Research ArticleSolving Volterra Integrodifferential Equations via DiagonallyImplicit Multistep Block Method

Nur Auni Baharum1 Zanariah Abdul Majid 12 and Norazak Senu 12

1 Institute for Mathematical Research Universiti Putra Malaysia 43400 Serdang Selangor Malaysia2Mathematics Department Faculty of Science Universiti Putra Malaysia 43400 Serdang Selangor Malaysia

Correspondence should be addressed to Zanariah Abdul Majid zana majid99yahoocom

Received 15 June 2017 Revised 13 October 2017 Accepted 23 November 2017 Published 1 January 2018

Academic Editor Theodore E Simos

Copyright copy 2018 Nur Auni Baharum et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

The performance of the numerical computation based on the diagonally implicit multistep block method for solving Volterraintegrodifferential equations (VIDE) of the second kind has been analyzed The numerical solutions of VIDE will be computed attwo points concurrently using the proposed numerical method and executed in the predictor-corrector (PECE)modeThe strategyto obtain the numerical solution of an integral part is discussed and the stability analysis of the diagonally implicit multistep blockmethod was investigated Numerical results showed the competence of diagonally implicit multistep block method when solvingVolterra integrodifferential equations compared to the existing methods

1 Introduction

Consider the Volterra integrodifferential equation of thesecond kind

1199101015840 (119909) = 119865 (119909 119910 (119909) 119911 (119909)) 119910 (1199090) = 1199100 (1)

where

119911 (119909) = int1199090119870 (119909 119904) 119910 (119904) 119889119904 0 le 119904 le 119909 (2)

The numerical methods are generated to solve (1) whichis a standard algorithm for ordinary differential equationsand Newton-Cote integration formulae are required forsolving the integral part since it cannot be solved explicitlyThese equations usually appeared in physics biology andengineering applications such as biological models neutrondiffusion wind ripple in the desert heat transfer and manymore

For many years several methods had been applied tosolve first-order problemofVIDEDay [1] proposedNewton-Cotes integration formula of the trapezoidal rule for thesolutions of outer and inner integral to obtain approximate

solutions of integrodifferential equations A comparisonbetween the variational iteration method and trapezoidalrule revealing that the variational iteration method is moreefficient and convenient to solve linear VIDE was discussedby Saadati et al [2]

In [3] finite difference method is used for solving linearVIDE by Raftari He transforms the Volterra integrodifferen-tial equation in a matrix form and solved it by using finitedifference method based on Simpsonrsquos rule and trapezoidalrule A fourth-order robust numerical method was presentedby Filiz [4] with a combination of the trapezoidal rule andSimpsonrsquos 13 rule to evaluate the solution of VIDE for kernelequal to oneThen he extended his workwith a Runge-Kutta-Verner method in [5] and used higher rules of numericalintegration method for solving the integral part

The extended trapezoidal method [6] was proposed forthe numerical solution of VIDE of the second kind andimplemented the method in 119875119864119862119864 scheme Mohamed andMajid [7] had solved the second kind of VIDE using one-stepblock method and the Newton-Cotes quadrature formulawas employed for finding the solution of the integral partThe multistep block method in [8] had implemented twoapproaches for solving VIDE for119870(119909 119904) = 1 and119870(119909 119904) = 1

HindawiInternational Journal of Mathematics and Mathematical SciencesVolume 2018 Article ID 7392452 10 pageshttpsdoiorg10115520187392452

2 International Journal of Mathematics and Mathematical Sciences

ℎℎℎ

xnminus1 xn xn+1 xn+2

Figure 1 Two-point multistep block method

2 Numerical Method

The proposed numerical method is in the form of blockmethod and it will generate two ormore solutions at the sametime The proposed method is a two-point block methodhence it will generate two solutions in one block

In Figure 1 the two approximate values of 119910119899+1 and 119910119899+2will be computed simultaneously in a blockThe approximatevalues of 119910119899+1 can be developed by integrating (1) overthe interval [119909119899 119909119899+1] while the interval for values 119910119899+2 is[119909119899 119909119899+2] Hence the formulae of 119910119899+1 and 119910119899+2 can beobtained as

int119909119899+119903119909119899

1199101015840 (119909) 119889119909 = int119909119899+119903119909119899

119865 (119909 119910 119911) 119889119909 (3)

Therefore

119910119899+119903 minus 119910119899 = int119909119899+119903119909119899

119865 (119909 119910 119911) 119889119909 (4)

where 119903 = 1 2 Then function 119865(119909 119910 119911) in (4) will beapproximated using Lagrange interpolating polynomial andthe interpolation points involved in obtaining the correctorformula of 119910119899+1 are 119909119899minus1 119909119899 119909119899+1 Taking 119909 = 119909119899+1 + 119904ℎ119889119909 = ℎ119889119904 and replacing into (4) the limit of integration in(4) will be minus1 to 0

The formulation of 119910119899+2 can be obtained when threepoints are involved in the interpolation polynomial that is119909119899minus1 119909119899+1 119909119899+2 Considering 119909 = 119909119899+2 + 119904ℎ 119889119909 = ℎ119889119904 in(4) and the limit of integration will be changed from minus2 to0 The corrector formulae of 119910119899+1 and 119910119899+2 will be obtainedusingMAPLE softwareThe predictor formulae are one orderless than the corrector formulae and the same process ofderivation is applied

Diagonally Implicit Multistep Block Method

Predictor

119910119901119899+1 = 119910119899 + ℎ [32119865119899 minus 12119865119899minus1] 119910119901119899+2 = 119910119899 + ℎ [4119865119899 minus 2119865119899minus1]

(5)

Corrector

119910119888119899+1 = 119910119899 + ℎ [ 512119865119899+1 +23119865119899 minus

112119865119899minus1]

119910119888119899+2 = 119910119899 + ℎ [29119865119899+2 + 53119865119899+1 + 19119865119899minus1] (6)

The matrix form of the corrector formulae is

[1 00 1] [

119910119899+1119910119899+2] = [

0 minus10 minus1] [

119910119899minus1119910119899 ]

+ ℎ[[[minus 112

231

953]]][119865119899minus1119865119899 ]

+ ℎ[[[

512 053

29]]][ 119865119899+1119865119899+2 ]

(7)

which is equivalent to the difference equations

1198600119884119898 = 1198601119884119898minus1 + ℎ (1198610119865119898minus1 + 1198611119865119898) (8)

where 1198600 1198601 1198610 and 1198611 are the coefficients with 119898-vectors119884119898 119884119898minus1 119865119898minus1 and 119865119898 defined as

119884119898 = [119910119899+1119910119899+2]

119884119898minus1 = [119910119899minus1119910119899 ]

119865119898minus1 = [119865119899minus1119865119899 ]

119865119898 = [119865119899+1119865119899+2]

(9)

3 Analysis of Diagonally ImplicitMultistep Block Method

31 Order and Convergence of the Method The order of themethod can be obtained by referring [9]

119896sum119895=0

[120572119895119910 (119909 + 119895ℎ) minus ℎ1205731198951199101015840 (119909 + 119895ℎ)]= 119862119901119910119901 + 119874 (ℎ119901+1)

(10)

where 119901 is the order of the linear multistep method 119874(ℎ119901+1)is the local truncation error and 119862119901 is defined as

119862119901 =119896sum119895=0

119895119901120572119895119901 minus 119895(119901minus1)120573119895(119901 minus 1) (11)

Definition 1 The numerical method is said to be in order 119901 ifthe linear operator of numerical method is

1198620 = 1198621 = 1198622 = sdot sdot sdot = 119862119901 = 0 119862119901+1 = 0 (12)

where 119862119901+1 is called as an error constant of the method

International Journal of Mathematics and Mathematical Sciences 3

The order of diagonally implicit multistep block methodin (7) can be determined by applying the formula in (11)hence the values of 120572 and 120573 are obtained as follows

1205720 = [00]

1205721 = [minus1minus1]

1205722 = [10]

1205723 = [01]

1205730 = [[[minus 11219]]]

1205731 = [[230]]

1205732 = [[[

51253]]]

1205733 = [[029]]

(13)

Substitute the values of 120572 and 120573 into (11) and obtain

1198620 =3sum119895=0

10 1198950120572119895 = [

00]

1198621 =3sum119895=0

11 1198951120572119895 minus

3sum119895=0

101198950120573119895 = [

00]

1198622 =3sum119895=0

12 1198952120572119895 minus

3sum119895=0

111198951120573119895 = [

00]

1198623 =3sum119895=0

13 1198953120572119895 minus

3sum119895=0

121198952120573119895 = [

00]

1198624 =3sum119895=0

14 1198954120572119895 minus

3sum119895=0

131198953120573119895 = [[[

minus 12419]]]

(14)

Therefore the diagonally implicit multistep block method isthird-order where the coefficient of error constant is

119862119901+1 = 1198624 = [minus 12419]119879 = [0 0]119879 (15)

Definition 2 The local truncation error at 119909119899+119896 of themethodis defined to be expression 119871[119910(119909119899) ℎ] when 119910(119909) is thetheoretical solution of the initial value problem

119871 [119910 (119909119899) ℎ] =119896sum119895=0

[120572119895119910 (119909 + 119895ℎ) minus ℎ1205731198951199101015840 (119909 + 119895ℎ)] (16)

For the formula 119910119888119899+1119910119888119899+1 = 119910119899 + ℎ [ 512119865119899+1 +

23119865119899 minus

112119865119899minus1] (17)

Since 119865119899+1 = 1199101015840(119909119899+1) Taylor expansion will be applied to thederivatives 1199101015840(119909119899+1) and 1199101015840(119909119899minus1) where 119865119899minus1 = 1199101015840(119909119899minus1)

1199101015840 (119909119899+1) = 1199101015840 (119909119899) + ℎ11991010158401015840 (119909119899) + ℎ22 119910101584010158401015840 (119909119899)+ 119874 (ℎ3)

1199101015840 (119909119899minus1) = 1199101015840 (119909119899) minus ℎ11991010158401015840 (119909119899) + ℎ22 119910101584010158401015840 (119909119899)+ 119874 (ℎ3)

(18)

Then since 119865119899 = 1199101015840(119909119899) we have119910119899+1 = 119910119899 + ℎ1199101015840119899 + ℎ

2

2 11991010158401015840119899 + ℎ3

6 119910101584010158401015840119899 + 119874 (ℎ4) (19)

So the local truncation error for 119910119899+1 is 119874(ℎ4) For theformula 119910119888119899+2

119910119888119899+2 = 119910119899 + ℎ [29119865119899+2 +53119865119899+1 +

19119865119899minus1] (20)

The Taylor expansion for 1199101015840(119909119899+2) = 119865119899+2 is given as

1199101015840 (119909119899+2) = 1199101015840 (119909119899) + 2ℎ11991010158401015840 (119909119899) + (2ℎ)22 119910101584010158401015840 (119909119899)+ 119874 (ℎ3)

(21)

Then we will have

119910119899+2 = 119910119899 + 2ℎ1199101015840119899 + 2ℎ211991010158401015840119899 + 43ℎ3119910101584010158401015840119899 + 119874 (ℎ4) (22)

This shows that the local truncation error for 119910119899+2 is 119874(ℎ4)Definition 3 Thenumerical method is said to be consistent ifthe order of method is 119901 ge 1 and the method is consistent ifand only if

119896sum119895=0

120572119895 = 0119896sum119895=0

119895120572119895 =119896sum119895=0

120573119895(23)

4 International Journal of Mathematics and Mathematical Sciences

Proof (i) sum119896119895=0 120572119895 = 03sum119895=0

120572119895 = 1205720 + 1205721 + 1205722 + 1205723

= [00] + [minus1minus1] + [

10] + [

01] = [

00]

(24)

(ii) sum119896119895=0 119895 sdot 120572119895 = sum119896119895=0 1205731198953sum119895=0

119895 sdot 120572119895 = 0 sdot 1205720 + 1 sdot 1205721 + 2 sdot 1205722 + 3 sdot 1205723

= 0 [00] + 1 [minus1minus1] + 2 [

10] + 3 [

01] = [

12]

3sum119895=0

120573119895 = 1205730 + 1205731 + 1205732 + 1205733

= [[[minus 11219]]]+ [[230]]

+ [[[

51253]]]+ [[029]]= [12]

(25)

Therefore3sum119895=0

119895 sdot 120572119895 =3sum119895=0

120573119895 = [12] (26)

ByDefinitions 1 and 3 the diagonally implicit multistep blockmethod is consistent

Definition 4 A block method is said to be zero-stable if andonly if providing the roots of 119877119895 119895 = 1(1)119896 of the firstcharacteristic polynomial 120588(119877) specified as

120588 (119877) = det[[119896sum119895=0

119860(119894)119877(119896minus119894)]]= 0 (27)

satisfies |119877119895| le 1 and those roots with |119877119895| = 1Proof The values of 119860 can be obtained in (7)

120588 (119903) = det [1198771198600 minus 1198601] = det[119877[1 00 1] minus [

0 minus10 minus1]]

= det[119877 10 119877 + 1] = 119877 (119877 + 1)

(28)

The diagonally implicit multistep block method is zero stablesince |119877| le 1Theorem 5 The method is said to be convergent if and only ifthe method is consistent and zero-stable

Proof By Definitions 1 3 and 4 the diagonally implicitmultistep block method is convergent

32 Stability Region of the Method In this section thestability region of the diagonally implicit multistep blockmethod of order three is discussed for the numerical solutionof VIDEThe test equation for first-order VIDE of the secondkind [10] is

1199101015840 (119909) = 120585119910 (119909) + 120578int1199090119910 (119905) 119889119905 (29)

where 120585 and 120578 are real constants 120585 = 120582 + 120583 and 120578 = minus120582120583Therefore

1199101015840 (119909) = (120582 + 120583) 119910 (119909) minus 120582120583int1199090119910 (119905) 119889119905 (30)

Definition 6 The method is said to be 119860-stable if and only ifthe region of absolute stability contains at the quarter planeℎ120585 lt 0 ℎ2120578 lt 0

From the proposed method for the numerical solutionthe characteristics polynomials 120588(119903) 120590(119903) 120588(119903) and (119903) canbe developed as follows

Corrector formula for 119910119899+1 is120588 (119903) = 1199032 minus 119903120590 (119903) = 5

121199032 + 23119903 minus 112 (31)

Corrector formula for 119910119899+2 is120588 (119903) = 1199033 minus 119903120590 (119903) = 2

91199033 + 531199032 + 19 (32)

Simpsonrsquos rule is

120588 (119903) = 1199032 minus 1 (119903) = 131199032 + 43119903 + 13

(33)

The stability polynomial of the diagonally implicit multistepblock method can be determined by substituting (31) (32)and (33) into this particular formula

120587 (119903 ℎ120585 ℎ2120578) = 120588 (119903) [120588 (119903) minus 1198671120590 (119903)]minus 1198672 (119903) 120590 (119903)

(34)

where1198671 = ℎ120585 and1198672 = ℎ2120583 Thus the stability polynomialof the proposed method is obtained

International Journal of Mathematics and Mathematical Sciences 5

minus15H1

H2

minus05

minus1

minus15

minus2

minus25

0minus05minus1

Figure 2 Stability region in11986711198672 plane

minus 7361198671 + 71081198672 minus 22711986721 minus 224311986722 + 554119867211199034

minus 7154119867211199033 +4118119867211199032 minus

535411986721119903 +

3211986711199032

minus 791198671119903 minus233611986711199034 +

1911986711199033 +

48111986711198672

+ 5486119867221199034 minus 131486119867221199033 + 9554119867221199032 minus 548611986722119903minus 2310811986721199034 minus 20554 11986721199033 + 11627 11986721199032 minus 19541198672119903minus 6181119867111986721199033 +

618111986711198672119903 minus

19119867111986721199032

+ 581119867111986721199034 + 1199034 minus 31199033 + 31199032 minus 119903 = 0

(35)

The region of stability polynomial can be illustrated inFigure 2

Regarding Definition 6 the third-order of diagonallyimplicitmultistep blockmethod in Figure 2 is119860-stablewithinthe shaded region

4 Implementation

The one-step methods are required for finding the first start-ing point at 119909119899+1 since Volterra integrodifferential equationsof the second kind have two types of kernels For119870(119909 119904) = 1Runge-Kuttamethod is involved in solving differential part ofVIDEwhilemidpointmethod is needed to solveVIDEwhen119870(119909 119904) = 1 Hence the predictor and the corrector formulacan be implemented until the end of the interval

119910119901119899+1 = 119910119899 + ℎ [32119865119899 minus12119865119899minus1]

119910119901119899+2 = 119910119899 + ℎ [4119865119899 minus 2119865119899minus1]

119910119888119899+1 = 119910119899 + ℎ [ 512119865119899+1 +23119865119899 minus

112119865119899minus1]

119910119888119899+2 = 119910119899 + ℎ [29119865119899+2 +53119865119899+1 +

19119865119899minus1]

(36)

Since 119911(119909) in 119865(119909 119910(119909) 119911(119909)) is the integral term in VIDEand cannot be solved explicitly therefore Simpsonrsquos rule isadapted for solving the integral part

(i) For 119870(119909 119904) = 1 Simpsonrsquos 13 rule is applied to solvethe integral term in VIDE

119911119899+2 = 119911119899 + ℎ [13119910119899 +43119910119899+1 +

13119910119899+2] (37)

(ii) For119870(119909 119904) = 1 composite Simpsonrsquos rule is employedfor solving the integral part

119911119899+2 = ℎ3119899+2sum119894=0

120596119904119894119870(119909119899+2 119909119894 119910119894)

119911119899+3 = ℎ3119899+2sum119894=0

120596119904119894119870(119909119899+3 119909119894 119910119894)

+ ℎ6 [119870 (119909119899+3 119909119899+2 119910119899+2)+ 4119870 (119909119899+2 119909119899+52 119910119899+52) + 119870 (119909119899+3 119909119899+3 119910119899+3)]

(38)

where 120596119904119894 are Simpsonrsquos rule weights 1 4 2 4 2 4 1 Theunknown value of 119910119899+52 in (38) can be estimated by usingLagrange interpolation at the point 119909119899 119909119899+1 119909119899+2 119909119899+3

119910119899+52 = 116119910119899 minus 516119910119899+1 + 1516119910119899+2 + 516119910119899+3 (39)

41 Algorithm of the Method The input of the programmingis the endpoints of 119886 and 119887 and the integer119873 The developedalgorithm for the method is given as follows

Step 1 Set

1199090 = 1198861199100 = 1205721199110 = 0ℎ = (119887 minus 119886)119873OUTPUT (1199090 1199100 1199110)

Step 2 For 119894 = 1When 119870(119909 119904) = 1 using RK3 to evaluate the value of119910When 119870(119909 119904) = 1 applying Midpoint Method

Step 3 For 119894 = 2 (1198732) do Steps 4ndash6Step 4 Set 119909 = 119886 + 119894ℎ

6 International Journal of Mathematics and Mathematical Sciences

Table 1 Numerical results for Example 1

ℎ 0025 00125 000625 0003125MAXE

RK3 83229(minus08) 10910(minus08) 13953(minus09) 17637(minus10)ABM3 19468(minus06) 24548(minus07) 30813(minus08) 38593(minus09)2PMBM 42079(minus07) 49165(minus08) 59272(minus09) 72715(minus10)DIMBM 41354(minus07) 47815(minus08) 58390(minus09) 72152(minus10)

TFCRK3 120 240 480 960ABM3 82 162 322 6422PMBM 43 83 163 323DIMBM 42 82 162 322

TSRK3 40 80 160 320ABM3 40 80 160 3202PMBM 21 41 81 161DIMBM 21 41 81 161

TimeRK3 01035 01840 02970 04527ABM3 00860 01410 02264 034542PMBM 00670 01200 02024 03204DIMBM 00450 01120 01600 02660

Step 5 Calculate for 119910119901119899+1 and 119911119901119899+1 119910119901119899+2 and 119911119901119899+2Step 6 Compute the solution for119910119888119899+1 and 119911119888119899+1119910119888119899+2 and 119911119888119899+2Step 7 Calculate the error

Step 8 OUTPUT (119909 119910 119911) and the absolute error

Step 9 STOP

5 Numerical Results

Four tested problems of first-order Volterra integrodifferen-tial equations were considered in order to study the perfor-mance of the diagonally implicit multistep block method

Example 1 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = 1 minus int119909

0119910 (119904) 119889119904 119910 (0) = 0 0 le 119909 le 1 (40)

Exact solution is 119910(119909) = sin(119909)Source [4]

Example 2 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = minus sin (119909) minus cos (119909) + int119909

02 cos (119909 minus 119904) 119910 (119904) 119889119904119910 (0) = 1 0 le 119909 le 5

(41)

Exact solution is 119910(119909) = exp(minus119909)Source [11]

Example 3 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 119909 exp (1 minus 119910 (119909)) minus 1

(1 + 119909)2 minus 119909minus int1199090

119909(1 + 119904)2 exp (1 minus 119910 (119904)) 119889119904

119910 (0) = 1 0 le 119909 le 4(42)

Exact solution is 119910(119909) = 1(1 + 119909)Source [12]

Example 4 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 2119909 minus 12 sin (1199094) + int

119909

01199092119904 cos (1199092119910 (119904)) 119889119904119910 (0) = 0 0 le 119909 le 2

(43)

Exact solution is 119910(119909) = 1199092Source [13]

Notations used in Tables 1ndash4 are as follows

ℎ step sizeMAXE maximum error

International Journal of Mathematics and Mathematical Sciences 7

Table 2 Numerical results for Example 2

ℎ 025 0125 00625 003125MAXE

BVMs 21607(minus01) 28411(minus02) 36378(minus03) 46011(minus04)ABM3 15401(minus02) 31321(minus03) 53778(minus04) 79127(minus05)2PMBM 57923(minus02) 12209(minus03) 37500(minus04) 84172(minus05)DIMBM 18211(minus02) 34489(minus03) 46803(minus04) 60807(minus05)

TFCBVMs - - - -ABM3 84 164 324 6442PMBM 50 90 170 330DIMBM 22 42 82 162

TSBVMs - - - -ABM3 20 40 80 1602PMBM 11 21 41 81DIMBM 11 21 41 81

TimeBVMs - - - -ABM3 00715 01546 02433 035462PMBM 00460 00780 01710 02738DIMBM 00140 00410 01170 01480

Table 3 Numerical results for Example 3

ℎ 0025 00125 000625 0003125MAXE

ABM3 23797(minus06) 32252(minus07) 42061(minus08) 53727(minus09)2PMBM 80020(minus06) 97319(minus07) 12000(minus07) 14862(minus08)DIMBM 36545(minus06) 46274(minus07) 58216(minus08) 72999(minus09)

TFCABM3 644 1284 2564 51242PMBM 330 650 1290 2570DIMBM 322 642 1282 2562

TimeABM3 03543 05677 11650 198552PMBM 02810 04210 07496 13570DIMBM 02610 03010 06740 11050

TS total steps

TFC total functions call

Time the execution time taken

RK3 Runge-Kutta method of order 3 with Simpsonrsquos13 rule by Filiz [4]ABM3 AdamBashforthMoultonOrder 3 with Simp-sonrsquos rule

BVMs Combination of BVMs and third-order Gen-eralized Adams Method by Chen and Zhang [11]

2PMBM Two points Multistep Block Method ofOrder 3 with Simpsonrsquos rule by Mohamed and Majid[8]DIMBM Diagonally implicit multistep blockmethodwith Simpsonrsquos rule proposed in this paper

Tables 1ndash4 display the numerical results for the four testedproblems when solved using the proposed block method andthe code was written in C language

The numerical results for Examples 1ndash4 displayed inTables 1ndash4 are solved numerically using the proposed numer-ical method with Simpsonrsquos rule In Table 1 the numerical

8 International Journal of Mathematics and Mathematical Sciences

Table 4 Numerical results for Example 4

ℎ 29 217 233 265MAXE

ABM3 50218(minus02) 18761(minus03) 11046(minus04) 58996(minus06)2PMBM 27425(minus02) 65258(minus04) 17781(minus04) 66888(minus06)DIMBM 88008(minus03) 86068(minus04) 17703(minus04) 66994(minus06)

TFCABM3 40 72 136 2462PMBM 22 38 70 134DIMBM 10 18 34 66

TimeABM3 00670 00723 01291 020302PMBM 00140 00352 00662 01336DIMBM 00050 00280 00350 00420

TFC900800700600500400300200100

2PMBMDIMBM

ABM3

RK3

minus9

minus8

minus7

minus6

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus9

minus8

minus7

minus6

02 03 0401

Time

ABM3

RK3

DIMBM2PMBM

MAXE

(ＦＩＡ1

0)

(b) Graph of Time versus MAXE

Figure 3 Graph of the numerical results when solving Example 1

results will be obtained when the step size ℎ = 002500125 000625 and 0003125 for the case when 119870(119909 119904) = 1The maximum error of DIMBM is comparable compared to2PMBMat all tested ℎ but the order of accuracy is the same orone order less compared to RK3 andABM3The performanceof DIMBM is better in terms of total functions call and totalnumber of steps compared to RK3 and ABM3

In Tables 2 3 and 4 the numerical results are solved usingthe proposed numerical method via composite Simpsonrsquos forthe integral part when 119870(119909 119904) = 1 In Table 2 the maximumerror of DIMBM is comparable compared to 2PMBM and

ABM3 The DIMBM manage to obtain less total functionscall compared to ABM3 and 2PMBM For the nonlinearExamples 3 and 4 we could observe that ABM3 and 2PMBMare expensive in terms of total functions call respectivelyFigures 3ndash6 display the numerical results of maximum errorversus total functions call when solving the tested problemsThis has shown the advantage of DIMBM in the form ofa standard multistep method because the cost per step ischeaper and the numerical results aremore accuratewhen thestep size is reduced In terms of timing DIMBM gave fasterresults compared to ABM3 and 2PMBM

International Journal of Mathematics and Mathematical Sciences 9

TFC600500400300200100

minus4

minus35

minus3

minus25

minus2

minus15

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

Time035030025020015010005

DIMBM2PMBM

minus4

minus35

minus3

minus25

minus2

minus15

ABM3MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 4 Graph of the numerical results when solving Example 2

TFC50004000300020001000

minus8

minus75

minus7

minus65

minus6

minus55

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

MAXE

(ＦＩＡ1

0)

DIMBM2PMBMABM3

06 08 10 12 14 1604 18

Time

minus8

minus75

minus7

minus65

minus6

minus55

(b) Graph of Time versus MAXE

Figure 5 Graph of the numerical results when solving Example 3

6 Conclusion

In this research we proposed the diagonally implicit multi-step block method for solving linear and nonlinear Volterra

integrodifferential equations and comparisons were madewith the existingmethod Comparisons with existingmethodreveal that the diagonally implicit multistep block method ismore efficient and cheaper

10 International Journal of Mathematics and Mathematical Sciences

TFC200150100150

DIMBM2PMBMABM3

minus5

minus4

minus3

minus2

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus5

minus4

minus3

minus2

DIMBM2PMBMABM3

008 018 020010 012 014 016004 006002

Time

MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 6 Graph of the numerical results when solving Example 4

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledged the financial supportof Fundamental Research Grant Scheme (FRGS5524973)and Graduate Research Fund (GRF) from Universiti PutraMalaysia

References

[1] J T Day ldquoNote on the Numerical Solution of Integro-Differential EquationsrdquoThe Computer Journal vol 9 no 4 pp394-395 1967

[2] R Saadati B Raftari H Adibi S M Vaezpour and S ShakerildquoA comparison between the Variational Iteration method andTrapezoidal rule for solving linear integro-differential equa-tionsrdquoWorld Applied Sciences Journal vol 4 no 3 pp 321ndash3252008

[3] B Raftari ldquoNumerical solutions of the linear volterra integro-differential equations homotopy perturbation method andfinite Difference methodrdquo World Applied Sciences Journal vol9 pp 7ndash12 2010

[4] A Filiz ldquoFourth-Order Robust Numerical Method for Integro-differential Equationsrdquo Asian Journal of Fuzzy and AppliedMathematics vol 1 no 1 pp 28ndash33 2013

[5] A Filiz ldquoNumerical Solution of a Non-linear Volterra Integro-differential Equation via Runge-Kutta-Verner Methodrdquo Inter-national Journal of Scientific and Research Publications vol 3article 9 2013

[6] F Ishak and S N Ahmad ldquoDevelopment of Extended Trape-zoidal Method for Numerical Solution of Volterra Integro-Differential Equationsrdquo International Journal of MathematicsComputational Physical Electrical and Computer Engineeringvol 10 no 11 article 52856 2016

[7] N A BMohamed and Z AMajid ldquoOne-step blockmethod forsolving Volterra integro-differential equationsrdquo AIP ConferenceProceedings vol 1682 no 1 Article ID 020018 2015

[8] N A Mohamed and Z A Majid ldquoMultistep block methodfor solving volterra integro-differential equationsrdquo MalaysianJournal of Mathematical Sciences vol 10 pp 33ndash48 2016

[9] J D Lambert Computational Methods in Ordinary DifferentialEquations John Wiley amp Sons New York NY USA 1973

[10] H Brunner and J D Lambert ldquoStability of numerical methodsfor volterra integro-differential equationsrdquo Computing vol 12no 1 pp 75ndash89 1974

[11] H Chen and C Zhang ldquoBoundary value methods for Volterraintegral and integro-differential equationsrdquo Applied Mathemat-ics and Computation vol 218 no 6 pp 2619ndash2630 2011

[12] R E Shaw ldquoA parallel algorithm for nonlinear volterra integro-differential equationsrdquo in Proceedings of the 2000 ACM Sympo-sium on Applied Computing SAC 2000 pp 86ndash88 Como ItalyMarch 2000

[13] 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

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

International Journal of Mathematics and Mathematical Sciences 3

The order of diagonally implicit multistep block methodin (7) can be determined by applying the formula in (11)hence the values of 120572 and 120573 are obtained as follows

1205720 = [00]

1205721 = [minus1minus1]

1205722 = [10]

1205723 = [01]

1205730 = [[[minus 11219]]]

1205731 = [[230]]

1205732 = [[[

51253]]]

1205733 = [[029]]

(13)

Substitute the values of 120572 and 120573 into (11) and obtain

1198620 =3sum119895=0

10 1198950120572119895 = [

00]

1198621 =3sum119895=0

11 1198951120572119895 minus

3sum119895=0

101198950120573119895 = [

00]

1198622 =3sum119895=0

12 1198952120572119895 minus

3sum119895=0

111198951120573119895 = [

00]

1198623 =3sum119895=0

13 1198953120572119895 minus

3sum119895=0

121198952120573119895 = [

00]

1198624 =3sum119895=0

14 1198954120572119895 minus

3sum119895=0

131198953120573119895 = [[[

minus 12419]]]

(14)

Therefore the diagonally implicit multistep block method isthird-order where the coefficient of error constant is

119862119901+1 = 1198624 = [minus 12419]119879 = [0 0]119879 (15)

Definition 2 The local truncation error at 119909119899+119896 of themethodis defined to be expression 119871[119910(119909119899) ℎ] when 119910(119909) is thetheoretical solution of the initial value problem

119871 [119910 (119909119899) ℎ] =119896sum119895=0

[120572119895119910 (119909 + 119895ℎ) minus ℎ1205731198951199101015840 (119909 + 119895ℎ)] (16)

For the formula 119910119888119899+1119910119888119899+1 = 119910119899 + ℎ [ 512119865119899+1 +

23119865119899 minus

112119865119899minus1] (17)

Since 119865119899+1 = 1199101015840(119909119899+1) Taylor expansion will be applied to thederivatives 1199101015840(119909119899+1) and 1199101015840(119909119899minus1) where 119865119899minus1 = 1199101015840(119909119899minus1)

1199101015840 (119909119899+1) = 1199101015840 (119909119899) + ℎ11991010158401015840 (119909119899) + ℎ22 119910101584010158401015840 (119909119899)+ 119874 (ℎ3)

1199101015840 (119909119899minus1) = 1199101015840 (119909119899) minus ℎ11991010158401015840 (119909119899) + ℎ22 119910101584010158401015840 (119909119899)+ 119874 (ℎ3)

(18)

Then since 119865119899 = 1199101015840(119909119899) we have119910119899+1 = 119910119899 + ℎ1199101015840119899 + ℎ

2

2 11991010158401015840119899 + ℎ3

6 119910101584010158401015840119899 + 119874 (ℎ4) (19)

So the local truncation error for 119910119899+1 is 119874(ℎ4) For theformula 119910119888119899+2

119910119888119899+2 = 119910119899 + ℎ [29119865119899+2 +53119865119899+1 +

19119865119899minus1] (20)

The Taylor expansion for 1199101015840(119909119899+2) = 119865119899+2 is given as

1199101015840 (119909119899+2) = 1199101015840 (119909119899) + 2ℎ11991010158401015840 (119909119899) + (2ℎ)22 119910101584010158401015840 (119909119899)+ 119874 (ℎ3)

(21)

Then we will have

119910119899+2 = 119910119899 + 2ℎ1199101015840119899 + 2ℎ211991010158401015840119899 + 43ℎ3119910101584010158401015840119899 + 119874 (ℎ4) (22)

This shows that the local truncation error for 119910119899+2 is 119874(ℎ4)Definition 3 Thenumerical method is said to be consistent ifthe order of method is 119901 ge 1 and the method is consistent ifand only if

119896sum119895=0

120572119895 = 0119896sum119895=0

119895120572119895 =119896sum119895=0

120573119895(23)

4 International Journal of Mathematics and Mathematical Sciences

Proof (i) sum119896119895=0 120572119895 = 03sum119895=0

120572119895 = 1205720 + 1205721 + 1205722 + 1205723

= [00] + [minus1minus1] + [

10] + [

01] = [

00]

(24)

(ii) sum119896119895=0 119895 sdot 120572119895 = sum119896119895=0 1205731198953sum119895=0

119895 sdot 120572119895 = 0 sdot 1205720 + 1 sdot 1205721 + 2 sdot 1205722 + 3 sdot 1205723

= 0 [00] + 1 [minus1minus1] + 2 [

10] + 3 [

01] = [

12]

3sum119895=0

120573119895 = 1205730 + 1205731 + 1205732 + 1205733

= [[[minus 11219]]]+ [[230]]

+ [[[

51253]]]+ [[029]]= [12]

(25)

Therefore3sum119895=0

119895 sdot 120572119895 =3sum119895=0

120573119895 = [12] (26)

ByDefinitions 1 and 3 the diagonally implicit multistep blockmethod is consistent

Definition 4 A block method is said to be zero-stable if andonly if providing the roots of 119877119895 119895 = 1(1)119896 of the firstcharacteristic polynomial 120588(119877) specified as

120588 (119877) = det[[119896sum119895=0

119860(119894)119877(119896minus119894)]]= 0 (27)

satisfies |119877119895| le 1 and those roots with |119877119895| = 1Proof The values of 119860 can be obtained in (7)

120588 (119903) = det [1198771198600 minus 1198601] = det[119877[1 00 1] minus [

0 minus10 minus1]]

= det[119877 10 119877 + 1] = 119877 (119877 + 1)

(28)

The diagonally implicit multistep block method is zero stablesince |119877| le 1Theorem 5 The method is said to be convergent if and only ifthe method is consistent and zero-stable

Proof By Definitions 1 3 and 4 the diagonally implicitmultistep block method is convergent

32 Stability Region of the Method In this section thestability region of the diagonally implicit multistep blockmethod of order three is discussed for the numerical solutionof VIDEThe test equation for first-order VIDE of the secondkind [10] is

1199101015840 (119909) = 120585119910 (119909) + 120578int1199090119910 (119905) 119889119905 (29)

where 120585 and 120578 are real constants 120585 = 120582 + 120583 and 120578 = minus120582120583Therefore

1199101015840 (119909) = (120582 + 120583) 119910 (119909) minus 120582120583int1199090119910 (119905) 119889119905 (30)

Definition 6 The method is said to be 119860-stable if and only ifthe region of absolute stability contains at the quarter planeℎ120585 lt 0 ℎ2120578 lt 0

From the proposed method for the numerical solutionthe characteristics polynomials 120588(119903) 120590(119903) 120588(119903) and (119903) canbe developed as follows

Corrector formula for 119910119899+1 is120588 (119903) = 1199032 minus 119903120590 (119903) = 5

121199032 + 23119903 minus 112 (31)

Corrector formula for 119910119899+2 is120588 (119903) = 1199033 minus 119903120590 (119903) = 2

91199033 + 531199032 + 19 (32)

Simpsonrsquos rule is

120588 (119903) = 1199032 minus 1 (119903) = 131199032 + 43119903 + 13

(33)

The stability polynomial of the diagonally implicit multistepblock method can be determined by substituting (31) (32)and (33) into this particular formula

120587 (119903 ℎ120585 ℎ2120578) = 120588 (119903) [120588 (119903) minus 1198671120590 (119903)]minus 1198672 (119903) 120590 (119903)

(34)

where1198671 = ℎ120585 and1198672 = ℎ2120583 Thus the stability polynomialof the proposed method is obtained

International Journal of Mathematics and Mathematical Sciences 5

minus15H1

H2

minus05

minus1

minus15

minus2

minus25

0minus05minus1

Figure 2 Stability region in11986711198672 plane

minus 7361198671 + 71081198672 minus 22711986721 minus 224311986722 + 554119867211199034

minus 7154119867211199033 +4118119867211199032 minus

535411986721119903 +

3211986711199032

minus 791198671119903 minus233611986711199034 +

1911986711199033 +

48111986711198672

+ 5486119867221199034 minus 131486119867221199033 + 9554119867221199032 minus 548611986722119903minus 2310811986721199034 minus 20554 11986721199033 + 11627 11986721199032 minus 19541198672119903minus 6181119867111986721199033 +

618111986711198672119903 minus

19119867111986721199032

+ 581119867111986721199034 + 1199034 minus 31199033 + 31199032 minus 119903 = 0

(35)

The region of stability polynomial can be illustrated inFigure 2

Regarding Definition 6 the third-order of diagonallyimplicitmultistep blockmethod in Figure 2 is119860-stablewithinthe shaded region

4 Implementation

The one-step methods are required for finding the first start-ing point at 119909119899+1 since Volterra integrodifferential equationsof the second kind have two types of kernels For119870(119909 119904) = 1Runge-Kuttamethod is involved in solving differential part ofVIDEwhilemidpointmethod is needed to solveVIDEwhen119870(119909 119904) = 1 Hence the predictor and the corrector formulacan be implemented until the end of the interval

119910119901119899+1 = 119910119899 + ℎ [32119865119899 minus12119865119899minus1]

119910119901119899+2 = 119910119899 + ℎ [4119865119899 minus 2119865119899minus1]

119910119888119899+1 = 119910119899 + ℎ [ 512119865119899+1 +23119865119899 minus

112119865119899minus1]

119910119888119899+2 = 119910119899 + ℎ [29119865119899+2 +53119865119899+1 +

19119865119899minus1]

(36)

Since 119911(119909) in 119865(119909 119910(119909) 119911(119909)) is the integral term in VIDEand cannot be solved explicitly therefore Simpsonrsquos rule isadapted for solving the integral part

(i) For 119870(119909 119904) = 1 Simpsonrsquos 13 rule is applied to solvethe integral term in VIDE

119911119899+2 = 119911119899 + ℎ [13119910119899 +43119910119899+1 +

13119910119899+2] (37)

(ii) For119870(119909 119904) = 1 composite Simpsonrsquos rule is employedfor solving the integral part

119911119899+2 = ℎ3119899+2sum119894=0

120596119904119894119870(119909119899+2 119909119894 119910119894)

119911119899+3 = ℎ3119899+2sum119894=0

120596119904119894119870(119909119899+3 119909119894 119910119894)

+ ℎ6 [119870 (119909119899+3 119909119899+2 119910119899+2)+ 4119870 (119909119899+2 119909119899+52 119910119899+52) + 119870 (119909119899+3 119909119899+3 119910119899+3)]

(38)

where 120596119904119894 are Simpsonrsquos rule weights 1 4 2 4 2 4 1 Theunknown value of 119910119899+52 in (38) can be estimated by usingLagrange interpolation at the point 119909119899 119909119899+1 119909119899+2 119909119899+3

119910119899+52 = 116119910119899 minus 516119910119899+1 + 1516119910119899+2 + 516119910119899+3 (39)

41 Algorithm of the Method The input of the programmingis the endpoints of 119886 and 119887 and the integer119873 The developedalgorithm for the method is given as follows

Step 1 Set

1199090 = 1198861199100 = 1205721199110 = 0ℎ = (119887 minus 119886)119873OUTPUT (1199090 1199100 1199110)

Step 2 For 119894 = 1When 119870(119909 119904) = 1 using RK3 to evaluate the value of119910When 119870(119909 119904) = 1 applying Midpoint Method

Step 3 For 119894 = 2 (1198732) do Steps 4ndash6Step 4 Set 119909 = 119886 + 119894ℎ

6 International Journal of Mathematics and Mathematical Sciences

Table 1 Numerical results for Example 1

ℎ 0025 00125 000625 0003125MAXE

RK3 83229(minus08) 10910(minus08) 13953(minus09) 17637(minus10)ABM3 19468(minus06) 24548(minus07) 30813(minus08) 38593(minus09)2PMBM 42079(minus07) 49165(minus08) 59272(minus09) 72715(minus10)DIMBM 41354(minus07) 47815(minus08) 58390(minus09) 72152(minus10)

TFCRK3 120 240 480 960ABM3 82 162 322 6422PMBM 43 83 163 323DIMBM 42 82 162 322

TSRK3 40 80 160 320ABM3 40 80 160 3202PMBM 21 41 81 161DIMBM 21 41 81 161

TimeRK3 01035 01840 02970 04527ABM3 00860 01410 02264 034542PMBM 00670 01200 02024 03204DIMBM 00450 01120 01600 02660

Step 5 Calculate for 119910119901119899+1 and 119911119901119899+1 119910119901119899+2 and 119911119901119899+2Step 6 Compute the solution for119910119888119899+1 and 119911119888119899+1119910119888119899+2 and 119911119888119899+2Step 7 Calculate the error

Step 8 OUTPUT (119909 119910 119911) and the absolute error

Step 9 STOP

5 Numerical Results

Four tested problems of first-order Volterra integrodifferen-tial equations were considered in order to study the perfor-mance of the diagonally implicit multistep block method

Example 1 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = 1 minus int119909

0119910 (119904) 119889119904 119910 (0) = 0 0 le 119909 le 1 (40)

Exact solution is 119910(119909) = sin(119909)Source [4]

Example 2 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = minus sin (119909) minus cos (119909) + int119909

02 cos (119909 minus 119904) 119910 (119904) 119889119904119910 (0) = 1 0 le 119909 le 5

(41)

Exact solution is 119910(119909) = exp(minus119909)Source [11]

Example 3 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 119909 exp (1 minus 119910 (119909)) minus 1

(1 + 119909)2 minus 119909minus int1199090

119909(1 + 119904)2 exp (1 minus 119910 (119904)) 119889119904

119910 (0) = 1 0 le 119909 le 4(42)

Exact solution is 119910(119909) = 1(1 + 119909)Source [12]

Example 4 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 2119909 minus 12 sin (1199094) + int

119909

01199092119904 cos (1199092119910 (119904)) 119889119904119910 (0) = 0 0 le 119909 le 2

(43)

Exact solution is 119910(119909) = 1199092Source [13]

Notations used in Tables 1ndash4 are as follows

ℎ step sizeMAXE maximum error

International Journal of Mathematics and Mathematical Sciences 7

Table 2 Numerical results for Example 2

ℎ 025 0125 00625 003125MAXE

BVMs 21607(minus01) 28411(minus02) 36378(minus03) 46011(minus04)ABM3 15401(minus02) 31321(minus03) 53778(minus04) 79127(minus05)2PMBM 57923(minus02) 12209(minus03) 37500(minus04) 84172(minus05)DIMBM 18211(minus02) 34489(minus03) 46803(minus04) 60807(minus05)

TFCBVMs - - - -ABM3 84 164 324 6442PMBM 50 90 170 330DIMBM 22 42 82 162

TSBVMs - - - -ABM3 20 40 80 1602PMBM 11 21 41 81DIMBM 11 21 41 81

TimeBVMs - - - -ABM3 00715 01546 02433 035462PMBM 00460 00780 01710 02738DIMBM 00140 00410 01170 01480

Table 3 Numerical results for Example 3

ℎ 0025 00125 000625 0003125MAXE

ABM3 23797(minus06) 32252(minus07) 42061(minus08) 53727(minus09)2PMBM 80020(minus06) 97319(minus07) 12000(minus07) 14862(minus08)DIMBM 36545(minus06) 46274(minus07) 58216(minus08) 72999(minus09)

TFCABM3 644 1284 2564 51242PMBM 330 650 1290 2570DIMBM 322 642 1282 2562

TimeABM3 03543 05677 11650 198552PMBM 02810 04210 07496 13570DIMBM 02610 03010 06740 11050

TS total steps

TFC total functions call

Time the execution time taken

RK3 Runge-Kutta method of order 3 with Simpsonrsquos13 rule by Filiz [4]ABM3 AdamBashforthMoultonOrder 3 with Simp-sonrsquos rule

BVMs Combination of BVMs and third-order Gen-eralized Adams Method by Chen and Zhang [11]

2PMBM Two points Multistep Block Method ofOrder 3 with Simpsonrsquos rule by Mohamed and Majid[8]DIMBM Diagonally implicit multistep blockmethodwith Simpsonrsquos rule proposed in this paper

Tables 1ndash4 display the numerical results for the four testedproblems when solved using the proposed block method andthe code was written in C language

The numerical results for Examples 1ndash4 displayed inTables 1ndash4 are solved numerically using the proposed numer-ical method with Simpsonrsquos rule In Table 1 the numerical

8 International Journal of Mathematics and Mathematical Sciences

Table 4 Numerical results for Example 4

ℎ 29 217 233 265MAXE

ABM3 50218(minus02) 18761(minus03) 11046(minus04) 58996(minus06)2PMBM 27425(minus02) 65258(minus04) 17781(minus04) 66888(minus06)DIMBM 88008(minus03) 86068(minus04) 17703(minus04) 66994(minus06)

TFCABM3 40 72 136 2462PMBM 22 38 70 134DIMBM 10 18 34 66

TimeABM3 00670 00723 01291 020302PMBM 00140 00352 00662 01336DIMBM 00050 00280 00350 00420

TFC900800700600500400300200100

2PMBMDIMBM

ABM3

RK3

minus9

minus8

minus7

minus6

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus9

minus8

minus7

minus6

02 03 0401

Time

ABM3

RK3

DIMBM2PMBM

MAXE

(ＦＩＡ1

0)

(b) Graph of Time versus MAXE

Figure 3 Graph of the numerical results when solving Example 1

results will be obtained when the step size ℎ = 002500125 000625 and 0003125 for the case when 119870(119909 119904) = 1The maximum error of DIMBM is comparable compared to2PMBMat all tested ℎ but the order of accuracy is the same orone order less compared to RK3 andABM3The performanceof DIMBM is better in terms of total functions call and totalnumber of steps compared to RK3 and ABM3

In Tables 2 3 and 4 the numerical results are solved usingthe proposed numerical method via composite Simpsonrsquos forthe integral part when 119870(119909 119904) = 1 In Table 2 the maximumerror of DIMBM is comparable compared to 2PMBM and

ABM3 The DIMBM manage to obtain less total functionscall compared to ABM3 and 2PMBM For the nonlinearExamples 3 and 4 we could observe that ABM3 and 2PMBMare expensive in terms of total functions call respectivelyFigures 3ndash6 display the numerical results of maximum errorversus total functions call when solving the tested problemsThis has shown the advantage of DIMBM in the form ofa standard multistep method because the cost per step ischeaper and the numerical results aremore accuratewhen thestep size is reduced In terms of timing DIMBM gave fasterresults compared to ABM3 and 2PMBM

International Journal of Mathematics and Mathematical Sciences 9

TFC600500400300200100

minus4

minus35

minus3

minus25

minus2

minus15

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

Time035030025020015010005

DIMBM2PMBM

minus4

minus35

minus3

minus25

minus2

minus15

ABM3MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 4 Graph of the numerical results when solving Example 2

TFC50004000300020001000

minus8

minus75

minus7

minus65

minus6

minus55

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

MAXE

(ＦＩＡ1

0)

DIMBM2PMBMABM3

06 08 10 12 14 1604 18

Time

minus8

minus75

minus7

minus65

minus6

minus55

(b) Graph of Time versus MAXE

Figure 5 Graph of the numerical results when solving Example 3

6 Conclusion

In this research we proposed the diagonally implicit multi-step block method for solving linear and nonlinear Volterra

integrodifferential equations and comparisons were madewith the existingmethod Comparisons with existingmethodreveal that the diagonally implicit multistep block method ismore efficient and cheaper

10 International Journal of Mathematics and Mathematical Sciences

TFC200150100150

DIMBM2PMBMABM3

minus5

minus4

minus3

minus2

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus5

minus4

minus3

minus2

DIMBM2PMBMABM3

008 018 020010 012 014 016004 006002

Time

MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 6 Graph of the numerical results when solving Example 4

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledged the financial supportof Fundamental Research Grant Scheme (FRGS5524973)and Graduate Research Fund (GRF) from Universiti PutraMalaysia

References

[1] J T Day ldquoNote on the Numerical Solution of Integro-Differential EquationsrdquoThe Computer Journal vol 9 no 4 pp394-395 1967

[2] R Saadati B Raftari H Adibi S M Vaezpour and S ShakerildquoA comparison between the Variational Iteration method andTrapezoidal rule for solving linear integro-differential equa-tionsrdquoWorld Applied Sciences Journal vol 4 no 3 pp 321ndash3252008

[3] B Raftari ldquoNumerical solutions of the linear volterra integro-differential equations homotopy perturbation method andfinite Difference methodrdquo World Applied Sciences Journal vol9 pp 7ndash12 2010

[4] A Filiz ldquoFourth-Order Robust Numerical Method for Integro-differential Equationsrdquo Asian Journal of Fuzzy and AppliedMathematics vol 1 no 1 pp 28ndash33 2013

[5] A Filiz ldquoNumerical Solution of a Non-linear Volterra Integro-differential Equation via Runge-Kutta-Verner Methodrdquo Inter-national Journal of Scientific and Research Publications vol 3article 9 2013

[6] F Ishak and S N Ahmad ldquoDevelopment of Extended Trape-zoidal Method for Numerical Solution of Volterra Integro-Differential Equationsrdquo International Journal of MathematicsComputational Physical Electrical and Computer Engineeringvol 10 no 11 article 52856 2016

[7] N A BMohamed and Z AMajid ldquoOne-step blockmethod forsolving Volterra integro-differential equationsrdquo AIP ConferenceProceedings vol 1682 no 1 Article ID 020018 2015

[8] N A Mohamed and Z A Majid ldquoMultistep block methodfor solving volterra integro-differential equationsrdquo MalaysianJournal of Mathematical Sciences vol 10 pp 33ndash48 2016

[9] J D Lambert Computational Methods in Ordinary DifferentialEquations John Wiley amp Sons New York NY USA 1973

[10] H Brunner and J D Lambert ldquoStability of numerical methodsfor volterra integro-differential equationsrdquo Computing vol 12no 1 pp 75ndash89 1974

[11] H Chen and C Zhang ldquoBoundary value methods for Volterraintegral and integro-differential equationsrdquo Applied Mathemat-ics and Computation vol 218 no 6 pp 2619ndash2630 2011

[12] R E Shaw ldquoA parallel algorithm for nonlinear volterra integro-differential equationsrdquo in Proceedings of the 2000 ACM Sympo-sium on Applied Computing SAC 2000 pp 86ndash88 Como ItalyMarch 2000

[13] 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

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

4 International Journal of Mathematics and Mathematical Sciences

Proof (i) sum119896119895=0 120572119895 = 03sum119895=0

120572119895 = 1205720 + 1205721 + 1205722 + 1205723

= [00] + [minus1minus1] + [

10] + [

01] = [

00]

(24)

(ii) sum119896119895=0 119895 sdot 120572119895 = sum119896119895=0 1205731198953sum119895=0

119895 sdot 120572119895 = 0 sdot 1205720 + 1 sdot 1205721 + 2 sdot 1205722 + 3 sdot 1205723

= 0 [00] + 1 [minus1minus1] + 2 [

10] + 3 [

01] = [

12]

3sum119895=0

120573119895 = 1205730 + 1205731 + 1205732 + 1205733

= [[[minus 11219]]]+ [[230]]

+ [[[

51253]]]+ [[029]]= [12]

(25)

Therefore3sum119895=0

119895 sdot 120572119895 =3sum119895=0

120573119895 = [12] (26)

ByDefinitions 1 and 3 the diagonally implicit multistep blockmethod is consistent

Definition 4 A block method is said to be zero-stable if andonly if providing the roots of 119877119895 119895 = 1(1)119896 of the firstcharacteristic polynomial 120588(119877) specified as

120588 (119877) = det[[119896sum119895=0

119860(119894)119877(119896minus119894)]]= 0 (27)

satisfies |119877119895| le 1 and those roots with |119877119895| = 1Proof The values of 119860 can be obtained in (7)

120588 (119903) = det [1198771198600 minus 1198601] = det[119877[1 00 1] minus [

0 minus10 minus1]]

= det[119877 10 119877 + 1] = 119877 (119877 + 1)

(28)

The diagonally implicit multistep block method is zero stablesince |119877| le 1Theorem 5 The method is said to be convergent if and only ifthe method is consistent and zero-stable

Proof By Definitions 1 3 and 4 the diagonally implicitmultistep block method is convergent

32 Stability Region of the Method In this section thestability region of the diagonally implicit multistep blockmethod of order three is discussed for the numerical solutionof VIDEThe test equation for first-order VIDE of the secondkind [10] is

1199101015840 (119909) = 120585119910 (119909) + 120578int1199090119910 (119905) 119889119905 (29)

where 120585 and 120578 are real constants 120585 = 120582 + 120583 and 120578 = minus120582120583Therefore

1199101015840 (119909) = (120582 + 120583) 119910 (119909) minus 120582120583int1199090119910 (119905) 119889119905 (30)

Definition 6 The method is said to be 119860-stable if and only ifthe region of absolute stability contains at the quarter planeℎ120585 lt 0 ℎ2120578 lt 0

From the proposed method for the numerical solutionthe characteristics polynomials 120588(119903) 120590(119903) 120588(119903) and (119903) canbe developed as follows

Corrector formula for 119910119899+1 is120588 (119903) = 1199032 minus 119903120590 (119903) = 5

121199032 + 23119903 minus 112 (31)

Corrector formula for 119910119899+2 is120588 (119903) = 1199033 minus 119903120590 (119903) = 2

91199033 + 531199032 + 19 (32)

Simpsonrsquos rule is

120588 (119903) = 1199032 minus 1 (119903) = 131199032 + 43119903 + 13

(33)

The stability polynomial of the diagonally implicit multistepblock method can be determined by substituting (31) (32)and (33) into this particular formula

120587 (119903 ℎ120585 ℎ2120578) = 120588 (119903) [120588 (119903) minus 1198671120590 (119903)]minus 1198672 (119903) 120590 (119903)

(34)

where1198671 = ℎ120585 and1198672 = ℎ2120583 Thus the stability polynomialof the proposed method is obtained

International Journal of Mathematics and Mathematical Sciences 5

minus15H1

H2

minus05

minus1

minus15

minus2

minus25

0minus05minus1

Figure 2 Stability region in11986711198672 plane

minus 7361198671 + 71081198672 minus 22711986721 minus 224311986722 + 554119867211199034

minus 7154119867211199033 +4118119867211199032 minus

535411986721119903 +

3211986711199032

minus 791198671119903 minus233611986711199034 +

1911986711199033 +

48111986711198672

+ 5486119867221199034 minus 131486119867221199033 + 9554119867221199032 minus 548611986722119903minus 2310811986721199034 minus 20554 11986721199033 + 11627 11986721199032 minus 19541198672119903minus 6181119867111986721199033 +

618111986711198672119903 minus

19119867111986721199032

+ 581119867111986721199034 + 1199034 minus 31199033 + 31199032 minus 119903 = 0

(35)

The region of stability polynomial can be illustrated inFigure 2

Regarding Definition 6 the third-order of diagonallyimplicitmultistep blockmethod in Figure 2 is119860-stablewithinthe shaded region

4 Implementation

The one-step methods are required for finding the first start-ing point at 119909119899+1 since Volterra integrodifferential equationsof the second kind have two types of kernels For119870(119909 119904) = 1Runge-Kuttamethod is involved in solving differential part ofVIDEwhilemidpointmethod is needed to solveVIDEwhen119870(119909 119904) = 1 Hence the predictor and the corrector formulacan be implemented until the end of the interval

119910119901119899+1 = 119910119899 + ℎ [32119865119899 minus12119865119899minus1]

119910119901119899+2 = 119910119899 + ℎ [4119865119899 minus 2119865119899minus1]

119910119888119899+1 = 119910119899 + ℎ [ 512119865119899+1 +23119865119899 minus

112119865119899minus1]

119910119888119899+2 = 119910119899 + ℎ [29119865119899+2 +53119865119899+1 +

19119865119899minus1]

(36)

Since 119911(119909) in 119865(119909 119910(119909) 119911(119909)) is the integral term in VIDEand cannot be solved explicitly therefore Simpsonrsquos rule isadapted for solving the integral part

(i) For 119870(119909 119904) = 1 Simpsonrsquos 13 rule is applied to solvethe integral term in VIDE

119911119899+2 = 119911119899 + ℎ [13119910119899 +43119910119899+1 +

13119910119899+2] (37)

(ii) For119870(119909 119904) = 1 composite Simpsonrsquos rule is employedfor solving the integral part

119911119899+2 = ℎ3119899+2sum119894=0

120596119904119894119870(119909119899+2 119909119894 119910119894)

119911119899+3 = ℎ3119899+2sum119894=0

120596119904119894119870(119909119899+3 119909119894 119910119894)

+ ℎ6 [119870 (119909119899+3 119909119899+2 119910119899+2)+ 4119870 (119909119899+2 119909119899+52 119910119899+52) + 119870 (119909119899+3 119909119899+3 119910119899+3)]

(38)

where 120596119904119894 are Simpsonrsquos rule weights 1 4 2 4 2 4 1 Theunknown value of 119910119899+52 in (38) can be estimated by usingLagrange interpolation at the point 119909119899 119909119899+1 119909119899+2 119909119899+3

119910119899+52 = 116119910119899 minus 516119910119899+1 + 1516119910119899+2 + 516119910119899+3 (39)

41 Algorithm of the Method The input of the programmingis the endpoints of 119886 and 119887 and the integer119873 The developedalgorithm for the method is given as follows

Step 1 Set

1199090 = 1198861199100 = 1205721199110 = 0ℎ = (119887 minus 119886)119873OUTPUT (1199090 1199100 1199110)

Step 2 For 119894 = 1When 119870(119909 119904) = 1 using RK3 to evaluate the value of119910When 119870(119909 119904) = 1 applying Midpoint Method

Step 3 For 119894 = 2 (1198732) do Steps 4ndash6Step 4 Set 119909 = 119886 + 119894ℎ

6 International Journal of Mathematics and Mathematical Sciences

Table 1 Numerical results for Example 1

ℎ 0025 00125 000625 0003125MAXE

RK3 83229(minus08) 10910(minus08) 13953(minus09) 17637(minus10)ABM3 19468(minus06) 24548(minus07) 30813(minus08) 38593(minus09)2PMBM 42079(minus07) 49165(minus08) 59272(minus09) 72715(minus10)DIMBM 41354(minus07) 47815(minus08) 58390(minus09) 72152(minus10)

TFCRK3 120 240 480 960ABM3 82 162 322 6422PMBM 43 83 163 323DIMBM 42 82 162 322

TSRK3 40 80 160 320ABM3 40 80 160 3202PMBM 21 41 81 161DIMBM 21 41 81 161

TimeRK3 01035 01840 02970 04527ABM3 00860 01410 02264 034542PMBM 00670 01200 02024 03204DIMBM 00450 01120 01600 02660

Step 5 Calculate for 119910119901119899+1 and 119911119901119899+1 119910119901119899+2 and 119911119901119899+2Step 6 Compute the solution for119910119888119899+1 and 119911119888119899+1119910119888119899+2 and 119911119888119899+2Step 7 Calculate the error

Step 8 OUTPUT (119909 119910 119911) and the absolute error

Step 9 STOP

5 Numerical Results

Four tested problems of first-order Volterra integrodifferen-tial equations were considered in order to study the perfor-mance of the diagonally implicit multistep block method

Example 1 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = 1 minus int119909

0119910 (119904) 119889119904 119910 (0) = 0 0 le 119909 le 1 (40)

Exact solution is 119910(119909) = sin(119909)Source [4]

Example 2 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = minus sin (119909) minus cos (119909) + int119909

02 cos (119909 minus 119904) 119910 (119904) 119889119904119910 (0) = 1 0 le 119909 le 5

(41)

Exact solution is 119910(119909) = exp(minus119909)Source [11]

Example 3 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 119909 exp (1 minus 119910 (119909)) minus 1

(1 + 119909)2 minus 119909minus int1199090

119909(1 + 119904)2 exp (1 minus 119910 (119904)) 119889119904

119910 (0) = 1 0 le 119909 le 4(42)

Exact solution is 119910(119909) = 1(1 + 119909)Source [12]

Example 4 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 2119909 minus 12 sin (1199094) + int

119909

01199092119904 cos (1199092119910 (119904)) 119889119904119910 (0) = 0 0 le 119909 le 2

(43)

Exact solution is 119910(119909) = 1199092Source [13]

Notations used in Tables 1ndash4 are as follows

ℎ step sizeMAXE maximum error

International Journal of Mathematics and Mathematical Sciences 7

Table 2 Numerical results for Example 2

ℎ 025 0125 00625 003125MAXE

BVMs 21607(minus01) 28411(minus02) 36378(minus03) 46011(minus04)ABM3 15401(minus02) 31321(minus03) 53778(minus04) 79127(minus05)2PMBM 57923(minus02) 12209(minus03) 37500(minus04) 84172(minus05)DIMBM 18211(minus02) 34489(minus03) 46803(minus04) 60807(minus05)

TFCBVMs - - - -ABM3 84 164 324 6442PMBM 50 90 170 330DIMBM 22 42 82 162

TSBVMs - - - -ABM3 20 40 80 1602PMBM 11 21 41 81DIMBM 11 21 41 81

TimeBVMs - - - -ABM3 00715 01546 02433 035462PMBM 00460 00780 01710 02738DIMBM 00140 00410 01170 01480

Table 3 Numerical results for Example 3

ℎ 0025 00125 000625 0003125MAXE

ABM3 23797(minus06) 32252(minus07) 42061(minus08) 53727(minus09)2PMBM 80020(minus06) 97319(minus07) 12000(minus07) 14862(minus08)DIMBM 36545(minus06) 46274(minus07) 58216(minus08) 72999(minus09)

TFCABM3 644 1284 2564 51242PMBM 330 650 1290 2570DIMBM 322 642 1282 2562

TimeABM3 03543 05677 11650 198552PMBM 02810 04210 07496 13570DIMBM 02610 03010 06740 11050

TS total steps

TFC total functions call

Time the execution time taken

RK3 Runge-Kutta method of order 3 with Simpsonrsquos13 rule by Filiz [4]ABM3 AdamBashforthMoultonOrder 3 with Simp-sonrsquos rule

BVMs Combination of BVMs and third-order Gen-eralized Adams Method by Chen and Zhang [11]

2PMBM Two points Multistep Block Method ofOrder 3 with Simpsonrsquos rule by Mohamed and Majid[8]DIMBM Diagonally implicit multistep blockmethodwith Simpsonrsquos rule proposed in this paper

Tables 1ndash4 display the numerical results for the four testedproblems when solved using the proposed block method andthe code was written in C language

The numerical results for Examples 1ndash4 displayed inTables 1ndash4 are solved numerically using the proposed numer-ical method with Simpsonrsquos rule In Table 1 the numerical

8 International Journal of Mathematics and Mathematical Sciences

Table 4 Numerical results for Example 4

ℎ 29 217 233 265MAXE

ABM3 50218(minus02) 18761(minus03) 11046(minus04) 58996(minus06)2PMBM 27425(minus02) 65258(minus04) 17781(minus04) 66888(minus06)DIMBM 88008(minus03) 86068(minus04) 17703(minus04) 66994(minus06)

TFCABM3 40 72 136 2462PMBM 22 38 70 134DIMBM 10 18 34 66

TimeABM3 00670 00723 01291 020302PMBM 00140 00352 00662 01336DIMBM 00050 00280 00350 00420

TFC900800700600500400300200100

2PMBMDIMBM

ABM3

RK3

minus9

minus8

minus7

minus6

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus9

minus8

minus7

minus6

02 03 0401

Time

ABM3

RK3

DIMBM2PMBM

MAXE

(ＦＩＡ1

0)

(b) Graph of Time versus MAXE

Figure 3 Graph of the numerical results when solving Example 1

results will be obtained when the step size ℎ = 002500125 000625 and 0003125 for the case when 119870(119909 119904) = 1The maximum error of DIMBM is comparable compared to2PMBMat all tested ℎ but the order of accuracy is the same orone order less compared to RK3 andABM3The performanceof DIMBM is better in terms of total functions call and totalnumber of steps compared to RK3 and ABM3

In Tables 2 3 and 4 the numerical results are solved usingthe proposed numerical method via composite Simpsonrsquos forthe integral part when 119870(119909 119904) = 1 In Table 2 the maximumerror of DIMBM is comparable compared to 2PMBM and

ABM3 The DIMBM manage to obtain less total functionscall compared to ABM3 and 2PMBM For the nonlinearExamples 3 and 4 we could observe that ABM3 and 2PMBMare expensive in terms of total functions call respectivelyFigures 3ndash6 display the numerical results of maximum errorversus total functions call when solving the tested problemsThis has shown the advantage of DIMBM in the form ofa standard multistep method because the cost per step ischeaper and the numerical results aremore accuratewhen thestep size is reduced In terms of timing DIMBM gave fasterresults compared to ABM3 and 2PMBM

International Journal of Mathematics and Mathematical Sciences 9

TFC600500400300200100

minus4

minus35

minus3

minus25

minus2

minus15

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

Time035030025020015010005

DIMBM2PMBM

minus4

minus35

minus3

minus25

minus2

minus15

ABM3MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 4 Graph of the numerical results when solving Example 2

TFC50004000300020001000

minus8

minus75

minus7

minus65

minus6

minus55

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

MAXE

(ＦＩＡ1

0)

DIMBM2PMBMABM3

06 08 10 12 14 1604 18

Time

minus8

minus75

minus7

minus65

minus6

minus55

(b) Graph of Time versus MAXE

Figure 5 Graph of the numerical results when solving Example 3

6 Conclusion

In this research we proposed the diagonally implicit multi-step block method for solving linear and nonlinear Volterra

integrodifferential equations and comparisons were madewith the existingmethod Comparisons with existingmethodreveal that the diagonally implicit multistep block method ismore efficient and cheaper

10 International Journal of Mathematics and Mathematical Sciences

TFC200150100150

DIMBM2PMBMABM3

minus5

minus4

minus3

minus2

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus5

minus4

minus3

minus2

DIMBM2PMBMABM3

008 018 020010 012 014 016004 006002

Time

MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 6 Graph of the numerical results when solving Example 4

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledged the financial supportof Fundamental Research Grant Scheme (FRGS5524973)and Graduate Research Fund (GRF) from Universiti PutraMalaysia

References

[1] J T Day ldquoNote on the Numerical Solution of Integro-Differential EquationsrdquoThe Computer Journal vol 9 no 4 pp394-395 1967

[2] R Saadati B Raftari H Adibi S M Vaezpour and S ShakerildquoA comparison between the Variational Iteration method andTrapezoidal rule for solving linear integro-differential equa-tionsrdquoWorld Applied Sciences Journal vol 4 no 3 pp 321ndash3252008

[3] B Raftari ldquoNumerical solutions of the linear volterra integro-differential equations homotopy perturbation method andfinite Difference methodrdquo World Applied Sciences Journal vol9 pp 7ndash12 2010

[4] A Filiz ldquoFourth-Order Robust Numerical Method for Integro-differential Equationsrdquo Asian Journal of Fuzzy and AppliedMathematics vol 1 no 1 pp 28ndash33 2013

[5] A Filiz ldquoNumerical Solution of a Non-linear Volterra Integro-differential Equation via Runge-Kutta-Verner Methodrdquo Inter-national Journal of Scientific and Research Publications vol 3article 9 2013

[6] F Ishak and S N Ahmad ldquoDevelopment of Extended Trape-zoidal Method for Numerical Solution of Volterra Integro-Differential Equationsrdquo International Journal of MathematicsComputational Physical Electrical and Computer Engineeringvol 10 no 11 article 52856 2016

[7] N A BMohamed and Z AMajid ldquoOne-step blockmethod forsolving Volterra integro-differential equationsrdquo AIP ConferenceProceedings vol 1682 no 1 Article ID 020018 2015

[8] N A Mohamed and Z A Majid ldquoMultistep block methodfor solving volterra integro-differential equationsrdquo MalaysianJournal of Mathematical Sciences vol 10 pp 33ndash48 2016

[9] J D Lambert Computational Methods in Ordinary DifferentialEquations John Wiley amp Sons New York NY USA 1973

[10] H Brunner and J D Lambert ldquoStability of numerical methodsfor volterra integro-differential equationsrdquo Computing vol 12no 1 pp 75ndash89 1974

[11] H Chen and C Zhang ldquoBoundary value methods for Volterraintegral and integro-differential equationsrdquo Applied Mathemat-ics and Computation vol 218 no 6 pp 2619ndash2630 2011

[12] R E Shaw ldquoA parallel algorithm for nonlinear volterra integro-differential equationsrdquo in Proceedings of the 2000 ACM Sympo-sium on Applied Computing SAC 2000 pp 86ndash88 Como ItalyMarch 2000

[13] 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

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

International Journal of Mathematics and Mathematical Sciences 5

minus15H1

H2

minus05

minus1

minus15

minus2

minus25

0minus05minus1

Figure 2 Stability region in11986711198672 plane

minus 7361198671 + 71081198672 minus 22711986721 minus 224311986722 + 554119867211199034

minus 7154119867211199033 +4118119867211199032 minus

535411986721119903 +

3211986711199032

minus 791198671119903 minus233611986711199034 +

1911986711199033 +

48111986711198672

+ 5486119867221199034 minus 131486119867221199033 + 9554119867221199032 minus 548611986722119903minus 2310811986721199034 minus 20554 11986721199033 + 11627 11986721199032 minus 19541198672119903minus 6181119867111986721199033 +

618111986711198672119903 minus

19119867111986721199032

+ 581119867111986721199034 + 1199034 minus 31199033 + 31199032 minus 119903 = 0

(35)

The region of stability polynomial can be illustrated inFigure 2

Regarding Definition 6 the third-order of diagonallyimplicitmultistep blockmethod in Figure 2 is119860-stablewithinthe shaded region

4 Implementation

The one-step methods are required for finding the first start-ing point at 119909119899+1 since Volterra integrodifferential equationsof the second kind have two types of kernels For119870(119909 119904) = 1Runge-Kuttamethod is involved in solving differential part ofVIDEwhilemidpointmethod is needed to solveVIDEwhen119870(119909 119904) = 1 Hence the predictor and the corrector formulacan be implemented until the end of the interval

119910119901119899+1 = 119910119899 + ℎ [32119865119899 minus12119865119899minus1]

119910119901119899+2 = 119910119899 + ℎ [4119865119899 minus 2119865119899minus1]

119910119888119899+1 = 119910119899 + ℎ [ 512119865119899+1 +23119865119899 minus

112119865119899minus1]

119910119888119899+2 = 119910119899 + ℎ [29119865119899+2 +53119865119899+1 +

19119865119899minus1]

(36)

Since 119911(119909) in 119865(119909 119910(119909) 119911(119909)) is the integral term in VIDEand cannot be solved explicitly therefore Simpsonrsquos rule isadapted for solving the integral part

(i) For 119870(119909 119904) = 1 Simpsonrsquos 13 rule is applied to solvethe integral term in VIDE

119911119899+2 = 119911119899 + ℎ [13119910119899 +43119910119899+1 +

13119910119899+2] (37)

(ii) For119870(119909 119904) = 1 composite Simpsonrsquos rule is employedfor solving the integral part

119911119899+2 = ℎ3119899+2sum119894=0

120596119904119894119870(119909119899+2 119909119894 119910119894)

119911119899+3 = ℎ3119899+2sum119894=0

120596119904119894119870(119909119899+3 119909119894 119910119894)

+ ℎ6 [119870 (119909119899+3 119909119899+2 119910119899+2)+ 4119870 (119909119899+2 119909119899+52 119910119899+52) + 119870 (119909119899+3 119909119899+3 119910119899+3)]

(38)

where 120596119904119894 are Simpsonrsquos rule weights 1 4 2 4 2 4 1 Theunknown value of 119910119899+52 in (38) can be estimated by usingLagrange interpolation at the point 119909119899 119909119899+1 119909119899+2 119909119899+3

119910119899+52 = 116119910119899 minus 516119910119899+1 + 1516119910119899+2 + 516119910119899+3 (39)

41 Algorithm of the Method The input of the programmingis the endpoints of 119886 and 119887 and the integer119873 The developedalgorithm for the method is given as follows

Step 1 Set

1199090 = 1198861199100 = 1205721199110 = 0ℎ = (119887 minus 119886)119873OUTPUT (1199090 1199100 1199110)

Step 2 For 119894 = 1When 119870(119909 119904) = 1 using RK3 to evaluate the value of119910When 119870(119909 119904) = 1 applying Midpoint Method

Step 3 For 119894 = 2 (1198732) do Steps 4ndash6Step 4 Set 119909 = 119886 + 119894ℎ

6 International Journal of Mathematics and Mathematical Sciences

Table 1 Numerical results for Example 1

ℎ 0025 00125 000625 0003125MAXE

RK3 83229(minus08) 10910(minus08) 13953(minus09) 17637(minus10)ABM3 19468(minus06) 24548(minus07) 30813(minus08) 38593(minus09)2PMBM 42079(minus07) 49165(minus08) 59272(minus09) 72715(minus10)DIMBM 41354(minus07) 47815(minus08) 58390(minus09) 72152(minus10)

TFCRK3 120 240 480 960ABM3 82 162 322 6422PMBM 43 83 163 323DIMBM 42 82 162 322

TSRK3 40 80 160 320ABM3 40 80 160 3202PMBM 21 41 81 161DIMBM 21 41 81 161

TimeRK3 01035 01840 02970 04527ABM3 00860 01410 02264 034542PMBM 00670 01200 02024 03204DIMBM 00450 01120 01600 02660

Step 5 Calculate for 119910119901119899+1 and 119911119901119899+1 119910119901119899+2 and 119911119901119899+2Step 6 Compute the solution for119910119888119899+1 and 119911119888119899+1119910119888119899+2 and 119911119888119899+2Step 7 Calculate the error

Step 8 OUTPUT (119909 119910 119911) and the absolute error

Step 9 STOP

5 Numerical Results

Four tested problems of first-order Volterra integrodifferen-tial equations were considered in order to study the perfor-mance of the diagonally implicit multistep block method

Example 1 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = 1 minus int119909

0119910 (119904) 119889119904 119910 (0) = 0 0 le 119909 le 1 (40)

Exact solution is 119910(119909) = sin(119909)Source [4]

Example 2 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = minus sin (119909) minus cos (119909) + int119909

02 cos (119909 minus 119904) 119910 (119904) 119889119904119910 (0) = 1 0 le 119909 le 5

(41)

Exact solution is 119910(119909) = exp(minus119909)Source [11]

Example 3 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 119909 exp (1 minus 119910 (119909)) minus 1

(1 + 119909)2 minus 119909minus int1199090

119909(1 + 119904)2 exp (1 minus 119910 (119904)) 119889119904

119910 (0) = 1 0 le 119909 le 4(42)

Exact solution is 119910(119909) = 1(1 + 119909)Source [12]

Example 4 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 2119909 minus 12 sin (1199094) + int

119909

01199092119904 cos (1199092119910 (119904)) 119889119904119910 (0) = 0 0 le 119909 le 2

(43)

Exact solution is 119910(119909) = 1199092Source [13]

Notations used in Tables 1ndash4 are as follows

ℎ step sizeMAXE maximum error

International Journal of Mathematics and Mathematical Sciences 7

Table 2 Numerical results for Example 2

ℎ 025 0125 00625 003125MAXE

BVMs 21607(minus01) 28411(minus02) 36378(minus03) 46011(minus04)ABM3 15401(minus02) 31321(minus03) 53778(minus04) 79127(minus05)2PMBM 57923(minus02) 12209(minus03) 37500(minus04) 84172(minus05)DIMBM 18211(minus02) 34489(minus03) 46803(minus04) 60807(minus05)

TFCBVMs - - - -ABM3 84 164 324 6442PMBM 50 90 170 330DIMBM 22 42 82 162

TSBVMs - - - -ABM3 20 40 80 1602PMBM 11 21 41 81DIMBM 11 21 41 81

TimeBVMs - - - -ABM3 00715 01546 02433 035462PMBM 00460 00780 01710 02738DIMBM 00140 00410 01170 01480

Table 3 Numerical results for Example 3

ℎ 0025 00125 000625 0003125MAXE

ABM3 23797(minus06) 32252(minus07) 42061(minus08) 53727(minus09)2PMBM 80020(minus06) 97319(minus07) 12000(minus07) 14862(minus08)DIMBM 36545(minus06) 46274(minus07) 58216(minus08) 72999(minus09)

TFCABM3 644 1284 2564 51242PMBM 330 650 1290 2570DIMBM 322 642 1282 2562

TimeABM3 03543 05677 11650 198552PMBM 02810 04210 07496 13570DIMBM 02610 03010 06740 11050

TS total steps

TFC total functions call

Time the execution time taken

RK3 Runge-Kutta method of order 3 with Simpsonrsquos13 rule by Filiz [4]ABM3 AdamBashforthMoultonOrder 3 with Simp-sonrsquos rule

BVMs Combination of BVMs and third-order Gen-eralized Adams Method by Chen and Zhang [11]

2PMBM Two points Multistep Block Method ofOrder 3 with Simpsonrsquos rule by Mohamed and Majid[8]DIMBM Diagonally implicit multistep blockmethodwith Simpsonrsquos rule proposed in this paper

Tables 1ndash4 display the numerical results for the four testedproblems when solved using the proposed block method andthe code was written in C language

The numerical results for Examples 1ndash4 displayed inTables 1ndash4 are solved numerically using the proposed numer-ical method with Simpsonrsquos rule In Table 1 the numerical

8 International Journal of Mathematics and Mathematical Sciences

Table 4 Numerical results for Example 4

ℎ 29 217 233 265MAXE

ABM3 50218(minus02) 18761(minus03) 11046(minus04) 58996(minus06)2PMBM 27425(minus02) 65258(minus04) 17781(minus04) 66888(minus06)DIMBM 88008(minus03) 86068(minus04) 17703(minus04) 66994(minus06)

TFCABM3 40 72 136 2462PMBM 22 38 70 134DIMBM 10 18 34 66

TimeABM3 00670 00723 01291 020302PMBM 00140 00352 00662 01336DIMBM 00050 00280 00350 00420

TFC900800700600500400300200100

2PMBMDIMBM

ABM3

RK3

minus9

minus8

minus7

minus6

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus9

minus8

minus7

minus6

02 03 0401

Time

ABM3

RK3

DIMBM2PMBM

MAXE

(ＦＩＡ1

0)

(b) Graph of Time versus MAXE

Figure 3 Graph of the numerical results when solving Example 1

results will be obtained when the step size ℎ = 002500125 000625 and 0003125 for the case when 119870(119909 119904) = 1The maximum error of DIMBM is comparable compared to2PMBMat all tested ℎ but the order of accuracy is the same orone order less compared to RK3 andABM3The performanceof DIMBM is better in terms of total functions call and totalnumber of steps compared to RK3 and ABM3

In Tables 2 3 and 4 the numerical results are solved usingthe proposed numerical method via composite Simpsonrsquos forthe integral part when 119870(119909 119904) = 1 In Table 2 the maximumerror of DIMBM is comparable compared to 2PMBM and

ABM3 The DIMBM manage to obtain less total functionscall compared to ABM3 and 2PMBM For the nonlinearExamples 3 and 4 we could observe that ABM3 and 2PMBMare expensive in terms of total functions call respectivelyFigures 3ndash6 display the numerical results of maximum errorversus total functions call when solving the tested problemsThis has shown the advantage of DIMBM in the form ofa standard multistep method because the cost per step ischeaper and the numerical results aremore accuratewhen thestep size is reduced In terms of timing DIMBM gave fasterresults compared to ABM3 and 2PMBM

International Journal of Mathematics and Mathematical Sciences 9

TFC600500400300200100

minus4

minus35

minus3

minus25

minus2

minus15

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

Time035030025020015010005

DIMBM2PMBM

minus4

minus35

minus3

minus25

minus2

minus15

ABM3MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 4 Graph of the numerical results when solving Example 2

TFC50004000300020001000

minus8

minus75

minus7

minus65

minus6

minus55

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

MAXE

(ＦＩＡ1

0)

DIMBM2PMBMABM3

06 08 10 12 14 1604 18

Time

minus8

minus75

minus7

minus65

minus6

minus55

(b) Graph of Time versus MAXE

Figure 5 Graph of the numerical results when solving Example 3

6 Conclusion

In this research we proposed the diagonally implicit multi-step block method for solving linear and nonlinear Volterra

integrodifferential equations and comparisons were madewith the existingmethod Comparisons with existingmethodreveal that the diagonally implicit multistep block method ismore efficient and cheaper

10 International Journal of Mathematics and Mathematical Sciences

TFC200150100150

DIMBM2PMBMABM3

minus5

minus4

minus3

minus2

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus5

minus4

minus3

minus2

DIMBM2PMBMABM3

008 018 020010 012 014 016004 006002

Time

MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 6 Graph of the numerical results when solving Example 4

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledged the financial supportof Fundamental Research Grant Scheme (FRGS5524973)and Graduate Research Fund (GRF) from Universiti PutraMalaysia

References

[1] J T Day ldquoNote on the Numerical Solution of Integro-Differential EquationsrdquoThe Computer Journal vol 9 no 4 pp394-395 1967

[2] R Saadati B Raftari H Adibi S M Vaezpour and S ShakerildquoA comparison between the Variational Iteration method andTrapezoidal rule for solving linear integro-differential equa-tionsrdquoWorld Applied Sciences Journal vol 4 no 3 pp 321ndash3252008

[3] B Raftari ldquoNumerical solutions of the linear volterra integro-differential equations homotopy perturbation method andfinite Difference methodrdquo World Applied Sciences Journal vol9 pp 7ndash12 2010

[4] A Filiz ldquoFourth-Order Robust Numerical Method for Integro-differential Equationsrdquo Asian Journal of Fuzzy and AppliedMathematics vol 1 no 1 pp 28ndash33 2013

[5] A Filiz ldquoNumerical Solution of a Non-linear Volterra Integro-differential Equation via Runge-Kutta-Verner Methodrdquo Inter-national Journal of Scientific and Research Publications vol 3article 9 2013

[6] F Ishak and S N Ahmad ldquoDevelopment of Extended Trape-zoidal Method for Numerical Solution of Volterra Integro-Differential Equationsrdquo International Journal of MathematicsComputational Physical Electrical and Computer Engineeringvol 10 no 11 article 52856 2016

[7] N A BMohamed and Z AMajid ldquoOne-step blockmethod forsolving Volterra integro-differential equationsrdquo AIP ConferenceProceedings vol 1682 no 1 Article ID 020018 2015

[8] N A Mohamed and Z A Majid ldquoMultistep block methodfor solving volterra integro-differential equationsrdquo MalaysianJournal of Mathematical Sciences vol 10 pp 33ndash48 2016

[9] J D Lambert Computational Methods in Ordinary DifferentialEquations John Wiley amp Sons New York NY USA 1973

[10] H Brunner and J D Lambert ldquoStability of numerical methodsfor volterra integro-differential equationsrdquo Computing vol 12no 1 pp 75ndash89 1974

[11] H Chen and C Zhang ldquoBoundary value methods for Volterraintegral and integro-differential equationsrdquo Applied Mathemat-ics and Computation vol 218 no 6 pp 2619ndash2630 2011

[12] R E Shaw ldquoA parallel algorithm for nonlinear volterra integro-differential equationsrdquo in Proceedings of the 2000 ACM Sympo-sium on Applied Computing SAC 2000 pp 86ndash88 Como ItalyMarch 2000

[13] 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

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

6 International Journal of Mathematics and Mathematical Sciences

Table 1 Numerical results for Example 1

ℎ 0025 00125 000625 0003125MAXE

RK3 83229(minus08) 10910(minus08) 13953(minus09) 17637(minus10)ABM3 19468(minus06) 24548(minus07) 30813(minus08) 38593(minus09)2PMBM 42079(minus07) 49165(minus08) 59272(minus09) 72715(minus10)DIMBM 41354(minus07) 47815(minus08) 58390(minus09) 72152(minus10)

TFCRK3 120 240 480 960ABM3 82 162 322 6422PMBM 43 83 163 323DIMBM 42 82 162 322

TSRK3 40 80 160 320ABM3 40 80 160 3202PMBM 21 41 81 161DIMBM 21 41 81 161

TimeRK3 01035 01840 02970 04527ABM3 00860 01410 02264 034542PMBM 00670 01200 02024 03204DIMBM 00450 01120 01600 02660

Step 5 Calculate for 119910119901119899+1 and 119911119901119899+1 119910119901119899+2 and 119911119901119899+2Step 6 Compute the solution for119910119888119899+1 and 119911119888119899+1119910119888119899+2 and 119911119888119899+2Step 7 Calculate the error

Step 8 OUTPUT (119909 119910 119911) and the absolute error

Step 9 STOP

5 Numerical Results

Four tested problems of first-order Volterra integrodifferen-tial equations were considered in order to study the perfor-mance of the diagonally implicit multistep block method

Example 1 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = 1 minus int119909

0119910 (119904) 119889119904 119910 (0) = 0 0 le 119909 le 1 (40)

Exact solution is 119910(119909) = sin(119909)Source [4]

Example 2 ((119870(119909 119904) = 1) linear VIDE)1199101015840 (119909) = minus sin (119909) minus cos (119909) + int119909

02 cos (119909 minus 119904) 119910 (119904) 119889119904119910 (0) = 1 0 le 119909 le 5

(41)

Exact solution is 119910(119909) = exp(minus119909)Source [11]

Example 3 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 119909 exp (1 minus 119910 (119909)) minus 1

(1 + 119909)2 minus 119909minus int1199090

119909(1 + 119904)2 exp (1 minus 119910 (119904)) 119889119904

119910 (0) = 1 0 le 119909 le 4(42)

Exact solution is 119910(119909) = 1(1 + 119909)Source [12]

Example 4 ((119870(119909 119904) = 1) nonlinear VIDE)1199101015840 (119909) = 2119909 minus 12 sin (1199094) + int

119909

01199092119904 cos (1199092119910 (119904)) 119889119904119910 (0) = 0 0 le 119909 le 2

(43)

Exact solution is 119910(119909) = 1199092Source [13]

Notations used in Tables 1ndash4 are as follows

ℎ step sizeMAXE maximum error

International Journal of Mathematics and Mathematical Sciences 7

Table 2 Numerical results for Example 2

ℎ 025 0125 00625 003125MAXE

BVMs 21607(minus01) 28411(minus02) 36378(minus03) 46011(minus04)ABM3 15401(minus02) 31321(minus03) 53778(minus04) 79127(minus05)2PMBM 57923(minus02) 12209(minus03) 37500(minus04) 84172(minus05)DIMBM 18211(minus02) 34489(minus03) 46803(minus04) 60807(minus05)

TFCBVMs - - - -ABM3 84 164 324 6442PMBM 50 90 170 330DIMBM 22 42 82 162

TSBVMs - - - -ABM3 20 40 80 1602PMBM 11 21 41 81DIMBM 11 21 41 81

TimeBVMs - - - -ABM3 00715 01546 02433 035462PMBM 00460 00780 01710 02738DIMBM 00140 00410 01170 01480

Table 3 Numerical results for Example 3

ℎ 0025 00125 000625 0003125MAXE

ABM3 23797(minus06) 32252(minus07) 42061(minus08) 53727(minus09)2PMBM 80020(minus06) 97319(minus07) 12000(minus07) 14862(minus08)DIMBM 36545(minus06) 46274(minus07) 58216(minus08) 72999(minus09)

TFCABM3 644 1284 2564 51242PMBM 330 650 1290 2570DIMBM 322 642 1282 2562

TimeABM3 03543 05677 11650 198552PMBM 02810 04210 07496 13570DIMBM 02610 03010 06740 11050

TS total steps

TFC total functions call

Time the execution time taken

RK3 Runge-Kutta method of order 3 with Simpsonrsquos13 rule by Filiz [4]ABM3 AdamBashforthMoultonOrder 3 with Simp-sonrsquos rule

BVMs Combination of BVMs and third-order Gen-eralized Adams Method by Chen and Zhang [11]

2PMBM Two points Multistep Block Method ofOrder 3 with Simpsonrsquos rule by Mohamed and Majid[8]DIMBM Diagonally implicit multistep blockmethodwith Simpsonrsquos rule proposed in this paper

Tables 1ndash4 display the numerical results for the four testedproblems when solved using the proposed block method andthe code was written in C language

The numerical results for Examples 1ndash4 displayed inTables 1ndash4 are solved numerically using the proposed numer-ical method with Simpsonrsquos rule In Table 1 the numerical

8 International Journal of Mathematics and Mathematical Sciences

Table 4 Numerical results for Example 4

ℎ 29 217 233 265MAXE

ABM3 50218(minus02) 18761(minus03) 11046(minus04) 58996(minus06)2PMBM 27425(minus02) 65258(minus04) 17781(minus04) 66888(minus06)DIMBM 88008(minus03) 86068(minus04) 17703(minus04) 66994(minus06)

TFCABM3 40 72 136 2462PMBM 22 38 70 134DIMBM 10 18 34 66

TimeABM3 00670 00723 01291 020302PMBM 00140 00352 00662 01336DIMBM 00050 00280 00350 00420

TFC900800700600500400300200100

2PMBMDIMBM

ABM3

RK3

minus9

minus8

minus7

minus6

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus9

minus8

minus7

minus6

02 03 0401

Time

ABM3

RK3

DIMBM2PMBM

MAXE

(ＦＩＡ1

0)

(b) Graph of Time versus MAXE

Figure 3 Graph of the numerical results when solving Example 1

results will be obtained when the step size ℎ = 002500125 000625 and 0003125 for the case when 119870(119909 119904) = 1The maximum error of DIMBM is comparable compared to2PMBMat all tested ℎ but the order of accuracy is the same orone order less compared to RK3 andABM3The performanceof DIMBM is better in terms of total functions call and totalnumber of steps compared to RK3 and ABM3

In Tables 2 3 and 4 the numerical results are solved usingthe proposed numerical method via composite Simpsonrsquos forthe integral part when 119870(119909 119904) = 1 In Table 2 the maximumerror of DIMBM is comparable compared to 2PMBM and

ABM3 The DIMBM manage to obtain less total functionscall compared to ABM3 and 2PMBM For the nonlinearExamples 3 and 4 we could observe that ABM3 and 2PMBMare expensive in terms of total functions call respectivelyFigures 3ndash6 display the numerical results of maximum errorversus total functions call when solving the tested problemsThis has shown the advantage of DIMBM in the form ofa standard multistep method because the cost per step ischeaper and the numerical results aremore accuratewhen thestep size is reduced In terms of timing DIMBM gave fasterresults compared to ABM3 and 2PMBM

International Journal of Mathematics and Mathematical Sciences 9

TFC600500400300200100

minus4

minus35

minus3

minus25

minus2

minus15

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

Time035030025020015010005

DIMBM2PMBM

minus4

minus35

minus3

minus25

minus2

minus15

ABM3MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 4 Graph of the numerical results when solving Example 2

TFC50004000300020001000

minus8

minus75

minus7

minus65

minus6

minus55

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

MAXE

(ＦＩＡ1

0)

DIMBM2PMBMABM3

06 08 10 12 14 1604 18

Time

minus8

minus75

minus7

minus65

minus6

minus55

(b) Graph of Time versus MAXE

Figure 5 Graph of the numerical results when solving Example 3

6 Conclusion

In this research we proposed the diagonally implicit multi-step block method for solving linear and nonlinear Volterra

integrodifferential equations and comparisons were madewith the existingmethod Comparisons with existingmethodreveal that the diagonally implicit multistep block method ismore efficient and cheaper

10 International Journal of Mathematics and Mathematical Sciences

TFC200150100150

DIMBM2PMBMABM3

minus5

minus4

minus3

minus2

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus5

minus4

minus3

minus2

DIMBM2PMBMABM3

008 018 020010 012 014 016004 006002

Time

MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 6 Graph of the numerical results when solving Example 4

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledged the financial supportof Fundamental Research Grant Scheme (FRGS5524973)and Graduate Research Fund (GRF) from Universiti PutraMalaysia

References

[1] J T Day ldquoNote on the Numerical Solution of Integro-Differential EquationsrdquoThe Computer Journal vol 9 no 4 pp394-395 1967

[2] R Saadati B Raftari H Adibi S M Vaezpour and S ShakerildquoA comparison between the Variational Iteration method andTrapezoidal rule for solving linear integro-differential equa-tionsrdquoWorld Applied Sciences Journal vol 4 no 3 pp 321ndash3252008

[3] B Raftari ldquoNumerical solutions of the linear volterra integro-differential equations homotopy perturbation method andfinite Difference methodrdquo World Applied Sciences Journal vol9 pp 7ndash12 2010

[4] A Filiz ldquoFourth-Order Robust Numerical Method for Integro-differential Equationsrdquo Asian Journal of Fuzzy and AppliedMathematics vol 1 no 1 pp 28ndash33 2013

[5] A Filiz ldquoNumerical Solution of a Non-linear Volterra Integro-differential Equation via Runge-Kutta-Verner Methodrdquo Inter-national Journal of Scientific and Research Publications vol 3article 9 2013

[6] F Ishak and S N Ahmad ldquoDevelopment of Extended Trape-zoidal Method for Numerical Solution of Volterra Integro-Differential Equationsrdquo International Journal of MathematicsComputational Physical Electrical and Computer Engineeringvol 10 no 11 article 52856 2016

[7] N A BMohamed and Z AMajid ldquoOne-step blockmethod forsolving Volterra integro-differential equationsrdquo AIP ConferenceProceedings vol 1682 no 1 Article ID 020018 2015

[8] N A Mohamed and Z A Majid ldquoMultistep block methodfor solving volterra integro-differential equationsrdquo MalaysianJournal of Mathematical Sciences vol 10 pp 33ndash48 2016

[9] J D Lambert Computational Methods in Ordinary DifferentialEquations John Wiley amp Sons New York NY USA 1973

[10] H Brunner and J D Lambert ldquoStability of numerical methodsfor volterra integro-differential equationsrdquo Computing vol 12no 1 pp 75ndash89 1974

[11] H Chen and C Zhang ldquoBoundary value methods for Volterraintegral and integro-differential equationsrdquo Applied Mathemat-ics and Computation vol 218 no 6 pp 2619ndash2630 2011

[12] R E Shaw ldquoA parallel algorithm for nonlinear volterra integro-differential equationsrdquo in Proceedings of the 2000 ACM Sympo-sium on Applied Computing SAC 2000 pp 86ndash88 Como ItalyMarch 2000

[13] 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

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

International Journal of Mathematics and Mathematical Sciences 7

Table 2 Numerical results for Example 2

ℎ 025 0125 00625 003125MAXE

BVMs 21607(minus01) 28411(minus02) 36378(minus03) 46011(minus04)ABM3 15401(minus02) 31321(minus03) 53778(minus04) 79127(minus05)2PMBM 57923(minus02) 12209(minus03) 37500(minus04) 84172(minus05)DIMBM 18211(minus02) 34489(minus03) 46803(minus04) 60807(minus05)

TFCBVMs - - - -ABM3 84 164 324 6442PMBM 50 90 170 330DIMBM 22 42 82 162

TSBVMs - - - -ABM3 20 40 80 1602PMBM 11 21 41 81DIMBM 11 21 41 81

TimeBVMs - - - -ABM3 00715 01546 02433 035462PMBM 00460 00780 01710 02738DIMBM 00140 00410 01170 01480

Table 3 Numerical results for Example 3

ℎ 0025 00125 000625 0003125MAXE

ABM3 23797(minus06) 32252(minus07) 42061(minus08) 53727(minus09)2PMBM 80020(minus06) 97319(minus07) 12000(minus07) 14862(minus08)DIMBM 36545(minus06) 46274(minus07) 58216(minus08) 72999(minus09)

TFCABM3 644 1284 2564 51242PMBM 330 650 1290 2570DIMBM 322 642 1282 2562

TimeABM3 03543 05677 11650 198552PMBM 02810 04210 07496 13570DIMBM 02610 03010 06740 11050

TS total steps

TFC total functions call

Time the execution time taken

RK3 Runge-Kutta method of order 3 with Simpsonrsquos13 rule by Filiz [4]ABM3 AdamBashforthMoultonOrder 3 with Simp-sonrsquos rule

BVMs Combination of BVMs and third-order Gen-eralized Adams Method by Chen and Zhang [11]

2PMBM Two points Multistep Block Method ofOrder 3 with Simpsonrsquos rule by Mohamed and Majid[8]DIMBM Diagonally implicit multistep blockmethodwith Simpsonrsquos rule proposed in this paper

Tables 1ndash4 display the numerical results for the four testedproblems when solved using the proposed block method andthe code was written in C language

The numerical results for Examples 1ndash4 displayed inTables 1ndash4 are solved numerically using the proposed numer-ical method with Simpsonrsquos rule In Table 1 the numerical

8 International Journal of Mathematics and Mathematical Sciences

Table 4 Numerical results for Example 4

ℎ 29 217 233 265MAXE

ABM3 50218(minus02) 18761(minus03) 11046(minus04) 58996(minus06)2PMBM 27425(minus02) 65258(minus04) 17781(minus04) 66888(minus06)DIMBM 88008(minus03) 86068(minus04) 17703(minus04) 66994(minus06)

TFCABM3 40 72 136 2462PMBM 22 38 70 134DIMBM 10 18 34 66

TimeABM3 00670 00723 01291 020302PMBM 00140 00352 00662 01336DIMBM 00050 00280 00350 00420

TFC900800700600500400300200100

2PMBMDIMBM

ABM3

RK3

minus9

minus8

minus7

minus6

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus9

minus8

minus7

minus6

02 03 0401

Time

ABM3

RK3

DIMBM2PMBM

MAXE

(ＦＩＡ1

0)

(b) Graph of Time versus MAXE

Figure 3 Graph of the numerical results when solving Example 1

results will be obtained when the step size ℎ = 002500125 000625 and 0003125 for the case when 119870(119909 119904) = 1The maximum error of DIMBM is comparable compared to2PMBMat all tested ℎ but the order of accuracy is the same orone order less compared to RK3 andABM3The performanceof DIMBM is better in terms of total functions call and totalnumber of steps compared to RK3 and ABM3

In Tables 2 3 and 4 the numerical results are solved usingthe proposed numerical method via composite Simpsonrsquos forthe integral part when 119870(119909 119904) = 1 In Table 2 the maximumerror of DIMBM is comparable compared to 2PMBM and

ABM3 The DIMBM manage to obtain less total functionscall compared to ABM3 and 2PMBM For the nonlinearExamples 3 and 4 we could observe that ABM3 and 2PMBMare expensive in terms of total functions call respectivelyFigures 3ndash6 display the numerical results of maximum errorversus total functions call when solving the tested problemsThis has shown the advantage of DIMBM in the form ofa standard multistep method because the cost per step ischeaper and the numerical results aremore accuratewhen thestep size is reduced In terms of timing DIMBM gave fasterresults compared to ABM3 and 2PMBM

International Journal of Mathematics and Mathematical Sciences 9

TFC600500400300200100

minus4

minus35

minus3

minus25

minus2

minus15

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

Time035030025020015010005

DIMBM2PMBM

minus4

minus35

minus3

minus25

minus2

minus15

ABM3MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 4 Graph of the numerical results when solving Example 2

TFC50004000300020001000

minus8

minus75

minus7

minus65

minus6

minus55

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

MAXE

(ＦＩＡ1

0)

DIMBM2PMBMABM3

06 08 10 12 14 1604 18

Time

minus8

minus75

minus7

minus65

minus6

minus55

(b) Graph of Time versus MAXE

Figure 5 Graph of the numerical results when solving Example 3

6 Conclusion

In this research we proposed the diagonally implicit multi-step block method for solving linear and nonlinear Volterra

integrodifferential equations and comparisons were madewith the existingmethod Comparisons with existingmethodreveal that the diagonally implicit multistep block method ismore efficient and cheaper

10 International Journal of Mathematics and Mathematical Sciences

TFC200150100150

DIMBM2PMBMABM3

minus5

minus4

minus3

minus2

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus5

minus4

minus3

minus2

DIMBM2PMBMABM3

008 018 020010 012 014 016004 006002

Time

MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 6 Graph of the numerical results when solving Example 4

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledged the financial supportof Fundamental Research Grant Scheme (FRGS5524973)and Graduate Research Fund (GRF) from Universiti PutraMalaysia

References

[1] J T Day ldquoNote on the Numerical Solution of Integro-Differential EquationsrdquoThe Computer Journal vol 9 no 4 pp394-395 1967

[2] R Saadati B Raftari H Adibi S M Vaezpour and S ShakerildquoA comparison between the Variational Iteration method andTrapezoidal rule for solving linear integro-differential equa-tionsrdquoWorld Applied Sciences Journal vol 4 no 3 pp 321ndash3252008

[3] B Raftari ldquoNumerical solutions of the linear volterra integro-differential equations homotopy perturbation method andfinite Difference methodrdquo World Applied Sciences Journal vol9 pp 7ndash12 2010

[4] A Filiz ldquoFourth-Order Robust Numerical Method for Integro-differential Equationsrdquo Asian Journal of Fuzzy and AppliedMathematics vol 1 no 1 pp 28ndash33 2013

[5] A Filiz ldquoNumerical Solution of a Non-linear Volterra Integro-differential Equation via Runge-Kutta-Verner Methodrdquo Inter-national Journal of Scientific and Research Publications vol 3article 9 2013

[6] F Ishak and S N Ahmad ldquoDevelopment of Extended Trape-zoidal Method for Numerical Solution of Volterra Integro-Differential Equationsrdquo International Journal of MathematicsComputational Physical Electrical and Computer Engineeringvol 10 no 11 article 52856 2016

[7] N A BMohamed and Z AMajid ldquoOne-step blockmethod forsolving Volterra integro-differential equationsrdquo AIP ConferenceProceedings vol 1682 no 1 Article ID 020018 2015

[8] N A Mohamed and Z A Majid ldquoMultistep block methodfor solving volterra integro-differential equationsrdquo MalaysianJournal of Mathematical Sciences vol 10 pp 33ndash48 2016

[9] J D Lambert Computational Methods in Ordinary DifferentialEquations John Wiley amp Sons New York NY USA 1973

[10] H Brunner and J D Lambert ldquoStability of numerical methodsfor volterra integro-differential equationsrdquo Computing vol 12no 1 pp 75ndash89 1974

[11] H Chen and C Zhang ldquoBoundary value methods for Volterraintegral and integro-differential equationsrdquo Applied Mathemat-ics and Computation vol 218 no 6 pp 2619ndash2630 2011

[12] R E Shaw ldquoA parallel algorithm for nonlinear volterra integro-differential equationsrdquo in Proceedings of the 2000 ACM Sympo-sium on Applied Computing SAC 2000 pp 86ndash88 Como ItalyMarch 2000

[13] 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

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

8 International Journal of Mathematics and Mathematical Sciences

Table 4 Numerical results for Example 4

ℎ 29 217 233 265MAXE

ABM3 50218(minus02) 18761(minus03) 11046(minus04) 58996(minus06)2PMBM 27425(minus02) 65258(minus04) 17781(minus04) 66888(minus06)DIMBM 88008(minus03) 86068(minus04) 17703(minus04) 66994(minus06)

TFCABM3 40 72 136 2462PMBM 22 38 70 134DIMBM 10 18 34 66

TimeABM3 00670 00723 01291 020302PMBM 00140 00352 00662 01336DIMBM 00050 00280 00350 00420

TFC900800700600500400300200100

2PMBMDIMBM

ABM3

RK3

minus9

minus8

minus7

minus6

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus9

minus8

minus7

minus6

02 03 0401

Time

ABM3

RK3

DIMBM2PMBM

MAXE

(ＦＩＡ1

0)

(b) Graph of Time versus MAXE

Figure 3 Graph of the numerical results when solving Example 1

results will be obtained when the step size ℎ = 002500125 000625 and 0003125 for the case when 119870(119909 119904) = 1The maximum error of DIMBM is comparable compared to2PMBMat all tested ℎ but the order of accuracy is the same orone order less compared to RK3 andABM3The performanceof DIMBM is better in terms of total functions call and totalnumber of steps compared to RK3 and ABM3

In Tables 2 3 and 4 the numerical results are solved usingthe proposed numerical method via composite Simpsonrsquos forthe integral part when 119870(119909 119904) = 1 In Table 2 the maximumerror of DIMBM is comparable compared to 2PMBM and

ABM3 The DIMBM manage to obtain less total functionscall compared to ABM3 and 2PMBM For the nonlinearExamples 3 and 4 we could observe that ABM3 and 2PMBMare expensive in terms of total functions call respectivelyFigures 3ndash6 display the numerical results of maximum errorversus total functions call when solving the tested problemsThis has shown the advantage of DIMBM in the form ofa standard multistep method because the cost per step ischeaper and the numerical results aremore accuratewhen thestep size is reduced In terms of timing DIMBM gave fasterresults compared to ABM3 and 2PMBM

International Journal of Mathematics and Mathematical Sciences 9

TFC600500400300200100

minus4

minus35

minus3

minus25

minus2

minus15

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

Time035030025020015010005

DIMBM2PMBM

minus4

minus35

minus3

minus25

minus2

minus15

ABM3MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 4 Graph of the numerical results when solving Example 2

TFC50004000300020001000

minus8

minus75

minus7

minus65

minus6

minus55

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

MAXE

(ＦＩＡ1

0)

DIMBM2PMBMABM3

06 08 10 12 14 1604 18

Time

minus8

minus75

minus7

minus65

minus6

minus55

(b) Graph of Time versus MAXE

Figure 5 Graph of the numerical results when solving Example 3

6 Conclusion

In this research we proposed the diagonally implicit multi-step block method for solving linear and nonlinear Volterra

integrodifferential equations and comparisons were madewith the existingmethod Comparisons with existingmethodreveal that the diagonally implicit multistep block method ismore efficient and cheaper

10 International Journal of Mathematics and Mathematical Sciences

TFC200150100150

DIMBM2PMBMABM3

minus5

minus4

minus3

minus2

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus5

minus4

minus3

minus2

DIMBM2PMBMABM3

008 018 020010 012 014 016004 006002

Time

MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 6 Graph of the numerical results when solving Example 4

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledged the financial supportof Fundamental Research Grant Scheme (FRGS5524973)and Graduate Research Fund (GRF) from Universiti PutraMalaysia

References

[1] J T Day ldquoNote on the Numerical Solution of Integro-Differential EquationsrdquoThe Computer Journal vol 9 no 4 pp394-395 1967

[2] R Saadati B Raftari H Adibi S M Vaezpour and S ShakerildquoA comparison between the Variational Iteration method andTrapezoidal rule for solving linear integro-differential equa-tionsrdquoWorld Applied Sciences Journal vol 4 no 3 pp 321ndash3252008

[3] B Raftari ldquoNumerical solutions of the linear volterra integro-differential equations homotopy perturbation method andfinite Difference methodrdquo World Applied Sciences Journal vol9 pp 7ndash12 2010

[4] A Filiz ldquoFourth-Order Robust Numerical Method for Integro-differential Equationsrdquo Asian Journal of Fuzzy and AppliedMathematics vol 1 no 1 pp 28ndash33 2013

[5] A Filiz ldquoNumerical Solution of a Non-linear Volterra Integro-differential Equation via Runge-Kutta-Verner Methodrdquo Inter-national Journal of Scientific and Research Publications vol 3article 9 2013

[6] F Ishak and S N Ahmad ldquoDevelopment of Extended Trape-zoidal Method for Numerical Solution of Volterra Integro-Differential Equationsrdquo International Journal of MathematicsComputational Physical Electrical and Computer Engineeringvol 10 no 11 article 52856 2016

[7] N A BMohamed and Z AMajid ldquoOne-step blockmethod forsolving Volterra integro-differential equationsrdquo AIP ConferenceProceedings vol 1682 no 1 Article ID 020018 2015

[8] N A Mohamed and Z A Majid ldquoMultistep block methodfor solving volterra integro-differential equationsrdquo MalaysianJournal of Mathematical Sciences vol 10 pp 33ndash48 2016

[9] J D Lambert Computational Methods in Ordinary DifferentialEquations John Wiley amp Sons New York NY USA 1973

[10] H Brunner and J D Lambert ldquoStability of numerical methodsfor volterra integro-differential equationsrdquo Computing vol 12no 1 pp 75ndash89 1974

[11] H Chen and C Zhang ldquoBoundary value methods for Volterraintegral and integro-differential equationsrdquo Applied Mathemat-ics and Computation vol 218 no 6 pp 2619ndash2630 2011

[12] R E Shaw ldquoA parallel algorithm for nonlinear volterra integro-differential equationsrdquo in Proceedings of the 2000 ACM Sympo-sium on Applied Computing SAC 2000 pp 86ndash88 Como ItalyMarch 2000

[13] 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

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

International Journal of Mathematics and Mathematical Sciences 9

TFC600500400300200100

minus4

minus35

minus3

minus25

minus2

minus15

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

Time035030025020015010005

DIMBM2PMBM

minus4

minus35

minus3

minus25

minus2

minus15

ABM3MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 4 Graph of the numerical results when solving Example 2

TFC50004000300020001000

minus8

minus75

minus7

minus65

minus6

minus55

DIMBM2PMBMABM3

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

MAXE

(ＦＩＡ1

0)

DIMBM2PMBMABM3

06 08 10 12 14 1604 18

Time

minus8

minus75

minus7

minus65

minus6

minus55

(b) Graph of Time versus MAXE

Figure 5 Graph of the numerical results when solving Example 3

6 Conclusion

In this research we proposed the diagonally implicit multi-step block method for solving linear and nonlinear Volterra

integrodifferential equations and comparisons were madewith the existingmethod Comparisons with existingmethodreveal that the diagonally implicit multistep block method ismore efficient and cheaper

10 International Journal of Mathematics and Mathematical Sciences

TFC200150100150

DIMBM2PMBMABM3

minus5

minus4

minus3

minus2

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus5

minus4

minus3

minus2

DIMBM2PMBMABM3

008 018 020010 012 014 016004 006002

Time

MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 6 Graph of the numerical results when solving Example 4

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledged the financial supportof Fundamental Research Grant Scheme (FRGS5524973)and Graduate Research Fund (GRF) from Universiti PutraMalaysia

References

[1] J T Day ldquoNote on the Numerical Solution of Integro-Differential EquationsrdquoThe Computer Journal vol 9 no 4 pp394-395 1967

[2] R Saadati B Raftari H Adibi S M Vaezpour and S ShakerildquoA comparison between the Variational Iteration method andTrapezoidal rule for solving linear integro-differential equa-tionsrdquoWorld Applied Sciences Journal vol 4 no 3 pp 321ndash3252008

[3] B Raftari ldquoNumerical solutions of the linear volterra integro-differential equations homotopy perturbation method andfinite Difference methodrdquo World Applied Sciences Journal vol9 pp 7ndash12 2010

[4] A Filiz ldquoFourth-Order Robust Numerical Method for Integro-differential Equationsrdquo Asian Journal of Fuzzy and AppliedMathematics vol 1 no 1 pp 28ndash33 2013

[5] A Filiz ldquoNumerical Solution of a Non-linear Volterra Integro-differential Equation via Runge-Kutta-Verner Methodrdquo Inter-national Journal of Scientific and Research Publications vol 3article 9 2013

[6] F Ishak and S N Ahmad ldquoDevelopment of Extended Trape-zoidal Method for Numerical Solution of Volterra Integro-Differential Equationsrdquo International Journal of MathematicsComputational Physical Electrical and Computer Engineeringvol 10 no 11 article 52856 2016

[7] N A BMohamed and Z AMajid ldquoOne-step blockmethod forsolving Volterra integro-differential equationsrdquo AIP ConferenceProceedings vol 1682 no 1 Article ID 020018 2015

[8] N A Mohamed and Z A Majid ldquoMultistep block methodfor solving volterra integro-differential equationsrdquo MalaysianJournal of Mathematical Sciences vol 10 pp 33ndash48 2016

[9] J D Lambert Computational Methods in Ordinary DifferentialEquations John Wiley amp Sons New York NY USA 1973

[10] H Brunner and J D Lambert ldquoStability of numerical methodsfor volterra integro-differential equationsrdquo Computing vol 12no 1 pp 75ndash89 1974

[11] H Chen and C Zhang ldquoBoundary value methods for Volterraintegral and integro-differential equationsrdquo Applied Mathemat-ics and Computation vol 218 no 6 pp 2619ndash2630 2011

[12] R E Shaw ldquoA parallel algorithm for nonlinear volterra integro-differential equationsrdquo in Proceedings of the 2000 ACM Sympo-sium on Applied Computing SAC 2000 pp 86ndash88 Como ItalyMarch 2000

[13] 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

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

10 International Journal of Mathematics and Mathematical Sciences

TFC200150100150

DIMBM2PMBMABM3

minus5

minus4

minus3

minus2

MAXE

(ＦＩＡ1

0)

(a) Graph of TFC versus MAXE

minus5

minus4

minus3

minus2

DIMBM2PMBMABM3

008 018 020010 012 014 016004 006002

Time

MAXE

(ＦＩＡ1

0)

(b) Graph of time versus MAXE

Figure 6 Graph of the numerical results when solving Example 4

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

The authors gratefully acknowledged the financial supportof Fundamental Research Grant Scheme (FRGS5524973)and Graduate Research Fund (GRF) from Universiti PutraMalaysia

References

[1] J T Day ldquoNote on the Numerical Solution of Integro-Differential EquationsrdquoThe Computer Journal vol 9 no 4 pp394-395 1967

[2] R Saadati B Raftari H Adibi S M Vaezpour and S ShakerildquoA comparison between the Variational Iteration method andTrapezoidal rule for solving linear integro-differential equa-tionsrdquoWorld Applied Sciences Journal vol 4 no 3 pp 321ndash3252008

[3] B Raftari ldquoNumerical solutions of the linear volterra integro-differential equations homotopy perturbation method andfinite Difference methodrdquo World Applied Sciences Journal vol9 pp 7ndash12 2010

[4] A Filiz ldquoFourth-Order Robust Numerical Method for Integro-differential Equationsrdquo Asian Journal of Fuzzy and AppliedMathematics vol 1 no 1 pp 28ndash33 2013

[5] A Filiz ldquoNumerical Solution of a Non-linear Volterra Integro-differential Equation via Runge-Kutta-Verner Methodrdquo Inter-national Journal of Scientific and Research Publications vol 3article 9 2013

[6] F Ishak and S N Ahmad ldquoDevelopment of Extended Trape-zoidal Method for Numerical Solution of Volterra Integro-Differential Equationsrdquo International Journal of MathematicsComputational Physical Electrical and Computer Engineeringvol 10 no 11 article 52856 2016

[7] N A BMohamed and Z AMajid ldquoOne-step blockmethod forsolving Volterra integro-differential equationsrdquo AIP ConferenceProceedings vol 1682 no 1 Article ID 020018 2015

[8] N A Mohamed and Z A Majid ldquoMultistep block methodfor solving volterra integro-differential equationsrdquo MalaysianJournal of Mathematical Sciences vol 10 pp 33ndash48 2016

[9] J D Lambert Computational Methods in Ordinary DifferentialEquations John Wiley amp Sons New York NY USA 1973

[10] H Brunner and J D Lambert ldquoStability of numerical methodsfor volterra integro-differential equationsrdquo Computing vol 12no 1 pp 75ndash89 1974

[11] H Chen and C Zhang ldquoBoundary value methods for Volterraintegral and integro-differential equationsrdquo Applied Mathemat-ics and Computation vol 218 no 6 pp 2619ndash2630 2011

[12] R E Shaw ldquoA parallel algorithm for nonlinear volterra integro-differential equationsrdquo in Proceedings of the 2000 ACM Sympo-sium on Applied Computing SAC 2000 pp 86ndash88 Como ItalyMarch 2000

[13] 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

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom