research article a three-stage fifth-order runge...

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 795397 7 pageshttpdxdoiorg1011552013795397

Research ArticleA Three-Stage Fifth-Order Runge-Kutta Method forDirectly Solving Special Third-Order Differential Equation withApplication to Thin Film Flow Problem

M Mechee12 N Senu3 F Ismail3 B Nikouravan4 and Z Siri1

1 Institute of Mathematical Sciences University of Malaya 50603 Kuala Lumpur Malaysia2 Department of Mathematics Faculty of Mathematics and Computer Sciences University of Kufa Najaf Iraq3 Department of Mathematics and Institute for Mathematical Research Universiti Putra Malaysia43400 Serdang Selangor Malaysia

4Department of Physics Faculty of Science University of Malaya 50603 Kuala Lumpur Malaysia

Correspondence should be addressed to N Senu norazakscienceupmedumy

Received 23 March 2013 Accepted 23 May 2013

Academic Editor Ebrahim Momoniat

Copyright copy 2013 M Mechee et al This 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

In this paper a three-stage fifth-order Runge-Kuttamethod for the integration of a special third-order ordinary differential equation(ODE) is constructed The zero stability of the method is proven The numerical study of a third-order ODE arising in thin filmflow of viscous fluid in physics is discussed The mathematical model of thin film flow has been solved using a new method andnumerical comparisons are made when the same problem is reduced to a first-order system of equations which are solved usingthe existing Runge-Kutta methods Numerical results have clearly shown the advantage and the efficiency of the new method

1 Introduction

A special third-order differential equation (ODE) of the form

119910101584010158401015840

(119909) = 119891 (119909 119910 (119909)) 119910 (1199090) = 120572

1199101015840(1199090) = 120573 119910

10158401015840(1199090) = 120574 119909 ge 119909

0

(1)

which is not explicitly dependent on the first derivative 1199101015840(119909)and the second derivative 11991010158401015840(119909) of the solution is frequentlyfound in many physical problems such as electromagneticwaves thin filmflow and gravity driven flowsThe solution to(1) can be obtained by reducing it to an equivalent first-ordersystem which is three times the dimension and can be solvedusing a standard Runge-Kuttamethod or amultistepmethod

Most researchers scientists and engineers solve problem(1) by converting the problem to a system of first-order equa-tions However there are also studies on numerical methodswhich solve (1) directly Such work can be seen in Awoyemi[1] Waeleh et al [2] Zainuddin [3] and Jator [4] Awoyemiand Idowu [5] and Jator [6] proposed a class of hybridcollocation methods for the direct solution of higher-order

ordinary differential equations (ODEs) Samat and Ismail[7] developed an embedded hybrid method for solving spe-cial second-order ODEs Waeleh et al [2] developed a blockmultistep method which can directly solve general third-order equations on the other hand Ibrahim et al [8]developed a multistep method that can directly solve stiffthird-order differential equations All of the methods dis-cussed above are multistep methods hence they need start-ing values when used to solve ODEs such as (1) Senu et al [9]derived the Runge-Kutta-Nystrom method for solving sec-ond-order ODEs directly Mechee et al [10] constructed newRunge-Kutta methods for solving (1)

In this paper we are concerned with a one-step methodparticularly the three-stage fifth-order Runge-Kutta methodfor directly solving special third-order ODEs Accordinglywe have developed a direct Runge-Kutta method (RKD)which can be directly used to solve (1) The advantage of thenewmethod overmultistepmethods is that it initialises itselfThe method produces 119910

119899+1 1199101015840119899+1

and 11991010158401015840

119899+1to approximate

119910(119909119899+1

) 1199101015840(119909119899+1

) and 11991010158401015840(119909119899+1

) where 119910119899+1

is the computedsolution and 119910(119909

119899+1) is the exact solution

2 Mathematical Problems in Engineering

2 Derivation of RKD Method

The general form of RKD method with 119904 stages for solvinginitial value problem (1) can be written as

119910119899+1

= 119910119899+ ℎ1199101015840

119899+

ℎ2

211991010158401015840

119899+ ℎ3

119904

sum

119894=1

119887119894119896119894

1199101015840

119899+1= 1199101015840

119899+ ℎ11991010158401015840

119899+ ℎ2

119904

sum

119894=1

1198871015840

119894119896119894

11991010158401015840

119899+1= 11991010158401015840

119899+ ℎ

119904

sum

119894=1

11988710158401015840

119894119896119894

(2)

where

1198961= 119891 (119909

119899 119910119899) (3)

119896119894= 119891(119909

119899+ 119888119894ℎ 119910119899+ ℎ1198881198941199101015840

119899+

ℎ2

21198882

11989411991010158401015840

119899+ ℎ3

119894minus1

sum

119895=1

119886119894119895119896119895) (4)

for 119894 = 2 3 119904The parameters of the RKD method are 119888

119894 119886119894119895 119887119894 1198871015840

119894 11988710158401015840

119894

for 119894 = 1 2 119904 and 119895 = 1 2 119904 are assumed to be realIf 119886119894119895= 0 for 119894 ⩽ 119895 it is an explicit method and otherwise it

is an implicit method The RKD method can be expressed inthe Butcher notation using the table of coefficients as follows

119888 119860

119887119879

1198871015840119879

11988710158401015840119879

(5)

To determine the coefficients of the method given by (2)ndash(4) the RKD method expression is expanded using Taylorrsquosseries expansion After some algebraic manipulations thisexpansion is equated to the true solution that is given byTaylorrsquos series expansion General order conditions for theRKD method can be found from the direct expansion of thelocal truncation error This idea is based on the derivation oforder conditions for the Runge-Kutta method introduced inDormand [11] The RKD formulae in (2) may be expressed as

119910119899+1

= 119910119899+ ℎΦ (119909

119899 119910119899)

1199101015840

119899+1= 1199101015840

119899+ ℎΦ1015840(119909119899 119910119899)

11991010158401015840

119899+1= 11991010158401015840

119899+ ℎΦ10158401015840(119909119899 119910119899)

(6)

where the increment functions are

Φ(119909119899 119910119899) = 1199101015840

119899+

ℎ

211991010158401015840

119899+ ℎ2

119904

sum

119894=1

119887119894119896119894

Φ1015840(119909119899 119910119899) = 11991010158401015840

119899+ ℎ

119904

sum

119894=1

1198871015840

119894119896119894

Φ10158401015840(119909119899 119910119899) =

119904

sum

119894=1

11988710158401015840

119894119896119894

(7)

and 119896119894is defined in formula (4) If Δ is the Taylor series

increment function then the local truncation errors of thesolution the first derivative and the second derivative maybe obtained by substituting the true solution 119910(119909) of (1) intothe RKD increment function This gives

119905119899+1

= ℎ [Φ minus Δ] 1199051015840

119899+1= ℎ [Φ

1015840minus Δ1015840]

11990510158401015840

119899+1= ℎ [Φ

10158401015840minus Δ10158401015840]

(8)

These expressions are best given in terms of elementarydifferentials and the Taylor series increment may be writtenas

Δ = 1199101015840+

1

2ℎ11991010158401015840+

1

6ℎ2119865(3)

1+

1

24ℎ3119865(4)

1+ 119874 (ℎ

4)

Δ1015840= 11991010158401015840+

1

2ℎ119865(3)

1+

1

6ℎ2119865(4)

1+

1

24ℎ3119865(5)

1+ 119874 (ℎ

4)

Δ10158401015840= 119865(3)

1+

1

2ℎ119865(4)

1+

1

6ℎ2119865(5)

1+ 119874 (ℎ

3)

(9)

where for the scalar case the first few elementary differentialsare

119865(3)

1= 119891

119865(4)

1= 119891119909+ 1198911199101199101015840

119865(5)

1= 119891119909119909

+ 21198911199091199101199101015840+ 11989111991011991010158401015840+ 119891119910119910(1199101015840)2

(10)

Using the above terms the increment functions Φ Φ1015840

and Φ10158401015840 for the RKD formula become

119904

sum

119894=1

119887119894119896119894=

119904

sum

119894=1

119887119894119891 +

119904

sum

119894=1

119887119894119888119894(119891119909+ 1198911199101199101015840) ℎ

+1

2

119904

sum

119894=1

1198871198941198882

119894(119891119909119909

+ 21198911199091199101199101015840+ 11989111991011991010158401015840+ 119891119910119910(1199101015840)2

) ℎ2

+ 119874 (ℎ3)

119904

sum

119894=1

1198871015840

119894119896119894=

119904

sum

119894=1

1198871015840

119894119891 +

119904

sum

119894=1

1198871015840

119894119888119894(119891119909+ 1198911199101199101015840) ℎ

+1

2

119904

sum

119894=1

1198871015840

1198941198882

119894(119891119909119909

+ 21198911199091199101199101015840+ 11989111991011991010158401015840+ 119891119910119910(1199101015840)2

) ℎ2

+ 119874 (ℎ3)

119904

sum

119894=1

11988710158401015840

119894119896119894=

119904

sum

119894=1

11988710158401015840

119894119891 +

119904

sum

119894=1

11988710158401015840

119894119888119894(119891119909+ 1198911199101199101015840) ℎ

+1

2

119904

sum

119894=1

11988710158401015840

1198941198882

119894(119891119909119909

+ 21198911199091199101199101015840+ 11989111991011991010158401015840+ 119891119910119910(1199101015840)2

) ℎ2

+ 119874 (ℎ3)

(11)

Mathematical Problems in Engineering 3

The expressions for the local truncation errors in the 119910

solution the first derivative and the second derivative are

119905119899+1

= ℎ3[

119904

sum

119894=1

119887119894119896119894minus (

1

6119865(3)

1+

1

24119865(4)

1+ sdot sdot sdot )]

1199051015840

119899+1= ℎ2[

119904

sum

119894=1

1198871015840

119894119896119894minus (

1

2119865(3)

1+

1

6119865(4)


11990510158401015840

119899+1= ℎ[

119904

sum

119894=1

11988710158401015840

119894119896119894minus (119865(3)

1+

1

2119865(4)

1+

1

6ℎ2119865(5)


(12)

Substituting (11) into (12) and expanding as a Taylorexpansion usingMAPLE 4 software as introduced by Ganderand Gruntz [12] the error equations or the order conditionsfor 119904-stage six-order RKD method can be written as follows

Order conditions for 119910

Order 3

sum119887119894=

1

6 (13)

Order 4

sum119887119894119888119894=

1

24 (14)

Order 5

sum1198871198941198882

119894=

1

60 (15)

Order 6

sum1198871198941198883

119894=

1

120 sum119887

119894119886119894119895=

1

720 (16)


Order 2

sum1198871015840

119894=

1

2 (17)

Order 3

sum1198871015840

119894119888119894=

1

6 (18)

Order 4

sum1198871015840

1198941198882

119894=

1

12 (19)

Order 5

sum1198871015840

1198941198883

119894=

1

20 sum119887

1015840

119894119886119894119895=

1

120 (20)

Order 6

sum1198871015840

1198941198884

119894=

1

30 sum119887

1015840

119894119886119894119895119888119895=

1

720

sum1198871015840

119894119888119894119886119894119895=

1

180

(21)


Order 1

sum11988710158401015840

119894= 1 (22)

Order 2

sum11988710158401015840

119894119888119894=

1

2 (23)

Order 3

sum11988710158401015840

1198941198882

119894=

1

3 (24)

Order 4

sum11988710158401015840

1198941198883

119894=

1

4 sum119887

10158401015840

119894119886119894119895=

1

24 (25)

Order 5

sum11988710158401015840

1198941198884

119894=

1

5 sum119887

10158401015840

119894119886119894119895119888119895=

1

120 sum119887

10158401015840

119894119888119894119886119894119895=

1

30

(26)

Order 6

sum11988710158401015840

1198941198882

119894119886119894119895=

1

36 sum119887

10158401015840

1198941198861198941198951198882

119895+ sum119887

10158401015840

119894119888119894119886119894119895119888119895=

7

720

sum11988710158401015840

1198941198885

119894=

1

6

sum11988710158401015840

1198941198861198941198951198882

119895=

1

360 sum119887

10158401015840

119894119888119894119886119894119895119888119895=

1

144

1

2sum11988710158401015840

1198941198861198941198951198882

119895+ sum119887

10158401015840

119894119888119894119886119894119895119888119895=

1

120

(27)

All indices are from 1 to 119904 To obtain the fifth-orderRKD method the following simplifying assumption is usedin order to reduce the number of equations to be solved

1198871015840

119894= 11988710158401015840

119894(1 minus 119888119894) 119894 = 1 119904 (28)

To derive the three-stage fifth-order RKDmethod we usethe algebraic conditions up to order five (13)ndash(15) (17)ndash(20)and (22)ndash(26) The resulting system of equations consists of16 nonlinear equations with 13 unknown variables for whichwe need to solve Consequently there is no solution since thenumber of equations exceeds the number of unknowns to besolved for To overcome this the simplifying assumption (28)is imposed This will reduce the number of equations to 11with 11 unknowns making the system solvable This systemhas no free parameter but is consistent and yields a uniquesolution The coefficients of the method are given below (seeThe RKD5 method in (29)) The error norms for 119910

1198991199101015840119899 and

11991010158401015840

119899are given by 120591

(6)2

= 4392052306 times 10minus4 1205911015840(6)

2=

4108422255 times 10minus3 12059110158401015840(6)

2= 4925724188 times 10

minus3 respec-tively where 120591(6) 1205911015840(6) and 120591

10158401015840(6) are error terms of the sixth-order conditions for 119910

1198991199101015840119899 and 119910

10158401015840

119899 respectively


The RKD5 Method Consider the following

0 0

3

5+

radic6

10

27

500+

19radic6

10000

3

5minus

radic6

10

33

2500minus

51radic6

5000

51

1250minus

11radic6

12500

1

18

1

18minus

radic6

48

1

18+

radic6

48

1

9

7

36minus

radic6

18

7

36+

radic6

18

1

9

4

9minus

radic6

36

4

9+

radic6

36

(29)

Next we will discuss the zero stability of the methodwhich is one of the criteria for the method to be convergentZero stability is an important tool for proving the stabilityand convergence of linear multistep methods The interestedreader is referred to the textbooks by Lambert [13] andButcher [14] in which zero stability is discussed Zero stabilityhas been discussed in Hairer et al [15] to determine anupper bound on the order of convergence of linear multistepmethods In studying the zero stability of the RKD methodwe can write method (2) as follows

(

1 0 0

0 1 0

0 0 1

)(

119910119899+1

ℎ1199101015840

119899

ℎ211991010158401015840

119899+1

) = (

1 11

2

0 1 1

0 0 1

)(

119910119899+1

ℎ1199101015840

119899

ℎ211991010158401015840

119899+1

)

119901 (120585) = det [119868120585 minus 119860] = det(120585 minus 1 minus1 minus

1

2

0 120585 minus 1 minus1

0 0 120585 minus 1

)

(30)

Thus the characteristic polynomial is

119901 (120585) = (120585 minus 1)3 (31)

Hence the method is zero stable since the roots 120585 = 1 areless than or equal to one

3 Problems Tested and Numerical Results

In this section we will apply the new method to some third-order differential equation problems The following explicitRK methods are selected for the numerical comparisons

(i) RKD5 the three-stage fifth-order RKD method de-rived in this paper

(ii) RK4 the four-stage fourth-order RK method as inButcher [14]

(iii) DOPRI the seven-stage fifth-order Runge-Kutta de-rived by Dormand [11]

Problem 1 Consider

1198893119910 (119909)

1198891199093=

3

8radic(1 + 119909)5

119910 (0) = 1 1199101015840

(0) =1

2

11991010158401015840

(0) = minus1

4

(32)

Exact solution 119910(119909) = radic1 + 119909

Problem 2 Consider

1198893119910 (119909)

1198891199093= 1199102

(119909) + cos2 (119909) minus cos (119909) minus 1 119910 (0) = 0

1199101015840

(0) = 1 11991010158401015840

(0) = 0

(33)

Exact solution 119910(119909) = sin(119909)

4 An Application to a Problem inThin Film Flow

In this section we will apply the proposed method to a well-known problem in physics regarding the thin film flow of aliquid Many problems have discussed this problem see forexample [16ndash23] In a survey paper Tuck and Schwartz [16]discussed the flow of a thin film of viscous fluid over a solidsurface Tension and gravity as well as viscosity are taken intoaccountThe problemwas formulated using third-order ordi-nary differential equations (ODEs) as follows

1198893119910

1198891199093= 119891 (119910) (34)

for the film profile 119910(119909) in a coordinate frame moving withthe fluid The form of 119891(119910) varies according to the physicalcontext Different forms of the function 119891 are studied inTuck and Schwartz [16] For drainage down a dry surfacethe form of 119891(119910) was given as

1198893119910

1198891199093= minus1 +

1

1199102 (35)

When the surface is prewetted by a thin film withthickness 120575 gt 0 (where 120575 gt 0 is very small) the function119891 is given by

119891 (119910) = minus1 +1 + 120575 + 120575

2

1199102minus

120575 + 1205752

1199103 (36)

Problems concerning the flow of thin films of viscousfluid with a free surface in which surface tension effects playa role typically lead to third-order ordinary differential equa-tions governing the shape of the free surface of the fluid 119910 =

119910(119909) According to Tuck and Schwartz [16] one such equa-tion is

119910101584010158401015840

= 119910minus119896 119909 ge 119909

0(37)


Table 1 Table comparing values of the numerical solution a fifth-order Runge-Kutta method (DOPRI) fourth-order Runge-Kuttamethod (RK4) and our new RKD5 method at 119909 isin [0 02 04 06

08 10] taking ℎ = 001 and 119896 = 2 with the initial conditions119910(0) = 119910

1015840(0) = 119910

10158401015840(0) = 1

119909 Exact solution DOPRI RK4 RKD500 1000000000 10000000000 10000000000 10000000000

02 1221211030 12212100045 12212100046 12212100045

04 1488834893 14888347799 14888347800 14888347799

06 1807361404 18073613977 18073613978 18073613977

08 2179819234 21798192339 21798192341 21798192339

10 2608275822 26082748676 26082748678 26082748676



1015840(0) = 119910

10158401015840(0) = 1


02 1221211030 12212100041 12212105060 12212099592

04 1488834893 14888347796 14888356990 14888344801

06 1807361404 18073613979 18073626884 18073605532

08 2179819234 21798192349 21798208831 21798175255

10 2608275822 26082748696 26082768844 26082719667

with initial conditions

119910 (1199090) = 120572 119910

1015840(1199090) = 120573 119910

10158401015840(1199090) = 120574 (38)

where 120572 120573 and 120574 are constants is of particular importancebecause it describes the dynamic balance between surface andviscous forces in a thin fluid layer in the absence (or neglect)of gravity

For comparison purposes wewill use Runge-Kuttameth-ods which are fourth-order (RK4) and fifth-order (DOPRI)methods respectively To use Runge-Kutta methods we write(1) as a system of three first-order equations Following Biazaret al [17] we can write (37) as the following system

1198891199101

119889119909= 1199102(119909)

1198891199102

119889119909= 1199103(119909)

1198891199103

119889119909= 119910minus119896

1(119909)

(39)

where

1199101(0) = 1 119910

2(0) = 1 119910

3(0) = 1 (40)

we have taken 1199090= 0 and 120572 = 120573 = 120574 = 1

Unfortunately for general 119896 (37) cannot be solved analyt-ically We can however use these reductions to determine anefficient way to solve (1) numerically We focus on the cases119896 = 2 and 119896 = 3 The results are displayed in Tables 1 and 2for the case 119896 = 2 and Tables 3 and 4 for the case 119896 = 3

For comparison purposes we present efficiency curvewhere the common logarithm of the maximum global error

Table 3 Table comparing values of the numerical solution usinga fifth-order Runge-Kutta method (DOPRI) fourth-order Runge-Kutta method (RK4) and our new RKD5 method at 119909 isin [0 02 04

06 08 10] taking ℎ = 001 and 119896 = 3 with the initial conditions119910(0) = 119910

1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211551425 12211551424 12211551423

04 14881052844 14881052842 14881052838

06 18042625484 18042625481 18042625471

08 21715227984 21715227981 21715227960

10 25909582594 25909582591 25909582556



1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211559590 12211551421 12211550887

04 14881067401 14881052848 14881049238

06 18042645823 18042625503 18042615558

08 21715254210 21715228023 21715208324

10 25909615178 25909582657 25909549758

along the integration versus the function evaluations asshown in Figures 1 and 2 From Figures 1 and 2 we observedthat the method RKD5 performs better when integratingthird-order ODE compared to RK4 and DOPRI methods

From Tables 1 and 2 we observe that the numerical resultsusing RKD5 are correct to five decimal places Applyingthe fourth- and fifth-order RK methods (RK4 and DOPRI)to (39) for 119896 = 2 also yields five-decimal-place accuracyWhile from Tables 3 and 4 we observed the numericalresults for the new method RKD5 agrees to eight decimalplaces when compared with the fourth- and fifth-order RKmethods (RK4 and DOPRI) This is consistent with theresults displayed in Tables 1 and 2 In Figures 3 and 4 we plotthe numerical solution 119910

119894for 119896 = 2 and 119896 = 3 respectively

with ℎ = 001 Figure 5 shows that the new RKD5 methodrequires less function evaluations than the RK4 and DOPRImethods This is because when problem (37) is solved usingRK4 and DOPRI method it need to be reduced to a systemof first-order equations which is three times the dimen-sion

5 Discussion and Conclusion

In this paper we have derived the order conditions for aRunge-Kutta method which can be used to directly solvespecial third-order ordinary differential equations A three-stage fifth-order RKD method has been derived and appliedto the thin film flow problem Numerical results show that


2000150010005000

0

Function evaluations

minus2

minus4

minus6

minus8

minus10

RKD5

RK4DOPRI

log 1

0(M

AXE

RR)

Figure 1 The efficiency curves of the RKD5 method and itscomparisons for Problem 1 with 119909end = 10 and ℎ = 12

119894 119894 = 1 5

7006005004003002001000

minus2

minus3

minus4

minus5

minus6

minus7

minus8

minus9

minus10


RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

101

yi

yi

xi

262422

218161412

08060402

Figure 3 Plot of the solution 119910119894for problem (37) for 119896 = 2 ℎ = 001

101

yi

yi

xi

26

24

22

2

18

16

14

12

08060402


5

4

3

2

1

0minus1minus15minus2minus25minus3minus35minus4

RKD5

RK4DOPRI

log 1

0(fu

nctio

n ev

alua

tions

)

log10(h)

Figure 5 Plot of graph for function evaluations against step size ℎfor 119896 = 3 ℎ = 110

119894 119894 = 1 4

the new method agrees very well with well-known existingmethods in the literature and required less function evalua-tions As such this method is more cost effective in terms ofcomputation time than other existing methods

References

[1] D O Awoyemi ldquoA P-stable linear multistep method for solvinggeneral third order ordinary differential equationsrdquo Interna-tional Journal of Computer Mathematics vol 80 no 8 pp 987ndash993 2003

[2] 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

[3] N Zainuddin 2-point block backward differentiation formulafor solving higher order ODEs [PhD thesis] Universiti PutraMalaysia

[4] 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


[5] 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

[6] S N Jator ldquoOn a class of hybrid methods for 11991010158401015840

= 119891(119909 119910 1199101015840)rdquo

International Journal of Pure and Applied Mathematics vol 59no 4 pp 381ndash395 2010

[7] F Samat and F Ismail ldquoAn embedded explicit hybrid for ordi-nary differential Equationsrdquo Journal of Mathematics and Statis-tics vol 8 pp 32ndash36 2012

[8] Z B Ibrahim K I Othman andM Suleiman ldquoImplicit r-pointblock backward differentiation formula for solving first-orderstiffODEsrdquoAppliedMathematics and Computation vol 186 no1 pp 558ndash565 2007

[9] N Senu M Suleiman and F Ismail ldquoAn embedded explicitrungekutta-nystrom method for solving oscillatory problemsrdquoPhysica Scripta vol 80 no 1 Article ID 015005 7 pages 2009

[10] M S Mechee F Ismail N Senu and Z Siri ldquoRunge-kuttamethods for solving special third order differential equations119910101584010158401015840

= 119891(119909 119910) directlyrdquo in Proceedings of the 8th Mathematicsand Physical ScienceGraduate Congress ChulalongkornUniver-sity Bangkok Thailand 2012

[11] J R Dormand Numerical Methods for Differential EquationsA Computational Approach CRC Press Boca Raton Fla USA1996

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

[13] J D Lambert Numerical Methods for Ordinary Differential Sys-tems The Initial Value Problem JohnWiley amp Sons ChichesterUK 1991

[14] J C ButcherNumericalMethods forOrdinaryDifferential Equa-tions John Wiley amp Sons Chichester UK 2nd edition 2008

[15] E Hairer S P Noslashrsett and G Wanner Solving Ordinary Dif-ferential Equations I Nonstiff Problems vol 8 Springer BerlinGermany 2nd edition 1993

[16] E O Tuck and L W Schwartz ldquoA numerical and asymptoticstudy of some third-order ordinary differential equations rele-vant to draining and coating flowsrdquo SIAM Review vol 32 no 3pp 453ndash469 1990

[17] J Biazar E Babolian and R Islam ldquoSolution of the systemof ordinary differential equations by Adomian decompositionmethodrdquo Applied Mathematics and Computation vol 147 no 3pp 713ndash719 2004

[18] E Momoniat and F M Mahomed ldquoSymmetry reduction andnumerical solution of a third-order ODE from thin film flowrdquoMathematical amp Computational Applications vol 15 no 4 pp709ndash719 2010

[19] F Bernis and L A Peletier ldquoTwo problems from draining flowsinvolving third-order ordinary differential equationsrdquo SIAMJournal on Mathematical Analysis vol 27 no 2 pp 515ndash5271996

[20] B R Duffy and S K Wilson ldquoA third-order differential equa-tion arising in thin-film flows and relevant to Tannerrsquos lawrdquoApplied Mathematics Letters vol 10 no 3 pp 63ndash68 1997

[21] T G Myers ldquoThin films with high surface tensionrdquo SIAM Re-view vol 40 no 3 pp 441ndash462 1998

[22] E Momoniat ldquoSymmetries first integrals and phase planes of athird-order ordinary differential equation from thin film flowrdquoMathematical and Computer Modelling vol 49 no 1-2 pp 215ndash225 2009

[23] E Momoniat ldquoNumerical investigation of a third-order ODEfrom thin film flowrdquoMeccanica vol 46 no 2 pp 313ndash323 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


2 Derivation of RKD Method

The general form of RKD method with 119904 stages for solvinginitial value problem (1) can be written as

119910119899+1

= 119910119899+ ℎ1199101015840

119899+

ℎ2

211991010158401015840

119899+ ℎ3

119904

sum

119894=1

119887119894119896119894

1199101015840

119899+1= 1199101015840

119899+ ℎ11991010158401015840

119899+ ℎ2

119904

sum

119894=1

1198871015840

119894119896119894

11991010158401015840

119899+1= 11991010158401015840

119899+ ℎ

119904

sum

119894=1

11988710158401015840

119894119896119894

(2)

where

1198961= 119891 (119909

119899 119910119899) (3)

119896119894= 119891(119909

119899+ 119888119894ℎ 119910119899+ ℎ1198881198941199101015840

119899+

ℎ2

21198882

11989411991010158401015840

119899+ ℎ3

119894minus1

sum

119895=1

119886119894119895119896119895) (4)

for 119894 = 2 3 119904The parameters of the RKD method are 119888

119894 119886119894119895 119887119894 1198871015840

119894 11988710158401015840

119894

for 119894 = 1 2 119904 and 119895 = 1 2 119904 are assumed to be realIf 119886119894119895= 0 for 119894 ⩽ 119895 it is an explicit method and otherwise it

is an implicit method The RKD method can be expressed inthe Butcher notation using the table of coefficients as follows

119888 119860

119887119879

1198871015840119879

11988710158401015840119879

(5)

To determine the coefficients of the method given by (2)ndash(4) the RKD method expression is expanded using Taylorrsquosseries expansion After some algebraic manipulations thisexpansion is equated to the true solution that is given byTaylorrsquos series expansion General order conditions for theRKD method can be found from the direct expansion of thelocal truncation error This idea is based on the derivation oforder conditions for the Runge-Kutta method introduced inDormand [11] The RKD formulae in (2) may be expressed as

119910119899+1

= 119910119899+ ℎΦ (119909

119899 119910119899)

1199101015840

119899+1= 1199101015840

119899+ ℎΦ1015840(119909119899 119910119899)

11991010158401015840

119899+1= 11991010158401015840

119899+ ℎΦ10158401015840(119909119899 119910119899)

(6)

where the increment functions are

Φ(119909119899 119910119899) = 1199101015840

119899+

ℎ

211991010158401015840

119899+ ℎ2

119904

sum

119894=1

119887119894119896119894

Φ1015840(119909119899 119910119899) = 11991010158401015840

119899+ ℎ

119904

sum

119894=1

1198871015840

119894119896119894

Φ10158401015840(119909119899 119910119899) =

119904

sum

119894=1

11988710158401015840

119894119896119894

(7)

and 119896119894is defined in formula (4) If Δ is the Taylor series

increment function then the local truncation errors of thesolution the first derivative and the second derivative maybe obtained by substituting the true solution 119910(119909) of (1) intothe RKD increment function This gives

119905119899+1

= ℎ [Φ minus Δ] 1199051015840

119899+1= ℎ [Φ

1015840minus Δ1015840]

11990510158401015840

119899+1= ℎ [Φ

10158401015840minus Δ10158401015840]

(8)

These expressions are best given in terms of elementarydifferentials and the Taylor series increment may be writtenas

Δ = 1199101015840+

1

2ℎ11991010158401015840+

1

6ℎ2119865(3)

1+

1

24ℎ3119865(4)

1+ 119874 (ℎ

4)

Δ1015840= 11991010158401015840+

1

2ℎ119865(3)

1+

1

6ℎ2119865(4)

1+

1

24ℎ3119865(5)

1+ 119874 (ℎ

4)

Δ10158401015840= 119865(3)

1+

1

2ℎ119865(4)

1+

1

6ℎ2119865(5)

1+ 119874 (ℎ

3)

(9)

where for the scalar case the first few elementary differentialsare

119865(3)

1= 119891

119865(4)

1= 119891119909+ 1198911199101199101015840

119865(5)

1= 119891119909119909

+ 21198911199091199101199101015840+ 11989111991011991010158401015840+ 119891119910119910(1199101015840)2

(10)

Using the above terms the increment functions Φ Φ1015840

and Φ10158401015840 for the RKD formula become

119904

sum

119894=1

119887119894119896119894=

119904

sum

119894=1

119887119894119891 +

119904

sum

119894=1

119887119894119888119894(119891119909+ 1198911199101199101015840) ℎ

+1

2

119904

sum

119894=1

1198871198941198882

119894(119891119909119909

+ 21198911199091199101199101015840+ 11989111991011991010158401015840+ 119891119910119910(1199101015840)2

) ℎ2

+ 119874 (ℎ3)

119904

sum

119894=1

1198871015840

119894119896119894=

119904

sum

119894=1

1198871015840

119894119891 +

119904

sum

119894=1

1198871015840

119894119888119894(119891119909+ 1198911199101199101015840) ℎ

+1

2

119904

sum

119894=1

1198871015840

1198941198882

119894(119891119909119909

+ 21198911199091199101199101015840+ 11989111991011991010158401015840+ 119891119910119910(1199101015840)2

) ℎ2

+ 119874 (ℎ3)

119904

sum

119894=1

11988710158401015840

119894119896119894=

119904

sum

119894=1

11988710158401015840

119894119891 +

119904

sum

119894=1

11988710158401015840

119894119888119894(119891119909+ 1198911199101199101015840) ℎ

+1

2

119904

sum

119894=1

11988710158401015840

1198941198882

119894(119891119909119909

+ 21198911199091199101199101015840+ 11989111991011991010158401015840+ 119891119910119910(1199101015840)2

) ℎ2

+ 119874 (ℎ3)

(11)




119905119899+1

= ℎ3[

119904

sum

119894=1

119887119894119896119894minus (

1

6119865(3)

1+

1

24119865(4)


1199051015840

119899+1= ℎ2[

119904

sum

119894=1

1198871015840

119894119896119894minus (

1

2119865(3)

1+

1

6119865(4)


11990510158401015840

119899+1= ℎ[

119904

sum

119894=1

11988710158401015840

119894119896119894minus (119865(3)

1+

1

2119865(4)

1+

1

6ℎ2119865(5)


(12)



Order 3

sum119887119894=

1

6 (13)

Order 4

sum119887119894119888119894=

1

24 (14)

Order 5

sum1198871198941198882

119894=

1

60 (15)

Order 6

sum1198871198941198883

119894=

1

120 sum119887

119894119886119894119895=

1

720 (16)


Order 2

sum1198871015840

119894=

1

2 (17)

Order 3

sum1198871015840

119894119888119894=

1

6 (18)

Order 4

sum1198871015840

1198941198882

119894=

1

12 (19)

Order 5

sum1198871015840

1198941198883

119894=

1

20 sum119887

1015840

119894119886119894119895=

1

120 (20)

Order 6

sum1198871015840

1198941198884

119894=

1

30 sum119887

1015840

119894119886119894119895119888119895=

1

720

sum1198871015840

119894119888119894119886119894119895=

1

180

(21)


Order 1

sum11988710158401015840

119894= 1 (22)

Order 2

sum11988710158401015840

119894119888119894=

1

2 (23)

Order 3

sum11988710158401015840

1198941198882

119894=

1

3 (24)

Order 4

sum11988710158401015840

1198941198883

119894=

1

4 sum119887

10158401015840

119894119886119894119895=

1

24 (25)

Order 5

sum11988710158401015840

1198941198884

119894=

1

5 sum119887

10158401015840

119894119886119894119895119888119895=

1

120 sum119887

10158401015840

119894119888119894119886119894119895=

1

30

(26)

Order 6

sum11988710158401015840

1198941198882

119894119886119894119895=

1

36 sum119887

10158401015840

1198941198861198941198951198882

119895+ sum119887

10158401015840

119894119888119894119886119894119895119888119895=

7

720

sum11988710158401015840

1198941198885

119894=

1

6

sum11988710158401015840

1198941198861198941198951198882

119895=

1

360 sum119887

10158401015840

119894119888119894119886119894119895119888119895=

1

144

1

2sum11988710158401015840

1198941198861198941198951198882

119895+ sum119887

10158401015840

119894119888119894119886119894119895119888119895=

1

120

(27)


1198871015840

119894= 11988710158401015840

119894(1 minus 119888119894) 119894 = 1 119904 (28)


1198991199101015840119899 and

11991010158401015840


(6)2

= 4392052306 times 10minus4 1205911015840(6)

2=

4108422255 times 10minus3 12059110158401015840(6)

2= 4925724188 times 10



1198991199101015840119899 and 119910

10158401015840

119899 respectively



0 0

3

5+

radic6

10

27

500+

19radic6

10000

3

5minus

radic6

10

33

2500minus

51radic6

5000

51

1250minus

11radic6

12500

1

18

1

18minus

radic6

48

1

18+

radic6

48

1

9

7

36minus

radic6

18

7

36+

radic6

18

1

9

4

9minus

radic6

36

4

9+

radic6

36

(29)


(

1 0 0

0 1 0

0 0 1

)(

119910119899+1

ℎ1199101015840

119899

ℎ211991010158401015840

119899+1

) = (

1 11

2

0 1 1

0 0 1

)(

119910119899+1

ℎ1199101015840

119899

ℎ211991010158401015840

119899+1

)


1

2


0 0 120585 minus 1

)

(30)


119901 (120585) = (120585 minus 1)3 (31)







Problem 1 Consider

1198893119910 (119909)

1198891199093=

3

8radic(1 + 119909)5

119910 (0) = 1 1199101015840

(0) =1

2

11991010158401015840

(0) = minus1

4

(32)


Problem 2 Consider

1198893119910 (119909)

1198891199093= 1199102


1199101015840

(0) = 1 11991010158401015840

(0) = 0

(33)




1198893119910

1198891199093= 119891 (119910) (34)


1198893119910

1198891199093= minus1 +

1

1199102 (35)


119891 (119910) = minus1 +1 + 120575 + 120575

2

1199102minus

120575 + 1205752

1199103 (36)



119910101584010158401015840

= 119910minus119896 119909 ge 119909

0(37)




1015840(0) = 119910

10158401015840(0) = 1


02 1221211030 12212100045 12212100046 12212100045

04 1488834893 14888347799 14888347800 14888347799

06 1807361404 18073613977 18073613978 18073613977

08 2179819234 21798192339 21798192341 21798192339

10 2608275822 26082748676 26082748678 26082748676



1015840(0) = 119910

10158401015840(0) = 1


02 1221211030 12212100041 12212105060 12212099592

04 1488834893 14888347796 14888356990 14888344801

06 1807361404 18073613979 18073626884 18073605532

08 2179819234 21798192349 21798208831 21798175255

10 2608275822 26082748696 26082768844 26082719667


119910 (1199090) = 120572 119910

1015840(1199090) = 120573 119910

10158401015840(1199090) = 120574 (38)



1198891199101

119889119909= 1199102(119909)

1198891199102

119889119909= 1199103(119909)

1198891199103

119889119909= 119910minus119896

1(119909)

(39)

where

1199101(0) = 1 119910

2(0) = 1 119910

3(0) = 1 (40)

we have taken 1199090= 0 and 120572 = 120573 = 120574 = 1





1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211551425 12211551424 12211551423

04 14881052844 14881052842 14881052838

06 18042625484 18042625481 18042625471

08 21715227984 21715227981 21715227960

10 25909582594 25909582591 25909582556



1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211559590 12211551421 12211550887

04 14881067401 14881052848 14881049238

06 18042645823 18042625503 18042615558

08 21715254210 21715228023 21715208324

10 25909615178 25909582657 25909549758








2000150010005000

0


minus2

minus4

minus6

minus8

minus10

RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

7006005004003002001000

minus2

minus3

minus4

minus5

minus6

minus7

minus8

minus9

minus10


RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

101

yi

yi

xi

262422

218161412

08060402


101

yi

yi

xi

26

24

22

2

18

16

14

12

08060402


5

4

3

2

1


RKD5

RK4DOPRI

log 1

0(fu

nctio

n ev

alua

tions

)

log10(h)


119894 119894 = 1 4


References








= 119891(119909 119910 1199101015840)rdquo



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119905119899+1

= ℎ3[

119904

sum

119894=1

119887119894119896119894minus (

1

6119865(3)

1+

1

24119865(4)


1199051015840

119899+1= ℎ2[

119904

sum

119894=1

1198871015840

119894119896119894minus (

1

2119865(3)

1+

1

6119865(4)


11990510158401015840

119899+1= ℎ[

119904

sum

119894=1

11988710158401015840

119894119896119894minus (119865(3)

1+

1

2119865(4)

1+

1

6ℎ2119865(5)


(12)



Order 3

sum119887119894=

1

6 (13)

Order 4

sum119887119894119888119894=

1

24 (14)

Order 5

sum1198871198941198882

119894=

1

60 (15)

Order 6

sum1198871198941198883

119894=

1

120 sum119887

119894119886119894119895=

1

720 (16)


Order 2

sum1198871015840

119894=

1

2 (17)

Order 3

sum1198871015840

119894119888119894=

1

6 (18)

Order 4

sum1198871015840

1198941198882

119894=

1

12 (19)

Order 5

sum1198871015840

1198941198883

119894=

1

20 sum119887

1015840

119894119886119894119895=

1

120 (20)

Order 6

sum1198871015840

1198941198884

119894=

1

30 sum119887

1015840

119894119886119894119895119888119895=

1

720

sum1198871015840

119894119888119894119886119894119895=

1

180

(21)


Order 1

sum11988710158401015840

119894= 1 (22)

Order 2

sum11988710158401015840

119894119888119894=

1

2 (23)

Order 3

sum11988710158401015840

1198941198882

119894=

1

3 (24)

Order 4

sum11988710158401015840

1198941198883

119894=

1

4 sum119887

10158401015840

119894119886119894119895=

1

24 (25)

Order 5

sum11988710158401015840

1198941198884

119894=

1

5 sum119887

10158401015840

119894119886119894119895119888119895=

1

120 sum119887

10158401015840

119894119888119894119886119894119895=

1

30

(26)

Order 6

sum11988710158401015840

1198941198882

119894119886119894119895=

1

36 sum119887

10158401015840

1198941198861198941198951198882

119895+ sum119887

10158401015840

119894119888119894119886119894119895119888119895=

7

720

sum11988710158401015840

1198941198885

119894=

1

6

sum11988710158401015840

1198941198861198941198951198882

119895=

1

360 sum119887

10158401015840

119894119888119894119886119894119895119888119895=

1

144

1

2sum11988710158401015840

1198941198861198941198951198882

119895+ sum119887

10158401015840

119894119888119894119886119894119895119888119895=

1

120

(27)


1198871015840

119894= 11988710158401015840

119894(1 minus 119888119894) 119894 = 1 119904 (28)


1198991199101015840119899 and

11991010158401015840


(6)2

= 4392052306 times 10minus4 1205911015840(6)

2=

4108422255 times 10minus3 12059110158401015840(6)

2= 4925724188 times 10



1198991199101015840119899 and 119910

10158401015840

119899 respectively



0 0

3

5+

radic6

10

27

500+

19radic6

10000

3

5minus

radic6

10

33

2500minus

51radic6

5000

51

1250minus

11radic6

12500

1

18

1

18minus

radic6

48

1

18+

radic6

48

1

9

7

36minus

radic6

18

7

36+

radic6

18

1

9

4

9minus

radic6

36

4

9+

radic6

36

(29)


(

1 0 0

0 1 0

0 0 1

)(

119910119899+1

ℎ1199101015840

119899

ℎ211991010158401015840

119899+1

) = (

1 11

2

0 1 1

0 0 1

)(

119910119899+1

ℎ1199101015840

119899

ℎ211991010158401015840

119899+1

)


1

2


0 0 120585 minus 1

)

(30)


119901 (120585) = (120585 minus 1)3 (31)







Problem 1 Consider

1198893119910 (119909)

1198891199093=

3

8radic(1 + 119909)5

119910 (0) = 1 1199101015840

(0) =1

2

11991010158401015840

(0) = minus1

4

(32)


Problem 2 Consider

1198893119910 (119909)

1198891199093= 1199102


1199101015840

(0) = 1 11991010158401015840

(0) = 0

(33)




1198893119910

1198891199093= 119891 (119910) (34)


1198893119910

1198891199093= minus1 +

1

1199102 (35)


119891 (119910) = minus1 +1 + 120575 + 120575

2

1199102minus

120575 + 1205752

1199103 (36)



119910101584010158401015840

= 119910minus119896 119909 ge 119909

0(37)




1015840(0) = 119910

10158401015840(0) = 1


02 1221211030 12212100045 12212100046 12212100045

04 1488834893 14888347799 14888347800 14888347799

06 1807361404 18073613977 18073613978 18073613977

08 2179819234 21798192339 21798192341 21798192339

10 2608275822 26082748676 26082748678 26082748676



1015840(0) = 119910

10158401015840(0) = 1


02 1221211030 12212100041 12212105060 12212099592

04 1488834893 14888347796 14888356990 14888344801

06 1807361404 18073613979 18073626884 18073605532

08 2179819234 21798192349 21798208831 21798175255

10 2608275822 26082748696 26082768844 26082719667


119910 (1199090) = 120572 119910

1015840(1199090) = 120573 119910

10158401015840(1199090) = 120574 (38)



1198891199101

119889119909= 1199102(119909)

1198891199102

119889119909= 1199103(119909)

1198891199103

119889119909= 119910minus119896

1(119909)

(39)

where

1199101(0) = 1 119910

2(0) = 1 119910

3(0) = 1 (40)

we have taken 1199090= 0 and 120572 = 120573 = 120574 = 1





1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211551425 12211551424 12211551423

04 14881052844 14881052842 14881052838

06 18042625484 18042625481 18042625471

08 21715227984 21715227981 21715227960

10 25909582594 25909582591 25909582556



1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211559590 12211551421 12211550887

04 14881067401 14881052848 14881049238

06 18042645823 18042625503 18042615558

08 21715254210 21715228023 21715208324

10 25909615178 25909582657 25909549758








2000150010005000

0


minus2

minus4

minus6

minus8

minus10

RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

7006005004003002001000

minus2

minus3

minus4

minus5

minus6

minus7

minus8

minus9

minus10


RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

101

yi

yi

xi

262422

218161412

08060402


101

yi

yi

xi

26

24

22

2

18

16

14

12

08060402


5

4

3

2

1


RKD5

RK4DOPRI

log 1

0(fu

nctio

n ev

alua

tions

)

log10(h)


119894 119894 = 1 4


References








= 119891(119909 119910 1199101015840)rdquo



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











0 0

3

5+

radic6

10

27

500+

19radic6

10000

3

5minus

radic6

10

33

2500minus

51radic6

5000

51

1250minus

11radic6

12500

1

18

1

18minus

radic6

48

1

18+

radic6

48

1

9

7

36minus

radic6

18

7

36+

radic6

18

1

9

4

9minus

radic6

36

4

9+

radic6

36

(29)


(

1 0 0

0 1 0

0 0 1

)(

119910119899+1

ℎ1199101015840

119899

ℎ211991010158401015840

119899+1

) = (

1 11

2

0 1 1

0 0 1

)(

119910119899+1

ℎ1199101015840

119899

ℎ211991010158401015840

119899+1

)


1

2


0 0 120585 minus 1

)

(30)


119901 (120585) = (120585 minus 1)3 (31)







Problem 1 Consider

1198893119910 (119909)

1198891199093=

3

8radic(1 + 119909)5

119910 (0) = 1 1199101015840

(0) =1

2

11991010158401015840

(0) = minus1

4

(32)


Problem 2 Consider

1198893119910 (119909)

1198891199093= 1199102


1199101015840

(0) = 1 11991010158401015840

(0) = 0

(33)




1198893119910

1198891199093= 119891 (119910) (34)


1198893119910

1198891199093= minus1 +

1

1199102 (35)


119891 (119910) = minus1 +1 + 120575 + 120575

2

1199102minus

120575 + 1205752

1199103 (36)



119910101584010158401015840

= 119910minus119896 119909 ge 119909

0(37)




1015840(0) = 119910

10158401015840(0) = 1


02 1221211030 12212100045 12212100046 12212100045

04 1488834893 14888347799 14888347800 14888347799

06 1807361404 18073613977 18073613978 18073613977

08 2179819234 21798192339 21798192341 21798192339

10 2608275822 26082748676 26082748678 26082748676



1015840(0) = 119910

10158401015840(0) = 1


02 1221211030 12212100041 12212105060 12212099592

04 1488834893 14888347796 14888356990 14888344801

06 1807361404 18073613979 18073626884 18073605532

08 2179819234 21798192349 21798208831 21798175255

10 2608275822 26082748696 26082768844 26082719667


119910 (1199090) = 120572 119910

1015840(1199090) = 120573 119910

10158401015840(1199090) = 120574 (38)



1198891199101

119889119909= 1199102(119909)

1198891199102

119889119909= 1199103(119909)

1198891199103

119889119909= 119910minus119896

1(119909)

(39)

where

1199101(0) = 1 119910

2(0) = 1 119910

3(0) = 1 (40)

we have taken 1199090= 0 and 120572 = 120573 = 120574 = 1





1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211551425 12211551424 12211551423

04 14881052844 14881052842 14881052838

06 18042625484 18042625481 18042625471

08 21715227984 21715227981 21715227960

10 25909582594 25909582591 25909582556



1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211559590 12211551421 12211550887

04 14881067401 14881052848 14881049238

06 18042645823 18042625503 18042615558

08 21715254210 21715228023 21715208324

10 25909615178 25909582657 25909549758








2000150010005000

0


minus2

minus4

minus6

minus8

minus10

RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

7006005004003002001000

minus2

minus3

minus4

minus5

minus6

minus7

minus8

minus9

minus10


RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

101

yi

yi

xi

262422

218161412

08060402


101

yi

yi

xi

26

24

22

2

18

16

14

12

08060402


5

4

3

2

1


RKD5

RK4DOPRI

log 1

0(fu

nctio

n ev

alua

tions

)

log10(h)


119894 119894 = 1 4


References








= 119891(119909 119910 1199101015840)rdquo



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












1015840(0) = 119910

10158401015840(0) = 1


02 1221211030 12212100045 12212100046 12212100045

04 1488834893 14888347799 14888347800 14888347799

06 1807361404 18073613977 18073613978 18073613977

08 2179819234 21798192339 21798192341 21798192339

10 2608275822 26082748676 26082748678 26082748676



1015840(0) = 119910

10158401015840(0) = 1


02 1221211030 12212100041 12212105060 12212099592

04 1488834893 14888347796 14888356990 14888344801

06 1807361404 18073613979 18073626884 18073605532

08 2179819234 21798192349 21798208831 21798175255

10 2608275822 26082748696 26082768844 26082719667


119910 (1199090) = 120572 119910

1015840(1199090) = 120573 119910

10158401015840(1199090) = 120574 (38)



1198891199101

119889119909= 1199102(119909)

1198891199102

119889119909= 1199103(119909)

1198891199103

119889119909= 119910minus119896

1(119909)

(39)

where

1199101(0) = 1 119910

2(0) = 1 119910

3(0) = 1 (40)

we have taken 1199090= 0 and 120572 = 120573 = 120574 = 1





1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211551425 12211551424 12211551423

04 14881052844 14881052842 14881052838

06 18042625484 18042625481 18042625471

08 21715227984 21715227981 21715227960

10 25909582594 25909582591 25909582556



1015840(0) = 119910

10158401015840(0) = 1

119909 RK4 DOPRI RKD500 10000000000 10000000000 10000000000

02 12211559590 12211551421 12211550887

04 14881067401 14881052848 14881049238

06 18042645823 18042625503 18042615558

08 21715254210 21715228023 21715208324

10 25909615178 25909582657 25909549758








2000150010005000

0


minus2

minus4

minus6

minus8

minus10

RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

7006005004003002001000

minus2

minus3

minus4

minus5

minus6

minus7

minus8

minus9

minus10


RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

101

yi

yi

xi

262422

218161412

08060402


101

yi

yi

xi

26

24

22

2

18

16

14

12

08060402


5

4

3

2

1


RKD5

RK4DOPRI

log 1

0(fu

nctio

n ev

alua

tions

)

log10(h)


119894 119894 = 1 4


References








= 119891(119909 119910 1199101015840)rdquo



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










2000150010005000

0


minus2

minus4

minus6

minus8

minus10

RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

7006005004003002001000

minus2

minus3

minus4

minus5

minus6

minus7

minus8

minus9

minus10


RKD5

RK4DOPRI

log 1

0(M

AXE

RR)


119894 119894 = 1 5

101

yi

yi

xi

262422

218161412

08060402


101

yi

yi

xi

26

24

22

2

18

16

14

12

08060402


5

4

3

2

1


RKD5

RK4DOPRI

log 1

0(fu

nctio

n ev

alua

tions

)

log10(h)


119894 119894 = 1 4


References








= 119891(119909 119910 1199101015840)rdquo



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












= 119891(119909 119910 1199101015840)rdquo



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article a three-stage fifth-order runge...

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