research article the higher accuracy fourth-order iade...

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 236548 13 pageshttpdxdoiorg1011552013236548

Research ArticleThe Higher Accuracy Fourth-Order IADE Algorithm

N Abu Mansor1 A K Zulkifle1 N Alias2 M K Hasan3 and M J N Boyce1

1 College of Engineering Universiti Tenaga Nasional Jalan Ikram-UNITEN 43000 Kajang Selangor Malaysia2 Ibnu Sina Institute of Fundamental Science Studies Universiti Teknologi Malaysia 81310 Skudai Johor Malaysia3 Faculty of Information Science and Technology Universiti Kebangsaan Malaysia 43600 Bangi Selangor Malaysia

Correspondence should be addressed to N Abu Mansor norelizaunitenedumy

Received 21 May 2013 Accepted 26 July 2013

Academic Editor Juan Manuel Pena

Copyright copy 2013 N Abu Mansor 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

This study develops the novel fourth-order iterative alternating decomposition explicit (IADE)method ofMitchell and Fairweather(IADEMF4) algorithm for the solution of the one-dimensional linear heat equationwithDirichlet boundary conditionsThehigher-order finite difference scheme is developed by representing the spatial derivative in the heat equation with the fourth-order finitedifference Crank-Nicolson approximationThis leads to the formation of pentadiagonal matrices in the systems of linear equationsThe algorithm also employs the higher accuracy of theMitchell and Fairweather variant Despite the schemersquos higher computationalcomplexity experimental results show that it is not only capable of enhancing the accuracy of the original correspondingmethod ofsecond-order (IADEMF2) but its solutions are also in very much agreement with the exact solutions Besides it is unconditionallystable and has proven to be convergent The IADEMF4 is also found to be more accurate more efficient and has better rate ofconvergence than the benchmarked fourth-order classical iterative methods namely the Jacobi (JAC4) the Gauss-Seidel (GS4)and the successive over-relaxation (SOR4) methods

1 Introduction

Numerical methods with accuracy of the order 119874(Δℎ)119899 119899 gt

2 are referred to as higher-order methods (ℎ = mesh size)Recent developments seem to desire methods of higher-order for achieving higher accuracy numerical solutionsfor problems involving partial differential equations Whenhigher-order methods are evaluated factors such as rate ofconvergence stability and boundary conditions have to beconsidered too There are mainly two categories of finitedifference higher-order schemes

The first category is the noncompact stencils that util-ize any grid point surrounding the grid points of interestwhere the difference schemes are implemented The methodinvolves larger matrix bandwidth but in many cases theincrease is not large [1] There will be a probable increasein execution time as more grid points are used Howeverincreasing the number of grid points provides advantagesin terms of enhancement in accuracy and improvement inresolution [2] To enhance accuracy the approach shouldconsider proper treatment of boundary conditions of com-parable accuracy

The second category is the compact stencils that useLax-Wendroff rsquos idea [3] proposed by MacKinnon and Carey[4] The method uses smaller number of stencils makingit computationally efficient and highly accurate However itrequires inversion of matrix to obtain spatial derivative ateach point Also the boundary stencil has a large effect onthe stability and accuracy of the scheme [5] Usually compactschemes of order higher than four require formation ofauxiliary equations due to boundary conditions This resultsin large bandwidth matrices which are not symmetric [6]Furthermore the schemes can get fairly complicated formorecomplex equations

Some of the current findings on higher-order methodsinclude the work by Sulaiman et al [7] who suggesteda fourth-order quarter sweep modified successive over-relaxation (QSMSOR) iterative method for solving a one-dimensional parabolic equation It is found to be superior interms of rate of convergence and execution time as comparedto other SOR methods Jha [8] formulated the six-orderaccurate quarter sweep alternating group explicit (QSAGE)iterative finite difference method for solving nonlinear sin-gular two-point boundary value problems The method can

2 Journal of Applied Mathematics

be implemented in parallel and is found to be superiorto the corresponding full sweep alternating group explicit(AGE) and SOR methods Jin et al [9] proposed an AGEiteration method of the fourth-order accuracy by integratingthe grouping explicit method with numerical boundary con-ditionsThemethod was used to solve initial-boundary valueproblem of convection equations Fu and Tan [10] showedthat an unconditionally stable split-step FDTD method withhigher-order spatial accuracy is more accurate than thelower-order methods The dispersion error of the proposedmethod is comparable with the higher-order ADI-FDTDmethod

Higher-order methods are also studied by Mohebbi andDeghan [11] who applied a compact finite difference approx-imation of fourth-order and the cubic C1 spline colloca-tion method to some one-dimensional heat and advection-diffusion equations The scheme has fourth-order accuracyin both space and time and unconditionally stable Liao[12] proposed an efficient and high-order accuracy of thefourth-order compact finite difference method to solve one-dimensional Burgersrsquo equation Tian and Ge [13] studied astable fourth-order compact ADI method for solving two-dimensional unsteady convection diffusion problems Themethod is temporally second-order and spatially fourth-order accurate which requires only a regular five-point 2Dstencil similar to that in the standard second-order methodsChun [14] solved some nonlinear equations by applying thefourth-order iterative methods containing the Kingrsquos fourth-order family It was observed that the proposed methodhas at least equal performance compared to other methodsof the same order Zhu et al [15] presented a high-orderparallel finite difference algorithm based on the domaindecomposition strategy The study used the classical explicitscheme to calculate the interface values between subdomainsand the fourth-order compact schemes for the interior valuesThe method has high accuracy and is convergent and stableGao and Xie [16] devised a fourth-order alternating directionimplicit compact finite difference schemes for the solutionof two-dimensional Schrodinger equations The method ishighly competitive as compared to other existing methodsand it achieves the expected convergence rate

This study develops a higher-order finite difference algo-rithm with noncompact stencils that is capable of deliveringhighly accurate solutions for a one-dimensional heat equa-tion with Dirichlet boundary conditions The study focuseson modifying the unconditionally stable and convergentsecond-order iterative alternating decomposition explicit(IADE) method of Mitchell and Fairweather (IADEMF2)The IADEMF2 which was originally proposed by Sahimiet al [17] employs the fractional splitting of the Mitchell andFairweather (MF) variant that has an accuracy of the order119874((Δ119905)

2

+ (Δ119909)4

) The scheme executes a two-stage processinvolving the solution of sets of tridiagonal equations alonglines parallel to the first and second time steps respectivelyIt is found to be unconditionally stable convergent andmoreaccurate than the classical AGE class of methods namely theAGE method based on the Peaceman and Rachford variant(AGE-PR) whose spatial accuracy is of second order and

the AGEmethod based on the Douglas and Rachford variant(AGE-DR) whose temporal accuracy is only of first orderThe detailed derivation of the IADEMF2 can be obtainedfrom [17]

In this paper a fourth-order Crank-Nicolson (CN) dif-ference approximation is applied to the spatial derivative inthe heat equation and the MF variant is employed leadingto the formation of the fourth-order IADEMF (IADEMF4)numerical algorithm The convergence of the IADEMF4is analyzed and proven Numerical experiments verify thepotential of the IADEMF4 in enhancing the accuracy ofthe IADEMF2 The results of the proposed higher-orderscheme are also compared with the benchmarked fourth-order classical iterative methods such as the fourth-orderGauss-Seidel (GS4) the fourth-order Jacobi (JAC4) and thefourth-order successive over-relaxation (SOR4) methods

This paper is organized as follows Section 2 discusses theimplementation and stability of the fourth-order CN approx-imation on the heat equation In Section 3 the IADEMF4algorithm is formulated Section 4 analyses the convergenceof the IADEMF4 Section 5 provides the equations for thebenchmarked fourth-order classical iterative methods Thecomputational complexity of the methods considered in thispaper is given in Section 6 while the pseudocode for theIADEMF4 sequential algorithm is presented in Section 7Theexperiments conducted are given in Section 8 Sections 9and 10 provide some discussion and conclusion based on theobtained numerical results

2 Fourth-Order Crank-NicolsonApproximation to the Spatial Derivativein the Heat Equation

Consider the following one-dimensional parabolic heat equa-tion (1) that has been suitably assumed to be nondimensiona-lised It models the flow of heat in a homogeneous unchang-ing medium of finite extent in the absence of heat source

120597119880

120597119905=1205972

119880

1205971199092(1)

subject to given initial and Dirichlet boundary conditions

119880 (119909 0) = 119891 (119909) 0 le 119909 le 1

119880 (0 119905) = 119892 (119905) 0 lt 119905 le 119879

119880 (1 119905) = ℎ (119905) 0 lt 119905 le 119879

(2)

For the problem in (1) the finite difference approach dis-cretizes the time-space domain by placing a rectangular gridover the domain with grid spacing ofΔ119905 andΔ119909 in the 119905- and119909-directions respectively The grid consists of the set of linesparallel to the 119905-axis given by 119909

119894= 119894Δ119909 119894 = 0 1 119898119898 + 1

and a set of lines parallel to the 119909-axis given by 119905119896= 119896Δ119905 119896 =

0 1 119899 119899+1 For simplicity the grid spacing is taken to beuniform so thatΔ119909 = 1(119898+1) andΔ119905 = 119879(119899+1) At a grid-point119875(119909

119894 119905119896) in the solution domain the dependent variable

119880(119909 119905) which represents the nondimensional temperature attime 119905 and at position 119909 is approximated by 119906119896

119894

Journal of Applied Mathematics 3

At the grid-point 119875(119909119894 119905119896+12

) the IADEMF4 schemereplaces the spatial derivative in the heat equation with ahigher-order particularly the fourth-order Crank-Nicolson(CN) difference approximation [18] This is shown in theexpression given in (3) with the central difference operatorsdefined as 1205752

119909119906119896

119894= 119906119896

119894minus1minus 2119906119896

119894+ 119906119896

119894+1and 120575

2

119909119906119896+1

119894= 119906119896+1

119894minus1minus

2119906119896+1

119894+ 119906119896+1

119894+1 The approach gives the fourth-order scheme a

spatial truncation error of the order 119874(Δ119909)4 It enhances theaccuracy of its second-order counterpart which has a largererror of the order 119874(Δ119909)2

1

Δ119905(119906119896+1

119894minus 119906119896

119894) =

1

2(Δ119909)2(1205752

119909minus

1

121205754

119909) (119906119896+1

119894+ 119906119896

119894) (3)

To determine the stability of (3) the von Neumann sta-bility analysis can be applied Let 120582 = Δ119905(Δ119909)

2 and since1205754

119909119906119896

119894= 1205752

119909(1205752

119909119906119896

119894) the discretization of (3) becomes

119906119896+1

119894minus 119906119896

119894=120582

2(minus

1

12119906119896+1

119894minus2+4

3119906119896+1

119894minus1

minus5

2119906119896+1

119894+4

3119906119896+1

119894+1minus

1

12119906119896+1

119894+2

minus1

12119906119896

119894minus2+4

3119906119896

119894minus1minus5

2119906119896

119894

+4

3119906119896

119894+1minus

1

12119906119896

119894+2)

(4)

The discretized equation in (4) is assumed to have a solu-tion in the form of a Fourier harmonic function that is 119906119896

119894=

120588119896

119890119904120573119894Δ119909 where 120588 is referred to as the amplification factor 120573 is

an arbitrary constant and 119904 = radicminus1 The amplification factorrepresents the time dependence of the solution If the Fourierfunction is substituted into (4) and then solve for 120588 the resultwill be

120588 =

1 minus (1205826) ((cos120573 minus 4)2

minus 9)

1 + (1205826) ((cos120573 minus 4)2

minus 9)

with 120573 = 120573Δ119909 (5)

Since |120588| le 1 then the fourth-order CN approximation isunconditionally stable for any choice of 120573 Δ119905 and Δ119909

3 The Formulation of the IADEMF4

The IADEMF4 is firstly developed based on the execution ofthe unconditionally stable fourth-order CN approximation(3) on the heat equation

Equation (4) can also be expressed as in (6) using thedefinitions of the constants given in (7)

119886119906119896+1

119894minus2+ 119887119906119896+1

119894minus1+ 119888119906119896+1

119894+ 119889119906119896+1

119894+1+ 119890119906119896+1

119894+2

= minus119886119906119896

119894minus2minus 119887119906119896

119894minus1+ 119888119906119896

119894minus 119889119906119896

119894+1minus 119890119906119896

119894+2

119894 = 2 3 119898 minus 1

(6)

119886 =120582

24 119887 = minus

2120582

3 119888 =

4 + 5120582

4

119889 = minus2120582

3 119890 =

120582

24 119888 =

4 minus 5120582

4

(7)

The approximation in (6) can be displayed in a matrix formsuch as 119860u = f (8) where 119860 is a sparse pentadiagonalcoefficient matrix and u = (119906

2 1199063 119906

119898minus2 119906119898minus1

)119879 is the

columnvector containing the unknownvalues of119906 at the timelevel 119896 + 1 The column vector f = (119891

2 1198913 119891

119898minus2 119891119898minus1

)119879

consists of boundary values and known 119906 values at theprevious time level 119896 The definitions for every entry in f aregiven in (9)

119860u = f

[[[[[[[[[

[

119888 119889 119890

119887 119888 119889 119890 O119886 119887 119888 119889 119890

d d d d119886 119887 119888 119889 119890

O 119886 119887 119888 119889

119886 119887 119888

]]]]]]]]]

](119898minus2)times(119898minus2)

[[[[[[

[

1199062

1199063

119906119898minus2

119906119898minus1

]]]]]]

]119896+1

=

[[[[[[

[

1198912

1198913

119891119898minus2

119891119898minus1

]]]]]]

]

(8)

1198912= minus119887 (119906

119896

1+ 119906119896+1

1) + 119888119906

119896

2minus 119889119906119896

3minus 119890119906119896

4

1198913= minus119886 (119906

119896

1+ 119906119896+1

1) minus 119887119906

119896

2+ 119888119906119896

3minus 119889119906119896

4minus 119890119906119896

5

119891119894= minus119886119906

119896

119894minus2minus 119887119906119896

119894minus1+ 119888119906119896

119894minus 119889119906119896

119894+1minus 119890119906119896

119894+2

for 119894 = 4 5 119898 minus 3

119891119898minus2

= minus119886119906119896

119898minus4minus 119887119906119896

119898minus3+ 119888119906119896

119898minus2

minus 119889119906119896

119898minus1minus 119890 (119906

119896

119898+ 119906119896+1

119898)

119891119898minus1

= minus119886119906119896

119898minus3minus 119887119906119896

119898minus2+ 119888119906119896

119898minus1minus 119889 (119906

119896

119898+ 119906119896+1

119898)

(9)

The evaluations of 1198912 1198913 119891119898minus2

and 119891119898minus1

require thevalues of 119906 at the boundaries 119894 = 1 and 119894 = 119898 Howeverthese values cannot be obtained numerically because theircomputations involve nodes at 119894 = minus1 and 119894 = 119898 + 2which are exterior to the considered solution domain Ifthe exact solutions of 119906 are available at 119894 = 1 and 119894 =

119898 then it is appropriate to consider them as the requiredboundary values Otherwise the boundary conditions haveto be formulated bearing in mind that they should be ofcomparable accuracy [1]

The IADEMF4 scheme secondly employs the higher-order accuracy formula of MF [19] The variant whose accu-racy is of the order 119874((Δ119905)2 + (Δ119909)

4

) is as given in (10) and(11)

(119903119868 + 1198661) u(119901+12) = (119903119868 minus 119892119866

2) u(119901) + f (10)

(119903119868 + 1198662) u(119901+1) = (119903119868 minus 119892119866

1) u(119901+12) + 119892f (11)

where 119903 119901 and 119868 represent an acceleration parameter theiteration index and an identity matrix respectively 119866

1and

1198662are two constituent matrices The vectors u(119901+1) and

u(119901+12) represent the required solution at the iteration level(119901 + 1) and at some intermediate level (119901 + 12) respectivelyThe relation of 119892 and 119903 is given by 119892 = (6 + 119903)6 119903 gt 0


Substitute the following expression obtained from (10)into (11)

u(119901+12) = (119903119868 + 1198661)minus1

[(119903119868 minus 1198921198662) u(119901) + f] (12)

and then simplify to obtain

(119903119868 + 1198662) u(119901+1) = (119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) u(119901)

+ [ (119903119868 minus 1198921198661) (119903119868 minus 119892119866

1)minus1

+ 119892] f (13)

As 119901 in (13) becomes sufficiently large the temperaturesolution reaches a steady state that is as 119901 rarr infin thenu(119901+1) rarr u and u(119901) rarr u Simplify the equation and thenmultiply by (119903119868+119866

1) After some algebraicmanipulations the

following expression is obtained

[119903 (1198661+ 1198662) (1 + 119892) + (1 minus 119892

2

)11986611198662] u = 119903 (1 + 119892) f (14)

Multiply (14) by 1119903(1 + 119892) and use the definition of 119892 tofinally obtain the following form

[ (1198661+ 1198662) minus

1

611986611198662] u = f (15)

The comparison between the expression 119860u = f in (8)and the form given in (15) suggests that the coefficient matrix119860 for the IADEMF4 can be decomposed into

119860 = 1198661+ 1198662minus1

611986611198662 (16)

The IADEMF4 requires the constituentmatrices1198661and119866

2in

(16) to be in the form of lower and upper tridiagonalmatricesrespectively in order to retain the pentadiagonal structure of119860 Thus

1198661=

[[[[[[[

[

1

1198971

1 119874

1

1198972

d2d dd 119897119898minus4

1

119874 119898minus4

119897119898minus3

1

]]]]]]]


1198662=

[[[[[[[

[

11989011

V1

1198902

2

V2

119874

d d d119874 119890

119898minus4119898minus4

V119898minus4

119890119898minus3

119898minus3

119890119898minus2

]]]]]]]


(17)

By substituting 1198661and 119866

2in (17) into the formula for the

decomposition of 119860 the entries of the resultant matrix arecompared with the entries of 119860 in (8) yielding the followingdefinitions

1198901=6 (119888 minus 1)

5

1=6119889

5 119897

1=

6119887

6 minus 1198901

1198902=6 (119888 minus 1) + 119897

11

5 V

119894=6119890

5for 119894 = 1 2 119898 minus 4

(18)

And for 119894 = 2 3 119898 minus 3

119894=6119889 + 119897

119894minus1V119894minus1

5

119894minus1=

6119886

6 minus 119890119894minus1

119897119894=6119887 +


6 minus 119890119894

119890119894+1

=6 (119888 minus 1) + 119897

119894119894+ 119894minus1

V119894minus1

5

(19)

Since1198661and119866

2are three bandedmatrices then (119903119868+119866

1)

and (119903119868+1198662) can be inverted easilyThe equations in (10) and

(11) are rearranged as in (20) and (21) respectively

u(119901+12) = (119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901)

+ (119903119868 + 1198661)minus1f

(20)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) u(119901+12)

+ 119892(119903119868 + 1198662)minus1f

(21)

The above two equations are computed and simplifiedleading to the computational formulae at each of the halfiteration levels as given in (22) and (23)

(i) At the (119901 + 12) iteration level

119906(119901+12)

119894=

1

119877(119864119894minus1

119906(119901)

119894+119882119894minus1

119906(119901)

119894+1+ 119881119894minus1

119906(119901)

119894+2

minus119894minus3

119906(119901+12)

119894minus2minus 119897119894minus2

119906(119901+12)

119894minus1+ 119891119894)

119894 = 2 3 119898 minus 2119898 minus 1

(22)

(ii) At the (119901 + 1) iteration level

119906(119901+1)

119894=

1

119885119894minus1

(119878119894minus3

119906(119901+12)

119894minus2+ 119876119894minus2

119906(119901+12)

119894minus1+ 119875119906(119901+12)

119894

minus119894minus1

119906(119901+1)

119894+1minus V119894minus1

119906(119901+1)

119894+2+ 119892119891119894)

119894 = 119898 minus 1 119898 minus 2 3 2

(23)

with

minus1

= 0= 1198970= 119881119898minus2

= 119881119898minus3

= 119882119898minus2

= 119898minus2

= V119898minus2

= V119898minus3

= 1198760= 119878minus1

= 1198780= 0

119877 = 1 + 119903 119875 = 119903 minus 119892

119864119894= 119903 minus 119892119890

119894 119885119894= 119903 + 119890

119894 119894 = 1 2 119898 minus 2

119882119894= minus119892

119894 119876119894= minus119892119897

119894 119894 = 1 2 119898 minus 3

119881119894=minus119892V119894 119878119894= minus119892

119894 119894 = 1 2 119898 minus 4

(24)

The IADEMF4 algorithm is regarded as a two-stageprocess involving two iteration levels (119901 + 12) and (119901 + 1)It is completed explicitly by using the required equations atthe two levels in alternate sweeps along all the grid pointsin the interval (0 1) until convergence is reached In (22)the calculation to determine the unknown 119906

(119901+12)

119894begins at

the left boundary and then moves to the right In a similar


manner 119906(119901+1)119894

in (23) is calculated by proceeding from theright boundary towards the left (Figure 1)

The computational molecules are depicted in Figures 2and 3

At each level of iteration the computational moleculesinvolve two known grid points at the new level and anotherthree known ones at the old level Clearly the method isexplicit

4 Convergence Analysis of the IADEMF4

This section proves the convergence of the IADEMF4 Since120582 gt 1will not guarantee an accurate approximation for 120597119880120597119905[18] the values of 120582 that are considered appropriate for theproof are 0 lt 120582 le 1

From the definitions in ((18) and (19)) the followingresults are obtained

1198901=6 (119888 minus 1)

5=3120582

2 implying that 0 lt 119890

1le3

2

1=6119889

5lt0 119897

1=

6119887

6 minus 1198901

lt0 V119894=6119890

5gt 0

for 119894 = 1 2 119898 minus 4

2=6119889 + 119897

1V1

5= 1+1198971V1

5lt 1 since 119897

1V1lt 0

1198902=6 (119888 minus 1) + 119897

11

5= 1198901+11989711

5gt 1198901 since 119897

11gt 0

(25)

Simple computation shows that 119897115 lt 1 Clearly 119890

2lt 6

Lemma 1 119890119894in ((18) and (19)) is such that 0 lt 119890

119894lt 6 for all

119894 = 1 2 3 119898 minus 2

Proof The results in (25) show that 6 gt 1198902gt 1198901gt 0 Assume

it is true that 6 gt 119890119896gt 119890119896minus1

gt 0 for all 119896 = 3 4 5 119898 minus 3Since the assumption implies that 6 gt 119890

119896minus1gt 119890119896minus2

gt 0 then119896minus1

= 6119886(6 minus 119890119896minus1

) gt 119896minus2

= 6119886(6 minus 119890119896minus2

) gt 0 Therefore119896minus1

V119896minus1

gt 119896minus2

V119896minus2

It has also been shown that

2lt 1lt 0 Assume it is true

that 119895lt 119895minus1

lt 0 for all 119895 = 3 4 119898 minus 4 The assumptionimplies that

119895minus1lt 119895minus2

lt 0 Thus 119895minus1

119895minus1

lt 119895minus2

119895minus2

lt 0

119895+1

=6119889 + 119897

119895V119895

5= 1+119897119895V119895

5= 1+

(6119887 + 119895minus1

119895minus1

) V119895

5 (6 minus 119890119895)

lt 1+

(6119887 + 119895minus2

119895minus2

) V119895minus1

5 (6 minus 119890119895minus1

)

= 1+119897119895minus1

V119895minus1

5= 119895lt 0

(26)

Since it is true that for 119895 + 1 119895+1

lt 119895lt 0 then by

induction the assumption 119895lt 119895minus1

lt 0 is true for all 119895 An

0

u(p+12)

i

u(p+1)

i

i iminus 1 i + 1 m + 1

Figure 1 The two-stage IADEMF4 algorithmThe directions of thesweeps at the (119901 + 12) and (119901 + 1) iteration levels

i iminus 1i minus 2 i + 1 i + 2

(p + 12)

(p)

Figure 2 Computational molecule of the IADEMF4 at the (119901+12)iteration level


(p + 12)

(p + 1)

Figure 3 Computational molecule of the IADEMF4 at the (119901 + 1)

iteration level

equivalent to the preceding statement would be 119896lt 119896minus1

lt 0

for all 119896 = 3 4 5 119898 minus 3 It follows that

119897119896119896=(6119887 +


) 119896

6 minus 119890119896

gt(6119887 +


) 119896minus1

6 minus 119890119896minus1

= 119897119896minus1

119896minus1

119890119896+1

=6 (119888 minus 1) + 119897

119896119896+ 119896minus1

V119896minus1

5= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

gt 1198901+119897119896minus1

119896minus1

+ 119896minus2

V119896minus2

5= 119890119896gt 0

(27)

Since it is true that for 119896+1 119890119896+1

gt 119890119896gt 0 then by induction

119890119896gt 119890119896minus1

gt 0 is also true for all 119896Suppose there is a 119896 such that 119890

119896gt 6 Then

119890119896+1

= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

= 1198901+

6119887 + 119896

5 (6 minus 119890119896)+119896minus1

5(119896minus1

119896

6 minus 119890119896

+ V119896minus1

) lt 1198901

(28)

This is a contradiction since 119890119896+1

gt 1198901 This verifies that 6 gt

119890119896gt 119890119896minus1

gt 0 119896 = 3 4 5 119898 minus 3 Let 119894 = 2 3 4 119898 minus 2and then 6 gt 119890

119894gt 119890119894minus1

gt 0 Therefore 0 lt 119890119894lt 6 119894 =

1 2 3 119898 minus 2 Lemma 1 is proved


Lemma 2 If 119903 gt 0 and 119892 = (6+ 119903)6 thenmax |(119903 minus119892119890119894)(119903 +

119890119894)| lt 1 i = 1 2 3 119898 minus 2

Proof Let max |(119903 minus 119892119890119894)(119903 + 119890

119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| for

some 119895Assume |(119903 minus 119892119890

119895)(119903 + 119890

119895)| ge 1 Since 119903 gt 0 and 119890

119895gt 0

then 119903 + 119890119895gt 0

If (119903 minus 119892119890119895)(119903 + 119890

119895) ge 1 then minus((6 + 119903)6)119890

119895ge 119890119895 which

implies that 119903 le minus12 This contradicts the fact that 119903 gt 0If (119903 minus 119892119890

119895)(119903 + 119890

119895) le minus1 then 2119903 minus ((6 + 119903)6)119890

119895le minus119890119895

which implies that 119890119895ge 12 This contradicts Lemma 1

So the assumption that |(119903 minus 119892119890119895)(119903 + 119890

119895)| ge 1 is false

Hence max |(119903 minus 119892119890119894)(119903 + 119890

119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| lt 1

Proposition 3 (119903119868 minus 1198921198661)(119903119868 + 119866

1)minus1

2lt 1

Proof Let 119865 = (119903119868 minus 1198921198661)(119903119868 + 119866

1)minus1 that is

119865 =

[[[[[[[[[[

[

119903 minus 119892

119889

d119903 minus 119892

119889119874

d d d

d d

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot119903 minus 119892

119889

]]]]]]]]]]

]

(29)

119865 is a lower triangular matrix with all the diagonal entries(whence the eigenvalues of 119865) equal to (119903 minus 119892)119889 where 119889 =

119903 + 1 Denote all the eigenvalues of 119865 by 120582119865 If 120588[119865] is defined

as the spectral radius of 119865 then

120588 [119865] = max 10038161003816100381610038161205821198651003816100381610038161003816 =

10038161003816100381610038161205821198651003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119903 minus 119892

119903 + 1

1003816100381610038161003816100381610038161003816le 119865

2 (30)

But by definition of 2-norm

1198652= maxx2 = 0

119865x2

x2

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

1003817100381710038171003817y10038171003817100381710038172

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

le maxw2=1

119865w2= maxw2=1

1003817100381710038171003817120582119865w10038171003817100381710038172

(31)

Since all the eigenvalues of 119865 are equal then

1198652le1003816100381610038161003816120582119865

1003816100381610038161003816 maxw2=1

w2=1003816100381610038161003816120582119865

1003816100381610038161003816 = 120588 [119865] (32)

Thus from (30) and (32) 1198652= |120582119865| = |(119903 minus 119892)(119903 + 1)|

By Lemma 2 with 119890119895replaced by 1 |(119903 minus 119892)(119903 + 1)| lt 1

is obtained leading to the result of Proposition 3 which is1198652= (119903119868 minus 119892119866

1)(119903119868 + 119866

1)minus1

2lt 1


2)minus1

2lt 1

Proof Let 119870 = (119903119868 minus 1198921198662)(119903119868 + 119866

2)minus1 that is

119870 =

[[[[[[[[[[[[[

[

119903 minus 1198921198901

1199111

d sdot sdot sdot sdot sdot sdot sdot sdot sdot

119903 minus 1198921198902

1199112

d sdot sdot sdot

d

119874 d d

119903 minus 119892119890119898

119911119898

]]]]]]]]]]]]]

]

(33)

119870 is an upper triangular matrix with all the diagonal entriesequal to (119903 minus 119892119890

119894)119911119894 119894 = 1 2 119898 minus 2 where 119911

119894= 119903 + 119890

119894

Since all the eigenvalues of119870 are distinct then119870 is similar toa diagonal matrix119863

By the Schur triangularization theorem [20] there is anorthogonal matrix 119874 such that 119874119879119870119874 = 119863 The diagonalentries of119863 are the eigenvalues of119870

1198702=10038171003817100381710038171003817119874119879

119870119874100381710038171003817100381710038172

= 1198632

= radicmaximum eigenvalue of 119863119879119863


= radic120588 (1198632) = radic120588 (1198702) = 120588 (119870)

(34)

By Lemma 2

1198702= 120588 (119870) = max

10038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119894

119903 + 119890119894

10038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119895

119903 + 119890119895

1003816100381610038161003816100381610038161003816100381610038161003816

lt 1 (35)

for some 119895 Hence this proves Proposition 4

From (20) and (21)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661)

times [(119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901) + (119903119868 + 119866

1)minus1f]

+ 119892 (119903119868 + 1198662)minus1f

(36)

Let

119872(119903) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662)

(37)

And let

119902 (119903) = [(119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

+119892(119903119868 + 1198662)minus1

] f (38)

Then u(119901+1) = 119872(119903)u(119901) + 119902(119903)

Theorem 5 The IADEMF4 is convergent if 120588[119872(119903)] lt 1 for119903 gt 0


Proof Define (119903) = (119903119868 + 1198662)119872(119903)(119903119868 + 119866

2)minus1 then

(119903) = (119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus1

(39)

Thus by similarity119872(119903) and (119903) have the same set of eigen-values Therefore

120588 [119872 (119903)]

= 120588 [ (119903)] le10038171003817100381710038171003817 (119903)

10038171003817100381710038171003817 2

=10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

le10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1100381710038171003817100381710038172

10038171003817100381710038171003817(119903119868 minus 119892119866

2) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

lt 1

(40)

The last inequality is due to Propositions 3 and 4The provenTheorem 5 assures the convergence of the IADEMF4

5 The Fourth-Order ClassicalIterative Methods

The system of linear equations in (8) may also be solvedby using the classical iterative methods of the fourth orderThey include the JAC4 the GS4 and the SOR4 methodsThese iterative methods are capable of exploiting the sparsestructure of the pentadiagonal matrix

The JAC4 algorithm can be represented by

119906(119901+1)

119894=

(119891119894minus119886119906(119901)

119894minus2minus 119887119906(119901)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(41)

The approximation of 119906(119901+1)119894

at the (119901 + 1)th iteration level iscomputed using the relevant values in the 119901th iteration

The GS4 method uses the most recent values of 119906(119901+1)119894minus2

and 119906(119901+1)

119894minus1to update the approximation value of 119906(119901+1)

119894 The

algorithm for the GS4 is as expressed in (42)

119906(119901+1)

119894=

(119891119894minus119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(42)

The SOR4 iterative method accelerates the convergencerate of the GS4 If 120596 is a relaxation parameter then for any120596 = 0 (42) can be rewritten as

119906(119901+1)

119894= (1 minus 120596) 119906

(119901)

119894

+

120596 (119891119894minus 119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(43)


(p + 1)

(p)

Figure 4 Computational molecule of the JAC4


(p + 1)

(p)

Figure 5 Computational molecule of the GS4SOR4

Table 1 Sequential arithmetic operations per iteration (119898 problemsize)

Method Number ofadditions

Number ofmultiplications

Total operationcount

IADEMF4 10(119898 minus 2) 13(119898 minus 2) 23(119898 minus 2)

IADEMF2 6119898 9119898 15119898

JAC4GS4 4(119898 minus 2) 5(119898 minus 2) 9(119898 minus 2)

SOR4 5(119898 minus 2) 7(119898 minus 2) 12(119898 minus 2)

The SOR4 algorithm reduces to the GS4 if 120596 = 1 Exceptfor very special cases it is difficult to obtain the analyticexpression for 120596 According to Young [21] the precise deter-mination of the optimal 120596 in the SOR is only known for asmall class of matrices The value of 120596 generally lies in therange of 1 lt 120596 lt 2

The computational molecules for the JAC4 and the GS4are as illustrated in Figures 4 and 5 respectively The SOR4has the same form as the GS4

6 Computational Complexity

The cost of implementing an algorithm can be assessedby examining its computational complexity The number ofarithmetic operations such as additions (including subtrac-tions) andmultiplications (including divisions) that is neededto perform by the algorithm can be straightforwardly count-ed Table 1 gives the number of sequential arithmetic opera-tions per iteration that is required to evaluate the algorithms

For the higher-order schemes trade-offbetween accuracyand speed usually happens The computational complexityhas an effect on the efficiency of a particular scheme Thusthis factor will be taken into account when discussing theresults of the numerical experiments conducted in this study

7 The Pseudocode of the IADEMF4Sequential Algorithm

Algorithm 1 illustrates the pseudocode of the IADEMF4sequential algorithm This algorithm can also be generalizedand applied to other methods discussed in this paper


begindetermine the parameters119898 119903 Δ119905 Δ119909 120582 120576 119879

determine initial conditions 119906(119901)119894

determine exact solutions 119880119894at time level 119879

while (time level lt 119879)determine boundary conditionsfor 119894 = 2 to 119894 = 119898 minus 1

compute 119891119894(refer to (9))

end-forset iteration = 0while (convergence conditions are not satisfied)

for 119894 = 2 to 119894 = 119898 minus 1

compute 119906(119901+12)119894

(refer to (22))end-forfor 119894 = 2 to 119894 = 119898 minus 1

compute 119906(119901+1)119894

(refer to (23))end-fortest for convergencecompute 119890

119894larr

10038161003816100381610038161003816119906(119901+1)

119894minus 119906(119901)

119894

10038161003816100381610038161003816if (max1003816100381610038161003816119890119894

1003816100381610038161003816 lt 120576)then 119906

(119901)

119894larr 119906(119901+1)

119894

add 1 to iteration (if necessary)end-while

end-whilefor 119894 = 2 to 119894 = 119898 minus 1

determine average absolute error root meansquare error and maximum error

end-forend

Algorithm 1 The IADEMF4 sequential algorithm

8 Numerical Experiments

Two experiments were conducted to test the sequentialnumerical performance of the proposed IADEMF4 methodagainst those of the benchmarked classical iterative methodsnamely the JAC4 theGS4 and the SOR4 Comparison is alsomade with the corresponding IADE method of the secondorder

Experiment 1 This problem was taken from Saulev [22]

120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (44)

subject to the initial condition

119880 (119909 0) = 4119909 (1 minus 119909) 0 le 119909 le 1 (45)

and the boundary conditions

119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (46)

The exact solution to the given problem is given by

119880 (119909 119905) =32

1205873

infin

sum

119896=1(2)

1

1198963119890minus12058721198962119905 sin (119896120587119909) (47)

Experiment 2 This problem was taken from Johnson andReiss [23]

120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (48)


119880 (119909 0) = sin (120587119909) (1 + 6 cos (120587119909)) 0 le 119909 le 1 (49)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (50)


119880 (119909 119905) = sin (120587119909) 119890minus1205872119905

+ 3 sin (2120587119909) 119890minus41205872119905

(51)


9 Results and Discussion

In each experiment the exact solutions at 119894 = 1 and 119894 = 119898

were taken as boundary values for the IADEMF4 and theother fourth-order classical iterative methods The conver-gence criterion used in the testing of each method was takenas 119906(119901+1) minus 119906

(119901)

infin

le 120576 where 120576 is the convergence toleranceThe selections of the optimum 119903 or 120596 were determined byexperiments As for the IADEMF2 the CN scheme with 120579 =

12 was employedFigures 6 7 8 and 9 visualize the behavior of the one-

dimensional parabolic heat solutions for both experimentsBy using119898 = 10 and a tolerance requirement of 120576 = 10

minus4 twodifferent mesh sizes 120582 = 05 and 120582 = 10 were consideredin each experiment The exact solutions are compared withthe IADEMF4 and the IADEMF2 numerical solutions Theoptimum value of 119903 chosen for each method as well as theircorresponding outcome of the number of iterations 119899 isstated in the legend of each figure Every figure reveals thatthe IADEMF4 is more accurate than the IADEMF2 andthe former converges with fewer numbers of iterations incomparison to the latter The numerical solutions of theIADEMF4 seem to be in very good agreement with theexact solutions For example in Figure 7 at 119909 = 05 thedifference between the exact solution and the numericalsolution using the IADEMF2 and using the IADEMF4 isabout 36 and 0 respectively These results imply thatthe accuracy and convergence rate of the second-order IADEmethod are enhanced by the implementation of the fourth-order CN approximation that leads to the formation of thecorresponding fourth-order IADE scheme

Figure 10 displays the graph of log (RMSE) versus log(Δ119909)for decreasing values of Δ119909 implemented on Experiment 2By considering 120576 = 10

minus4 and fixing Δ119905 = 00001 the value ofΔ119909 was initially taken as Δ119909 = 0125 It was then successivelyhalved into Δ1199092 Δ1199094 and Δ1199098

The figure shows that amongst the tested methods theroot mean square error (RMSE) of the IADEMF4 and theIADEMF2 decreases linearly as the values of Δ119909 decreasesThe slope of the IADEMF4 is approximately equal to 4which corresponds to its fourth-order spatial accuracy TheIADEMF2 with a second-order spatial accuracy has a slopethat is approximately equal to 2 It is clear that the IADEMF4is always more accurate than the IADEMF2 for the differentconsidered mesh sizes The graphs of the SOR4 (120596 = 105)the GS4 and the JAC4 show that their accuracies of fourthorder tend to lack as Δ119909 decreases largely due to the effect ofincreasing round-off errors as the value of119898 increases

Tables 2 3 4 5 6 7 8 and 9 provide numerical results interms of the average absolute error (AAE) root mean squareerror (RMSE) maximum error (ME) number of iterations(119899) and execution time (ET) measured in seconds (s) Theresults are obtained from both experiments for two differentvalues of 119898 and mesh size 120582 It is generally observed thatwhen 119898 = 700 120582 = 05 and 120576 = 10

minus6 the IADEMF4has the least average absolute error root mean square errorand maximum error in comparison with the other methodsunder consideration (Tables 2ndash5) When the size of 119898 was

0 01 02 03 04 05 06 07 08 09 1x

Exact solutionIADEMF4 (r = 08 n = 172)IADEMF2 (r = 03 n = 185)

00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)

Figure 6 Numerical and exact solutions for Experiment 1 120582 = 05Δ119909 = 01 Δ119905 = 0005 119905 = 025 and 120576 = 10

minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

90E minus 03

U(xt)


minus4

ten times bigger (119898 = 7000) and a more stringent tolerancecriterion was set (120576 = 10

minus10

) the accuracy of the IADEMF4clearly outperforms the other methods for a mesh size of120582 = 05 (Tables 6 and 8) However the difference in the errorsamongst the tested methods is not so obvious for the case of120582 = 10 (Tables 7 and 9)The achievement of the fourth-orderIADEmethod in Experiments 1 and 2 can be clearly seen fromthe results in Tables 2 and 8 respectively where it has caused ahuge 94 reduction of RMSE from its corresponding second-order IADE method

The IADEMF4 has the advantage of featuring higheraccuracy due to the execution of the fourth-order CNapprox-imation coupled with the fourth-order accurate MF variant


0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4

The IADEMF2 is only derived from the second-order CNapproximation but its combination with its correspondingfourth-order MF variant managed to produce errors whichare better than theGS4 JAC4 and the SOR4 Even though theclassical iterative methods are also derived from the fourth-order accurate CN type approximation they are lackingin accuracy due to the round-off errors that have beenaccumulated from the time the execution starts till it ends

With regards to the rate of convergence the resultsfrom each table demonstrate that the number of iterationsproduced by the IADEMF2 and the fourth-order classicaliterative methods is greater than or at least equal to that of

0

minus2

minus208 minus277 minus347 minus416

minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2

Figure 10 Log(RMSE) versus Log(Δ119909) Experiment 2 Δ119905 = 00001119905 = 0002 and 120576 = 10

minus4

Table 2 Experiment 1 119898 = 700 120582 = 05 Δ119909 = 143 times 10minus3 Δ119905 =

102 times 10minus6 119905 = 509 times 10

minus5 and 120576 = 10minus6

Method AAE RMSE ME 119899 ET (s)IADEMF4(119903 = 07) 267119890 minus 9 330119890 minus 9 212119890 minus 8 100 440119890 minus 2

IADEMF2(119903 = 08) 188119890 minus 8 495119890 minus 8 323119890 minus 7 100 635119890 minus 3

SOR4(120596 = 12) 540119890 minus 7 544119890 minus 7 552119890 minus 7 108 460119890 minus 2

GS4 540119890 minus 6 543119890 minus 6 332119890 minus 6 150 472119890 minus 2

JAC4 228119890 minus 5 229119890 minus 5 232119890 minus 5 150 473119890 minus 2









the IADEMF4 The convergence rate of the latter surpassesthe others in the case of 119898 = 7000 and 120576 = 10

minus10 (Tables6ndash8) Even though the operational count of the IADEMF4is relatively quite large (Table 1) due to its higher level ofaccuracy its increasing number of correct digits at eachiteration causes it to converge at a faster rate The IADEMF2has the advantage of having less mathematical operationsthus it is competitive in terms of convergence rate but atthe expense of accuracy The application of the fourth-orderCN approximation on the heat equation proves that thethree classical iterative methods also converge with JAC4appearing to be the slowest and the least accurate amongst all





Method AAE RMSE ME n ET (s)IADEMF4(119903 = 07) 207119890 minus 8 231119890 minus 8 345119890 minus 8 100 903119890 minus 3





























It is found that the convergence rate of the fourth-orderand second-order methods improves with the application oflarger mesh size that is from 120582 = 05 to 120582 = 10 Thecoarser meshes cause a reduction in the computational oper-ations thus giving better rate of convergence However a

















more accurate solution is obtained by using finer mesh Forexample Table 6 shows that for 120582 = 05 the IADEMF4 hasRMSE = 433119890 minus 10 and 119899 = 173 whereas Table 7 shows thatfor 120582 = 10 its RMSE = 163119890 minus 7 and 119899 = 112 In generalamongst all the testedmethods the IADEMF4 still maintainsits greater accuracy characteristic even with coarser meshes

In terms of execution time the results from every tabledisplay shorter execution time for the IADEMF2 in compari-son to the IADEMF4 This is expected since the IADEMF2has lower computational complexity (Table 1) Despite theachievement in accuracy the IADEMF4 has to performmorecomputational work than the IADEMF2 since the former hasto utilize values of 119906 at two grid points on either side of thepoint (119894Δ119909 119896Δ119905) along the 119896th time level Thus if accuracyis desired then the preferred sequential numerical algorithmwould be the IADEMF4 On the other hand if execution timematters then the choice would be the IADEMF2

Amongst the fourth-order methods the IADEMF4 exe-cutes in the least amount of time Even though its computa-tional complexity is relatively large it operates with the leastnumber of iterations thus enabling it to be the most efficient

10 Conclusion

This paper proposes the development of the novel fourth-order IADEMF4 finite difference scheme The higher-orderscheme is developed by representing the spatial derivative inthe heat equation with the fourth-order finite difference CNapproximation This leads to the formation of pentadiagonalmatrices in the systems of linear equations with larger com-putational stencils The algorithm also employs the higheraccuracy of the Mitchell and Fairweather variant Despite


the fourth-order IADE schemersquos higher computational com-plexity this technique is proved to be valuable becauseit enhances the accuracy of its second-order counterpartnamely the IADEMF2 In addition the higher accuracy ofIADEMF4 is verified as a convergent and unconditionallystable scheme and is superior in terms of rate of convergenceIt has also been proven to execute more efficiently in compar-ison to the other benchmarked fourth-order classical iterativemethods such as the GS4 the SOR4 and the JAC4 The in-creasing number of correct digits at each iteration serves as anadvantage for the IADEMF4 thus yielding faster rate of con-vergence with higher level of accuracy

In conclusion the proposed IADEMF4 scheme affordsusers many advantages with respect to higher accuracy sta-bility and rate of convergence and it serves as an alternativeefficient technique for the solution of a one-dimensional heatequation with Dirichlet boundary conditions

The proposed fourth-order scheme can be modified andadapted to more general multidimensional linear and non-linear parabolic elliptic and hyperbolic partial differentialequations using different types of boundary conditionsBesides it can also be considered for applications in problemsthat require higher-order accuracy with high resolution suchas problems in nanocomputing that require the solution ofvery large sparse systems of equation

The explicit and high accuracy feature of the IADEMF4can be exploited in two- and three-dimensional heat prob-lems by applying the scheme as pentadiagonal solvers forthe system of linear equations arising in the sweeps of ahigher-order alternating direction implicit (ADI) schemeThe lower-order ADI scheme was initially proposed byPeaceman and Rachford [24] The higher space dimensionsare expected not to cause unsatisfactory performance for theADI-IADEMF4 as long as the proposed scheme is stableconvergent and efficient as the process of its implementa-tion continues to advance from one-time step to anotherPathirana et al [25] implemented the ADI in developing atwo-dimensional model for incorporating flood damage inurban drainage planningThe proposed model is found to bestable numerically accurate and computationally efficientMirzavand et al [26] proposed the ADI-FDTD methodfor the physical modeling of high-frequency semiconductordevices The approach is able to reduce significantly the full-wave simulation time

It is possible to parallelize the IADEMF4 algorithm sincethe calculation of the new iteration only depends on theknown values from the last iteration (Figures 2 and 3) Futurework is to exploit the explicit computational properties andefficiency of the IADEMF4 for parallelization and executionon distributed parallel computing systems The idea is tospeed up the execution time without compromising its accu-racy especially on problems involving very large linear sys-tems of equations

References

[1] J Noye Computational Techniques for Differential Equationsvol 83 Elsevier Science North-Holland Publishing Amster-dam The Netherlands 1984

[2] G W Recktenwald ldquoFinite difference approximations to theheat equationrdquoMechanical Engineering vol 10 pp 1ndash27 2004

[3] P Lax and B Wendroff ldquoSystems of conservation lawsrdquo Com-munications on Pure and Applied Mathematics vol 13 pp 217ndash237 1960

[4] R J MacKinnon and G F Carey ldquoAnalysis of material interfacediscontinuities and superconvergent fluxes in finite differencetheoryrdquo Journal of Computational Physics vol 75 no 1 pp 151ndash167 1988

[5] R Hixon and E Turkel High Accuracy Compact MacCormack-Type Schemes For Computational Aerocoustics NASACentre forAerospace Information Lewis Research Centre 1998

[6] Y Lin Higher-order finite difference methods for solving heatequations [PhD thesis] Southern Illinois University 2008

[7] J Sulaiman M K Hasan M Othman M and S A AbdulKarim ldquoFourth-order QSMOR iterative method for the solu-tion of one-dimensional parabolic PDErsquosrdquo in Proceedings of theInternational Conference on Industrial and AppliedMathematics(CIAM rsquo10) pp 34ndash39 ITB Bandung Indonesia 2010

[8] N Jha ldquoThe application of sixth order accurate parallel quar-ter sweep alternating group explicit algorithm for nonlinearboundary value problems with singularityrdquo in Proceedings ofthe 2nd International Conference on Methods and Models inComputer Science (ICM2CS rsquo10) pp 76ndash80 December 2010

[9] Y Jin G Jin and J Li ldquoA class of high-precision finite differenceparallel algorithms for convection equationsrdquo Journal of Conver-gence Information Technology vol 6 no 1 pp 79ndash82 2011

[10] W Fu and E L Tan ldquoDevelopment of split-step FDTDmethodwith higher-order spatial accuracyrdquo Electronics Letters vol 40no 20 pp 1252ndash1253 2004

[11] A Mohebbi andM Dehghan ldquoHigh-order compact solution ofthe one-dimensional heat and advection-diffusion equationsrdquoApplied Mathematical Modelling vol 34 no 10 pp 3071ndash30842010

[12] W Liao ldquoAn implicit fourth-order compact finite differencescheme for one-dimensional Burgersrsquo equationrdquo Applied Math-ematics and Computation vol 206 no 2 pp 755ndash764 2008

[13] Z F Tian and Y B Ge ldquoA fourth-order compact ADI methodfor solving two-dimensional unsteady convection-diffusionproblemsrdquo Journal of Computational and Applied Mathematicsvol 198 no 1 pp 268ndash286 2007

[14] C Chun ldquoSome fourth-order iterativemethods for solving non-linear equationsrdquo Applied Mathematics and Computation vol195 no 2 pp 454ndash459 2008

[15] S Zhu Z Yu and J Zhao ldquoAhigh-order parallel finite differencealgorithmrdquo Applied Mathematics and Computation vol 183 no1 pp 365ndash372 2006

[16] Z Gao and S Xie ldquoFourth-order alternating direction implicitcompact finite difference schemes for two-dimensionalSchrodinger equationsrdquo Applied Numerical Mathematics vol61 no 4 pp 593ndash614 2011

[17] M S Sahimi A Ahmad and A A Bakar ldquoThe Iterative Alter-nating Decomposition Explicit (IADE) method to solve theheat conduction equationrdquo International Journal of ComputerMathematics vol 47 pp 219ndash229 1993

[18] G D Smith Numerical Solution of Partial Differential Equa-tions Finite Difference Methods Oxford University Press 2ndedition 1978

[19] A RMitchell andG Fairweather ldquoImproved forms of the alter-nating direction methods of Douglas Peaceman and Rachfordfor solving parabolic and elliptic equationsrdquo Numerische Math-ematik vol 6 no 1 pp 285ndash292 1964


[20] B N DattaNumerical Linear Algebra and Applications BrooksCole Publishing Pacific Grove Calif USA 1995

[21] D M Young Iterative Solution of Large Linear Systems Aca-demic Press New York NY USA 1971

[22] V K Saulev Integration of Equations of Parabolic Type by theMethod of Nets Pergamon Press Oxford UK 1964

[23] L W Johnson and R D Riess Numerical Analysis Addison-Wesley Reading Mass USA 2nd edition 1982

[24] D W Peaceman and H H Rachford Jr ldquoThe numerical solu-tion of parabolic and elliptic differential equationsrdquo Journal ofthe Society for Industrial and Applied Mathematics vol 3 pp28ndash41 1955

[25] A Pathirana S Tsegaye B Gersonius and K VairavamoorthyldquoA simple 2-D inundation model for incorporating flood dam-age in urban drainage planningrdquo Hydrology and Earth SystemSciences vol 15 no 8 pp 2747ndash2761 2011

[26] R Mirzavand A Abdipour G Moradi and M MovahhedildquoFull-wave semiconductor devices simulation using ADI-FDTD methodrdquo Progress in Electromagnetics Research vol 11pp 191ndash202 2010

Submit your manuscripts athttpwwwhindawicom

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


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Algebra

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Stochastic AnalysisInternational Journal of


be implemented in parallel and is found to be superiorto the corresponding full sweep alternating group explicit(AGE) and SOR methods Jin et al [9] proposed an AGEiteration method of the fourth-order accuracy by integratingthe grouping explicit method with numerical boundary con-ditionsThemethod was used to solve initial-boundary valueproblem of convection equations Fu and Tan [10] showedthat an unconditionally stable split-step FDTD method withhigher-order spatial accuracy is more accurate than thelower-order methods The dispersion error of the proposedmethod is comparable with the higher-order ADI-FDTDmethod

Higher-order methods are also studied by Mohebbi andDeghan [11] who applied a compact finite difference approx-imation of fourth-order and the cubic C1 spline colloca-tion method to some one-dimensional heat and advection-diffusion equations The scheme has fourth-order accuracyin both space and time and unconditionally stable Liao[12] proposed an efficient and high-order accuracy of thefourth-order compact finite difference method to solve one-dimensional Burgersrsquo equation Tian and Ge [13] studied astable fourth-order compact ADI method for solving two-dimensional unsteady convection diffusion problems Themethod is temporally second-order and spatially fourth-order accurate which requires only a regular five-point 2Dstencil similar to that in the standard second-order methodsChun [14] solved some nonlinear equations by applying thefourth-order iterative methods containing the Kingrsquos fourth-order family It was observed that the proposed methodhas at least equal performance compared to other methodsof the same order Zhu et al [15] presented a high-orderparallel finite difference algorithm based on the domaindecomposition strategy The study used the classical explicitscheme to calculate the interface values between subdomainsand the fourth-order compact schemes for the interior valuesThe method has high accuracy and is convergent and stableGao and Xie [16] devised a fourth-order alternating directionimplicit compact finite difference schemes for the solutionof two-dimensional Schrodinger equations The method ishighly competitive as compared to other existing methodsand it achieves the expected convergence rate

This study develops a higher-order finite difference algo-rithm with noncompact stencils that is capable of deliveringhighly accurate solutions for a one-dimensional heat equa-tion with Dirichlet boundary conditions The study focuseson modifying the unconditionally stable and convergentsecond-order iterative alternating decomposition explicit(IADE) method of Mitchell and Fairweather (IADEMF2)The IADEMF2 which was originally proposed by Sahimiet al [17] employs the fractional splitting of the Mitchell andFairweather (MF) variant that has an accuracy of the order119874((Δ119905)

2

+ (Δ119909)4

) The scheme executes a two-stage processinvolving the solution of sets of tridiagonal equations alonglines parallel to the first and second time steps respectivelyIt is found to be unconditionally stable convergent andmoreaccurate than the classical AGE class of methods namely theAGE method based on the Peaceman and Rachford variant(AGE-PR) whose spatial accuracy is of second order and

the AGEmethod based on the Douglas and Rachford variant(AGE-DR) whose temporal accuracy is only of first orderThe detailed derivation of the IADEMF2 can be obtainedfrom [17]

In this paper a fourth-order Crank-Nicolson (CN) dif-ference approximation is applied to the spatial derivative inthe heat equation and the MF variant is employed leadingto the formation of the fourth-order IADEMF (IADEMF4)numerical algorithm The convergence of the IADEMF4is analyzed and proven Numerical experiments verify thepotential of the IADEMF4 in enhancing the accuracy ofthe IADEMF2 The results of the proposed higher-orderscheme are also compared with the benchmarked fourth-order classical iterative methods such as the fourth-orderGauss-Seidel (GS4) the fourth-order Jacobi (JAC4) and thefourth-order successive over-relaxation (SOR4) methods

This paper is organized as follows Section 2 discusses theimplementation and stability of the fourth-order CN approx-imation on the heat equation In Section 3 the IADEMF4algorithm is formulated Section 4 analyses the convergenceof the IADEMF4 Section 5 provides the equations for thebenchmarked fourth-order classical iterative methods Thecomputational complexity of the methods considered in thispaper is given in Section 6 while the pseudocode for theIADEMF4 sequential algorithm is presented in Section 7Theexperiments conducted are given in Section 8 Sections 9and 10 provide some discussion and conclusion based on theobtained numerical results

2 Fourth-Order Crank-NicolsonApproximation to the Spatial Derivativein the Heat Equation

Consider the following one-dimensional parabolic heat equa-tion (1) that has been suitably assumed to be nondimensiona-lised It models the flow of heat in a homogeneous unchang-ing medium of finite extent in the absence of heat source

120597119880

120597119905=1205972

119880

1205971199092(1)

subject to given initial and Dirichlet boundary conditions

119880 (119909 0) = 119891 (119909) 0 le 119909 le 1

119880 (0 119905) = 119892 (119905) 0 lt 119905 le 119879

119880 (1 119905) = ℎ (119905) 0 lt 119905 le 119879

(2)

For the problem in (1) the finite difference approach dis-cretizes the time-space domain by placing a rectangular gridover the domain with grid spacing ofΔ119905 andΔ119909 in the 119905- and119909-directions respectively The grid consists of the set of linesparallel to the 119905-axis given by 119909

119894= 119894Δ119909 119894 = 0 1 119898119898 + 1

and a set of lines parallel to the 119909-axis given by 119905119896= 119896Δ119905 119896 =

0 1 119899 119899+1 For simplicity the grid spacing is taken to beuniform so thatΔ119909 = 1(119898+1) andΔ119905 = 119879(119899+1) At a grid-point119875(119909

119894 119905119896) in the solution domain the dependent variable

119880(119909 119905) which represents the nondimensional temperature attime 119905 and at position 119909 is approximated by 119906119896

119894


At the grid-point 119875(119909119894 119905119896+12


119909119906119896

119894= 119906119896

119894minus1minus 2119906119896

119894+ 119906119896

119894+1and 120575

2

119909119906119896+1

119894= 119906119896+1

119894minus1minus

2119906119896+1

119894+ 119906119896+1



1

Δ119905(119906119896+1

119894minus 119906119896

119894) =

1

2(Δ119909)2(1205752

119909minus

1

121205754

119909) (119906119896+1

119894+ 119906119896

119894) (3)


2 and since1205754

119909119906119896

119894= 1205752

119909(1205752

119909119906119896


119906119896+1

119894minus 119906119896

119894=120582

2(minus

1

12119906119896+1

119894minus2+4

3119906119896+1

119894minus1

minus5

2119906119896+1

119894+4

3119906119896+1

119894+1minus

1

12119906119896+1

119894+2

minus1

12119906119896

119894minus2+4

3119906119896

119894minus1minus5

2119906119896

119894

+4

3119906119896

119894+1minus

1

12119906119896

119894+2)

(4)


119894=

120588119896



120588 =

1 minus (1205826) ((cos120573 minus 4)2

minus 9)

1 + (1205826) ((cos120573 minus 4)2

minus 9)

with 120573 = 120573Δ119909 (5)





119886119906119896+1

119894minus2+ 119887119906119896+1

119894minus1+ 119888119906119896+1

119894+ 119889119906119896+1

119894+1+ 119890119906119896+1

119894+2

= minus119886119906119896

119894minus2minus 119887119906119896

119894minus1+ 119888119906119896

119894minus 119889119906119896

119894+1minus 119890119906119896

119894+2

119894 = 2 3 119898 minus 1

(6)

119886 =120582

24 119887 = minus

2120582

3 119888 =

4 + 5120582

4

119889 = minus2120582

3 119890 =

120582

24 119888 =

4 minus 5120582

4

(7)


2 1199063 119906

119898minus2 119906119898minus1

)119879 is the


2 1198913 119891

119898minus2 119891119898minus1

)119879


119860u = f

[[[[[[[[[

[

119888 119889 119890

119887 119888 119889 119890 O119886 119887 119888 119889 119890

d d d d119886 119887 119888 119889 119890

O 119886 119887 119888 119889

119886 119887 119888

]]]]]]]]]


[[[[[[

[

1199062

1199063

119906119898minus2

119906119898minus1

]]]]]]

]119896+1

=

[[[[[[

[

1198912

1198913

119891119898minus2

119891119898minus1

]]]]]]

]

(8)

1198912= minus119887 (119906

119896

1+ 119906119896+1

1) + 119888119906

119896

2minus 119889119906119896

3minus 119890119906119896

4

1198913= minus119886 (119906

119896

1+ 119906119896+1

1) minus 119887119906

119896

2+ 119888119906119896

3minus 119889119906119896

4minus 119890119906119896

5

119891119894= minus119886119906

119896

119894minus2minus 119887119906119896

119894minus1+ 119888119906119896

119894minus 119889119906119896

119894+1minus 119890119906119896

119894+2

for 119894 = 4 5 119898 minus 3

119891119898minus2

= minus119886119906119896

119898minus4minus 119887119906119896

119898minus3+ 119888119906119896

119898minus2

minus 119889119906119896

119898minus1minus 119890 (119906

119896

119898+ 119906119896+1

119898)

119891119898minus1

= minus119886119906119896

119898minus3minus 119887119906119896

119898minus2+ 119888119906119896

119898minus1minus 119889 (119906

119896

119898+ 119906119896+1

119898)

(9)


and 119891119898minus1




4


(119903119868 + 1198661) u(119901+12) = (119903119868 minus 119892119866

2) u(119901) + f (10)

(119903119868 + 1198662) u(119901+1) = (119903119868 minus 119892119866

1) u(119901+12) + 119892f (11)


1and





u(119901+12) = (119903119868 + 1198661)minus1

[(119903119868 minus 1198921198662) u(119901) + f] (12)


(119903119868 + 1198662) u(119901+1) = (119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) u(119901)

+ [ (119903119868 minus 1198921198661) (119903119868 minus 119892119866

1)minus1

+ 119892] f (13)




[119903 (1198661+ 1198662) (1 + 119892) + (1 minus 119892

2

)11986611198662] u = 119903 (1 + 119892) f (14)


[ (1198661+ 1198662) minus

1

611986611198662] u = f (15)


119860 = 1198661+ 1198662minus1

611986611198662 (16)


2in


1198661=

[[[[[[[

[

1

1198971

1 119874

1

1198972

d2d dd 119897119898minus4

1

119874 119898minus4

119897119898minus3

1

]]]]]]]


1198662=

[[[[[[[

[

11989011

V1

1198902

2

V2

119874

d d d119874 119890


V119898minus4

119890119898minus3

119898minus3

119890119898minus2

]]]]]]]


(17)




1198901=6 (119888 minus 1)

5

1=6119889

5 119897

1=

6119887

6 minus 1198901

1198902=6 (119888 minus 1) + 119897

11

5 V

119894=6119890

5for 119894 = 1 2 119898 minus 4

(18)

And for 119894 = 2 3 119898 minus 3

119894=6119889 + 119897


5

119894minus1=

6119886

6 minus 119890119894minus1

119897119894=6119887 +


6 minus 119890119894

119890119894+1

=6 (119888 minus 1) + 119897

119894119894+ 119894minus1

V119894minus1

5

(19)

Since1198661and119866


1)



u(119901+12) = (119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901)

+ (119903119868 + 1198661)minus1f

(20)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) u(119901+12)

+ 119892(119903119868 + 1198662)minus1f

(21)



119906(119901+12)

119894=

1

119877(119864119894minus1

119906(119901)

119894+119882119894minus1

119906(119901)

119894+1+ 119881119894minus1

119906(119901)

119894+2

minus119894minus3

119906(119901+12)


119906(119901+12)

119894minus1+ 119891119894)

119894 = 2 3 119898 minus 2119898 minus 1

(22)


119906(119901+1)

119894=

1

119885119894minus1

(119878119894minus3

119906(119901+12)

119894minus2+ 119876119894minus2

119906(119901+12)

119894minus1+ 119875119906(119901+12)

119894

minus119894minus1

119906(119901+1)


119906(119901+1)

119894+2+ 119892119891119894)

119894 = 119898 minus 1 119898 minus 2 3 2

(23)

with

minus1

= 0= 1198970= 119881119898minus2

= 119881119898minus3

= 119882119898minus2

= 119898minus2

= V119898minus2

= V119898minus3

= 1198760= 119878minus1

= 1198780= 0

119877 = 1 + 119903 119875 = 119903 minus 119892

119864119894= 119903 minus 119892119890

119894 119885119894= 119903 + 119890

119894 119894 = 1 2 119898 minus 2

119882119894= minus119892

119894 119876119894= minus119892119897

119894 119894 = 1 2 119898 minus 3

119881119894=minus119892V119894 119878119894= minus119892

119894 119894 = 1 2 119898 minus 4

(24)


(119901+12)

119894begins at



manner 119906(119901+1)119894







1198901=6 (119888 minus 1)

5=3120582


1le3

2

1=6119889

5lt0 119897

1=

6119887

6 minus 1198901

lt0 V119894=6119890

5gt 0

for 119894 = 1 2 119898 minus 4

2=6119889 + 119897

1V1

5= 1+1198971V1

5lt 1 since 119897

1V1lt 0

1198902=6 (119888 minus 1) + 119897

11

5= 1198901+11989711

5gt 1198901 since 119897

11gt 0

(25)


2lt 6


119894lt 6 for all

119894 = 1 2 3 119898 minus 2




119896minus1gt 119890119896minus2


= 6119886(6 minus 119890119896minus1

) gt 119896minus2

= 6119886(6 minus 119890119896minus2


V119896minus1

gt 119896minus2

V119896minus2







119895minus1

lt 119895minus2

119895minus2

lt 0

119895+1

=6119889 + 119897

119895V119895

5= 1+119897119895V119895

5= 1+

(6119887 + 119895minus1

119895minus1

) V119895

5 (6 minus 119890119895)

lt 1+

(6119887 + 119895minus2

119895minus2

) V119895minus1

5 (6 minus 119890119895minus1

)

= 1+119897119895minus1

V119895minus1

5= 119895lt 0

(26)





0

u(p+12)

i

u(p+1)

i




(p + 12)

(p)



(p + 12)

(p + 1)


iteration level


lt 0


119897119896119896=(6119887 +


) 119896

6 minus 119890119896

gt(6119887 +


) 119896minus1

6 minus 119890119896minus1

= 119897119896minus1

119896minus1

119890119896+1

=6 (119888 minus 1) + 119897

119896119896+ 119896minus1

V119896minus1

5= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

gt 1198901+119897119896minus1

119896minus1

+ 119896minus2

V119896minus2

5= 119890119896gt 0

(27)



119890119896gt 119890119896minus1


119896gt 6 Then

119890119896+1

= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

= 1198901+

6119887 + 119896

5 (6 minus 119890119896)+119896minus1

5(119896minus1

119896

6 minus 119890119896

+ V119896minus1

) lt 1198901

(28)



119890119896gt 119890119896minus1


119894gt 119890119894minus1





119890119894)| lt 1 i = 1 2 3 119898 minus 2


119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| for


119895)(119903 + 119890

119895)| ge 1 Since 119903 gt 0 and 119890

119895gt 0

then 119903 + 119890119895gt 0

If (119903 minus 119892119890119895)(119903 + 119890

119895) ge 1 then minus((6 + 119903)6)119890

119895ge 119890119895 which


119895)(119903 + 119890


119895le minus119890119895




Hence max |(119903 minus 119892119890119894)(119903 + 119890

119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| lt 1


1)minus1

2lt 1

Proof Let 119865 = (119903119868 minus 1198921198661)(119903119868 + 119866

1)minus1 that is

119865 =

[[[[[[[[[[

[

119903 minus 119892

119889

d119903 minus 119892

119889119874

d d d

d d


119889

]]]]]]]]]]

]

(29)




120588 [119865] = max 10038161003816100381610038161205821198651003816100381610038161003816 =

10038161003816100381610038161205821198651003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119903 minus 119892

119903 + 1

1003816100381610038161003816100381610038161003816le 119865

2 (30)


1198652= maxx2 = 0

119865x2

x2

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

1003817100381710038171003817y10038171003817100381710038172

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

le maxw2=1

119865w2= maxw2=1

1003817100381710038171003817120582119865w10038171003817100381710038172

(31)


1198652le1003816100381610038161003816120582119865

1003816100381610038161003816 maxw2=1

w2=1003816100381610038161003816120582119865

1003816100381610038161003816 = 120588 [119865] (32)




1)(119903119868 + 119866

1)minus1

2lt 1


2)minus1

2lt 1

Proof Let 119870 = (119903119868 minus 1198921198662)(119903119868 + 119866

2)minus1 that is

119870 =

[[[[[[[[[[[[[

[

119903 minus 1198921198901

1199111


119903 minus 1198921198902

1199112

d sdot sdot sdot

d

119874 d d

119903 minus 119892119890119898

119911119898

]]]]]]]]]]]]]

]

(33)


119894)119911119894 119894 = 1 2 119898 minus 2 where 119911

119894= 119903 + 119890

119894



1198702=10038171003817100381710038171003817119874119879

119870119874100381710038171003817100381710038172

= 1198632



= radic120588 (1198632) = radic120588 (1198702) = 120588 (119870)

(34)

By Lemma 2

1198702= 120588 (119870) = max

10038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119894

119903 + 119890119894

10038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119895

119903 + 119890119895

1003816100381610038161003816100381610038161003816100381610038161003816

lt 1 (35)


From (20) and (21)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661)

times [(119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901) + (119903119868 + 119866

1)minus1f]

+ 119892 (119903119868 + 1198662)minus1f

(36)

Let

119872(119903) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662)

(37)

And let

119902 (119903) = [(119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

+119892(119903119868 + 1198662)minus1

] f (38)

Then u(119901+1) = 119872(119903)u(119901) + 119902(119903)



Proof Define (119903) = (119903119868 + 1198662)119872(119903)(119903119868 + 119866

2)minus1 then

(119903) = (119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus1

(39)


120588 [119872 (119903)]

= 120588 [ (119903)] le10038171003817100381710038171003817 (119903)

10038171003817100381710038171003817 2

=10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

le10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1100381710038171003817100381710038172

10038171003817100381710038171003817(119903119868 minus 119892119866

2) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

lt 1

(40)





119906(119901+1)

119894=

(119891119894minus119886119906(119901)

119894minus2minus 119887119906(119901)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(41)




and 119906(119901+1)


119894 The


119906(119901+1)

119894=

(119891119894minus119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(42)


119906(119901+1)

119894= (1 minus 120596) 119906

(119901)

119894

+

120596 (119891119894minus 119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(43)


(p + 1)

(p)



(p + 1)

(p)







IADEMF2 6119898 9119898 15119898

















for 119894 = 2 to 119894 = 119898 minus 1

compute 119906(119901+12)119894


compute 119906(119901+1)119894


119894larr

10038161003816100381610038161003816119906(119901+1)

119894minus 119906(119901)

119894

10038161003816100381610038161003816if (max1003816100381610038161003816119890119894

1003816100381610038161003816 lt 120576)then 119906

(119901)

119894larr 119906(119901+1)

119894




end-forend





120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (44)


119880 (119909 0) = 4119909 (1 minus 119909) 0 le 119909 le 1 (45)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (46)


119880 (119909 119905) =32

1205873

infin

sum

119896=1(2)

1

1198963119890minus12058721198962119905 sin (119896120587119909) (47)


120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (48)


119880 (119909 0) = sin (120587119909) (1 + 6 cos (120587119909)) 0 le 119909 le 1 (49)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (50)


119880 (119909 119905) = sin (120587119909) 119890minus1205872119905

+ 3 sin (2120587119909) 119890minus41205872119905

(51)





(119901)

infin










0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

90E minus 03

U(xt)


minus4


minus10




0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4



0

minus2


minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2


minus4








































































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










At the grid-point 119875(119909119894 119905119896+12


119909119906119896

119894= 119906119896

119894minus1minus 2119906119896

119894+ 119906119896

119894+1and 120575

2

119909119906119896+1

119894= 119906119896+1

119894minus1minus

2119906119896+1

119894+ 119906119896+1



1

Δ119905(119906119896+1

119894minus 119906119896

119894) =

1

2(Δ119909)2(1205752

119909minus

1

121205754

119909) (119906119896+1

119894+ 119906119896

119894) (3)


2 and since1205754

119909119906119896

119894= 1205752

119909(1205752

119909119906119896


119906119896+1

119894minus 119906119896

119894=120582

2(minus

1

12119906119896+1

119894minus2+4

3119906119896+1

119894minus1

minus5

2119906119896+1

119894+4

3119906119896+1

119894+1minus

1

12119906119896+1

119894+2

minus1

12119906119896

119894minus2+4

3119906119896

119894minus1minus5

2119906119896

119894

+4

3119906119896

119894+1minus

1

12119906119896

119894+2)

(4)


119894=

120588119896



120588 =

1 minus (1205826) ((cos120573 minus 4)2

minus 9)

1 + (1205826) ((cos120573 minus 4)2

minus 9)

with 120573 = 120573Δ119909 (5)





119886119906119896+1

119894minus2+ 119887119906119896+1

119894minus1+ 119888119906119896+1

119894+ 119889119906119896+1

119894+1+ 119890119906119896+1

119894+2

= minus119886119906119896

119894minus2minus 119887119906119896

119894minus1+ 119888119906119896

119894minus 119889119906119896

119894+1minus 119890119906119896

119894+2

119894 = 2 3 119898 minus 1

(6)

119886 =120582

24 119887 = minus

2120582

3 119888 =

4 + 5120582

4

119889 = minus2120582

3 119890 =

120582

24 119888 =

4 minus 5120582

4

(7)


2 1199063 119906

119898minus2 119906119898minus1

)119879 is the


2 1198913 119891

119898minus2 119891119898minus1

)119879


119860u = f

[[[[[[[[[

[

119888 119889 119890

119887 119888 119889 119890 O119886 119887 119888 119889 119890

d d d d119886 119887 119888 119889 119890

O 119886 119887 119888 119889

119886 119887 119888

]]]]]]]]]


[[[[[[

[

1199062

1199063

119906119898minus2

119906119898minus1

]]]]]]

]119896+1

=

[[[[[[

[

1198912

1198913

119891119898minus2

119891119898minus1

]]]]]]

]

(8)

1198912= minus119887 (119906

119896

1+ 119906119896+1

1) + 119888119906

119896

2minus 119889119906119896

3minus 119890119906119896

4

1198913= minus119886 (119906

119896

1+ 119906119896+1

1) minus 119887119906

119896

2+ 119888119906119896

3minus 119889119906119896

4minus 119890119906119896

5

119891119894= minus119886119906

119896

119894minus2minus 119887119906119896

119894minus1+ 119888119906119896

119894minus 119889119906119896

119894+1minus 119890119906119896

119894+2

for 119894 = 4 5 119898 minus 3

119891119898minus2

= minus119886119906119896

119898minus4minus 119887119906119896

119898minus3+ 119888119906119896

119898minus2

minus 119889119906119896

119898minus1minus 119890 (119906

119896

119898+ 119906119896+1

119898)

119891119898minus1

= minus119886119906119896

119898minus3minus 119887119906119896

119898minus2+ 119888119906119896

119898minus1minus 119889 (119906

119896

119898+ 119906119896+1

119898)

(9)


and 119891119898minus1




4


(119903119868 + 1198661) u(119901+12) = (119903119868 minus 119892119866

2) u(119901) + f (10)

(119903119868 + 1198662) u(119901+1) = (119903119868 minus 119892119866

1) u(119901+12) + 119892f (11)


1and





u(119901+12) = (119903119868 + 1198661)minus1

[(119903119868 minus 1198921198662) u(119901) + f] (12)


(119903119868 + 1198662) u(119901+1) = (119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) u(119901)

+ [ (119903119868 minus 1198921198661) (119903119868 minus 119892119866

1)minus1

+ 119892] f (13)




[119903 (1198661+ 1198662) (1 + 119892) + (1 minus 119892

2

)11986611198662] u = 119903 (1 + 119892) f (14)


[ (1198661+ 1198662) minus

1

611986611198662] u = f (15)


119860 = 1198661+ 1198662minus1

611986611198662 (16)


2in


1198661=

[[[[[[[

[

1

1198971

1 119874

1

1198972

d2d dd 119897119898minus4

1

119874 119898minus4

119897119898minus3

1

]]]]]]]


1198662=

[[[[[[[

[

11989011

V1

1198902

2

V2

119874

d d d119874 119890


V119898minus4

119890119898minus3

119898minus3

119890119898minus2

]]]]]]]


(17)




1198901=6 (119888 minus 1)

5

1=6119889

5 119897

1=

6119887

6 minus 1198901

1198902=6 (119888 minus 1) + 119897

11

5 V

119894=6119890

5for 119894 = 1 2 119898 minus 4

(18)

And for 119894 = 2 3 119898 minus 3

119894=6119889 + 119897


5

119894minus1=

6119886

6 minus 119890119894minus1

119897119894=6119887 +


6 minus 119890119894

119890119894+1

=6 (119888 minus 1) + 119897

119894119894+ 119894minus1

V119894minus1

5

(19)

Since1198661and119866


1)



u(119901+12) = (119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901)

+ (119903119868 + 1198661)minus1f

(20)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) u(119901+12)

+ 119892(119903119868 + 1198662)minus1f

(21)



119906(119901+12)

119894=

1

119877(119864119894minus1

119906(119901)

119894+119882119894minus1

119906(119901)

119894+1+ 119881119894minus1

119906(119901)

119894+2

minus119894minus3

119906(119901+12)


119906(119901+12)

119894minus1+ 119891119894)

119894 = 2 3 119898 minus 2119898 minus 1

(22)


119906(119901+1)

119894=

1

119885119894minus1

(119878119894minus3

119906(119901+12)

119894minus2+ 119876119894minus2

119906(119901+12)

119894minus1+ 119875119906(119901+12)

119894

minus119894minus1

119906(119901+1)


119906(119901+1)

119894+2+ 119892119891119894)

119894 = 119898 minus 1 119898 minus 2 3 2

(23)

with

minus1

= 0= 1198970= 119881119898minus2

= 119881119898minus3

= 119882119898minus2

= 119898minus2

= V119898minus2

= V119898minus3

= 1198760= 119878minus1

= 1198780= 0

119877 = 1 + 119903 119875 = 119903 minus 119892

119864119894= 119903 minus 119892119890

119894 119885119894= 119903 + 119890

119894 119894 = 1 2 119898 minus 2

119882119894= minus119892

119894 119876119894= minus119892119897

119894 119894 = 1 2 119898 minus 3

119881119894=minus119892V119894 119878119894= minus119892

119894 119894 = 1 2 119898 minus 4

(24)


(119901+12)

119894begins at



manner 119906(119901+1)119894







1198901=6 (119888 minus 1)

5=3120582


1le3

2

1=6119889

5lt0 119897

1=

6119887

6 minus 1198901

lt0 V119894=6119890

5gt 0

for 119894 = 1 2 119898 minus 4

2=6119889 + 119897

1V1

5= 1+1198971V1

5lt 1 since 119897

1V1lt 0

1198902=6 (119888 minus 1) + 119897

11

5= 1198901+11989711

5gt 1198901 since 119897

11gt 0

(25)


2lt 6


119894lt 6 for all

119894 = 1 2 3 119898 minus 2




119896minus1gt 119890119896minus2


= 6119886(6 minus 119890119896minus1

) gt 119896minus2

= 6119886(6 minus 119890119896minus2


V119896minus1

gt 119896minus2

V119896minus2







119895minus1

lt 119895minus2

119895minus2

lt 0

119895+1

=6119889 + 119897

119895V119895

5= 1+119897119895V119895

5= 1+

(6119887 + 119895minus1

119895minus1

) V119895

5 (6 minus 119890119895)

lt 1+

(6119887 + 119895minus2

119895minus2

) V119895minus1

5 (6 minus 119890119895minus1

)

= 1+119897119895minus1

V119895minus1

5= 119895lt 0

(26)





0

u(p+12)

i

u(p+1)

i




(p + 12)

(p)



(p + 12)

(p + 1)


iteration level


lt 0


119897119896119896=(6119887 +


) 119896

6 minus 119890119896

gt(6119887 +


) 119896minus1

6 minus 119890119896minus1

= 119897119896minus1

119896minus1

119890119896+1

=6 (119888 minus 1) + 119897

119896119896+ 119896minus1

V119896minus1

5= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

gt 1198901+119897119896minus1

119896minus1

+ 119896minus2

V119896minus2

5= 119890119896gt 0

(27)



119890119896gt 119890119896minus1


119896gt 6 Then

119890119896+1

= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

= 1198901+

6119887 + 119896

5 (6 minus 119890119896)+119896minus1

5(119896minus1

119896

6 minus 119890119896

+ V119896minus1

) lt 1198901

(28)



119890119896gt 119890119896minus1


119894gt 119890119894minus1





119890119894)| lt 1 i = 1 2 3 119898 minus 2


119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| for


119895)(119903 + 119890

119895)| ge 1 Since 119903 gt 0 and 119890

119895gt 0

then 119903 + 119890119895gt 0

If (119903 minus 119892119890119895)(119903 + 119890

119895) ge 1 then minus((6 + 119903)6)119890

119895ge 119890119895 which


119895)(119903 + 119890


119895le minus119890119895




Hence max |(119903 minus 119892119890119894)(119903 + 119890

119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| lt 1


1)minus1

2lt 1

Proof Let 119865 = (119903119868 minus 1198921198661)(119903119868 + 119866

1)minus1 that is

119865 =

[[[[[[[[[[

[

119903 minus 119892

119889

d119903 minus 119892

119889119874

d d d

d d


119889

]]]]]]]]]]

]

(29)




120588 [119865] = max 10038161003816100381610038161205821198651003816100381610038161003816 =

10038161003816100381610038161205821198651003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119903 minus 119892

119903 + 1

1003816100381610038161003816100381610038161003816le 119865

2 (30)


1198652= maxx2 = 0

119865x2

x2

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

1003817100381710038171003817y10038171003817100381710038172

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

le maxw2=1

119865w2= maxw2=1

1003817100381710038171003817120582119865w10038171003817100381710038172

(31)


1198652le1003816100381610038161003816120582119865

1003816100381610038161003816 maxw2=1

w2=1003816100381610038161003816120582119865

1003816100381610038161003816 = 120588 [119865] (32)




1)(119903119868 + 119866

1)minus1

2lt 1


2)minus1

2lt 1

Proof Let 119870 = (119903119868 minus 1198921198662)(119903119868 + 119866

2)minus1 that is

119870 =

[[[[[[[[[[[[[

[

119903 minus 1198921198901

1199111


119903 minus 1198921198902

1199112

d sdot sdot sdot

d

119874 d d

119903 minus 119892119890119898

119911119898

]]]]]]]]]]]]]

]

(33)


119894)119911119894 119894 = 1 2 119898 minus 2 where 119911

119894= 119903 + 119890

119894



1198702=10038171003817100381710038171003817119874119879

119870119874100381710038171003817100381710038172

= 1198632



= radic120588 (1198632) = radic120588 (1198702) = 120588 (119870)

(34)

By Lemma 2

1198702= 120588 (119870) = max

10038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119894

119903 + 119890119894

10038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119895

119903 + 119890119895

1003816100381610038161003816100381610038161003816100381610038161003816

lt 1 (35)


From (20) and (21)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661)

times [(119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901) + (119903119868 + 119866

1)minus1f]

+ 119892 (119903119868 + 1198662)minus1f

(36)

Let

119872(119903) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662)

(37)

And let

119902 (119903) = [(119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

+119892(119903119868 + 1198662)minus1

] f (38)

Then u(119901+1) = 119872(119903)u(119901) + 119902(119903)



Proof Define (119903) = (119903119868 + 1198662)119872(119903)(119903119868 + 119866

2)minus1 then

(119903) = (119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus1

(39)


120588 [119872 (119903)]

= 120588 [ (119903)] le10038171003817100381710038171003817 (119903)

10038171003817100381710038171003817 2

=10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

le10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1100381710038171003817100381710038172

10038171003817100381710038171003817(119903119868 minus 119892119866

2) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

lt 1

(40)





119906(119901+1)

119894=

(119891119894minus119886119906(119901)

119894minus2minus 119887119906(119901)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(41)




and 119906(119901+1)


119894 The


119906(119901+1)

119894=

(119891119894minus119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(42)


119906(119901+1)

119894= (1 minus 120596) 119906

(119901)

119894

+

120596 (119891119894minus 119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(43)


(p + 1)

(p)



(p + 1)

(p)







IADEMF2 6119898 9119898 15119898

















for 119894 = 2 to 119894 = 119898 minus 1

compute 119906(119901+12)119894


compute 119906(119901+1)119894


119894larr

10038161003816100381610038161003816119906(119901+1)

119894minus 119906(119901)

119894

10038161003816100381610038161003816if (max1003816100381610038161003816119890119894

1003816100381610038161003816 lt 120576)then 119906

(119901)

119894larr 119906(119901+1)

119894




end-forend





120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (44)


119880 (119909 0) = 4119909 (1 minus 119909) 0 le 119909 le 1 (45)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (46)


119880 (119909 119905) =32

1205873

infin

sum

119896=1(2)

1

1198963119890minus12058721198962119905 sin (119896120587119909) (47)


120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (48)


119880 (119909 0) = sin (120587119909) (1 + 6 cos (120587119909)) 0 le 119909 le 1 (49)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (50)


119880 (119909 119905) = sin (120587119909) 119890minus1205872119905

+ 3 sin (2120587119909) 119890minus41205872119905

(51)





(119901)

infin










0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

90E minus 03

U(xt)


minus4


minus10




0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4



0

minus2


minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2


minus4








































































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











u(119901+12) = (119903119868 + 1198661)minus1

[(119903119868 minus 1198921198662) u(119901) + f] (12)


(119903119868 + 1198662) u(119901+1) = (119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) u(119901)

+ [ (119903119868 minus 1198921198661) (119903119868 minus 119892119866

1)minus1

+ 119892] f (13)




[119903 (1198661+ 1198662) (1 + 119892) + (1 minus 119892

2

)11986611198662] u = 119903 (1 + 119892) f (14)


[ (1198661+ 1198662) minus

1

611986611198662] u = f (15)


119860 = 1198661+ 1198662minus1

611986611198662 (16)


2in


1198661=

[[[[[[[

[

1

1198971

1 119874

1

1198972

d2d dd 119897119898minus4

1

119874 119898minus4

119897119898minus3

1

]]]]]]]


1198662=

[[[[[[[

[

11989011

V1

1198902

2

V2

119874

d d d119874 119890


V119898minus4

119890119898minus3

119898minus3

119890119898minus2

]]]]]]]


(17)




1198901=6 (119888 minus 1)

5

1=6119889

5 119897

1=

6119887

6 minus 1198901

1198902=6 (119888 minus 1) + 119897

11

5 V

119894=6119890

5for 119894 = 1 2 119898 minus 4

(18)

And for 119894 = 2 3 119898 minus 3

119894=6119889 + 119897


5

119894minus1=

6119886

6 minus 119890119894minus1

119897119894=6119887 +


6 minus 119890119894

119890119894+1

=6 (119888 minus 1) + 119897

119894119894+ 119894minus1

V119894minus1

5

(19)

Since1198661and119866


1)



u(119901+12) = (119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901)

+ (119903119868 + 1198661)minus1f

(20)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) u(119901+12)

+ 119892(119903119868 + 1198662)minus1f

(21)



119906(119901+12)

119894=

1

119877(119864119894minus1

119906(119901)

119894+119882119894minus1

119906(119901)

119894+1+ 119881119894minus1

119906(119901)

119894+2

minus119894minus3

119906(119901+12)


119906(119901+12)

119894minus1+ 119891119894)

119894 = 2 3 119898 minus 2119898 minus 1

(22)


119906(119901+1)

119894=

1

119885119894minus1

(119878119894minus3

119906(119901+12)

119894minus2+ 119876119894minus2

119906(119901+12)

119894minus1+ 119875119906(119901+12)

119894

minus119894minus1

119906(119901+1)


119906(119901+1)

119894+2+ 119892119891119894)

119894 = 119898 minus 1 119898 minus 2 3 2

(23)

with

minus1

= 0= 1198970= 119881119898minus2

= 119881119898minus3

= 119882119898minus2

= 119898minus2

= V119898minus2

= V119898minus3

= 1198760= 119878minus1

= 1198780= 0

119877 = 1 + 119903 119875 = 119903 minus 119892

119864119894= 119903 minus 119892119890

119894 119885119894= 119903 + 119890

119894 119894 = 1 2 119898 minus 2

119882119894= minus119892

119894 119876119894= minus119892119897

119894 119894 = 1 2 119898 minus 3

119881119894=minus119892V119894 119878119894= minus119892

119894 119894 = 1 2 119898 minus 4

(24)


(119901+12)

119894begins at



manner 119906(119901+1)119894







1198901=6 (119888 minus 1)

5=3120582


1le3

2

1=6119889

5lt0 119897

1=

6119887

6 minus 1198901

lt0 V119894=6119890

5gt 0

for 119894 = 1 2 119898 minus 4

2=6119889 + 119897

1V1

5= 1+1198971V1

5lt 1 since 119897

1V1lt 0

1198902=6 (119888 minus 1) + 119897

11

5= 1198901+11989711

5gt 1198901 since 119897

11gt 0

(25)


2lt 6


119894lt 6 for all

119894 = 1 2 3 119898 minus 2




119896minus1gt 119890119896minus2


= 6119886(6 minus 119890119896minus1

) gt 119896minus2

= 6119886(6 minus 119890119896minus2


V119896minus1

gt 119896minus2

V119896minus2







119895minus1

lt 119895minus2

119895minus2

lt 0

119895+1

=6119889 + 119897

119895V119895

5= 1+119897119895V119895

5= 1+

(6119887 + 119895minus1

119895minus1

) V119895

5 (6 minus 119890119895)

lt 1+

(6119887 + 119895minus2

119895minus2

) V119895minus1

5 (6 minus 119890119895minus1

)

= 1+119897119895minus1

V119895minus1

5= 119895lt 0

(26)





0

u(p+12)

i

u(p+1)

i




(p + 12)

(p)



(p + 12)

(p + 1)


iteration level


lt 0


119897119896119896=(6119887 +


) 119896

6 minus 119890119896

gt(6119887 +


) 119896minus1

6 minus 119890119896minus1

= 119897119896minus1

119896minus1

119890119896+1

=6 (119888 minus 1) + 119897

119896119896+ 119896minus1

V119896minus1

5= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

gt 1198901+119897119896minus1

119896minus1

+ 119896minus2

V119896minus2

5= 119890119896gt 0

(27)



119890119896gt 119890119896minus1


119896gt 6 Then

119890119896+1

= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

= 1198901+

6119887 + 119896

5 (6 minus 119890119896)+119896minus1

5(119896minus1

119896

6 minus 119890119896

+ V119896minus1

) lt 1198901

(28)



119890119896gt 119890119896minus1


119894gt 119890119894minus1





119890119894)| lt 1 i = 1 2 3 119898 minus 2


119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| for


119895)(119903 + 119890

119895)| ge 1 Since 119903 gt 0 and 119890

119895gt 0

then 119903 + 119890119895gt 0

If (119903 minus 119892119890119895)(119903 + 119890

119895) ge 1 then minus((6 + 119903)6)119890

119895ge 119890119895 which


119895)(119903 + 119890


119895le minus119890119895




Hence max |(119903 minus 119892119890119894)(119903 + 119890

119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| lt 1


1)minus1

2lt 1

Proof Let 119865 = (119903119868 minus 1198921198661)(119903119868 + 119866

1)minus1 that is

119865 =

[[[[[[[[[[

[

119903 minus 119892

119889

d119903 minus 119892

119889119874

d d d

d d


119889

]]]]]]]]]]

]

(29)




120588 [119865] = max 10038161003816100381610038161205821198651003816100381610038161003816 =

10038161003816100381610038161205821198651003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119903 minus 119892

119903 + 1

1003816100381610038161003816100381610038161003816le 119865

2 (30)


1198652= maxx2 = 0

119865x2

x2

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

1003817100381710038171003817y10038171003817100381710038172

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

le maxw2=1

119865w2= maxw2=1

1003817100381710038171003817120582119865w10038171003817100381710038172

(31)


1198652le1003816100381610038161003816120582119865

1003816100381610038161003816 maxw2=1

w2=1003816100381610038161003816120582119865

1003816100381610038161003816 = 120588 [119865] (32)




1)(119903119868 + 119866

1)minus1

2lt 1


2)minus1

2lt 1

Proof Let 119870 = (119903119868 minus 1198921198662)(119903119868 + 119866

2)minus1 that is

119870 =

[[[[[[[[[[[[[

[

119903 minus 1198921198901

1199111


119903 minus 1198921198902

1199112

d sdot sdot sdot

d

119874 d d

119903 minus 119892119890119898

119911119898

]]]]]]]]]]]]]

]

(33)


119894)119911119894 119894 = 1 2 119898 minus 2 where 119911

119894= 119903 + 119890

119894



1198702=10038171003817100381710038171003817119874119879

119870119874100381710038171003817100381710038172

= 1198632



= radic120588 (1198632) = radic120588 (1198702) = 120588 (119870)

(34)

By Lemma 2

1198702= 120588 (119870) = max

10038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119894

119903 + 119890119894

10038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119895

119903 + 119890119895

1003816100381610038161003816100381610038161003816100381610038161003816

lt 1 (35)


From (20) and (21)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661)

times [(119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901) + (119903119868 + 119866

1)minus1f]

+ 119892 (119903119868 + 1198662)minus1f

(36)

Let

119872(119903) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662)

(37)

And let

119902 (119903) = [(119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

+119892(119903119868 + 1198662)minus1

] f (38)

Then u(119901+1) = 119872(119903)u(119901) + 119902(119903)



Proof Define (119903) = (119903119868 + 1198662)119872(119903)(119903119868 + 119866

2)minus1 then

(119903) = (119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus1

(39)


120588 [119872 (119903)]

= 120588 [ (119903)] le10038171003817100381710038171003817 (119903)

10038171003817100381710038171003817 2

=10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

le10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1100381710038171003817100381710038172

10038171003817100381710038171003817(119903119868 minus 119892119866

2) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

lt 1

(40)





119906(119901+1)

119894=

(119891119894minus119886119906(119901)

119894minus2minus 119887119906(119901)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(41)




and 119906(119901+1)


119894 The


119906(119901+1)

119894=

(119891119894minus119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(42)


119906(119901+1)

119894= (1 minus 120596) 119906

(119901)

119894

+

120596 (119891119894minus 119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(43)


(p + 1)

(p)



(p + 1)

(p)







IADEMF2 6119898 9119898 15119898

















for 119894 = 2 to 119894 = 119898 minus 1

compute 119906(119901+12)119894


compute 119906(119901+1)119894


119894larr

10038161003816100381610038161003816119906(119901+1)

119894minus 119906(119901)

119894

10038161003816100381610038161003816if (max1003816100381610038161003816119890119894

1003816100381610038161003816 lt 120576)then 119906

(119901)

119894larr 119906(119901+1)

119894




end-forend





120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (44)


119880 (119909 0) = 4119909 (1 minus 119909) 0 le 119909 le 1 (45)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (46)


119880 (119909 119905) =32

1205873

infin

sum

119896=1(2)

1

1198963119890minus12058721198962119905 sin (119896120587119909) (47)


120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (48)


119880 (119909 0) = sin (120587119909) (1 + 6 cos (120587119909)) 0 le 119909 le 1 (49)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (50)


119880 (119909 119905) = sin (120587119909) 119890minus1205872119905

+ 3 sin (2120587119909) 119890minus41205872119905

(51)





(119901)

infin










0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

90E minus 03

U(xt)


minus4


minus10




0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4



0

minus2


minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2


minus4








































































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










manner 119906(119901+1)119894







1198901=6 (119888 minus 1)

5=3120582


1le3

2

1=6119889

5lt0 119897

1=

6119887

6 minus 1198901

lt0 V119894=6119890

5gt 0

for 119894 = 1 2 119898 minus 4

2=6119889 + 119897

1V1

5= 1+1198971V1

5lt 1 since 119897

1V1lt 0

1198902=6 (119888 minus 1) + 119897

11

5= 1198901+11989711

5gt 1198901 since 119897

11gt 0

(25)


2lt 6


119894lt 6 for all

119894 = 1 2 3 119898 minus 2




119896minus1gt 119890119896minus2


= 6119886(6 minus 119890119896minus1

) gt 119896minus2

= 6119886(6 minus 119890119896minus2


V119896minus1

gt 119896minus2

V119896minus2







119895minus1

lt 119895minus2

119895minus2

lt 0

119895+1

=6119889 + 119897

119895V119895

5= 1+119897119895V119895

5= 1+

(6119887 + 119895minus1

119895minus1

) V119895

5 (6 minus 119890119895)

lt 1+

(6119887 + 119895minus2

119895minus2

) V119895minus1

5 (6 minus 119890119895minus1

)

= 1+119897119895minus1

V119895minus1

5= 119895lt 0

(26)





0

u(p+12)

i

u(p+1)

i




(p + 12)

(p)



(p + 12)

(p + 1)


iteration level


lt 0


119897119896119896=(6119887 +


) 119896

6 minus 119890119896

gt(6119887 +


) 119896minus1

6 minus 119890119896minus1

= 119897119896minus1

119896minus1

119890119896+1

=6 (119888 minus 1) + 119897

119896119896+ 119896minus1

V119896minus1

5= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

gt 1198901+119897119896minus1

119896minus1

+ 119896minus2

V119896minus2

5= 119890119896gt 0

(27)



119890119896gt 119890119896minus1


119896gt 6 Then

119890119896+1

= 1198901+119897119896119896+ 119896minus1

V119896minus1

5

= 1198901+

6119887 + 119896

5 (6 minus 119890119896)+119896minus1

5(119896minus1

119896

6 minus 119890119896

+ V119896minus1

) lt 1198901

(28)



119890119896gt 119890119896minus1


119894gt 119890119894minus1





119890119894)| lt 1 i = 1 2 3 119898 minus 2


119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| for


119895)(119903 + 119890

119895)| ge 1 Since 119903 gt 0 and 119890

119895gt 0

then 119903 + 119890119895gt 0

If (119903 minus 119892119890119895)(119903 + 119890

119895) ge 1 then minus((6 + 119903)6)119890

119895ge 119890119895 which


119895)(119903 + 119890


119895le minus119890119895




Hence max |(119903 minus 119892119890119894)(119903 + 119890

119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| lt 1


1)minus1

2lt 1

Proof Let 119865 = (119903119868 minus 1198921198661)(119903119868 + 119866

1)minus1 that is

119865 =

[[[[[[[[[[

[

119903 minus 119892

119889

d119903 minus 119892

119889119874

d d d

d d


119889

]]]]]]]]]]

]

(29)




120588 [119865] = max 10038161003816100381610038161205821198651003816100381610038161003816 =

10038161003816100381610038161205821198651003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119903 minus 119892

119903 + 1

1003816100381610038161003816100381610038161003816le 119865

2 (30)


1198652= maxx2 = 0

119865x2

x2

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

1003817100381710038171003817y10038171003817100381710038172

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

le maxw2=1

119865w2= maxw2=1

1003817100381710038171003817120582119865w10038171003817100381710038172

(31)


1198652le1003816100381610038161003816120582119865

1003816100381610038161003816 maxw2=1

w2=1003816100381610038161003816120582119865

1003816100381610038161003816 = 120588 [119865] (32)




1)(119903119868 + 119866

1)minus1

2lt 1


2)minus1

2lt 1

Proof Let 119870 = (119903119868 minus 1198921198662)(119903119868 + 119866

2)minus1 that is

119870 =

[[[[[[[[[[[[[

[

119903 minus 1198921198901

1199111


119903 minus 1198921198902

1199112

d sdot sdot sdot

d

119874 d d

119903 minus 119892119890119898

119911119898

]]]]]]]]]]]]]

]

(33)


119894)119911119894 119894 = 1 2 119898 minus 2 where 119911

119894= 119903 + 119890

119894



1198702=10038171003817100381710038171003817119874119879

119870119874100381710038171003817100381710038172

= 1198632



= radic120588 (1198632) = radic120588 (1198702) = 120588 (119870)

(34)

By Lemma 2

1198702= 120588 (119870) = max

10038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119894

119903 + 119890119894

10038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119895

119903 + 119890119895

1003816100381610038161003816100381610038161003816100381610038161003816

lt 1 (35)


From (20) and (21)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661)

times [(119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901) + (119903119868 + 119866

1)minus1f]

+ 119892 (119903119868 + 1198662)minus1f

(36)

Let

119872(119903) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662)

(37)

And let

119902 (119903) = [(119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

+119892(119903119868 + 1198662)minus1

] f (38)

Then u(119901+1) = 119872(119903)u(119901) + 119902(119903)



Proof Define (119903) = (119903119868 + 1198662)119872(119903)(119903119868 + 119866

2)minus1 then

(119903) = (119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus1

(39)


120588 [119872 (119903)]

= 120588 [ (119903)] le10038171003817100381710038171003817 (119903)

10038171003817100381710038171003817 2

=10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

le10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1100381710038171003817100381710038172

10038171003817100381710038171003817(119903119868 minus 119892119866

2) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

lt 1

(40)





119906(119901+1)

119894=

(119891119894minus119886119906(119901)

119894minus2minus 119887119906(119901)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(41)




and 119906(119901+1)


119894 The


119906(119901+1)

119894=

(119891119894minus119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(42)


119906(119901+1)

119894= (1 minus 120596) 119906

(119901)

119894

+

120596 (119891119894minus 119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(43)


(p + 1)

(p)



(p + 1)

(p)







IADEMF2 6119898 9119898 15119898

















for 119894 = 2 to 119894 = 119898 minus 1

compute 119906(119901+12)119894


compute 119906(119901+1)119894


119894larr

10038161003816100381610038161003816119906(119901+1)

119894minus 119906(119901)

119894

10038161003816100381610038161003816if (max1003816100381610038161003816119890119894

1003816100381610038161003816 lt 120576)then 119906

(119901)

119894larr 119906(119901+1)

119894




end-forend





120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (44)


119880 (119909 0) = 4119909 (1 minus 119909) 0 le 119909 le 1 (45)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (46)


119880 (119909 119905) =32

1205873

infin

sum

119896=1(2)

1

1198963119890minus12058721198962119905 sin (119896120587119909) (47)


120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (48)


119880 (119909 0) = sin (120587119909) (1 + 6 cos (120587119909)) 0 le 119909 le 1 (49)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (50)


119880 (119909 119905) = sin (120587119909) 119890minus1205872119905

+ 3 sin (2120587119909) 119890minus41205872119905

(51)





(119901)

infin










0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

90E minus 03

U(xt)


minus4


minus10




0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4



0

minus2


minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2


minus4








































































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119890119894)| lt 1 i = 1 2 3 119898 minus 2


119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| for


119895)(119903 + 119890

119895)| ge 1 Since 119903 gt 0 and 119890

119895gt 0

then 119903 + 119890119895gt 0

If (119903 minus 119892119890119895)(119903 + 119890

119895) ge 1 then minus((6 + 119903)6)119890

119895ge 119890119895 which


119895)(119903 + 119890


119895le minus119890119895




Hence max |(119903 minus 119892119890119894)(119903 + 119890

119894)| = |(119903 minus 119892119890

119895)(119903 + 119890

119895)| lt 1


1)minus1

2lt 1

Proof Let 119865 = (119903119868 minus 1198921198661)(119903119868 + 119866

1)minus1 that is

119865 =

[[[[[[[[[[

[

119903 minus 119892

119889

d119903 minus 119892

119889119874

d d d

d d


119889

]]]]]]]]]]

]

(29)




120588 [119865] = max 10038161003816100381610038161205821198651003816100381610038161003816 =

10038161003816100381610038161205821198651003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119903 minus 119892

119903 + 1

1003816100381610038161003816100381610038161003816le 119865

2 (30)


1198652= maxx2 = 0

119865x2

x2

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

1003817100381710038171003817y10038171003817100381710038172

= maxy2=1

1003817100381710038171003817119865y10038171003817100381710038172

le maxw2=1

119865w2= maxw2=1

1003817100381710038171003817120582119865w10038171003817100381710038172

(31)


1198652le1003816100381610038161003816120582119865

1003816100381610038161003816 maxw2=1

w2=1003816100381610038161003816120582119865

1003816100381610038161003816 = 120588 [119865] (32)




1)(119903119868 + 119866

1)minus1

2lt 1


2)minus1

2lt 1

Proof Let 119870 = (119903119868 minus 1198921198662)(119903119868 + 119866

2)minus1 that is

119870 =

[[[[[[[[[[[[[

[

119903 minus 1198921198901

1199111


119903 minus 1198921198902

1199112

d sdot sdot sdot

d

119874 d d

119903 minus 119892119890119898

119911119898

]]]]]]]]]]]]]

]

(33)


119894)119911119894 119894 = 1 2 119898 minus 2 where 119911

119894= 119903 + 119890

119894



1198702=10038171003817100381710038171003817119874119879

119870119874100381710038171003817100381710038172

= 1198632



= radic120588 (1198632) = radic120588 (1198702) = 120588 (119870)

(34)

By Lemma 2

1198702= 120588 (119870) = max

10038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119894

119903 + 119890119894

10038161003816100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

119903 minus 119892119890119895

119903 + 119890119895

1003816100381610038161003816100381610038161003816100381610038161003816

lt 1 (35)


From (20) and (21)

u(119901+1) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661)

times [(119903119868 + 1198661)minus1

(119903119868 minus 1198921198662) u(119901) + (119903119868 + 119866

1)minus1f]

+ 119892 (119903119868 + 1198662)minus1f

(36)

Let

119872(119903) = (119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662)

(37)

And let

119902 (119903) = [(119903119868 + 1198662)minus1

(119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

+119892(119903119868 + 1198662)minus1

] f (38)

Then u(119901+1) = 119872(119903)u(119901) + 119902(119903)



Proof Define (119903) = (119903119868 + 1198662)119872(119903)(119903119868 + 119866

2)minus1 then

(119903) = (119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus1

(39)


120588 [119872 (119903)]

= 120588 [ (119903)] le10038171003817100381710038171003817 (119903)

10038171003817100381710038171003817 2

=10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

le10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1100381710038171003817100381710038172

10038171003817100381710038171003817(119903119868 minus 119892119866

2) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

lt 1

(40)





119906(119901+1)

119894=

(119891119894minus119886119906(119901)

119894minus2minus 119887119906(119901)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(41)




and 119906(119901+1)


119894 The


119906(119901+1)

119894=

(119891119894minus119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(42)


119906(119901+1)

119894= (1 minus 120596) 119906

(119901)

119894

+

120596 (119891119894minus 119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(43)


(p + 1)

(p)



(p + 1)

(p)







IADEMF2 6119898 9119898 15119898

















for 119894 = 2 to 119894 = 119898 minus 1

compute 119906(119901+12)119894


compute 119906(119901+1)119894


119894larr

10038161003816100381610038161003816119906(119901+1)

119894minus 119906(119901)

119894

10038161003816100381610038161003816if (max1003816100381610038161003816119890119894

1003816100381610038161003816 lt 120576)then 119906

(119901)

119894larr 119906(119901+1)

119894




end-forend





120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (44)


119880 (119909 0) = 4119909 (1 minus 119909) 0 le 119909 le 1 (45)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (46)


119880 (119909 119905) =32

1205873

infin

sum

119896=1(2)

1

1198963119890minus12058721198962119905 sin (119896120587119909) (47)


120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (48)


119880 (119909 0) = sin (120587119909) (1 + 6 cos (120587119909)) 0 le 119909 le 1 (49)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (50)


119880 (119909 119905) = sin (120587119909) 119890minus1205872119905

+ 3 sin (2120587119909) 119890minus41205872119905

(51)





(119901)

infin










0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

90E minus 03

U(xt)


minus4


minus10




0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4



0

minus2


minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2


minus4








































































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Proof Define (119903) = (119903119868 + 1198662)119872(119903)(119903119868 + 119866

2)minus1 then

(119903) = (119903119868 minus 1198921198661) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus1

(39)


120588 [119872 (119903)]

= 120588 [ (119903)] le10038171003817100381710038171003817 (119903)

10038171003817100381710038171003817 2

=10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1

(119903119868 minus 1198921198662) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

le10038171003817100381710038171003817(119903119868 minus 119892119866

1) (119903119868 + 119866

1)minus1100381710038171003817100381710038172

10038171003817100381710038171003817(119903119868 minus 119892119866

2) (119903119868 + 119866

2)minus110038171003817100381710038171003817 2

lt 1

(40)





119906(119901+1)

119894=

(119891119894minus119886119906(119901)

119894minus2minus 119887119906(119901)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(41)




and 119906(119901+1)


119894 The


119906(119901+1)

119894=

(119891119894minus119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(42)


119906(119901+1)

119894= (1 minus 120596) 119906

(119901)

119894

+

120596 (119891119894minus 119886119906(119901+1)

119894minus2minus 119887119906(119901+1)

119894minus1minus 119889119906(119901)

119894+1minus 119890119906(119901)

119894+2)

119888

119894 = 2 3 119898 minus 1

(43)


(p + 1)

(p)



(p + 1)

(p)







IADEMF2 6119898 9119898 15119898

















for 119894 = 2 to 119894 = 119898 minus 1

compute 119906(119901+12)119894


compute 119906(119901+1)119894


119894larr

10038161003816100381610038161003816119906(119901+1)

119894minus 119906(119901)

119894

10038161003816100381610038161003816if (max1003816100381610038161003816119890119894

1003816100381610038161003816 lt 120576)then 119906

(119901)

119894larr 119906(119901+1)

119894




end-forend





120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (44)


119880 (119909 0) = 4119909 (1 minus 119909) 0 le 119909 le 1 (45)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (46)


119880 (119909 119905) =32

1205873

infin

sum

119896=1(2)

1

1198963119890minus12058721198962119905 sin (119896120587119909) (47)


120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (48)


119880 (119909 0) = sin (120587119909) (1 + 6 cos (120587119909)) 0 le 119909 le 1 (49)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (50)


119880 (119909 119905) = sin (120587119909) 119890minus1205872119905

+ 3 sin (2120587119909) 119890minus41205872119905

(51)





(119901)

infin










0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

90E minus 03

U(xt)


minus4


minus10




0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4



0

minus2


minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2


minus4








































































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















for 119894 = 2 to 119894 = 119898 minus 1

compute 119906(119901+12)119894


compute 119906(119901+1)119894


119894larr

10038161003816100381610038161003816119906(119901+1)

119894minus 119906(119901)

119894

10038161003816100381610038161003816if (max1003816100381610038161003816119890119894

1003816100381610038161003816 lt 120576)then 119906

(119901)

119894larr 119906(119901+1)

119894




end-forend





120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (44)


119880 (119909 0) = 4119909 (1 minus 119909) 0 le 119909 le 1 (45)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (46)


119880 (119909 119905) =32

1205873

infin

sum

119896=1(2)

1

1198963119890minus12058721198962119905 sin (119896120587119909) (47)


120597119880

120597119905=1205972

119880

1205971199092 0 le 119909 le 1 (48)


119880 (119909 0) = sin (120587119909) (1 + 6 cos (120587119909)) 0 le 119909 le 1 (49)


119880 (0 119905) = 119880 (1 119905) = 0 119905 ge 0 (50)


119880 (119909 119905) = sin (120587119909) 119890minus1205872119905

+ 3 sin (2120587119909) 119890minus41205872119905

(51)





(119901)

infin










0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

90E minus 03

U(xt)


minus4


minus10




0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4



0

minus2


minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2


minus4








































































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













(119901)

infin










0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

90E minus 03

U(xt)


minus4


minus10




0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4



0

minus2


minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2


minus4








































































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 02

20E minus 02

30E minus 02

40E minus 02

50E minus 02

60E minus 02

70E minus 02

80E minus 02

90E minus 02

10E minus 01

U(xt)


minus4

0 01 02 03 04 05 06 07 08 09 1x


00E + 00

10E minus 03

20E minus 03

30E minus 03

40E minus 03

50E minus 03

60E minus 03

70E minus 03

80E minus 03

U(xt)


minus4



0

minus2


minus4

minus8

minus6

minus10

minus12

minus14

minus16

minus18

GS4JAC4SOR4

log

(RM

SE)

log(Δx)

IADEMF4

IADEMF2


minus4








































































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra






























































10 Conclusion








References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra















References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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