Research ArticleNumerical Solution of Two-Point Boundary ValueProblems by Interpolating Subdivision Schemes

Ghulam Mustafa and Syeda Tehmina Ejaz

Department of Mathematics The Islamia University of Bahawalpur Punjab 63100 Pakistan

Correspondence should be addressed to Ghulam Mustafa ghulammustafaiubedupk

Received 23 January 2014 Revised 9 June 2014 Accepted 22 June 2014 Published 17 July 2014

Academic Editor Chun-Gang Zhu

Copyright copy 2014 G Mustafa and S T Ejaz This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

A numerical interpolating algorithmof collocation is formulated based on 8-point binary interpolating subdivision schemes for thegeneration of curves to solve the two-point third order boundary value problems It is observed that the algorithmproduces smoothcontinuous solutions of the problems Numerical examples are given to illustrate the algorithm and its convergence Moreover theapproximation properties of the collocation algorithm have also been discussed

1 Introduction

Subdivision plays an important role in computer aidedgeometric design Particularly subdivision schemes for curvedesign consist of repeated refinement of control polygons Inparticular the linear schemes are well studied with rules fordefining the control points at the finer level as finite linearcombinations of the control points in the coarser level

The concept of subdivision has been initiated by de Rham[1] Later on in [2] Dyn et al studied a family of schemeswithmask of size four indexed by a tension parameter In [3]Dubuc and Deslauriers studied a family of schemes indexedby the size of the mask 2119898 and the arity (or base) 119887 that isat each subdivision step the scheme (119898 119887) uses masks of 2119898points to compute 119887 new points corresponding to each oldedge Mustafa and Rehman [4] unified all existing even-pointinterpolating and approximating schemes by offering generalformula for the mask of (2119898 + 4)-point even-ary subdivisionscheme Aslam et al [5] presented an explicit formula whichunified the mask of (2119898 minus 1)-point interpolating as wellas approximating schemes Mustafa et al [6] presented anexplicit formula for the mask of odd-points 119887-ary (for anyodd 119898 ge 3) interpolating subdivision schemes Higherarity schemes have been introduced in [7 8] Mustafa andIvrissimitzis [9] have showed that when subdivision is usedas a data modelling tool large support or large arity schemes

which generally produce smoother curves it may outperformsimpler schemes

An interpolating subdivision scheme approach to theconstruction of approximate solutions of two-point secondorder boundary value problems was first time introduced in[10] In this approach amethod of collocation was formulatedfor linear two-point second order boundary value problemsIt is proved that the algorithms produce smooth continuoussolutions provided the algorithms are chosen appropriatelyLater on in [11] they reformulated the collocationmethod bysubdivision scheme in order to compute numerical solutionsfor two-point boundary-value problems of differential equa-tions with deviating argumentsThey demonstrated that theirapproach has further development to their previousworks forsolving various types of two-point second order boundary-value problems

They approximated the derivative boundary conditionsby forward difference operator so it is difficult (with no flex-ibility to improve the solution) to generalize their approachto solve third or higher order boundary value problems Weexpress derivative boundary conditions at end points by usinginterpolatory subdivision algorithm therefore it is easy togeneralize our approach to deal with higher order boundaryvalue problems

In this paper we have reformulated the collocationalgorithm by using 8-point interpolating subdivision scheme

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 721314 13 pageshttpdxdoiorg1011552014721314

2 Abstract and Applied Analysis

to compute the approximate solution of two-point third orderboundary value problems Particularly collocation algorithmwith septic precision treatments at the endpoints has thepower of approximation119874(ℎ2) Our reformulated collocationalgorithm treats the following type of two-point third orderboundary value problem

119910101584010158401015840(119909) = 119886 (119909) 119910 (119909) + 119887 (119909) 0 le 119909 le 1

119910 (0) = 119910119903 119910

1015840(0) = 0 119910 (1) = 119910

119897

(1)

where 119886(119909) and 119887(119909) are continuous and 119886(119909) ge 0 on [0 1]The outline of the paper is as follows In Section 2 we

rewrite general form of interpolating subdivision scheme forcurve design [12] and some related resultsThe 8-point binaryinterpolating scheme and derivatives of its basis functionhave also been discussed in this section In Section 3 anumerical interpolating algorithm of collocation to solve (1)is formulated and its boundary treatments are discussedIn Section 4 approximation properties of this algorithmare given In Section 5 numerical examples are presentedSection 6 is devoted for conclusion and the possible futureresearch directions

2 Interpolating Schemes for Curve Design

A general compact form of symmetric univariate binaryinterpolating subdivision scheme [12] which maps polygon119901119896= 119901119896

119894119894isinZ to a refined polygon 119901119896+1 = 119901119896+1

119894119894isinZ is defined

by

119901119896+1

2119894= 119901119896

119894

119901119896+1

2119894+1=

119899

sum

119895=0

119871119899119895(119901119896

119894minus119895+ 119901119896

119894+119895+1)

(2)

where 119899 is called the degree of the scheme and the constantsare given by

119871119899119895=

((2119899 + 1))2

2 (4119899) (2119899 + 1)

(minus1)119895

(2119895 + 1)(2119899 + 1

119899 minus 119895)

119895 = 0 1 2 119899

(3)

where ( 2119899+1119899minus119895 ) denotes the binomial coefficientThe boundary treatments are necessary to produce

smooth curve segments by scheme (2) Normally higherorder approximation formulae should be used near the endsof the segments and thus Lagrange formulae of order 2119899 + 1are recommended

Remark 1 Let 120601(119909) be the limit curve generated from thecardinal data 119901

119894= (119894 120575

0)119879 that is 120601(119909) is the fundamental

solution of the subdivision scheme (2) then

120601 (119894) = 1 119894 = 0

0 119894 = 0(4)

Furthermore 120601(119909) satisfies the following two-scale equation

120601 (119909) = 120601119899(119909)

= 120601 (2119909) +

119899

sum

119895=minus119899

119871119899|119895|

120601 (2119909 minus 2119895 + 1) 119909 isin R(5)

Lemma 2 (see [12 13]) The support of the fundamentalsolution 120601

119899(119909) to scheme (2) is finite Explicitly supp120601

119899(119909) =

(minus2119899 minus 1 2119899 + 1)

Lemma 3 (see [10]) Given a square matrix 119860 of order 119899 letthe normalized left and right (generalized) eigenvectors of119860 bedenoted by 120578

119894 120585119894 Then for any vector 119891 isin R119899 there exists

following Fourier expansion

119891 =

119899

sum

119894=1

(119891119879120578119894) 120585119894 (6)

Lemma4 (see [10]) Suppose119865(119905) 119905 isin R is a regular and1198622119899+2curve inR119898119898 ge 2 Let119875(119905) 119905 isin R be the limit curve generatedby (2) from the initial data 119875

119894= 119865(119894ℎ) 119894 isin Z 0 lt ℎ lt 1 Then

on any finite interval [119886 119887] the following estimates hold

1003817100381710038171003817119865 (ℎ119905) minus 119901 (119905)1003817100381710038171003817infin

le1198722119899+2

(119865)

(2119899 + 2)ℎ2119899+2

= 119874 (ℎ2119899+2

)

10038171003817100381710038171003817ℎ119895119865119895(ℎ119905) minus 119901

119895(119905)10038171003817100381710038171003817infin

= 119874 (ℎ2119899+2minus119895

)

119895 = 0 1 2 119899 + 2

2

(7)

where the number1198722119899+2

(119865) depends only on the derivatives of119865(119905) and 119899

21 8-Point Interpolating Scheme For 119899 = 3 by (2) and (3) wehave the following 8-point binary interpolating subdivisionscheme for curve design

119901119896+1

2119894= 119901119896

119894

119901119896+1

2119894+1=1225

2048(119901119896

119894minus 119901119896

119894+1) minus

245

2048(119901119896

119894minus1minus 119901119896

119894+2)

+49

2048(119901119896

119894minus2minus 119901119896

119894+3) minus

5

2048(119901119896

119894minus3minus 119901119896

119894+4)

(8)

This scheme is 1198623-continuous in [14] and reproduces poly-nomial curve of degree seven in [15] The local subdivisionmatrix of (8) is denoted and defined by

Abstract and Applied Analysis 3

119878 =

((((((((((((((

(

0 0 0 1 0 0 0 0 0 0 0 0 0

11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0 0

0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0

0 0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0

0 0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0 0 0 0 0 0 0 1 0 0 0

))))))))))))))

)

(9)

where 11987130

= 12252048 11987131

= minus2452048 11987132

= 492048and 119871

33= minus52048 The some of its eigenvalues is

120582 = 11

21

41

81

161

321

641

128 (10)

For an eigenvalue 120582 the eigenvectors 120585 and 120578 that satisfy 119878120585 =120582120585 and 120578119878119879 = 120578120582 are called right and left eigenvectors of thematrix 119878 respectively Some of the normalized left and righteigenvectors corresponding to eigenvalues are

1205850= (1 1 1 1 1 1 1 1 1 1 1 1 1)

119879

1205780= (0 0 0 0 0 0 1 0 0 0 0 0 0)

119879

1205851= (minus6 minus5 minus4 minus3 minus2 minus1 0 1 2 3 4 5 6)

119879

1205781= (

1

5946360) (minus5 1024 13225 minus199680 1141695

minus 4715520 0 4715520 minus1141695

199680 minus13225 minus1024 5)119879

1205852= (36 25 16 9 4 1 0 1 4 9 16 25 36)

119879

1205782= (

1

34546860) (275 minus28160 minus182613 2607616

minus 12053651 45634048 minus71955030

45634048 minus12053651 2607616

minus182613 minus28160 275)119879

1205853= (minus216 minus125 minus64 minus27 minus8 minus1 0 1 8 27 64 125 216)

119879

1205783= (

1

15039360) (225 minus11520 10952 476928

minus 3047987 4677632 0 minus4677632

3047987 minus476928 minus10952

11520 minus225)119879

(11)

Since 120585119879119894120578119895= 1 for 119894 = 119895 and 0 otherwise then by using

Lemmas 2 and 3 we get the following result

Lemma 5 The fundamental solution (Cardinal basis) 120601(119909) isthrice continuously differentiable and supported on (minus7 7) andits derivatives at integers are given by

1206011015840(119894) = 2 sign (119894) 119890119879

|119894|1205781

12060110158401015840(119894) = 2

2119890119879

|119894|1205782

120601101584010158401015840(119894) = 2

3 sign (119894) 119890119879|119894|1205783

minus 6 le 119894 le 6

(12)

where

1198900= (0 0 0 0 0 0 1 0 0 0 0 0 0)

119879

1198901= (0 0 0 0 0 1 0 0 0 0 0 0 0)

119879

1198902= (0 0 0 0 1 0 0 0 0 0 0 0 0)

119879

1198903= (0 0 0 1 0 0 0 0 0 0 0 0 0)

119879

1198904= (0 0 1 0 0 0 0 0 0 0 0 0 0)

119879

1198905= (0 1 0 0 0 0 0 0 0 0 0 0 0)

119879

1198906= (1 0 0 0 0 0 0 0 0 0 0 0 0)

119879

1206011015840(0) = 0 120601

1015840(plusmn1) = ∓

78592

49553

1206011015840(plusmn2) = plusmn

76113

198212 120601

1015840(plusmn3) = ∓

3328

49553

1206011015840(plusmn4) = plusmn

2645

594636 120601

1015840(plusmn5) = plusmn

256

743295

1206011015840(plusmn6) =

1

594636

4 Abstract and Applied Analysis

12060110158401015840(0) = minus

342643

41124 120601

10158401015840(plusmn1) =

5704256

1079505

12060110158401015840(plusmn2) = minus

12053651

8636040 120601

10158401015840(plusmn3) =

325952

1079505

12060110158401015840(plusmn4) = minus

60871

2878680 120601

10158401015840(plusmn5) = minus

704

215901

12060110158401015840(plusmn6) =

55

1727208

120601101584010158401015840(0) = 0 120601

101584010158401015840(plusmn1) = ∓

292352

117495

120601101584010158401015840(plusmn2) = plusmn

3047987

1879920 120601

101584010158401015840(plusmn3) = ∓

3312

13055

120601101584010158401015840(plusmn4) = ∓

1369

234990 120601

101584010158401015840(plusmn5) = plusmn

16

2611

120601101584010158401015840(plusmn6) = ∓

5

41776

(13)

The graphical representations of the basis limit functiondefined on cardinal data and its derivatives up to order threefor 119899 = 3 are shown in Figure 1 Figure 1(a) represents thebasis limit function defined in (4) Graphical representationsof first second and third derivatives of basis limit functionsobtained from (5) for 119899 = 3 are shown in Figures 1(b) 1(c)and 1(d) at 119894 = 0 1 minus1 respectively The numeric values offirst second and third derivative are given in (13)

3 Numerical InterpolatingCollocation Algorithm

In this section first we formulate a numerical interpolat-ing collocation algorithm for linear third order two-pointboundary value problemsThenwe settle down the boundaryconditions to get unique solution

31 The Collocation Algorithm Let 119873 be a positive integer(119873 ge 6) ℎ = 1119873 and 119909

119894= 119894119873 = 119894ℎ 119894 = 0 1 2 119873

and set 119886119894= 119886(119909

119894) 119887119894= 119887(119909119894) Let

119885 (119909) =

119873+6

sum

119894=minus6

119911119894120601(

119909 minus 119909119894

ℎ) 0 le 119909 le 1 (14)

be the approximate solution to (1) where 119911119894 are the

unknown to be determined by (1)The collocation algorithmtogether with the boundary conditions to be discussed isgiven by setting

119885101584010158401015840(119909119895) = 119886 (119909

119895)119885 (119909

119895) + 119887 (119909

119895) 119895 = 0 1 2 119873

(15)

where

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ)

11988510158401015840(119909119895) =

1

ℎ2

119873+6

sum

119894=minus6

11991111989412060110158401015840(

119909119895minus 119909119894

ℎ)

119885101584010158401015840(119909119895) =

1

ℎ3

119873+6

sum

119894=minus6

119911119894120601101584010158401015840(

119909119895minus 119909119894

ℎ)

(16)

Using (14) and (16) in (15) we get following119873 + 1 system ofequations

119873+6

sum

119894=minus6

119911119894120601101584010158401015840(

119909119895minus 119909119894

ℎ) minus ℎ3119886119895

119873+6

sum

119894=minus6

119911119894120601(

119909119895minus 119909119894

ℎ) = ℎ

3119887119895

119895 = 0 1 2 119873

(17)

Now we simplify the above system in the followingtheorems

Theorem 6 For 119895 = 0 by (17) one gets

119911minus6120601101584010158401015840

6+ 119911minus5120601101584010158401015840

5+ 119911minus4120601101584010158401015840

4+ 119911minus3120601101584010158401015840

3+ 119911minus2120601101584010158401015840

2

+ 119911minus1120601101584010158401015840

1+ 11991101199020+ 1199111120601101584010158401015840

minus1+ 1199112120601101584010158401015840

minus2+ 1199113120601101584010158401015840

minus3

+ 1199114120601101584010158401015840

minus4+ 1199115120601101584010158401015840

minus5+ 1199116120601101584010158401015840

minus6= ℎ31198870

(18)

where 120601101584010158401015840119895= 120601101584010158401015840(119895) and 119902

0= 120601101584010158401015840

0minus 1198860ℎ3

Proof Substituting 119895 = 0 in (17) we get

119911minus6120601101584010158401015840(1199090minus 119909minus6

ℎ) + 119911minus5120601101584010158401015840(1199090minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601101584010158401015840(1199090minus 119909119873+5

ℎ) + 119911119873+6

120601101584010158401015840(1199090minus 119909119873+6

ℎ)

minus 1198860ℎ3119911minus6120601(

1199090minus 119909minus6

ℎ) + 119911minus5120601(

1199090minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601(1199090minus 119909119873+5

ℎ) + 119911119873+6

120601(1199090minus 119909119873+6

ℎ)

= ℎ31198870

(19)

For 119909119895= 119895ℎ 119895 = 0 1 2 119873 this implies

119911minus6120601101584010158401015840(6) + 119911

minus5120601101584010158401015840(5) + sdot sdot sdot + 119911

119873+5120601101584010158401015840(minus119873 minus 5)

+ 119911119873+6

120601101584010158401015840(minus119873 minus 6)

minus 1198860ℎ3119911minus6120601 (6) + 119911

minus5120601 (5) + sdot sdot sdot

+119911119873+5

120601 (minus119873 minus 5) + 119911119873+6

120601 (minus119873 minus 6) = ℎ31198870

(20)

Abstract and Applied Analysis 5

1

08

06

04

02

0

minus4 minus2 0 2 4 6

(a)

minus4minus6 minus2 0 2 4 6

15

1

05

0

minus05

minus1

minus15

(b)

1

0

minus1

minus2

minus4minus6 minus2 0 2 4 6

(c)

minus4minus2 0 2 4 6

03

02

01

0

minus01

minus02

minus03

(d)

Figure 1 Interpolatory basis function 1206013(119894) is shown in (a) first derivative of 120601

3(119894) i-e 1206011015840

3(119894) shown in (b) second derivative of 120601

3(119894) i-e 12060110158401015840

3(119894)

shown in (c) and third derivative of 1206013(119894) i-e 120601101584010158401015840

3(119894) shown in (d) respectively

Since the support of basis function 120601(119909) is (minus7 7) 1206011015840(119909)12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval (minus7 7) also

by (4) and (13) we get

119911minus6120601101584010158401015840(6) + 119911

minus5120601101584010158401015840(5) + 119911

minus4120601101584010158401015840(4) + 119911

minus3120601101584010158401015840(3)

+ 119911minus2120601101584010158401015840(2) + 119911

minus1120601101584010158401015840(1) + 119911

0120601101584010158401015840(0)

+ 1199111120601101584010158401015840(minus1) + 119911

2120601101584010158401015840(minus2) + 119911

3120601101584010158401015840(minus3)

+ 1199114120601101584010158401015840(minus4) + 119911

5120601101584010158401015840(minus5) + 119911

6120601101584010158401015840(minus6)

minus 1198860ℎ31199110120601 (0) = ℎ

31198870

(21)

If 120601101584010158401015840119895= 120601101584010158401015840(119895) then

119911minus6120601101584010158401015840

6+ 119911minus5120601101584010158401015840

5+ 119911minus4120601101584010158401015840

4+ 119911minus3120601101584010158401015840

3+ 119911minus2120601101584010158401015840

2

+ 119911minus1120601101584010158401015840

1+ 1199110(120601101584010158401015840

0minus 1198860ℎ3) + 1199111120601101584010158401015840

minus1+ 1199112120601101584010158401015840

minus2

+ 1199113120601101584010158401015840

minus3+ 1199114120601101584010158401015840

minus4+ 1199115120601101584010158401015840

minus5+ 1199116120601101584010158401015840

minus6= ℎ31198870

(22)

For 1199020= 120601101584010158401015840

0minus1198860ℎ3 we get (18)This completes the proof

Theorem 7 For 119895 = 1 2 3 119873 the system (17) is equivalentto

119911minus6120601101584010158401015840

119895+6+ 119911minus5120601101584010158401015840

119895+5+ sdot sdot sdot + 119911

0120601101584010158401015840

119895

+ 1199111(120601101584010158401015840

119895minus1minus 119886119895ℎ3120601119895minus1) + 1199112(120601101584010158401015840

119895minus2minus 119886119895ℎ3120601119895minus2) + sdot sdot sdot

+ 119911119873minus1

(120601101584010158401015840

119895minus119873+1minus 119886119895ℎ3120601119895minus119873+1

) + 119911119873(120601101584010158401015840

119895minus119873minus 119886119895ℎ3120601119895minus119873

)

+ 119911119873+1

120601101584010158401015840

119895minus119873minus1+ sdot sdot sdot + 119911

119873+6120601101584010158401015840

119895minus119873minus6= ℎ3119887119895

(23)

Proof By expanding (17) we get

119911minus6120601101584010158401015840(

119909119895minus 119909minus6

ℎ) + 119911minus5120601101584010158401015840(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601101584010158401015840(

119909119895minus 119909119873+6

ℎ)

minus 119886119895ℎ3119911minus6120601(

119909119895minus 119909minus6

ℎ) + 119911minus5120601(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601(

119909119895minus 119909119873+6

ℎ)

= ℎ3119887119895

(24)

For 119909119895= 119895ℎ 119895 = 1 2 119873 we get

119911minus6120601101584010158401015840(119895 + 6) + 119911

minus5120601101584010158401015840(119895 + 5) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(119895 minus 119873 minus 5) + 119911

119873+6120601101584010158401015840(119895 minus 119873 minus 6)

minus 119886119895ℎ3119911minus6120601 (119895 + 6) + 119911

minus5120601 (119895 + 5) + sdot sdot sdot

+119911119873+5

120601 (119895 minus 119873 minus 5) + 119911119873+6

120601 (119895 minus 119873 minus 6)

= ℎ3119887119895

(25)

6 Abstract and Applied Analysis

If 120601101584010158401015840119895= 120601101584010158401015840(119895) for 119895 = 1 2 119873 then

119911minus6(120601101584010158401015840

119895+6minus 119886119895ℎ3120601119895+6) + 119911minus5(120601101584010158401015840

119895+5minus 119886119895ℎ3120601119895+5)

+ sdot sdot sdot + 119911119873+5

(120601101584010158401015840

119895minus119873minus5minus 119886119895ℎ3120601119895minus119873minus5

)

+ 119911119873+6

(120601101584010158401015840

119895minus119873minus6minus 119886119895ℎ3120601119895minus119873minus6

) = ℎ3119887119895

(26)

Since 1206011015840(119909) 12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval(minus7 7) then by (4) and (13) we get (23)

From (18) and (23) we get following undeterminedsystem of (119873 + 1) equations with (119873 + 13) unknowns 119911

119894

119860119885 = 119863 (27)

where the matrices 119860 119885 and119863 of orders (119873 + 1) times (119873 + 13)119873 + 13 and119873 + 1 respectively are given by

119860 =

(((

(

120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199020

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199021

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199022

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

0 0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199023

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus5120601101584010158401015840

minus60

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6

)))

)

(28)

119885 = (119911minus6 119911minus5 119911minus4 119911minus3 119911minus2 119911

119873+6)119879 and 119863 = (119887

0ℎ3 1198871ℎ3

1198872ℎ3 1198873ℎ3 119887

119873ℎ3)119879 where120601101584010158401015840

119895= 120601101584010158401015840(119895) and 119902

119895= 120601101584010158401015840

0minus119886119895ℎ3

32 Adjustment of Boundary Conditions The order of thecoefficient matrix (28) is (119873 + 1) times (119873 + 13) In orderto get unique solution of the system we need twelve moreconditions Here we consider only two different cases Incoming section we will show that the approximate solutioncan be improved by adjusting different boundary conditions

Case 1 If we assume 11991110158400= 0 (equivalently 119911

0= 119910119903= finite)

then two conditions can be achieved by using following givenboundary conditions i-e

1199110= 119910119903 119911

1015840

0= 0 119911

119873= 119910119897 (29)

Still we need ten more conditions to get stable system Sincesubdivision scheme reproduces seven degree polynomials wedefine boundary conditions of order eight for solution of (27)For simplicity only the left end points are discussed and thevalues of right end points 119911

119873+1 119911119873+2

119911119873+3

119911119873+4

119911119873+5

can betreated similarly

The values 119911minus5 119911minus4 119911minus3 119911minus2 119911minus1

can be determined bythe septic polynomial 119902(119909) interpolating at (119909

119894 119911119894) 0 le 119894 le 7

Precisely we have

119911minus119894= 119902 (minus119909

119894) 119894 = 1 2 3 4 5 (30)

where

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119885(119909119894minus119895) (31)

Since by (14) 119885(119909119894) = 119911119894for 119894 = 1 2 3 4 5 then by replacing

119909119894by minus119909

119894 we have

119902 (minus119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911minus119894+119895

(32)

Hence the following boundary conditions can be employed atthe left end

8

sum

119895=0

(8

119895) (minus1)

119895119911minus119894+119895

= 0 119894 = 5 4 3 2 1 (33)

Similarly for the right end we can define 119911119894= 119902(minus119909

119894) 119894 =

119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5 and

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911119894minus119895 (34)

So we have the following boundary conditions at the rightend

8

sum

119895=0

(8

119895) (minus1)

119895119911119894minus119895

= 0

119894 = 119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5

(35)

Finally we get the following new system of (119873 + 13) linearequations with (119873 + 13) unknowns 119911

119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) and (35)

119861119885 = 119877 (36)

where the coefficients matrix 119861 = (1198611198790 119860119879 119861119879

1)119879 119860 is defined

by (28) and 1198610and 119861

1are formed by (29) (33) and (35)

Abstract and Applied Analysis 7

1198610=

(((((((

(

0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 sdot sdot sdot 0 0

0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 sdot sdot sdot 0 0

0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 sdot sdot sdot 0 0

0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 sdot sdot sdot 0 0

0 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 sdot sdot sdot 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0 0 sdot sdot sdot 0 0

)))))))

)

(37)

where the first five rows of 1198610come from (33) and the sixth

row comes from (29) at 1199110= 119910119903 Consider

1198611=

(((((((

(

0 0 sdot sdot sdot 0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 sdot sdot sdot 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 0

0 0 sdot sdot sdot 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0

0 0 sdot sdot sdot 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0

0 0 sdot sdot sdot 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0

0 0 sdot sdot sdot 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0

)))))))

)

(38)

where first row of 1198611comes from (29) at 119911

119873= 119910119897remaining

rows come from (35) and the matrices119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 0 119910119903 119863119879 119910119897 0 0 0 0 0)

119879

(39)

Case 2 In this case we express the given boundary condition1199111015840

0= 0 in the following wayBy using (16) we have

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ) (40)

As we defined earlier 119909119895= 119895ℎ if we put 119895 = 0 we get 119909

0= 0

the above equation can be written as

1198851015840(0) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) (41)

Since by boundary condition 11991110158400= 1198851015840(0) = 0

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) = 0 (42)

By using (13) we can express above equation as

1

594636119911minus6+

256

743295119911minus5+

2645

594636119911minus4minus3328

49553119911minus3

+76113

198212119911minus2minus78592

49553119911minus1+78592

495531199111minus76113

1982121199112

+3328

495531199113minus

2645

5946361199114minus

256

7432951199115minus

1

5946361199116= 0

(43)Finally we get a following new system of (119873 + 13) linear

equations with (119873 + 13) unknowns 119911119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) for 119894 = 5 4 3 2 (35) and (43)

119861119885 = 119877 (44)

where the coefficients matrix 119861 = (B1198790 119860119879 119861119879

1)119879 119860 is defined

by (28) andB0and 119861

1formed by (29) (33) (35) and (43) are

defined as

8 Abstract and Applied Analysis

B0= (

(

0 1 minus8 28 minus56 70 minus56 28 minus8

0 0 1 minus8 28 minus56 70 minus56 28

0 0 0 1 minus8 28 minus56 70 minus56

0 0 0 0 1 minus8 28 minus56 70

1

594636

256

743295

2645

594636minus3328

49553

76113

198212minus78592

495530

78592

49553

76113

198212

0 0 0 0 0 0 1 0 0

1 0 0 0 0 sdot sdot sdot 0 0

minus8 1 0 0 0 sdot sdot sdot 0 0

28 minus8 1 0 0 sdot sdot sdot 0 0

minus56 28 minus8 1 0 sdot sdot sdot 0 0

3328

49553minus2645

594636minus

256

743295minus

1

5946360 sdot sdot sdot 0 0

0 0 0 0 0 sdot sdot sdot 0 0

)

)

(45)

in B first four rows come from (33) for 119894 = 5 4 3 2 fifth rowcomes from (43) and the last row comes from (29)

The matrix 1198611is same as defined in Case 1 and the

matrices 119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 1199101015840(0) 119910

119903 119863119879 119910119897 0 0 0 0 0)

119879

(46)

The nonsingularity of the coefficients matrix 119861 has beendiscussed in next section

33 Existence of the Solution In this section we discuss thenonsingularity of the coefficients matrix 119861 We observe thatthe coefficients matrix 119861 is neither symmetric nor diagonallydominant However it can be shown that 119861 is a nonsingularSince 119861 is almost a band matrix with half band width 7numerical complexity for solving the linear system usingGaussian elimination is about 49(119873 + 9)multiplications Forlarge 119873 the matrix is almost symmetric except the first andlast six rows and columns due to the boundary conditionsTherefore we first consider the symmetric part of it that issquare symmetric matrix 119862 of order119873 + 3 defined as

119862 =(

(

120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1

)

)

(47)

It can be shown that 119862 is always nonsingular for each valueof 119873 However 119861 is nonsingular for 119873 ⩽ 1000 We havechecked the nonsingularity of matrix 119861 by different methodsIn first method we observe that the determinants of matrix 119861increase as 119873 increases and it is not zero for 119873 ⩽ 1000 So119861 is nonsingular The determinants of 119861 at some values of119873are shown in Table 1 In second method we observe that for119873 ⩽ 1000 the eigenvalues of matrix 119861 are nonzero so by [16]matrix119861 is nonsingular However for119873 gt 1000matrix119861mayor may not be nonsingular Therefore we claim that systems(36) and (44) are stable for119873 ⩽ 1000

4 Error Estimation

In this section we discuss the approximation properties ofthe numerical interpolating collocation algorithm Since thescheme (8) reproduces polynomial curve of degree seven soby Dyn [14] scheme has approximation order eight So thecollocation algorithm (14) with septic precision treatments at

the endpoints has the power of approximation 119874(ℎ2) Herewe present our main result for error estimation

Proposition 8 Suppose the exact solution 119910(119909) isin 1198628[0 1]

and 119911119894 are obtained by solving either (36) or (44) with 8th

order boundary condition at the end points then

err (119909)infin =10038171003817100381710038171003817119885119895minus 11991011989510038171003817100381710038171003817infin

= 119874 (ℎ2minus119895) 119895 = 0 1 2 (48)

where 119895 denotes the order of derivative

Proof Since the order of approximation of subdivisionscheme (8) is eight so by using (13) we can write for smoothfunction 119910(119909) and small ℎ as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910 (119909119895minus 6ℎ) minus 11520119910 (119909

119895minus 5ℎ)

+ 10952119910 (119909119895minus 4ℎ) + 476928119910 (119909

119895minus 3ℎ)

Abstract and Applied Analysis 9

minus 3047987119910 (119909119895minus 2ℎ) + 4677632119910 (119909

119895minus ℎ)

minus 4677632119910 (119909119895+ ℎ) + 3047987119910 (119909

119895+ 2ℎ)

minus 476928119910 (119909119895+ 3ℎ) minus 10952119910 (119909

119895+ 4ℎ)

+ 11520119910 (119909119895+ 5ℎ) minus225119910 (119909

119895+ 6ℎ)

(49)

This can be written as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910119895minus6

minus 11520119910119895minus5

+ 10952119910119895minus4

+ 476928119910119895minus3

minus 3047987119910119895minus2

+ 4677632119910119895minus1

minus 4677632119910119895+1

+ 3047987119910119895+2

minus 476928119910119895+3

minus10952119910119895+4

+ 11520119910119895+5

minus 225119910119895+6

(50)

Similarly we have

119885101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119911119895minus6

minus 11520119911119895minus5

+ 10952119911119895minus4

+ 476928119911119895minus3

minus 3047987119911119895minus2

+ 4677632119911119895minus1

minus 4677632119911119895+1

+ 3047987119911119895+2

minus 476928119911119895+3

minus10952119911119895+4

+ 11520119911119895+5

minus 225119911119895+6

(51)

If we define error function 119890(119909) = 119885(119909) minus 119910(119909) and errorvectors at the nodes by

119890 (119909119895) = 119885 (119909

119895) minus 119910 (119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 (52)

or equivalently 119890119895= 119885119895minus 119910(119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 then

this implies

1198901015840

119895= 1198851015840

119895minus 1199101015840(119909 + 119895ℎ)

11989010158401015840

119895= 11988510158401015840

119895minus 11991010158401015840(119909 + 119895ℎ)

119890101584010158401015840

119895= 119885101584010158401015840

119895minus 119910101584010158401015840(119909 + 119895ℎ)

(53)

Table 1 Determinants of the matrices

119873 119862 119861

10 minus866756 11048870018371741

50 minus177183 14981270309

100 minus552709050 11964492

500 minus5033491471916955 times 1036

4728852755761116 times 1021

1000 minus4477989536166907 times 1071

42069711017699999 times 1040

By subtracting (50) from (51) we get

119885101584010158401015840(119909119895) minus 119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225 (119911119895minus6

minus 119910119895minus6) minus 11520 (119911

119895minus5minus 119910119895minus5)

+ 10952 (119911119895minus4

minus 119910119895minus4) + 476928 (119911

119895minus3minus 119910119895minus3)

minus 3047987 (119911119895minus2

minus 119910119895minus2) + 4677632 (119911

119895minus1minus 119910119895minus1)

minus 4677632 (119911119895+1

minus 119910119895+1) + 3047987 (119911

119895+2minus 119910119895+2)

minus 476928 (119911119895+3

minus 119910119895+3) minus 10952 (119911

119895+4minus 119910119895+4)

+11520 (119911119895+5

minus 119910119895+5) minus 225 (119911

119895+6minus 119910119895+6)

(54)

This implies

119890101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119890119895minus6

minus 11520119890119895minus5

+ 10952119890119895minus4

+ 476928119890119895minus3

minus 3047987119890119895minus2

+ 4677632119890119895minus1

minus 4677632119890119895+1

+ 3047987119890119895+2

minus 476928119890119895+3

minus10952119890119895+4

+ 11520119890119895+5

minus 225119890119895+6

(55)

By Lemma 5 we get the following expression

119890101584010158401015840

119895=1

ℎ3120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 120601101584010158401015840

0119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6 + 119874 (ℎ

8)

(56)

where 119895 = 0 1 2 119873By subtracting (1) from (15) we get

119885101584010158401015840

119895minus 119910101584010158401015840

119895= 119886119895(119885119894minus 119884119895) (57)

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

to compute the approximate solution of two-point third orderboundary value problems Particularly collocation algorithmwith septic precision treatments at the endpoints has thepower of approximation119874(ℎ2) Our reformulated collocationalgorithm treats the following type of two-point third orderboundary value problem

119910101584010158401015840(119909) = 119886 (119909) 119910 (119909) + 119887 (119909) 0 le 119909 le 1

119910 (0) = 119910119903 119910

1015840(0) = 0 119910 (1) = 119910

119897

(1)

where 119886(119909) and 119887(119909) are continuous and 119886(119909) ge 0 on [0 1]The outline of the paper is as follows In Section 2 we

rewrite general form of interpolating subdivision scheme forcurve design [12] and some related resultsThe 8-point binaryinterpolating scheme and derivatives of its basis functionhave also been discussed in this section In Section 3 anumerical interpolating algorithm of collocation to solve (1)is formulated and its boundary treatments are discussedIn Section 4 approximation properties of this algorithmare given In Section 5 numerical examples are presentedSection 6 is devoted for conclusion and the possible futureresearch directions

2 Interpolating Schemes for Curve Design

A general compact form of symmetric univariate binaryinterpolating subdivision scheme [12] which maps polygon119901119896= 119901119896

119894119894isinZ to a refined polygon 119901119896+1 = 119901119896+1

119894119894isinZ is defined

by

119901119896+1

2119894= 119901119896

119894

119901119896+1

2119894+1=

119899

sum

119895=0

119871119899119895(119901119896

119894minus119895+ 119901119896

119894+119895+1)

(2)

where 119899 is called the degree of the scheme and the constantsare given by

119871119899119895=

((2119899 + 1))2

2 (4119899) (2119899 + 1)

(minus1)119895

(2119895 + 1)(2119899 + 1

119899 minus 119895)

119895 = 0 1 2 119899

(3)

where ( 2119899+1119899minus119895 ) denotes the binomial coefficientThe boundary treatments are necessary to produce

smooth curve segments by scheme (2) Normally higherorder approximation formulae should be used near the endsof the segments and thus Lagrange formulae of order 2119899 + 1are recommended

Remark 1 Let 120601(119909) be the limit curve generated from thecardinal data 119901

119894= (119894 120575

0)119879 that is 120601(119909) is the fundamental

solution of the subdivision scheme (2) then

120601 (119894) = 1 119894 = 0

0 119894 = 0(4)

Furthermore 120601(119909) satisfies the following two-scale equation

120601 (119909) = 120601119899(119909)

= 120601 (2119909) +

119899

sum

119895=minus119899

119871119899|119895|

120601 (2119909 minus 2119895 + 1) 119909 isin R(5)

Lemma 2 (see [12 13]) The support of the fundamentalsolution 120601

119899(119909) to scheme (2) is finite Explicitly supp120601

119899(119909) =

(minus2119899 minus 1 2119899 + 1)

Lemma 3 (see [10]) Given a square matrix 119860 of order 119899 letthe normalized left and right (generalized) eigenvectors of119860 bedenoted by 120578

119894 120585119894 Then for any vector 119891 isin R119899 there exists

following Fourier expansion

119891 =

119899

sum

119894=1

(119891119879120578119894) 120585119894 (6)

Lemma4 (see [10]) Suppose119865(119905) 119905 isin R is a regular and1198622119899+2curve inR119898119898 ge 2 Let119875(119905) 119905 isin R be the limit curve generatedby (2) from the initial data 119875

119894= 119865(119894ℎ) 119894 isin Z 0 lt ℎ lt 1 Then

on any finite interval [119886 119887] the following estimates hold

1003817100381710038171003817119865 (ℎ119905) minus 119901 (119905)1003817100381710038171003817infin

le1198722119899+2

(119865)

(2119899 + 2)ℎ2119899+2

= 119874 (ℎ2119899+2

)

10038171003817100381710038171003817ℎ119895119865119895(ℎ119905) minus 119901

119895(119905)10038171003817100381710038171003817infin

= 119874 (ℎ2119899+2minus119895

)

119895 = 0 1 2 119899 + 2

2

(7)

where the number1198722119899+2

(119865) depends only on the derivatives of119865(119905) and 119899

21 8-Point Interpolating Scheme For 119899 = 3 by (2) and (3) wehave the following 8-point binary interpolating subdivisionscheme for curve design

119901119896+1

2119894= 119901119896

119894

119901119896+1

2119894+1=1225

2048(119901119896

119894minus 119901119896

119894+1) minus

245

2048(119901119896

119894minus1minus 119901119896

119894+2)

+49

2048(119901119896

119894minus2minus 119901119896

119894+3) minus

5

2048(119901119896

119894minus3minus 119901119896

119894+4)

(8)

This scheme is 1198623-continuous in [14] and reproduces poly-nomial curve of degree seven in [15] The local subdivisionmatrix of (8) is denoted and defined by

Abstract and Applied Analysis 3

119878 =

((((((((((((((

(

0 0 0 1 0 0 0 0 0 0 0 0 0

11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0 0

0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0

0 0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0

0 0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0 0 0 0 0 0 0 1 0 0 0

))))))))))))))

)

(9)

where 11987130

= 12252048 11987131

= minus2452048 11987132

= 492048and 119871

33= minus52048 The some of its eigenvalues is

120582 = 11

21

41

81

161

321

641

128 (10)

For an eigenvalue 120582 the eigenvectors 120585 and 120578 that satisfy 119878120585 =120582120585 and 120578119878119879 = 120578120582 are called right and left eigenvectors of thematrix 119878 respectively Some of the normalized left and righteigenvectors corresponding to eigenvalues are

1205850= (1 1 1 1 1 1 1 1 1 1 1 1 1)

119879

1205780= (0 0 0 0 0 0 1 0 0 0 0 0 0)

119879

1205851= (minus6 minus5 minus4 minus3 minus2 minus1 0 1 2 3 4 5 6)

119879

1205781= (

1

5946360) (minus5 1024 13225 minus199680 1141695

minus 4715520 0 4715520 minus1141695

199680 minus13225 minus1024 5)119879

1205852= (36 25 16 9 4 1 0 1 4 9 16 25 36)

119879

1205782= (

1

34546860) (275 minus28160 minus182613 2607616

minus 12053651 45634048 minus71955030

45634048 minus12053651 2607616

minus182613 minus28160 275)119879

1205853= (minus216 minus125 minus64 minus27 minus8 minus1 0 1 8 27 64 125 216)

119879

1205783= (

1

15039360) (225 minus11520 10952 476928

minus 3047987 4677632 0 minus4677632

3047987 minus476928 minus10952

11520 minus225)119879

(11)

Since 120585119879119894120578119895= 1 for 119894 = 119895 and 0 otherwise then by using

Lemmas 2 and 3 we get the following result

Lemma 5 The fundamental solution (Cardinal basis) 120601(119909) isthrice continuously differentiable and supported on (minus7 7) andits derivatives at integers are given by

1206011015840(119894) = 2 sign (119894) 119890119879

|119894|1205781

12060110158401015840(119894) = 2

2119890119879

|119894|1205782

120601101584010158401015840(119894) = 2

3 sign (119894) 119890119879|119894|1205783

minus 6 le 119894 le 6

(12)

where

1198900= (0 0 0 0 0 0 1 0 0 0 0 0 0)

119879

1198901= (0 0 0 0 0 1 0 0 0 0 0 0 0)

119879

1198902= (0 0 0 0 1 0 0 0 0 0 0 0 0)

119879

1198903= (0 0 0 1 0 0 0 0 0 0 0 0 0)

119879

1198904= (0 0 1 0 0 0 0 0 0 0 0 0 0)

119879

1198905= (0 1 0 0 0 0 0 0 0 0 0 0 0)

119879

1198906= (1 0 0 0 0 0 0 0 0 0 0 0 0)

119879

1206011015840(0) = 0 120601

1015840(plusmn1) = ∓

78592

49553

1206011015840(plusmn2) = plusmn

76113

198212 120601

1015840(plusmn3) = ∓

3328

49553

1206011015840(plusmn4) = plusmn

2645

594636 120601

1015840(plusmn5) = plusmn

256

743295

1206011015840(plusmn6) =

1

594636

4 Abstract and Applied Analysis

12060110158401015840(0) = minus

342643

41124 120601

10158401015840(plusmn1) =

5704256

1079505

12060110158401015840(plusmn2) = minus

12053651

8636040 120601

10158401015840(plusmn3) =

325952

1079505

12060110158401015840(plusmn4) = minus

60871

2878680 120601

10158401015840(plusmn5) = minus

704

215901

12060110158401015840(plusmn6) =

55

1727208

120601101584010158401015840(0) = 0 120601

101584010158401015840(plusmn1) = ∓

292352

117495

120601101584010158401015840(plusmn2) = plusmn

3047987

1879920 120601

101584010158401015840(plusmn3) = ∓

3312

13055

120601101584010158401015840(plusmn4) = ∓

1369

234990 120601

101584010158401015840(plusmn5) = plusmn

16

2611

120601101584010158401015840(plusmn6) = ∓

5

41776

(13)

The graphical representations of the basis limit functiondefined on cardinal data and its derivatives up to order threefor 119899 = 3 are shown in Figure 1 Figure 1(a) represents thebasis limit function defined in (4) Graphical representationsof first second and third derivatives of basis limit functionsobtained from (5) for 119899 = 3 are shown in Figures 1(b) 1(c)and 1(d) at 119894 = 0 1 minus1 respectively The numeric values offirst second and third derivative are given in (13)

3 Numerical InterpolatingCollocation Algorithm

In this section first we formulate a numerical interpolat-ing collocation algorithm for linear third order two-pointboundary value problemsThenwe settle down the boundaryconditions to get unique solution

31 The Collocation Algorithm Let 119873 be a positive integer(119873 ge 6) ℎ = 1119873 and 119909

119894= 119894119873 = 119894ℎ 119894 = 0 1 2 119873

and set 119886119894= 119886(119909

119894) 119887119894= 119887(119909119894) Let

119885 (119909) =

119873+6

sum

119894=minus6

119911119894120601(

119909 minus 119909119894

ℎ) 0 le 119909 le 1 (14)

be the approximate solution to (1) where 119911119894 are the

unknown to be determined by (1)The collocation algorithmtogether with the boundary conditions to be discussed isgiven by setting

119885101584010158401015840(119909119895) = 119886 (119909

119895)119885 (119909

119895) + 119887 (119909

119895) 119895 = 0 1 2 119873

(15)

where

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ)

11988510158401015840(119909119895) =

1

ℎ2

119873+6

sum

119894=minus6

11991111989412060110158401015840(

119909119895minus 119909119894

ℎ)

119885101584010158401015840(119909119895) =

1

ℎ3

119873+6

sum

119894=minus6

119911119894120601101584010158401015840(

119909119895minus 119909119894

ℎ)

(16)

Using (14) and (16) in (15) we get following119873 + 1 system ofequations

119873+6

sum

119894=minus6

119911119894120601101584010158401015840(

119909119895minus 119909119894

ℎ) minus ℎ3119886119895

119873+6

sum

119894=minus6

119911119894120601(

119909119895minus 119909119894

ℎ) = ℎ

3119887119895

119895 = 0 1 2 119873

(17)

Now we simplify the above system in the followingtheorems

Theorem 6 For 119895 = 0 by (17) one gets

119911minus6120601101584010158401015840

6+ 119911minus5120601101584010158401015840

5+ 119911minus4120601101584010158401015840

4+ 119911minus3120601101584010158401015840

3+ 119911minus2120601101584010158401015840

2

+ 119911minus1120601101584010158401015840

1+ 11991101199020+ 1199111120601101584010158401015840

minus1+ 1199112120601101584010158401015840

minus2+ 1199113120601101584010158401015840

minus3

+ 1199114120601101584010158401015840

minus4+ 1199115120601101584010158401015840

minus5+ 1199116120601101584010158401015840

minus6= ℎ31198870

(18)

where 120601101584010158401015840119895= 120601101584010158401015840(119895) and 119902

0= 120601101584010158401015840

0minus 1198860ℎ3

Proof Substituting 119895 = 0 in (17) we get

119911minus6120601101584010158401015840(1199090minus 119909minus6

ℎ) + 119911minus5120601101584010158401015840(1199090minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601101584010158401015840(1199090minus 119909119873+5

ℎ) + 119911119873+6

120601101584010158401015840(1199090minus 119909119873+6

ℎ)

minus 1198860ℎ3119911minus6120601(

1199090minus 119909minus6

ℎ) + 119911minus5120601(

1199090minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601(1199090minus 119909119873+5

ℎ) + 119911119873+6

120601(1199090minus 119909119873+6

ℎ)

= ℎ31198870

(19)

For 119909119895= 119895ℎ 119895 = 0 1 2 119873 this implies

119911minus6120601101584010158401015840(6) + 119911

minus5120601101584010158401015840(5) + sdot sdot sdot + 119911

119873+5120601101584010158401015840(minus119873 minus 5)

+ 119911119873+6

120601101584010158401015840(minus119873 minus 6)

minus 1198860ℎ3119911minus6120601 (6) + 119911

minus5120601 (5) + sdot sdot sdot

+119911119873+5

120601 (minus119873 minus 5) + 119911119873+6

120601 (minus119873 minus 6) = ℎ31198870

(20)

Abstract and Applied Analysis 5

1

08

06

04

02

0

minus4 minus2 0 2 4 6

(a)

minus4minus6 minus2 0 2 4 6

15

1

05

0

minus05

minus1

minus15

(b)

1

0

minus1

minus2

minus4minus6 minus2 0 2 4 6

(c)

minus4minus2 0 2 4 6

03

02

01

0

minus01

minus02

minus03

(d)

Figure 1 Interpolatory basis function 1206013(119894) is shown in (a) first derivative of 120601

3(119894) i-e 1206011015840

3(119894) shown in (b) second derivative of 120601

3(119894) i-e 12060110158401015840

3(119894)

shown in (c) and third derivative of 1206013(119894) i-e 120601101584010158401015840

3(119894) shown in (d) respectively

Since the support of basis function 120601(119909) is (minus7 7) 1206011015840(119909)12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval (minus7 7) also

by (4) and (13) we get

119911minus6120601101584010158401015840(6) + 119911

minus5120601101584010158401015840(5) + 119911

minus4120601101584010158401015840(4) + 119911

minus3120601101584010158401015840(3)

+ 119911minus2120601101584010158401015840(2) + 119911

minus1120601101584010158401015840(1) + 119911

0120601101584010158401015840(0)

+ 1199111120601101584010158401015840(minus1) + 119911

2120601101584010158401015840(minus2) + 119911

3120601101584010158401015840(minus3)

+ 1199114120601101584010158401015840(minus4) + 119911

5120601101584010158401015840(minus5) + 119911

6120601101584010158401015840(minus6)

minus 1198860ℎ31199110120601 (0) = ℎ

31198870

(21)

If 120601101584010158401015840119895= 120601101584010158401015840(119895) then

119911minus6120601101584010158401015840

6+ 119911minus5120601101584010158401015840

5+ 119911minus4120601101584010158401015840

4+ 119911minus3120601101584010158401015840

3+ 119911minus2120601101584010158401015840

2

+ 119911minus1120601101584010158401015840

1+ 1199110(120601101584010158401015840

0minus 1198860ℎ3) + 1199111120601101584010158401015840

minus1+ 1199112120601101584010158401015840

minus2

+ 1199113120601101584010158401015840

minus3+ 1199114120601101584010158401015840

minus4+ 1199115120601101584010158401015840

minus5+ 1199116120601101584010158401015840

minus6= ℎ31198870

(22)

For 1199020= 120601101584010158401015840

0minus1198860ℎ3 we get (18)This completes the proof

Theorem 7 For 119895 = 1 2 3 119873 the system (17) is equivalentto

119911minus6120601101584010158401015840

119895+6+ 119911minus5120601101584010158401015840

119895+5+ sdot sdot sdot + 119911

0120601101584010158401015840

119895

+ 1199111(120601101584010158401015840

119895minus1minus 119886119895ℎ3120601119895minus1) + 1199112(120601101584010158401015840

119895minus2minus 119886119895ℎ3120601119895minus2) + sdot sdot sdot

+ 119911119873minus1

(120601101584010158401015840

119895minus119873+1minus 119886119895ℎ3120601119895minus119873+1

) + 119911119873(120601101584010158401015840

119895minus119873minus 119886119895ℎ3120601119895minus119873

)

+ 119911119873+1

120601101584010158401015840

119895minus119873minus1+ sdot sdot sdot + 119911

119873+6120601101584010158401015840

119895minus119873minus6= ℎ3119887119895

(23)

Proof By expanding (17) we get

119911minus6120601101584010158401015840(

119909119895minus 119909minus6

ℎ) + 119911minus5120601101584010158401015840(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601101584010158401015840(

119909119895minus 119909119873+6

ℎ)

minus 119886119895ℎ3119911minus6120601(

119909119895minus 119909minus6

ℎ) + 119911minus5120601(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601(

119909119895minus 119909119873+6

ℎ)

= ℎ3119887119895

(24)

For 119909119895= 119895ℎ 119895 = 1 2 119873 we get

119911minus6120601101584010158401015840(119895 + 6) + 119911

minus5120601101584010158401015840(119895 + 5) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(119895 minus 119873 minus 5) + 119911

119873+6120601101584010158401015840(119895 minus 119873 minus 6)

minus 119886119895ℎ3119911minus6120601 (119895 + 6) + 119911

minus5120601 (119895 + 5) + sdot sdot sdot

+119911119873+5

120601 (119895 minus 119873 minus 5) + 119911119873+6

120601 (119895 minus 119873 minus 6)

= ℎ3119887119895

(25)

6 Abstract and Applied Analysis

If 120601101584010158401015840119895= 120601101584010158401015840(119895) for 119895 = 1 2 119873 then

119911minus6(120601101584010158401015840

119895+6minus 119886119895ℎ3120601119895+6) + 119911minus5(120601101584010158401015840

119895+5minus 119886119895ℎ3120601119895+5)

+ sdot sdot sdot + 119911119873+5

(120601101584010158401015840

119895minus119873minus5minus 119886119895ℎ3120601119895minus119873minus5

)

+ 119911119873+6

(120601101584010158401015840

119895minus119873minus6minus 119886119895ℎ3120601119895minus119873minus6

) = ℎ3119887119895

(26)

Since 1206011015840(119909) 12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval(minus7 7) then by (4) and (13) we get (23)

From (18) and (23) we get following undeterminedsystem of (119873 + 1) equations with (119873 + 13) unknowns 119911

119894

119860119885 = 119863 (27)

where the matrices 119860 119885 and119863 of orders (119873 + 1) times (119873 + 13)119873 + 13 and119873 + 1 respectively are given by

119860 =

(((

(

120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199020

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199021

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199022

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

0 0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199023

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus5120601101584010158401015840

minus60

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6

)))

)

(28)

119885 = (119911minus6 119911minus5 119911minus4 119911minus3 119911minus2 119911

119873+6)119879 and 119863 = (119887

0ℎ3 1198871ℎ3

1198872ℎ3 1198873ℎ3 119887

119873ℎ3)119879 where120601101584010158401015840

119895= 120601101584010158401015840(119895) and 119902

119895= 120601101584010158401015840

0minus119886119895ℎ3

32 Adjustment of Boundary Conditions The order of thecoefficient matrix (28) is (119873 + 1) times (119873 + 13) In orderto get unique solution of the system we need twelve moreconditions Here we consider only two different cases Incoming section we will show that the approximate solutioncan be improved by adjusting different boundary conditions

Case 1 If we assume 11991110158400= 0 (equivalently 119911

0= 119910119903= finite)

then two conditions can be achieved by using following givenboundary conditions i-e

1199110= 119910119903 119911

1015840

0= 0 119911

119873= 119910119897 (29)

Still we need ten more conditions to get stable system Sincesubdivision scheme reproduces seven degree polynomials wedefine boundary conditions of order eight for solution of (27)For simplicity only the left end points are discussed and thevalues of right end points 119911

119873+1 119911119873+2

119911119873+3

119911119873+4

119911119873+5

can betreated similarly

The values 119911minus5 119911minus4 119911minus3 119911minus2 119911minus1

can be determined bythe septic polynomial 119902(119909) interpolating at (119909

119894 119911119894) 0 le 119894 le 7

Precisely we have

119911minus119894= 119902 (minus119909

119894) 119894 = 1 2 3 4 5 (30)

where

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119885(119909119894minus119895) (31)

Since by (14) 119885(119909119894) = 119911119894for 119894 = 1 2 3 4 5 then by replacing

119909119894by minus119909

119894 we have

119902 (minus119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911minus119894+119895

(32)

Hence the following boundary conditions can be employed atthe left end

8

sum

119895=0

(8

119895) (minus1)

119895119911minus119894+119895

= 0 119894 = 5 4 3 2 1 (33)

Similarly for the right end we can define 119911119894= 119902(minus119909

119894) 119894 =

119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5 and

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911119894minus119895 (34)

So we have the following boundary conditions at the rightend

8

sum

119895=0

(8

119895) (minus1)

119895119911119894minus119895

= 0

119894 = 119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5

(35)

Finally we get the following new system of (119873 + 13) linearequations with (119873 + 13) unknowns 119911

119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) and (35)

119861119885 = 119877 (36)

where the coefficients matrix 119861 = (1198611198790 119860119879 119861119879

1)119879 119860 is defined

by (28) and 1198610and 119861

1are formed by (29) (33) and (35)

Abstract and Applied Analysis 7

1198610=

(((((((

(

0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 sdot sdot sdot 0 0

0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 sdot sdot sdot 0 0

0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 sdot sdot sdot 0 0

0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 sdot sdot sdot 0 0

0 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 sdot sdot sdot 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0 0 sdot sdot sdot 0 0

)))))))

)

(37)

where the first five rows of 1198610come from (33) and the sixth

row comes from (29) at 1199110= 119910119903 Consider

1198611=

(((((((

(

0 0 sdot sdot sdot 0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 sdot sdot sdot 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 0

0 0 sdot sdot sdot 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0

0 0 sdot sdot sdot 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0

0 0 sdot sdot sdot 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0

0 0 sdot sdot sdot 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0

)))))))

)

(38)

where first row of 1198611comes from (29) at 119911

119873= 119910119897remaining

rows come from (35) and the matrices119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 0 119910119903 119863119879 119910119897 0 0 0 0 0)

119879

(39)

Case 2 In this case we express the given boundary condition1199111015840

0= 0 in the following wayBy using (16) we have

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ) (40)

As we defined earlier 119909119895= 119895ℎ if we put 119895 = 0 we get 119909

0= 0

the above equation can be written as

1198851015840(0) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) (41)

Since by boundary condition 11991110158400= 1198851015840(0) = 0

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) = 0 (42)

By using (13) we can express above equation as

1

594636119911minus6+

256

743295119911minus5+

2645

594636119911minus4minus3328

49553119911minus3

+76113

198212119911minus2minus78592

49553119911minus1+78592

495531199111minus76113

1982121199112

+3328

495531199113minus

2645

5946361199114minus

256

7432951199115minus

1

5946361199116= 0

(43)Finally we get a following new system of (119873 + 13) linear

equations with (119873 + 13) unknowns 119911119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) for 119894 = 5 4 3 2 (35) and (43)

119861119885 = 119877 (44)

where the coefficients matrix 119861 = (B1198790 119860119879 119861119879

1)119879 119860 is defined

by (28) andB0and 119861

1formed by (29) (33) (35) and (43) are

defined as

8 Abstract and Applied Analysis

B0= (

(

0 1 minus8 28 minus56 70 minus56 28 minus8

0 0 1 minus8 28 minus56 70 minus56 28

0 0 0 1 minus8 28 minus56 70 minus56

0 0 0 0 1 minus8 28 minus56 70

1

594636

256

743295

2645

594636minus3328

49553

76113

198212minus78592

495530

78592

49553

76113

198212

0 0 0 0 0 0 1 0 0

1 0 0 0 0 sdot sdot sdot 0 0

minus8 1 0 0 0 sdot sdot sdot 0 0

28 minus8 1 0 0 sdot sdot sdot 0 0

minus56 28 minus8 1 0 sdot sdot sdot 0 0

3328

49553minus2645

594636minus

256

743295minus

1

5946360 sdot sdot sdot 0 0

0 0 0 0 0 sdot sdot sdot 0 0

)

)

(45)

in B first four rows come from (33) for 119894 = 5 4 3 2 fifth rowcomes from (43) and the last row comes from (29)

The matrix 1198611is same as defined in Case 1 and the

matrices 119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 1199101015840(0) 119910

119903 119863119879 119910119897 0 0 0 0 0)

119879

(46)

The nonsingularity of the coefficients matrix 119861 has beendiscussed in next section

33 Existence of the Solution In this section we discuss thenonsingularity of the coefficients matrix 119861 We observe thatthe coefficients matrix 119861 is neither symmetric nor diagonallydominant However it can be shown that 119861 is a nonsingularSince 119861 is almost a band matrix with half band width 7numerical complexity for solving the linear system usingGaussian elimination is about 49(119873 + 9)multiplications Forlarge 119873 the matrix is almost symmetric except the first andlast six rows and columns due to the boundary conditionsTherefore we first consider the symmetric part of it that issquare symmetric matrix 119862 of order119873 + 3 defined as

119862 =(

(

120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1

)

)

(47)

It can be shown that 119862 is always nonsingular for each valueof 119873 However 119861 is nonsingular for 119873 ⩽ 1000 We havechecked the nonsingularity of matrix 119861 by different methodsIn first method we observe that the determinants of matrix 119861increase as 119873 increases and it is not zero for 119873 ⩽ 1000 So119861 is nonsingular The determinants of 119861 at some values of119873are shown in Table 1 In second method we observe that for119873 ⩽ 1000 the eigenvalues of matrix 119861 are nonzero so by [16]matrix119861 is nonsingular However for119873 gt 1000matrix119861mayor may not be nonsingular Therefore we claim that systems(36) and (44) are stable for119873 ⩽ 1000

4 Error Estimation

In this section we discuss the approximation properties ofthe numerical interpolating collocation algorithm Since thescheme (8) reproduces polynomial curve of degree seven soby Dyn [14] scheme has approximation order eight So thecollocation algorithm (14) with septic precision treatments at

the endpoints has the power of approximation 119874(ℎ2) Herewe present our main result for error estimation

Proposition 8 Suppose the exact solution 119910(119909) isin 1198628[0 1]

and 119911119894 are obtained by solving either (36) or (44) with 8th

order boundary condition at the end points then

err (119909)infin =10038171003817100381710038171003817119885119895minus 11991011989510038171003817100381710038171003817infin

= 119874 (ℎ2minus119895) 119895 = 0 1 2 (48)

where 119895 denotes the order of derivative

Proof Since the order of approximation of subdivisionscheme (8) is eight so by using (13) we can write for smoothfunction 119910(119909) and small ℎ as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910 (119909119895minus 6ℎ) minus 11520119910 (119909

119895minus 5ℎ)

+ 10952119910 (119909119895minus 4ℎ) + 476928119910 (119909

119895minus 3ℎ)

Abstract and Applied Analysis 9

minus 3047987119910 (119909119895minus 2ℎ) + 4677632119910 (119909

119895minus ℎ)

minus 4677632119910 (119909119895+ ℎ) + 3047987119910 (119909

119895+ 2ℎ)

minus 476928119910 (119909119895+ 3ℎ) minus 10952119910 (119909

119895+ 4ℎ)

+ 11520119910 (119909119895+ 5ℎ) minus225119910 (119909

119895+ 6ℎ)

(49)

This can be written as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910119895minus6

minus 11520119910119895minus5

+ 10952119910119895minus4

+ 476928119910119895minus3

minus 3047987119910119895minus2

+ 4677632119910119895minus1

minus 4677632119910119895+1

+ 3047987119910119895+2

minus 476928119910119895+3

minus10952119910119895+4

+ 11520119910119895+5

minus 225119910119895+6

(50)

Similarly we have

119885101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119911119895minus6

minus 11520119911119895minus5

+ 10952119911119895minus4

+ 476928119911119895minus3

minus 3047987119911119895minus2

+ 4677632119911119895minus1

minus 4677632119911119895+1

+ 3047987119911119895+2

minus 476928119911119895+3

minus10952119911119895+4

+ 11520119911119895+5

minus 225119911119895+6

(51)

If we define error function 119890(119909) = 119885(119909) minus 119910(119909) and errorvectors at the nodes by

119890 (119909119895) = 119885 (119909

119895) minus 119910 (119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 (52)

or equivalently 119890119895= 119885119895minus 119910(119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 then

this implies

1198901015840

119895= 1198851015840

119895minus 1199101015840(119909 + 119895ℎ)

11989010158401015840

119895= 11988510158401015840

119895minus 11991010158401015840(119909 + 119895ℎ)

119890101584010158401015840

119895= 119885101584010158401015840

119895minus 119910101584010158401015840(119909 + 119895ℎ)

(53)

Table 1 Determinants of the matrices

119873 119862 119861

10 minus866756 11048870018371741

50 minus177183 14981270309

100 minus552709050 11964492

500 minus5033491471916955 times 1036

4728852755761116 times 1021

1000 minus4477989536166907 times 1071

42069711017699999 times 1040

By subtracting (50) from (51) we get

119885101584010158401015840(119909119895) minus 119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225 (119911119895minus6

minus 119910119895minus6) minus 11520 (119911

119895minus5minus 119910119895minus5)

+ 10952 (119911119895minus4

minus 119910119895minus4) + 476928 (119911

119895minus3minus 119910119895minus3)

minus 3047987 (119911119895minus2

minus 119910119895minus2) + 4677632 (119911

119895minus1minus 119910119895minus1)

minus 4677632 (119911119895+1

minus 119910119895+1) + 3047987 (119911

119895+2minus 119910119895+2)

minus 476928 (119911119895+3

minus 119910119895+3) minus 10952 (119911

119895+4minus 119910119895+4)

+11520 (119911119895+5

minus 119910119895+5) minus 225 (119911

119895+6minus 119910119895+6)

(54)

This implies

119890101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119890119895minus6

minus 11520119890119895minus5

+ 10952119890119895minus4

+ 476928119890119895minus3

minus 3047987119890119895minus2

+ 4677632119890119895minus1

minus 4677632119890119895+1

+ 3047987119890119895+2

minus 476928119890119895+3

minus10952119890119895+4

+ 11520119890119895+5

minus 225119890119895+6

(55)

By Lemma 5 we get the following expression

119890101584010158401015840

119895=1

ℎ3120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 120601101584010158401015840

0119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6 + 119874 (ℎ

8)

(56)

where 119895 = 0 1 2 119873By subtracting (1) from (15) we get

119885101584010158401015840

119895minus 119910101584010158401015840

119895= 119886119895(119885119894minus 119884119895) (57)

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

119878 =

((((((((((((((

(

0 0 0 1 0 0 0 0 0 0 0 0 0

11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0 0

0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0

0 0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0

0 0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 11987133

11987132

11987131

11987130

11987130

11987131

11987132

11987133

0 0 0 0 0 0 0 0 0 1 0 0 0

))))))))))))))

)

(9)

where 11987130

= 12252048 11987131

= minus2452048 11987132

= 492048and 119871

33= minus52048 The some of its eigenvalues is

120582 = 11

21

41

81

161

321

641

128 (10)

For an eigenvalue 120582 the eigenvectors 120585 and 120578 that satisfy 119878120585 =120582120585 and 120578119878119879 = 120578120582 are called right and left eigenvectors of thematrix 119878 respectively Some of the normalized left and righteigenvectors corresponding to eigenvalues are

1205850= (1 1 1 1 1 1 1 1 1 1 1 1 1)

119879

1205780= (0 0 0 0 0 0 1 0 0 0 0 0 0)

119879

1205851= (minus6 minus5 minus4 minus3 minus2 minus1 0 1 2 3 4 5 6)

119879

1205781= (

1

5946360) (minus5 1024 13225 minus199680 1141695

minus 4715520 0 4715520 minus1141695

199680 minus13225 minus1024 5)119879

1205852= (36 25 16 9 4 1 0 1 4 9 16 25 36)

119879

1205782= (

1

34546860) (275 minus28160 minus182613 2607616

minus 12053651 45634048 minus71955030

45634048 minus12053651 2607616

minus182613 minus28160 275)119879

1205853= (minus216 minus125 minus64 minus27 minus8 minus1 0 1 8 27 64 125 216)

119879

1205783= (

1

15039360) (225 minus11520 10952 476928

minus 3047987 4677632 0 minus4677632

3047987 minus476928 minus10952

11520 minus225)119879

(11)

Since 120585119879119894120578119895= 1 for 119894 = 119895 and 0 otherwise then by using

Lemmas 2 and 3 we get the following result

Lemma 5 The fundamental solution (Cardinal basis) 120601(119909) isthrice continuously differentiable and supported on (minus7 7) andits derivatives at integers are given by

1206011015840(119894) = 2 sign (119894) 119890119879

|119894|1205781

12060110158401015840(119894) = 2

2119890119879

|119894|1205782

120601101584010158401015840(119894) = 2

3 sign (119894) 119890119879|119894|1205783

minus 6 le 119894 le 6

(12)

where

1198900= (0 0 0 0 0 0 1 0 0 0 0 0 0)

119879

1198901= (0 0 0 0 0 1 0 0 0 0 0 0 0)

119879

1198902= (0 0 0 0 1 0 0 0 0 0 0 0 0)

119879

1198903= (0 0 0 1 0 0 0 0 0 0 0 0 0)

119879

1198904= (0 0 1 0 0 0 0 0 0 0 0 0 0)

119879

1198905= (0 1 0 0 0 0 0 0 0 0 0 0 0)

119879

1198906= (1 0 0 0 0 0 0 0 0 0 0 0 0)

119879

1206011015840(0) = 0 120601

1015840(plusmn1) = ∓

78592

49553

1206011015840(plusmn2) = plusmn

76113

198212 120601

1015840(plusmn3) = ∓

3328

49553

1206011015840(plusmn4) = plusmn

2645

594636 120601

1015840(plusmn5) = plusmn

256

743295

1206011015840(plusmn6) =

1

594636

4 Abstract and Applied Analysis

12060110158401015840(0) = minus

342643

41124 120601

10158401015840(plusmn1) =

5704256

1079505

12060110158401015840(plusmn2) = minus

12053651

8636040 120601

10158401015840(plusmn3) =

325952

1079505

12060110158401015840(plusmn4) = minus

60871

2878680 120601

10158401015840(plusmn5) = minus

704

215901

12060110158401015840(plusmn6) =

55

1727208

120601101584010158401015840(0) = 0 120601

101584010158401015840(plusmn1) = ∓

292352

117495

120601101584010158401015840(plusmn2) = plusmn

3047987

1879920 120601

101584010158401015840(plusmn3) = ∓

3312

13055

120601101584010158401015840(plusmn4) = ∓

1369

234990 120601

101584010158401015840(plusmn5) = plusmn

16

2611

120601101584010158401015840(plusmn6) = ∓

5

41776

(13)

The graphical representations of the basis limit functiondefined on cardinal data and its derivatives up to order threefor 119899 = 3 are shown in Figure 1 Figure 1(a) represents thebasis limit function defined in (4) Graphical representationsof first second and third derivatives of basis limit functionsobtained from (5) for 119899 = 3 are shown in Figures 1(b) 1(c)and 1(d) at 119894 = 0 1 minus1 respectively The numeric values offirst second and third derivative are given in (13)

3 Numerical InterpolatingCollocation Algorithm

In this section first we formulate a numerical interpolat-ing collocation algorithm for linear third order two-pointboundary value problemsThenwe settle down the boundaryconditions to get unique solution

31 The Collocation Algorithm Let 119873 be a positive integer(119873 ge 6) ℎ = 1119873 and 119909

119894= 119894119873 = 119894ℎ 119894 = 0 1 2 119873

and set 119886119894= 119886(119909

119894) 119887119894= 119887(119909119894) Let

119885 (119909) =

119873+6

sum

119894=minus6

119911119894120601(

119909 minus 119909119894

ℎ) 0 le 119909 le 1 (14)

be the approximate solution to (1) where 119911119894 are the

unknown to be determined by (1)The collocation algorithmtogether with the boundary conditions to be discussed isgiven by setting

119885101584010158401015840(119909119895) = 119886 (119909

119895)119885 (119909

119895) + 119887 (119909

119895) 119895 = 0 1 2 119873

(15)

where

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ)

11988510158401015840(119909119895) =

1

ℎ2

119873+6

sum

119894=minus6

11991111989412060110158401015840(

119909119895minus 119909119894

ℎ)

119885101584010158401015840(119909119895) =

1

ℎ3

119873+6

sum

119894=minus6

119911119894120601101584010158401015840(

119909119895minus 119909119894

ℎ)

(16)

Using (14) and (16) in (15) we get following119873 + 1 system ofequations

119873+6

sum

119894=minus6

119911119894120601101584010158401015840(

119909119895minus 119909119894

ℎ) minus ℎ3119886119895

119873+6

sum

119894=minus6

119911119894120601(

119909119895minus 119909119894

ℎ) = ℎ

3119887119895

119895 = 0 1 2 119873

(17)

Now we simplify the above system in the followingtheorems

Theorem 6 For 119895 = 0 by (17) one gets

119911minus6120601101584010158401015840

6+ 119911minus5120601101584010158401015840

5+ 119911minus4120601101584010158401015840

4+ 119911minus3120601101584010158401015840

3+ 119911minus2120601101584010158401015840

2

+ 119911minus1120601101584010158401015840

1+ 11991101199020+ 1199111120601101584010158401015840

minus1+ 1199112120601101584010158401015840

minus2+ 1199113120601101584010158401015840

minus3

+ 1199114120601101584010158401015840

minus4+ 1199115120601101584010158401015840

minus5+ 1199116120601101584010158401015840

minus6= ℎ31198870

(18)

where 120601101584010158401015840119895= 120601101584010158401015840(119895) and 119902

0= 120601101584010158401015840

0minus 1198860ℎ3

Proof Substituting 119895 = 0 in (17) we get

119911minus6120601101584010158401015840(1199090minus 119909minus6

ℎ) + 119911minus5120601101584010158401015840(1199090minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601101584010158401015840(1199090minus 119909119873+5

ℎ) + 119911119873+6

120601101584010158401015840(1199090minus 119909119873+6

ℎ)

minus 1198860ℎ3119911minus6120601(

1199090minus 119909minus6

ℎ) + 119911minus5120601(

1199090minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601(1199090minus 119909119873+5

ℎ) + 119911119873+6

120601(1199090minus 119909119873+6

ℎ)

= ℎ31198870

(19)

For 119909119895= 119895ℎ 119895 = 0 1 2 119873 this implies

119911minus6120601101584010158401015840(6) + 119911

minus5120601101584010158401015840(5) + sdot sdot sdot + 119911

119873+5120601101584010158401015840(minus119873 minus 5)

+ 119911119873+6

120601101584010158401015840(minus119873 minus 6)

minus 1198860ℎ3119911minus6120601 (6) + 119911

minus5120601 (5) + sdot sdot sdot

+119911119873+5

120601 (minus119873 minus 5) + 119911119873+6

120601 (minus119873 minus 6) = ℎ31198870

(20)

Abstract and Applied Analysis 5

1

08

06

04

02

0

minus4 minus2 0 2 4 6

(a)

minus4minus6 minus2 0 2 4 6

15

1

05

0

minus05

minus1

minus15

(b)

1

0

minus1

minus2

minus4minus6 minus2 0 2 4 6

(c)

minus4minus2 0 2 4 6

03

02

01

0

minus01

minus02

minus03

(d)

Figure 1 Interpolatory basis function 1206013(119894) is shown in (a) first derivative of 120601

3(119894) i-e 1206011015840

3(119894) shown in (b) second derivative of 120601

3(119894) i-e 12060110158401015840

3(119894)

shown in (c) and third derivative of 1206013(119894) i-e 120601101584010158401015840

3(119894) shown in (d) respectively

Since the support of basis function 120601(119909) is (minus7 7) 1206011015840(119909)12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval (minus7 7) also

by (4) and (13) we get

119911minus6120601101584010158401015840(6) + 119911

minus5120601101584010158401015840(5) + 119911

minus4120601101584010158401015840(4) + 119911

minus3120601101584010158401015840(3)

+ 119911minus2120601101584010158401015840(2) + 119911

minus1120601101584010158401015840(1) + 119911

0120601101584010158401015840(0)

+ 1199111120601101584010158401015840(minus1) + 119911

2120601101584010158401015840(minus2) + 119911

3120601101584010158401015840(minus3)

+ 1199114120601101584010158401015840(minus4) + 119911

5120601101584010158401015840(minus5) + 119911

6120601101584010158401015840(minus6)

minus 1198860ℎ31199110120601 (0) = ℎ

31198870

(21)

If 120601101584010158401015840119895= 120601101584010158401015840(119895) then

119911minus6120601101584010158401015840

6+ 119911minus5120601101584010158401015840

5+ 119911minus4120601101584010158401015840

4+ 119911minus3120601101584010158401015840

3+ 119911minus2120601101584010158401015840

2

+ 119911minus1120601101584010158401015840

1+ 1199110(120601101584010158401015840

0minus 1198860ℎ3) + 1199111120601101584010158401015840

minus1+ 1199112120601101584010158401015840

minus2

+ 1199113120601101584010158401015840

minus3+ 1199114120601101584010158401015840

minus4+ 1199115120601101584010158401015840

minus5+ 1199116120601101584010158401015840

minus6= ℎ31198870

(22)

For 1199020= 120601101584010158401015840

0minus1198860ℎ3 we get (18)This completes the proof

Theorem 7 For 119895 = 1 2 3 119873 the system (17) is equivalentto

119911minus6120601101584010158401015840

119895+6+ 119911minus5120601101584010158401015840

119895+5+ sdot sdot sdot + 119911

0120601101584010158401015840

119895

+ 1199111(120601101584010158401015840

119895minus1minus 119886119895ℎ3120601119895minus1) + 1199112(120601101584010158401015840

119895minus2minus 119886119895ℎ3120601119895minus2) + sdot sdot sdot

+ 119911119873minus1

(120601101584010158401015840

119895minus119873+1minus 119886119895ℎ3120601119895minus119873+1

) + 119911119873(120601101584010158401015840

119895minus119873minus 119886119895ℎ3120601119895minus119873

)

+ 119911119873+1

120601101584010158401015840

119895minus119873minus1+ sdot sdot sdot + 119911

119873+6120601101584010158401015840

119895minus119873minus6= ℎ3119887119895

(23)

Proof By expanding (17) we get

119911minus6120601101584010158401015840(

119909119895minus 119909minus6

ℎ) + 119911minus5120601101584010158401015840(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601101584010158401015840(

119909119895minus 119909119873+6

ℎ)

minus 119886119895ℎ3119911minus6120601(

119909119895minus 119909minus6

ℎ) + 119911minus5120601(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601(

119909119895minus 119909119873+6

ℎ)

= ℎ3119887119895

(24)

For 119909119895= 119895ℎ 119895 = 1 2 119873 we get

119911minus6120601101584010158401015840(119895 + 6) + 119911

minus5120601101584010158401015840(119895 + 5) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(119895 minus 119873 minus 5) + 119911

119873+6120601101584010158401015840(119895 minus 119873 minus 6)

minus 119886119895ℎ3119911minus6120601 (119895 + 6) + 119911

minus5120601 (119895 + 5) + sdot sdot sdot

+119911119873+5

120601 (119895 minus 119873 minus 5) + 119911119873+6

120601 (119895 minus 119873 minus 6)

= ℎ3119887119895

(25)

6 Abstract and Applied Analysis

If 120601101584010158401015840119895= 120601101584010158401015840(119895) for 119895 = 1 2 119873 then

119911minus6(120601101584010158401015840

119895+6minus 119886119895ℎ3120601119895+6) + 119911minus5(120601101584010158401015840

119895+5minus 119886119895ℎ3120601119895+5)

+ sdot sdot sdot + 119911119873+5

(120601101584010158401015840

119895minus119873minus5minus 119886119895ℎ3120601119895minus119873minus5

)

+ 119911119873+6

(120601101584010158401015840

119895minus119873minus6minus 119886119895ℎ3120601119895minus119873minus6

) = ℎ3119887119895

(26)

Since 1206011015840(119909) 12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval(minus7 7) then by (4) and (13) we get (23)

From (18) and (23) we get following undeterminedsystem of (119873 + 1) equations with (119873 + 13) unknowns 119911

119894

119860119885 = 119863 (27)

where the matrices 119860 119885 and119863 of orders (119873 + 1) times (119873 + 13)119873 + 13 and119873 + 1 respectively are given by

119860 =

(((

(

120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199020

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199021

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199022

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

0 0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199023

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus5120601101584010158401015840

minus60

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6

)))

)

(28)

119885 = (119911minus6 119911minus5 119911minus4 119911minus3 119911minus2 119911

119873+6)119879 and 119863 = (119887

0ℎ3 1198871ℎ3

1198872ℎ3 1198873ℎ3 119887

119873ℎ3)119879 where120601101584010158401015840

119895= 120601101584010158401015840(119895) and 119902

119895= 120601101584010158401015840

0minus119886119895ℎ3

32 Adjustment of Boundary Conditions The order of thecoefficient matrix (28) is (119873 + 1) times (119873 + 13) In orderto get unique solution of the system we need twelve moreconditions Here we consider only two different cases Incoming section we will show that the approximate solutioncan be improved by adjusting different boundary conditions

Case 1 If we assume 11991110158400= 0 (equivalently 119911

0= 119910119903= finite)

then two conditions can be achieved by using following givenboundary conditions i-e

1199110= 119910119903 119911

1015840

0= 0 119911

119873= 119910119897 (29)

Still we need ten more conditions to get stable system Sincesubdivision scheme reproduces seven degree polynomials wedefine boundary conditions of order eight for solution of (27)For simplicity only the left end points are discussed and thevalues of right end points 119911

119873+1 119911119873+2

119911119873+3

119911119873+4

119911119873+5

can betreated similarly

The values 119911minus5 119911minus4 119911minus3 119911minus2 119911minus1

can be determined bythe septic polynomial 119902(119909) interpolating at (119909

119894 119911119894) 0 le 119894 le 7

Precisely we have

119911minus119894= 119902 (minus119909

119894) 119894 = 1 2 3 4 5 (30)

where

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119885(119909119894minus119895) (31)

Since by (14) 119885(119909119894) = 119911119894for 119894 = 1 2 3 4 5 then by replacing

119909119894by minus119909

119894 we have

119902 (minus119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911minus119894+119895

(32)

Hence the following boundary conditions can be employed atthe left end

8

sum

119895=0

(8

119895) (minus1)

119895119911minus119894+119895

= 0 119894 = 5 4 3 2 1 (33)

Similarly for the right end we can define 119911119894= 119902(minus119909

119894) 119894 =

119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5 and

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911119894minus119895 (34)

So we have the following boundary conditions at the rightend

8

sum

119895=0

(8

119895) (minus1)

119895119911119894minus119895

= 0

119894 = 119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5

(35)

Finally we get the following new system of (119873 + 13) linearequations with (119873 + 13) unknowns 119911

119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) and (35)

119861119885 = 119877 (36)

where the coefficients matrix 119861 = (1198611198790 119860119879 119861119879

1)119879 119860 is defined

by (28) and 1198610and 119861

1are formed by (29) (33) and (35)

Abstract and Applied Analysis 7

1198610=

(((((((

(

0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 sdot sdot sdot 0 0

0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 sdot sdot sdot 0 0

0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 sdot sdot sdot 0 0

0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 sdot sdot sdot 0 0

0 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 sdot sdot sdot 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0 0 sdot sdot sdot 0 0

)))))))

)

(37)

where the first five rows of 1198610come from (33) and the sixth

row comes from (29) at 1199110= 119910119903 Consider

1198611=

(((((((

(

0 0 sdot sdot sdot 0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 sdot sdot sdot 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 0

0 0 sdot sdot sdot 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0

0 0 sdot sdot sdot 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0

0 0 sdot sdot sdot 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0

0 0 sdot sdot sdot 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0

)))))))

)

(38)

where first row of 1198611comes from (29) at 119911

119873= 119910119897remaining

rows come from (35) and the matrices119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 0 119910119903 119863119879 119910119897 0 0 0 0 0)

119879

(39)

Case 2 In this case we express the given boundary condition1199111015840

0= 0 in the following wayBy using (16) we have

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ) (40)

As we defined earlier 119909119895= 119895ℎ if we put 119895 = 0 we get 119909

0= 0

the above equation can be written as

1198851015840(0) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) (41)

Since by boundary condition 11991110158400= 1198851015840(0) = 0

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) = 0 (42)

By using (13) we can express above equation as

1

594636119911minus6+

256

743295119911minus5+

2645

594636119911minus4minus3328

49553119911minus3

+76113

198212119911minus2minus78592

49553119911minus1+78592

495531199111minus76113

1982121199112

+3328

495531199113minus

2645

5946361199114minus

256

7432951199115minus

1

5946361199116= 0

(43)Finally we get a following new system of (119873 + 13) linear

equations with (119873 + 13) unknowns 119911119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) for 119894 = 5 4 3 2 (35) and (43)

119861119885 = 119877 (44)

where the coefficients matrix 119861 = (B1198790 119860119879 119861119879

1)119879 119860 is defined

by (28) andB0and 119861

1formed by (29) (33) (35) and (43) are

defined as

8 Abstract and Applied Analysis

B0= (

(

0 1 minus8 28 minus56 70 minus56 28 minus8

0 0 1 minus8 28 minus56 70 minus56 28

0 0 0 1 minus8 28 minus56 70 minus56

0 0 0 0 1 minus8 28 minus56 70

1

594636

256

743295

2645

594636minus3328

49553

76113

198212minus78592

495530

78592

49553

76113

198212

0 0 0 0 0 0 1 0 0

1 0 0 0 0 sdot sdot sdot 0 0

minus8 1 0 0 0 sdot sdot sdot 0 0

28 minus8 1 0 0 sdot sdot sdot 0 0

minus56 28 minus8 1 0 sdot sdot sdot 0 0

3328

49553minus2645

594636minus

256

743295minus

1

5946360 sdot sdot sdot 0 0

0 0 0 0 0 sdot sdot sdot 0 0

)

)

(45)

in B first four rows come from (33) for 119894 = 5 4 3 2 fifth rowcomes from (43) and the last row comes from (29)

The matrix 1198611is same as defined in Case 1 and the

matrices 119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 1199101015840(0) 119910

119903 119863119879 119910119897 0 0 0 0 0)

119879

(46)

The nonsingularity of the coefficients matrix 119861 has beendiscussed in next section

33 Existence of the Solution In this section we discuss thenonsingularity of the coefficients matrix 119861 We observe thatthe coefficients matrix 119861 is neither symmetric nor diagonallydominant However it can be shown that 119861 is a nonsingularSince 119861 is almost a band matrix with half band width 7numerical complexity for solving the linear system usingGaussian elimination is about 49(119873 + 9)multiplications Forlarge 119873 the matrix is almost symmetric except the first andlast six rows and columns due to the boundary conditionsTherefore we first consider the symmetric part of it that issquare symmetric matrix 119862 of order119873 + 3 defined as

119862 =(

(

120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1

)

)

(47)

It can be shown that 119862 is always nonsingular for each valueof 119873 However 119861 is nonsingular for 119873 ⩽ 1000 We havechecked the nonsingularity of matrix 119861 by different methodsIn first method we observe that the determinants of matrix 119861increase as 119873 increases and it is not zero for 119873 ⩽ 1000 So119861 is nonsingular The determinants of 119861 at some values of119873are shown in Table 1 In second method we observe that for119873 ⩽ 1000 the eigenvalues of matrix 119861 are nonzero so by [16]matrix119861 is nonsingular However for119873 gt 1000matrix119861mayor may not be nonsingular Therefore we claim that systems(36) and (44) are stable for119873 ⩽ 1000

4 Error Estimation

In this section we discuss the approximation properties ofthe numerical interpolating collocation algorithm Since thescheme (8) reproduces polynomial curve of degree seven soby Dyn [14] scheme has approximation order eight So thecollocation algorithm (14) with septic precision treatments at

the endpoints has the power of approximation 119874(ℎ2) Herewe present our main result for error estimation

Proposition 8 Suppose the exact solution 119910(119909) isin 1198628[0 1]

and 119911119894 are obtained by solving either (36) or (44) with 8th

order boundary condition at the end points then

err (119909)infin =10038171003817100381710038171003817119885119895minus 11991011989510038171003817100381710038171003817infin

= 119874 (ℎ2minus119895) 119895 = 0 1 2 (48)

where 119895 denotes the order of derivative

Proof Since the order of approximation of subdivisionscheme (8) is eight so by using (13) we can write for smoothfunction 119910(119909) and small ℎ as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910 (119909119895minus 6ℎ) minus 11520119910 (119909

119895minus 5ℎ)

+ 10952119910 (119909119895minus 4ℎ) + 476928119910 (119909

119895minus 3ℎ)

Abstract and Applied Analysis 9

minus 3047987119910 (119909119895minus 2ℎ) + 4677632119910 (119909

119895minus ℎ)

minus 4677632119910 (119909119895+ ℎ) + 3047987119910 (119909

119895+ 2ℎ)

minus 476928119910 (119909119895+ 3ℎ) minus 10952119910 (119909

119895+ 4ℎ)

+ 11520119910 (119909119895+ 5ℎ) minus225119910 (119909

119895+ 6ℎ)

(49)

This can be written as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910119895minus6

minus 11520119910119895minus5

+ 10952119910119895minus4

+ 476928119910119895minus3

minus 3047987119910119895minus2

+ 4677632119910119895minus1

minus 4677632119910119895+1

+ 3047987119910119895+2

minus 476928119910119895+3

minus10952119910119895+4

+ 11520119910119895+5

minus 225119910119895+6

(50)

Similarly we have

119885101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119911119895minus6

minus 11520119911119895minus5

+ 10952119911119895minus4

+ 476928119911119895minus3

minus 3047987119911119895minus2

+ 4677632119911119895minus1

minus 4677632119911119895+1

+ 3047987119911119895+2

minus 476928119911119895+3

minus10952119911119895+4

+ 11520119911119895+5

minus 225119911119895+6

(51)

If we define error function 119890(119909) = 119885(119909) minus 119910(119909) and errorvectors at the nodes by

119890 (119909119895) = 119885 (119909

119895) minus 119910 (119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 (52)

or equivalently 119890119895= 119885119895minus 119910(119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 then

this implies

1198901015840

119895= 1198851015840

119895minus 1199101015840(119909 + 119895ℎ)

11989010158401015840

119895= 11988510158401015840

119895minus 11991010158401015840(119909 + 119895ℎ)

119890101584010158401015840

119895= 119885101584010158401015840

119895minus 119910101584010158401015840(119909 + 119895ℎ)

(53)

Table 1 Determinants of the matrices

119873 119862 119861

10 minus866756 11048870018371741

50 minus177183 14981270309

100 minus552709050 11964492

500 minus5033491471916955 times 1036

4728852755761116 times 1021

1000 minus4477989536166907 times 1071

42069711017699999 times 1040

By subtracting (50) from (51) we get

119885101584010158401015840(119909119895) minus 119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225 (119911119895minus6

minus 119910119895minus6) minus 11520 (119911

119895minus5minus 119910119895minus5)

+ 10952 (119911119895minus4

minus 119910119895minus4) + 476928 (119911

119895minus3minus 119910119895minus3)

minus 3047987 (119911119895minus2

minus 119910119895minus2) + 4677632 (119911

119895minus1minus 119910119895minus1)

minus 4677632 (119911119895+1

minus 119910119895+1) + 3047987 (119911

119895+2minus 119910119895+2)

minus 476928 (119911119895+3

minus 119910119895+3) minus 10952 (119911

119895+4minus 119910119895+4)

+11520 (119911119895+5

minus 119910119895+5) minus 225 (119911

119895+6minus 119910119895+6)

(54)

This implies

119890101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119890119895minus6

minus 11520119890119895minus5

+ 10952119890119895minus4

+ 476928119890119895minus3

minus 3047987119890119895minus2

+ 4677632119890119895minus1

minus 4677632119890119895+1

+ 3047987119890119895+2

minus 476928119890119895+3

minus10952119890119895+4

+ 11520119890119895+5

minus 225119890119895+6

(55)

By Lemma 5 we get the following expression

119890101584010158401015840

119895=1

ℎ3120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 120601101584010158401015840

0119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6 + 119874 (ℎ

8)

(56)

where 119895 = 0 1 2 119873By subtracting (1) from (15) we get

119885101584010158401015840

119895minus 119910101584010158401015840

119895= 119886119895(119885119894minus 119884119895) (57)

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

12060110158401015840(0) = minus

342643

41124 120601

10158401015840(plusmn1) =

5704256

1079505

12060110158401015840(plusmn2) = minus

12053651

8636040 120601

10158401015840(plusmn3) =

325952

1079505

12060110158401015840(plusmn4) = minus

60871

2878680 120601

10158401015840(plusmn5) = minus

704

215901

12060110158401015840(plusmn6) =

55

1727208

120601101584010158401015840(0) = 0 120601

101584010158401015840(plusmn1) = ∓

292352

117495

120601101584010158401015840(plusmn2) = plusmn

3047987

1879920 120601

101584010158401015840(plusmn3) = ∓

3312

13055

120601101584010158401015840(plusmn4) = ∓

1369

234990 120601

101584010158401015840(plusmn5) = plusmn

16

2611

120601101584010158401015840(plusmn6) = ∓

5

41776

(13)

The graphical representations of the basis limit functiondefined on cardinal data and its derivatives up to order threefor 119899 = 3 are shown in Figure 1 Figure 1(a) represents thebasis limit function defined in (4) Graphical representationsof first second and third derivatives of basis limit functionsobtained from (5) for 119899 = 3 are shown in Figures 1(b) 1(c)and 1(d) at 119894 = 0 1 minus1 respectively The numeric values offirst second and third derivative are given in (13)

3 Numerical InterpolatingCollocation Algorithm

In this section first we formulate a numerical interpolat-ing collocation algorithm for linear third order two-pointboundary value problemsThenwe settle down the boundaryconditions to get unique solution

31 The Collocation Algorithm Let 119873 be a positive integer(119873 ge 6) ℎ = 1119873 and 119909

119894= 119894119873 = 119894ℎ 119894 = 0 1 2 119873

and set 119886119894= 119886(119909

119894) 119887119894= 119887(119909119894) Let

119885 (119909) =

119873+6

sum

119894=minus6

119911119894120601(

119909 minus 119909119894

ℎ) 0 le 119909 le 1 (14)

be the approximate solution to (1) where 119911119894 are the

unknown to be determined by (1)The collocation algorithmtogether with the boundary conditions to be discussed isgiven by setting

119885101584010158401015840(119909119895) = 119886 (119909

119895)119885 (119909

119895) + 119887 (119909

119895) 119895 = 0 1 2 119873

(15)

where

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ)

11988510158401015840(119909119895) =

1

ℎ2

119873+6

sum

119894=minus6

11991111989412060110158401015840(

119909119895minus 119909119894

ℎ)

119885101584010158401015840(119909119895) =

1

ℎ3

119873+6

sum

119894=minus6

119911119894120601101584010158401015840(

119909119895minus 119909119894

ℎ)

(16)

Using (14) and (16) in (15) we get following119873 + 1 system ofequations

119873+6

sum

119894=minus6

119911119894120601101584010158401015840(

119909119895minus 119909119894

ℎ) minus ℎ3119886119895

119873+6

sum

119894=minus6

119911119894120601(

119909119895minus 119909119894

ℎ) = ℎ

3119887119895

119895 = 0 1 2 119873

(17)

Now we simplify the above system in the followingtheorems

Theorem 6 For 119895 = 0 by (17) one gets

119911minus6120601101584010158401015840

6+ 119911minus5120601101584010158401015840

5+ 119911minus4120601101584010158401015840

4+ 119911minus3120601101584010158401015840

3+ 119911minus2120601101584010158401015840

2

+ 119911minus1120601101584010158401015840

1+ 11991101199020+ 1199111120601101584010158401015840

minus1+ 1199112120601101584010158401015840

minus2+ 1199113120601101584010158401015840

minus3

+ 1199114120601101584010158401015840

minus4+ 1199115120601101584010158401015840

minus5+ 1199116120601101584010158401015840

minus6= ℎ31198870

(18)

where 120601101584010158401015840119895= 120601101584010158401015840(119895) and 119902

0= 120601101584010158401015840

0minus 1198860ℎ3

Proof Substituting 119895 = 0 in (17) we get

119911minus6120601101584010158401015840(1199090minus 119909minus6

ℎ) + 119911minus5120601101584010158401015840(1199090minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601101584010158401015840(1199090minus 119909119873+5

ℎ) + 119911119873+6

120601101584010158401015840(1199090minus 119909119873+6

ℎ)

minus 1198860ℎ3119911minus6120601(

1199090minus 119909minus6

ℎ) + 119911minus5120601(

1199090minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601(1199090minus 119909119873+5

ℎ) + 119911119873+6

120601(1199090minus 119909119873+6

ℎ)

= ℎ31198870

(19)

For 119909119895= 119895ℎ 119895 = 0 1 2 119873 this implies

119911minus6120601101584010158401015840(6) + 119911

minus5120601101584010158401015840(5) + sdot sdot sdot + 119911

119873+5120601101584010158401015840(minus119873 minus 5)

+ 119911119873+6

120601101584010158401015840(minus119873 minus 6)

minus 1198860ℎ3119911minus6120601 (6) + 119911

minus5120601 (5) + sdot sdot sdot

+119911119873+5

120601 (minus119873 minus 5) + 119911119873+6

120601 (minus119873 minus 6) = ℎ31198870

(20)

Abstract and Applied Analysis 5

1

08

06

04

02

0

minus4 minus2 0 2 4 6

(a)

minus4minus6 minus2 0 2 4 6

15

1

05

0

minus05

minus1

minus15

(b)

1

0

minus1

minus2

minus4minus6 minus2 0 2 4 6

(c)

minus4minus2 0 2 4 6

03

02

01

0

minus01

minus02

minus03

(d)

Figure 1 Interpolatory basis function 1206013(119894) is shown in (a) first derivative of 120601

3(119894) i-e 1206011015840

3(119894) shown in (b) second derivative of 120601

3(119894) i-e 12060110158401015840

3(119894)

shown in (c) and third derivative of 1206013(119894) i-e 120601101584010158401015840

3(119894) shown in (d) respectively

Since the support of basis function 120601(119909) is (minus7 7) 1206011015840(119909)12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval (minus7 7) also

by (4) and (13) we get

119911minus6120601101584010158401015840(6) + 119911

minus5120601101584010158401015840(5) + 119911

minus4120601101584010158401015840(4) + 119911

minus3120601101584010158401015840(3)

+ 119911minus2120601101584010158401015840(2) + 119911

minus1120601101584010158401015840(1) + 119911

0120601101584010158401015840(0)

+ 1199111120601101584010158401015840(minus1) + 119911

2120601101584010158401015840(minus2) + 119911

3120601101584010158401015840(minus3)

+ 1199114120601101584010158401015840(minus4) + 119911

5120601101584010158401015840(minus5) + 119911

6120601101584010158401015840(minus6)

minus 1198860ℎ31199110120601 (0) = ℎ

31198870

(21)

If 120601101584010158401015840119895= 120601101584010158401015840(119895) then

119911minus6120601101584010158401015840

6+ 119911minus5120601101584010158401015840

5+ 119911minus4120601101584010158401015840

4+ 119911minus3120601101584010158401015840

3+ 119911minus2120601101584010158401015840

2

+ 119911minus1120601101584010158401015840

1+ 1199110(120601101584010158401015840

0minus 1198860ℎ3) + 1199111120601101584010158401015840

minus1+ 1199112120601101584010158401015840

minus2

+ 1199113120601101584010158401015840

minus3+ 1199114120601101584010158401015840

minus4+ 1199115120601101584010158401015840

minus5+ 1199116120601101584010158401015840

minus6= ℎ31198870

(22)

For 1199020= 120601101584010158401015840

0minus1198860ℎ3 we get (18)This completes the proof

Theorem 7 For 119895 = 1 2 3 119873 the system (17) is equivalentto

119911minus6120601101584010158401015840

119895+6+ 119911minus5120601101584010158401015840

119895+5+ sdot sdot sdot + 119911

0120601101584010158401015840

119895

+ 1199111(120601101584010158401015840

119895minus1minus 119886119895ℎ3120601119895minus1) + 1199112(120601101584010158401015840

119895minus2minus 119886119895ℎ3120601119895minus2) + sdot sdot sdot

+ 119911119873minus1

(120601101584010158401015840

119895minus119873+1minus 119886119895ℎ3120601119895minus119873+1

) + 119911119873(120601101584010158401015840

119895minus119873minus 119886119895ℎ3120601119895minus119873

)

+ 119911119873+1

120601101584010158401015840

119895minus119873minus1+ sdot sdot sdot + 119911

119873+6120601101584010158401015840

119895minus119873minus6= ℎ3119887119895

(23)

Proof By expanding (17) we get

119911minus6120601101584010158401015840(

119909119895minus 119909minus6

ℎ) + 119911minus5120601101584010158401015840(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601101584010158401015840(

119909119895minus 119909119873+6

ℎ)

minus 119886119895ℎ3119911minus6120601(

119909119895minus 119909minus6

ℎ) + 119911minus5120601(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601(

119909119895minus 119909119873+6

ℎ)

= ℎ3119887119895

(24)

For 119909119895= 119895ℎ 119895 = 1 2 119873 we get

119911minus6120601101584010158401015840(119895 + 6) + 119911

minus5120601101584010158401015840(119895 + 5) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(119895 minus 119873 minus 5) + 119911

119873+6120601101584010158401015840(119895 minus 119873 minus 6)

minus 119886119895ℎ3119911minus6120601 (119895 + 6) + 119911

minus5120601 (119895 + 5) + sdot sdot sdot

+119911119873+5

120601 (119895 minus 119873 minus 5) + 119911119873+6

120601 (119895 minus 119873 minus 6)

= ℎ3119887119895

(25)

6 Abstract and Applied Analysis

If 120601101584010158401015840119895= 120601101584010158401015840(119895) for 119895 = 1 2 119873 then

119911minus6(120601101584010158401015840

119895+6minus 119886119895ℎ3120601119895+6) + 119911minus5(120601101584010158401015840

119895+5minus 119886119895ℎ3120601119895+5)

+ sdot sdot sdot + 119911119873+5

(120601101584010158401015840

119895minus119873minus5minus 119886119895ℎ3120601119895minus119873minus5

)

+ 119911119873+6

(120601101584010158401015840

119895minus119873minus6minus 119886119895ℎ3120601119895minus119873minus6

) = ℎ3119887119895

(26)

Since 1206011015840(119909) 12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval(minus7 7) then by (4) and (13) we get (23)

From (18) and (23) we get following undeterminedsystem of (119873 + 1) equations with (119873 + 13) unknowns 119911

119894

119860119885 = 119863 (27)

where the matrices 119860 119885 and119863 of orders (119873 + 1) times (119873 + 13)119873 + 13 and119873 + 1 respectively are given by

119860 =

(((

(

120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199020

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199021

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199022

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

0 0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199023

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus5120601101584010158401015840

minus60

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6

)))

)

(28)

119885 = (119911minus6 119911minus5 119911minus4 119911minus3 119911minus2 119911

119873+6)119879 and 119863 = (119887

0ℎ3 1198871ℎ3

1198872ℎ3 1198873ℎ3 119887

119873ℎ3)119879 where120601101584010158401015840

119895= 120601101584010158401015840(119895) and 119902

119895= 120601101584010158401015840

0minus119886119895ℎ3

32 Adjustment of Boundary Conditions The order of thecoefficient matrix (28) is (119873 + 1) times (119873 + 13) In orderto get unique solution of the system we need twelve moreconditions Here we consider only two different cases Incoming section we will show that the approximate solutioncan be improved by adjusting different boundary conditions

Case 1 If we assume 11991110158400= 0 (equivalently 119911

0= 119910119903= finite)

then two conditions can be achieved by using following givenboundary conditions i-e

1199110= 119910119903 119911

1015840

0= 0 119911

119873= 119910119897 (29)

Still we need ten more conditions to get stable system Sincesubdivision scheme reproduces seven degree polynomials wedefine boundary conditions of order eight for solution of (27)For simplicity only the left end points are discussed and thevalues of right end points 119911

119873+1 119911119873+2

119911119873+3

119911119873+4

119911119873+5

can betreated similarly

The values 119911minus5 119911minus4 119911minus3 119911minus2 119911minus1

can be determined bythe septic polynomial 119902(119909) interpolating at (119909

119894 119911119894) 0 le 119894 le 7

Precisely we have

119911minus119894= 119902 (minus119909

119894) 119894 = 1 2 3 4 5 (30)

where

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119885(119909119894minus119895) (31)

Since by (14) 119885(119909119894) = 119911119894for 119894 = 1 2 3 4 5 then by replacing

119909119894by minus119909

119894 we have

119902 (minus119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911minus119894+119895

(32)

Hence the following boundary conditions can be employed atthe left end

8

sum

119895=0

(8

119895) (minus1)

119895119911minus119894+119895

= 0 119894 = 5 4 3 2 1 (33)

Similarly for the right end we can define 119911119894= 119902(minus119909

119894) 119894 =

119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5 and

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911119894minus119895 (34)

So we have the following boundary conditions at the rightend

8

sum

119895=0

(8

119895) (minus1)

119895119911119894minus119895

= 0

119894 = 119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5

(35)

Finally we get the following new system of (119873 + 13) linearequations with (119873 + 13) unknowns 119911

119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) and (35)

119861119885 = 119877 (36)

where the coefficients matrix 119861 = (1198611198790 119860119879 119861119879

1)119879 119860 is defined

by (28) and 1198610and 119861

1are formed by (29) (33) and (35)

Abstract and Applied Analysis 7

1198610=

(((((((

(

0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 sdot sdot sdot 0 0

0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 sdot sdot sdot 0 0

0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 sdot sdot sdot 0 0

0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 sdot sdot sdot 0 0

0 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 sdot sdot sdot 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0 0 sdot sdot sdot 0 0

)))))))

)

(37)

where the first five rows of 1198610come from (33) and the sixth

row comes from (29) at 1199110= 119910119903 Consider

1198611=

(((((((

(

0 0 sdot sdot sdot 0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 sdot sdot sdot 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 0

0 0 sdot sdot sdot 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0

0 0 sdot sdot sdot 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0

0 0 sdot sdot sdot 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0

0 0 sdot sdot sdot 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0

)))))))

)

(38)

where first row of 1198611comes from (29) at 119911

119873= 119910119897remaining

rows come from (35) and the matrices119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 0 119910119903 119863119879 119910119897 0 0 0 0 0)

119879

(39)

Case 2 In this case we express the given boundary condition1199111015840

0= 0 in the following wayBy using (16) we have

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ) (40)

As we defined earlier 119909119895= 119895ℎ if we put 119895 = 0 we get 119909

0= 0

the above equation can be written as

1198851015840(0) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) (41)

Since by boundary condition 11991110158400= 1198851015840(0) = 0

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) = 0 (42)

By using (13) we can express above equation as

1

594636119911minus6+

256

743295119911minus5+

2645

594636119911minus4minus3328

49553119911minus3

+76113

198212119911minus2minus78592

49553119911minus1+78592

495531199111minus76113

1982121199112

+3328

495531199113minus

2645

5946361199114minus

256

7432951199115minus

1

5946361199116= 0

(43)Finally we get a following new system of (119873 + 13) linear

equations with (119873 + 13) unknowns 119911119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) for 119894 = 5 4 3 2 (35) and (43)

119861119885 = 119877 (44)

where the coefficients matrix 119861 = (B1198790 119860119879 119861119879

1)119879 119860 is defined

by (28) andB0and 119861

1formed by (29) (33) (35) and (43) are

defined as

8 Abstract and Applied Analysis

B0= (

(

0 1 minus8 28 minus56 70 minus56 28 minus8

0 0 1 minus8 28 minus56 70 minus56 28

0 0 0 1 minus8 28 minus56 70 minus56

0 0 0 0 1 minus8 28 minus56 70

1

594636

256

743295

2645

594636minus3328

49553

76113

198212minus78592

495530

78592

49553

76113

198212

0 0 0 0 0 0 1 0 0

1 0 0 0 0 sdot sdot sdot 0 0

minus8 1 0 0 0 sdot sdot sdot 0 0

28 minus8 1 0 0 sdot sdot sdot 0 0

minus56 28 minus8 1 0 sdot sdot sdot 0 0

3328

49553minus2645

594636minus

256

743295minus

1

5946360 sdot sdot sdot 0 0

0 0 0 0 0 sdot sdot sdot 0 0

)

)

(45)

in B first four rows come from (33) for 119894 = 5 4 3 2 fifth rowcomes from (43) and the last row comes from (29)

The matrix 1198611is same as defined in Case 1 and the

matrices 119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 1199101015840(0) 119910

119903 119863119879 119910119897 0 0 0 0 0)

119879

(46)

The nonsingularity of the coefficients matrix 119861 has beendiscussed in next section

33 Existence of the Solution In this section we discuss thenonsingularity of the coefficients matrix 119861 We observe thatthe coefficients matrix 119861 is neither symmetric nor diagonallydominant However it can be shown that 119861 is a nonsingularSince 119861 is almost a band matrix with half band width 7numerical complexity for solving the linear system usingGaussian elimination is about 49(119873 + 9)multiplications Forlarge 119873 the matrix is almost symmetric except the first andlast six rows and columns due to the boundary conditionsTherefore we first consider the symmetric part of it that issquare symmetric matrix 119862 of order119873 + 3 defined as

119862 =(

(

120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1

)

)

(47)

It can be shown that 119862 is always nonsingular for each valueof 119873 However 119861 is nonsingular for 119873 ⩽ 1000 We havechecked the nonsingularity of matrix 119861 by different methodsIn first method we observe that the determinants of matrix 119861increase as 119873 increases and it is not zero for 119873 ⩽ 1000 So119861 is nonsingular The determinants of 119861 at some values of119873are shown in Table 1 In second method we observe that for119873 ⩽ 1000 the eigenvalues of matrix 119861 are nonzero so by [16]matrix119861 is nonsingular However for119873 gt 1000matrix119861mayor may not be nonsingular Therefore we claim that systems(36) and (44) are stable for119873 ⩽ 1000

4 Error Estimation

In this section we discuss the approximation properties ofthe numerical interpolating collocation algorithm Since thescheme (8) reproduces polynomial curve of degree seven soby Dyn [14] scheme has approximation order eight So thecollocation algorithm (14) with septic precision treatments at

the endpoints has the power of approximation 119874(ℎ2) Herewe present our main result for error estimation

Proposition 8 Suppose the exact solution 119910(119909) isin 1198628[0 1]

and 119911119894 are obtained by solving either (36) or (44) with 8th

order boundary condition at the end points then

err (119909)infin =10038171003817100381710038171003817119885119895minus 11991011989510038171003817100381710038171003817infin

= 119874 (ℎ2minus119895) 119895 = 0 1 2 (48)

where 119895 denotes the order of derivative

Proof Since the order of approximation of subdivisionscheme (8) is eight so by using (13) we can write for smoothfunction 119910(119909) and small ℎ as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910 (119909119895minus 6ℎ) minus 11520119910 (119909

119895minus 5ℎ)

+ 10952119910 (119909119895minus 4ℎ) + 476928119910 (119909

119895minus 3ℎ)

Abstract and Applied Analysis 9

minus 3047987119910 (119909119895minus 2ℎ) + 4677632119910 (119909

119895minus ℎ)

minus 4677632119910 (119909119895+ ℎ) + 3047987119910 (119909

119895+ 2ℎ)

minus 476928119910 (119909119895+ 3ℎ) minus 10952119910 (119909

119895+ 4ℎ)

+ 11520119910 (119909119895+ 5ℎ) minus225119910 (119909

119895+ 6ℎ)

(49)

This can be written as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910119895minus6

minus 11520119910119895minus5

+ 10952119910119895minus4

+ 476928119910119895minus3

minus 3047987119910119895minus2

+ 4677632119910119895minus1

minus 4677632119910119895+1

+ 3047987119910119895+2

minus 476928119910119895+3

minus10952119910119895+4

+ 11520119910119895+5

minus 225119910119895+6

(50)

Similarly we have

119885101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119911119895minus6

minus 11520119911119895minus5

+ 10952119911119895minus4

+ 476928119911119895minus3

minus 3047987119911119895minus2

+ 4677632119911119895minus1

minus 4677632119911119895+1

+ 3047987119911119895+2

minus 476928119911119895+3

minus10952119911119895+4

+ 11520119911119895+5

minus 225119911119895+6

(51)

If we define error function 119890(119909) = 119885(119909) minus 119910(119909) and errorvectors at the nodes by

119890 (119909119895) = 119885 (119909

119895) minus 119910 (119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 (52)

or equivalently 119890119895= 119885119895minus 119910(119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 then

this implies

1198901015840

119895= 1198851015840

119895minus 1199101015840(119909 + 119895ℎ)

11989010158401015840

119895= 11988510158401015840

119895minus 11991010158401015840(119909 + 119895ℎ)

119890101584010158401015840

119895= 119885101584010158401015840

119895minus 119910101584010158401015840(119909 + 119895ℎ)

(53)

Table 1 Determinants of the matrices

119873 119862 119861

10 minus866756 11048870018371741

50 minus177183 14981270309

100 minus552709050 11964492

500 minus5033491471916955 times 1036

4728852755761116 times 1021

1000 minus4477989536166907 times 1071

42069711017699999 times 1040

By subtracting (50) from (51) we get

119885101584010158401015840(119909119895) minus 119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225 (119911119895minus6

minus 119910119895minus6) minus 11520 (119911

119895minus5minus 119910119895minus5)

+ 10952 (119911119895minus4

minus 119910119895minus4) + 476928 (119911

119895minus3minus 119910119895minus3)

minus 3047987 (119911119895minus2

minus 119910119895minus2) + 4677632 (119911

119895minus1minus 119910119895minus1)

minus 4677632 (119911119895+1

minus 119910119895+1) + 3047987 (119911

119895+2minus 119910119895+2)

minus 476928 (119911119895+3

minus 119910119895+3) minus 10952 (119911

119895+4minus 119910119895+4)

+11520 (119911119895+5

minus 119910119895+5) minus 225 (119911

119895+6minus 119910119895+6)

(54)

This implies

119890101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119890119895minus6

minus 11520119890119895minus5

+ 10952119890119895minus4

+ 476928119890119895minus3

minus 3047987119890119895minus2

+ 4677632119890119895minus1

minus 4677632119890119895+1

+ 3047987119890119895+2

minus 476928119890119895+3

minus10952119890119895+4

+ 11520119890119895+5

minus 225119890119895+6

(55)

By Lemma 5 we get the following expression

119890101584010158401015840

119895=1

ℎ3120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 120601101584010158401015840

0119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6 + 119874 (ℎ

8)

(56)

where 119895 = 0 1 2 119873By subtracting (1) from (15) we get

119885101584010158401015840

119895minus 119910101584010158401015840

119895= 119886119895(119885119894minus 119884119895) (57)

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

1

08

06

04

02

0

minus4 minus2 0 2 4 6

(a)

minus4minus6 minus2 0 2 4 6

15

1

05

0

minus05

minus1

minus15

(b)

1

0

minus1

minus2

minus4minus6 minus2 0 2 4 6

(c)

minus4minus2 0 2 4 6

03

02

01

0

minus01

minus02

minus03

(d)

Figure 1 Interpolatory basis function 1206013(119894) is shown in (a) first derivative of 120601

3(119894) i-e 1206011015840

3(119894) shown in (b) second derivative of 120601

3(119894) i-e 12060110158401015840

3(119894)

shown in (c) and third derivative of 1206013(119894) i-e 120601101584010158401015840

3(119894) shown in (d) respectively

Since the support of basis function 120601(119909) is (minus7 7) 1206011015840(119909)12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval (minus7 7) also

by (4) and (13) we get

119911minus6120601101584010158401015840(6) + 119911

minus5120601101584010158401015840(5) + 119911

minus4120601101584010158401015840(4) + 119911

minus3120601101584010158401015840(3)

+ 119911minus2120601101584010158401015840(2) + 119911

minus1120601101584010158401015840(1) + 119911

0120601101584010158401015840(0)

+ 1199111120601101584010158401015840(minus1) + 119911

2120601101584010158401015840(minus2) + 119911

3120601101584010158401015840(minus3)

+ 1199114120601101584010158401015840(minus4) + 119911

5120601101584010158401015840(minus5) + 119911

6120601101584010158401015840(minus6)

minus 1198860ℎ31199110120601 (0) = ℎ

31198870

(21)

If 120601101584010158401015840119895= 120601101584010158401015840(119895) then

119911minus6120601101584010158401015840

6+ 119911minus5120601101584010158401015840

5+ 119911minus4120601101584010158401015840

4+ 119911minus3120601101584010158401015840

3+ 119911minus2120601101584010158401015840

2

+ 119911minus1120601101584010158401015840

1+ 1199110(120601101584010158401015840

0minus 1198860ℎ3) + 1199111120601101584010158401015840

minus1+ 1199112120601101584010158401015840

minus2

+ 1199113120601101584010158401015840

minus3+ 1199114120601101584010158401015840

minus4+ 1199115120601101584010158401015840

minus5+ 1199116120601101584010158401015840

minus6= ℎ31198870

(22)

For 1199020= 120601101584010158401015840

0minus1198860ℎ3 we get (18)This completes the proof

Theorem 7 For 119895 = 1 2 3 119873 the system (17) is equivalentto

119911minus6120601101584010158401015840

119895+6+ 119911minus5120601101584010158401015840

119895+5+ sdot sdot sdot + 119911

0120601101584010158401015840

119895

+ 1199111(120601101584010158401015840

119895minus1minus 119886119895ℎ3120601119895minus1) + 1199112(120601101584010158401015840

119895minus2minus 119886119895ℎ3120601119895minus2) + sdot sdot sdot

+ 119911119873minus1

(120601101584010158401015840

119895minus119873+1minus 119886119895ℎ3120601119895minus119873+1

) + 119911119873(120601101584010158401015840

119895minus119873minus 119886119895ℎ3120601119895minus119873

)

+ 119911119873+1

120601101584010158401015840

119895minus119873minus1+ sdot sdot sdot + 119911

119873+6120601101584010158401015840

119895minus119873minus6= ℎ3119887119895

(23)

Proof By expanding (17) we get

119911minus6120601101584010158401015840(

119909119895minus 119909minus6

ℎ) + 119911minus5120601101584010158401015840(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601101584010158401015840(

119909119895minus 119909119873+6

ℎ)

minus 119886119895ℎ3119911minus6120601(

119909119895minus 119909minus6

ℎ) + 119911minus5120601(

119909119895minus 119909minus5

ℎ) + sdot sdot sdot

+119911119873+5

120601(

119909119895minus 119909119873+5

ℎ) + 119911119873+6

120601(

119909119895minus 119909119873+6

ℎ)

= ℎ3119887119895

(24)

For 119909119895= 119895ℎ 119895 = 1 2 119873 we get

119911minus6120601101584010158401015840(119895 + 6) + 119911

minus5120601101584010158401015840(119895 + 5) + sdot sdot sdot

+ 119911119873+5

120601101584010158401015840(119895 minus 119873 minus 5) + 119911

119873+6120601101584010158401015840(119895 minus 119873 minus 6)

minus 119886119895ℎ3119911minus6120601 (119895 + 6) + 119911

minus5120601 (119895 + 5) + sdot sdot sdot

+119911119873+5

120601 (119895 minus 119873 minus 5) + 119911119873+6

120601 (119895 minus 119873 minus 6)

= ℎ3119887119895

(25)

6 Abstract and Applied Analysis

If 120601101584010158401015840119895= 120601101584010158401015840(119895) for 119895 = 1 2 119873 then

119911minus6(120601101584010158401015840

119895+6minus 119886119895ℎ3120601119895+6) + 119911minus5(120601101584010158401015840

119895+5minus 119886119895ℎ3120601119895+5)

+ sdot sdot sdot + 119911119873+5

(120601101584010158401015840

119895minus119873minus5minus 119886119895ℎ3120601119895minus119873minus5

)

+ 119911119873+6

(120601101584010158401015840

119895minus119873minus6minus 119886119895ℎ3120601119895minus119873minus6

) = ℎ3119887119895

(26)

Since 1206011015840(119909) 12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval(minus7 7) then by (4) and (13) we get (23)

From (18) and (23) we get following undeterminedsystem of (119873 + 1) equations with (119873 + 13) unknowns 119911

119894

119860119885 = 119863 (27)

where the matrices 119860 119885 and119863 of orders (119873 + 1) times (119873 + 13)119873 + 13 and119873 + 1 respectively are given by

119860 =

(((

(

120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199020

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199021

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199022

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

0 0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199023

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus5120601101584010158401015840

minus60

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6

)))

)

(28)

119885 = (119911minus6 119911minus5 119911minus4 119911minus3 119911minus2 119911

119873+6)119879 and 119863 = (119887

0ℎ3 1198871ℎ3

1198872ℎ3 1198873ℎ3 119887

119873ℎ3)119879 where120601101584010158401015840

119895= 120601101584010158401015840(119895) and 119902

119895= 120601101584010158401015840

0minus119886119895ℎ3

32 Adjustment of Boundary Conditions The order of thecoefficient matrix (28) is (119873 + 1) times (119873 + 13) In orderto get unique solution of the system we need twelve moreconditions Here we consider only two different cases Incoming section we will show that the approximate solutioncan be improved by adjusting different boundary conditions

Case 1 If we assume 11991110158400= 0 (equivalently 119911

0= 119910119903= finite)

then two conditions can be achieved by using following givenboundary conditions i-e

1199110= 119910119903 119911

1015840

0= 0 119911

119873= 119910119897 (29)

Still we need ten more conditions to get stable system Sincesubdivision scheme reproduces seven degree polynomials wedefine boundary conditions of order eight for solution of (27)For simplicity only the left end points are discussed and thevalues of right end points 119911

119873+1 119911119873+2

119911119873+3

119911119873+4

119911119873+5

can betreated similarly

The values 119911minus5 119911minus4 119911minus3 119911minus2 119911minus1

can be determined bythe septic polynomial 119902(119909) interpolating at (119909

119894 119911119894) 0 le 119894 le 7

Precisely we have

119911minus119894= 119902 (minus119909

119894) 119894 = 1 2 3 4 5 (30)

where

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119885(119909119894minus119895) (31)

Since by (14) 119885(119909119894) = 119911119894for 119894 = 1 2 3 4 5 then by replacing

119909119894by minus119909

119894 we have

119902 (minus119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911minus119894+119895

(32)

Hence the following boundary conditions can be employed atthe left end

8

sum

119895=0

(8

119895) (minus1)

119895119911minus119894+119895

= 0 119894 = 5 4 3 2 1 (33)

Similarly for the right end we can define 119911119894= 119902(minus119909

119894) 119894 =

119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5 and

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911119894minus119895 (34)

So we have the following boundary conditions at the rightend

8

sum

119895=0

(8

119895) (minus1)

119895119911119894minus119895

= 0

119894 = 119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5

(35)

Finally we get the following new system of (119873 + 13) linearequations with (119873 + 13) unknowns 119911

119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) and (35)

119861119885 = 119877 (36)

where the coefficients matrix 119861 = (1198611198790 119860119879 119861119879

1)119879 119860 is defined

by (28) and 1198610and 119861

1are formed by (29) (33) and (35)

Abstract and Applied Analysis 7

1198610=

(((((((

(

0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 sdot sdot sdot 0 0

0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 sdot sdot sdot 0 0

0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 sdot sdot sdot 0 0

0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 sdot sdot sdot 0 0

0 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 sdot sdot sdot 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0 0 sdot sdot sdot 0 0

)))))))

)

(37)

where the first five rows of 1198610come from (33) and the sixth

row comes from (29) at 1199110= 119910119903 Consider

1198611=

(((((((

(

0 0 sdot sdot sdot 0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 sdot sdot sdot 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 0

0 0 sdot sdot sdot 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0

0 0 sdot sdot sdot 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0

0 0 sdot sdot sdot 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0

0 0 sdot sdot sdot 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0

)))))))

)

(38)

where first row of 1198611comes from (29) at 119911

119873= 119910119897remaining

rows come from (35) and the matrices119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 0 119910119903 119863119879 119910119897 0 0 0 0 0)

119879

(39)

Case 2 In this case we express the given boundary condition1199111015840

0= 0 in the following wayBy using (16) we have

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ) (40)

As we defined earlier 119909119895= 119895ℎ if we put 119895 = 0 we get 119909

0= 0

the above equation can be written as

1198851015840(0) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) (41)

Since by boundary condition 11991110158400= 1198851015840(0) = 0

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) = 0 (42)

By using (13) we can express above equation as

1

594636119911minus6+

256

743295119911minus5+

2645

594636119911minus4minus3328

49553119911minus3

+76113

198212119911minus2minus78592

49553119911minus1+78592

495531199111minus76113

1982121199112

+3328

495531199113minus

2645

5946361199114minus

256

7432951199115minus

1

5946361199116= 0

(43)Finally we get a following new system of (119873 + 13) linear

equations with (119873 + 13) unknowns 119911119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) for 119894 = 5 4 3 2 (35) and (43)

119861119885 = 119877 (44)

where the coefficients matrix 119861 = (B1198790 119860119879 119861119879

1)119879 119860 is defined

by (28) andB0and 119861

1formed by (29) (33) (35) and (43) are

defined as

8 Abstract and Applied Analysis

B0= (

(

0 1 minus8 28 minus56 70 minus56 28 minus8

0 0 1 minus8 28 minus56 70 minus56 28

0 0 0 1 minus8 28 minus56 70 minus56

0 0 0 0 1 minus8 28 minus56 70

1

594636

256

743295

2645

594636minus3328

49553

76113

198212minus78592

495530

78592

49553

76113

198212

0 0 0 0 0 0 1 0 0

1 0 0 0 0 sdot sdot sdot 0 0

minus8 1 0 0 0 sdot sdot sdot 0 0

28 minus8 1 0 0 sdot sdot sdot 0 0

minus56 28 minus8 1 0 sdot sdot sdot 0 0

3328

49553minus2645

594636minus

256

743295minus

1

5946360 sdot sdot sdot 0 0

0 0 0 0 0 sdot sdot sdot 0 0

)

)

(45)

in B first four rows come from (33) for 119894 = 5 4 3 2 fifth rowcomes from (43) and the last row comes from (29)

The matrix 1198611is same as defined in Case 1 and the

matrices 119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 1199101015840(0) 119910

119903 119863119879 119910119897 0 0 0 0 0)

119879

(46)

The nonsingularity of the coefficients matrix 119861 has beendiscussed in next section

33 Existence of the Solution In this section we discuss thenonsingularity of the coefficients matrix 119861 We observe thatthe coefficients matrix 119861 is neither symmetric nor diagonallydominant However it can be shown that 119861 is a nonsingularSince 119861 is almost a band matrix with half band width 7numerical complexity for solving the linear system usingGaussian elimination is about 49(119873 + 9)multiplications Forlarge 119873 the matrix is almost symmetric except the first andlast six rows and columns due to the boundary conditionsTherefore we first consider the symmetric part of it that issquare symmetric matrix 119862 of order119873 + 3 defined as

119862 =(

(

120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1

)

)

(47)

It can be shown that 119862 is always nonsingular for each valueof 119873 However 119861 is nonsingular for 119873 ⩽ 1000 We havechecked the nonsingularity of matrix 119861 by different methodsIn first method we observe that the determinants of matrix 119861increase as 119873 increases and it is not zero for 119873 ⩽ 1000 So119861 is nonsingular The determinants of 119861 at some values of119873are shown in Table 1 In second method we observe that for119873 ⩽ 1000 the eigenvalues of matrix 119861 are nonzero so by [16]matrix119861 is nonsingular However for119873 gt 1000matrix119861mayor may not be nonsingular Therefore we claim that systems(36) and (44) are stable for119873 ⩽ 1000

4 Error Estimation

In this section we discuss the approximation properties ofthe numerical interpolating collocation algorithm Since thescheme (8) reproduces polynomial curve of degree seven soby Dyn [14] scheme has approximation order eight So thecollocation algorithm (14) with septic precision treatments at

the endpoints has the power of approximation 119874(ℎ2) Herewe present our main result for error estimation

Proposition 8 Suppose the exact solution 119910(119909) isin 1198628[0 1]

and 119911119894 are obtained by solving either (36) or (44) with 8th

order boundary condition at the end points then

err (119909)infin =10038171003817100381710038171003817119885119895minus 11991011989510038171003817100381710038171003817infin

= 119874 (ℎ2minus119895) 119895 = 0 1 2 (48)

where 119895 denotes the order of derivative

Proof Since the order of approximation of subdivisionscheme (8) is eight so by using (13) we can write for smoothfunction 119910(119909) and small ℎ as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910 (119909119895minus 6ℎ) minus 11520119910 (119909

119895minus 5ℎ)

+ 10952119910 (119909119895minus 4ℎ) + 476928119910 (119909

119895minus 3ℎ)

Abstract and Applied Analysis 9

minus 3047987119910 (119909119895minus 2ℎ) + 4677632119910 (119909

119895minus ℎ)

minus 4677632119910 (119909119895+ ℎ) + 3047987119910 (119909

119895+ 2ℎ)

minus 476928119910 (119909119895+ 3ℎ) minus 10952119910 (119909

119895+ 4ℎ)

+ 11520119910 (119909119895+ 5ℎ) minus225119910 (119909

119895+ 6ℎ)

(49)

This can be written as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910119895minus6

minus 11520119910119895minus5

+ 10952119910119895minus4

+ 476928119910119895minus3

minus 3047987119910119895minus2

+ 4677632119910119895minus1

minus 4677632119910119895+1

+ 3047987119910119895+2

minus 476928119910119895+3

minus10952119910119895+4

+ 11520119910119895+5

minus 225119910119895+6

(50)

Similarly we have

119885101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119911119895minus6

minus 11520119911119895minus5

+ 10952119911119895minus4

+ 476928119911119895minus3

minus 3047987119911119895minus2

+ 4677632119911119895minus1

minus 4677632119911119895+1

+ 3047987119911119895+2

minus 476928119911119895+3

minus10952119911119895+4

+ 11520119911119895+5

minus 225119911119895+6

(51)

If we define error function 119890(119909) = 119885(119909) minus 119910(119909) and errorvectors at the nodes by

119890 (119909119895) = 119885 (119909

119895) minus 119910 (119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 (52)

or equivalently 119890119895= 119885119895minus 119910(119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 then

this implies

1198901015840

119895= 1198851015840

119895minus 1199101015840(119909 + 119895ℎ)

11989010158401015840

119895= 11988510158401015840

119895minus 11991010158401015840(119909 + 119895ℎ)

119890101584010158401015840

119895= 119885101584010158401015840

119895minus 119910101584010158401015840(119909 + 119895ℎ)

(53)

Table 1 Determinants of the matrices

119873 119862 119861

10 minus866756 11048870018371741

50 minus177183 14981270309

100 minus552709050 11964492

500 minus5033491471916955 times 1036

4728852755761116 times 1021

1000 minus4477989536166907 times 1071

42069711017699999 times 1040

By subtracting (50) from (51) we get

119885101584010158401015840(119909119895) minus 119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225 (119911119895minus6

minus 119910119895minus6) minus 11520 (119911

119895minus5minus 119910119895minus5)

+ 10952 (119911119895minus4

minus 119910119895minus4) + 476928 (119911

119895minus3minus 119910119895minus3)

minus 3047987 (119911119895minus2

minus 119910119895minus2) + 4677632 (119911

119895minus1minus 119910119895minus1)

minus 4677632 (119911119895+1

minus 119910119895+1) + 3047987 (119911

119895+2minus 119910119895+2)

minus 476928 (119911119895+3

minus 119910119895+3) minus 10952 (119911

119895+4minus 119910119895+4)

+11520 (119911119895+5

minus 119910119895+5) minus 225 (119911

119895+6minus 119910119895+6)

(54)

This implies

119890101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119890119895minus6

minus 11520119890119895minus5

+ 10952119890119895minus4

+ 476928119890119895minus3

minus 3047987119890119895minus2

+ 4677632119890119895minus1

minus 4677632119890119895+1

+ 3047987119890119895+2

minus 476928119890119895+3

minus10952119890119895+4

+ 11520119890119895+5

minus 225119890119895+6

(55)

By Lemma 5 we get the following expression

119890101584010158401015840

119895=1

ℎ3120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 120601101584010158401015840

0119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6 + 119874 (ℎ

8)

(56)

where 119895 = 0 1 2 119873By subtracting (1) from (15) we get

119885101584010158401015840

119895minus 119910101584010158401015840

119895= 119886119895(119885119894minus 119884119895) (57)

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

If 120601101584010158401015840119895= 120601101584010158401015840(119895) for 119895 = 1 2 119873 then

119911minus6(120601101584010158401015840

119895+6minus 119886119895ℎ3120601119895+6) + 119911minus5(120601101584010158401015840

119895+5minus 119886119895ℎ3120601119895+5)

+ sdot sdot sdot + 119911119873+5

(120601101584010158401015840

119895minus119873minus5minus 119886119895ℎ3120601119895minus119873minus5

)

+ 119911119873+6

(120601101584010158401015840

119895minus119873minus6minus 119886119895ℎ3120601119895minus119873minus6

) = ℎ3119887119895

(26)

Since 1206011015840(119909) 12060110158401015840(119909) and 120601101584010158401015840(119909) are zero outside the interval(minus7 7) then by (4) and (13) we get (23)

From (18) and (23) we get following undeterminedsystem of (119873 + 1) equations with (119873 + 13) unknowns 119911

119894

119860119885 = 119863 (27)

where the matrices 119860 119885 and119863 of orders (119873 + 1) times (119873 + 13)119873 + 13 and119873 + 1 respectively are given by

119860 =

(((

(

120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199020

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199021

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199022

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

0 0 0 120601101584010158401015840

6120601101584010158401015840

5120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

11199023

120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus5120601101584010158401015840

minus60

0 0 0 0 0 0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6

)))

)

(28)

119885 = (119911minus6 119911minus5 119911minus4 119911minus3 119911minus2 119911

119873+6)119879 and 119863 = (119887

0ℎ3 1198871ℎ3

1198872ℎ3 1198873ℎ3 119887

119873ℎ3)119879 where120601101584010158401015840

119895= 120601101584010158401015840(119895) and 119902

119895= 120601101584010158401015840

0minus119886119895ℎ3

32 Adjustment of Boundary Conditions The order of thecoefficient matrix (28) is (119873 + 1) times (119873 + 13) In orderto get unique solution of the system we need twelve moreconditions Here we consider only two different cases Incoming section we will show that the approximate solutioncan be improved by adjusting different boundary conditions

Case 1 If we assume 11991110158400= 0 (equivalently 119911

0= 119910119903= finite)

then two conditions can be achieved by using following givenboundary conditions i-e

1199110= 119910119903 119911

1015840

0= 0 119911

119873= 119910119897 (29)

Still we need ten more conditions to get stable system Sincesubdivision scheme reproduces seven degree polynomials wedefine boundary conditions of order eight for solution of (27)For simplicity only the left end points are discussed and thevalues of right end points 119911

119873+1 119911119873+2

119911119873+3

119911119873+4

119911119873+5

can betreated similarly

The values 119911minus5 119911minus4 119911minus3 119911minus2 119911minus1

can be determined bythe septic polynomial 119902(119909) interpolating at (119909

119894 119911119894) 0 le 119894 le 7

Precisely we have

119911minus119894= 119902 (minus119909

119894) 119894 = 1 2 3 4 5 (30)

where

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119885(119909119894minus119895) (31)

Since by (14) 119885(119909119894) = 119911119894for 119894 = 1 2 3 4 5 then by replacing

119909119894by minus119909

119894 we have

119902 (minus119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911minus119894+119895

(32)

Hence the following boundary conditions can be employed atthe left end

8

sum

119895=0

(8

119895) (minus1)

119895119911minus119894+119895

= 0 119894 = 5 4 3 2 1 (33)

Similarly for the right end we can define 119911119894= 119902(minus119909

119894) 119894 =

119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5 and

119902 (119909119894) =

8

sum

119895=1

(8

119895) (minus1)

119895+1119911119894minus119895 (34)

So we have the following boundary conditions at the rightend

8

sum

119895=0

(8

119895) (minus1)

119895119911119894minus119895

= 0

119894 = 119873 + 1119873 + 2119873 + 3119873 + 4119873 + 5

(35)

Finally we get the following new system of (119873 + 13) linearequations with (119873 + 13) unknowns 119911

119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) and (35)

119861119885 = 119877 (36)

where the coefficients matrix 119861 = (1198611198790 119860119879 119861119879

1)119879 119860 is defined

by (28) and 1198610and 119861

1are formed by (29) (33) and (35)

Abstract and Applied Analysis 7

1198610=

(((((((

(

0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 sdot sdot sdot 0 0

0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 sdot sdot sdot 0 0

0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 sdot sdot sdot 0 0

0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 sdot sdot sdot 0 0

0 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 sdot sdot sdot 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0 0 sdot sdot sdot 0 0

)))))))

)

(37)

where the first five rows of 1198610come from (33) and the sixth

row comes from (29) at 1199110= 119910119903 Consider

1198611=

(((((((

(

0 0 sdot sdot sdot 0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 sdot sdot sdot 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 0

0 0 sdot sdot sdot 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0

0 0 sdot sdot sdot 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0

0 0 sdot sdot sdot 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0

0 0 sdot sdot sdot 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0

)))))))

)

(38)

where first row of 1198611comes from (29) at 119911

119873= 119910119897remaining

rows come from (35) and the matrices119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 0 119910119903 119863119879 119910119897 0 0 0 0 0)

119879

(39)

Case 2 In this case we express the given boundary condition1199111015840

0= 0 in the following wayBy using (16) we have

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ) (40)

As we defined earlier 119909119895= 119895ℎ if we put 119895 = 0 we get 119909

0= 0

the above equation can be written as

1198851015840(0) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) (41)

Since by boundary condition 11991110158400= 1198851015840(0) = 0

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) = 0 (42)

By using (13) we can express above equation as

1

594636119911minus6+

256

743295119911minus5+

2645

594636119911minus4minus3328

49553119911minus3

+76113

198212119911minus2minus78592

49553119911minus1+78592

495531199111minus76113

1982121199112

+3328

495531199113minus

2645

5946361199114minus

256

7432951199115minus

1

5946361199116= 0

(43)Finally we get a following new system of (119873 + 13) linear

equations with (119873 + 13) unknowns 119911119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) for 119894 = 5 4 3 2 (35) and (43)

119861119885 = 119877 (44)

where the coefficients matrix 119861 = (B1198790 119860119879 119861119879

1)119879 119860 is defined

by (28) andB0and 119861

1formed by (29) (33) (35) and (43) are

defined as

8 Abstract and Applied Analysis

B0= (

(

0 1 minus8 28 minus56 70 minus56 28 minus8

0 0 1 minus8 28 minus56 70 minus56 28

0 0 0 1 minus8 28 minus56 70 minus56

0 0 0 0 1 minus8 28 minus56 70

1

594636

256

743295

2645

594636minus3328

49553

76113

198212minus78592

495530

78592

49553

76113

198212

0 0 0 0 0 0 1 0 0

1 0 0 0 0 sdot sdot sdot 0 0

minus8 1 0 0 0 sdot sdot sdot 0 0

28 minus8 1 0 0 sdot sdot sdot 0 0

minus56 28 minus8 1 0 sdot sdot sdot 0 0

3328

49553minus2645

594636minus

256

743295minus

1

5946360 sdot sdot sdot 0 0

0 0 0 0 0 sdot sdot sdot 0 0

)

)

(45)

in B first four rows come from (33) for 119894 = 5 4 3 2 fifth rowcomes from (43) and the last row comes from (29)

The matrix 1198611is same as defined in Case 1 and the

matrices 119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 1199101015840(0) 119910

119903 119863119879 119910119897 0 0 0 0 0)

119879

(46)

The nonsingularity of the coefficients matrix 119861 has beendiscussed in next section

33 Existence of the Solution In this section we discuss thenonsingularity of the coefficients matrix 119861 We observe thatthe coefficients matrix 119861 is neither symmetric nor diagonallydominant However it can be shown that 119861 is a nonsingularSince 119861 is almost a band matrix with half band width 7numerical complexity for solving the linear system usingGaussian elimination is about 49(119873 + 9)multiplications Forlarge 119873 the matrix is almost symmetric except the first andlast six rows and columns due to the boundary conditionsTherefore we first consider the symmetric part of it that issquare symmetric matrix 119862 of order119873 + 3 defined as

119862 =(

(

120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1

)

)

(47)

It can be shown that 119862 is always nonsingular for each valueof 119873 However 119861 is nonsingular for 119873 ⩽ 1000 We havechecked the nonsingularity of matrix 119861 by different methodsIn first method we observe that the determinants of matrix 119861increase as 119873 increases and it is not zero for 119873 ⩽ 1000 So119861 is nonsingular The determinants of 119861 at some values of119873are shown in Table 1 In second method we observe that for119873 ⩽ 1000 the eigenvalues of matrix 119861 are nonzero so by [16]matrix119861 is nonsingular However for119873 gt 1000matrix119861mayor may not be nonsingular Therefore we claim that systems(36) and (44) are stable for119873 ⩽ 1000

4 Error Estimation

In this section we discuss the approximation properties ofthe numerical interpolating collocation algorithm Since thescheme (8) reproduces polynomial curve of degree seven soby Dyn [14] scheme has approximation order eight So thecollocation algorithm (14) with septic precision treatments at

the endpoints has the power of approximation 119874(ℎ2) Herewe present our main result for error estimation

Proposition 8 Suppose the exact solution 119910(119909) isin 1198628[0 1]

and 119911119894 are obtained by solving either (36) or (44) with 8th

order boundary condition at the end points then

err (119909)infin =10038171003817100381710038171003817119885119895minus 11991011989510038171003817100381710038171003817infin

= 119874 (ℎ2minus119895) 119895 = 0 1 2 (48)

where 119895 denotes the order of derivative

Proof Since the order of approximation of subdivisionscheme (8) is eight so by using (13) we can write for smoothfunction 119910(119909) and small ℎ as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910 (119909119895minus 6ℎ) minus 11520119910 (119909

119895minus 5ℎ)

+ 10952119910 (119909119895minus 4ℎ) + 476928119910 (119909

119895minus 3ℎ)

Abstract and Applied Analysis 9

minus 3047987119910 (119909119895minus 2ℎ) + 4677632119910 (119909

119895minus ℎ)

minus 4677632119910 (119909119895+ ℎ) + 3047987119910 (119909

119895+ 2ℎ)

minus 476928119910 (119909119895+ 3ℎ) minus 10952119910 (119909

119895+ 4ℎ)

+ 11520119910 (119909119895+ 5ℎ) minus225119910 (119909

119895+ 6ℎ)

(49)

This can be written as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910119895minus6

minus 11520119910119895minus5

+ 10952119910119895minus4

+ 476928119910119895minus3

minus 3047987119910119895minus2

+ 4677632119910119895minus1

minus 4677632119910119895+1

+ 3047987119910119895+2

minus 476928119910119895+3

minus10952119910119895+4

+ 11520119910119895+5

minus 225119910119895+6

(50)

Similarly we have

119885101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119911119895minus6

minus 11520119911119895minus5

+ 10952119911119895minus4

+ 476928119911119895minus3

minus 3047987119911119895minus2

+ 4677632119911119895minus1

minus 4677632119911119895+1

+ 3047987119911119895+2

minus 476928119911119895+3

minus10952119911119895+4

+ 11520119911119895+5

minus 225119911119895+6

(51)

If we define error function 119890(119909) = 119885(119909) minus 119910(119909) and errorvectors at the nodes by

119890 (119909119895) = 119885 (119909

119895) minus 119910 (119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 (52)

or equivalently 119890119895= 119885119895minus 119910(119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 then

this implies

1198901015840

119895= 1198851015840

119895minus 1199101015840(119909 + 119895ℎ)

11989010158401015840

119895= 11988510158401015840

119895minus 11991010158401015840(119909 + 119895ℎ)

119890101584010158401015840

119895= 119885101584010158401015840

119895minus 119910101584010158401015840(119909 + 119895ℎ)

(53)

Table 1 Determinants of the matrices

119873 119862 119861

10 minus866756 11048870018371741

50 minus177183 14981270309

100 minus552709050 11964492

500 minus5033491471916955 times 1036

4728852755761116 times 1021

1000 minus4477989536166907 times 1071

42069711017699999 times 1040

By subtracting (50) from (51) we get

119885101584010158401015840(119909119895) minus 119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225 (119911119895minus6

minus 119910119895minus6) minus 11520 (119911

119895minus5minus 119910119895minus5)

+ 10952 (119911119895minus4

minus 119910119895minus4) + 476928 (119911

119895minus3minus 119910119895minus3)

minus 3047987 (119911119895minus2

minus 119910119895minus2) + 4677632 (119911

119895minus1minus 119910119895minus1)

minus 4677632 (119911119895+1

minus 119910119895+1) + 3047987 (119911

119895+2minus 119910119895+2)

minus 476928 (119911119895+3

minus 119910119895+3) minus 10952 (119911

119895+4minus 119910119895+4)

+11520 (119911119895+5

minus 119910119895+5) minus 225 (119911

119895+6minus 119910119895+6)

(54)

This implies

119890101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119890119895minus6

minus 11520119890119895minus5

+ 10952119890119895minus4

+ 476928119890119895minus3

minus 3047987119890119895minus2

+ 4677632119890119895minus1

minus 4677632119890119895+1

+ 3047987119890119895+2

minus 476928119890119895+3

minus10952119890119895+4

+ 11520119890119895+5

minus 225119890119895+6

(55)

By Lemma 5 we get the following expression

119890101584010158401015840

119895=1

ℎ3120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 120601101584010158401015840

0119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6 + 119874 (ℎ

8)

(56)

where 119895 = 0 1 2 119873By subtracting (1) from (15) we get

119885101584010158401015840

119895minus 119910101584010158401015840

119895= 119886119895(119885119894minus 119884119895) (57)

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

1198610=

(((((((

(

0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 sdot sdot sdot 0 0

0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 sdot sdot sdot 0 0

0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 sdot sdot sdot 0 0

0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 sdot sdot sdot 0 0

0 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 sdot sdot sdot 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0 0 sdot sdot sdot 0 0

)))))))

)

(37)

where the first five rows of 1198610come from (33) and the sixth

row comes from (29) at 1199110= 119910119903 Consider

1198611=

(((((((

(

0 0 sdot sdot sdot 0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 sdot sdot sdot 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0 0

0 0 sdot sdot sdot 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0 0

0 0 sdot sdot sdot 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0 0

0 0 sdot sdot sdot 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0 0

0 0 sdot sdot sdot 0 0 0 0 1 minus8 28 minus56 70 minus56 28 minus8 1 0

)))))))

)

(38)

where first row of 1198611comes from (29) at 119911

119873= 119910119897remaining

rows come from (35) and the matrices119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 0 119910119903 119863119879 119910119897 0 0 0 0 0)

119879

(39)

Case 2 In this case we express the given boundary condition1199111015840

0= 0 in the following wayBy using (16) we have

1198851015840(119909119895) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(

119909119895minus 119909119894

ℎ) (40)

As we defined earlier 119909119895= 119895ℎ if we put 119895 = 0 we get 119909

0= 0

the above equation can be written as

1198851015840(0) =

1

ℎ

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) (41)

Since by boundary condition 11991110158400= 1198851015840(0) = 0

119873+6

sum

119894=minus6

1199111198941206011015840(minus119894) = 0 (42)

By using (13) we can express above equation as

1

594636119911minus6+

256

743295119911minus5+

2645

594636119911minus4minus3328

49553119911minus3

+76113

198212119911minus2minus78592

49553119911minus1+78592

495531199111minus76113

1982121199112

+3328

495531199113minus

2645

5946361199114minus

256

7432951199115minus

1

5946361199116= 0

(43)Finally we get a following new system of (119873 + 13) linear

equations with (119873 + 13) unknowns 119911119894 in which 119873 + 1

equations are obtained from (18) and (23) two equationsfrom boundary conditions (29) and ten from boundaryconditions (33) for 119894 = 5 4 3 2 (35) and (43)

119861119885 = 119877 (44)

where the coefficients matrix 119861 = (B1198790 119860119879 119861119879

1)119879 119860 is defined

by (28) andB0and 119861

1formed by (29) (33) (35) and (43) are

defined as

8 Abstract and Applied Analysis

B0= (

(

0 1 minus8 28 minus56 70 minus56 28 minus8

0 0 1 minus8 28 minus56 70 minus56 28

0 0 0 1 minus8 28 minus56 70 minus56

0 0 0 0 1 minus8 28 minus56 70

1

594636

256

743295

2645

594636minus3328

49553

76113

198212minus78592

495530

78592

49553

76113

198212

0 0 0 0 0 0 1 0 0

1 0 0 0 0 sdot sdot sdot 0 0

minus8 1 0 0 0 sdot sdot sdot 0 0

28 minus8 1 0 0 sdot sdot sdot 0 0

minus56 28 minus8 1 0 sdot sdot sdot 0 0

3328

49553minus2645

594636minus

256

743295minus

1

5946360 sdot sdot sdot 0 0

0 0 0 0 0 sdot sdot sdot 0 0

)

)

(45)

in B first four rows come from (33) for 119894 = 5 4 3 2 fifth rowcomes from (43) and the last row comes from (29)

The matrix 1198611is same as defined in Case 1 and the

matrices 119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 1199101015840(0) 119910

119903 119863119879 119910119897 0 0 0 0 0)

119879

(46)

The nonsingularity of the coefficients matrix 119861 has beendiscussed in next section

33 Existence of the Solution In this section we discuss thenonsingularity of the coefficients matrix 119861 We observe thatthe coefficients matrix 119861 is neither symmetric nor diagonallydominant However it can be shown that 119861 is a nonsingularSince 119861 is almost a band matrix with half band width 7numerical complexity for solving the linear system usingGaussian elimination is about 49(119873 + 9)multiplications Forlarge 119873 the matrix is almost symmetric except the first andlast six rows and columns due to the boundary conditionsTherefore we first consider the symmetric part of it that issquare symmetric matrix 119862 of order119873 + 3 defined as

119862 =(

(

120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1

)

)

(47)

It can be shown that 119862 is always nonsingular for each valueof 119873 However 119861 is nonsingular for 119873 ⩽ 1000 We havechecked the nonsingularity of matrix 119861 by different methodsIn first method we observe that the determinants of matrix 119861increase as 119873 increases and it is not zero for 119873 ⩽ 1000 So119861 is nonsingular The determinants of 119861 at some values of119873are shown in Table 1 In second method we observe that for119873 ⩽ 1000 the eigenvalues of matrix 119861 are nonzero so by [16]matrix119861 is nonsingular However for119873 gt 1000matrix119861mayor may not be nonsingular Therefore we claim that systems(36) and (44) are stable for119873 ⩽ 1000

4 Error Estimation

In this section we discuss the approximation properties ofthe numerical interpolating collocation algorithm Since thescheme (8) reproduces polynomial curve of degree seven soby Dyn [14] scheme has approximation order eight So thecollocation algorithm (14) with septic precision treatments at

the endpoints has the power of approximation 119874(ℎ2) Herewe present our main result for error estimation

Proposition 8 Suppose the exact solution 119910(119909) isin 1198628[0 1]

and 119911119894 are obtained by solving either (36) or (44) with 8th

order boundary condition at the end points then

err (119909)infin =10038171003817100381710038171003817119885119895minus 11991011989510038171003817100381710038171003817infin

= 119874 (ℎ2minus119895) 119895 = 0 1 2 (48)

where 119895 denotes the order of derivative

Proof Since the order of approximation of subdivisionscheme (8) is eight so by using (13) we can write for smoothfunction 119910(119909) and small ℎ as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910 (119909119895minus 6ℎ) minus 11520119910 (119909

119895minus 5ℎ)

+ 10952119910 (119909119895minus 4ℎ) + 476928119910 (119909

119895minus 3ℎ)

Abstract and Applied Analysis 9

minus 3047987119910 (119909119895minus 2ℎ) + 4677632119910 (119909

119895minus ℎ)

minus 4677632119910 (119909119895+ ℎ) + 3047987119910 (119909

119895+ 2ℎ)

minus 476928119910 (119909119895+ 3ℎ) minus 10952119910 (119909

119895+ 4ℎ)

+ 11520119910 (119909119895+ 5ℎ) minus225119910 (119909

119895+ 6ℎ)

(49)

This can be written as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910119895minus6

minus 11520119910119895minus5

+ 10952119910119895minus4

+ 476928119910119895minus3

minus 3047987119910119895minus2

+ 4677632119910119895minus1

minus 4677632119910119895+1

+ 3047987119910119895+2

minus 476928119910119895+3

minus10952119910119895+4

+ 11520119910119895+5

minus 225119910119895+6

(50)

Similarly we have

119885101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119911119895minus6

minus 11520119911119895minus5

+ 10952119911119895minus4

+ 476928119911119895minus3

minus 3047987119911119895minus2

+ 4677632119911119895minus1

minus 4677632119911119895+1

+ 3047987119911119895+2

minus 476928119911119895+3

minus10952119911119895+4

+ 11520119911119895+5

minus 225119911119895+6

(51)

If we define error function 119890(119909) = 119885(119909) minus 119910(119909) and errorvectors at the nodes by

119890 (119909119895) = 119885 (119909

119895) minus 119910 (119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 (52)

or equivalently 119890119895= 119885119895minus 119910(119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 then

this implies

1198901015840

119895= 1198851015840

119895minus 1199101015840(119909 + 119895ℎ)

11989010158401015840

119895= 11988510158401015840

119895minus 11991010158401015840(119909 + 119895ℎ)

119890101584010158401015840

119895= 119885101584010158401015840

119895minus 119910101584010158401015840(119909 + 119895ℎ)

(53)

Table 1 Determinants of the matrices

119873 119862 119861

10 minus866756 11048870018371741

50 minus177183 14981270309

100 minus552709050 11964492

500 minus5033491471916955 times 1036

4728852755761116 times 1021

1000 minus4477989536166907 times 1071

42069711017699999 times 1040

By subtracting (50) from (51) we get

119885101584010158401015840(119909119895) minus 119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225 (119911119895minus6

minus 119910119895minus6) minus 11520 (119911

119895minus5minus 119910119895minus5)

+ 10952 (119911119895minus4

minus 119910119895minus4) + 476928 (119911

119895minus3minus 119910119895minus3)

minus 3047987 (119911119895minus2

minus 119910119895minus2) + 4677632 (119911

119895minus1minus 119910119895minus1)

minus 4677632 (119911119895+1

minus 119910119895+1) + 3047987 (119911

119895+2minus 119910119895+2)

minus 476928 (119911119895+3

minus 119910119895+3) minus 10952 (119911

119895+4minus 119910119895+4)

+11520 (119911119895+5

minus 119910119895+5) minus 225 (119911

119895+6minus 119910119895+6)

(54)

This implies

119890101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119890119895minus6

minus 11520119890119895minus5

+ 10952119890119895minus4

+ 476928119890119895minus3

minus 3047987119890119895minus2

+ 4677632119890119895minus1

minus 4677632119890119895+1

+ 3047987119890119895+2

minus 476928119890119895+3

minus10952119890119895+4

+ 11520119890119895+5

minus 225119890119895+6

(55)

By Lemma 5 we get the following expression

119890101584010158401015840

119895=1

ℎ3120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 120601101584010158401015840

0119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6 + 119874 (ℎ

8)

(56)

where 119895 = 0 1 2 119873By subtracting (1) from (15) we get

119885101584010158401015840

119895minus 119910101584010158401015840

119895= 119886119895(119885119894minus 119884119895) (57)

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

B0= (

(

0 1 minus8 28 minus56 70 minus56 28 minus8

0 0 1 minus8 28 minus56 70 minus56 28

0 0 0 1 minus8 28 minus56 70 minus56

0 0 0 0 1 minus8 28 minus56 70

1

594636

256

743295

2645

594636minus3328

49553

76113

198212minus78592

495530

78592

49553

76113

198212

0 0 0 0 0 0 1 0 0

1 0 0 0 0 sdot sdot sdot 0 0

minus8 1 0 0 0 sdot sdot sdot 0 0

28 minus8 1 0 0 sdot sdot sdot 0 0

minus56 28 minus8 1 0 sdot sdot sdot 0 0

3328

49553minus2645

594636minus

256

743295minus

1

5946360 sdot sdot sdot 0 0

0 0 0 0 0 sdot sdot sdot 0 0

)

)

(45)

in B first four rows come from (33) for 119894 = 5 4 3 2 fifth rowcomes from (43) and the last row comes from (29)

The matrix 1198611is same as defined in Case 1 and the

matrices 119885 and 119877 are defined as

119885 = (119911minus6 119911minus5 119911

119873+5 119911119873+6

)119879

119877 = (0 0 0 0 1199101015840(0) 119910

119903 119863119879 119910119897 0 0 0 0 0)

119879

(46)

The nonsingularity of the coefficients matrix 119861 has beendiscussed in next section

33 Existence of the Solution In this section we discuss thenonsingularity of the coefficients matrix 119861 We observe thatthe coefficients matrix 119861 is neither symmetric nor diagonallydominant However it can be shown that 119861 is a nonsingularSince 119861 is almost a band matrix with half band width 7numerical complexity for solving the linear system usingGaussian elimination is about 49(119873 + 9)multiplications Forlarge 119873 the matrix is almost symmetric except the first andlast six rows and columns due to the boundary conditionsTherefore we first consider the symmetric part of it that issquare symmetric matrix 119862 of order119873 + 3 defined as

119862 =(

(

120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5120601101584010158401015840

minus6sdot sdot sdot 0 0 0

120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4120601101584010158401015840

minus5sdot sdot sdot 0 0 0

120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3120601101584010158401015840

minus4sdot sdot sdot 0 0 0

120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1120601101584010158401015840

0120601101584010158401015840

minus1120601101584010158401015840

minus2120601101584010158401015840

minus3sdot sdot sdot 0 0 0

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

4120601101584010158401015840

3120601101584010158401015840

2

0 0 0 0 0 0 0 0 sdot sdot sdot 120601101584010158401015840

3120601101584010158401015840

2120601101584010158401015840

1

)

)

(47)

It can be shown that 119862 is always nonsingular for each valueof 119873 However 119861 is nonsingular for 119873 ⩽ 1000 We havechecked the nonsingularity of matrix 119861 by different methodsIn first method we observe that the determinants of matrix 119861increase as 119873 increases and it is not zero for 119873 ⩽ 1000 So119861 is nonsingular The determinants of 119861 at some values of119873are shown in Table 1 In second method we observe that for119873 ⩽ 1000 the eigenvalues of matrix 119861 are nonzero so by [16]matrix119861 is nonsingular However for119873 gt 1000matrix119861mayor may not be nonsingular Therefore we claim that systems(36) and (44) are stable for119873 ⩽ 1000

4 Error Estimation

In this section we discuss the approximation properties ofthe numerical interpolating collocation algorithm Since thescheme (8) reproduces polynomial curve of degree seven soby Dyn [14] scheme has approximation order eight So thecollocation algorithm (14) with septic precision treatments at

the endpoints has the power of approximation 119874(ℎ2) Herewe present our main result for error estimation

Proposition 8 Suppose the exact solution 119910(119909) isin 1198628[0 1]

and 119911119894 are obtained by solving either (36) or (44) with 8th

order boundary condition at the end points then

err (119909)infin =10038171003817100381710038171003817119885119895minus 11991011989510038171003817100381710038171003817infin

= 119874 (ℎ2minus119895) 119895 = 0 1 2 (48)

where 119895 denotes the order of derivative

Proof Since the order of approximation of subdivisionscheme (8) is eight so by using (13) we can write for smoothfunction 119910(119909) and small ℎ as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910 (119909119895minus 6ℎ) minus 11520119910 (119909

119895minus 5ℎ)

+ 10952119910 (119909119895minus 4ℎ) + 476928119910 (119909

119895minus 3ℎ)

Abstract and Applied Analysis 9

minus 3047987119910 (119909119895minus 2ℎ) + 4677632119910 (119909

119895minus ℎ)

minus 4677632119910 (119909119895+ ℎ) + 3047987119910 (119909

119895+ 2ℎ)

minus 476928119910 (119909119895+ 3ℎ) minus 10952119910 (119909

119895+ 4ℎ)

+ 11520119910 (119909119895+ 5ℎ) minus225119910 (119909

119895+ 6ℎ)

(49)

This can be written as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910119895minus6

minus 11520119910119895minus5

+ 10952119910119895minus4

+ 476928119910119895minus3

minus 3047987119910119895minus2

+ 4677632119910119895minus1

minus 4677632119910119895+1

+ 3047987119910119895+2

minus 476928119910119895+3

minus10952119910119895+4

+ 11520119910119895+5

minus 225119910119895+6

(50)

Similarly we have

119885101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119911119895minus6

minus 11520119911119895minus5

+ 10952119911119895minus4

+ 476928119911119895minus3

minus 3047987119911119895minus2

+ 4677632119911119895minus1

minus 4677632119911119895+1

+ 3047987119911119895+2

minus 476928119911119895+3

minus10952119911119895+4

+ 11520119911119895+5

minus 225119911119895+6

(51)

If we define error function 119890(119909) = 119885(119909) minus 119910(119909) and errorvectors at the nodes by

119890 (119909119895) = 119885 (119909

119895) minus 119910 (119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 (52)

or equivalently 119890119895= 119885119895minus 119910(119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 then

this implies

1198901015840

119895= 1198851015840

119895minus 1199101015840(119909 + 119895ℎ)

11989010158401015840

119895= 11988510158401015840

119895minus 11991010158401015840(119909 + 119895ℎ)

119890101584010158401015840

119895= 119885101584010158401015840

119895minus 119910101584010158401015840(119909 + 119895ℎ)

(53)

Table 1 Determinants of the matrices

119873 119862 119861

10 minus866756 11048870018371741

50 minus177183 14981270309

100 minus552709050 11964492

500 minus5033491471916955 times 1036

4728852755761116 times 1021

1000 minus4477989536166907 times 1071

42069711017699999 times 1040

By subtracting (50) from (51) we get

119885101584010158401015840(119909119895) minus 119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225 (119911119895minus6

minus 119910119895minus6) minus 11520 (119911

119895minus5minus 119910119895minus5)

+ 10952 (119911119895minus4

minus 119910119895minus4) + 476928 (119911

119895minus3minus 119910119895minus3)

minus 3047987 (119911119895minus2

minus 119910119895minus2) + 4677632 (119911

119895minus1minus 119910119895minus1)

minus 4677632 (119911119895+1

minus 119910119895+1) + 3047987 (119911

119895+2minus 119910119895+2)

minus 476928 (119911119895+3

minus 119910119895+3) minus 10952 (119911

119895+4minus 119910119895+4)

+11520 (119911119895+5

minus 119910119895+5) minus 225 (119911

119895+6minus 119910119895+6)

(54)

This implies

119890101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119890119895minus6

minus 11520119890119895minus5

+ 10952119890119895minus4

+ 476928119890119895minus3

minus 3047987119890119895minus2

+ 4677632119890119895minus1

minus 4677632119890119895+1

+ 3047987119890119895+2

minus 476928119890119895+3

minus10952119890119895+4

+ 11520119890119895+5

minus 225119890119895+6

(55)

By Lemma 5 we get the following expression

119890101584010158401015840

119895=1

ℎ3120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 120601101584010158401015840

0119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6 + 119874 (ℎ

8)

(56)

where 119895 = 0 1 2 119873By subtracting (1) from (15) we get

119885101584010158401015840

119895minus 119910101584010158401015840

119895= 119886119895(119885119894minus 119884119895) (57)

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 9

minus 3047987119910 (119909119895minus 2ℎ) + 4677632119910 (119909

119895minus ℎ)

minus 4677632119910 (119909119895+ ℎ) + 3047987119910 (119909

119895+ 2ℎ)

minus 476928119910 (119909119895+ 3ℎ) minus 10952119910 (119909

119895+ 4ℎ)

+ 11520119910 (119909119895+ 5ℎ) minus225119910 (119909

119895+ 6ℎ)

(49)

This can be written as

119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119910119895minus6

minus 11520119910119895minus5

+ 10952119910119895minus4

+ 476928119910119895minus3

minus 3047987119910119895minus2

+ 4677632119910119895minus1

minus 4677632119910119895+1

+ 3047987119910119895+2

minus 476928119910119895+3

minus10952119910119895+4

+ 11520119910119895+5

minus 225119910119895+6

(50)

Similarly we have

119885101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119911119895minus6

minus 11520119911119895minus5

+ 10952119911119895minus4

+ 476928119911119895minus3

minus 3047987119911119895minus2

+ 4677632119911119895minus1

minus 4677632119911119895+1

+ 3047987119911119895+2

minus 476928119911119895+3

minus10952119911119895+4

+ 11520119911119895+5

minus 225119911119895+6

(51)

If we define error function 119890(119909) = 119885(119909) minus 119910(119909) and errorvectors at the nodes by

119890 (119909119895) = 119885 (119909

119895) minus 119910 (119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 (52)

or equivalently 119890119895= 119885119895minus 119910(119909

119895+ 119895ℎ) minus6 le 119895 le 119873 + 6 then

this implies

1198901015840

119895= 1198851015840

119895minus 1199101015840(119909 + 119895ℎ)

11989010158401015840

119895= 11988510158401015840

119895minus 11991010158401015840(119909 + 119895ℎ)

119890101584010158401015840

119895= 119885101584010158401015840

119895minus 119910101584010158401015840(119909 + 119895ℎ)

(53)

Table 1 Determinants of the matrices

119873 119862 119861

10 minus866756 11048870018371741

50 minus177183 14981270309

100 minus552709050 11964492

500 minus5033491471916955 times 1036

4728852755761116 times 1021

1000 minus4477989536166907 times 1071

42069711017699999 times 1040

By subtracting (50) from (51) we get

119885101584010158401015840(119909119895) minus 119910101584010158401015840(119909119895)

=23

15039360ℎ3

times 225 (119911119895minus6

minus 119910119895minus6) minus 11520 (119911

119895minus5minus 119910119895minus5)

+ 10952 (119911119895minus4

minus 119910119895minus4) + 476928 (119911

119895minus3minus 119910119895minus3)

minus 3047987 (119911119895minus2

minus 119910119895minus2) + 4677632 (119911

119895minus1minus 119910119895minus1)

minus 4677632 (119911119895+1

minus 119910119895+1) + 3047987 (119911

119895+2minus 119910119895+2)

minus 476928 (119911119895+3

minus 119910119895+3) minus 10952 (119911

119895+4minus 119910119895+4)

+11520 (119911119895+5

minus 119910119895+5) minus 225 (119911

119895+6minus 119910119895+6)

(54)

This implies

119890101584010158401015840(119909119895)

=23

15039360ℎ3

times 225119890119895minus6

minus 11520119890119895minus5

+ 10952119890119895minus4

+ 476928119890119895minus3

minus 3047987119890119895minus2

+ 4677632119890119895minus1

minus 4677632119890119895+1

+ 3047987119890119895+2

minus 476928119890119895+3

minus10952119890119895+4

+ 11520119890119895+5

minus 225119890119895+6

(55)

By Lemma 5 we get the following expression

119890101584010158401015840

119895=1

ℎ3120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 120601101584010158401015840

0119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6 + 119874 (ℎ

8)

(56)

where 119895 = 0 1 2 119873By subtracting (1) from (15) we get

119885101584010158401015840

119895minus 119910101584010158401015840

119895= 119886119895(119885119894minus 119884119895) (57)

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Abstract and Applied Analysis

This implies

119890101584010158401015840

119895= 119886119895119890119895 0 le 119894 le 119873 (58)

Using (56) we get

120601101584010158401015840

6119890119895minus6

+ 120601101584010158401015840

5119890119895minus5

+ 120601101584010158401015840

4119890119895minus4

+ 120601101584010158401015840

3119890119895minus3

+ 120601101584010158401015840

2119890119895minus2

+ 120601101584010158401015840

1119890119895minus1

+ 119902119895119890119895+ 120601101584010158401015840

minus1119890119895+1

+ 120601101584010158401015840

minus2119890119895+2

+ 120601101584010158401015840

minus3119890119895+3

+ 120601101584010158401015840

minus4119890119895+4

+ 120601101584010158401015840

minus5119890119895+5

+ 120601101584010158401015840

minus6119890119895+6

= 0

(59)

where 119902119895= 120601101584010158401015840

0minus ℎ3119886119895and 119895 = 0 1 2 119873

As 0 le 119909 le 1 and 119909119895= 119895ℎ 119895 = 0 1 2 119873

so 1198900 1198901 119890

119873are nonzero while 119890

minus6 119890minus5 119890

minus1and

119890119873+1

119890119873+2

119890119873+6

are zero because they lie outside theinterval [0 1] Let us define these (the left and right end) errorvalues as

119890119895=

max0le119896le7

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) minus6 le 119895 le 0

max119873minus7le119896le119873

10038161003816100381610038161198901198961003816100381610038161003816 119874 (ℎ

8) 119873 le 119895 le 119873 + 6

(60)

Thus system (59) is equivalent to

(119861 + 119874 (ℎ6)) 119864 = 0 (61)

where 119861 + 119874(ℎ6) is the matrix obtained by deleting the firstand last six rows and columns of the matrix 119861 where

119864 = (119890minus6 119890minus5 119890minus4 119890

119873+4 119890119873+5

119890119873+6

)119879

(62)

By using (7)

(119861 + 119874 (ℎ6)) 119864 = 119874 (ℎ

8)10038171003817100381710038171003817119885 (119909119895) minus 119910 (119909

119895)10038171003817100381710038171003817

= 119874 (ℎ8) 119864 = 119874 (ℎ

8)

(63)

Hence for small ℎ the coefficients matrix 119861 + 119874(ℎ6) will

be invertible and thus using the standard result from linearalgebra we have

119864 le (

10038171003817100381710038171003817119861minus110038171003817100381710038171003817

1 minus 119874 (ℎ6)119874 (ℎ8)) = 119874 (ℎ

2) (64)

This completes the proof

The above discussion suggests that the approximationsof the solution computed by the method developed inpervious section are second order accurate approximationsThis suggestion is supported by the numerical experimentsgiven in the next section

5 Numerical Examples and Discussions

In this section the numerical collocation algorithm basedon 8-point interpolating subdivision scheme described inSection 3 with the 8th order boundary conditions at the endpoints is tested on the two-point third order boundary valueproblems Absolute errors in the analytical solutions are alsocalculated For the sake of comparisons we also tabulated theresults in this section

Example 1 Consider the boundary value problem

119910101584010158401015840(119909) = 119910 (119909) minus 3119890

119909 0 lt 119909 lt 1 (65)

with boundary conditions 1199101015840(0) = 0 119910(1) = 0 119910(0) = 1The analytical solution of this problem is

119910 (119909) = (1 minus 119909) 119890119909 (66)

By using the collocation algorithm for 119873 = 10 we getfollowing solution of the above problem119885

119895= sum16

119894=minus6119911119894120601(119895minus119894)

where the values of 119911minus6 119911minus5 119911

5 11991116 by using (36) are

245275284967525 0690493373832105

0811531107670859

0901997043612006 0962835576958380

0995101658947808 1 0978925359857389

0933503956247400 0865636023693632

0797539554103408 0771795250759200

0651392727341382 0519777983609805

0380902186129123 0199271785305208 0

minus 0131140299007903 minus0247662275449558

minus 0342309352611581 minus0406997945734322

minus0432827137019005 minus0413625213339484

(67)

and by using (44) are

08188703235427 08701325409827 09139995186898

09498836119009 09768946175522 09940000846665

1 09934991816499 09728768806892 09362530908960

08814510666459 08059555490326 07068662002854

05808456281783 04240613657868 02321210447258 0

minus 02780394553965 minus06085378480019

minus 09989382153934 minus1457696393548

1993136716176 minus254764506053343

(68)

By using two different boundary treatments presented inSection 3 we obtained two different solutions which arepresented in Table 2 along with their absolute errors Thegraphical representations of the analytic and approximatesolutions of the above problem are shown in Figure 2Figure 2(a) represents the comparison of analytic and approx-imate solutions obtained by (36) while analytic and approx-imate solutions obtained by (44) are shown in Figure 2(b)From this table and figure we observe that the solutionobtained by (44) is significantly better than the solution

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 11

Table 2 Solutions and error estimation of Example 1

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 1 1 00000 0 001 09946538262 0978925359857389 09934991878 00157284663 00011546383702 09771222064 0933503956247400 09728768936 00436182502 00042453128003 09449011656 0865636023693632 09362531105 00792651419 00086480551304 08950948188 0797539554103408 08814510903 00975552647 00136437285005 08243606355 0771795250759200 08059555784 00525653847 0018405057106 07288475200 0651392727341382 07068662336 00774547927 0021981286407 06041258121 0519777983609805 05808456679 00843478285 0023280144108 04451081856 0380902186129123 042406141196 006420599959 0021046773609 02459603111 0199271785305208 023212109670 00466885258 0013839214410 0 0 00000 0000 0000

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

1

08

06

04

02

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 2 Comparison between analytic and approximating solutions

obtained by (36) So our claim that is the approximatesolution can be improved by adjusting boundary treatmentis justified The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 9755 times 10minus2 and2328 times 10

minus2 respectively

Example 2 Consider the following third order boundaryvalues problem

119910101584010158401015840(119909) = 119909119910 (119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909 0 lt 119909 lt 1

(69)

with boundary conditions 119910(0) = 0 = 119910(1) 1199101015840(0) = 1 Itsexact solution is

119910 (119909) = 119909 (1 minus 119909) 119890119909 (70)

By the homogeneous process of the boundary conditionlet 119906(119909) = 119910(119909) minus 119909(1 minus 119909) then above problem can betransformed into its equivalent form

119906101584010158401015840(119909) = 119909119906 (119909) + 119909

2(1 minus 119909) + (119909

3minus 21199092minus 5119909 minus 3) 119890

119909

0 lt 119909 lt 1 119906 (0) = 119906 (1) = 1199061015840(0) = 0

(71)

The solutions of this problem and their absolute errorsobtained by two different boundary treatments are shownin Table 3 The graphical representations of the analytic andapproximate solutions are shown in Figure 3 Figure 3(a)represents the comparison of analytic and approximate

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Abstract and Applied Analysis

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(a)

04

03

02

01

0

0 02 04 06 08 1

Analytic solution Y

Approximate solution Z

(b)

Figure 3 Comparison between analytic and approximating solutions

Table 3 Solutions and error estimation of Example 2

119909119895 Analytic solution 119884 Approximate solution

119885 by (36)Approximate solution

119885 by (44)Absolute error

by (36)Absolute error

by (44)00 0 0 00000 0000 000001 009946538262 01142745006 009710744670 001480911798 00023579359202 01954244413 02168561360 01876024920 00214316947 0007821949303 02834703497 03045203260 02767792681 00210499763 0006691081604 03580379275 03732002884 03437107723 00151623609 0014327155205 04121803178 04177996505 04018316295 00056193327 0010348688306 04373085120 04319808639 04229368864 00053276481 0014371625607 04228880685 04079258312 04199575084 00149622373 0002930560108 03560865485 03360647276 03473442367 00200218209 0008742311809 02213642800 02047684968 02069998108 00165957832 0014364469210 0 0 00000 0000 0000

solutions obtained by (36) Figure 3(b) represents the com-parison of analytic and approximate solutions obtained by(44) We observe that the solution obtained by (44) has lessabsolute error than that of the solution obtained by (36)Again this supports our claim

Comparison The maximum absolute errors in the solutionsobtained by (36) and (44) at step size 10 are 2143 times 10

minus2

and 1437 times 10minus2 respectively Caglar et al [17] obtained thesame maximum absolute errors but at the step size 32 and 50respectively Therefore we conclude that our method is moreefficient than that of Caglar et al

6 Conclusion and Future Work

In this work we present an interpolatory symmetric subdivi-sion algorithm for the numerical solution of third order linearproblems Septic polynomials were used for the adjustmentof boundary conditions at the end points We establishedcollocation method and obtained stable system of linearequations which can be solved by any well-known numericalmethod The numerical result showed that the adjustmentof boundary conditions at the end points influence theaccuracy of approximate solution That is the accuracy ofthe solution can be improved by the proper adjustmentof boundary conditions So our algorithm has flexibility

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 13

to improve the results by adjusting boundary conditionsThe automatic selection and adjustment of the boundaryconditions which improve the approximation order of thesolution is possible future research direction

Conflict of Interests

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

Acknowledgments

The authors thank the anonymous reviewer whose valuablecomments greatly improved this work This work is sup-ported by Indigenous PhD Scholarship Scheme of HigherEducation Commission (HEC) Pakistan

References

[1] G de Rham ldquoSur une courbe planerdquo Journal de MathematiquesPures et Appliquees vol 35 pp 25ndash42 1956

[2] N Dyn D Levin and J A Gregory ldquoA 4-point interpolatorysubdivision scheme for curve designrdquoComputer Aided Geomet-ric Design vol 4 no 4 pp 257ndash268 1987

[3] G Deslauriers and S Dubuc ldquoSymmetric iterative interpolationprocessesrdquo Constructive Approximation vol 5 no 1 pp 49ndash681989

[4] G Mustafa and N A Rehman ldquoThe mask of (2119887 + 4)-pointn-ary subdivision schemerdquo Computing Archives for ScientificComputing vol 90 no 1-2 pp 1ndash14 2010

[5] M Aslam G Mustafa and A Ghaffar ldquo(2119899 minus 1)-point ternaryapproximating and interpolating subdivision schemesrdquo Journalof Applied Mathematics vol 2011 Article ID 832630 12 pages2011

[6] G Mustafa J Deng P Ashraf and N A Rehman ldquoThe maskof odd points 119899-ary interpolating subdivision schemerdquo Journalof Applied Mathematics vol 2012 Article ID 205863 20 pages2012

[7] GMustafa and F Khan ldquoA new 4-point C3 quaternary approxi-mating subdivision schemerdquo Abstract and Applied Analysis vol2009 Article ID 301967 14 pages 2009

[8] G Mustafa P Ashraf and J Deng ldquoGeneralized and unifiedfamilies of inter-polating schemesrdquo Numerical Mathematics-Theory Methods and Applications vol 7 no 2014 pp 193ndash2132014

[9] G Mustafa and I P Ivrissimitzis ldquoModel selection for theDubuc-Deslauriers family of subdivision schemesrdquo in Proceed-ings of the 14th IMA Conference on Mathematics of SurfacesUniversity of Birmingham Birmingham UK September 2013

[10] R Qu and R P Agarwal ldquoSolving two point boundary valueproblems by interpolatory subdivision algorithmsrdquo Interna-tional Journal of Computer Mathematics vol 60 no 3-4 pp279ndash294 1996

[11] R Qu and R P Agarwal ldquoA subdivision approach to theconstruction of approximate solutions of boundary-value prob-lems with deviating argumentsrdquo Journal of Computers ampMathematics with Applications vol 35 no 11 pp 121ndash135 1998

[12] R Qu ldquoCurve and surface interpolation by subdivision algo-rithmsrdquo Computer Aided Drafting Design and Manufacturingvol 4 no 2 pp 28ndash39 1994

[13] R B Qu and R P Agarwal ldquoA cross difference approach to theanalysis of subdivision algorithmsrdquo Neural Parallel amp ScientificComputations vol 3 no 3 pp 393ndash416 1995

[14] N Dyn ldquoTutorial on multiresolution in geometric modellingsummer school lecture notes seriesrdquo in Mathematics amp Visual-ization I Armin A Ewald and S F Michael Eds Springer2002

[15] C Conti and K Hormann ldquoPolynomial reproduction forunivariate subdivision schemes of any arityrdquo Journal of Approx-imation Theory vol 163 no 4 pp 413ndash437 2011

[16] G Strang Linear Algebra and Its Applications Cengage Learn-ing India Private Limited 4th edition 2011

[17] H N Caglar S H Caglar and E H Twizell ldquoThe numericalsolution of third-order boundary-value problems with fourth-degree 119861-spline functionsrdquo International Journal of ComputerMathematics vol 71 no 3 pp 373ndash381 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of