Research Article Numerical Solution of Nonlinear Fredholm ...

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 674364 9 pageshttpdxdoiorg1011552013674364

Research ArticleNumerical Solution of Nonlinear Fredholm Integro-DifferentialEquations Using Spectral Homotopy Analysis Method

Z Pashazadeh Atabakan A Kazemi Nasab A KJlJccedilman and Zainidin K Eshkuvatov

Department of Mathematics University Putra Malaysia (UPM) 43400 Serdang Selangor Malaysia

Correspondence should be addressed to A Kılıcman akilicmanputraupmedumy

Received 25 February 2013 Revised 18 April 2013 Accepted 20 April 2013

Academic Editor Fazal M Mahomed

Copyright copy 2013 Z Pashazadeh Atabakan et alThis is an open access article distributed under the Creative CommonsAttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited

Spectral homotopy analysis method (SHAM) as a modification of homotopy analysis method (HAM) is applied to obtain solutionof high-order nonlinear Fredholm integro-differential problems The existence and uniqueness of the solution and convergence ofthe proposed method are proved Some examples are given to approve the efficiency and the accuracy of the proposed methodThe SHAM results show that the proposed approach is quite reasonable when compared to homotopy analysis method Lagrangeinterpolation solutions and exact solutions

1 Introduction

The integro-differential equations stem from the mathemat-ical modeling of many complex real-life problems Manyscientific phenomena have been formulated using integro-differential equations [1 2] Solving nonlinear integro-differential equation is much more difficult than linear oneanalytically So different types of numerical methods havebeen used to obtain an efficient approximation solution [3 4]In 1992 Liao [5] proposed the homotopy analysis method(HAM) concept in topology for solving nonlinear differentialequations Liao [6 7] found that the convergence of seriessolutions of nonlinear equations cannot be guaranteed by theearly HAM Further Liao [6] introduced a nonzero auxiliaryparameter to solve this limitation Unlike the special cases ofHAM such as Lyapunoversquos artificial small parameter method[8] Adomian decomposition method [9ndash12] and the 120575-expansion method [13] this method need not a small pertur-bation parameter In the HAM the perturbation techniques[14] need not be converted a nonlinear problem to infinitenumber of linear problems The homotopy analysis methodis applicable for solving problems having strong nonlinearity[15] even if they do not have any small or large parametersso it is more powerful than traditional perturbationmethods

The convergence region and the rate of approximationin series can been adjusted by this method Also it can

give us freedom to use different base function to approxi-mate a non linear problem The convergence of HAM forsolving Volterra-Fredholm integro-differential equations ispresented in [16]

In 2010 Motsa et al [17] suggested the so-called spectralhomotopy analysis method (SHAM) using the Chebyshevpseudospectral method to solve the linear high-order defor-mation equations Since the SHAM combines the HAMwith the numerical techniques it provides us larger freedomto choose auxiliary linear operators Thus one can choosemore complicated auxiliary linear operators in the frameof the SHAM In theory any continuous function in abounded interval can be best approximated using Chebyshevpolynomial So the SHAMprovides larger freedom to choosethe auxiliary linear operator 119871 and initial guess Further it iseasy to employ the optimal convergence-control parameter inthe frame of the SHAMThus the SHAM has great potentialto solve more complicated nonlinear problems in scienceand engineering although further modifications in theoryand more applications are needed Chebyshev polynomial isconsidered a kind of special function There are many otherspecial functions such as Hermite polynomial Legendrepolynomial Airy function Bessel function Riemann zetafunction and hypergeometric functions Since the HAMprovides us extremely large freedom to choose the auxiliarylinear operator 119871 and the initial guess it should be possible

2 Mathematical Problems in Engineering

to develop a generalized spectral HAM which can use aproper special function for a given nonlinear problem Thespectral homotopy analysis method has been used for solvingpartial and ordinary differential equations [18ndash20] Spectralhomotopy analysis method and its convergency for solvinga class of optimal control problems are presented in [21]Motsa et al [17ndash19] found that the spectral homotopy analysismethod is more efficient than the homotopy analysis methodas it does not depend on the rule of solution expressionand the rule of ergodicity This method is more flexiblethan homotopy analysis method since it allows for a widerrange of linear and nonlinear operators and one is notrestricted to use the method of higher-order differentialmapping for solving boundary value problems in boundeddomains unlike the homotopy analysis method The rangeof admissible ℎ values is much wider in spectral homotopyanalysis method than in homotopy analysis method Themain restriction of HAM in solving integral equations isto choose the best initial guess as the series solution isconvergent In SHAM the initial approximation is taken to bethe solution of the nonhomogeneous linear part of the givenequation In 2012 Pashazadeh Atabakan et al solved linearVolterra and Fredholm integro-differential equations usingspectral homotopy analysis method see [22]

In this paper we apply spectral homotopy analysismethod (SHAM) to solve higher-order nonlinear Fredholmtype of integro-differential equations Fredholm integro-differential equation is given by

2

sum

119895=0

119886119895(119909) 119910(119895)(119909) = 119891 (119909) + 120583int

1

minus1

119896 (119909 119905) [119910 (119905)]119903

119889119905

119910 (minus1) = 119910 (1) = 0

(1)

where120583 is constant value119891(119909) 119896(119909 119905) [119910(119905)]119903 and 119886119895(119909)119903 ge 1

are functions that have suitable derivatives on interval minus1 le119905 le 119909 le 1 and 119886

2(119909) = 0

The paper is organized in the following way Section 2includes a brief introduction in homotopy analysis methodSpectral homotopy analysis method for solving nonlinearFredholm integral equations is presented in Section 3 Theexistence and uniqueness of the solution and convergence ofthe proposed method are proved in Section 4 In Section 5numerical examples are presented In Section 6 concludingremarks are given

2 Homotopy Analysis Solution

In this section we give a brief introduction to HAM Weconsider the following differential equation in a general formas follows

119873[119910 (120578)] = 0 (2)

where 119873 is nonlinear operator 120578 denotes independent vari-ables and 119910(120578) is an unknown function respectively Forsimplicity we disregard all initial and all boundary conditions

which can be dealt in similar way The so-called zero-orderdeformation equation was constructed by Liao as follows

(1 minus 119901) 119871 [120595 (120578 119901) minus 1199100(120578)] = 119901ℎ119867 (120578) (119873 [120595 (120578 119901)])

(3)

where 119901 isin [0 1] is the embedding parameter ℎ is a nonzeroconvergence-parameter 119867(120578) is an auxiliary function 119910

0(120578)

is called an initial guess of 119910(120578) and 120595(120578 119901) is an unknownfunction In addition 119871 is an auxiliary linear operator and119873is nonlinear operator as follows

119871 (120595 (119909 119901)) = 119886119896(119909)

1205972120595 (119909 119901)

1205971199092(4)

with the property 119871(sum2119895=0119888119895119905119895) = 0 where 119888

119895 are constants

and

119873[120595 (119909 119901)] =

2

sum

119895=0

119886119895(119909)

120597119895120595 (119909 119901)

120597119909119895minus 119891 (119909)

minus 120583int

1

minus1

119896 (119909 119905) 120595119903(119905) 119889119905

(5)

is a nonlinear operator Obviously when 119901 = 0 and 119901 = 1it holds 120595(120578 0) = 119910

0(120578) and 120595(120578 1) = 119910(120578) In this way as

119901 increase from 0 to 1 120595(120578 119901) alter from initial guess 1199100(120578)

to the solution 119910(120578) and 120595(120578 119901) is expanded in Taylor serieswith respect to 119901 as follows

120595 (120578 119901) = 1199100(120578) +

+infin

sum

119898=1

119910119898(120578) 119901119898 (6)

where

119910119898(120578) = 119863

119898[120595 (120578 119901)]

119863119898120595 =

1

119898

120597119898120595

120597119901119898

10038161003816100381610038161003816100381610038161003816119901=0

(7)

The series (6) converges at 119901 = 1 if the auxiliary linearoperator the initial guess the convergence-parameter andthe auxiliary function are properly selected as follows

120595 (120578) = 1199100(120578) +

+infin

sum

119898=1

119910119898(120578) (8)

The admissible and valid values of the convergence-parameter ℎ are found from the horizontal portion of theℎ-curves Liao proved that 119910(120578) is one of the solutions oforiginal nonlinear equation As119867(120578) = 1 so (3) becomes

(1 minus 119901) 119871 [120595 (120578 119901) minus 1199100(120578)] = 119901ℎ (119873 [120595 (120578 119901)]) (9)

Define the vector 119910119898= 1199100(120578) 1199101(120578) 119910

119898(120578) Operating

on both side of (9) with119863119898 we have the so called119898th-order

deformation equation as follows

119871 [119910119898(120578) minus 120594

119898119910119898minus1

(120578)] = ℎ119867 (120578) 119877119898(119910119898minus1

(120578)) (10)

Mathematical Problems in Engineering 3

where

119877119898(119910119898minus1) =

1

(119898 minus 1)

120597119898minus1119873[120595 (120578 119901)]

120597119901119898minus1

100381610038161003816100381610038161003816100381610038161003816119901=0

120594119898= 0 119898 le 1

1 otherwise

(11)

119910119898(120578) for 119898 ge 0 that is governed by the linear equation

(10) can be solved by symbolic computation software such asMAPLE MATLAB and similar CAS

3 Spectral-Homotopy Analysis Solution

Consider the non linear Fredholm integro-differential equa-tion2

sum

119895=0

119886119895(119909) 119910(119895)(119909) = 119891 (119909) + 120583int

1

minus1

119896 (119909 119905) [119910 (119905)]119903

119889119905

119910 (minus1) = 119910 (1) = 0

(12)

We begin by defining the following linear operator

119871 (120595 (119909 119901)) =

2

sum

119895=0

119886119895(119909)

120597119895120595 (119909 119901)

120597119909119895 (13)

where119901 isin [0 1] is the embedding parameter and120595(119909 119901) is anunknown function The zeroth-order deformation equationis given by

(1 minus 119901) 119871 [120595 (120578 119901) minus 1199100(120578)] = 119901ℎ (119873 [120595 (120578 119901)] minus 119891 (120578))

(14)

where ℎ is the nonzero convergence controlling auxiliaryparameter and119873 is a nonlinear operator given by

119873[120595 (120578 119901)] =

2

sum

119895=0

119886119895(120578)

120597119895120595 (120578 119901)

120597120578119895minus 119891 (120578)

minus 120583int

1

minus1

119896 (120578 119905) 120595119903(119905) 119889119905

(15)

Differentiating (14) 119898 times with respect to the embeddingparameter 119901 setting 119901 = 0 and finally dividing them by 119898we have the so called119898th-order deformation equation

119871 [119910119898(120578) minus 120594

119898119910119898minus1

(120578)] = ℎ119877119898 (16)

subject to boundary conditions

119910119898(minus1) = 119910

119898(1) = 0 (17)

where

119877119898(120578) =

2

sum

119895=0

119886119895(120578)

120597119895120595 (120578 119901)

120597120578119895minus 119891 (120578) (1 minus 120594

119898)

minus 120583int

1

minus1

119896 (120578 119905) 120595119903(119905) 119889119905

(18)

The initial approximation that is used in the higher-orderequation (18) is obtained on solving the following equation

2

sum

119895=0

119886119895(119909) 119910(119895)

0(119909) = 119891 (119909) (19)


1199100(minus1) = 119910

0(1) = 0 (20)

where we use the Chebyshev pseudospectral method to solve(19)-(20)

We first approximate 1199100(120578) by a truncated series of

Chebyshev polynomial of the following form

1199100(120578) asymp 119910

119873

0(120578119895) =

119873

sum

119896=0

119910119896119879119896(120578119895) 119895 = 0 119873 (21)

where 119879119896is the 119896th Chebyshev polynomials 119910

119896are coef-

ficients and Gauss-Lobatto collocation points 1205780 1205781 120578

119873

which are the extrema of the 119873th-order Chebyshev polyno-mial defined by

120578119895= cos(

120587119895

119873) (22)

Derivatives of the functions1199100(120578) at the collocation points

are represented as

1198891199041199100(120578119896)

119889120578119904=

119873

sum

119895=0

119863119904

1198961198951199100(120578119895) 119896 = 0 119873 (23)

where 119904 is the order of differentiation and119863 is the Chebyshevspectral differentiationmatrix Following [23] we express theentries of the differentiation matrix119863 as

119863119896119895= (

minus1

2)119888119896

119888119895

times(minus1)119896+119895

sin (120587 (119895 + 119896) 2119873) sin (120587 (119895 minus 119896) 2119873) 119895 = 119896

119863119896119895= (

minus1

2)

119909119896

sin2 (120587119896119873) 119896 = 0119873 119896 = 119895

11986300= minus119863119873119873=21198732+ 1

6

(24)

Substituting (21)ndash(23) into (19) will result in

AY0= F (25)

subject to the boundary conditions

1199100(1205780) = 1199100(120578119873) = 0 (26)


where

A =2

sum

119895=0

a119895D119895

Y0= [1199100(1205780) 1199100(1205781) 119910

0(120578119873)]119879

F = [119891 (1205780) 119891 (120578

1) 119891 (120578

119873)]119879

a119903= diag (119886

119903(1205780) 119886119903(1205781) 119886

119903(120578119873))

(27)

The values of 1199100(120578119894) 119894 = 0 119873 are determined from the

equation

Y0= Aminus1F (28)

which is the initial approximation for the SHAM solu-tion of the governing equation (12) Apply the Chebyshevpseudospectral transformation on (16)ndash(18) to obtain thefollowing result

AY119898= (120594119898+ ℎ)AY

119898minus1minus ℎ [S

119898minus1minus (1 minus 120594

119898) F] (29)


119910119898(1205780) = 119910119898(120578119873) = 0 (30)

where A and F were defined in and

Y119898= [119910119898(1205780) 119910119898(1205781) 119910

119898(120578119873)]119879

s119898= int

1

minus1

119896 (120591 119905) [Y119898]119903

119889119905

(31)

To implement the boundary condition (30) we delete the firstand the last rows of S

119898minus1 F and the first and the last rows and

columns ofA Finally this recursive formula can be written asfollows

Y119898= (120594119898+ ℎ)Y

119898minus1minus ℎAminus1 [S


119898) F119898minus1]

(32)

with starting from the initial approximation we can obtainhigher-order approximation Y

119898for 119898 ge 1 recursively To

compute the integral in (32) we use the Clenshaw-Curtisquadrature formula as follows

S119898(120578) = int

1

minus1

119896 (120578 119905 Y119898) 119889119905 =

119873

sum

119895=0

119908119895119896 (120578 120578

119895 Y119898) (33)

where the nodes 120578119895are given by (22) and the weights 119908

119895are

given by

1199080= 119908119873=

1

1198732 119873 odd1

1198732 minus 1 119873 even

(34)

119908119897=2

119873120574119897

[1 minus

lfloor1198732rfloor

sum

119896=1

2

1205742119896(41198962 minus 1)

cos 2119896119897120587119873]

119897 = 1 119873 minus 1

(35)

where 1205740= 120574119873= 2 and 120574

119897= 1 for 119897 = 1 119873 minus 1 Y is

a column vector of the elements of the vector Y that iscomputed as follows

Y119898=

119898

sum

1198991=0

119910119898minus1198991

1198991

sum

1198992=0

1199101198991minus1198992

sdot sdot sdot

119899119903minus2

sum

119899119903minus1=0

119910119899119903minus2minus119899119903minus1

119910119899119903minus1

(36)

where119898 119903 ge 0 are positive integers [24]Regarding to accuracy the stability and the error of

previous quadrature formula at the Gauss-Lobatto points werefer the reader to [25]

4 Convergence Analysis

Following the authors in [7 16 26] we present the con-vergence of spectral homotopy analysis method for solvingFredholm integro-differential equations

In view of (13) and (27) (12) can be written as follows

AY = F + 120583int1

minus1

119896 (119909 119905)G (Y) 119889119905 (37)

where Y F and G(Y) are vector functionsWe obtain

Y = Aminus1F + 120583int1

minus1

119896 (119909 119905)Aminus1G (Y) 119889119905 (38)

By substituting F = Aminus1F and G(Y) = Aminus1G(Y) in (38) weobtain

Y = F + 120583int1

minus1

119896 (119909 119905) G (Y) 119889119905 (39)

In (39) we assume that F is bounded for all 119905 in 119862 = [minus1 1]and

|119896 (119909 119905)| le 119872 (40)

Also we suppose that the non linear term G(Y) is Lipschitzcontinuous with

10038171003817100381710038171003817G (Y) minus G (Ylowast)10038171003817100381710038171003817 le 119871

1003817100381710038171003817Y minus Ylowast1003817100381710038171003817 (41)

If we set 120572 = 2120583119871119872 then the following can be proved byusing the previous assumptions

Theorem 1 Thenonlinear Fredholm integro-differential equa-tion in (32) has a unique solution whenever 0 lt 120572 lt 1

Proof Let Y and Ylowast be two different solutions of (39) then

1003817100381710038171003817Y minus Ylowast1003817100381710038171003817 =

100381710038171003817100381710038171003817100381710038171003817

120583int

1

minus1

119896 (119909 119905) [G (Y) minus G (Ylowast)] 119889119905100381710038171003817100381710038171003817100381710038171003817

le 120583int

1

minus1

|119896 (119909 119905)|10038171003817100381710038171003817G (Y) minus G (Ylowast)10038171003817100381710038171003817 119889119905

le 21205831198711198721003817100381710038171003817Y minus Y

lowast1003817100381710038171003817

(42)

Sowe get (1minus120572)YminusYlowast le 0 Since 0 lt 120572 lt 1 so YminusYlowast = 0therefore Y = Ylowast and this completes the proof


60000004

60000003

60000002

60000001

6

59999999

59999998

59999997

59999996

minus3 minus2

ℎ

0minus1 1

(a)

minus3 minus2

ℎ

0minus1 1 2 3

100

50

minus50

0

minus100

(b)

Figure 1 The ℎ-curve 11991010158401015840(0) and 119910101584010158401015840(0) for 10th-order (a) SHAM (b) HAM

Theorem 2 If the series solution Y = suminfin119898=0

Y119898obtained from

(32) is convergent then it converges to the exact solution of theproblem (39)

Proof We assume

Y =infin

sum

119898=0

Y119898 V (Y) =

infin

sum

119898=0

G (Y119898) (43)

where lim119898rarrinfin

Y119898= 0 We can write

119899

sum

119898=1

[Y119898minus 120594119898Y119898minus1]

= Y1+ (Y2minus Y1) + sdot sdot sdot + (Y

119899minus Y119899minus1) = Y119899

(44)

Hence from (44)infin

sum

119898=1

[Y119898minus 120594119898Y119898minus1] = 0 (45)

so using (45) and the definition of the linear operator 119871 wehaveinfin

sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = 119871 [

infin

sum

119898=1

Y119898minus 120594119898Y119898minus1] = 0 (46)

Therefore from (16) we can obtain thatinfin

sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = ℎ

infin

sum

119898=1

119877119898(Y119898minus1) = 0 (47)

Since ℎ = 0 we haveinfin

sum

119898=1

119877119898(Y119898minus1) = 0 (48)

By applying (39) and (43)

infin

sum

119898=1

119877119898(Y119898minus1)

=

infin

sum

119898=1

[Y119898minus1

minus (1 minus 120594119898minus1) F minus 120583int

1

minus1

119896 (119909 119905) G (Y119898minus1) 119889119905]

= Y minus F minus 120583int1

minus1

119896 (119909 119905)V (Y) 119889119905(49)

Therefore Ymust be the exact solution of (39)

5 Numerical Examples

In this section we apply the technique described in Section 3to some illustrative examples of higher-order nonlinear Fred-holm integro-differential equations

Example 1 Consider the second-order Fredholm integro-differential equation

11991010158401015840(119909) = 6119909 + int

1

minus1

119909119905(1199101015840(119905))2

(119910 (119905))2

119889119905 (50)

subject to 119910(minus1) = 119910(1) = 0 with the exact solution 119910(119909) =1199093minus 119909 We employ SHAM and HAM to solve this example

From the ℎ-curves (Figure 1) it is found that when minus15 leℎ le 15 and minus1 le ℎ le 0 the SHAM solution andHAM solution converge to the exact solution respectivelyA numerical results of Example 1 against different order ofSHAM approximate solutions is shown in Table 1


Table 1 The numerical results of Example 1 against different order of SHAM approximate solutions with ℎ = minus001

119909SHAM Numerical

2nd order 4th order100000 0 0 0099965 minus001162119 minus001162119 minus001162119099861 minus004513180 minus004513187 minus004513187099687 minus016001177 minus016001177 minus016001177099443 minus022774902 minus022774902 minus022774902099130 minus029155781 minus029155781 minus029155781098748 minus034334545 minus034334545 minus034334545098297 minus037606083 minus037606087 minus03760608097778 minus038445192 minus038445192 minus038445192097191 minus036563660 minus036563661 minus036563661

Example 2 Consider the second order Fredholm integro-differential equation

11990911991010158401015840(119909) + 119909

21199101015840(119909) + 2119910 (119909)

= (minus1205872119909 + 2) sin (120587119909) + 1205871199092 cos (120587119909)

+ int

1

minus1

cos (120587119905) 1199104 (119905) 119889119905

(51)

subject to 119910(minus1) = 119910(1) = 0 with the exact solution 119910(119909) =sin(120587119909) We employ HAM and SHAM to solve this exampleThe numerical results of Example 2 against different order ofSHAM approximate solutions with ℎ = minus001 is shown inTable 2 In Table 3 there is a comparison of the numericalresult against the HAM and SHAM approximation solutionsat different orders with ℎ = minus0001 It is worth noting that theSHAM results become very highly accurate only with a fewiterations and fifth-order solutions are very close to the exactsolution Comparison of the numerical solution with the 4th-order SHAM solution for ℎ = minus001 is made in Figure 2 Asit is shown in Figure 3 the rate of convergency in SHAM isfaster than HAM In Figure 4 it is found that when minus25 leℎ le 05 and minus1 le ℎ le 1 the SHAM solution and HAMsolution converge to the exact solution respectively In HAMwe choose 119910

0(119909) = 1 minus 119909

2 as initial guess

Example 3 Consider the first-order Fredholm integro-differential equation [27 28]

1199101015840(119909) = minus

1

2119890119909+2+3

2119890119909+ int

1

0

119890119909minus1199051199103(119905) 119889119905 (52)

subject to the boundary condition 119910(0) = 1 In order toapply the SHAM for solving the given problem we shouldtransform using an appropriate change of variables as

119909 =120577 + 1

2 120577 isin [minus1 1] (53)

Then we use the following transformation

119910 (119909) = 119884 (120577) + 119890(119909+1)2

(54)

1050

05

119909

minus1

minus05

minus1 minus05

1

Figure 2 Comparison of the numerical solution of Example 2 withthe 4th-order SHAM solution for ℎ = minus001

We make the governing boundary condition homogeneousSubstituting (54) into the governing equation and boundarycondition results in

1198841015840(120577) =

1

4int

1

minus1

119890(120577minus119905)2

(1198843(119905) + 3119890

119905+1119884 (119905) + 3119890

(119905+1)21198842(119905)) 119889119905

(55)

subject to the boundary condition 119884(minus1) = 0 A comparisonbetween absolute errors in solutions by SHAM Lagrangeinterpolation and Rationalized Haar functions is tabulatedin Table 4 It is also worth noting that the SHAM results arevery close to exact solutions only with two iterations

6 Conclusion

In this paper we presented the application of spectralhomotopy analysis method (SHAM) for solving nonlinearFredholm integro-differential equations A comparison wasmade between exact analytical solutions and numerical



119909 2nd order 3rd order 4th order Numerical100000 0 0 0 0099965 000437807 000437807 000437807 000437807099861 000109471 000109471 000109471 000109471099687 000984768 000984768 000984768 000984768099443 001749926 001749926 001749926 001749926099130 002732631 002732631 002732631 002732631098748 003931949 003931949 003931950 003931950098297 005346606 005346607 005346607 006974900097778 006974898 006974899 006974899 006974900097191 00881459 008814599 008814599 008814600

Table 3 Numerical result of Example 2 against the HAM and the SHAM solutions with ℎ = minus0001

119909SHAM HAM Numerical

5th order 6th order 7th order 3rd order 4th orderminus097191 minus00881460 minus00881460 minus00881460 minus005395836 minus005794467 minus00881460minus097778 minus006974902 minus006974902 minus006974902 minus004280765 minus004597139 minus006974902minus098297 minus005346609 minus005346609 minus005346609 minus003289259 minus003532441 minus005346607minus098748 minus003931951 minus003931951 minus003931951 minus002424140 minus002603420 minus003931950minus099130 minus002732631 minus002732631 minus002732631 minus001687877 minus001812740 minus002732630minus099443 001749926 001749926 001749926 minus000609972 minus001162680 minus001749926minus099687 minus000984768 minus000984768 minus000984768 minus000609972 minus001162680 minus000984768minus099861 minus000437807 minus000437807 minus000437807 minus000271424 minus000655115 minus000437807minus099965 minus000109471 000109471 minus000109471 minus000067905 minus000072931 minus000109471minus100000 0 0 0 0 0 0

0

2

times10minus8

4

6

8

10

minus1 minus05 0 05 1119909

(a)

0

02

04

06

08

1

12

14

16

minus1 minus05 0 05 1119909

(b)

Figure 3 Comparison of the absolute error of third-order (a) SHAM (b) HAM


minus100

minus200

minus300

minus400

minus500

minus600

minus700

minus800

minus900

minus5 minus4 minus3 minus2 minus1 0 1 2ℎ

(a)

minus4 minus2 0 2 4ℎ

minus15

minus1

minus05

0

05

1

15times10

6

(b)

Figure 4 The ℎ-curve 11991010158401015840(minus1) and 119910101584010158401015840(1) for 6th-order (a) SHAM (b) HAM

Table 4 A comparison of absolute errors between SHAM LIM and RHFS

119909SHAM LIM RHFS

2nd order (ℎ = minus1) 6th order 119896 = 32

00 0 0 80 times 10minus5

01 0 10 times 10minus7

20 times 10minus5

02 20 times 10minus19

70 times 10minus7

50 times 10minus5


10 times 10minus6

10 times 10minus5


20 times 10minus5


40 times 10minus6

70 times 10minus5

results obtained by the spectral homotopy analysis methodRationalized Haar functions and Lagrange interpolationsolutions In Example 1 the numerical results indicate thatthe rate of convergency in SHAM is faster than HAM Inthis example we found that the forth-order SHAM approx-imation sufficiently gives a match with the numerical resultsup to eight decimal places In contrast HAM solutions havea good agreement with the numerical results in 20th orderwith six decimal places As we can see in Table 4 the spectralhomotopy analysis results are more accurate and efficientthan Lagrange interpolation solutions and rationalized Haarfunctions solutions [27 28] As it is shown in Figures 1 and4 the rang of admissible values of ℎ is much wider in SHAMthan HAM

In this paper we employed the spectral homotopy analy-sis method to solve nonlinear Fredholm integro-difflerentialequations however it remains to be generalized and verifiedformore complicated integral equations that we consider it asfuture works

Acknowledgment

Theauthors express their sincere thanks to the referees for thecareful and details reading of the earlier version of the paperand very helpful suggestions The authors also gratefullyacknowledge that this research was partially supported bythe University PutraMalaysia under the ERGSGrant Schemehaving Project no 5527068

References

[1] L K Forbes S Crozier and D M Doddrell ldquoCalculatingcurrent densities and fields produced by shielded magnetic res-onance imaging probesrdquo SIAM Journal on AppliedMathematicsvol 57 no 2 pp 401ndash425 1997

[2] K Parand S Abbasbandy S Kazem and J A Rad ldquoA novelapplication of radial basis functions for solving a model of first-order integro-ordinary differential equationrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 16 no 11pp 4250ndash4258 2011


[3] P Darania and A Ebadian ldquoA method for the numerical solu-tion of the integro-differential equationsrdquo Applied Mathematicsand Computation vol 188 no 1 pp 657ndash668 2007

[4] A Karamete andM Sezer ldquoA Taylor collocationmethod for thesolution of linear integro-differential equationsrdquo InternationalJournal of Computer Mathematics vol 79 no 9 pp 987ndash10002002

[5] S J Liao The proposed homotopy analysis technique for thesolution ofnonlinear problems [PhD thesis] Shanghai Jiao TongUniversity Shanghai China 1992

[6] S J Liao The proposed homotopy analysis technique for thesolutionof non linear problems [PhD dissertation] Shanghai JiaoTong University Shanghai China 1992

[7] S Liao Beyond Perturbation Introduction to the HomotopyAnalysis Method vol 2 of CRC Series Modern Mechanics andMathematics Chapman amp HallCRC Boca Raton Fla USA2004

[8] A M LyapunovThe General Problem of the Stability of MotionTaylor amp Francis London UK 1992

[9] G Adomian ldquoA review of the decomposition method andsome recent results for nonlinear equationsrdquoMathematical andComputer Modelling vol 13 no 7 pp 17ndash43 1990

[10] G Adomian and R Rach ldquoNoise terms in decompositionsolution seriesrdquo Computers amp Mathematics with Applicationsvol 24 no 11 pp 61ndash64 1992

[11] G Adomian and R Rach ldquoAnalytic solution of nonlinearboundary value problems in several dimensions by decompo-sitionrdquo Journal of Mathematical Analysis and Applications vol174 no 1 pp 118ndash137 1993

[12] G Adomian Solving Frontier Problems of Physics The Decom-position Method vol 60 of Fundamental Theories of PhysicsKluwer Academic Publishers Dordrecht The Netherlands1994

[13] P K Bera and J Datta ldquoLinear delta expansion technique forthe solution of anharmonic oscillationsrdquo PRAMANA Journal ofPhysics vol 68 no 1 pp 117ndash122 2007

[14] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[15] J H He ldquoThe homotopy perturbation method for nonlinearoscillator with discontinuitiesrdquo Applied Mathematics and Com-putation vol 5 pp 287ndash292 2004

[16] Sh S Behzadi S Abbasbandy T Allahviranlo and A YildirimldquoApplication of Homotopy analysis method for solving a classof nonlinear Volterra-Fredholm integro-differential equationsrdquoJournal of Applied Analysis and Computation vol 1 no 1 pp1ndash14 2012

[17] S S Motsa P Sibanda and S Shateyi ldquoA new spectral-homotopy analysismethod for solving a nonlinear second orderBVPrdquo Communications in Nonlinear Science and NumericalSimulation vol 15 no 9 pp 2293ndash2302 2010

[18] S S Motsa P Sibanda F G Awad and S Shateyi ldquoA newspectral-homotopy analysis method for the MHD Jeffery-Hamel problemrdquo Computers amp Fluids vol 39 no 7 pp 1219ndash1225 2010

[19] S S Motsa and P Sibanda ldquoA new algorithm for solvingsingular IVPsof Lane-Emden typerdquo in Proceedings of the 4thInternational Conferenceon Applied Mathematics SimulationModelling (WSEAS rsquo10) pp 176ndash180 Corfu Island Greece July2010

[20] S S Motsa S Shateyi G T Marewo and P Sibanda ldquoAnimproved spectral homotopy analysis method for MHD flowin a semi-porous channelrdquo Numerical Algorithms vol 60 no3 pp 463ndash481 2012

[21] H Saberi Nik S Effati S S Motsa and M Shirazian ldquoSpectralhomotopy analysismethod and its convergence for solving aclass of nonlinear optimalcontrol problemsrdquo Numerical Algo-rithms 2013

[22] Z Pashazadeh Atabakan A Kılıcman and A Kazemi NasabldquoOn spectralhomotopy analysismethod for solvingVolterra andFredholm typeof integro-differential equationsrdquo Abstract andApplied Analysis vol 2012 Article ID 960289 16 pages 2012

[23] W S Don and A Solomonoff ldquoAccuracy and speed in com-puting the Chebyshev collocation derivativerdquo SIAM Journal onScientific Computing vol 16 no 6 pp 1253ndash1268 1995

[24] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[25] P J Davis and P Rabinowits Method of Numerical IntegrationAcademic Press London UK 2nd edition 1970

[26] H Jafari M Alipour and H Tajadodi ldquoConvergence of homo-topy perturbation method for solving integral equationsrdquo ThaiJournal of Mathematics vol 8 no 3 pp 511ndash520 2010

[27] A Shahsavaran and A Shahsavaran ldquoApplication of Lagrangeinterpolation for nonlinear integro differential equationsrdquoApplied Mathematical Sciences vol 6 no 17ndash20 pp 887ndash8922012

[28] F Mirzaee ldquoThe RHFs for solution of nonlinear Fredholmintegro-differential equationsrdquo Applied Mathematical Sciencesvol 5 no 69ndash72 pp 3453ndash3464 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


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Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

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Applied MathematicsJournal of


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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


to develop a generalized spectral HAM which can use aproper special function for a given nonlinear problem Thespectral homotopy analysis method has been used for solvingpartial and ordinary differential equations [18ndash20] Spectralhomotopy analysis method and its convergency for solvinga class of optimal control problems are presented in [21]Motsa et al [17ndash19] found that the spectral homotopy analysismethod is more efficient than the homotopy analysis methodas it does not depend on the rule of solution expressionand the rule of ergodicity This method is more flexiblethan homotopy analysis method since it allows for a widerrange of linear and nonlinear operators and one is notrestricted to use the method of higher-order differentialmapping for solving boundary value problems in boundeddomains unlike the homotopy analysis method The rangeof admissible ℎ values is much wider in spectral homotopyanalysis method than in homotopy analysis method Themain restriction of HAM in solving integral equations isto choose the best initial guess as the series solution isconvergent In SHAM the initial approximation is taken to bethe solution of the nonhomogeneous linear part of the givenequation In 2012 Pashazadeh Atabakan et al solved linearVolterra and Fredholm integro-differential equations usingspectral homotopy analysis method see [22]

In this paper we apply spectral homotopy analysismethod (SHAM) to solve higher-order nonlinear Fredholmtype of integro-differential equations Fredholm integro-differential equation is given by

2

sum

119895=0

119886119895(119909) 119910(119895)(119909) = 119891 (119909) + 120583int

1

minus1

119896 (119909 119905) [119910 (119905)]119903

119889119905

119910 (minus1) = 119910 (1) = 0

(1)

where120583 is constant value119891(119909) 119896(119909 119905) [119910(119905)]119903 and 119886119895(119909)119903 ge 1

are functions that have suitable derivatives on interval minus1 le119905 le 119909 le 1 and 119886

2(119909) = 0

The paper is organized in the following way Section 2includes a brief introduction in homotopy analysis methodSpectral homotopy analysis method for solving nonlinearFredholm integral equations is presented in Section 3 Theexistence and uniqueness of the solution and convergence ofthe proposed method are proved in Section 4 In Section 5numerical examples are presented In Section 6 concludingremarks are given

2 Homotopy Analysis Solution

In this section we give a brief introduction to HAM Weconsider the following differential equation in a general formas follows

119873[119910 (120578)] = 0 (2)

where 119873 is nonlinear operator 120578 denotes independent vari-ables and 119910(120578) is an unknown function respectively Forsimplicity we disregard all initial and all boundary conditions

which can be dealt in similar way The so-called zero-orderdeformation equation was constructed by Liao as follows

(1 minus 119901) 119871 [120595 (120578 119901) minus 1199100(120578)] = 119901ℎ119867 (120578) (119873 [120595 (120578 119901)])

(3)

where 119901 isin [0 1] is the embedding parameter ℎ is a nonzeroconvergence-parameter 119867(120578) is an auxiliary function 119910

0(120578)

is called an initial guess of 119910(120578) and 120595(120578 119901) is an unknownfunction In addition 119871 is an auxiliary linear operator and119873is nonlinear operator as follows

119871 (120595 (119909 119901)) = 119886119896(119909)

1205972120595 (119909 119901)

1205971199092(4)

with the property 119871(sum2119895=0119888119895119905119895) = 0 where 119888

119895 are constants

and

119873[120595 (119909 119901)] =

2

sum

119895=0

119886119895(119909)

120597119895120595 (119909 119901)

120597119909119895minus 119891 (119909)

minus 120583int

1

minus1

119896 (119909 119905) 120595119903(119905) 119889119905

(5)

is a nonlinear operator Obviously when 119901 = 0 and 119901 = 1it holds 120595(120578 0) = 119910

0(120578) and 120595(120578 1) = 119910(120578) In this way as

119901 increase from 0 to 1 120595(120578 119901) alter from initial guess 1199100(120578)

to the solution 119910(120578) and 120595(120578 119901) is expanded in Taylor serieswith respect to 119901 as follows

120595 (120578 119901) = 1199100(120578) +

+infin

sum

119898=1

119910119898(120578) 119901119898 (6)

where

119910119898(120578) = 119863

119898[120595 (120578 119901)]

119863119898120595 =

1

119898

120597119898120595

120597119901119898

10038161003816100381610038161003816100381610038161003816119901=0

(7)

The series (6) converges at 119901 = 1 if the auxiliary linearoperator the initial guess the convergence-parameter andthe auxiliary function are properly selected as follows

120595 (120578) = 1199100(120578) +

+infin

sum

119898=1

119910119898(120578) (8)

The admissible and valid values of the convergence-parameter ℎ are found from the horizontal portion of theℎ-curves Liao proved that 119910(120578) is one of the solutions oforiginal nonlinear equation As119867(120578) = 1 so (3) becomes

(1 minus 119901) 119871 [120595 (120578 119901) minus 1199100(120578)] = 119901ℎ (119873 [120595 (120578 119901)]) (9)

Define the vector 119910119898= 1199100(120578) 1199101(120578) 119910

119898(120578) Operating

on both side of (9) with119863119898 we have the so called119898th-order

deformation equation as follows

119871 [119910119898(120578) minus 120594

119898119910119898minus1

(120578)] = ℎ119867 (120578) 119877119898(119910119898minus1

(120578)) (10)


where

119877119898(119910119898minus1) =

1

(119898 minus 1)

120597119898minus1119873[120595 (120578 119901)]

120597119901119898minus1

100381610038161003816100381610038161003816100381610038161003816119901=0

120594119898= 0 119898 le 1

1 otherwise

(11)





sum

119895=0

119886119895(119909) 119910(119895)(119909) = 119891 (119909) + 120583int

1

minus1

119896 (119909 119905) [119910 (119905)]119903

119889119905

119910 (minus1) = 119910 (1) = 0

(12)


119871 (120595 (119909 119901)) =

2

sum

119895=0

119886119895(119909)

120597119895120595 (119909 119901)

120597119909119895 (13)



(14)


119873[120595 (120578 119901)] =

2

sum

119895=0

119886119895(120578)

120597119895120595 (120578 119901)

120597120578119895minus 119891 (120578)

minus 120583int

1

minus1

119896 (120578 119905) 120595119903(119905) 119889119905

(15)


119871 [119910119898(120578) minus 120594

119898119910119898minus1

(120578)] = ℎ119877119898 (16)


119910119898(minus1) = 119910

119898(1) = 0 (17)

where

119877119898(120578) =

2

sum

119895=0

119886119895(120578)

120597119895120595 (120578 119901)

120597120578119895minus 119891 (120578) (1 minus 120594

119898)

minus 120583int

1

minus1

119896 (120578 119905) 120595119903(119905) 119889119905

(18)


2

sum

119895=0

119886119895(119909) 119910(119895)

0(119909) = 119891 (119909) (19)


1199100(minus1) = 119910

0(1) = 0 (20)




1199100(120578) asymp 119910

119873

0(120578119895) =

119873

sum

119896=0

119910119896119879119896(120578119895) 119895 = 0 119873 (21)


119896are coef-


119873


120578119895= cos(

120587119895

119873) (22)


are represented as

1198891199041199100(120578119896)

119889120578119904=

119873

sum

119895=0

119863119904

1198961198951199100(120578119895) 119896 = 0 119873 (23)


119863119896119895= (

minus1

2)119888119896

119888119895

times(minus1)119896+119895

sin (120587 (119895 + 119896) 2119873) sin (120587 (119895 minus 119896) 2119873) 119895 = 119896

119863119896119895= (

minus1

2)

119909119896

sin2 (120587119896119873) 119896 = 0119873 119896 = 119895

11986300= minus119863119873119873=21198732+ 1

6

(24)


AY0= F (25)


1199100(1205780) = 1199100(120578119873) = 0 (26)


where

A =2

sum

119895=0

a119895D119895

Y0= [1199100(1205780) 1199100(1205781) 119910

0(120578119873)]119879

F = [119891 (1205780) 119891 (120578

1) 119891 (120578

119873)]119879

a119903= diag (119886

119903(1205780) 119886119903(1205781) 119886

119903(120578119873))

(27)


equation

Y0= Aminus1F (28)


AY119898= (120594119898+ ℎ)AY



119898) F] (29)


119910119898(1205780) = 119910119898(120578119873) = 0 (30)


Y119898= [119910119898(1205780) 119910119898(1205781) 119910

119898(120578119873)]119879

s119898= int

1

minus1

119896 (120591 119905) [Y119898]119903

119889119905

(31)




Y119898= (120594119898+ ℎ)Y



119898) F119898minus1]

(32)




S119898(120578) = int

1

minus1

119896 (120578 119905 Y119898) 119889119905 =

119873

sum

119895=0

119908119895119896 (120578 120578

119895 Y119898) (33)


119895are

given by

1199080= 119908119873=

1

1198732 119873 odd1

1198732 minus 1 119873 even

(34)

119908119897=2

119873120574119897

[1 minus

lfloor1198732rfloor

sum

119896=1

2

1205742119896(41198962 minus 1)

cos 2119896119897120587119873]

119897 = 1 119873 minus 1

(35)

where 1205740= 120574119873= 2 and 120574

119897= 1 for 119897 = 1 119873 minus 1 Y is


Y119898=

119898

sum

1198991=0

119910119898minus1198991

1198991

sum

1198992=0

1199101198991minus1198992

sdot sdot sdot

119899119903minus2

sum

119899119903minus1=0


119910119899119903minus1

(36)






AY = F + 120583int1

minus1

119896 (119909 119905)G (Y) 119889119905 (37)



minus1

119896 (119909 119905)Aminus1G (Y) 119889119905 (38)


Y = F + 120583int1

minus1

119896 (119909 119905) G (Y) 119889119905 (39)


|119896 (119909 119905)| le 119872 (40)



1003817100381710038171003817Y minus Ylowast1003817100381710038171003817 (41)




1003817100381710038171003817Y minus Ylowast1003817100381710038171003817 =

100381710038171003817100381710038171003817100381710038171003817

120583int

1

minus1


le 120583int

1

minus1


le 21205831198711198721003817100381710038171003817Y minus Y

lowast1003817100381710038171003817

(42)



60000004

60000003

60000002

60000001

6

59999999

59999998

59999997

59999996

minus3 minus2

ℎ

0minus1 1

(a)

minus3 minus2

ℎ

0minus1 1 2 3

100

50

minus50

0

minus100

(b)





Proof We assume

Y =infin

sum

119898=0

Y119898 V (Y) =

infin

sum

119898=0

G (Y119898) (43)



119899

sum

119898=1

[Y119898minus 120594119898Y119898minus1]


119899minus Y119899minus1) = Y119899

(44)


sum

119898=1

[Y119898minus 120594119898Y119898minus1] = 0 (45)


sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = 119871 [

infin

sum

119898=1

Y119898minus 120594119898Y119898minus1] = 0 (46)


sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = ℎ

infin

sum

119898=1

119877119898(Y119898minus1) = 0 (47)


sum

119898=1

119877119898(Y119898minus1) = 0 (48)


infin

sum

119898=1

119877119898(Y119898minus1)

=

infin

sum

119898=1

[Y119898minus1


1

minus1

119896 (119909 119905) G (Y119898minus1) 119889119905]


minus1

119896 (119909 119905)V (Y) 119889119905(49)





11991010158401015840(119909) = 6119909 + int

1

minus1

119909119905(1199101015840(119905))2

(119910 (119905))2

119889119905 (50)








11990911991010158401015840(119909) + 119909

21199101015840(119909) + 2119910 (119909)

= (minus1205872119909 + 2) sin (120587119909) + 1205871199092 cos (120587119909)

+ int

1

minus1

cos (120587119905) 1199104 (119905) 119889119905

(51)


0(119909) = 1 minus 119909

2 as initial guess


1199101015840(119909) = minus

1

2119890119909+2+3

2119890119909+ int

1

0

119890119909minus1199051199103(119905) 119889119905 (52)


119909 =120577 + 1

2 120577 isin [minus1 1] (53)


119910 (119909) = 119884 (120577) + 119890(119909+1)2

(54)

1050

05

119909

minus1

minus05

minus1 minus05

1



1198841015840(120577) =

1

4int

1

minus1

119890(120577minus119905)2

(1198843(119905) + 3119890

119905+1119884 (119905) + 3119890

(119905+1)21198842(119905)) 119889119905

(55)


6 Conclusion








0

2

times10minus8

4

6

8

10

minus1 minus05 0 05 1119909

(a)

0

02

04

06

08

1

12

14

16

minus1 minus05 0 05 1119909

(b)



minus100

minus200

minus300

minus400

minus500

minus600

minus700

minus800

minus900


(a)


minus15

minus1

minus05

0

05

1

15times10

6

(b)



119909SHAM LIM RHFS




20 times 10minus5


70 times 10minus7

50 times 10minus5


10 times 10minus6

10 times 10minus5


20 times 10minus5


40 times 10minus6

70 times 10minus5



Acknowledgment


References





































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










where

119877119898(119910119898minus1) =

1

(119898 minus 1)

120597119898minus1119873[120595 (120578 119901)]

120597119901119898minus1

100381610038161003816100381610038161003816100381610038161003816119901=0

120594119898= 0 119898 le 1

1 otherwise

(11)





sum

119895=0

119886119895(119909) 119910(119895)(119909) = 119891 (119909) + 120583int

1

minus1

119896 (119909 119905) [119910 (119905)]119903

119889119905

119910 (minus1) = 119910 (1) = 0

(12)


119871 (120595 (119909 119901)) =

2

sum

119895=0

119886119895(119909)

120597119895120595 (119909 119901)

120597119909119895 (13)



(14)


119873[120595 (120578 119901)] =

2

sum

119895=0

119886119895(120578)

120597119895120595 (120578 119901)

120597120578119895minus 119891 (120578)

minus 120583int

1

minus1

119896 (120578 119905) 120595119903(119905) 119889119905

(15)


119871 [119910119898(120578) minus 120594

119898119910119898minus1

(120578)] = ℎ119877119898 (16)


119910119898(minus1) = 119910

119898(1) = 0 (17)

where

119877119898(120578) =

2

sum

119895=0

119886119895(120578)

120597119895120595 (120578 119901)

120597120578119895minus 119891 (120578) (1 minus 120594

119898)

minus 120583int

1

minus1

119896 (120578 119905) 120595119903(119905) 119889119905

(18)


2

sum

119895=0

119886119895(119909) 119910(119895)

0(119909) = 119891 (119909) (19)


1199100(minus1) = 119910

0(1) = 0 (20)




1199100(120578) asymp 119910

119873

0(120578119895) =

119873

sum

119896=0

119910119896119879119896(120578119895) 119895 = 0 119873 (21)


119896are coef-


119873


120578119895= cos(

120587119895

119873) (22)


are represented as

1198891199041199100(120578119896)

119889120578119904=

119873

sum

119895=0

119863119904

1198961198951199100(120578119895) 119896 = 0 119873 (23)


119863119896119895= (

minus1

2)119888119896

119888119895

times(minus1)119896+119895

sin (120587 (119895 + 119896) 2119873) sin (120587 (119895 minus 119896) 2119873) 119895 = 119896

119863119896119895= (

minus1

2)

119909119896

sin2 (120587119896119873) 119896 = 0119873 119896 = 119895

11986300= minus119863119873119873=21198732+ 1

6

(24)


AY0= F (25)


1199100(1205780) = 1199100(120578119873) = 0 (26)


where

A =2

sum

119895=0

a119895D119895

Y0= [1199100(1205780) 1199100(1205781) 119910

0(120578119873)]119879

F = [119891 (1205780) 119891 (120578

1) 119891 (120578

119873)]119879

a119903= diag (119886

119903(1205780) 119886119903(1205781) 119886

119903(120578119873))

(27)


equation

Y0= Aminus1F (28)


AY119898= (120594119898+ ℎ)AY



119898) F] (29)


119910119898(1205780) = 119910119898(120578119873) = 0 (30)


Y119898= [119910119898(1205780) 119910119898(1205781) 119910

119898(120578119873)]119879

s119898= int

1

minus1

119896 (120591 119905) [Y119898]119903

119889119905

(31)




Y119898= (120594119898+ ℎ)Y



119898) F119898minus1]

(32)




S119898(120578) = int

1

minus1

119896 (120578 119905 Y119898) 119889119905 =

119873

sum

119895=0

119908119895119896 (120578 120578

119895 Y119898) (33)


119895are

given by

1199080= 119908119873=

1

1198732 119873 odd1

1198732 minus 1 119873 even

(34)

119908119897=2

119873120574119897

[1 minus

lfloor1198732rfloor

sum

119896=1

2

1205742119896(41198962 minus 1)

cos 2119896119897120587119873]

119897 = 1 119873 minus 1

(35)

where 1205740= 120574119873= 2 and 120574

119897= 1 for 119897 = 1 119873 minus 1 Y is


Y119898=

119898

sum

1198991=0

119910119898minus1198991

1198991

sum

1198992=0

1199101198991minus1198992

sdot sdot sdot

119899119903minus2

sum

119899119903minus1=0


119910119899119903minus1

(36)






AY = F + 120583int1

minus1

119896 (119909 119905)G (Y) 119889119905 (37)



minus1

119896 (119909 119905)Aminus1G (Y) 119889119905 (38)


Y = F + 120583int1

minus1

119896 (119909 119905) G (Y) 119889119905 (39)


|119896 (119909 119905)| le 119872 (40)



1003817100381710038171003817Y minus Ylowast1003817100381710038171003817 (41)




1003817100381710038171003817Y minus Ylowast1003817100381710038171003817 =

100381710038171003817100381710038171003817100381710038171003817

120583int

1

minus1


le 120583int

1

minus1


le 21205831198711198721003817100381710038171003817Y minus Y

lowast1003817100381710038171003817

(42)



60000004

60000003

60000002

60000001

6

59999999

59999998

59999997

59999996

minus3 minus2

ℎ

0minus1 1

(a)

minus3 minus2

ℎ

0minus1 1 2 3

100

50

minus50

0

minus100

(b)





Proof We assume

Y =infin

sum

119898=0

Y119898 V (Y) =

infin

sum

119898=0

G (Y119898) (43)



119899

sum

119898=1

[Y119898minus 120594119898Y119898minus1]


119899minus Y119899minus1) = Y119899

(44)


sum

119898=1

[Y119898minus 120594119898Y119898minus1] = 0 (45)


sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = 119871 [

infin

sum

119898=1

Y119898minus 120594119898Y119898minus1] = 0 (46)


sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = ℎ

infin

sum

119898=1

119877119898(Y119898minus1) = 0 (47)


sum

119898=1

119877119898(Y119898minus1) = 0 (48)


infin

sum

119898=1

119877119898(Y119898minus1)

=

infin

sum

119898=1

[Y119898minus1


1

minus1

119896 (119909 119905) G (Y119898minus1) 119889119905]


minus1

119896 (119909 119905)V (Y) 119889119905(49)





11991010158401015840(119909) = 6119909 + int

1

minus1

119909119905(1199101015840(119905))2

(119910 (119905))2

119889119905 (50)








11990911991010158401015840(119909) + 119909

21199101015840(119909) + 2119910 (119909)

= (minus1205872119909 + 2) sin (120587119909) + 1205871199092 cos (120587119909)

+ int

1

minus1

cos (120587119905) 1199104 (119905) 119889119905

(51)


0(119909) = 1 minus 119909

2 as initial guess


1199101015840(119909) = minus

1

2119890119909+2+3

2119890119909+ int

1

0

119890119909minus1199051199103(119905) 119889119905 (52)


119909 =120577 + 1

2 120577 isin [minus1 1] (53)


119910 (119909) = 119884 (120577) + 119890(119909+1)2

(54)

1050

05

119909

minus1

minus05

minus1 minus05

1



1198841015840(120577) =

1

4int

1

minus1

119890(120577minus119905)2

(1198843(119905) + 3119890

119905+1119884 (119905) + 3119890

(119905+1)21198842(119905)) 119889119905

(55)


6 Conclusion








0

2

times10minus8

4

6

8

10

minus1 minus05 0 05 1119909

(a)

0

02

04

06

08

1

12

14

16

minus1 minus05 0 05 1119909

(b)



minus100

minus200

minus300

minus400

minus500

minus600

minus700

minus800

minus900


(a)


minus15

minus1

minus05

0

05

1

15times10

6

(b)



119909SHAM LIM RHFS




20 times 10minus5


70 times 10minus7

50 times 10minus5


10 times 10minus6

10 times 10minus5


20 times 10minus5


40 times 10minus6

70 times 10minus5



Acknowledgment


References





































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










where

A =2

sum

119895=0

a119895D119895

Y0= [1199100(1205780) 1199100(1205781) 119910

0(120578119873)]119879

F = [119891 (1205780) 119891 (120578

1) 119891 (120578

119873)]119879

a119903= diag (119886

119903(1205780) 119886119903(1205781) 119886

119903(120578119873))

(27)


equation

Y0= Aminus1F (28)


AY119898= (120594119898+ ℎ)AY



119898) F] (29)


119910119898(1205780) = 119910119898(120578119873) = 0 (30)


Y119898= [119910119898(1205780) 119910119898(1205781) 119910

119898(120578119873)]119879

s119898= int

1

minus1

119896 (120591 119905) [Y119898]119903

119889119905

(31)




Y119898= (120594119898+ ℎ)Y



119898) F119898minus1]

(32)




S119898(120578) = int

1

minus1

119896 (120578 119905 Y119898) 119889119905 =

119873

sum

119895=0

119908119895119896 (120578 120578

119895 Y119898) (33)


119895are

given by

1199080= 119908119873=

1

1198732 119873 odd1

1198732 minus 1 119873 even

(34)

119908119897=2

119873120574119897

[1 minus

lfloor1198732rfloor

sum

119896=1

2

1205742119896(41198962 minus 1)

cos 2119896119897120587119873]

119897 = 1 119873 minus 1

(35)

where 1205740= 120574119873= 2 and 120574

119897= 1 for 119897 = 1 119873 minus 1 Y is


Y119898=

119898

sum

1198991=0

119910119898minus1198991

1198991

sum

1198992=0

1199101198991minus1198992

sdot sdot sdot

119899119903minus2

sum

119899119903minus1=0


119910119899119903minus1

(36)






AY = F + 120583int1

minus1

119896 (119909 119905)G (Y) 119889119905 (37)



minus1

119896 (119909 119905)Aminus1G (Y) 119889119905 (38)


Y = F + 120583int1

minus1

119896 (119909 119905) G (Y) 119889119905 (39)


|119896 (119909 119905)| le 119872 (40)



1003817100381710038171003817Y minus Ylowast1003817100381710038171003817 (41)




1003817100381710038171003817Y minus Ylowast1003817100381710038171003817 =

100381710038171003817100381710038171003817100381710038171003817

120583int

1

minus1


le 120583int

1

minus1


le 21205831198711198721003817100381710038171003817Y minus Y

lowast1003817100381710038171003817

(42)



60000004

60000003

60000002

60000001

6

59999999

59999998

59999997

59999996

minus3 minus2

ℎ

0minus1 1

(a)

minus3 minus2

ℎ

0minus1 1 2 3

100

50

minus50

0

minus100

(b)





Proof We assume

Y =infin

sum

119898=0

Y119898 V (Y) =

infin

sum

119898=0

G (Y119898) (43)



119899

sum

119898=1

[Y119898minus 120594119898Y119898minus1]


119899minus Y119899minus1) = Y119899

(44)


sum

119898=1

[Y119898minus 120594119898Y119898minus1] = 0 (45)


sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = 119871 [

infin

sum

119898=1

Y119898minus 120594119898Y119898minus1] = 0 (46)


sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = ℎ

infin

sum

119898=1

119877119898(Y119898minus1) = 0 (47)


sum

119898=1

119877119898(Y119898minus1) = 0 (48)


infin

sum

119898=1

119877119898(Y119898minus1)

=

infin

sum

119898=1

[Y119898minus1


1

minus1

119896 (119909 119905) G (Y119898minus1) 119889119905]


minus1

119896 (119909 119905)V (Y) 119889119905(49)





11991010158401015840(119909) = 6119909 + int

1

minus1

119909119905(1199101015840(119905))2

(119910 (119905))2

119889119905 (50)








11990911991010158401015840(119909) + 119909

21199101015840(119909) + 2119910 (119909)

= (minus1205872119909 + 2) sin (120587119909) + 1205871199092 cos (120587119909)

+ int

1

minus1

cos (120587119905) 1199104 (119905) 119889119905

(51)


0(119909) = 1 minus 119909

2 as initial guess


1199101015840(119909) = minus

1

2119890119909+2+3

2119890119909+ int

1

0

119890119909minus1199051199103(119905) 119889119905 (52)


119909 =120577 + 1

2 120577 isin [minus1 1] (53)


119910 (119909) = 119884 (120577) + 119890(119909+1)2

(54)

1050

05

119909

minus1

minus05

minus1 minus05

1



1198841015840(120577) =

1

4int

1

minus1

119890(120577minus119905)2

(1198843(119905) + 3119890

119905+1119884 (119905) + 3119890

(119905+1)21198842(119905)) 119889119905

(55)


6 Conclusion








0

2

times10minus8

4

6

8

10

minus1 minus05 0 05 1119909

(a)

0

02

04

06

08

1

12

14

16

minus1 minus05 0 05 1119909

(b)



minus100

minus200

minus300

minus400

minus500

minus600

minus700

minus800

minus900


(a)


minus15

minus1

minus05

0

05

1

15times10

6

(b)



119909SHAM LIM RHFS




20 times 10minus5


70 times 10minus7

50 times 10minus5


10 times 10minus6

10 times 10minus5


20 times 10minus5


40 times 10minus6

70 times 10minus5



Acknowledgment


References





































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










60000004

60000003

60000002

60000001

6

59999999

59999998

59999997

59999996

minus3 minus2

ℎ

0minus1 1

(a)

minus3 minus2

ℎ

0minus1 1 2 3

100

50

minus50

0

minus100

(b)





Proof We assume

Y =infin

sum

119898=0

Y119898 V (Y) =

infin

sum

119898=0

G (Y119898) (43)



119899

sum

119898=1

[Y119898minus 120594119898Y119898minus1]


119899minus Y119899minus1) = Y119899

(44)


sum

119898=1

[Y119898minus 120594119898Y119898minus1] = 0 (45)


sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = 119871 [

infin

sum

119898=1

Y119898minus 120594119898Y119898minus1] = 0 (46)


sum

119898=1

119871 [Y119898minus 120594119898Y119898minus1] = ℎ

infin

sum

119898=1

119877119898(Y119898minus1) = 0 (47)


sum

119898=1

119877119898(Y119898minus1) = 0 (48)


infin

sum

119898=1

119877119898(Y119898minus1)

=

infin

sum

119898=1

[Y119898minus1


1

minus1

119896 (119909 119905) G (Y119898minus1) 119889119905]


minus1

119896 (119909 119905)V (Y) 119889119905(49)





11991010158401015840(119909) = 6119909 + int

1

minus1

119909119905(1199101015840(119905))2

(119910 (119905))2

119889119905 (50)








11990911991010158401015840(119909) + 119909

21199101015840(119909) + 2119910 (119909)

= (minus1205872119909 + 2) sin (120587119909) + 1205871199092 cos (120587119909)

+ int

1

minus1

cos (120587119905) 1199104 (119905) 119889119905

(51)


0(119909) = 1 minus 119909

2 as initial guess


1199101015840(119909) = minus

1

2119890119909+2+3

2119890119909+ int

1

0

119890119909minus1199051199103(119905) 119889119905 (52)


119909 =120577 + 1

2 120577 isin [minus1 1] (53)


119910 (119909) = 119884 (120577) + 119890(119909+1)2

(54)

1050

05

119909

minus1

minus05

minus1 minus05

1



1198841015840(120577) =

1

4int

1

minus1

119890(120577minus119905)2

(1198843(119905) + 3119890

119905+1119884 (119905) + 3119890

(119905+1)21198842(119905)) 119889119905

(55)


6 Conclusion








0

2

times10minus8

4

6

8

10

minus1 minus05 0 05 1119909

(a)

0

02

04

06

08

1

12

14

16

minus1 minus05 0 05 1119909

(b)



minus100

minus200

minus300

minus400

minus500

minus600

minus700

minus800

minus900


(a)


minus15

minus1

minus05

0

05

1

15times10

6

(b)



119909SHAM LIM RHFS




20 times 10minus5


70 times 10minus7

50 times 10minus5


10 times 10minus6

10 times 10minus5


20 times 10minus5


40 times 10minus6

70 times 10minus5



Acknowledgment


References





































Volume 2014




Journal of











Journal of


Function Spaces






Algebra














11990911991010158401015840(119909) + 119909

21199101015840(119909) + 2119910 (119909)

= (minus1205872119909 + 2) sin (120587119909) + 1205871199092 cos (120587119909)

+ int

1

minus1

cos (120587119905) 1199104 (119905) 119889119905

(51)


0(119909) = 1 minus 119909

2 as initial guess


1199101015840(119909) = minus

1

2119890119909+2+3

2119890119909+ int

1

0

119890119909minus1199051199103(119905) 119889119905 (52)


119909 =120577 + 1

2 120577 isin [minus1 1] (53)


119910 (119909) = 119884 (120577) + 119890(119909+1)2

(54)

1050

05

119909

minus1

minus05

minus1 minus05

1



1198841015840(120577) =

1

4int

1

minus1

119890(120577minus119905)2

(1198843(119905) + 3119890

119905+1119884 (119905) + 3119890

(119905+1)21198842(119905)) 119889119905

(55)


6 Conclusion








0

2

times10minus8

4

6

8

10

minus1 minus05 0 05 1119909

(a)

0

02

04

06

08

1

12

14

16

minus1 minus05 0 05 1119909

(b)



minus100

minus200

minus300

minus400

minus500

minus600

minus700

minus800

minus900


(a)


minus15

minus1

minus05

0

05

1

15times10

6

(b)



119909SHAM LIM RHFS




20 times 10minus5


70 times 10minus7

50 times 10minus5


10 times 10minus6

10 times 10minus5


20 times 10minus5


40 times 10minus6

70 times 10minus5



Acknowledgment


References





































Volume 2014




Journal of











Journal of


Function Spaces






Algebra















0

2

times10minus8

4

6

8

10

minus1 minus05 0 05 1119909

(a)

0

02

04

06

08

1

12

14

16

minus1 minus05 0 05 1119909

(b)



minus100

minus200

minus300

minus400

minus500

minus600

minus700

minus800

minus900


(a)


minus15

minus1

minus05

0

05

1

15times10

6

(b)



119909SHAM LIM RHFS




20 times 10minus5


70 times 10minus7

50 times 10minus5


10 times 10minus6

10 times 10minus5


20 times 10minus5


40 times 10minus6

70 times 10minus5



Acknowledgment


References





































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










minus100

minus200

minus300

minus400

minus500

minus600

minus700

minus800

minus900


(a)


minus15

minus1

minus05

0

05

1

15times10

6

(b)



119909SHAM LIM RHFS




20 times 10minus5


70 times 10minus7

50 times 10minus5


10 times 10minus6

10 times 10minus5


20 times 10minus5


40 times 10minus6

70 times 10minus5



Acknowledgment


References





































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











































Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Research Article Numerical Solution of Nonlinear Fredholm ...

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