 Springer New York NY USA 1993

[17] L Zhang ldquoBoundary value problem for first order impulsivefunctional integro-differential equationsrdquo Journal of Computa-tional and Applied Mathematics vol 235 no 8 pp 2442ndash24502011

[18] G Wang L Zhang and G Song ldquoExtremal solutions for thefirst order impulsive functional differential equations with

8 Abstract and Applied Analysis

upper and lower solutions in reversed orderrdquo Journal of Com-putational and AppliedMathematics vol 235 no 1 pp 325ndash3332010

[19] M Dehghan and R Salehi ldquoThe numerical solution of thenon-linear integro-differential equations based on the meshlessmethodrdquo Journal of Computational and Applied Mathematicsvol 236 no 9 pp 2367ndash2377 2012

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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OptimizationJournal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

V119899(119909) = int

119909

0

119890minusint119909

1199031198721(119904)119889119904

times [119892 (119903 V119899minus1

(119903) V119899minus1

(1205741(119903))

119880V119899minus1

(119903) 119881V119899minus1

(119903)) + 1198721(119903) 119911119899minus1

(119903)

minus 1198722(119903) (V119899minus V119899minus1

) (1205742(119903))

minus 1198723(119903) 119880 (V

119899minus V119899minus1

) (119903)

minus1198724(119903) 119881 (V

119899minus V119899minus1

) (119903)] 119889119903

+ 119890minusint119909

01198721(119904)119889119904

[1205781V119899minus1

(120581)

+ 1205782int

119887

0

120596 (119903 V119899minus1

(119903)) 119889119903 + 119886]

forall119909 isin 119868 119899 = 1 2 3

1199060le 1199061le sdot sdot sdot le 119906

119899le sdot sdot sdot le 119906 le V le sdot sdot sdot le V

119899

le sdot sdot sdot le V1le V0

(9)

Proof See [18]

Theorem 3 Consider that assumptions of Theorem 2 holdMoreover consider the following

(1198714) there are nonnegative bounded functions 120572

1(119909) 120572

2(119909)

1205723(119909) 120572

4(119909) on 119868 such that

119892 (119909 1206011 1206012 1198801206011 1198811206011) minus 119892 (119909 120595

1 1205952 1198801205951 1198811205951)

le 1205721(119909) (120601

1minus 1205951) + 1205722(119909) (120601

2minus 1205952)

+ 1205723(119909)119880 (120601

1minus 1205951) + 1205724(119909)119881 (120601

1minus 1205951)

(10)

if 1199060(119909) le 120595

1le 1206011

le V0(119909) 119906

0(1205741(119909)) le 120595

2le 1206012

le

V0(1205741(119909))

(1198715) there is120573(119909) isin C(119868R+) such that120596(119909 120595)minus120596(119909 120595

minus) le

120573(119909)(120595 minus 120595minus) if 1199060(119905) le 120595

minusle 120595 le V

0(119909)

Then problem (1) has a unique solution 119906minus

isin [1199060 V0]

In addition there are sequences 119906119899(119909) V119899(119909) sub [119906

0 V0]

that these are monotone and 119906119899

rarr 119906minus V119899

rarr 119906minus

for 119899 rarr +infin This convergence is uniformly on 119909 isin

119868 where 119906119899(119909) V

119899(119909) are defined as (9) such that

119906119899minus 119906minus119888le 119871119899V0minus 1199060119888 119899 isin N where

119871 = 1205781+ int

119887

0

[1205782120573 (119909) + 120572

1(119909) + 119872

1(119909) + 120572

2(119909) + 119872

2(119909)

+ (1205723(119909) + 119872

3(119909)) int

1205742(119909)

0

119870 (119909 119903) 119889119903

+ (1205724(119909) + 119872

4(119909)) int

119887

0

ℎ (119909 119903) 119889119903] 119889119909 lt 1

(11)

Proof See [18]

3 Sinc Function

In this section we recall the basis function and some ofits applicabilities In here definition of sinc(119909) function isfollowed by

Sinc (119909) =

sin (120587119909)

120587119909 119909 = 0

1 119909 = 0

(12)

Now for ℎ gt 0 and integer 119895 we define 119895th Sinc functionwith step size ℎ by

119878 (119895 ℎ) (119909) =sin (120587 (119909 minus 119895ℎ) ℎ)

120587 (119909 minus 119895ℎ) ℎ (13)

31 Sinc Estimation on [119886 119887] Let 119909 = 120593(119908) be a transforma-tion that denotes a conformal transformation which transfersthe simply connected domain 119860 onto a strip region 119860

119889such

that

120593 ((119886 119887)) = (minusinfininfin) lim119909rarr119886

120593 (119909) = minusinfin

lim119909rarr119887

120593 (119909) = infin

(14)

In here 120597119860 is boundary of 119860 and in order to have the Sincestimation on (119886 119887) conformal transformation is applied asfollows

120593 (119905) = ln(119905 minus 119886

119887 minus 119905) (15)

This function transfers the eye-shaped complex region

119908 = 119909 + 119894119910

1003816100381610038161003816100381610038161003816arg(

119908 minus 119886

119887 minus 119908)

1003816100381610038161003816100381610038161003816lt 119889 le

120587

2 (16)

onto 119860119889that it is a strip region

119860119889

= 120590 = 120572 + 120573119894 1003816100381610038161003816120573

1003816100381610038161003816 lt 119889 lt120587

2 (17)

The basis functions on finite interval (119886 119887) are given by

119878 (119895 ℎ) ∘ 120593 (119909) =sin (120587 (120593 (119909) minus 119895ℎ) ℎ)

120587 (120593 (119909) minus 119895ℎ) ℎ (18)

and also Sinc function for interpolation points 119909119895

= 119895ℎ isgiven by

119878 (119895 ℎ) (119896ℎ) = 120575(0)

119895119896=

1 119895 = 119896

0 119895 = 119896(19)

Then 119878(119895 ℎ) ∘ 120593(119909) shows behavior of Kronecker deltafunction on the network points

119909119895= 120593minus1

(119895ℎ) =119886 + 119887119890

119895ℎ

1 + 119890119895ℎ (20)

4 Abstract and Applied Analysis

The approximation of 119892(119909) by interpolation and quadratureformulas for int

119887

119886119892(119909)119889119909 is

119892 (119909) asymp

119872

sum

119895=minus119872

119892 (119909119895) 119878 (119895 ℎ) ∘ 120593 (119909)

int

119887

119886

119892 (119909) 119889119909 asymp ℎ

119872

sum

119895=minus119872

119892 (119909119895)

1205931015840 (119909119895)

(21)

Theorem 4 Consider that for a map 119908 = 120593minus1

(120585) the map119892(120593minus1

(120585)) satisfies

(1) 119892 isin 1198671(119863119889) for 119889 gt 0

(2) 119892 decays exponentially on the real line such that

1003816100381610038161003816119892 (119909)1003816100381610038161003816 le 120572 exp (minus120573 |119909|) forall119909 isin R 120572 120573 gt 0 (22)

with some 120572 120573 and 119889 Then one has

sup119886lt119909lt119887

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119909) minus

119872

sum

119895=minus119872

119892 (120593minus1

(119895ℎ)) 119878 (119895 ℎ) ∘ 120593 (119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

le 119862radic119872 exp(minusradic120587119889120573119872)

(23)

That there is some 119862 and ℎ is ℎ = radic120587119889120573119872

Proof See [16]

Definition 5 Let 119871120572(119860) be the set of all analytic functions 119892

for which there exists a constant 119862 such that

1003816100381610038161003816119892 (119911)1003816100381610038161003816 le 119862

|119890120593(119911)

|120572

(1 + |119890120593(119911)|)2120572

119911 isin 119860 0 lt 120572 le 1 (24)

Theorem6 Let1198921205931015840isin 119871120572(119860) with 0 lt 120572 le 1 and 0 lt 119889 le 120587

also let ℎ = radic120587119889120572119872 Then there exists a constant 1198621 which

is independent of 119872 such that

10038161003816100381610038161003816100381610038161003816100381610038161003816

int

119909119895

119886

119892 (119905) 119889119905 minus ℎ

119872

sum

119896=minus119872

120575(minus1)

119895119896

119892 (119909119895)

1205931015840 (119909119896)

10038161003816100381610038161003816100381610038161003816100381610038161003816

le 1198621119890minusradic120587119889120572119872

(25)

where

120575(minus1)

119895119896=

1

2+ int

119895minus119896

0

sin (120587119905)

120587119905119889119905 (26)

and 120593(119909) 119909119896are defined as above

Proof See [16]

32 Sinc-Qudrature Method In this section for solvingequation

119910 (119909) = 1205821119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892(119904 119910 (119904) 119910 (1205741(119904))

int

1205742(119904)

0

119896 (119904 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910 (119863 (119904)) 119889119903) 119889119904

(27)

we try to discrete integral equation by quadrature formula as

int

119887

119886

119892 (119904) 119889119904 = ℎ

119872

sum

119895=minus119872

119892 (119904119895)

1205931015840 (119904119895)

+ 119874(exp(minus2120587119889119872

log (2120587119889119872120573)))

int

119904

119886

119891 (119909) 119889119909 = ℎ

119872

sum

119895=minus119872

119892 (119909119895)

1205931015840 (119909119895)120578ℎ119895

(119904)

+ 119874(log119872

119872exp(minus

120587119889119872

log (120587119889119872120573)))

(28)

where

120578ℎ119895

(119904) =1

2+

1

120587119903119894(120587

119903 minus 119895ℎ

ℎ) 119903

119894= int

119909

0

sin (119905)

119905119889119905

(29)

with 119909119895

= 119904119895

= (119886 + 119887119890119895ℎ

)(1 + 119895ℎ) 119895 = minus119872 119872 and ℎ =

(1119872) log(120587119889119872120573) (see [16])Now by substituting quadrature formulas in the integral

equation (4) we have

119910119872

(119909) = 1205781119910 (120581) + 120578

2ℎ

119872

sum

119895=minus119872

120596 (119904119895 119910 (119904119895))

1205931015840 (119904119895)

+ 119886

+ ℎ

119872

sum

119895=minus119872

119892(119904119895 119910 (119904119895) 119910 (120572 (119904

119895))

ℎ

119872

sum

119894=minus119872

119870(119904119895 119903119894) 119910 (119861 (119903

119894))

1205931015840 (119903119894)

120578ℎ119894

(119905)

ℎ

119872

sum

119894=minus119872

ℎ (119904119895 119903119894) 119910 (119863 (119903

119894))

1205931015840 (119903119894)

)

times (1205931015840(119904119895))minus1

120578ℎ119895

(119909)

(30)

Abstract and Applied Analysis 5

Now for 119899 = 1 2 3 let

1199101119872

(119909) = 1205781119910 (120581)

119910119899+1119872

(119909)

= 1205781119910119899119872

(120581) + 1205782ℎ

119872

sum

119895=minus119872

120596 (119904119895 119910119899119872

(119904119895))

1205931015840 (119904119895)

+ 119886

+ ℎ

119872

sum

119895=minus119872

119892(119904119895 119910119899119872

(119904119895) 119910119899119872

(120572 (119904119895))

ℎ

119872

sum

119894=minus119872

119886 (119904119895 119903119894) 119910119899119872

(119861 (119903119894))

1205931015840 (119903119894)

120578ℎ119894

(119909)

ℎ

119872

sum

119894=minus119872

ℎ (119904119895 119903119894) 119910119899119872

(120575 (119903119894))

1205931015840 (119903119894)

)

times (1205931015840(119904119895))minus1

120578ℎ119895

(119909)

(31)

4 Convergence of Method

In this section we present a theorem that shows a boundfor 119910(119909) minus 119910

119899(119909) with the real norm where 119910(119909) is the exact

solution of problem (4) and 119910119899(119909) is an estimation for 119910(119909) by

using Sinc function in interpolaion and quadrature formulaThe result is shown as follows

Theorem 7 Under the assumptions (1198711)ndash(1198715) iterative

approximation approach (31) is convergent to exact solution if1199101(119909) is closed enough to the it and

1198732le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

(32)

where 1198731

= sup120573(119909) 119909 isin 119868 1198732

= max120572119894(119909) 119894 = 1 2 3 4

119909 isin 119868 1198733

= sup119870(119909 119903) 119909 119903 isin 119868 and 1198734

= supℎ(119909 119903)

119909 119903 isin 119868

Proof For a fixed 119872 let

119910119899+1119872

(119909) = 1205781119909119899119872

(120591) + 1205782int

119887

0

120596 (119903 119910119899119872

(119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119910119899119872

(119904) 119910119899119872

(1205741(119904)) 119880119910

119899119872(119904)

119881119910119899119872

(119904)) 119889119904

119910 (119909) = 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119909 (119904) 119910 (1205741(119904)) 119880119910 (119904) 119881119910 (119904)) 119889119904

(33)

Now1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816

+ 1205782

100381610038161003816100381610038161003816100381610038161003816

int

119887

0

(120596 (119903 119910119899(119903)) minus 120596 (119903 119910 (119903))) 119889119903

100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

int

119909

0

[119892(119904 119910119899(119904) 119910119899(1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910119899(119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910119899(119863 (119903)) 119889119903)

minus 119891(119904 119910 (119904) 119910 (1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910 (119863 (119903)) 119889119903)]119889119904

100381610038161003816100381610038161003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816 + 12057821198871198731

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816

+ 1198871198732

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816 + 119887119873

2

1003816100381610038161003816119910119899 (1205741 (119909)) minus 119910 (1205741(119909))

1003816100381610038161003816

+ 11988711987321198733

1003816100381610038161003816119909119899 (119861 (119903)) minus 119910 (119861 (119903))1003816100381610038161003816

+ 11988711987321198724

1003816100381610038161003816119910119899 (119863 (119903)) minus 119910 (119863 (119903))1003816100381610038161003816

le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))

1003816100381610038161003816119910119899119872 (119909) minus 119910 (119909)1003816100381610038161003816

(34)

Then1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816 le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))119899

times10038161003816100381610038161199101119872 (119909) minus 119910 (119909)

1003816100381610038161003816

(35)

Now by assumption in theorem we have 1205781+12057821198871198731+1198871198732(2+

1198733+ 1198734) lt 1

Because 1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734) lt 1205781+ 12057821198871198731+

119887((1 minus (1205781+ 12057821198871198731))119887(2 + 119873

3+ 1198734))(2 + 119873

3+ 1198734) = 1 and

then lim119899rarr+infin

119910119899+1119872

(119909) = 119910(119909) 0 le 119909 le 119887

5 Illustrative Examples

In this section for showing efficiency of the iterative schemeand in order to show the facts of the exact solution wegive some examples below All routines have been written inMathematica 7 and a Dual-Core CPU 200GHz is used torun the programs Also about efficiency and accuracy of theproposed numerical method we present absolute errors fordifferent examples Therefore numerical results are shown infigures to illustrate the efficiency of this scheme

6 Abstract and Applied Analysis

5 10 15 20 25 30

0001

01

n

En60

10minus5

10minus7

10minus9

Figure 1 Absolute errors of Example 1 For different values ofiteration 119899

Example 1 Consider the following problem

119889119910

119889119909=

1

151199093[119909 minus 119910 (119909)] minus

1

5119909119910 (1205741(119909)) +

1

10011990921199102(1205741(119909))

+1

500119909[1199093minus int

(12)119909

0

8119909119903119910 (1199032) 119889119903]

5

minus1

6001199092[1199093minus int

(12)119909

0

8119909119903119910(1199032)119889119903]

6

+1

7001199092[1199092minus int

1

0

21199092119903119910 (1199032) 119889119903]

7

minus1

8001199093[1199092minus int

1

0

21199092119903119910(1199032)119889119903]

8

equiv 119866119910 (119909)

119909 isin 119868 = [0 1]

119910 (0) =1

4119910 (120581) +

1

4int

1

0

(2119903 + 119910 (119903)) 119889119903 +1

4

(36)

where 120581 isin (0 1] 1205741isin 119862(119868 119868) and 120574

1(119909) le 119909 on 119868

Now we want to show that assumptions (1198711)ndash(1198715) and

convergence criterion which was prepared in Theorem 7 aresatisfied for integral equation (4)

Noting that 1205742(119909) = (12)119909 119861(119909) = 119863(119909) = 119909

2 forall119909 isin 119868then (119871

1)ndash(1198715) is true For details see [6]

In addition we have 120573(119909) = 1 1205721(119909) = 0 120572

2(119909) =

(150)1199092 1205723(119909) = 120572

4(119909) = (1100)119909

17 119870(119909 119903) = 8119909119903ℎ(119909 119903) = 2119909

2119903 and 120578

1= 1205782

= 14 then 1198731

= 1 1198732

= 1501198733= 8 and 119873

4= 2 and then

1198732=

1

50le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

=1

24 (37)

02 04 06 08 10

090

092

094

096

t

x(t)

Figure 2 Approximate solution for Example 1

Therefore the iterative method (31) converges to the exactsolution of this equation Now based on Sinc quadraturescheme we can have some successive approximations forsolution of this example for 120581 = 12 Absolute error for 119899thapproximation and 119873 points quadrature method is definedby

119890119899119872

= max119909isin[01]

1003816100381610038161003816119910119899+1119872 (119909) minus 119910

119899119872(119909)

1003816100381610038161003816 119899 = 1 2 3

(38)

For different values of 119899 absolute errors are depicted inFigure 1 and approximate solution for 119899 = 30 is depicted inFigure 2

Example 2 In Example 1 of [19] authors considered theintegrodifferential equation

119889119910

119889119909= 1 minus

1

31199093+ int

1

0

11990931199102(119911) 119889119911

119910 (0) = 0

(39)

The exact solution is 119910(119909) = 119909 Maximum absolute error foreach iteration and different values of quadrature points aredepicted in Figures 3 and 4 By comparing these results withthe numerical results given in [19] in Table 2 efficiency andaccuracy of current approach are guaranteed

6 Conclusion

In this paper we apply a numerical approach by Sinc functionfor reaching the estimated solution of integrodifferentialequation with enteral boundary condition and with delayparameter To reach this aim we change this problem to afunctional enteral equation The Sinc estimation has expo-nential convergence rate such as119874(minus119888119890

11987212

) that this propertyis an advantage so we applied it to solve our problem byusing collocation method Finally some examples are solvedby this numericalmethod to show the efficiency and accuracyof Sinc estimation It is worthy to note that this method can

Abstract and Applied Analysis 7

2 4 6 8 10 12 14

01

n

En300

10minus4

10minus7

10minus10

Figure 3 Maximum absolute error related to each iteration inExample 2

100 150 200 250 300

0001

N

E15N

10minus5

10minus7

10minus9

10minus11

Figure 4 Maximum error related to different quadrature points inExample 2

be used for solving integrodifferential equations with integralboundary conditions with deviating arguments arising in allsciences such as chemistry physics and other fields of appliedmathematics

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authors thank the referee for hisher careful reading ofthe paper and useful suggestion

References

[1] R P Agarwal D OrsquoRegan and P J Y Wong Positive Solutionsof Differential Difference and Integral Equations Kluwer Aca-demic Publishers Dordrecht The Netherlands 1999

[2] T A Burton ldquoDifferential inequalities for integral and delaydifferential equationsrdquo in Comparison Methods and StabilityTheory vol 162 of Lecture Notes in Pure and Applied Mathemat-ics pp 35ndash56 Dekker New York NY USA 1994

[3] B Ahmad and B S Alghamdi ldquoApproximation of solutionsof the nonlinear Duffing equation involving both integral andnon-integral forcing terms with separated boundary condi-tionsrdquo Computer Physics Communications vol 179 no 6 pp409ndash416 2008

[4] B Ahmad ldquoOn the existence of p-periodic solutions forDuffing type integro-differential equations with p-LaplacianrdquoLobachevskii Journal ofMathematics vol 29 no 1 pp 1ndash4 2008

[5] Y K Chang and J J Nieto ldquoExistence of solutions for impulsiveneutral integro-differential inclusions with nonlocal initial con-ditions via fractional operatorsrdquo Numerical Functional Analysisand Optimization vol 30 no 3-4 pp 227ndash244 2009

[6] Z Luo and J J Nieto ldquoNew results for the periodic boundaryvalue problem for impulsive integro-differential equationsrdquoNonlinear Analysis Theory Methods amp Applications vol 70 no6 pp 2248ndash2260 2009

[7] T Jankowski ldquoDifferential equations with integral boundaryconditionsrdquo Journal of Computational andAppliedMathematicsvol 147 no 1 pp 1ndash8 2002

[8] G Wang G Song and L Zhang ldquoIntegral boundary valueproblems for first order integro-differential equations withdeviating argumentsrdquo Journal of Computational and AppliedMathematics vol 225 no 2 pp 602ndash611 2009

[9] K Maleknejad P Torabi and R Mollapourasl ldquoFixed pointmethod for solving nonlinear quadratic Volterra integral equa-tionsrdquo Computers amp Mathematics with Applications vol 62 no6 pp 2555ndash2566 2011

[10] K Maleknejad and M N Sahlan ldquoThe method of momentsfor solution of second kind Fredholm integral equations basedon B-spline waveletsrdquo International Journal of Computer Math-ematics vol 87 no 7 pp 1602ndash1616 2010

[11] K Maleknejad K Nouri and R Mollapourasl ldquoExistence ofsolutions for some nonlinear integral equationsrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no6 pp 2559ndash2564 2009

[12] W Volk ldquoThe iterated Galerkin method for linear integro-differential equationsrdquo Journal of Computational and AppliedMathematics vol 21 no 1 pp 63ndash74 1988

[13] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces vol 60 Marcel Dekker New York NY USA 1980

[14] A Mohsen and M El-Gamel ldquoA sinc-collocation method forthe linear Fredholm integro-differential equationsrdquo Journal ofApplied Mathematics and Physics vol 58 no 3 pp 380ndash3902007

[15] A Avudainayagam and C Vani ldquoWavelet-Galerkin method forintegro-differential equationsrdquoAppliedNumericalMathematicsvol 32 no 3 pp 247ndash254 2000

[16] F StengerNumerical Methods Based on Sinc and Analytic Func-tions vol 20 Springer New York NY USA 1993

[17] L Zhang ldquoBoundary value problem for first order impulsivefunctional integro-differential equationsrdquo Journal of Computa-tional and Applied Mathematics vol 235 no 8 pp 2442ndash24502011

[18] G Wang L Zhang and G Song ldquoExtremal solutions for thefirst order impulsive functional differential equations with

8 Abstract and Applied Analysis

upper and lower solutions in reversed orderrdquo Journal of Com-putational and AppliedMathematics vol 235 no 1 pp 325ndash3332010

[19] M Dehghan and R Salehi ldquoThe numerical solution of thenon-linear integro-differential equations based on the meshlessmethodrdquo Journal of Computational and Applied Mathematicsvol 236 no 9 pp 2367ndash2377 2012

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

The approximation of 119892(119909) by interpolation and quadratureformulas for int

119887

119886119892(119909)119889119909 is

119892 (119909) asymp

119872

sum

119895=minus119872

119892 (119909119895) 119878 (119895 ℎ) ∘ 120593 (119909)

int

119887

119886

119892 (119909) 119889119909 asymp ℎ

119872

sum

119895=minus119872

119892 (119909119895)

1205931015840 (119909119895)

(21)

Theorem 4 Consider that for a map 119908 = 120593minus1

(120585) the map119892(120593minus1

(120585)) satisfies

(1) 119892 isin 1198671(119863119889) for 119889 gt 0

(2) 119892 decays exponentially on the real line such that

1003816100381610038161003816119892 (119909)1003816100381610038161003816 le 120572 exp (minus120573 |119909|) forall119909 isin R 120572 120573 gt 0 (22)

with some 120572 120573 and 119889 Then one has

sup119886lt119909lt119887

10038161003816100381610038161003816100381610038161003816100381610038161003816

119892 (119909) minus

119872

sum

119895=minus119872

119892 (120593minus1

(119895ℎ)) 119878 (119895 ℎ) ∘ 120593 (119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

le 119862radic119872 exp(minusradic120587119889120573119872)

(23)

That there is some 119862 and ℎ is ℎ = radic120587119889120573119872

Proof See [16]

Definition 5 Let 119871120572(119860) be the set of all analytic functions 119892

for which there exists a constant 119862 such that

1003816100381610038161003816119892 (119911)1003816100381610038161003816 le 119862

|119890120593(119911)

|120572

(1 + |119890120593(119911)|)2120572

119911 isin 119860 0 lt 120572 le 1 (24)

Theorem6 Let1198921205931015840isin 119871120572(119860) with 0 lt 120572 le 1 and 0 lt 119889 le 120587

also let ℎ = radic120587119889120572119872 Then there exists a constant 1198621 which

is independent of 119872 such that

10038161003816100381610038161003816100381610038161003816100381610038161003816

int

119909119895

119886

119892 (119905) 119889119905 minus ℎ

119872

sum

119896=minus119872

120575(minus1)

119895119896

119892 (119909119895)

1205931015840 (119909119896)

10038161003816100381610038161003816100381610038161003816100381610038161003816

le 1198621119890minusradic120587119889120572119872

(25)

where

120575(minus1)

119895119896=

1

2+ int

119895minus119896

0

sin (120587119905)

120587119905119889119905 (26)

and 120593(119909) 119909119896are defined as above

Proof See [16]

32 Sinc-Qudrature Method In this section for solvingequation

119910 (119909) = 1205821119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892(119904 119910 (119904) 119910 (1205741(119904))

int

1205742(119904)

0

119896 (119904 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910 (119863 (119904)) 119889119903) 119889119904

(27)

we try to discrete integral equation by quadrature formula as

int

119887

119886

119892 (119904) 119889119904 = ℎ

119872

sum

119895=minus119872

119892 (119904119895)

1205931015840 (119904119895)

+ 119874(exp(minus2120587119889119872

log (2120587119889119872120573)))

int

119904

119886

119891 (119909) 119889119909 = ℎ

119872

sum

119895=minus119872

119892 (119909119895)

1205931015840 (119909119895)120578ℎ119895

(119904)

+ 119874(log119872

119872exp(minus

120587119889119872

log (120587119889119872120573)))

(28)

where

120578ℎ119895

(119904) =1

2+

1

120587119903119894(120587

119903 minus 119895ℎ

ℎ) 119903

119894= int

119909

0

sin (119905)

119905119889119905

(29)

with 119909119895

= 119904119895

= (119886 + 119887119890119895ℎ

)(1 + 119895ℎ) 119895 = minus119872 119872 and ℎ =

(1119872) log(120587119889119872120573) (see [16])Now by substituting quadrature formulas in the integral

equation (4) we have

119910119872

(119909) = 1205781119910 (120581) + 120578

2ℎ

119872

sum

119895=minus119872

120596 (119904119895 119910 (119904119895))

1205931015840 (119904119895)

+ 119886

+ ℎ

119872

sum

119895=minus119872

119892(119904119895 119910 (119904119895) 119910 (120572 (119904

119895))

ℎ

119872

sum

119894=minus119872

119870(119904119895 119903119894) 119910 (119861 (119903

119894))

1205931015840 (119903119894)

120578ℎ119894

(119905)

ℎ

119872

sum

119894=minus119872

ℎ (119904119895 119903119894) 119910 (119863 (119903

119894))

1205931015840 (119903119894)

)

times (1205931015840(119904119895))minus1

120578ℎ119895

(119909)

(30)

Abstract and Applied Analysis 5

Now for 119899 = 1 2 3 let

1199101119872

(119909) = 1205781119910 (120581)

119910119899+1119872

(119909)

= 1205781119910119899119872

(120581) + 1205782ℎ

119872

sum

119895=minus119872

120596 (119904119895 119910119899119872

(119904119895))

1205931015840 (119904119895)

+ 119886

+ ℎ

119872

sum

119895=minus119872

119892(119904119895 119910119899119872

(119904119895) 119910119899119872

(120572 (119904119895))

ℎ

119872

sum

119894=minus119872

119886 (119904119895 119903119894) 119910119899119872

(119861 (119903119894))

1205931015840 (119903119894)

120578ℎ119894

(119909)

ℎ

119872

sum

119894=minus119872

ℎ (119904119895 119903119894) 119910119899119872

(120575 (119903119894))

1205931015840 (119903119894)

)

times (1205931015840(119904119895))minus1

120578ℎ119895

(119909)

(31)

4 Convergence of Method

In this section we present a theorem that shows a boundfor 119910(119909) minus 119910

119899(119909) with the real norm where 119910(119909) is the exact

solution of problem (4) and 119910119899(119909) is an estimation for 119910(119909) by

using Sinc function in interpolaion and quadrature formulaThe result is shown as follows

Theorem 7 Under the assumptions (1198711)ndash(1198715) iterative

approximation approach (31) is convergent to exact solution if1199101(119909) is closed enough to the it and

1198732le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

(32)

where 1198731

= sup120573(119909) 119909 isin 119868 1198732

= max120572119894(119909) 119894 = 1 2 3 4

119909 isin 119868 1198733

= sup119870(119909 119903) 119909 119903 isin 119868 and 1198734

= supℎ(119909 119903)

119909 119903 isin 119868

Proof For a fixed 119872 let

119910119899+1119872

(119909) = 1205781119909119899119872

(120591) + 1205782int

119887

0

120596 (119903 119910119899119872

(119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119910119899119872

(119904) 119910119899119872

(1205741(119904)) 119880119910

119899119872(119904)

119881119910119899119872

(119904)) 119889119904

119910 (119909) = 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119909 (119904) 119910 (1205741(119904)) 119880119910 (119904) 119881119910 (119904)) 119889119904

(33)

Now1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816

+ 1205782

100381610038161003816100381610038161003816100381610038161003816

int

119887

0

(120596 (119903 119910119899(119903)) minus 120596 (119903 119910 (119903))) 119889119903

100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

int

119909

0

[119892(119904 119910119899(119904) 119910119899(1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910119899(119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910119899(119863 (119903)) 119889119903)

minus 119891(119904 119910 (119904) 119910 (1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910 (119863 (119903)) 119889119903)]119889119904

100381610038161003816100381610038161003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816 + 12057821198871198731

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816

+ 1198871198732

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816 + 119887119873

2

1003816100381610038161003816119910119899 (1205741 (119909)) minus 119910 (1205741(119909))

1003816100381610038161003816

+ 11988711987321198733

1003816100381610038161003816119909119899 (119861 (119903)) minus 119910 (119861 (119903))1003816100381610038161003816

+ 11988711987321198724

1003816100381610038161003816119910119899 (119863 (119903)) minus 119910 (119863 (119903))1003816100381610038161003816

le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))

1003816100381610038161003816119910119899119872 (119909) minus 119910 (119909)1003816100381610038161003816

(34)

Then1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816 le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))119899

times10038161003816100381610038161199101119872 (119909) minus 119910 (119909)

1003816100381610038161003816

(35)

Now by assumption in theorem we have 1205781+12057821198871198731+1198871198732(2+

1198733+ 1198734) lt 1

Because 1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734) lt 1205781+ 12057821198871198731+

119887((1 minus (1205781+ 12057821198871198731))119887(2 + 119873

3+ 1198734))(2 + 119873

3+ 1198734) = 1 and

then lim119899rarr+infin

119910119899+1119872

(119909) = 119910(119909) 0 le 119909 le 119887

5 Illustrative Examples

In this section for showing efficiency of the iterative schemeand in order to show the facts of the exact solution wegive some examples below All routines have been written inMathematica 7 and a Dual-Core CPU 200GHz is used torun the programs Also about efficiency and accuracy of theproposed numerical method we present absolute errors fordifferent examples Therefore numerical results are shown infigures to illustrate the efficiency of this scheme

6 Abstract and Applied Analysis

5 10 15 20 25 30

0001

01

n

En60

10minus5

10minus7

10minus9

Figure 1 Absolute errors of Example 1 For different values ofiteration 119899

Example 1 Consider the following problem

119889119910

119889119909=

1

151199093[119909 minus 119910 (119909)] minus

1

5119909119910 (1205741(119909)) +

1

10011990921199102(1205741(119909))

+1

500119909[1199093minus int

(12)119909

0

8119909119903119910 (1199032) 119889119903]

5

minus1

6001199092[1199093minus int

(12)119909

0

8119909119903119910(1199032)119889119903]

6

+1

7001199092[1199092minus int

1

0

21199092119903119910 (1199032) 119889119903]

7

minus1

8001199093[1199092minus int

1

0

21199092119903119910(1199032)119889119903]

8

equiv 119866119910 (119909)

119909 isin 119868 = [0 1]

119910 (0) =1

4119910 (120581) +

1

4int

1

0

(2119903 + 119910 (119903)) 119889119903 +1

4

(36)

where 120581 isin (0 1] 1205741isin 119862(119868 119868) and 120574

1(119909) le 119909 on 119868

Now we want to show that assumptions (1198711)ndash(1198715) and

convergence criterion which was prepared in Theorem 7 aresatisfied for integral equation (4)

Noting that 1205742(119909) = (12)119909 119861(119909) = 119863(119909) = 119909

2 forall119909 isin 119868then (119871

1)ndash(1198715) is true For details see [6]

In addition we have 120573(119909) = 1 1205721(119909) = 0 120572

2(119909) =

(150)1199092 1205723(119909) = 120572

4(119909) = (1100)119909

17 119870(119909 119903) = 8119909119903ℎ(119909 119903) = 2119909

2119903 and 120578

1= 1205782

= 14 then 1198731

= 1 1198732

= 1501198733= 8 and 119873

4= 2 and then

1198732=

1

50le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

=1

24 (37)

02 04 06 08 10

090

092

094

096

t

x(t)

Figure 2 Approximate solution for Example 1

Therefore the iterative method (31) converges to the exactsolution of this equation Now based on Sinc quadraturescheme we can have some successive approximations forsolution of this example for 120581 = 12 Absolute error for 119899thapproximation and 119873 points quadrature method is definedby

119890119899119872

= max119909isin[01]

1003816100381610038161003816119910119899+1119872 (119909) minus 119910

119899119872(119909)

1003816100381610038161003816 119899 = 1 2 3

(38)

For different values of 119899 absolute errors are depicted inFigure 1 and approximate solution for 119899 = 30 is depicted inFigure 2

Example 2 In Example 1 of [19] authors considered theintegrodifferential equation

119889119910

119889119909= 1 minus

1

31199093+ int

1

0

11990931199102(119911) 119889119911

119910 (0) = 0

(39)

The exact solution is 119910(119909) = 119909 Maximum absolute error foreach iteration and different values of quadrature points aredepicted in Figures 3 and 4 By comparing these results withthe numerical results given in [19] in Table 2 efficiency andaccuracy of current approach are guaranteed

6 Conclusion

In this paper we apply a numerical approach by Sinc functionfor reaching the estimated solution of integrodifferentialequation with enteral boundary condition and with delayparameter To reach this aim we change this problem to afunctional enteral equation The Sinc estimation has expo-nential convergence rate such as119874(minus119888119890

11987212

) that this propertyis an advantage so we applied it to solve our problem byusing collocation method Finally some examples are solvedby this numericalmethod to show the efficiency and accuracyof Sinc estimation It is worthy to note that this method can

Abstract and Applied Analysis 7

2 4 6 8 10 12 14

01

n

En300

10minus4

10minus7

10minus10

Figure 3 Maximum absolute error related to each iteration inExample 2

100 150 200 250 300

0001

N

E15N

10minus5

10minus7

10minus9

10minus11

Figure 4 Maximum error related to different quadrature points inExample 2

be used for solving integrodifferential equations with integralboundary conditions with deviating arguments arising in allsciences such as chemistry physics and other fields of appliedmathematics

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authors thank the referee for hisher careful reading ofthe paper and useful suggestion

References

[1] R P Agarwal D OrsquoRegan and P J Y Wong Positive Solutionsof Differential Difference and Integral Equations Kluwer Aca-demic Publishers Dordrecht The Netherlands 1999

[2] T A Burton ldquoDifferential inequalities for integral and delaydifferential equationsrdquo in Comparison Methods and StabilityTheory vol 162 of Lecture Notes in Pure and Applied Mathemat-ics pp 35ndash56 Dekker New York NY USA 1994

[3] B Ahmad and B S Alghamdi ldquoApproximation of solutionsof the nonlinear Duffing equation involving both integral andnon-integral forcing terms with separated boundary condi-tionsrdquo Computer Physics Communications vol 179 no 6 pp409ndash416 2008

[4] B Ahmad ldquoOn the existence of p-periodic solutions forDuffing type integro-differential equations with p-LaplacianrdquoLobachevskii Journal ofMathematics vol 29 no 1 pp 1ndash4 2008

[5] Y K Chang and J J Nieto ldquoExistence of solutions for impulsiveneutral integro-differential inclusions with nonlocal initial con-ditions via fractional operatorsrdquo Numerical Functional Analysisand Optimization vol 30 no 3-4 pp 227ndash244 2009

[6] Z Luo and J J Nieto ldquoNew results for the periodic boundaryvalue problem for impulsive integro-differential equationsrdquoNonlinear Analysis Theory Methods amp Applications vol 70 no6 pp 2248ndash2260 2009

[7] T Jankowski ldquoDifferential equations with integral boundaryconditionsrdquo Journal of Computational andAppliedMathematicsvol 147 no 1 pp 1ndash8 2002

[8] G Wang G Song and L Zhang ldquoIntegral boundary valueproblems for first order integro-differential equations withdeviating argumentsrdquo Journal of Computational and AppliedMathematics vol 225 no 2 pp 602ndash611 2009

[9] K Maleknejad P Torabi and R Mollapourasl ldquoFixed pointmethod for solving nonlinear quadratic Volterra integral equa-tionsrdquo Computers amp Mathematics with Applications vol 62 no6 pp 2555ndash2566 2011

[10] K Maleknejad and M N Sahlan ldquoThe method of momentsfor solution of second kind Fredholm integral equations basedon B-spline waveletsrdquo International Journal of Computer Math-ematics vol 87 no 7 pp 1602ndash1616 2010

[11] K Maleknejad K Nouri and R Mollapourasl ldquoExistence ofsolutions for some nonlinear integral equationsrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no6 pp 2559ndash2564 2009

[12] W Volk ldquoThe iterated Galerkin method for linear integro-differential equationsrdquo Journal of Computational and AppliedMathematics vol 21 no 1 pp 63ndash74 1988

[13] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces vol 60 Marcel Dekker New York NY USA 1980

[14] A Mohsen and M El-Gamel ldquoA sinc-collocation method forthe linear Fredholm integro-differential equationsrdquo Journal ofApplied Mathematics and Physics vol 58 no 3 pp 380ndash3902007

[15] A Avudainayagam and C Vani ldquoWavelet-Galerkin method forintegro-differential equationsrdquoAppliedNumericalMathematicsvol 32 no 3 pp 247ndash254 2000

[16] F StengerNumerical Methods Based on Sinc and Analytic Func-tions vol 20 Springer New York NY USA 1993

[17] L Zhang ldquoBoundary value problem for first order impulsivefunctional integro-differential equationsrdquo Journal of Computa-tional and Applied Mathematics vol 235 no 8 pp 2442ndash24502011

[18] G Wang L Zhang and G Song ldquoExtremal solutions for thefirst order impulsive functional differential equations with

8 Abstract and Applied Analysis

upper and lower solutions in reversed orderrdquo Journal of Com-putational and AppliedMathematics vol 235 no 1 pp 325ndash3332010

[19] M Dehghan and R Salehi ldquoThe numerical solution of thenon-linear integro-differential equations based on the meshlessmethodrdquo Journal of Computational and Applied Mathematicsvol 236 no 9 pp 2367ndash2377 2012

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

Now for 119899 = 1 2 3 let

1199101119872

(119909) = 1205781119910 (120581)

119910119899+1119872

(119909)

= 1205781119910119899119872

(120581) + 1205782ℎ

119872

sum

119895=minus119872

120596 (119904119895 119910119899119872

(119904119895))

1205931015840 (119904119895)

+ 119886

+ ℎ

119872

sum

119895=minus119872

119892(119904119895 119910119899119872

(119904119895) 119910119899119872

(120572 (119904119895))

ℎ

119872

sum

119894=minus119872

119886 (119904119895 119903119894) 119910119899119872

(119861 (119903119894))

1205931015840 (119903119894)

120578ℎ119894

(119909)

ℎ

119872

sum

119894=minus119872

ℎ (119904119895 119903119894) 119910119899119872

(120575 (119903119894))

1205931015840 (119903119894)

)

times (1205931015840(119904119895))minus1

120578ℎ119895

(119909)

(31)

4 Convergence of Method

In this section we present a theorem that shows a boundfor 119910(119909) minus 119910

119899(119909) with the real norm where 119910(119909) is the exact

solution of problem (4) and 119910119899(119909) is an estimation for 119910(119909) by

using Sinc function in interpolaion and quadrature formulaThe result is shown as follows

Theorem 7 Under the assumptions (1198711)ndash(1198715) iterative

approximation approach (31) is convergent to exact solution if1199101(119909) is closed enough to the it and

1198732le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

(32)

where 1198731

= sup120573(119909) 119909 isin 119868 1198732

= max120572119894(119909) 119894 = 1 2 3 4

119909 isin 119868 1198733

= sup119870(119909 119903) 119909 119903 isin 119868 and 1198734

= supℎ(119909 119903)

119909 119903 isin 119868

Proof For a fixed 119872 let

119910119899+1119872

(119909) = 1205781119909119899119872

(120591) + 1205782int

119887

0

120596 (119903 119910119899119872

(119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119910119899119872

(119904) 119910119899119872

(1205741(119904)) 119880119910

119899119872(119904)

119881119910119899119872

(119904)) 119889119904

119910 (119909) = 1205781119910 (120581) + 120578

2int

119887

0

120596 (119903 119910 (119903)) 119889119903 + 119886

+ int

119909

0

119892 (119904 119909 (119904) 119910 (1205741(119904)) 119880119910 (119904) 119881119910 (119904)) 119889119904

(33)

Now1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816

+ 1205782

100381610038161003816100381610038161003816100381610038161003816

int

119887

0

(120596 (119903 119910119899(119903)) minus 120596 (119903 119910 (119903))) 119889119903

100381610038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

int

119909

0

[119892(119904 119910119899(119904) 119910119899(1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910119899(119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910119899(119863 (119903)) 119889119903)

minus 119891(119904 119910 (119904) 119910 (1205741(119904))

int

1205742(119904)

0

119870 (119904 119903) 119910 (119861 (119903)) 119889119903

int

119887

0

ℎ (119904 119903) 119910 (119863 (119903)) 119889119903)]119889119904

100381610038161003816100381610038161003816100381610038161003816

le 1205781

1003816100381610038161003816119910119899 (120581) minus 119910 (120581)1003816100381610038161003816 + 12057821198871198731

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816

+ 1198871198732

1003816100381610038161003816119910119899 (119909) minus 119910 (119909)1003816100381610038161003816 + 119887119873

2

1003816100381610038161003816119910119899 (1205741 (119909)) minus 119910 (1205741(119909))

1003816100381610038161003816

+ 11988711987321198733

1003816100381610038161003816119909119899 (119861 (119903)) minus 119910 (119861 (119903))1003816100381610038161003816

+ 11988711987321198724

1003816100381610038161003816119910119899 (119863 (119903)) minus 119910 (119863 (119903))1003816100381610038161003816

le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))

1003816100381610038161003816119910119899119872 (119909) minus 119910 (119909)1003816100381610038161003816

(34)

Then1003816100381610038161003816119910119899+1119872 (119909) minus 119910 (119909)

1003816100381610038161003816 le (1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734))119899

times10038161003816100381610038161199101119872 (119909) minus 119910 (119909)

1003816100381610038161003816

(35)

Now by assumption in theorem we have 1205781+12057821198871198731+1198871198732(2+

1198733+ 1198734) lt 1

Because 1205781+ 12057821198871198731+ 1198871198732(2 + 119873

3+ 1198734) lt 1205781+ 12057821198871198731+

119887((1 minus (1205781+ 12057821198871198731))119887(2 + 119873

3+ 1198734))(2 + 119873

3+ 1198734) = 1 and

then lim119899rarr+infin

119910119899+1119872

(119909) = 119910(119909) 0 le 119909 le 119887

5 Illustrative Examples

In this section for showing efficiency of the iterative schemeand in order to show the facts of the exact solution wegive some examples below All routines have been written inMathematica 7 and a Dual-Core CPU 200GHz is used torun the programs Also about efficiency and accuracy of theproposed numerical method we present absolute errors fordifferent examples Therefore numerical results are shown infigures to illustrate the efficiency of this scheme

6 Abstract and Applied Analysis

5 10 15 20 25 30

0001

01

n

En60

10minus5

10minus7

10minus9

Figure 1 Absolute errors of Example 1 For different values ofiteration 119899

Example 1 Consider the following problem

119889119910

119889119909=

1

151199093[119909 minus 119910 (119909)] minus

1

5119909119910 (1205741(119909)) +

1

10011990921199102(1205741(119909))

+1

500119909[1199093minus int

(12)119909

0

8119909119903119910 (1199032) 119889119903]

5

minus1

6001199092[1199093minus int

(12)119909

0

8119909119903119910(1199032)119889119903]

6

+1

7001199092[1199092minus int

1

0

21199092119903119910 (1199032) 119889119903]

7

minus1

8001199093[1199092minus int

1

0

21199092119903119910(1199032)119889119903]

8

equiv 119866119910 (119909)

119909 isin 119868 = [0 1]

119910 (0) =1

4119910 (120581) +

1

4int

1

0

(2119903 + 119910 (119903)) 119889119903 +1

4

(36)

where 120581 isin (0 1] 1205741isin 119862(119868 119868) and 120574

1(119909) le 119909 on 119868

Now we want to show that assumptions (1198711)ndash(1198715) and

convergence criterion which was prepared in Theorem 7 aresatisfied for integral equation (4)

Noting that 1205742(119909) = (12)119909 119861(119909) = 119863(119909) = 119909

2 forall119909 isin 119868then (119871

1)ndash(1198715) is true For details see [6]

In addition we have 120573(119909) = 1 1205721(119909) = 0 120572

2(119909) =

(150)1199092 1205723(119909) = 120572

4(119909) = (1100)119909

17 119870(119909 119903) = 8119909119903ℎ(119909 119903) = 2119909

2119903 and 120578

1= 1205782

= 14 then 1198731

= 1 1198732

= 1501198733= 8 and 119873

4= 2 and then

1198732=

1

50le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

=1

24 (37)

02 04 06 08 10

090

092

094

096

t

x(t)

Figure 2 Approximate solution for Example 1

Therefore the iterative method (31) converges to the exactsolution of this equation Now based on Sinc quadraturescheme we can have some successive approximations forsolution of this example for 120581 = 12 Absolute error for 119899thapproximation and 119873 points quadrature method is definedby

119890119899119872

= max119909isin[01]

1003816100381610038161003816119910119899+1119872 (119909) minus 119910

119899119872(119909)

1003816100381610038161003816 119899 = 1 2 3

(38)

For different values of 119899 absolute errors are depicted inFigure 1 and approximate solution for 119899 = 30 is depicted inFigure 2

Example 2 In Example 1 of [19] authors considered theintegrodifferential equation

119889119910

119889119909= 1 minus

1

31199093+ int

1

0

11990931199102(119911) 119889119911

119910 (0) = 0

(39)

The exact solution is 119910(119909) = 119909 Maximum absolute error foreach iteration and different values of quadrature points aredepicted in Figures 3 and 4 By comparing these results withthe numerical results given in [19] in Table 2 efficiency andaccuracy of current approach are guaranteed

6 Conclusion

In this paper we apply a numerical approach by Sinc functionfor reaching the estimated solution of integrodifferentialequation with enteral boundary condition and with delayparameter To reach this aim we change this problem to afunctional enteral equation The Sinc estimation has expo-nential convergence rate such as119874(minus119888119890

11987212

) that this propertyis an advantage so we applied it to solve our problem byusing collocation method Finally some examples are solvedby this numericalmethod to show the efficiency and accuracyof Sinc estimation It is worthy to note that this method can

Abstract and Applied Analysis 7

2 4 6 8 10 12 14

01

n

En300

10minus4

10minus7

10minus10

Figure 3 Maximum absolute error related to each iteration inExample 2

100 150 200 250 300

0001

N

E15N

10minus5

10minus7

10minus9

10minus11

Figure 4 Maximum error related to different quadrature points inExample 2

be used for solving integrodifferential equations with integralboundary conditions with deviating arguments arising in allsciences such as chemistry physics and other fields of appliedmathematics

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authors thank the referee for hisher careful reading ofthe paper and useful suggestion

References

[1] R P Agarwal D OrsquoRegan and P J Y Wong Positive Solutionsof Differential Difference and Integral Equations Kluwer Aca-demic Publishers Dordrecht The Netherlands 1999

[2] T A Burton ldquoDifferential inequalities for integral and delaydifferential equationsrdquo in Comparison Methods and StabilityTheory vol 162 of Lecture Notes in Pure and Applied Mathemat-ics pp 35ndash56 Dekker New York NY USA 1994

[3] B Ahmad and B S Alghamdi ldquoApproximation of solutionsof the nonlinear Duffing equation involving both integral andnon-integral forcing terms with separated boundary condi-tionsrdquo Computer Physics Communications vol 179 no 6 pp409ndash416 2008

[4] B Ahmad ldquoOn the existence of p-periodic solutions forDuffing type integro-differential equations with p-LaplacianrdquoLobachevskii Journal ofMathematics vol 29 no 1 pp 1ndash4 2008

[5] Y K Chang and J J Nieto ldquoExistence of solutions for impulsiveneutral integro-differential inclusions with nonlocal initial con-ditions via fractional operatorsrdquo Numerical Functional Analysisand Optimization vol 30 no 3-4 pp 227ndash244 2009

[6] Z Luo and J J Nieto ldquoNew results for the periodic boundaryvalue problem for impulsive integro-differential equationsrdquoNonlinear Analysis Theory Methods amp Applications vol 70 no6 pp 2248ndash2260 2009

[7] T Jankowski ldquoDifferential equations with integral boundaryconditionsrdquo Journal of Computational andAppliedMathematicsvol 147 no 1 pp 1ndash8 2002

[8] G Wang G Song and L Zhang ldquoIntegral boundary valueproblems for first order integro-differential equations withdeviating argumentsrdquo Journal of Computational and AppliedMathematics vol 225 no 2 pp 602ndash611 2009

[9] K Maleknejad P Torabi and R Mollapourasl ldquoFixed pointmethod for solving nonlinear quadratic Volterra integral equa-tionsrdquo Computers amp Mathematics with Applications vol 62 no6 pp 2555ndash2566 2011

[10] K Maleknejad and M N Sahlan ldquoThe method of momentsfor solution of second kind Fredholm integral equations basedon B-spline waveletsrdquo International Journal of Computer Math-ematics vol 87 no 7 pp 1602ndash1616 2010

[11] K Maleknejad K Nouri and R Mollapourasl ldquoExistence ofsolutions for some nonlinear integral equationsrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no6 pp 2559ndash2564 2009

[12] W Volk ldquoThe iterated Galerkin method for linear integro-differential equationsrdquo Journal of Computational and AppliedMathematics vol 21 no 1 pp 63ndash74 1988

[13] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces vol 60 Marcel Dekker New York NY USA 1980

[14] A Mohsen and M El-Gamel ldquoA sinc-collocation method forthe linear Fredholm integro-differential equationsrdquo Journal ofApplied Mathematics and Physics vol 58 no 3 pp 380ndash3902007

[15] A Avudainayagam and C Vani ldquoWavelet-Galerkin method forintegro-differential equationsrdquoAppliedNumericalMathematicsvol 32 no 3 pp 247ndash254 2000

[16] F StengerNumerical Methods Based on Sinc and Analytic Func-tions vol 20 Springer New York NY USA 1993

[17] L Zhang ldquoBoundary value problem for first order impulsivefunctional integro-differential equationsrdquo Journal of Computa-tional and Applied Mathematics vol 235 no 8 pp 2442ndash24502011

[18] G Wang L Zhang and G Song ldquoExtremal solutions for thefirst order impulsive functional differential equations with

8 Abstract and Applied Analysis

upper and lower solutions in reversed orderrdquo Journal of Com-putational and AppliedMathematics vol 235 no 1 pp 325ndash3332010

[19] M Dehghan and R Salehi ldquoThe numerical solution of thenon-linear integro-differential equations based on the meshlessmethodrdquo Journal of Computational and Applied Mathematicsvol 236 no 9 pp 2367ndash2377 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

5 10 15 20 25 30

0001

01

n

En60

10minus5

10minus7

10minus9

Figure 1 Absolute errors of Example 1 For different values ofiteration 119899

Example 1 Consider the following problem

119889119910

119889119909=

1

151199093[119909 minus 119910 (119909)] minus

1

5119909119910 (1205741(119909)) +

1

10011990921199102(1205741(119909))

+1

500119909[1199093minus int

(12)119909

0

8119909119903119910 (1199032) 119889119903]

5

minus1

6001199092[1199093minus int

(12)119909

0

8119909119903119910(1199032)119889119903]

6

+1

7001199092[1199092minus int

1

0

21199092119903119910 (1199032) 119889119903]

7

minus1

8001199093[1199092minus int

1

0

21199092119903119910(1199032)119889119903]

8

equiv 119866119910 (119909)

119909 isin 119868 = [0 1]

119910 (0) =1

4119910 (120581) +

1

4int

1

0

(2119903 + 119910 (119903)) 119889119903 +1

4

(36)

where 120581 isin (0 1] 1205741isin 119862(119868 119868) and 120574

1(119909) le 119909 on 119868

Now we want to show that assumptions (1198711)ndash(1198715) and

convergence criterion which was prepared in Theorem 7 aresatisfied for integral equation (4)

Noting that 1205742(119909) = (12)119909 119861(119909) = 119863(119909) = 119909

2 forall119909 isin 119868then (119871

1)ndash(1198715) is true For details see [6]

In addition we have 120573(119909) = 1 1205721(119909) = 0 120572

2(119909) =

(150)1199092 1205723(119909) = 120572

4(119909) = (1100)119909

17 119870(119909 119903) = 8119909119903ℎ(119909 119903) = 2119909

2119903 and 120578

1= 1205782

= 14 then 1198731

= 1 1198732

= 1501198733= 8 and 119873

4= 2 and then

1198732=

1

50le

1 minus (1205781+ 12057821198871198731)

119887 (2 + 1198733+ 1198734)

=1

24 (37)

02 04 06 08 10

090

092

094

096

t

x(t)

Figure 2 Approximate solution for Example 1

Therefore the iterative method (31) converges to the exactsolution of this equation Now based on Sinc quadraturescheme we can have some successive approximations forsolution of this example for 120581 = 12 Absolute error for 119899thapproximation and 119873 points quadrature method is definedby

119890119899119872

= max119909isin[01]

1003816100381610038161003816119910119899+1119872 (119909) minus 119910

119899119872(119909)

1003816100381610038161003816 119899 = 1 2 3

(38)

For different values of 119899 absolute errors are depicted inFigure 1 and approximate solution for 119899 = 30 is depicted inFigure 2

Example 2 In Example 1 of [19] authors considered theintegrodifferential equation

119889119910

119889119909= 1 minus

1

31199093+ int

1

0

11990931199102(119911) 119889119911

119910 (0) = 0

(39)

The exact solution is 119910(119909) = 119909 Maximum absolute error foreach iteration and different values of quadrature points aredepicted in Figures 3 and 4 By comparing these results withthe numerical results given in [19] in Table 2 efficiency andaccuracy of current approach are guaranteed

6 Conclusion

In this paper we apply a numerical approach by Sinc functionfor reaching the estimated solution of integrodifferentialequation with enteral boundary condition and with delayparameter To reach this aim we change this problem to afunctional enteral equation The Sinc estimation has expo-nential convergence rate such as119874(minus119888119890

11987212

) that this propertyis an advantage so we applied it to solve our problem byusing collocation method Finally some examples are solvedby this numericalmethod to show the efficiency and accuracyof Sinc estimation It is worthy to note that this method can

Abstract and Applied Analysis 7

2 4 6 8 10 12 14

01

n

En300

10minus4

10minus7

10minus10

Figure 3 Maximum absolute error related to each iteration inExample 2

100 150 200 250 300

0001

N

E15N

10minus5

10minus7

10minus9

10minus11

Figure 4 Maximum error related to different quadrature points inExample 2

be used for solving integrodifferential equations with integralboundary conditions with deviating arguments arising in allsciences such as chemistry physics and other fields of appliedmathematics

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authors thank the referee for hisher careful reading ofthe paper and useful suggestion

References

[1] R P Agarwal D OrsquoRegan and P J Y Wong Positive Solutionsof Differential Difference and Integral Equations Kluwer Aca-demic Publishers Dordrecht The Netherlands 1999

[2] T A Burton ldquoDifferential inequalities for integral and delaydifferential equationsrdquo in Comparison Methods and StabilityTheory vol 162 of Lecture Notes in Pure and Applied Mathemat-ics pp 35ndash56 Dekker New York NY USA 1994

[3] B Ahmad and B S Alghamdi ldquoApproximation of solutionsof the nonlinear Duffing equation involving both integral andnon-integral forcing terms with separated boundary condi-tionsrdquo Computer Physics Communications vol 179 no 6 pp409ndash416 2008

[4] B Ahmad ldquoOn the existence of p-periodic solutions forDuffing type integro-differential equations with p-LaplacianrdquoLobachevskii Journal ofMathematics vol 29 no 1 pp 1ndash4 2008

[5] Y K Chang and J J Nieto ldquoExistence of solutions for impulsiveneutral integro-differential inclusions with nonlocal initial con-ditions via fractional operatorsrdquo Numerical Functional Analysisand Optimization vol 30 no 3-4 pp 227ndash244 2009

[6] Z Luo and J J Nieto ldquoNew results for the periodic boundaryvalue problem for impulsive integro-differential equationsrdquoNonlinear Analysis Theory Methods amp Applications vol 70 no6 pp 2248ndash2260 2009

[7] T Jankowski ldquoDifferential equations with integral boundaryconditionsrdquo Journal of Computational andAppliedMathematicsvol 147 no 1 pp 1ndash8 2002

[8] G Wang G Song and L Zhang ldquoIntegral boundary valueproblems for first order integro-differential equations withdeviating argumentsrdquo Journal of Computational and AppliedMathematics vol 225 no 2 pp 602ndash611 2009

[9] K Maleknejad P Torabi and R Mollapourasl ldquoFixed pointmethod for solving nonlinear quadratic Volterra integral equa-tionsrdquo Computers amp Mathematics with Applications vol 62 no6 pp 2555ndash2566 2011

[10] K Maleknejad and M N Sahlan ldquoThe method of momentsfor solution of second kind Fredholm integral equations basedon B-spline waveletsrdquo International Journal of Computer Math-ematics vol 87 no 7 pp 1602ndash1616 2010

[11] K Maleknejad K Nouri and R Mollapourasl ldquoExistence ofsolutions for some nonlinear integral equationsrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no6 pp 2559ndash2564 2009

[12] W Volk ldquoThe iterated Galerkin method for linear integro-differential equationsrdquo Journal of Computational and AppliedMathematics vol 21 no 1 pp 63ndash74 1988

[13] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces vol 60 Marcel Dekker New York NY USA 1980

[14] A Mohsen and M El-Gamel ldquoA sinc-collocation method forthe linear Fredholm integro-differential equationsrdquo Journal ofApplied Mathematics and Physics vol 58 no 3 pp 380ndash3902007

[15] A Avudainayagam and C Vani ldquoWavelet-Galerkin method forintegro-differential equationsrdquoAppliedNumericalMathematicsvol 32 no 3 pp 247ndash254 2000

[16] F StengerNumerical Methods Based on Sinc and Analytic Func-tions vol 20 Springer New York NY USA 1993

[17] L Zhang ldquoBoundary value problem for first order impulsivefunctional integro-differential equationsrdquo Journal of Computa-tional and Applied Mathematics vol 235 no 8 pp 2442ndash24502011

[18] G Wang L Zhang and G Song ldquoExtremal solutions for thefirst order impulsive functional differential equations with

8 Abstract and Applied Analysis

upper and lower solutions in reversed orderrdquo Journal of Com-putational and AppliedMathematics vol 235 no 1 pp 325ndash3332010

[19] M Dehghan and R Salehi ldquoThe numerical solution of thenon-linear integro-differential equations based on the meshlessmethodrdquo Journal of Computational and Applied Mathematicsvol 236 no 9 pp 2367ndash2377 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

2 4 6 8 10 12 14

01

n

En300

10minus4

10minus7

10minus10

Figure 3 Maximum absolute error related to each iteration inExample 2

100 150 200 250 300

0001

N

E15N

10minus5

10minus7

10minus9

10minus11

Figure 4 Maximum error related to different quadrature points inExample 2

be used for solving integrodifferential equations with integralboundary conditions with deviating arguments arising in allsciences such as chemistry physics and other fields of appliedmathematics

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The authors thank the referee for hisher careful reading ofthe paper and useful suggestion

References

[1] R P Agarwal D OrsquoRegan and P J Y Wong Positive Solutionsof Differential Difference and Integral Equations Kluwer Aca-demic Publishers Dordrecht The Netherlands 1999

[2] T A Burton ldquoDifferential inequalities for integral and delaydifferential equationsrdquo in Comparison Methods and StabilityTheory vol 162 of Lecture Notes in Pure and Applied Mathemat-ics pp 35ndash56 Dekker New York NY USA 1994

[3] B Ahmad and B S Alghamdi ldquoApproximation of solutionsof the nonlinear Duffing equation involving both integral andnon-integral forcing terms with separated boundary condi-tionsrdquo Computer Physics Communications vol 179 no 6 pp409ndash416 2008

[4] B Ahmad ldquoOn the existence of p-periodic solutions forDuffing type integro-differential equations with p-LaplacianrdquoLobachevskii Journal ofMathematics vol 29 no 1 pp 1ndash4 2008

[5] Y K Chang and J J Nieto ldquoExistence of solutions for impulsiveneutral integro-differential inclusions with nonlocal initial con-ditions via fractional operatorsrdquo Numerical Functional Analysisand Optimization vol 30 no 3-4 pp 227ndash244 2009

[6] Z Luo and J J Nieto ldquoNew results for the periodic boundaryvalue problem for impulsive integro-differential equationsrdquoNonlinear Analysis Theory Methods amp Applications vol 70 no6 pp 2248ndash2260 2009

[7] T Jankowski ldquoDifferential equations with integral boundaryconditionsrdquo Journal of Computational andAppliedMathematicsvol 147 no 1 pp 1ndash8 2002

[8] G Wang G Song and L Zhang ldquoIntegral boundary valueproblems for first order integro-differential equations withdeviating argumentsrdquo Journal of Computational and AppliedMathematics vol 225 no 2 pp 602ndash611 2009

[9] K Maleknejad P Torabi and R Mollapourasl ldquoFixed pointmethod for solving nonlinear quadratic Volterra integral equa-tionsrdquo Computers amp Mathematics with Applications vol 62 no6 pp 2555ndash2566 2011

[10] K Maleknejad and M N Sahlan ldquoThe method of momentsfor solution of second kind Fredholm integral equations basedon B-spline waveletsrdquo International Journal of Computer Math-ematics vol 87 no 7 pp 1602ndash1616 2010

[11] K Maleknejad K Nouri and R Mollapourasl ldquoExistence ofsolutions for some nonlinear integral equationsrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 14 no6 pp 2559ndash2564 2009

[12] W Volk ldquoThe iterated Galerkin method for linear integro-differential equationsrdquo Journal of Computational and AppliedMathematics vol 21 no 1 pp 63ndash74 1988

[13] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces vol 60 Marcel Dekker New York NY USA 1980

[14] A Mohsen and M El-Gamel ldquoA sinc-collocation method forthe linear Fredholm integro-differential equationsrdquo Journal ofApplied Mathematics and Physics vol 58 no 3 pp 380ndash3902007

[15] A Avudainayagam and C Vani ldquoWavelet-Galerkin method forintegro-differential equationsrdquoAppliedNumericalMathematicsvol 32 no 3 pp 247ndash254 2000

[16] F StengerNumerical Methods Based on Sinc and Analytic Func-tions vol 20 Springer New York NY USA 1993

[17] L Zhang ldquoBoundary value problem for first order impulsivefunctional integro-differential equationsrdquo Journal of Computa-tional and Applied Mathematics vol 235 no 8 pp 2442ndash24502011

[18] G Wang L Zhang and G Song ldquoExtremal solutions for thefirst order impulsive functional differential equations with

8 Abstract and Applied Analysis

upper and lower solutions in reversed orderrdquo Journal of Com-putational and AppliedMathematics vol 235 no 1 pp 325ndash3332010

[19] M Dehghan and R Salehi ldquoThe numerical solution of thenon-linear integro-differential equations based on the meshlessmethodrdquo Journal of Computational and Applied Mathematicsvol 236 no 9 pp 2367ndash2377 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

upper and lower solutions in reversed orderrdquo Journal of Com-putational and AppliedMathematics vol 235 no 1 pp 325ndash3332010

[19] M Dehghan and R Salehi ldquoThe numerical solution of thenon-linear integro-differential equations based on the meshlessmethodrdquo Journal of Computational and Applied Mathematicsvol 236 no 9 pp 2367ndash2377 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of