Research ArticleRunge-Kutta Type Methods for Directly Solving SpecialFourth-Order Ordinary Differential Equations

Kasim Hussain12 Fudziah Ismail13 and Norazak Senu13

1Department of Mathematics Faculty of Science Universiti Putra Malaysia (UPM) 43400 Serdang Selangor Malaysia2Department of Mathematics College of Science Al-Mustansiriya University Baghdad Iraq3Institute for Mathematical Research Universiti Putra Malaysia (UPM) 43400 Serdang Selangor Malaysia

Correspondence should be addressed to Fudziah Ismail fudziah iyahoocommy

Received 25 May 2015 Revised 14 July 2015 Accepted 21 July 2015

Academic Editor Yan-Jun Liu

Copyright copy 2015 Kasim Hussain et alThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

A Runge-Kutta type method for directly solving special fourth-order ordinary differential equations (ODEs) which is denoted byRKFD method is constructed The order conditions of RKFD method up to order five are derived based on the order conditionsthree-stage fourth- and fifth-order Runge-Kutta type methods are constructed Zero-stability of the RKFD method is provenNumerical results obtained are compared with the existing Runge-Kutta methods in the scientific literature after reducing theproblems into a system of first-order ODEs and solving them Numerical results are presented to illustrate the robustness andcompetency of the new methods in terms of accuracy and number of function evaluations

1 Introduction

This paper deals with the numerical integration of the specialfourth-order ordinary differential equations (ODEs) of theform

119910(119894V)

(119909) = 119891 (119909 119910) (1)

with initial conditions

119910 (1199090) = 1199100

1199101015840(1199090) = 119910

1015840

0

11991010158401015840(1199090) = 119910

10158401015840

0

119910101584010158401015840(1199090) = 119910

101584010158401015840

0

(2)

in which the first second and third derivatives do notappear explicitly This type of problems often arises in manyfields of applied science such as mechanics astrophysicsquantum chemistry and electronic and control engineeringThe general approach for solving the higher-order ordinarydifferential equation (ODE) is by transforming it into anequivalent first-order system of differential equations and

then applying the appropriate numerical methods to solvethe resulting system (see [1ndash5]) However the applicationof such technique takes a lot of computational time (see[6 7]) Direct integration method is proposed to avoid suchcomputational encumbrance and increase the efficiency ofthe method Many authors have proposed several numer-ical methods for directly approximating the solutions forthe higher-order ODEs for example Kayode [8] proposedzero-stable predictor-corrector methods for solving fourth-order ordinary differential equations Majid and Suleiman[9] derived one point method to solve system of higher-order ODEs Jain et al [10] constructed finite differencemethod for solving fourth-order ODEs Waeleh et al [11]constructed a new block method for solving directly higher-order ODEs Awoyemi and Idowu [12] proposed a hybridcollocations method for solving third-order ODEs Hybridlinear multistep method with three steps to solve second-order ODEs was introduced by Jator [13] and all themethodsdiscussed above are multistep in nature

This paper primarily aims to construct a one-stepmethodof orders four and five to solve special fourth-order ODEsdirectly these new methods are self-starting in nature Thepaper is organized as follows In Section 2 the derivation

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 893763 11 pageshttpdxdoiorg1011552015893763

2 Mathematical Problems in Engineering

of the order conditions of RKFD method is presented InSection 3 the zero-stability of RKFD method is given InSection 4 three-stage RKFDmethods of order four and orderfive are constructed In Section 5 numerical examples aregiven to show the effectiveness and competency of the newRKFD methods as compared with the well known Runge-Kutta methods from the scientific literature Conclusions aregiven in Section 6

2 Derivation of the RKFD Method

The general form of RKFD method with 119904-stage for directlysolving special fourth-order ODEs (1) can be written asfollows

119910119899+1 = 119910119899 + ℎ1199101015840

119899 +ℎ2

211991010158401015840

119899 +ℎ3

6119910101584010158401015840

119899 + ℎ4119904

sum

119894=1119887119894119896119894 (3)

1199101015840

119899+1 = 1199101015840

119899 + ℎ211991010158401015840

119899 +ℎ2

2119910101584010158401015840

119899 + ℎ3119904

sum

119894=11198871015840

119894 119896119894 (4)

11991010158401015840

119899+1 = 11991010158401015840

119899 + ℎ119910101584010158401015840

119899 + ℎ2119904

sum

119894=111988710158401015840

119894 119896119894 (5)

119910101584010158401015840

119899+1 = 119910101584010158401015840

119899 + ℎ

119904

sum

119894=1119887101584010158401015840

119894 119896119894 (6)

where

1198961 = 119891 (119909119899 119910119899) (7)

119896119894 = 119891(119909119899 + 119888119894ℎ 119910119899 + ℎ1198881198941199101015840

119899 +ℎ2

21198882119894 11991010158401015840

119899 +ℎ3

61198883119894 119910101584010158401015840

119899

+ ℎ4119894minus1sum

119895=1119886119894119895119896119895) 119894 = 2 3 119904

(8)

All parameters 119887119894 1198871015840119894 11988710158401015840119894 119887101584010158401015840119894 119886119894119895 and 119888119894 of the RKFD method

are used for 119894 119895 = 1 2 119904 and are supposed to be real TheRKFD method is an explicit method if 119886119894119895 = 0 for 119894 le 119895 andis an implicit method if 119886119894119895 = 0 for 119894 le 119895 The RKFD methodcan be represented by Butcher tableau as follows

119888 119860

119887119879

1198871015840119879

11988710158401015840119879

119887101584010158401015840119879

(9)

To determine the parameters of the RKFD method givenin (3)ndash(8) the RKFD method expression is expanded usingthe Taylor series expansion After performing some algebraicmanipulations this expansion is equated to the true solutionthat is given by the Taylor series expansion The directexpansion of the local truncation error is used to derivethe general order conditions for the RKFD method This

approach depends on the derivation of order conditions forthe Runge-Kutta method proposed in Dormand [14] TheRKFD method in (3)ndash(6) can be written as follows

119910119899+1 = 119910119899 + ℎ120595 (119909119899 119910119899)

1199101015840

119899+1 = 1199101015840

119899 + ℎ1205951015840(119909119899 119910119899)

11991010158401015840

119899+1 = 11991010158401015840

119899 + ℎ12059510158401015840(119909119899 119910119899)

119910101584010158401015840

119899+1 = 119910101584010158401015840

119899 + ℎ120595101584010158401015840(119909119899 119910119899)

(10)

where the increment functions are

120595 (119909119899 119910119899) = 1199101015840

119899 +ℎ

211991010158401015840

119899 +ℎ2

6119910101584010158401015840

119899 + ℎ3119904

sum

119894=1119887119894119896119894

1205951015840(119909119899 119910119899) = 119910

10158401015840

119899 +ℎ

2119910101584010158401015840

119899 + ℎ2119904

sum

119894=11198871015840

119894 119896119894

12059510158401015840(119909119899 119910119899) = 119910

101584010158401015840

119899 + ℎ

119904

sum

119894=111988710158401015840

119894 119896119894

120595101584010158401015840(119909119899 119910119899) =

119904

sum

119894=1119887101584010158401015840

119894 119896119894

(11)

where 119896119894 is given in (8)The first few elementary differentials for the scalar equa-

tion are

119865(4)1 = 119910

(119894V)= 119891

119865(5)1 = 119891119909 +119891119910119910

1015840

119865(6)1 = 119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2

119865(7)1 = 119891119909119909119909 + 3119891119909119909119910119910

1015840+ 3119891119909119910119910 (119910

1015840)2+ 3119891119909119910119910

10158401015840

+ 3119891119910119910119910101584011991010158401015840+119891119910119910119910 (119910

1015840)3+119891119910119910

101584010158401015840

(12)

We assume that the Taylor series increment function is ΔThe local truncation errors of 119910(119909) 1199101015840(119909) 11991010158401015840(119909) and

119910101584010158401015840(119909) can be obtained after substituting the exact solution

of (1) into (11) as follows

120591119899+1 = ℎ [120595minusΔ]

1205911015840

119899+1 = ℎ [1205951015840minusΔ1015840]

12059110158401015840

119899+1 = ℎ [12059510158401015840minusΔ10158401015840]

120591101584010158401015840

119899+1 = ℎ [120595101584010158401015840minusΔ101584010158401015840]

(13)

Mathematical Problems in Engineering 3

The Taylor series increment functions of 119910(119909) 1199101015840(119909) 11991010158401015840(119909)and 119910

101584010158401015840(119909) can be expressed as follows

Δ = 1199101015840+12ℎ11991010158401015840+16ℎ2119910101584010158401015840+

124

ℎ3119865(4)1 +

1120

ℎ4119865(5)1

+1720

ℎ5119865(6)1 +119874 (ℎ

6)

Δ1015840= 11991010158401015840+12ℎ119910101584010158401015840+16ℎ2119865(4)1 +

124

ℎ3119865(5)1

+1120

ℎ4119865(6)1 +

1720

ℎ5119865(7)1 +119874 (ℎ

6)

Δ10158401015840= 119910101584010158401015840+12ℎ119865(4)1 +

16ℎ2119865(5)1 +

124

ℎ3119865(6)1

+1120

ℎ4119865(7)1 +

1720

ℎ5119865(8)1 +119874 (ℎ

6)

Δ101584010158401015840

= 119865(4)1 +

12ℎ119865(5)1 +

16ℎ2119865(6)1 +

124

ℎ3119865(7)1

+1720

ℎ4119865(8)1 +119874 (ℎ

5)

(14)

Substituting (12) into (11) the increment functions 120595 1205951015840 12059510158401015840and 120595

101584010158401015840 for RKFD method become as follows119904

sum

119894=1119887119894119896119894

=

119904

sum

119894=1119887119894119891+

119904

sum

119894=1119887119894119888119894 (119891119909 +119891119910119910

1015840) ℎ

+12

119904

sum

119894=1119887119894119888

2119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=1119887119894119896119894

=

119904

sum

119894=1119887119894119865(4)1 +

119904

sum

119894=1119887119894119888119894ℎ119865

(5)1 +

12

119904

sum

119894=1119887119894119888

2119894 ℎ

2119865(6)1 +119874 (ℎ

3)

119904

sum

119894=11198871015840

119894 119896119894

=

119904

sum

119894=11198871015840

119894119891+

119904

sum

119894=11198871015840

119894 119888119894 (119891119909 +1198911199101199101015840) ℎ

+12

119904

sum

119894=11198871015840

119894 1198882119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=11198871015840

119894 119896119894

=

119904

sum

119894=11198871015840

119894 119865(4)1 +

119904

sum

119894=11198871015840

119894 119888119894ℎ119865(5)1 +

12

119904

sum

119894=11198871015840

119894 1198882119894 ℎ

2119865(6)1 +119874 (ℎ

3)

119904

sum

119894=111988710158401015840

119894 119896119894

=

119904

sum

119894=111988710158401015840

119894 119891+

119904

sum

119894=111988710158401015840

119894 119888119894 (119891119909 +1198911199101199101015840) ℎ

+12

119904

sum

119894=111988710158401015840

119894 1198882119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=111988710158401015840

119894 119896119894

=

119904

sum

119894=111988710158401015840

119894 119865(4)1 +

119904

sum

119894=111988710158401015840

119894 119888119894ℎ119865(5)1 +

12

119904

sum

119894=111988710158401015840

119894 1198882119894 ℎ

2119865(6)1

+119874 (ℎ3)

119904

sum

119894=1119887101584010158401015840

119894 119896119894

=

119904

sum

119894=1119887101584010158401015840

119894 119891+

119904

sum

119894=1119887101584010158401015840

119894 119888119894 (119891119909 +1198911199101199101015840) ℎ

+12

119904

sum

119894=1119887101584010158401015840

119894 1198882119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=1119887101584010158401015840

119894 119896119894

=

119904

sum

119894=1119887101584010158401015840

119894 119865(4)1 +

119904

sum

119894=1119887101584010158401015840

119894 119888119894ℎ119865(5)1 +

12

119904

sum

119894=1119887101584010158401015840

119894 1198882119894 ℎ

2119865(6)1

+119874 (ℎ3)

(15)

Using (11) and (14) the local truncation errors (13) can bewritten as follows

120591119899+1 = ℎ4[

119904

sum

119894=1119887119894119896119894 minus(

124

119865(4)1 +

1120

ℎ119865(5)1 + sdot sdot sdot)]

1205911015840

119899+1 = ℎ3[

119904

sum

119894=11198871015840

119894 119896119894 minus(16119865(4)1 +

124

ℎ119865(5)1 + sdot sdot sdot)]

12059110158401015840

119899+1 = ℎ2[

119904

sum

119894=111988710158401015840

119894 119896119894 minus(12119865(4)1 +

16ℎ119865(5)1 + sdot sdot sdot)]

120591101584010158401015840

119899+1

= ℎ[

119904

sum

119894=1119887101584010158401015840

119894 119896119894 minus(119865(4)1 +

12ℎ119865(5)1 +

16ℎ2119865(6)1 + sdot sdot sdot)]

(16)

4 Mathematical Problems in Engineering

By offsetting (15) into (16) and expanding as a Taylor seriesexpansion using computer algebra package MAPLE (see[15]) the local truncation errors or the order conditions for 119904-stage fifth-order RKFD method can be written as follows

The order conditions for 119910 arefourth order

119904

sum

119894=1119887119894 =

124

(17)

fifth order119904

sum

119894=1119887119894119888119894 =

1120

(18)

The order conditions for 1199101015840 arethird order

119904

sum

119894=11198871015840

119894 =16 (19)

fourth order119904

sum

119894=11198871015840

119894 119888119894 =124

(20)

fifth order119904

sum

119894=11198871015840

119894 1198882119894 =

160

(21)

The order conditions for 11991010158401015840 aresecond order

119904

sum

119894=111988710158401015840

119894 =12 (22)

third order119904

sum

119894=111988710158401015840

119894 119888119894 =16 (23)

fourth order119904

sum

119894=111988710158401015840

119894 1198882119894 =

112

(24)

fifth order119904

sum

119894=111988710158401015840

119894 1198883119894 =

120

(25)

The order conditions for 119910101584010158401015840 arefirst order

119904

sum

119894=1119887101584010158401015840

119894 = 1 (26)

second order119904

sum

119894=1119887101584010158401015840

119894 119888119894 =12 (27)

third order119904

sum

119894=1119887101584010158401015840

119894 1198882119894 =

13 (28)

fourth order119904

sum

119894=1119887101584010158401015840

119894 1198883119894 =

14 (29)

fifth order119904

sum

119894=1119887101584010158401015840

119894 1198884119894 =

15

119904

sum

119894119895=1119887101584010158401015840

119894 119886119894119895 =1120

(30)

3 Zero-Stability of the RKFD Method

In this section we discuss the convergence of the RKFDmethod by introducing the concept of zero-stability of theRKFD method A good numerical method is a methodin which the numerical approximation to the solutionconverges and zero-stability is a significant criterion forconvergence The zero-stability concept for those numericalmethods that are used for solving first- and second-orderODEs can be seen in Lambert [16] Dormand [14] andButcher [4] The RKFD method (3)ndash(8) can be expressed inthe matrix form as follows

[[[[[

[

1 0 0 00 1 0 00 0 1 00 0 0 0

]]]]]

]

[[[[[[

[

119910119899+1

ℎ1199101015840119899+1

ℎ211991010158401015840119899+1

ℎ3119910101584010158401015840119899+1

]]]]]]

]

=

[[[[[[[[

[

1 1 12

16

0 1 1 12

0 0 1 10 0 0 1

]]]]]]]]

]

[[[[[[

[

119910119899

ℎ1199101015840119899

ℎ211991010158401015840119899

ℎ3119910101584010158401015840119899

]]]]]]

]

(31)

where 119868 = [

1 0 0 00 1 0 00 0 1 00 0 0 0

] is the identity matrix coefficients of 119910119899+1

ℎ1199101015840119899+1 ℎ

211991010158401015840119899+1 and ℎ

3119910101584010158401015840119899+1 respectively and 119860 = [

1 1 12 160 1 1 120 0 1 10 0 0 1

]

is matrix coefficients of 119910119899 ℎ1199101015840119899 ℎ

211991010158401015840119899 and ℎ

3119910101584010158401015840119899 respectively

The characteristic polynomial of the RKFD method isdenoted by 984858(120577) which can be written as follows

984858 (120577) =1003816100381610038161003816119868120577 minus119860

1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

120577 minus 1 minus1 minus12

minus16

0 120577 minus 1 minus1 minus12

0 0 120577 minus 1 minus10 0 0 120577 minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(32)

Hence 984858(120577) = (120577 minus 1)4We find that all the roots are 120577 = 1 1 1 1 Generalizing

the theorem proposed by Henrici [17] for solving specialsecond-order ODEs therefore the RKFD method is zero-stable since all roots are less than or equal to the value of 1

Mathematical Problems in Engineering 5

4 Construction of RKFD Methods

In this section we proceed to construct explicit RKFDmethods based on the order conditions which we havederived in Section 2

41 A Three-Stage RKFD Method of Order Four This sectionwill focus on the derivation of a three-stage RKFDmethod oforder four where we use the algebraic order conditions (17)(19)-(20) (22)ndash(24) and (26)ndash(29) respectivelyThe resultingsystem of equations consists of 10 nonlinear equations with14 unknown variables to be solved solving the systemsimultaneously yields a solution with four free parameters 119888311988710158401 1198871 and 1198873 as follows

1198882 = minus3 minus 41198883minus4 + 61198883

119887101584010158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

119887101584010158401015840

2 = minus16 minus 721198883 + 10811988823 minus 54119888331081198883 minus 15011988823 minus 27 + 7211988833

119887101584010158401015840

3 =1

3611988833 minus 4811988823 + 181198883

11988710158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

11988710158401015840

2 = minusminus201198883 minus 1811988833 + 3311988823 + 41081198883 minus 15011988823 minus 27 + 7211988833

11988710158401015840

3 = minusminus1 + 1198883

3611988833 minus 4811988823 + 181198883

1198871015840

2 = minusminus48119887101584011198883 + 7211988710158401119888

23 minus 2 + 111198883 minus 1211988823

12 (3 minus 81198883 + 611988823 )

1198871015840

3 = minusminus4 + 3611988710158401 + 51198883 minus 48119887101584011198883

minus961198883 + 7211988823 + 36

1198872 = minus 1198871 minus 1198873 +124

(33)

Thus these free parameters can be chosen by minimizingthe local truncation error norms of the fifth-order conditionsaccording to Dormand et al [18] However we have anotherthree free parameters 11988621 11988631 and 11988632 that do not appear infourth-order conditions but they appear in the minimizationof error equations for fifth-order conditions of 119910101584010158401015840

The error norms and global error of fifth-order conditionsare defined as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(5)119894)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(5)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

(34)

where 120591(5) 1205911015840(5) 12059110158401015840(5) and 120591101584010158401015840(5) are the local truncation errornorms for119910119910101584011991010158401015840 and119910101584010158401015840 respectively and 120591(5)119892 is the globalerror

Consequently we find the error norms of 119910 1199101015840 and 11991010158401015840

respectively as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1240

radic(minus3601198873 + 11 minus 720119888231198873 + 48011988711198883 minus 3601198871 + 96011988731198883 minus 141198883)

2

(minus2 + 31198883)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

1240

radic(minus5011988823 + 48011988710158401119888

23 minus 360119887101584011198883 + 481198883 minus 7)

2

(minus2 + 31198883)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

1120

radic(3 minus 121198883 + 1011988823 )

2

(minus2 + 31198883)2

(35)

6 Mathematical Problems in Engineering

Our goal is to choose the free parameters 1198883 11988710158401 1198871 and 1198873 such

that the error norms of fifth-order conditions have minimalvalue By plotting the graph of 12059110158401015840(5)2 versus 1198883 and choosinga small value of 1198883 in the interval [07 3] we find that 1198883 =

1720 is the optimal value which yields a minimum value for12059110158401015840(5)

2 = 3787878788 times 10minus4 Substituting the value of 1198883 =1720 into 120591(5)2 and 120591

1015840(5)2 we get

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1440

radic(3 minus 1601198871 + 2141198873)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

11760

radic(minus31 + 54411988710158401)2

(36)

Also through plotting the graph of 120591(5)2 against 1198871 and 1198873in the interval [minus01 05] and choosing a small value of 1198873we get that 1198873 = 120 is the optimal value which gives 1198871 =

17200 and 120591(5)2 = 2272727273 times 10minus4

Now utilizing the same technique where we draw thegraph of 1205911015840(5)2 versus 119887

10158401 in the interval [minus1 1] we find that

11988710158401 = 118 is the best choice and with this value of 11988710158401 we get1205911015840(5)

2 = 4419191919 times 10minus4Therefore the error equation of the fifth-order condition

of 119910101584010158401015840 is as follows

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

14802160

(1604749085

+ 6194971660900119886221 + 87611744001198862111988631

+ 87611744001198862111988632 minus 19920720292011988621

+ 3097600119886231 + 61952001198863111988632 minus 140863360011988631

+ 3097600119886232 minus 140863360011988632)12

(37)

Consequently the global error is

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172=

1144064800

(1452377332189

+ 55754744948100119886221 + 788505696001198862111988631

+ 788505696001198862111988632 minus 1792864826280011988621

+ 27878400119886231 + 557568001198863111988632

minus 1267770240011988631 + 27878400119886232

minus 1267770240011988632)12

(38)

Byminimizing the error norm in (37) and global error in (38)with respect to the free parameters 11988621 11988631 and 11988632 we get11988621 = minus15 11988631 = 19125 and 11988632 = 19125 which produces120591101584010158401015840(5)

2 = 37878787879times10minus4 and 120591(5)119892 2 = 73068870183times10minus4 Finally all the coefficients of three-stage fourth-order

RKFDmethod are written in Butcher tableau and denoted byRKFD4 method as follows

411

minus15

1720

19125

19125

17200

minus775

120

118

2091926

51926

47408

8472568

1001819

47408

13312568

20005457

(39)

42 A Three-Stage RKFD Method of Order Five In thissection a three-stage RKFD method of order five will bederivedThe algebraic order conditions up to order five ((17)-(18) (19)ndash(21) (22)ndash(25) and (26)ndash(30)) need to be solvedThe resulting system of equations consists of fifteen nonlinearequations solving the system simultaneously which resultsin a solution with three free parameters 1198871 11988621 and 11988631 asfollows

1198882 =35+radic610

1198883 =35minusradic610

119887101584010158401015840

1 =19

119887101584010158401015840

2 =49minusradic636

119887101584010158401015840

3 =49+radic636

11988710158401015840

1 =19

11988710158401015840

2 =736

minusradic618

11988710158401015840

3 =736

+radic618

1198871015840

1 =118

1198871015840

2 =118

minusradic648

1198871015840

3 =118

+radic648

Mathematical Problems in Engineering 7

1198872 =148

minusradic672

+(minus12+radic62)1198871

1198873 =148

+radic672

minus(12+radic62)1198871

11988632 = (minus131125

+16radic6125

)11988621 minus 11988631 +12625

minus3radic62500

(40)

Thus these free parameters can be chosen by minimizing thelocal truncation error norms of the sixth-order conditionsThe error norms and the global error of the sixth-orderconditions are given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(6)119894)2

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(6)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

(41)

where 120591(6) 1205911015840(6) 12059110158401015840(6) and 120591

101584010158401015840(6) are the local truncationerror norms for 119910 1199101015840 11991010158401015840 and 119910

101584010158401015840 of the RKFD methodrespectively 120591(6)119892 is the global error The error equation ofsixth-order condition for 119910with respect to the free parameter1198871 is as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

13600

radic(minus19 + 10801198871)2 (42)

The error equation 120591(6)2 has a minimum value equal to

zero at 1198871 = 191080 asymp 001759259259 which leads to 1198872 =

131080 minus 11radic62160 and 1198873 = 131080 + 11radic62160 Thetruncation error norms of the sixth-order condition of 119910 119910101584011991010158401015840 and 119910

101584010158401015840 are calculated as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

11200

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

13600

(139+ 628800119886221 minus 76800119886221radic6

minus 984011988621 minus 376011988621radic6+ 42radic6)12

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

13600

(362minus 1176011988621radic6+ 103200119886221radic6

+ 60000011988621radic611988631 minus 3684011988621 minus 3444011988631

+ 120radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 1498800119886221)12

(43)

Also the global error can be written as

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172=

13600

(minus1552011988621radic6+ 26400119886221radic6

+ 60000011988621radic611988631 minus 4668011988621 minus 3444011988631 + 511

+ 162radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 2127600119886221)

(44)

Now minimizing the error coefficients in (43) and (44) withrespect to the free parameters 11988621 11988631 we obtain 11988621 =

4059187793 and 11988631 = minus1502532215 which gives 11988632 =

1826569317Thus the error equations for 119910 1199101015840 11991010158401015840 and 119910101584010158401015840are computed and given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 = 8333333333times 10minus4

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 = 1666666668times 10minus3

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 = 1666666667times 10minus3

(45)

and global error norm is10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172= 2499999999times 10minus3 (46)

Therefore the parameters of the three-stage fifth-order RKFDmethod denoted by RKFD5 can be represented in Butchertableau as follows

35+radic610

4059187793

35minusradic610

minus1502532215

1826569317

191080

131080

minus11radic62160

131080

+11radic62160

118

118

minusradic648

118

+radic648

19

736

minusradic618

736

+radic618

19

49minusradic636

49+radic636

(47)

5 Numerical Examples

In this section some numerical examples will be solved toshow the efficiency of the new RKFD methods of order four

8 Mathematical Problems in Engineering

25 3 35 4 45

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

0

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 1 The efficiency curves for Example 1 with ℎ = 012119894 119894 =0 2 3 4

25 3 35 4 45

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 2 The efficiency curves for Example 2 with ℎ = 012119894 119894 =0 2 3 4

and order five which are denoted by RKFD4 and RKFD5respectively The comparison is made with the well knownmethods in the scientific literature We use in the numericalcomparisons the criteria based on computing the maximumerror in the solution (max error = max(|119910(119905119899) minus 119910119899|)) whichis equal to the maximum between absolute errors of thetrue solutions and the computed solutions Figures 1ndash7 showthe efficiency curves of Log10 (max error) against the com-putational effort measured by Log10 (function evaluations)required by eachmethodThe followingmethods are used forcomparison

2 22 24 26 28 3 32 34 36 38 4

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 3 The efficiency curves for Example 3 with ℎ = 012119894 119894 =0 2 3 4

14 16 18 2 22 24 26 28 3 32 34

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 4 The efficiency curves for Example 4 with ℎ = 012119894 119894 =0 2 3 4

(i) RKFD5 the three-stage fifth-order RKFD methodderived in this paper

(ii) RKFD4 the three-stage fourth-order RKFD methodderived in this paper

(iii) RK5Bs6 the six-stage fifth-order Runge-Kuttamethod given in Butcher [4]

(iv) RK5Ns6 the six-stage fifth-order Runge-Kuttamethod given in Hairer [5]

(v) RK4s4 the four-stage fourth-order Runge-Kuttamethod given in Dormand [14]

Mathematical Problems in Engineering 9

18 2 22 24 26 28 3 32 34 36 38

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 5 The efficiency curves for Example 5 with ℎ = 012119894 119894 =0 2 3 4

18 2 22 24 26 28 3 32 34 36 38

minus12

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 6 The efficiency curves for Example 6 with ℎ = 012119894 119894 =0 2 3 4

(vi) RK38 the four-stage fourth-order 38 rule Runge-Kutta method given in Butcher [4]

Example 1 The homogeneous linear problem is as follows

119910(119894V)

= minus 4119910

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 2 119910101584010158401015840 (0) = 2(48)

The exact solution is given by 119910(119909) = e119909 sin(119909) The problemis integrated in the interval [0 10]

24 26 28 3 32 34 36 38 4 42minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 7 The efficiency curves for Example 7 with ℎ = 00252119894119894 = 0 1 2 3

Example 2 The nonhomogeneous nonlinear problem is asfollows

119910(119894V)

= 1199102+ cos2 (119909) + sin (119909) minus 1

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = minus1(49)

The exact solution is given by 119910(119909) = sin(119909) The problem isintegrated in the interval [0 10]

Example 3 The homogeneous linear problem with noncon-stant coefficients is as follows

119910(119894V)

= (161199094 minus 481199092 + 12) 119910

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus2 119910101584010158401015840 (0) = 0(50)

The exact solution is given by 119910(119909) = eminus1199092 The problem is

integrated in the interval [0 3]

Example 4 The nonlinear problem is as follows

119910(119894V)

=3 sin (119910) (3 + 2sin2 (119910))

cos7 (119910)

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = 1

(51)

The exact solution is given by 119910(119909) = arcsin(119909) The problemis integrated in the interval [0 1205874]

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

of the order conditions of RKFD method is presented InSection 3 the zero-stability of RKFD method is given InSection 4 three-stage RKFDmethods of order four and orderfive are constructed In Section 5 numerical examples aregiven to show the effectiveness and competency of the newRKFD methods as compared with the well known Runge-Kutta methods from the scientific literature Conclusions aregiven in Section 6

2 Derivation of the RKFD Method

The general form of RKFD method with 119904-stage for directlysolving special fourth-order ODEs (1) can be written asfollows

119910119899+1 = 119910119899 + ℎ1199101015840

119899 +ℎ2

211991010158401015840

119899 +ℎ3

6119910101584010158401015840

119899 + ℎ4119904

sum

119894=1119887119894119896119894 (3)

1199101015840

119899+1 = 1199101015840

119899 + ℎ211991010158401015840

119899 +ℎ2

2119910101584010158401015840

119899 + ℎ3119904

sum

119894=11198871015840

119894 119896119894 (4)

11991010158401015840

119899+1 = 11991010158401015840

119899 + ℎ119910101584010158401015840

119899 + ℎ2119904

sum

119894=111988710158401015840

119894 119896119894 (5)

119910101584010158401015840

119899+1 = 119910101584010158401015840

119899 + ℎ

119904

sum

119894=1119887101584010158401015840

119894 119896119894 (6)

where

1198961 = 119891 (119909119899 119910119899) (7)

119896119894 = 119891(119909119899 + 119888119894ℎ 119910119899 + ℎ1198881198941199101015840

119899 +ℎ2

21198882119894 11991010158401015840

119899 +ℎ3

61198883119894 119910101584010158401015840

119899

+ ℎ4119894minus1sum

119895=1119886119894119895119896119895) 119894 = 2 3 119904

(8)

All parameters 119887119894 1198871015840119894 11988710158401015840119894 119887101584010158401015840119894 119886119894119895 and 119888119894 of the RKFD method

are used for 119894 119895 = 1 2 119904 and are supposed to be real TheRKFD method is an explicit method if 119886119894119895 = 0 for 119894 le 119895 andis an implicit method if 119886119894119895 = 0 for 119894 le 119895 The RKFD methodcan be represented by Butcher tableau as follows

119888 119860

119887119879

1198871015840119879

11988710158401015840119879

119887101584010158401015840119879

(9)

To determine the parameters of the RKFD method givenin (3)ndash(8) the RKFD method expression is expanded usingthe Taylor series expansion After performing some algebraicmanipulations this expansion is equated to the true solutionthat is given by the Taylor series expansion The directexpansion of the local truncation error is used to derivethe general order conditions for the RKFD method This

approach depends on the derivation of order conditions forthe Runge-Kutta method proposed in Dormand [14] TheRKFD method in (3)ndash(6) can be written as follows

119910119899+1 = 119910119899 + ℎ120595 (119909119899 119910119899)

1199101015840

119899+1 = 1199101015840

119899 + ℎ1205951015840(119909119899 119910119899)

11991010158401015840

119899+1 = 11991010158401015840

119899 + ℎ12059510158401015840(119909119899 119910119899)

119910101584010158401015840

119899+1 = 119910101584010158401015840

119899 + ℎ120595101584010158401015840(119909119899 119910119899)

(10)

where the increment functions are

120595 (119909119899 119910119899) = 1199101015840

119899 +ℎ

211991010158401015840

119899 +ℎ2

6119910101584010158401015840

119899 + ℎ3119904

sum

119894=1119887119894119896119894

1205951015840(119909119899 119910119899) = 119910

10158401015840

119899 +ℎ

2119910101584010158401015840

119899 + ℎ2119904

sum

119894=11198871015840

119894 119896119894

12059510158401015840(119909119899 119910119899) = 119910

101584010158401015840

119899 + ℎ

119904

sum

119894=111988710158401015840

119894 119896119894

120595101584010158401015840(119909119899 119910119899) =

119904

sum

119894=1119887101584010158401015840

119894 119896119894

(11)

where 119896119894 is given in (8)The first few elementary differentials for the scalar equa-

tion are

119865(4)1 = 119910

(119894V)= 119891

119865(5)1 = 119891119909 +119891119910119910

1015840

119865(6)1 = 119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2

119865(7)1 = 119891119909119909119909 + 3119891119909119909119910119910

1015840+ 3119891119909119910119910 (119910

1015840)2+ 3119891119909119910119910

10158401015840

+ 3119891119910119910119910101584011991010158401015840+119891119910119910119910 (119910

1015840)3+119891119910119910

101584010158401015840

(12)

We assume that the Taylor series increment function is ΔThe local truncation errors of 119910(119909) 1199101015840(119909) 11991010158401015840(119909) and

119910101584010158401015840(119909) can be obtained after substituting the exact solution

of (1) into (11) as follows

120591119899+1 = ℎ [120595minusΔ]

1205911015840

119899+1 = ℎ [1205951015840minusΔ1015840]

12059110158401015840

119899+1 = ℎ [12059510158401015840minusΔ10158401015840]

120591101584010158401015840

119899+1 = ℎ [120595101584010158401015840minusΔ101584010158401015840]

(13)

Mathematical Problems in Engineering 3

The Taylor series increment functions of 119910(119909) 1199101015840(119909) 11991010158401015840(119909)and 119910

101584010158401015840(119909) can be expressed as follows

Δ = 1199101015840+12ℎ11991010158401015840+16ℎ2119910101584010158401015840+

124

ℎ3119865(4)1 +

1120

ℎ4119865(5)1

+1720

ℎ5119865(6)1 +119874 (ℎ

6)

Δ1015840= 11991010158401015840+12ℎ119910101584010158401015840+16ℎ2119865(4)1 +

124

ℎ3119865(5)1

+1120

ℎ4119865(6)1 +

1720

ℎ5119865(7)1 +119874 (ℎ

6)

Δ10158401015840= 119910101584010158401015840+12ℎ119865(4)1 +

16ℎ2119865(5)1 +

124

ℎ3119865(6)1

+1120

ℎ4119865(7)1 +

1720

ℎ5119865(8)1 +119874 (ℎ

6)

Δ101584010158401015840

= 119865(4)1 +

12ℎ119865(5)1 +

16ℎ2119865(6)1 +

124

ℎ3119865(7)1

+1720

ℎ4119865(8)1 +119874 (ℎ

5)

(14)

Substituting (12) into (11) the increment functions 120595 1205951015840 12059510158401015840and 120595

101584010158401015840 for RKFD method become as follows119904

sum

119894=1119887119894119896119894

=

119904

sum

119894=1119887119894119891+

119904

sum

119894=1119887119894119888119894 (119891119909 +119891119910119910

1015840) ℎ

+12

119904

sum

119894=1119887119894119888

2119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=1119887119894119896119894

=

119904

sum

119894=1119887119894119865(4)1 +

119904

sum

119894=1119887119894119888119894ℎ119865

(5)1 +

12

119904

sum

119894=1119887119894119888

2119894 ℎ

2119865(6)1 +119874 (ℎ

3)

119904

sum

119894=11198871015840

119894 119896119894

=

119904

sum

119894=11198871015840

119894119891+

119904

sum

119894=11198871015840

119894 119888119894 (119891119909 +1198911199101199101015840) ℎ

+12

119904

sum

119894=11198871015840

119894 1198882119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=11198871015840

119894 119896119894

=

119904

sum

119894=11198871015840

119894 119865(4)1 +

119904

sum

119894=11198871015840

119894 119888119894ℎ119865(5)1 +

12

119904

sum

119894=11198871015840

119894 1198882119894 ℎ

2119865(6)1 +119874 (ℎ

3)

119904

sum

119894=111988710158401015840

119894 119896119894

=

119904

sum

119894=111988710158401015840

119894 119891+

119904

sum

119894=111988710158401015840

119894 119888119894 (119891119909 +1198911199101199101015840) ℎ

+12

119904

sum

119894=111988710158401015840

119894 1198882119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=111988710158401015840

119894 119896119894

=

119904

sum

119894=111988710158401015840

119894 119865(4)1 +

119904

sum

119894=111988710158401015840

119894 119888119894ℎ119865(5)1 +

12

119904

sum

119894=111988710158401015840

119894 1198882119894 ℎ

2119865(6)1

+119874 (ℎ3)

119904

sum

119894=1119887101584010158401015840

119894 119896119894

=

119904

sum

119894=1119887101584010158401015840

119894 119891+

119904

sum

119894=1119887101584010158401015840

119894 119888119894 (119891119909 +1198911199101199101015840) ℎ

+12

119904

sum

119894=1119887101584010158401015840

119894 1198882119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=1119887101584010158401015840

119894 119896119894

=

119904

sum

119894=1119887101584010158401015840

119894 119865(4)1 +

119904

sum

119894=1119887101584010158401015840

119894 119888119894ℎ119865(5)1 +

12

119904

sum

119894=1119887101584010158401015840

119894 1198882119894 ℎ

2119865(6)1

+119874 (ℎ3)

(15)

Using (11) and (14) the local truncation errors (13) can bewritten as follows

120591119899+1 = ℎ4[

119904

sum

119894=1119887119894119896119894 minus(

124

119865(4)1 +

1120

ℎ119865(5)1 + sdot sdot sdot)]

1205911015840

119899+1 = ℎ3[

119904

sum

119894=11198871015840

119894 119896119894 minus(16119865(4)1 +

124

ℎ119865(5)1 + sdot sdot sdot)]

12059110158401015840

119899+1 = ℎ2[

119904

sum

119894=111988710158401015840

119894 119896119894 minus(12119865(4)1 +

16ℎ119865(5)1 + sdot sdot sdot)]

120591101584010158401015840

119899+1

= ℎ[

119904

sum

119894=1119887101584010158401015840

119894 119896119894 minus(119865(4)1 +

12ℎ119865(5)1 +

16ℎ2119865(6)1 + sdot sdot sdot)]

(16)

4 Mathematical Problems in Engineering

By offsetting (15) into (16) and expanding as a Taylor seriesexpansion using computer algebra package MAPLE (see[15]) the local truncation errors or the order conditions for 119904-stage fifth-order RKFD method can be written as follows

The order conditions for 119910 arefourth order

119904

sum

119894=1119887119894 =

124

(17)

fifth order119904

sum

119894=1119887119894119888119894 =

1120

(18)

The order conditions for 1199101015840 arethird order

119904

sum

119894=11198871015840

119894 =16 (19)

fourth order119904

sum

119894=11198871015840

119894 119888119894 =124

(20)

fifth order119904

sum

119894=11198871015840

119894 1198882119894 =

160

(21)

The order conditions for 11991010158401015840 aresecond order

119904

sum

119894=111988710158401015840

119894 =12 (22)

third order119904

sum

119894=111988710158401015840

119894 119888119894 =16 (23)

fourth order119904

sum

119894=111988710158401015840

119894 1198882119894 =

112

(24)

fifth order119904

sum

119894=111988710158401015840

119894 1198883119894 =

120

(25)

The order conditions for 119910101584010158401015840 arefirst order

119904

sum

119894=1119887101584010158401015840

119894 = 1 (26)

second order119904

sum

119894=1119887101584010158401015840

119894 119888119894 =12 (27)

third order119904

sum

119894=1119887101584010158401015840

119894 1198882119894 =

13 (28)

fourth order119904

sum

119894=1119887101584010158401015840

119894 1198883119894 =

14 (29)

fifth order119904

sum

119894=1119887101584010158401015840

119894 1198884119894 =

15

119904

sum

119894119895=1119887101584010158401015840

119894 119886119894119895 =1120

(30)

3 Zero-Stability of the RKFD Method

In this section we discuss the convergence of the RKFDmethod by introducing the concept of zero-stability of theRKFD method A good numerical method is a methodin which the numerical approximation to the solutionconverges and zero-stability is a significant criterion forconvergence The zero-stability concept for those numericalmethods that are used for solving first- and second-orderODEs can be seen in Lambert [16] Dormand [14] andButcher [4] The RKFD method (3)ndash(8) can be expressed inthe matrix form as follows

[[[[[

[

1 0 0 00 1 0 00 0 1 00 0 0 0

]]]]]

]

[[[[[[

[

119910119899+1

ℎ1199101015840119899+1

ℎ211991010158401015840119899+1

ℎ3119910101584010158401015840119899+1

]]]]]]

]

=

[[[[[[[[

[

1 1 12

16

0 1 1 12

0 0 1 10 0 0 1

]]]]]]]]

]

[[[[[[

[

119910119899

ℎ1199101015840119899

ℎ211991010158401015840119899

ℎ3119910101584010158401015840119899

]]]]]]

]

(31)

where 119868 = [

1 0 0 00 1 0 00 0 1 00 0 0 0

] is the identity matrix coefficients of 119910119899+1

ℎ1199101015840119899+1 ℎ

211991010158401015840119899+1 and ℎ

3119910101584010158401015840119899+1 respectively and 119860 = [

1 1 12 160 1 1 120 0 1 10 0 0 1

]

is matrix coefficients of 119910119899 ℎ1199101015840119899 ℎ

211991010158401015840119899 and ℎ

3119910101584010158401015840119899 respectively

The characteristic polynomial of the RKFD method isdenoted by 984858(120577) which can be written as follows

984858 (120577) =1003816100381610038161003816119868120577 minus119860

1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

120577 minus 1 minus1 minus12

minus16

0 120577 minus 1 minus1 minus12

0 0 120577 minus 1 minus10 0 0 120577 minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(32)

Hence 984858(120577) = (120577 minus 1)4We find that all the roots are 120577 = 1 1 1 1 Generalizing

the theorem proposed by Henrici [17] for solving specialsecond-order ODEs therefore the RKFD method is zero-stable since all roots are less than or equal to the value of 1

Mathematical Problems in Engineering 5

4 Construction of RKFD Methods

In this section we proceed to construct explicit RKFDmethods based on the order conditions which we havederived in Section 2

41 A Three-Stage RKFD Method of Order Four This sectionwill focus on the derivation of a three-stage RKFDmethod oforder four where we use the algebraic order conditions (17)(19)-(20) (22)ndash(24) and (26)ndash(29) respectivelyThe resultingsystem of equations consists of 10 nonlinear equations with14 unknown variables to be solved solving the systemsimultaneously yields a solution with four free parameters 119888311988710158401 1198871 and 1198873 as follows

1198882 = minus3 minus 41198883minus4 + 61198883

119887101584010158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

119887101584010158401015840

2 = minus16 minus 721198883 + 10811988823 minus 54119888331081198883 minus 15011988823 minus 27 + 7211988833

119887101584010158401015840

3 =1

3611988833 minus 4811988823 + 181198883

11988710158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

11988710158401015840

2 = minusminus201198883 minus 1811988833 + 3311988823 + 41081198883 minus 15011988823 minus 27 + 7211988833

11988710158401015840

3 = minusminus1 + 1198883

3611988833 minus 4811988823 + 181198883

1198871015840

2 = minusminus48119887101584011198883 + 7211988710158401119888

23 minus 2 + 111198883 minus 1211988823

12 (3 minus 81198883 + 611988823 )

1198871015840

3 = minusminus4 + 3611988710158401 + 51198883 minus 48119887101584011198883

minus961198883 + 7211988823 + 36

1198872 = minus 1198871 minus 1198873 +124

(33)

Thus these free parameters can be chosen by minimizingthe local truncation error norms of the fifth-order conditionsaccording to Dormand et al [18] However we have anotherthree free parameters 11988621 11988631 and 11988632 that do not appear infourth-order conditions but they appear in the minimizationof error equations for fifth-order conditions of 119910101584010158401015840

The error norms and global error of fifth-order conditionsare defined as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(5)119894)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(5)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

(34)

where 120591(5) 1205911015840(5) 12059110158401015840(5) and 120591101584010158401015840(5) are the local truncation errornorms for119910119910101584011991010158401015840 and119910101584010158401015840 respectively and 120591(5)119892 is the globalerror

Consequently we find the error norms of 119910 1199101015840 and 11991010158401015840

respectively as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1240

radic(minus3601198873 + 11 minus 720119888231198873 + 48011988711198883 minus 3601198871 + 96011988731198883 minus 141198883)

2

(minus2 + 31198883)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

1240

radic(minus5011988823 + 48011988710158401119888

23 minus 360119887101584011198883 + 481198883 minus 7)

2

(minus2 + 31198883)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

1120

radic(3 minus 121198883 + 1011988823 )

2

(minus2 + 31198883)2

(35)

6 Mathematical Problems in Engineering

Our goal is to choose the free parameters 1198883 11988710158401 1198871 and 1198873 such

that the error norms of fifth-order conditions have minimalvalue By plotting the graph of 12059110158401015840(5)2 versus 1198883 and choosinga small value of 1198883 in the interval [07 3] we find that 1198883 =

1720 is the optimal value which yields a minimum value for12059110158401015840(5)

2 = 3787878788 times 10minus4 Substituting the value of 1198883 =1720 into 120591(5)2 and 120591

1015840(5)2 we get

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1440

radic(3 minus 1601198871 + 2141198873)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

11760

radic(minus31 + 54411988710158401)2

(36)

Also through plotting the graph of 120591(5)2 against 1198871 and 1198873in the interval [minus01 05] and choosing a small value of 1198873we get that 1198873 = 120 is the optimal value which gives 1198871 =

17200 and 120591(5)2 = 2272727273 times 10minus4

Now utilizing the same technique where we draw thegraph of 1205911015840(5)2 versus 119887

10158401 in the interval [minus1 1] we find that

11988710158401 = 118 is the best choice and with this value of 11988710158401 we get1205911015840(5)

2 = 4419191919 times 10minus4Therefore the error equation of the fifth-order condition

of 119910101584010158401015840 is as follows

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

14802160

(1604749085

+ 6194971660900119886221 + 87611744001198862111988631

+ 87611744001198862111988632 minus 19920720292011988621

+ 3097600119886231 + 61952001198863111988632 minus 140863360011988631

+ 3097600119886232 minus 140863360011988632)12

(37)

Consequently the global error is

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172=

1144064800

(1452377332189

+ 55754744948100119886221 + 788505696001198862111988631

+ 788505696001198862111988632 minus 1792864826280011988621

+ 27878400119886231 + 557568001198863111988632

minus 1267770240011988631 + 27878400119886232

minus 1267770240011988632)12

(38)

Byminimizing the error norm in (37) and global error in (38)with respect to the free parameters 11988621 11988631 and 11988632 we get11988621 = minus15 11988631 = 19125 and 11988632 = 19125 which produces120591101584010158401015840(5)

2 = 37878787879times10minus4 and 120591(5)119892 2 = 73068870183times10minus4 Finally all the coefficients of three-stage fourth-order

RKFDmethod are written in Butcher tableau and denoted byRKFD4 method as follows

411

minus15

1720

19125

19125

17200

minus775

120

118

2091926

51926

47408

8472568

1001819

47408

13312568

20005457

(39)

42 A Three-Stage RKFD Method of Order Five In thissection a three-stage RKFD method of order five will bederivedThe algebraic order conditions up to order five ((17)-(18) (19)ndash(21) (22)ndash(25) and (26)ndash(30)) need to be solvedThe resulting system of equations consists of fifteen nonlinearequations solving the system simultaneously which resultsin a solution with three free parameters 1198871 11988621 and 11988631 asfollows

1198882 =35+radic610

1198883 =35minusradic610

119887101584010158401015840

1 =19

119887101584010158401015840

2 =49minusradic636

119887101584010158401015840

3 =49+radic636

11988710158401015840

1 =19

11988710158401015840

2 =736

minusradic618

11988710158401015840

3 =736

+radic618

1198871015840

1 =118

1198871015840

2 =118

minusradic648

1198871015840

3 =118

+radic648

Mathematical Problems in Engineering 7

1198872 =148

minusradic672

+(minus12+radic62)1198871

1198873 =148

+radic672

minus(12+radic62)1198871

11988632 = (minus131125

+16radic6125

)11988621 minus 11988631 +12625

minus3radic62500

(40)

Thus these free parameters can be chosen by minimizing thelocal truncation error norms of the sixth-order conditionsThe error norms and the global error of the sixth-orderconditions are given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(6)119894)2

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(6)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

(41)

where 120591(6) 1205911015840(6) 12059110158401015840(6) and 120591

101584010158401015840(6) are the local truncationerror norms for 119910 1199101015840 11991010158401015840 and 119910

101584010158401015840 of the RKFD methodrespectively 120591(6)119892 is the global error The error equation ofsixth-order condition for 119910with respect to the free parameter1198871 is as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

13600

radic(minus19 + 10801198871)2 (42)

The error equation 120591(6)2 has a minimum value equal to

zero at 1198871 = 191080 asymp 001759259259 which leads to 1198872 =

131080 minus 11radic62160 and 1198873 = 131080 + 11radic62160 Thetruncation error norms of the sixth-order condition of 119910 119910101584011991010158401015840 and 119910

101584010158401015840 are calculated as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

11200

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

13600

(139+ 628800119886221 minus 76800119886221radic6

minus 984011988621 minus 376011988621radic6+ 42radic6)12

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

13600

(362minus 1176011988621radic6+ 103200119886221radic6

+ 60000011988621radic611988631 minus 3684011988621 minus 3444011988631

+ 120radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 1498800119886221)12

(43)

Also the global error can be written as

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172=

13600

(minus1552011988621radic6+ 26400119886221radic6

+ 60000011988621radic611988631 minus 4668011988621 minus 3444011988631 + 511

+ 162radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 2127600119886221)

(44)

Now minimizing the error coefficients in (43) and (44) withrespect to the free parameters 11988621 11988631 we obtain 11988621 =

4059187793 and 11988631 = minus1502532215 which gives 11988632 =

1826569317Thus the error equations for 119910 1199101015840 11991010158401015840 and 119910101584010158401015840are computed and given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 = 8333333333times 10minus4

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 = 1666666668times 10minus3

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 = 1666666667times 10minus3

(45)

and global error norm is10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172= 2499999999times 10minus3 (46)

Therefore the parameters of the three-stage fifth-order RKFDmethod denoted by RKFD5 can be represented in Butchertableau as follows

35+radic610

4059187793

35minusradic610

minus1502532215

1826569317

191080

131080

minus11radic62160

131080

+11radic62160

118

118

minusradic648

118

+radic648

19

736

minusradic618

736

+radic618

19

49minusradic636

49+radic636

(47)

5 Numerical Examples

In this section some numerical examples will be solved toshow the efficiency of the new RKFD methods of order four

8 Mathematical Problems in Engineering

25 3 35 4 45

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

0

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 1 The efficiency curves for Example 1 with ℎ = 012119894 119894 =0 2 3 4

25 3 35 4 45

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 2 The efficiency curves for Example 2 with ℎ = 012119894 119894 =0 2 3 4

and order five which are denoted by RKFD4 and RKFD5respectively The comparison is made with the well knownmethods in the scientific literature We use in the numericalcomparisons the criteria based on computing the maximumerror in the solution (max error = max(|119910(119905119899) minus 119910119899|)) whichis equal to the maximum between absolute errors of thetrue solutions and the computed solutions Figures 1ndash7 showthe efficiency curves of Log10 (max error) against the com-putational effort measured by Log10 (function evaluations)required by eachmethodThe followingmethods are used forcomparison

2 22 24 26 28 3 32 34 36 38 4

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 3 The efficiency curves for Example 3 with ℎ = 012119894 119894 =0 2 3 4

14 16 18 2 22 24 26 28 3 32 34

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 4 The efficiency curves for Example 4 with ℎ = 012119894 119894 =0 2 3 4

(i) RKFD5 the three-stage fifth-order RKFD methodderived in this paper

(ii) RKFD4 the three-stage fourth-order RKFD methodderived in this paper

(iii) RK5Bs6 the six-stage fifth-order Runge-Kuttamethod given in Butcher [4]

(iv) RK5Ns6 the six-stage fifth-order Runge-Kuttamethod given in Hairer [5]

(v) RK4s4 the four-stage fourth-order Runge-Kuttamethod given in Dormand [14]

Mathematical Problems in Engineering 9

18 2 22 24 26 28 3 32 34 36 38

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 5 The efficiency curves for Example 5 with ℎ = 012119894 119894 =0 2 3 4

18 2 22 24 26 28 3 32 34 36 38

minus12

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 6 The efficiency curves for Example 6 with ℎ = 012119894 119894 =0 2 3 4

(vi) RK38 the four-stage fourth-order 38 rule Runge-Kutta method given in Butcher [4]

Example 1 The homogeneous linear problem is as follows

119910(119894V)

= minus 4119910

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 2 119910101584010158401015840 (0) = 2(48)

The exact solution is given by 119910(119909) = e119909 sin(119909) The problemis integrated in the interval [0 10]

24 26 28 3 32 34 36 38 4 42minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 7 The efficiency curves for Example 7 with ℎ = 00252119894119894 = 0 1 2 3

Example 2 The nonhomogeneous nonlinear problem is asfollows

119910(119894V)

= 1199102+ cos2 (119909) + sin (119909) minus 1

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = minus1(49)

The exact solution is given by 119910(119909) = sin(119909) The problem isintegrated in the interval [0 10]

Example 3 The homogeneous linear problem with noncon-stant coefficients is as follows

119910(119894V)

= (161199094 minus 481199092 + 12) 119910

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus2 119910101584010158401015840 (0) = 0(50)

The exact solution is given by 119910(119909) = eminus1199092 The problem is

integrated in the interval [0 3]

Example 4 The nonlinear problem is as follows

119910(119894V)

=3 sin (119910) (3 + 2sin2 (119910))

cos7 (119910)

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = 1

(51)

The exact solution is given by 119910(119909) = arcsin(119909) The problemis integrated in the interval [0 1205874]

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

The Taylor series increment functions of 119910(119909) 1199101015840(119909) 11991010158401015840(119909)and 119910

101584010158401015840(119909) can be expressed as follows

Δ = 1199101015840+12ℎ11991010158401015840+16ℎ2119910101584010158401015840+

124

ℎ3119865(4)1 +

1120

ℎ4119865(5)1

+1720

ℎ5119865(6)1 +119874 (ℎ

6)

Δ1015840= 11991010158401015840+12ℎ119910101584010158401015840+16ℎ2119865(4)1 +

124

ℎ3119865(5)1

+1120

ℎ4119865(6)1 +

1720

ℎ5119865(7)1 +119874 (ℎ

6)

Δ10158401015840= 119910101584010158401015840+12ℎ119865(4)1 +

16ℎ2119865(5)1 +

124

ℎ3119865(6)1

+1120

ℎ4119865(7)1 +

1720

ℎ5119865(8)1 +119874 (ℎ

6)

Δ101584010158401015840

= 119865(4)1 +

12ℎ119865(5)1 +

16ℎ2119865(6)1 +

124

ℎ3119865(7)1

+1720

ℎ4119865(8)1 +119874 (ℎ

5)

(14)

Substituting (12) into (11) the increment functions 120595 1205951015840 12059510158401015840and 120595

101584010158401015840 for RKFD method become as follows119904

sum

119894=1119887119894119896119894

=

119904

sum

119894=1119887119894119891+

119904

sum

119894=1119887119894119888119894 (119891119909 +119891119910119910

1015840) ℎ

+12

119904

sum

119894=1119887119894119888

2119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=1119887119894119896119894

=

119904

sum

119894=1119887119894119865(4)1 +

119904

sum

119894=1119887119894119888119894ℎ119865

(5)1 +

12

119904

sum

119894=1119887119894119888

2119894 ℎ

2119865(6)1 +119874 (ℎ

3)

119904

sum

119894=11198871015840

119894 119896119894

=

119904

sum

119894=11198871015840

119894119891+

119904

sum

119894=11198871015840

119894 119888119894 (119891119909 +1198911199101199101015840) ℎ

+12

119904

sum

119894=11198871015840

119894 1198882119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=11198871015840

119894 119896119894

=

119904

sum

119894=11198871015840

119894 119865(4)1 +

119904

sum

119894=11198871015840

119894 119888119894ℎ119865(5)1 +

12

119904

sum

119894=11198871015840

119894 1198882119894 ℎ

2119865(6)1 +119874 (ℎ

3)

119904

sum

119894=111988710158401015840

119894 119896119894

=

119904

sum

119894=111988710158401015840

119894 119891+

119904

sum

119894=111988710158401015840

119894 119888119894 (119891119909 +1198911199101199101015840) ℎ

+12

119904

sum

119894=111988710158401015840

119894 1198882119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=111988710158401015840

119894 119896119894

=

119904

sum

119894=111988710158401015840

119894 119865(4)1 +

119904

sum

119894=111988710158401015840

119894 119888119894ℎ119865(5)1 +

12

119904

sum

119894=111988710158401015840

119894 1198882119894 ℎ

2119865(6)1

+119874 (ℎ3)

119904

sum

119894=1119887101584010158401015840

119894 119896119894

=

119904

sum

119894=1119887101584010158401015840

119894 119891+

119904

sum

119894=1119887101584010158401015840

119894 119888119894 (119891119909 +1198911199101199101015840) ℎ

+12

119904

sum

119894=1119887101584010158401015840

119894 1198882119894 (119891119909119909 + 2119891119909119910119910

1015840+119891119910119910

10158401015840+119891119910119910 (119910

1015840)2) ℎ

2

+119874 (ℎ3)

119904

sum

119894=1119887101584010158401015840

119894 119896119894

=

119904

sum

119894=1119887101584010158401015840

119894 119865(4)1 +

119904

sum

119894=1119887101584010158401015840

119894 119888119894ℎ119865(5)1 +

12

119904

sum

119894=1119887101584010158401015840

119894 1198882119894 ℎ

2119865(6)1

+119874 (ℎ3)

(15)

Using (11) and (14) the local truncation errors (13) can bewritten as follows

120591119899+1 = ℎ4[

119904

sum

119894=1119887119894119896119894 minus(

124

119865(4)1 +

1120

ℎ119865(5)1 + sdot sdot sdot)]

1205911015840

119899+1 = ℎ3[

119904

sum

119894=11198871015840

119894 119896119894 minus(16119865(4)1 +

124

ℎ119865(5)1 + sdot sdot sdot)]

12059110158401015840

119899+1 = ℎ2[

119904

sum

119894=111988710158401015840

119894 119896119894 minus(12119865(4)1 +

16ℎ119865(5)1 + sdot sdot sdot)]

120591101584010158401015840

119899+1

= ℎ[

119904

sum

119894=1119887101584010158401015840

119894 119896119894 minus(119865(4)1 +

12ℎ119865(5)1 +

16ℎ2119865(6)1 + sdot sdot sdot)]

(16)

4 Mathematical Problems in Engineering

By offsetting (15) into (16) and expanding as a Taylor seriesexpansion using computer algebra package MAPLE (see[15]) the local truncation errors or the order conditions for 119904-stage fifth-order RKFD method can be written as follows

The order conditions for 119910 arefourth order

119904

sum

119894=1119887119894 =

124

(17)

fifth order119904

sum

119894=1119887119894119888119894 =

1120

(18)

The order conditions for 1199101015840 arethird order

119904

sum

119894=11198871015840

119894 =16 (19)

fourth order119904

sum

119894=11198871015840

119894 119888119894 =124

(20)

fifth order119904

sum

119894=11198871015840

119894 1198882119894 =

160

(21)

The order conditions for 11991010158401015840 aresecond order

119904

sum

119894=111988710158401015840

119894 =12 (22)

third order119904

sum

119894=111988710158401015840

119894 119888119894 =16 (23)

fourth order119904

sum

119894=111988710158401015840

119894 1198882119894 =

112

(24)

fifth order119904

sum

119894=111988710158401015840

119894 1198883119894 =

120

(25)

The order conditions for 119910101584010158401015840 arefirst order

119904

sum

119894=1119887101584010158401015840

119894 = 1 (26)

second order119904

sum

119894=1119887101584010158401015840

119894 119888119894 =12 (27)

third order119904

sum

119894=1119887101584010158401015840

119894 1198882119894 =

13 (28)

fourth order119904

sum

119894=1119887101584010158401015840

119894 1198883119894 =

14 (29)

fifth order119904

sum

119894=1119887101584010158401015840

119894 1198884119894 =

15

119904

sum

119894119895=1119887101584010158401015840

119894 119886119894119895 =1120

(30)

3 Zero-Stability of the RKFD Method

In this section we discuss the convergence of the RKFDmethod by introducing the concept of zero-stability of theRKFD method A good numerical method is a methodin which the numerical approximation to the solutionconverges and zero-stability is a significant criterion forconvergence The zero-stability concept for those numericalmethods that are used for solving first- and second-orderODEs can be seen in Lambert [16] Dormand [14] andButcher [4] The RKFD method (3)ndash(8) can be expressed inthe matrix form as follows

[[[[[

[

1 0 0 00 1 0 00 0 1 00 0 0 0

]]]]]

]

[[[[[[

[

119910119899+1

ℎ1199101015840119899+1

ℎ211991010158401015840119899+1

ℎ3119910101584010158401015840119899+1

]]]]]]

]

=

[[[[[[[[

[

1 1 12

16

0 1 1 12

0 0 1 10 0 0 1

]]]]]]]]

]

[[[[[[

[

119910119899

ℎ1199101015840119899

ℎ211991010158401015840119899

ℎ3119910101584010158401015840119899

]]]]]]

]

(31)

where 119868 = [

1 0 0 00 1 0 00 0 1 00 0 0 0

] is the identity matrix coefficients of 119910119899+1

ℎ1199101015840119899+1 ℎ

211991010158401015840119899+1 and ℎ

3119910101584010158401015840119899+1 respectively and 119860 = [

1 1 12 160 1 1 120 0 1 10 0 0 1

]

is matrix coefficients of 119910119899 ℎ1199101015840119899 ℎ

211991010158401015840119899 and ℎ

3119910101584010158401015840119899 respectively

The characteristic polynomial of the RKFD method isdenoted by 984858(120577) which can be written as follows

984858 (120577) =1003816100381610038161003816119868120577 minus119860

1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

120577 minus 1 minus1 minus12

minus16

0 120577 minus 1 minus1 minus12

0 0 120577 minus 1 minus10 0 0 120577 minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(32)

Hence 984858(120577) = (120577 minus 1)4We find that all the roots are 120577 = 1 1 1 1 Generalizing

the theorem proposed by Henrici [17] for solving specialsecond-order ODEs therefore the RKFD method is zero-stable since all roots are less than or equal to the value of 1

Mathematical Problems in Engineering 5

4 Construction of RKFD Methods

In this section we proceed to construct explicit RKFDmethods based on the order conditions which we havederived in Section 2

41 A Three-Stage RKFD Method of Order Four This sectionwill focus on the derivation of a three-stage RKFDmethod oforder four where we use the algebraic order conditions (17)(19)-(20) (22)ndash(24) and (26)ndash(29) respectivelyThe resultingsystem of equations consists of 10 nonlinear equations with14 unknown variables to be solved solving the systemsimultaneously yields a solution with four free parameters 119888311988710158401 1198871 and 1198873 as follows

1198882 = minus3 minus 41198883minus4 + 61198883

119887101584010158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

119887101584010158401015840

2 = minus16 minus 721198883 + 10811988823 minus 54119888331081198883 minus 15011988823 minus 27 + 7211988833

119887101584010158401015840

3 =1

3611988833 minus 4811988823 + 181198883

11988710158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

11988710158401015840

2 = minusminus201198883 minus 1811988833 + 3311988823 + 41081198883 minus 15011988823 minus 27 + 7211988833

11988710158401015840

3 = minusminus1 + 1198883

3611988833 minus 4811988823 + 181198883

1198871015840

2 = minusminus48119887101584011198883 + 7211988710158401119888

23 minus 2 + 111198883 minus 1211988823

12 (3 minus 81198883 + 611988823 )

1198871015840

3 = minusminus4 + 3611988710158401 + 51198883 minus 48119887101584011198883

minus961198883 + 7211988823 + 36

1198872 = minus 1198871 minus 1198873 +124

(33)

Thus these free parameters can be chosen by minimizingthe local truncation error norms of the fifth-order conditionsaccording to Dormand et al [18] However we have anotherthree free parameters 11988621 11988631 and 11988632 that do not appear infourth-order conditions but they appear in the minimizationof error equations for fifth-order conditions of 119910101584010158401015840

The error norms and global error of fifth-order conditionsare defined as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(5)119894)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(5)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

(34)

where 120591(5) 1205911015840(5) 12059110158401015840(5) and 120591101584010158401015840(5) are the local truncation errornorms for119910119910101584011991010158401015840 and119910101584010158401015840 respectively and 120591(5)119892 is the globalerror

Consequently we find the error norms of 119910 1199101015840 and 11991010158401015840

respectively as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1240

radic(minus3601198873 + 11 minus 720119888231198873 + 48011988711198883 minus 3601198871 + 96011988731198883 minus 141198883)

2

(minus2 + 31198883)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

1240

radic(minus5011988823 + 48011988710158401119888

23 minus 360119887101584011198883 + 481198883 minus 7)

2

(minus2 + 31198883)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

1120

radic(3 minus 121198883 + 1011988823 )

2

(minus2 + 31198883)2

(35)

6 Mathematical Problems in Engineering

Our goal is to choose the free parameters 1198883 11988710158401 1198871 and 1198873 such

that the error norms of fifth-order conditions have minimalvalue By plotting the graph of 12059110158401015840(5)2 versus 1198883 and choosinga small value of 1198883 in the interval [07 3] we find that 1198883 =

1720 is the optimal value which yields a minimum value for12059110158401015840(5)

2 = 3787878788 times 10minus4 Substituting the value of 1198883 =1720 into 120591(5)2 and 120591

1015840(5)2 we get

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1440

radic(3 minus 1601198871 + 2141198873)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

11760

radic(minus31 + 54411988710158401)2

(36)

Also through plotting the graph of 120591(5)2 against 1198871 and 1198873in the interval [minus01 05] and choosing a small value of 1198873we get that 1198873 = 120 is the optimal value which gives 1198871 =

17200 and 120591(5)2 = 2272727273 times 10minus4

Now utilizing the same technique where we draw thegraph of 1205911015840(5)2 versus 119887

10158401 in the interval [minus1 1] we find that

11988710158401 = 118 is the best choice and with this value of 11988710158401 we get1205911015840(5)

2 = 4419191919 times 10minus4Therefore the error equation of the fifth-order condition

of 119910101584010158401015840 is as follows

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

14802160

(1604749085

+ 6194971660900119886221 + 87611744001198862111988631

+ 87611744001198862111988632 minus 19920720292011988621

+ 3097600119886231 + 61952001198863111988632 minus 140863360011988631

+ 3097600119886232 minus 140863360011988632)12

(37)

Consequently the global error is

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172=

1144064800

(1452377332189

+ 55754744948100119886221 + 788505696001198862111988631

+ 788505696001198862111988632 minus 1792864826280011988621

+ 27878400119886231 + 557568001198863111988632

minus 1267770240011988631 + 27878400119886232

minus 1267770240011988632)12

(38)

Byminimizing the error norm in (37) and global error in (38)with respect to the free parameters 11988621 11988631 and 11988632 we get11988621 = minus15 11988631 = 19125 and 11988632 = 19125 which produces120591101584010158401015840(5)

2 = 37878787879times10minus4 and 120591(5)119892 2 = 73068870183times10minus4 Finally all the coefficients of three-stage fourth-order

RKFDmethod are written in Butcher tableau and denoted byRKFD4 method as follows

411

minus15

1720

19125

19125

17200

minus775

120

118

2091926

51926

47408

8472568

1001819

47408

13312568

20005457

(39)

42 A Three-Stage RKFD Method of Order Five In thissection a three-stage RKFD method of order five will bederivedThe algebraic order conditions up to order five ((17)-(18) (19)ndash(21) (22)ndash(25) and (26)ndash(30)) need to be solvedThe resulting system of equations consists of fifteen nonlinearequations solving the system simultaneously which resultsin a solution with three free parameters 1198871 11988621 and 11988631 asfollows

1198882 =35+radic610

1198883 =35minusradic610

119887101584010158401015840

1 =19

119887101584010158401015840

2 =49minusradic636

119887101584010158401015840

3 =49+radic636

11988710158401015840

1 =19

11988710158401015840

2 =736

minusradic618

11988710158401015840

3 =736

+radic618

1198871015840

1 =118

1198871015840

2 =118

minusradic648

1198871015840

3 =118

+radic648

Mathematical Problems in Engineering 7

1198872 =148

minusradic672

+(minus12+radic62)1198871

1198873 =148

+radic672

minus(12+radic62)1198871

11988632 = (minus131125

+16radic6125

)11988621 minus 11988631 +12625

minus3radic62500

(40)

Thus these free parameters can be chosen by minimizing thelocal truncation error norms of the sixth-order conditionsThe error norms and the global error of the sixth-orderconditions are given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(6)119894)2

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(6)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

(41)

where 120591(6) 1205911015840(6) 12059110158401015840(6) and 120591

101584010158401015840(6) are the local truncationerror norms for 119910 1199101015840 11991010158401015840 and 119910

101584010158401015840 of the RKFD methodrespectively 120591(6)119892 is the global error The error equation ofsixth-order condition for 119910with respect to the free parameter1198871 is as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

13600

radic(minus19 + 10801198871)2 (42)

The error equation 120591(6)2 has a minimum value equal to

zero at 1198871 = 191080 asymp 001759259259 which leads to 1198872 =

131080 minus 11radic62160 and 1198873 = 131080 + 11radic62160 Thetruncation error norms of the sixth-order condition of 119910 119910101584011991010158401015840 and 119910

101584010158401015840 are calculated as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

11200

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

13600

(139+ 628800119886221 minus 76800119886221radic6

minus 984011988621 minus 376011988621radic6+ 42radic6)12

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

13600

(362minus 1176011988621radic6+ 103200119886221radic6

+ 60000011988621radic611988631 minus 3684011988621 minus 3444011988631

+ 120radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 1498800119886221)12

(43)

Also the global error can be written as

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172=

13600

(minus1552011988621radic6+ 26400119886221radic6

+ 60000011988621radic611988631 minus 4668011988621 minus 3444011988631 + 511

+ 162radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 2127600119886221)

(44)

Now minimizing the error coefficients in (43) and (44) withrespect to the free parameters 11988621 11988631 we obtain 11988621 =

4059187793 and 11988631 = minus1502532215 which gives 11988632 =

1826569317Thus the error equations for 119910 1199101015840 11991010158401015840 and 119910101584010158401015840are computed and given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 = 8333333333times 10minus4

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 = 1666666668times 10minus3

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 = 1666666667times 10minus3

(45)

and global error norm is10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172= 2499999999times 10minus3 (46)

Therefore the parameters of the three-stage fifth-order RKFDmethod denoted by RKFD5 can be represented in Butchertableau as follows

35+radic610

4059187793

35minusradic610

minus1502532215

1826569317

191080

131080

minus11radic62160

131080

+11radic62160

118

118

minusradic648

118

+radic648

19

736

minusradic618

736

+radic618

19

49minusradic636

49+radic636

(47)

5 Numerical Examples

In this section some numerical examples will be solved toshow the efficiency of the new RKFD methods of order four

8 Mathematical Problems in Engineering

25 3 35 4 45

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

0

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 1 The efficiency curves for Example 1 with ℎ = 012119894 119894 =0 2 3 4

25 3 35 4 45

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 2 The efficiency curves for Example 2 with ℎ = 012119894 119894 =0 2 3 4

and order five which are denoted by RKFD4 and RKFD5respectively The comparison is made with the well knownmethods in the scientific literature We use in the numericalcomparisons the criteria based on computing the maximumerror in the solution (max error = max(|119910(119905119899) minus 119910119899|)) whichis equal to the maximum between absolute errors of thetrue solutions and the computed solutions Figures 1ndash7 showthe efficiency curves of Log10 (max error) against the com-putational effort measured by Log10 (function evaluations)required by eachmethodThe followingmethods are used forcomparison

2 22 24 26 28 3 32 34 36 38 4

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 3 The efficiency curves for Example 3 with ℎ = 012119894 119894 =0 2 3 4

14 16 18 2 22 24 26 28 3 32 34

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 4 The efficiency curves for Example 4 with ℎ = 012119894 119894 =0 2 3 4

(i) RKFD5 the three-stage fifth-order RKFD methodderived in this paper

(ii) RKFD4 the three-stage fourth-order RKFD methodderived in this paper

(iii) RK5Bs6 the six-stage fifth-order Runge-Kuttamethod given in Butcher [4]

(iv) RK5Ns6 the six-stage fifth-order Runge-Kuttamethod given in Hairer [5]

(v) RK4s4 the four-stage fourth-order Runge-Kuttamethod given in Dormand [14]

Mathematical Problems in Engineering 9

18 2 22 24 26 28 3 32 34 36 38

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 5 The efficiency curves for Example 5 with ℎ = 012119894 119894 =0 2 3 4

18 2 22 24 26 28 3 32 34 36 38

minus12

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 6 The efficiency curves for Example 6 with ℎ = 012119894 119894 =0 2 3 4

(vi) RK38 the four-stage fourth-order 38 rule Runge-Kutta method given in Butcher [4]

Example 1 The homogeneous linear problem is as follows

119910(119894V)

= minus 4119910

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 2 119910101584010158401015840 (0) = 2(48)

The exact solution is given by 119910(119909) = e119909 sin(119909) The problemis integrated in the interval [0 10]

24 26 28 3 32 34 36 38 4 42minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 7 The efficiency curves for Example 7 with ℎ = 00252119894119894 = 0 1 2 3

Example 2 The nonhomogeneous nonlinear problem is asfollows

119910(119894V)

= 1199102+ cos2 (119909) + sin (119909) minus 1

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = minus1(49)

The exact solution is given by 119910(119909) = sin(119909) The problem isintegrated in the interval [0 10]

Example 3 The homogeneous linear problem with noncon-stant coefficients is as follows

119910(119894V)

= (161199094 minus 481199092 + 12) 119910

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus2 119910101584010158401015840 (0) = 0(50)

The exact solution is given by 119910(119909) = eminus1199092 The problem is

integrated in the interval [0 3]

Example 4 The nonlinear problem is as follows

119910(119894V)

=3 sin (119910) (3 + 2sin2 (119910))

cos7 (119910)

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = 1

(51)

The exact solution is given by 119910(119909) = arcsin(119909) The problemis integrated in the interval [0 1205874]

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

By offsetting (15) into (16) and expanding as a Taylor seriesexpansion using computer algebra package MAPLE (see[15]) the local truncation errors or the order conditions for 119904-stage fifth-order RKFD method can be written as follows

The order conditions for 119910 arefourth order

119904

sum

119894=1119887119894 =

124

(17)

fifth order119904

sum

119894=1119887119894119888119894 =

1120

(18)

The order conditions for 1199101015840 arethird order

119904

sum

119894=11198871015840

119894 =16 (19)

fourth order119904

sum

119894=11198871015840

119894 119888119894 =124

(20)

fifth order119904

sum

119894=11198871015840

119894 1198882119894 =

160

(21)

The order conditions for 11991010158401015840 aresecond order

119904

sum

119894=111988710158401015840

119894 =12 (22)

third order119904

sum

119894=111988710158401015840

119894 119888119894 =16 (23)

fourth order119904

sum

119894=111988710158401015840

119894 1198882119894 =

112

(24)

fifth order119904

sum

119894=111988710158401015840

119894 1198883119894 =

120

(25)

The order conditions for 119910101584010158401015840 arefirst order

119904

sum

119894=1119887101584010158401015840

119894 = 1 (26)

second order119904

sum

119894=1119887101584010158401015840

119894 119888119894 =12 (27)

third order119904

sum

119894=1119887101584010158401015840

119894 1198882119894 =

13 (28)

fourth order119904

sum

119894=1119887101584010158401015840

119894 1198883119894 =

14 (29)

fifth order119904

sum

119894=1119887101584010158401015840

119894 1198884119894 =

15

119904

sum

119894119895=1119887101584010158401015840

119894 119886119894119895 =1120

(30)

3 Zero-Stability of the RKFD Method

In this section we discuss the convergence of the RKFDmethod by introducing the concept of zero-stability of theRKFD method A good numerical method is a methodin which the numerical approximation to the solutionconverges and zero-stability is a significant criterion forconvergence The zero-stability concept for those numericalmethods that are used for solving first- and second-orderODEs can be seen in Lambert [16] Dormand [14] andButcher [4] The RKFD method (3)ndash(8) can be expressed inthe matrix form as follows

[[[[[

[

1 0 0 00 1 0 00 0 1 00 0 0 0

]]]]]

]

[[[[[[

[

119910119899+1

ℎ1199101015840119899+1

ℎ211991010158401015840119899+1

ℎ3119910101584010158401015840119899+1

]]]]]]

]

=

[[[[[[[[

[

1 1 12

16

0 1 1 12

0 0 1 10 0 0 1

]]]]]]]]

]

[[[[[[

[

119910119899

ℎ1199101015840119899

ℎ211991010158401015840119899

ℎ3119910101584010158401015840119899

]]]]]]

]

(31)

where 119868 = [

1 0 0 00 1 0 00 0 1 00 0 0 0

] is the identity matrix coefficients of 119910119899+1

ℎ1199101015840119899+1 ℎ

211991010158401015840119899+1 and ℎ

3119910101584010158401015840119899+1 respectively and 119860 = [

1 1 12 160 1 1 120 0 1 10 0 0 1

]

is matrix coefficients of 119910119899 ℎ1199101015840119899 ℎ

211991010158401015840119899 and ℎ

3119910101584010158401015840119899 respectively

The characteristic polynomial of the RKFD method isdenoted by 984858(120577) which can be written as follows

984858 (120577) =1003816100381610038161003816119868120577 minus119860

1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

120577 minus 1 minus1 minus12

minus16

0 120577 minus 1 minus1 minus12

0 0 120577 minus 1 minus10 0 0 120577 minus 1

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(32)

Hence 984858(120577) = (120577 minus 1)4We find that all the roots are 120577 = 1 1 1 1 Generalizing

the theorem proposed by Henrici [17] for solving specialsecond-order ODEs therefore the RKFD method is zero-stable since all roots are less than or equal to the value of 1

Mathematical Problems in Engineering 5

4 Construction of RKFD Methods

In this section we proceed to construct explicit RKFDmethods based on the order conditions which we havederived in Section 2

41 A Three-Stage RKFD Method of Order Four This sectionwill focus on the derivation of a three-stage RKFDmethod oforder four where we use the algebraic order conditions (17)(19)-(20) (22)ndash(24) and (26)ndash(29) respectivelyThe resultingsystem of equations consists of 10 nonlinear equations with14 unknown variables to be solved solving the systemsimultaneously yields a solution with four free parameters 119888311988710158401 1198871 and 1198873 as follows

1198882 = minus3 minus 41198883minus4 + 61198883

119887101584010158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

119887101584010158401015840

2 = minus16 minus 721198883 + 10811988823 minus 54119888331081198883 minus 15011988823 minus 27 + 7211988833

119887101584010158401015840

3 =1

3611988833 minus 4811988823 + 181198883

11988710158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

11988710158401015840

2 = minusminus201198883 minus 1811988833 + 3311988823 + 41081198883 minus 15011988823 minus 27 + 7211988833

11988710158401015840

3 = minusminus1 + 1198883

3611988833 minus 4811988823 + 181198883

1198871015840

2 = minusminus48119887101584011198883 + 7211988710158401119888

23 minus 2 + 111198883 minus 1211988823

12 (3 minus 81198883 + 611988823 )

1198871015840

3 = minusminus4 + 3611988710158401 + 51198883 minus 48119887101584011198883

minus961198883 + 7211988823 + 36

1198872 = minus 1198871 minus 1198873 +124

(33)

Thus these free parameters can be chosen by minimizingthe local truncation error norms of the fifth-order conditionsaccording to Dormand et al [18] However we have anotherthree free parameters 11988621 11988631 and 11988632 that do not appear infourth-order conditions but they appear in the minimizationof error equations for fifth-order conditions of 119910101584010158401015840

The error norms and global error of fifth-order conditionsare defined as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(5)119894)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(5)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

(34)

where 120591(5) 1205911015840(5) 12059110158401015840(5) and 120591101584010158401015840(5) are the local truncation errornorms for119910119910101584011991010158401015840 and119910101584010158401015840 respectively and 120591(5)119892 is the globalerror

Consequently we find the error norms of 119910 1199101015840 and 11991010158401015840

respectively as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1240

radic(minus3601198873 + 11 minus 720119888231198873 + 48011988711198883 minus 3601198871 + 96011988731198883 minus 141198883)

2

(minus2 + 31198883)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

1240

radic(minus5011988823 + 48011988710158401119888

23 minus 360119887101584011198883 + 481198883 minus 7)

2

(minus2 + 31198883)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

1120

radic(3 minus 121198883 + 1011988823 )

2

(minus2 + 31198883)2

(35)

6 Mathematical Problems in Engineering

Our goal is to choose the free parameters 1198883 11988710158401 1198871 and 1198873 such

that the error norms of fifth-order conditions have minimalvalue By plotting the graph of 12059110158401015840(5)2 versus 1198883 and choosinga small value of 1198883 in the interval [07 3] we find that 1198883 =

1720 is the optimal value which yields a minimum value for12059110158401015840(5)

2 = 3787878788 times 10minus4 Substituting the value of 1198883 =1720 into 120591(5)2 and 120591

1015840(5)2 we get

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1440

radic(3 minus 1601198871 + 2141198873)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

11760

radic(minus31 + 54411988710158401)2

(36)

Also through plotting the graph of 120591(5)2 against 1198871 and 1198873in the interval [minus01 05] and choosing a small value of 1198873we get that 1198873 = 120 is the optimal value which gives 1198871 =

17200 and 120591(5)2 = 2272727273 times 10minus4

Now utilizing the same technique where we draw thegraph of 1205911015840(5)2 versus 119887

10158401 in the interval [minus1 1] we find that

11988710158401 = 118 is the best choice and with this value of 11988710158401 we get1205911015840(5)

2 = 4419191919 times 10minus4Therefore the error equation of the fifth-order condition

of 119910101584010158401015840 is as follows

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

14802160

(1604749085

+ 6194971660900119886221 + 87611744001198862111988631

+ 87611744001198862111988632 minus 19920720292011988621

+ 3097600119886231 + 61952001198863111988632 minus 140863360011988631

+ 3097600119886232 minus 140863360011988632)12

(37)

Consequently the global error is

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172=

1144064800

(1452377332189

+ 55754744948100119886221 + 788505696001198862111988631

+ 788505696001198862111988632 minus 1792864826280011988621

+ 27878400119886231 + 557568001198863111988632

minus 1267770240011988631 + 27878400119886232

minus 1267770240011988632)12

(38)

Byminimizing the error norm in (37) and global error in (38)with respect to the free parameters 11988621 11988631 and 11988632 we get11988621 = minus15 11988631 = 19125 and 11988632 = 19125 which produces120591101584010158401015840(5)

2 = 37878787879times10minus4 and 120591(5)119892 2 = 73068870183times10minus4 Finally all the coefficients of three-stage fourth-order

RKFDmethod are written in Butcher tableau and denoted byRKFD4 method as follows

411

minus15

1720

19125

19125

17200

minus775

120

118

2091926

51926

47408

8472568

1001819

47408

13312568

20005457

(39)

42 A Three-Stage RKFD Method of Order Five In thissection a three-stage RKFD method of order five will bederivedThe algebraic order conditions up to order five ((17)-(18) (19)ndash(21) (22)ndash(25) and (26)ndash(30)) need to be solvedThe resulting system of equations consists of fifteen nonlinearequations solving the system simultaneously which resultsin a solution with three free parameters 1198871 11988621 and 11988631 asfollows

1198882 =35+radic610

1198883 =35minusradic610

119887101584010158401015840

1 =19

119887101584010158401015840

2 =49minusradic636

119887101584010158401015840

3 =49+radic636

11988710158401015840

1 =19

11988710158401015840

2 =736

minusradic618

11988710158401015840

3 =736

+radic618

1198871015840

1 =118

1198871015840

2 =118

minusradic648

1198871015840

3 =118

+radic648

Mathematical Problems in Engineering 7

1198872 =148

minusradic672

+(minus12+radic62)1198871

1198873 =148

+radic672

minus(12+radic62)1198871

11988632 = (minus131125

+16radic6125

)11988621 minus 11988631 +12625

minus3radic62500

(40)

Thus these free parameters can be chosen by minimizing thelocal truncation error norms of the sixth-order conditionsThe error norms and the global error of the sixth-orderconditions are given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(6)119894)2

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(6)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

(41)

where 120591(6) 1205911015840(6) 12059110158401015840(6) and 120591

101584010158401015840(6) are the local truncationerror norms for 119910 1199101015840 11991010158401015840 and 119910

101584010158401015840 of the RKFD methodrespectively 120591(6)119892 is the global error The error equation ofsixth-order condition for 119910with respect to the free parameter1198871 is as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

13600

radic(minus19 + 10801198871)2 (42)

The error equation 120591(6)2 has a minimum value equal to

zero at 1198871 = 191080 asymp 001759259259 which leads to 1198872 =

131080 minus 11radic62160 and 1198873 = 131080 + 11radic62160 Thetruncation error norms of the sixth-order condition of 119910 119910101584011991010158401015840 and 119910

101584010158401015840 are calculated as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

11200

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

13600

(139+ 628800119886221 minus 76800119886221radic6

minus 984011988621 minus 376011988621radic6+ 42radic6)12

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

13600

(362minus 1176011988621radic6+ 103200119886221radic6

+ 60000011988621radic611988631 minus 3684011988621 minus 3444011988631

+ 120radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 1498800119886221)12

(43)

Also the global error can be written as

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172=

13600

(minus1552011988621radic6+ 26400119886221radic6

+ 60000011988621radic611988631 minus 4668011988621 minus 3444011988631 + 511

+ 162radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 2127600119886221)

(44)

Now minimizing the error coefficients in (43) and (44) withrespect to the free parameters 11988621 11988631 we obtain 11988621 =

4059187793 and 11988631 = minus1502532215 which gives 11988632 =

1826569317Thus the error equations for 119910 1199101015840 11991010158401015840 and 119910101584010158401015840are computed and given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 = 8333333333times 10minus4

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 = 1666666668times 10minus3

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 = 1666666667times 10minus3

(45)

and global error norm is10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172= 2499999999times 10minus3 (46)

Therefore the parameters of the three-stage fifth-order RKFDmethod denoted by RKFD5 can be represented in Butchertableau as follows

35+radic610

4059187793

35minusradic610

minus1502532215

1826569317

191080

131080

minus11radic62160

131080

+11radic62160

118

118

minusradic648

118

+radic648

19

736

minusradic618

736

+radic618

19

49minusradic636

49+radic636

(47)

5 Numerical Examples

In this section some numerical examples will be solved toshow the efficiency of the new RKFD methods of order four

8 Mathematical Problems in Engineering

25 3 35 4 45

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

0

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 1 The efficiency curves for Example 1 with ℎ = 012119894 119894 =0 2 3 4

25 3 35 4 45

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 2 The efficiency curves for Example 2 with ℎ = 012119894 119894 =0 2 3 4

and order five which are denoted by RKFD4 and RKFD5respectively The comparison is made with the well knownmethods in the scientific literature We use in the numericalcomparisons the criteria based on computing the maximumerror in the solution (max error = max(|119910(119905119899) minus 119910119899|)) whichis equal to the maximum between absolute errors of thetrue solutions and the computed solutions Figures 1ndash7 showthe efficiency curves of Log10 (max error) against the com-putational effort measured by Log10 (function evaluations)required by eachmethodThe followingmethods are used forcomparison

2 22 24 26 28 3 32 34 36 38 4

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 3 The efficiency curves for Example 3 with ℎ = 012119894 119894 =0 2 3 4

14 16 18 2 22 24 26 28 3 32 34

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 4 The efficiency curves for Example 4 with ℎ = 012119894 119894 =0 2 3 4

(i) RKFD5 the three-stage fifth-order RKFD methodderived in this paper

(ii) RKFD4 the three-stage fourth-order RKFD methodderived in this paper

(iii) RK5Bs6 the six-stage fifth-order Runge-Kuttamethod given in Butcher [4]

(iv) RK5Ns6 the six-stage fifth-order Runge-Kuttamethod given in Hairer [5]

(v) RK4s4 the four-stage fourth-order Runge-Kuttamethod given in Dormand [14]

Mathematical Problems in Engineering 9

18 2 22 24 26 28 3 32 34 36 38

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 5 The efficiency curves for Example 5 with ℎ = 012119894 119894 =0 2 3 4

18 2 22 24 26 28 3 32 34 36 38

minus12

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 6 The efficiency curves for Example 6 with ℎ = 012119894 119894 =0 2 3 4

(vi) RK38 the four-stage fourth-order 38 rule Runge-Kutta method given in Butcher [4]

Example 1 The homogeneous linear problem is as follows

119910(119894V)

= minus 4119910

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 2 119910101584010158401015840 (0) = 2(48)

The exact solution is given by 119910(119909) = e119909 sin(119909) The problemis integrated in the interval [0 10]

24 26 28 3 32 34 36 38 4 42minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 7 The efficiency curves for Example 7 with ℎ = 00252119894119894 = 0 1 2 3

Example 2 The nonhomogeneous nonlinear problem is asfollows

119910(119894V)

= 1199102+ cos2 (119909) + sin (119909) minus 1

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = minus1(49)

The exact solution is given by 119910(119909) = sin(119909) The problem isintegrated in the interval [0 10]

Example 3 The homogeneous linear problem with noncon-stant coefficients is as follows

119910(119894V)

= (161199094 minus 481199092 + 12) 119910

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus2 119910101584010158401015840 (0) = 0(50)

The exact solution is given by 119910(119909) = eminus1199092 The problem is

integrated in the interval [0 3]

Example 4 The nonlinear problem is as follows

119910(119894V)

=3 sin (119910) (3 + 2sin2 (119910))

cos7 (119910)

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = 1

(51)

The exact solution is given by 119910(119909) = arcsin(119909) The problemis integrated in the interval [0 1205874]

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

4 Construction of RKFD Methods

In this section we proceed to construct explicit RKFDmethods based on the order conditions which we havederived in Section 2

41 A Three-Stage RKFD Method of Order Four This sectionwill focus on the derivation of a three-stage RKFDmethod oforder four where we use the algebraic order conditions (17)(19)-(20) (22)ndash(24) and (26)ndash(29) respectivelyThe resultingsystem of equations consists of 10 nonlinear equations with14 unknown variables to be solved solving the systemsimultaneously yields a solution with four free parameters 119888311988710158401 1198871 and 1198873 as follows

1198882 = minus3 minus 41198883minus4 + 61198883

119887101584010158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

119887101584010158401015840

2 = minus16 minus 721198883 + 10811988823 minus 54119888331081198883 minus 15011988823 minus 27 + 7211988833

119887101584010158401015840

3 =1

3611988833 minus 4811988823 + 181198883

11988710158401015840

1 =611988823 minus 61198883 + 161198883 (minus3 + 41198883)

11988710158401015840

2 = minusminus201198883 minus 1811988833 + 3311988823 + 41081198883 minus 15011988823 minus 27 + 7211988833

11988710158401015840

3 = minusminus1 + 1198883

3611988833 minus 4811988823 + 181198883

1198871015840

2 = minusminus48119887101584011198883 + 7211988710158401119888

23 minus 2 + 111198883 minus 1211988823

12 (3 minus 81198883 + 611988823 )

1198871015840

3 = minusminus4 + 3611988710158401 + 51198883 minus 48119887101584011198883

minus961198883 + 7211988823 + 36

1198872 = minus 1198871 minus 1198873 +124

(33)

Thus these free parameters can be chosen by minimizingthe local truncation error norms of the fifth-order conditionsaccording to Dormand et al [18] However we have anotherthree free parameters 11988621 11988631 and 11988632 that do not appear infourth-order conditions but they appear in the minimizationof error equations for fifth-order conditions of 119910101584010158401015840

The error norms and global error of fifth-order conditionsare defined as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(5)119894)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(5)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(5)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(5)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(5)119894

)2

(34)

where 120591(5) 1205911015840(5) 12059110158401015840(5) and 120591101584010158401015840(5) are the local truncation errornorms for119910119910101584011991010158401015840 and119910101584010158401015840 respectively and 120591(5)119892 is the globalerror

Consequently we find the error norms of 119910 1199101015840 and 11991010158401015840

respectively as follows

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1240

radic(minus3601198873 + 11 minus 720119888231198873 + 48011988711198883 minus 3601198871 + 96011988731198883 minus 141198883)

2

(minus2 + 31198883)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

1240

radic(minus5011988823 + 48011988710158401119888

23 minus 360119887101584011198883 + 481198883 minus 7)

2

(minus2 + 31198883)2

1003817100381710038171003817100381712059110158401015840(5)100381710038171003817100381710038172 =

1120

radic(3 minus 121198883 + 1011988823 )

2

(minus2 + 31198883)2

(35)

6 Mathematical Problems in Engineering

Our goal is to choose the free parameters 1198883 11988710158401 1198871 and 1198873 such

that the error norms of fifth-order conditions have minimalvalue By plotting the graph of 12059110158401015840(5)2 versus 1198883 and choosinga small value of 1198883 in the interval [07 3] we find that 1198883 =

1720 is the optimal value which yields a minimum value for12059110158401015840(5)

2 = 3787878788 times 10minus4 Substituting the value of 1198883 =1720 into 120591(5)2 and 120591

1015840(5)2 we get

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1440

radic(3 minus 1601198871 + 2141198873)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

11760

radic(minus31 + 54411988710158401)2

(36)

Also through plotting the graph of 120591(5)2 against 1198871 and 1198873in the interval [minus01 05] and choosing a small value of 1198873we get that 1198873 = 120 is the optimal value which gives 1198871 =

17200 and 120591(5)2 = 2272727273 times 10minus4

Now utilizing the same technique where we draw thegraph of 1205911015840(5)2 versus 119887

10158401 in the interval [minus1 1] we find that

11988710158401 = 118 is the best choice and with this value of 11988710158401 we get1205911015840(5)

2 = 4419191919 times 10minus4Therefore the error equation of the fifth-order condition

of 119910101584010158401015840 is as follows

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

14802160

(1604749085

+ 6194971660900119886221 + 87611744001198862111988631

+ 87611744001198862111988632 minus 19920720292011988621

+ 3097600119886231 + 61952001198863111988632 minus 140863360011988631

+ 3097600119886232 minus 140863360011988632)12

(37)

Consequently the global error is

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172=

1144064800

(1452377332189

+ 55754744948100119886221 + 788505696001198862111988631

+ 788505696001198862111988632 minus 1792864826280011988621

+ 27878400119886231 + 557568001198863111988632

minus 1267770240011988631 + 27878400119886232

minus 1267770240011988632)12

(38)

Byminimizing the error norm in (37) and global error in (38)with respect to the free parameters 11988621 11988631 and 11988632 we get11988621 = minus15 11988631 = 19125 and 11988632 = 19125 which produces120591101584010158401015840(5)

2 = 37878787879times10minus4 and 120591(5)119892 2 = 73068870183times10minus4 Finally all the coefficients of three-stage fourth-order

RKFDmethod are written in Butcher tableau and denoted byRKFD4 method as follows

411

minus15

1720

19125

19125

17200

minus775

120

118

2091926

51926

47408

8472568

1001819

47408

13312568

20005457

(39)

42 A Three-Stage RKFD Method of Order Five In thissection a three-stage RKFD method of order five will bederivedThe algebraic order conditions up to order five ((17)-(18) (19)ndash(21) (22)ndash(25) and (26)ndash(30)) need to be solvedThe resulting system of equations consists of fifteen nonlinearequations solving the system simultaneously which resultsin a solution with three free parameters 1198871 11988621 and 11988631 asfollows

1198882 =35+radic610

1198883 =35minusradic610

119887101584010158401015840

1 =19

119887101584010158401015840

2 =49minusradic636

119887101584010158401015840

3 =49+radic636

11988710158401015840

1 =19

11988710158401015840

2 =736

minusradic618

11988710158401015840

3 =736

+radic618

1198871015840

1 =118

1198871015840

2 =118

minusradic648

1198871015840

3 =118

+radic648

Mathematical Problems in Engineering 7

1198872 =148

minusradic672

+(minus12+radic62)1198871

1198873 =148

+radic672

minus(12+radic62)1198871

11988632 = (minus131125

+16radic6125

)11988621 minus 11988631 +12625

minus3radic62500

(40)

Thus these free parameters can be chosen by minimizing thelocal truncation error norms of the sixth-order conditionsThe error norms and the global error of the sixth-orderconditions are given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(6)119894)2

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(6)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

(41)

where 120591(6) 1205911015840(6) 12059110158401015840(6) and 120591

101584010158401015840(6) are the local truncationerror norms for 119910 1199101015840 11991010158401015840 and 119910

101584010158401015840 of the RKFD methodrespectively 120591(6)119892 is the global error The error equation ofsixth-order condition for 119910with respect to the free parameter1198871 is as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

13600

radic(minus19 + 10801198871)2 (42)

The error equation 120591(6)2 has a minimum value equal to

zero at 1198871 = 191080 asymp 001759259259 which leads to 1198872 =

131080 minus 11radic62160 and 1198873 = 131080 + 11radic62160 Thetruncation error norms of the sixth-order condition of 119910 119910101584011991010158401015840 and 119910

101584010158401015840 are calculated as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

11200

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

13600

(139+ 628800119886221 minus 76800119886221radic6

minus 984011988621 minus 376011988621radic6+ 42radic6)12

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

13600

(362minus 1176011988621radic6+ 103200119886221radic6

+ 60000011988621radic611988631 minus 3684011988621 minus 3444011988631

+ 120radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 1498800119886221)12

(43)

Also the global error can be written as

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172=

13600

(minus1552011988621radic6+ 26400119886221radic6

+ 60000011988621radic611988631 minus 4668011988621 minus 3444011988631 + 511

+ 162radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 2127600119886221)

(44)

Now minimizing the error coefficients in (43) and (44) withrespect to the free parameters 11988621 11988631 we obtain 11988621 =

4059187793 and 11988631 = minus1502532215 which gives 11988632 =

1826569317Thus the error equations for 119910 1199101015840 11991010158401015840 and 119910101584010158401015840are computed and given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 = 8333333333times 10minus4

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 = 1666666668times 10minus3

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 = 1666666667times 10minus3

(45)

and global error norm is10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172= 2499999999times 10minus3 (46)

Therefore the parameters of the three-stage fifth-order RKFDmethod denoted by RKFD5 can be represented in Butchertableau as follows

35+radic610

4059187793

35minusradic610

minus1502532215

1826569317

191080

131080

minus11radic62160

131080

+11radic62160

118

118

minusradic648

118

+radic648

19

736

minusradic618

736

+radic618

19

49minusradic636

49+radic636

(47)

5 Numerical Examples

In this section some numerical examples will be solved toshow the efficiency of the new RKFD methods of order four

8 Mathematical Problems in Engineering

25 3 35 4 45

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

0

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 1 The efficiency curves for Example 1 with ℎ = 012119894 119894 =0 2 3 4

25 3 35 4 45

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 2 The efficiency curves for Example 2 with ℎ = 012119894 119894 =0 2 3 4

and order five which are denoted by RKFD4 and RKFD5respectively The comparison is made with the well knownmethods in the scientific literature We use in the numericalcomparisons the criteria based on computing the maximumerror in the solution (max error = max(|119910(119905119899) minus 119910119899|)) whichis equal to the maximum between absolute errors of thetrue solutions and the computed solutions Figures 1ndash7 showthe efficiency curves of Log10 (max error) against the com-putational effort measured by Log10 (function evaluations)required by eachmethodThe followingmethods are used forcomparison

2 22 24 26 28 3 32 34 36 38 4

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 3 The efficiency curves for Example 3 with ℎ = 012119894 119894 =0 2 3 4

14 16 18 2 22 24 26 28 3 32 34

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 4 The efficiency curves for Example 4 with ℎ = 012119894 119894 =0 2 3 4

(i) RKFD5 the three-stage fifth-order RKFD methodderived in this paper

(ii) RKFD4 the three-stage fourth-order RKFD methodderived in this paper

(iii) RK5Bs6 the six-stage fifth-order Runge-Kuttamethod given in Butcher [4]

(iv) RK5Ns6 the six-stage fifth-order Runge-Kuttamethod given in Hairer [5]

(v) RK4s4 the four-stage fourth-order Runge-Kuttamethod given in Dormand [14]

Mathematical Problems in Engineering 9

18 2 22 24 26 28 3 32 34 36 38

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 5 The efficiency curves for Example 5 with ℎ = 012119894 119894 =0 2 3 4

18 2 22 24 26 28 3 32 34 36 38

minus12

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 6 The efficiency curves for Example 6 with ℎ = 012119894 119894 =0 2 3 4

(vi) RK38 the four-stage fourth-order 38 rule Runge-Kutta method given in Butcher [4]

Example 1 The homogeneous linear problem is as follows

119910(119894V)

= minus 4119910

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 2 119910101584010158401015840 (0) = 2(48)

The exact solution is given by 119910(119909) = e119909 sin(119909) The problemis integrated in the interval [0 10]

24 26 28 3 32 34 36 38 4 42minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 7 The efficiency curves for Example 7 with ℎ = 00252119894119894 = 0 1 2 3

Example 2 The nonhomogeneous nonlinear problem is asfollows

119910(119894V)

= 1199102+ cos2 (119909) + sin (119909) minus 1

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = minus1(49)

The exact solution is given by 119910(119909) = sin(119909) The problem isintegrated in the interval [0 10]

Example 3 The homogeneous linear problem with noncon-stant coefficients is as follows

119910(119894V)

= (161199094 minus 481199092 + 12) 119910

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus2 119910101584010158401015840 (0) = 0(50)

The exact solution is given by 119910(119909) = eminus1199092 The problem is

integrated in the interval [0 3]

Example 4 The nonlinear problem is as follows

119910(119894V)

=3 sin (119910) (3 + 2sin2 (119910))

cos7 (119910)

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = 1

(51)

The exact solution is given by 119910(119909) = arcsin(119909) The problemis integrated in the interval [0 1205874]

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

Our goal is to choose the free parameters 1198883 11988710158401 1198871 and 1198873 such

that the error norms of fifth-order conditions have minimalvalue By plotting the graph of 12059110158401015840(5)2 versus 1198883 and choosinga small value of 1198883 in the interval [07 3] we find that 1198883 =

1720 is the optimal value which yields a minimum value for12059110158401015840(5)

2 = 3787878788 times 10minus4 Substituting the value of 1198883 =1720 into 120591(5)2 and 120591

1015840(5)2 we get

10038171003817100381710038171003817120591(5)100381710038171003817100381710038172 =

1440

radic(3 minus 1601198871 + 2141198873)2

100381710038171003817100381710038171205911015840(5)100381710038171003817100381710038172 =

11760

radic(minus31 + 54411988710158401)2

(36)

Also through plotting the graph of 120591(5)2 against 1198871 and 1198873in the interval [minus01 05] and choosing a small value of 1198873we get that 1198873 = 120 is the optimal value which gives 1198871 =

17200 and 120591(5)2 = 2272727273 times 10minus4

Now utilizing the same technique where we draw thegraph of 1205911015840(5)2 versus 119887

10158401 in the interval [minus1 1] we find that

11988710158401 = 118 is the best choice and with this value of 11988710158401 we get1205911015840(5)

2 = 4419191919 times 10minus4Therefore the error equation of the fifth-order condition

of 119910101584010158401015840 is as follows

10038171003817100381710038171003817120591101584010158401015840(5)100381710038171003817100381710038172 =

14802160

(1604749085

+ 6194971660900119886221 + 87611744001198862111988631

+ 87611744001198862111988632 minus 19920720292011988621

+ 3097600119886231 + 61952001198863111988632 minus 140863360011988631

+ 3097600119886232 minus 140863360011988632)12

(37)

Consequently the global error is

10038171003817100381710038171003817120591(5)119892

100381710038171003817100381710038172=

1144064800

(1452377332189

+ 55754744948100119886221 + 788505696001198862111988631

+ 788505696001198862111988632 minus 1792864826280011988621

+ 27878400119886231 + 557568001198863111988632

minus 1267770240011988631 + 27878400119886232

minus 1267770240011988632)12

(38)

Byminimizing the error norm in (37) and global error in (38)with respect to the free parameters 11988621 11988631 and 11988632 we get11988621 = minus15 11988631 = 19125 and 11988632 = 19125 which produces120591101584010158401015840(5)

2 = 37878787879times10minus4 and 120591(5)119892 2 = 73068870183times10minus4 Finally all the coefficients of three-stage fourth-order

RKFDmethod are written in Butcher tableau and denoted byRKFD4 method as follows

411

minus15

1720

19125

19125

17200

minus775

120

118

2091926

51926

47408

8472568

1001819

47408

13312568

20005457

(39)

42 A Three-Stage RKFD Method of Order Five In thissection a three-stage RKFD method of order five will bederivedThe algebraic order conditions up to order five ((17)-(18) (19)ndash(21) (22)ndash(25) and (26)ndash(30)) need to be solvedThe resulting system of equations consists of fifteen nonlinearequations solving the system simultaneously which resultsin a solution with three free parameters 1198871 11988621 and 11988631 asfollows

1198882 =35+radic610

1198883 =35minusradic610

119887101584010158401015840

1 =19

119887101584010158401015840

2 =49minusradic636

119887101584010158401015840

3 =49+radic636

11988710158401015840

1 =19

11988710158401015840

2 =736

minusradic618

11988710158401015840

3 =736

+radic618

1198871015840

1 =118

1198871015840

2 =118

minusradic648

1198871015840

3 =118

+radic648

Mathematical Problems in Engineering 7

1198872 =148

minusradic672

+(minus12+radic62)1198871

1198873 =148

+radic672

minus(12+radic62)1198871

11988632 = (minus131125

+16radic6125

)11988621 minus 11988631 +12625

minus3radic62500

(40)

Thus these free parameters can be chosen by minimizing thelocal truncation error norms of the sixth-order conditionsThe error norms and the global error of the sixth-orderconditions are given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(6)119894)2

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(6)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

(41)

where 120591(6) 1205911015840(6) 12059110158401015840(6) and 120591

101584010158401015840(6) are the local truncationerror norms for 119910 1199101015840 11991010158401015840 and 119910

101584010158401015840 of the RKFD methodrespectively 120591(6)119892 is the global error The error equation ofsixth-order condition for 119910with respect to the free parameter1198871 is as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

13600

radic(minus19 + 10801198871)2 (42)

The error equation 120591(6)2 has a minimum value equal to

zero at 1198871 = 191080 asymp 001759259259 which leads to 1198872 =

131080 minus 11radic62160 and 1198873 = 131080 + 11radic62160 Thetruncation error norms of the sixth-order condition of 119910 119910101584011991010158401015840 and 119910

101584010158401015840 are calculated as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

11200

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

13600

(139+ 628800119886221 minus 76800119886221radic6

minus 984011988621 minus 376011988621radic6+ 42radic6)12

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

13600

(362minus 1176011988621radic6+ 103200119886221radic6

+ 60000011988621radic611988631 minus 3684011988621 minus 3444011988631

+ 120radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 1498800119886221)12

(43)

Also the global error can be written as

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172=

13600

(minus1552011988621radic6+ 26400119886221radic6

+ 60000011988621radic611988631 minus 4668011988621 minus 3444011988631 + 511

+ 162radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 2127600119886221)

(44)

Now minimizing the error coefficients in (43) and (44) withrespect to the free parameters 11988621 11988631 we obtain 11988621 =

4059187793 and 11988631 = minus1502532215 which gives 11988632 =

1826569317Thus the error equations for 119910 1199101015840 11991010158401015840 and 119910101584010158401015840are computed and given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 = 8333333333times 10minus4

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 = 1666666668times 10minus3

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 = 1666666667times 10minus3

(45)

and global error norm is10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172= 2499999999times 10minus3 (46)

Therefore the parameters of the three-stage fifth-order RKFDmethod denoted by RKFD5 can be represented in Butchertableau as follows

35+radic610

4059187793

35minusradic610

minus1502532215

1826569317

191080

131080

minus11radic62160

131080

+11radic62160

118

118

minusradic648

118

+radic648

19

736

minusradic618

736

+radic618

19

49minusradic636

49+radic636

(47)

5 Numerical Examples

In this section some numerical examples will be solved toshow the efficiency of the new RKFD methods of order four

8 Mathematical Problems in Engineering

25 3 35 4 45

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

0

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 1 The efficiency curves for Example 1 with ℎ = 012119894 119894 =0 2 3 4

25 3 35 4 45

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 2 The efficiency curves for Example 2 with ℎ = 012119894 119894 =0 2 3 4

and order five which are denoted by RKFD4 and RKFD5respectively The comparison is made with the well knownmethods in the scientific literature We use in the numericalcomparisons the criteria based on computing the maximumerror in the solution (max error = max(|119910(119905119899) minus 119910119899|)) whichis equal to the maximum between absolute errors of thetrue solutions and the computed solutions Figures 1ndash7 showthe efficiency curves of Log10 (max error) against the com-putational effort measured by Log10 (function evaluations)required by eachmethodThe followingmethods are used forcomparison

2 22 24 26 28 3 32 34 36 38 4

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 3 The efficiency curves for Example 3 with ℎ = 012119894 119894 =0 2 3 4

14 16 18 2 22 24 26 28 3 32 34

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 4 The efficiency curves for Example 4 with ℎ = 012119894 119894 =0 2 3 4

(i) RKFD5 the three-stage fifth-order RKFD methodderived in this paper

(ii) RKFD4 the three-stage fourth-order RKFD methodderived in this paper

(iii) RK5Bs6 the six-stage fifth-order Runge-Kuttamethod given in Butcher [4]

(iv) RK5Ns6 the six-stage fifth-order Runge-Kuttamethod given in Hairer [5]

(v) RK4s4 the four-stage fourth-order Runge-Kuttamethod given in Dormand [14]

Mathematical Problems in Engineering 9

18 2 22 24 26 28 3 32 34 36 38

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 5 The efficiency curves for Example 5 with ℎ = 012119894 119894 =0 2 3 4

18 2 22 24 26 28 3 32 34 36 38

minus12

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 6 The efficiency curves for Example 6 with ℎ = 012119894 119894 =0 2 3 4

(vi) RK38 the four-stage fourth-order 38 rule Runge-Kutta method given in Butcher [4]

Example 1 The homogeneous linear problem is as follows

119910(119894V)

= minus 4119910

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 2 119910101584010158401015840 (0) = 2(48)

The exact solution is given by 119910(119909) = e119909 sin(119909) The problemis integrated in the interval [0 10]

24 26 28 3 32 34 36 38 4 42minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 7 The efficiency curves for Example 7 with ℎ = 00252119894119894 = 0 1 2 3

Example 2 The nonhomogeneous nonlinear problem is asfollows

119910(119894V)

= 1199102+ cos2 (119909) + sin (119909) minus 1

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = minus1(49)

The exact solution is given by 119910(119909) = sin(119909) The problem isintegrated in the interval [0 10]

Example 3 The homogeneous linear problem with noncon-stant coefficients is as follows

119910(119894V)

= (161199094 minus 481199092 + 12) 119910

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus2 119910101584010158401015840 (0) = 0(50)

The exact solution is given by 119910(119909) = eminus1199092 The problem is

integrated in the interval [0 3]

Example 4 The nonlinear problem is as follows

119910(119894V)

=3 sin (119910) (3 + 2sin2 (119910))

cos7 (119910)

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = 1

(51)

The exact solution is given by 119910(119909) = arcsin(119909) The problemis integrated in the interval [0 1205874]

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

1198872 =148

minusradic672

+(minus12+radic62)1198871

1198873 =148

+radic672

minus(12+radic62)1198871

11988632 = (minus131125

+16radic6125

)11988621 minus 11988631 +12625

minus3radic62500

(40)

Thus these free parameters can be chosen by minimizing thelocal truncation error norms of the sixth-order conditionsThe error norms and the global error of the sixth-orderconditions are given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

radic

119899119901+1

sum

119894=1(120591(6)119894)2

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

radic

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

radic

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

radic

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172

= radic

119899119901+1

sum

119894=1(120591(6)119894)2+

1198991015840119901+1

sum

119894=1(1205911015840(6)119894

)2+

11989910158401015840119901+1

sum

119894=1(12059110158401015840(6)119894

)2+

119899101584010158401015840119901+1

sum

119894=1(120591101584010158401015840(6)119894

)2

(41)

where 120591(6) 1205911015840(6) 12059110158401015840(6) and 120591

101584010158401015840(6) are the local truncationerror norms for 119910 1199101015840 11991010158401015840 and 119910

101584010158401015840 of the RKFD methodrespectively 120591(6)119892 is the global error The error equation ofsixth-order condition for 119910with respect to the free parameter1198871 is as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 =

13600

radic(minus19 + 10801198871)2 (42)

The error equation 120591(6)2 has a minimum value equal to

zero at 1198871 = 191080 asymp 001759259259 which leads to 1198872 =

131080 minus 11radic62160 and 1198873 = 131080 + 11radic62160 Thetruncation error norms of the sixth-order condition of 119910 119910101584011991010158401015840 and 119910

101584010158401015840 are calculated as follows

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 =

11200

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 =

13600

(139+ 628800119886221 minus 76800119886221radic6

minus 984011988621 minus 376011988621radic6+ 42radic6)12

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 =

13600

(362minus 1176011988621radic6+ 103200119886221radic6

+ 60000011988621radic611988631 minus 3684011988621 minus 3444011988631

+ 120radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 1498800119886221)12

(43)

Also the global error can be written as

10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172=

13600

(minus1552011988621radic6+ 26400119886221radic6

+ 60000011988621radic611988631 minus 4668011988621 minus 3444011988631 + 511

+ 162radic6minus 11840radic611988631 + 21000001198862111988631

+ 1330800119886231 + 448800radic6119886231 + 2127600119886221)

(44)

Now minimizing the error coefficients in (43) and (44) withrespect to the free parameters 11988621 11988631 we obtain 11988621 =

4059187793 and 11988631 = minus1502532215 which gives 11988632 =

1826569317Thus the error equations for 119910 1199101015840 11991010158401015840 and 119910101584010158401015840are computed and given by

10038171003817100381710038171003817120591(6)100381710038171003817100381710038172 = 0

100381710038171003817100381710038171205911015840(6)100381710038171003817100381710038172 = 8333333333times 10minus4

1003817100381710038171003817100381712059110158401015840(6)100381710038171003817100381710038172 = 1666666668times 10minus3

10038171003817100381710038171003817120591101584010158401015840(6)100381710038171003817100381710038172 = 1666666667times 10minus3

(45)

and global error norm is10038171003817100381710038171003817120591(6)119892

100381710038171003817100381710038172= 2499999999times 10minus3 (46)

Therefore the parameters of the three-stage fifth-order RKFDmethod denoted by RKFD5 can be represented in Butchertableau as follows

35+radic610

4059187793

35minusradic610

minus1502532215

1826569317

191080

131080

minus11radic62160

131080

+11radic62160

118

118

minusradic648

118

+radic648

19

736

minusradic618

736

+radic618

19

49minusradic636

49+radic636

(47)

5 Numerical Examples

In this section some numerical examples will be solved toshow the efficiency of the new RKFD methods of order four

8 Mathematical Problems in Engineering

25 3 35 4 45

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

0

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 1 The efficiency curves for Example 1 with ℎ = 012119894 119894 =0 2 3 4

25 3 35 4 45

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 2 The efficiency curves for Example 2 with ℎ = 012119894 119894 =0 2 3 4

and order five which are denoted by RKFD4 and RKFD5respectively The comparison is made with the well knownmethods in the scientific literature We use in the numericalcomparisons the criteria based on computing the maximumerror in the solution (max error = max(|119910(119905119899) minus 119910119899|)) whichis equal to the maximum between absolute errors of thetrue solutions and the computed solutions Figures 1ndash7 showthe efficiency curves of Log10 (max error) against the com-putational effort measured by Log10 (function evaluations)required by eachmethodThe followingmethods are used forcomparison

2 22 24 26 28 3 32 34 36 38 4

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 3 The efficiency curves for Example 3 with ℎ = 012119894 119894 =0 2 3 4

14 16 18 2 22 24 26 28 3 32 34

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 4 The efficiency curves for Example 4 with ℎ = 012119894 119894 =0 2 3 4

(i) RKFD5 the three-stage fifth-order RKFD methodderived in this paper

(ii) RKFD4 the three-stage fourth-order RKFD methodderived in this paper

(iii) RK5Bs6 the six-stage fifth-order Runge-Kuttamethod given in Butcher [4]

(iv) RK5Ns6 the six-stage fifth-order Runge-Kuttamethod given in Hairer [5]

(v) RK4s4 the four-stage fourth-order Runge-Kuttamethod given in Dormand [14]

Mathematical Problems in Engineering 9

18 2 22 24 26 28 3 32 34 36 38

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 5 The efficiency curves for Example 5 with ℎ = 012119894 119894 =0 2 3 4

18 2 22 24 26 28 3 32 34 36 38

minus12

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 6 The efficiency curves for Example 6 with ℎ = 012119894 119894 =0 2 3 4

(vi) RK38 the four-stage fourth-order 38 rule Runge-Kutta method given in Butcher [4]

Example 1 The homogeneous linear problem is as follows

119910(119894V)

= minus 4119910

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 2 119910101584010158401015840 (0) = 2(48)

The exact solution is given by 119910(119909) = e119909 sin(119909) The problemis integrated in the interval [0 10]

24 26 28 3 32 34 36 38 4 42minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 7 The efficiency curves for Example 7 with ℎ = 00252119894119894 = 0 1 2 3

Example 2 The nonhomogeneous nonlinear problem is asfollows

119910(119894V)

= 1199102+ cos2 (119909) + sin (119909) minus 1

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = minus1(49)

The exact solution is given by 119910(119909) = sin(119909) The problem isintegrated in the interval [0 10]

Example 3 The homogeneous linear problem with noncon-stant coefficients is as follows

119910(119894V)

= (161199094 minus 481199092 + 12) 119910

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus2 119910101584010158401015840 (0) = 0(50)

The exact solution is given by 119910(119909) = eminus1199092 The problem is

integrated in the interval [0 3]

Example 4 The nonlinear problem is as follows

119910(119894V)

=3 sin (119910) (3 + 2sin2 (119910))

cos7 (119910)

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = 1

(51)

The exact solution is given by 119910(119909) = arcsin(119909) The problemis integrated in the interval [0 1205874]

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

25 3 35 4 45

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

0

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 1 The efficiency curves for Example 1 with ℎ = 012119894 119894 =0 2 3 4

25 3 35 4 45

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 2 The efficiency curves for Example 2 with ℎ = 012119894 119894 =0 2 3 4

and order five which are denoted by RKFD4 and RKFD5respectively The comparison is made with the well knownmethods in the scientific literature We use in the numericalcomparisons the criteria based on computing the maximumerror in the solution (max error = max(|119910(119905119899) minus 119910119899|)) whichis equal to the maximum between absolute errors of thetrue solutions and the computed solutions Figures 1ndash7 showthe efficiency curves of Log10 (max error) against the com-putational effort measured by Log10 (function evaluations)required by eachmethodThe followingmethods are used forcomparison

2 22 24 26 28 3 32 34 36 38 4

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 3 The efficiency curves for Example 3 with ℎ = 012119894 119894 =0 2 3 4

14 16 18 2 22 24 26 28 3 32 34

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 4 The efficiency curves for Example 4 with ℎ = 012119894 119894 =0 2 3 4

(i) RKFD5 the three-stage fifth-order RKFD methodderived in this paper

(ii) RKFD4 the three-stage fourth-order RKFD methodderived in this paper

(iii) RK5Bs6 the six-stage fifth-order Runge-Kuttamethod given in Butcher [4]

(iv) RK5Ns6 the six-stage fifth-order Runge-Kuttamethod given in Hairer [5]

(v) RK4s4 the four-stage fourth-order Runge-Kuttamethod given in Dormand [14]

Mathematical Problems in Engineering 9

18 2 22 24 26 28 3 32 34 36 38

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 5 The efficiency curves for Example 5 with ℎ = 012119894 119894 =0 2 3 4

18 2 22 24 26 28 3 32 34 36 38

minus12

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 6 The efficiency curves for Example 6 with ℎ = 012119894 119894 =0 2 3 4

(vi) RK38 the four-stage fourth-order 38 rule Runge-Kutta method given in Butcher [4]

Example 1 The homogeneous linear problem is as follows

119910(119894V)

= minus 4119910

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 2 119910101584010158401015840 (0) = 2(48)

The exact solution is given by 119910(119909) = e119909 sin(119909) The problemis integrated in the interval [0 10]

24 26 28 3 32 34 36 38 4 42minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 7 The efficiency curves for Example 7 with ℎ = 00252119894119894 = 0 1 2 3

Example 2 The nonhomogeneous nonlinear problem is asfollows

119910(119894V)

= 1199102+ cos2 (119909) + sin (119909) minus 1

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = minus1(49)

The exact solution is given by 119910(119909) = sin(119909) The problem isintegrated in the interval [0 10]

Example 3 The homogeneous linear problem with noncon-stant coefficients is as follows

119910(119894V)

= (161199094 minus 481199092 + 12) 119910

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus2 119910101584010158401015840 (0) = 0(50)

The exact solution is given by 119910(119909) = eminus1199092 The problem is

integrated in the interval [0 3]

Example 4 The nonlinear problem is as follows

119910(119894V)

=3 sin (119910) (3 + 2sin2 (119910))

cos7 (119910)

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = 1

(51)

The exact solution is given by 119910(119909) = arcsin(119909) The problemis integrated in the interval [0 1205874]

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

18 2 22 24 26 28 3 32 34 36 38

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 5 The efficiency curves for Example 5 with ℎ = 012119894 119894 =0 2 3 4

18 2 22 24 26 28 3 32 34 36 38

minus12

minus11

minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 6 The efficiency curves for Example 6 with ℎ = 012119894 119894 =0 2 3 4

(vi) RK38 the four-stage fourth-order 38 rule Runge-Kutta method given in Butcher [4]

Example 1 The homogeneous linear problem is as follows

119910(119894V)

= minus 4119910

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 2 119910101584010158401015840 (0) = 2(48)

The exact solution is given by 119910(119909) = e119909 sin(119909) The problemis integrated in the interval [0 10]

24 26 28 3 32 34 36 38 4 42minus10

minus9

minus8

minus7

minus6

minus5

minus4

minus3

minus2

minus1

RKFD5RKFD4RK5Bs6

RK5Ns6RK4s4RK38

Log10

(function evaluations)

Log 1

0

(max

erro

r)

Figure 7 The efficiency curves for Example 7 with ℎ = 00252119894119894 = 0 1 2 3

Example 2 The nonhomogeneous nonlinear problem is asfollows

119910(119894V)

= 1199102+ cos2 (119909) + sin (119909) minus 1

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = minus1(49)

The exact solution is given by 119910(119909) = sin(119909) The problem isintegrated in the interval [0 10]

Example 3 The homogeneous linear problem with noncon-stant coefficients is as follows

119910(119894V)

= (161199094 minus 481199092 + 12) 119910

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus2 119910101584010158401015840 (0) = 0(50)

The exact solution is given by 119910(119909) = eminus1199092 The problem is

integrated in the interval [0 3]

Example 4 The nonlinear problem is as follows

119910(119894V)

=3 sin (119910) (3 + 2sin2 (119910))

cos7 (119910)

119910 (0) = 0 1199101015840 (0) = 1 11991010158401015840 (0) = 0 119910101584010158401015840 (0) = 1

(51)

The exact solution is given by 119910(119909) = arcsin(119909) The problemis integrated in the interval [0 1205874]

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

Example 5 The linear system is as follows

119910(119894V)

= e3119909119906

119910 (0) = 1 1199101015840 (0) = minus1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = minus1

119911(119894V)

= 16eminus119909119910

119911 (0) = 1 1199111015840 (0) = minus2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = minus8

119908(119894V)

= 81eminus119909119911

119908 (0) = 1 1199081015840 (0) = minus3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = minus27

119906(119894V)

= 256eminus119909119908

119906 (0) = 1 1199061015840 (0) = minus4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = minus64

(52)

The exact solution is given by

119910 = eminus119909

119911 = eminus2119909

119908 = eminus3119909

119906 = eminus4119909

(53)

The problem is integrated in the interval [0 2]

Example 6 The nonlinear system is as follows

119910(119894V)

= 119910+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119910 (0) = 1 1199101015840 (0) = 0 11991010158401015840 (0) = minus1 119910101584010158401015840 (0) = 0

119911(119894V)

= 119911minus1

radic1199102 + 1199112+

1radic1199082 + 1199062

119911 (0) = 0 1199111015840 (0) = 1 11991110158401015840 (0) = 0 119911101584010158401015840 (0) = minus1

119908(119894V)

= 16119908+1

radic1199102 + 1199112minus

1radic1199082 + 1199062

119908 (0) = 1 1199081015840 (0) = 0 11990810158401015840 (0) = minus4 119908101584010158401015840 (0) = 0

119906(119894V)

= 16119906minus 1

radic1199102 + 1199112+

1radic1199082 + 1199062

119906 (0) = 0 1199061015840 (0) = 2 11990610158401015840 (0) = 0 119906101584010158401015840 (0) = minus8

(54)

The exact solution is given by

119910 = cos (119909)

119911 = sin (119909)

119908 = cos (2119909)

119906 = sin (2119909)

(55)

The problem is integrated in the interval [0 2]

Example 7 The nonlinear system is as follows

119910(119894V)

=1199112

119908

119910 (0) = 1 1199101015840 (0) = 1 11991010158401015840 (0) = 1 119910101584010158401015840 (0) = 1

119911(119894V)

= 161199082

119906

119911 (0) = 1 1199111015840 (0) = 2 11991110158401015840 (0) = 4 119911101584010158401015840 (0) = 8

119908(119894V)

= 811199062

1199105

119908 (0) = 1 1199081015840 (0) = 3 11990810158401015840 (0) = 9 119908101584010158401015840 (0) = 27

119906(119894V)

= 2561199104

119906 (0) = 1 1199061015840 (0) = 4 11990610158401015840 (0) = 16 119906101584010158401015840 (0) = 64

(56)

The exact solution is given by

119910 = e119909

119911 = e2119909

119908 = e3119909

119906 = e4119909

(57)

The problem is integrated in the interval [0 2]

6 Conclusion

This paper deals with Runge-Kutta type method denotedby RKFD method for directly solving special fourth-orderODEs of the form 119910

(119894V)(119909) = 119891(119909 119910) First we derived the

order conditions for RKFDmethod which were then used toconstruct three-stage fourth- and fifth-order RKFDmethodsThe methods are denoted by RKFD5 and RKFD4 respec-tively We also proved that the RKFD method is zero-stableFrom the numerical results we observed that the new RKFDmethods are more competent as compared with the existingRunge-Kutta methods in the scientific literature From thenumerical results we conclude that the new RKFD methodsare computationally more efficient in solving special fourth-order ODEs and outperformed the existingmethods in termsof error precision and number of function evaluations

Conflict of Interests

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

References

[1] P Onumanyi U W Sirisena and S N Jator ldquoContinuous finitedifference approximations for solving differential equationsrdquoInternational Journal of Computer Mathematics vol 72 no 1pp 15ndash27 1999

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 11

[2] D Sarafyan ldquoNew algorithms for the continuous approximatesolution of ordinary differential equations and the upgradingof the order of the processesrdquo Computers amp Mathematics withApplications vol 20 no 1 pp 77ndash100 1990

[3] GDahlquist ldquoOn accuracy and unconditional stability of linearmultistep methods for second order differential equationsrdquo BITNumerical Mathematics vol 18 no 2 pp 133ndash136 1978

[4] J C Butcher Numerical Methods for Ordinary DifferentialEquations JohnWileyamp Sons NewYorkNYUSA 2nd edition2008

[5] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 of Springer Seriesin Computational Mathematics Springer Berlin Germany 2ndedition 1993

[6] S N Jator and J Li ldquoA self-starting linear multistep methodfor a direct solution of the general second-order initial valueproblemrdquo International Journal of Computer Mathematics vol86 no 5 pp 827ndash836 2009

[7] D O Awoyemi ldquoA new sixth-order algorithm for general sec-ond order ordinary differential equationrdquo International Journalof Computer Mathematics vol 77 no 1 pp 117ndash124 2001

[8] S J Kayode ldquoAn efficient zero-stable numerical method forfourth-order differential equationsrdquo International Journal ofMathematics and Mathematical Sciences vol 2008 Article ID364021 10 pages 2008

[9] Z A Majid and M B Suleiman ldquoDirect integration implicitvariable steps method for solving higher order systems ofordinary differential equations directlyrdquo Sains Malaysiana vol35 no 2 pp 63ndash68 2006

[10] M-K Jain S R K Iyengar and J S V Saldanha ldquoNumericalsolution of a fourth-order ordinary differential equationrdquo Jour-nal of Engineering Mathematics vol 11 no 4 pp 373ndash380 1977

[11] N Waeleh Z A Majid and F Ismail ldquoA new algorithm forsolving higher order IVPs of ODEsrdquo Applied MathematicalSciences vol 5 no 53ndash56 pp 2795ndash2805 2011

[12] D O Awoyemi and O M Idowu ldquoA class of hybrid collocationmethods for third-order ordinary differential equationsrdquo Inter-national Journal of Computer Mathematics vol 82 no 10 pp1287ndash1293 2005

[13] S N Jator ldquoSolving second order initial value problems by ahybrid multistep method without predictorsrdquo Applied Mathe-matics and Computation vol 217 no 8 pp 4036ndash4046 2010

[14] J R DormandNumerical Methods for Differential Equations AComputational Approach Library of Engineering MathematicsCRC Press Boca Raton Fla USA 1996

[15] W Gander and D Gruntz ldquoDerivation of numerical methodsusing computer algebrardquo SIAM Review vol 41 no 3 pp 577ndash593 1999

[16] J D Lambert Numerical Methods for Ordinary DifferentialSystemsThe Initial Value Problem JohnWiley amp Sons LondonUK 1991

[17] P Henrici Elements of Numerical Analysis John Wiley amp SonsNew York NY USA 1964

[18] J R DormandM E A EL-Mikkawy and P J Prince ldquoFamiliesof Runge-Kutta Nystrom formulaerdquo IMA Journal of NumericalAnalysis vol 7 pp 235ndash250 1987

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of