Commun Nonlinear Sci Numer Simulat 14 (2009) 1132–1138

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Commun Nonlinear Sci Numer Simulat

journal homepage: www.elsevier .com/locate /cnsns

Numerical solution of special 12th-order boundary value problemsusing differential transform method

Siraj-Ul Islam a,*, Sirajul Haq b, Javid Ali c

a University of Engineering and Technology, Peshawar, NWFP, Pakistanb Faculty of Engineering Sciences, GIK Institute, Topi, NWFP, Pakistanc Kohat University of Science and Technology, Kohat, NWFP, Pakistan

a r t i c l e i n f o a b s t r a c t

Article history:Received 17 August 2007Accepted 29 February 2008Available online 14 March 2008

PACS:02.60.Lj

Keywords:Twelfth-order boundary value problemsBoundary value problemsDifferential transform method (DTM)

1007-5704/$ - see front matter � 2008 Elsevier B.Vdoi:10.1016/j.cnsns.2008.02.012

* Corresponding author. Tel.: +92 03339243749.E-mail addresses: siraj-ul-islam@nwfpuet.edu.pk

In this paper, a differential transform method (DTM) is used to find the numerical solutionof a special 12th-order boundary value problems with two point boundary conditions. Theanalysis is accompanied by testing differential transform method both on linear and non-linear problems from the literature [Wazwaz AM. Approximate solutions to boundaryvalue problems of higher-order by the modified decomposition method. Comput MathAppl 2000:40;679–91; Siddiqi SS, Ghazala Akram. Solutions of 12th order boundary valueproblems using non-polynomial spline technique. Appl Math Comput 2007. doi:10.1016/j.amc.2007.10.015; Siddiqi SS, Twizell EH. Spline solutions of linear 12th-order boundaryvalue problems. J Comput Appl Math 1997;78:371–90]. Numerical experiments and com-parison with existing methods are performed to demonstrate reliability and efficiency ofthe proposed method.

� 2008 Elsevier B.V. All rights reserved.

1. Introduction

In literature a variety of methods, exact, approximate and purely numerical, are available for solution of differential equa-tions. The most common numerical techniques used are: Adomian decomposition method [1], variational iteration [2],splines [3], homotopy perturbation [4] etc. In [5] Euler method, Taylor method, Runge–Kutta method serve as an introduc-tion to numerical method for solving initial value problems. However, the Taylor method requires the calculations of thehigh-order derivatives. Among these solution techniques, DTM is one of the semi-analytical numerical method for the solu-tion of differential equations. The basic idea of DTM was initially introduced by Zhou [6] in 1986. Its main application thereinwas to solve both linear and nonlinear initial value problems arising in electrical circuit analysis. DTM is a semi-analyticalnumerical technique that uses Taylor series for the solution of differential equations in the form of a polynomial. The DTM isan alternative procedure for getting Taylor series solution of the given differential equations. By using this method, we get aseries solution, in practice a truncated series solution. The series often coincides with the Taylor expansion of the true solu-tion at point x ¼ 0 in the case of initial value problems. This method reduces the size of computational domain and is appli-cable to many problems easily. The method is also used for the solution of differential-algebraic equations (DAEs) of index-1by Ayaz [7]. Liu et.al. [8] analyzed higher index differential-algebraic equations using this technique where they showed thatthe method is effective in case of index-2 DAEs but not suitable for DAEs of index-3. Application of the two-dimensional DTM

. All rights reserved.

, siraj_islam_pk@yahoo.com (S.-U. Islam).

S.-U. Islam et al. / Commun Nonlinear Sci Numer Simulat 14 (2009) 1132–1138 1133

was studied by Ayaz [9] for the solution of partial differential equations. Comparison of this method with adomian decom-position method was done by Hassan [10] to solve PDEs. The same author used this method to solve higher-order initial va-lue problems where he studied second and third-order initial value problems to show the efficiency of the method. Fourth-order boundary value problem was studied by Ertürk [11] in 2007.

Twelfth-order boundary value problems arise in the context when a uniform magnetic field is applied across the fluid inthe same direction as gravity (see [12]). When instability sets in as an ordinary convection, it is modeled by the 10th-orderboundary value problems; when instability sets in as overstability, it is modeled by the 12th-order boundary value problems.This type of linear and nonlinear problems were solved by Wazwaz [1] using modified decomposition method. Siddique andTwizell [14] and Siddique and Ghazala [13] investigated these type of problems using spline based approach. The literatureon the numerical solution of 12th-order boundary value problems and associated eigenvalue problems is sparse. Existenceand uniqueness of 12th-order boundary value problems are discussed by Agarwal [15].

In this paper, the DTM is applied to 12th-order boundary value problems. The method is tested on both linear and non-linear problems from the literature [1,13,14]. The numerical results are compared with [1,13,14] to show effectiveness andaccuracy of the method.

Rest of the paper is organized as follows. In Section 2, we introduce differential transform technique. In Section 3, we dis-cuss the method in the context of 12th-order boundary value problems. In Section 4, we present numerical examples thatdemonstrate how the proposed method works. In Section 5, we summarize the results.

2. Differential transform technique

If uðxÞ is a given function, then its differential transform is defined as

Table 1Fundam

Functio

f ðxÞ ¼ gf ðxÞ ¼ a

f ðxÞ ¼ g

f ðxÞ ¼ g

f ðxÞ ¼ x

f ðxÞ ¼ g

f ðxÞ ¼ s

f ðxÞ ¼ c

UðrÞ ¼ 1r!

druðxÞdxr

����x¼0

ð1Þ

and the inverse differential transform of UðrÞ is defined by

uðxÞ ¼X1r¼0

xrUðrÞ: ð2Þ

In actual application, the function uðxÞ is expressed by a finite series

uðxÞ ¼XN

r¼0

xrUðrÞ: ð3Þ

Eq. (2) implies thatP1

r¼Nþ1xrUðrÞ is negligibly small. In Table 1 the fundamental operations related to one-dimensional prob-lems are listed. These results are given in [11].

3. Analysis of the method

In this section, we give analysis of the method for 12th-order boundary value problem. In fact our target is to find solutionof the following boundary value problem:

uð12ÞðxÞ þ f ðxÞuðxÞ ¼ gðxÞ; a < x < b; ð4Þ

subject to the boundary conditions:

ental operations of one-dimensional DTM

n Transform function

ðxÞ � hðxÞ FðrÞ ¼ GðrÞ � HðrÞgðxÞ FðrÞ ¼ aGðrÞðmÞðxÞ FðrÞ ¼ ðmþrÞ!

r! Gðmþ rÞðxÞhðxÞ FðrÞ ¼

Prk¼0GðkÞHðr � kÞ

m FðrÞ ¼ dðr �mÞ; dðpÞ ¼ 1;p ¼ 00; p–0

�:

1ðxÞg2ðxÞ � � � gmðxÞ FðrÞ ¼Pr

km�1¼0 � � �Pk3

k2¼0

Pk2k1¼0fG1ðk1Þ

G2ðk2 � k1Þ � � �Gmðr � km�1Þginðxxþ aÞ FðrÞ ¼ xr

r! sinðpr2 þ aÞ

osðxxþ aÞ FðrÞ ¼ xr

r! cosðpr2 þ aÞ

1134 S.-U. Islam et al. / Commun Nonlinear Sci Numer Simulat 14 (2009) 1132–1138

uðaÞ ¼ a0; uðbÞ ¼ a1;

uð1ÞðaÞ ¼ c0; uð1ÞðbÞ ¼ c1;

uð2ÞðaÞ ¼ d0; uð2ÞðbÞ ¼ d1;

uð3ÞðaÞ ¼ m0; uð3ÞðbÞ ¼ m1;

uð4ÞðaÞ ¼ n0; uð4ÞðbÞ ¼ n1;

uð5ÞðaÞ ¼ x0; uð5ÞðbÞ ¼ x1;

ð5Þ

where ai; ci; di; mi; ni and xi, i ¼ 0;1 are finite real constants and f ðxÞ and gðxÞ are continuous functions on the interval ½a; b�.Differential transform of Eq. (4) gives

Uðr þ 12Þ ¼ r!

ðr þ 12Þ! GðrÞ �Xr

k¼0

FðkÞYðr � kÞ( )

; ð6Þ

where UðrÞ, FðrÞ and GðrÞ are the differential transformations of uðxÞ, f ðxÞ and gðxÞ, respectively.Taking differential transform of the boundary conditions given in Eq. (5), we get

XN

r¼0

UðrÞar ¼ a0;XN

r¼0

rUðrÞar�1 ¼ c0;

XN

r¼0

rðr � 1ÞUðrÞar�2 ¼ d0;

XN

r¼0

rðr � 1Þðr � 2ÞUðrÞar�3 ¼ m0;

XN

r¼0

rðr � 1Þðr � 2Þðr � 3ÞUðrÞar�4 ¼ n0;

XN

r¼0

rðr � 1Þðr � 2Þðr � 3Þðr � 4ÞUðrÞar�5 ¼ x0;

XN

r¼0

UðrÞbr ¼ a1;XN

r¼0

rUðrÞbr�1 ¼ c1;

XN

r¼0

rðr � 1ÞUðrÞbr�2 ¼ d1;

XN

r¼0

rðr � 1Þðr � 2ÞUðrÞbr�3 ¼ m1;

XN

r¼0

rðr � 1Þðr � 2Þðr � 3ÞUðrÞbr�4 ¼ n1;

XN

r¼0

rðr � 1Þðr � 2Þðr � 3Þðr � 4ÞUðrÞbr�5 ¼ x1:

ð7Þ

The values of UðiÞ; i ¼ 1;2;3; . . . ;N, can be obtained from Eqs. (6) and (7). Using these values in Eq. (3), we obtain solution ofthe boundary value problem given in Eqs. (4) and (5).

4. Numerical examples

In this section, we consider linear and nonlinear problems to test the performance of the method.

Example 1. Consider the following linear 12th-order boundary value problem:

uð12ÞðxÞ ¼ uðxÞ � 24x cos x� 132 sin x; �1 < x < 1 ð8Þ

with the boundary conditions:

uð�1Þ ¼ 0;uð1Þð�1Þ ¼ 2 sinð1Þ; uð2Þð�1Þ ¼ �4 cosð1Þ � 2 sinð1Þ;uð3Þð�1Þ ¼ 6 cosð1Þ � 6 sinð1Þ; uð4Þð�1Þ ¼ 8 cosð1Þ þ 12 sinð1Þ;uð5Þð�1Þ ¼ �20 cosð1Þ þ 10 sinð1Þ;uð1Þ ¼ 0;uð1Þð1Þ ¼ 2 sinð1Þ; uð2Þð1Þ ¼ 4 cosð1Þ þ 2 sinð1Þ;uð3Þð1Þ ¼ 6 cosð1Þ � 6 sinð1Þ; uð4Þð1Þ ¼ �8 cosð1Þ � 12 sinð1Þ;uð5Þð1Þ ¼ �20 cosð1Þ þ 10 sinð1Þ:

ð9Þ

S.-U. Islam et al. / Commun Nonlinear Sci Numer Simulat 14 (2009) 1132–1138 1135

The analytical solution of the above problem is given by,

uðxÞ ¼ ðx2 � 1Þ sin x: ð10Þ

This problem is reported in [14] with different set of the boundary conditions. Taking the differential transformation of Eq.(8) and boundary conditions given in Eq. (9), it follows that

Uðr þ 12Þ ¼ r!

ðr þ 12Þ! UðrÞ � 24Xr

l¼0

dðl� 1Þðr � lÞ! cosðr � lÞ p

2� 132

r!sin

rp2

!: ð11Þ

The transformed boundary conditions at x ¼ 1 and at x ¼ �1 are given by

XN

r¼0

UðrÞð�1Þr ¼ 0;XN

r¼0

UðrÞ ¼ 0;

XN

r¼0

rUðrÞð�1Þr�1 ¼ 2 sinð1Þ;XN

r¼0

rUðrÞ ¼ 2 sinð1Þ;

XN

r¼0

rðr � 1ÞUðrÞð�1Þr�2 ¼ �4 cosð1Þ � 2 sinð1Þ;

XN

r¼0

rðr � 1ÞUðrÞ ¼ 4 cosð1Þ þ 2 sinð1Þ;

XN

r¼0

rðr � 1Þðr � 2ÞUðrÞð�1Þr�3 ¼ 6 cosð1Þ � 6 sinð1Þ;

XN

r¼0

rðr � 1Þðr � 2ÞUðrÞ ¼ 6 cosð1Þ � 6 sinð1Þ;

XN

r¼0

rðr � 1Þðr � 2Þðr � 3ÞUðrÞð�1Þr�4 ¼ 8 cosð1Þ þ 12 sinð1Þ;

XN

r¼0

rðr � 1Þðr � 2Þðr � 3ÞUðrÞ ¼ �8 cosð1Þ � 12 sinð1Þ;

XN

r¼0

rðr � 1Þðr � 2Þðr � 3Þðr � 4ÞUðrÞð�1Þr�5 ¼ �20 cosð1Þ þ 10 sinð1Þ;

XN

r¼0

rðr � 1Þðr � 2Þðr � 3Þðr � 4ÞUðrÞ ¼ �20 cosð1Þ þ 10 sinð1Þ:

ð12Þ

Using Eqs. (11) and (12), the following series solution up to Oðx14Þ is obtained:

uðxÞ ¼ �1122163868037811600

� xþ 1890974887640530

x2 þ 76

x3 � 11479368910958320

x4

� 21120

x5 þ 11219019448010320

x6 þ 51127159925600

x7 � 13878698243674240

x8 � 1803289631360

x9

þ 126777904664896000

x10 þ 3411812254457600

x11 � 158516688252300616898560000

x12

� 1576227020800

x13 þ Oðx14Þ: ð13Þ

The numerical results corresponding to Example 1 are given in Table 2. We have not listed the results of the spline basedtechnique [13,14] in the Table as the results of the DTM given in the Table 2 are far superior than [14,13]. In [14] the max-imum error at the interior points of the interval is 2:07Eð�003Þ and near the boundary points the maximum error is1:06Eð013Þ for h ¼ 1

9. In [13] the maximum error is 4:69Eð�005Þ. In our case the maximum error is �8:7157Eð�009Þ, whichshows clear superiority of the DTM over the spline based technique [13,14].

Example 2. Consider another linear 12th-order boundary value problem reported in [13]

uð12ÞðxÞ þ xuðxÞ ¼ �ð120þ 23xþ x3Þex; �1 < x < 1 ð14Þ

with the boundary conditions:

uðkÞð0Þ ¼ kð2� kÞ; uðkÞð1Þ ¼ �k2e ð15Þ

where k ¼ 0ð1Þ5.

Table 2Numerical results for Example 1

x Analytic solution Solution by DTM Error*

0.0 0.00000000000000 0.00000000000000 0.00000.1 �0.09883508248036 �0.09883508248036 �1.6376E�0150.2 �0.19072255756326 �0.19072255756305 �2.0797E�0130.3 �0.26892338806182 �0.26892338805838 �3.4360E�0120.4 �0.32711140753927 �0.32711140751471 �2.4556E�0110.5 �0.35956915395315 �0.35956915384294 �1.1021E�0100.6 �0.36137118297282 �0.36137118260606 �3.6677E�0100.7 �0.32855102049122 �0.32855101950177 �9.8945E�0100.8 �0.25824819272383 �0.25824819043994 �2.2839E�0090.9 �0.14883211282922 �0.14883210815326 �4.6760E�0091.0 0.00000000000000 0.00000000871568 �8.7157E�009

*Error = analytical solution � numerical solution.

1136 S.-U. Islam et al. / Commun Nonlinear Sci Numer Simulat 14 (2009) 1132–1138

Using properties of the differential transform given in the Table 1, the transformed form of Eq. (14) is given by

Table 3Numeri

x

0.00.10.20.30.40.50.60.70.80.91.0

*Error =

Uðr þ 12Þ ¼ �r!

ðr þ 12Þ!120

r!þXr

l¼0

dðl� 1ÞUðr � lÞ þ 23dðl� 1Þðr � lÞ! þ

dðl� 3Þðr � lÞ!

� �" #ð16Þ

with the transformed boundary conditions given in Eq. (15) at x ¼ 0

UðkÞ ¼ kð2� kÞk!

; k ¼ 0ð1Þ5: ð17Þ

Using Eqs. (16) and (17), the transformed boundary conditions at x ¼ 1 are given by

XN

r¼0

UðrÞ ¼ 0;XN

r¼0

Yk

i¼0

ðr � iÞUðrÞ" #

¼ �ðkþ 1Þ2e; k ¼ 0ð1Þ4: ð18Þ

From Eqs. (3), (17) and (18), a series solution of order Oðx14Þ of the boundary value problem (14) and (15) is obtained inthe following form:

uðxÞ ¼ x� 12

x3 � 13

x4 � 18

x5 � 2675998028000

x6 � 114811653120

x7 � 68095725440

x8 � 691839553920

x9 � 6889322963200

x10

� 46551756339200

x11 � 120479001600

x12 � 1436227020800

x13 þ Oðx14Þ: ð19Þ

Numerical results obtained form the above equation are compared with exact solution in Table 3.The numerical results corresponding to Example 2 are given in Table 3. In [14] the maximum error at the interior points of

the interval is 2:07Eð�003Þ and near the boundary points the maximum error is 1:06Eð013Þ for h ¼ 19. In [13] the maximum

error is 7:38Eð�009Þ. In our case the maximum error is �1:39Eð�0010Þ, which shows clear superiority of the DTM over thespline based technique [13,14].

Example 3. Now we consider a nonlinear 12th-order boundary value problem:

uð12ÞðxÞ ¼ 2exu2ðxÞ þ uð3ÞðxÞ; 0 < x < 1: ð20Þ

We consider the following two sets of boundary conditions separately:

cal results for Example 2

Analytic solution Solution by DTM Error*

0.00000000000000 0.00000000000000 0.00000.09946538262681 0.09946538262688 �7.5065E�0140.19542444130563 0.19542444130840 �2.7686E�0120.28347034959096 0.28347034960823 �1.7271E�0110.35803792743390 0.35803792748414 �5.0232E�0110.41218031767503 0.41218031776843 �9.3401E�0110.43730851209372 0.43730851222163 �1.2791E�0100.42288806856880 0.42288806870797 �1.3917E�0100.35608654855879 0.35608654868158 �1.2278E�0100.22136428000413 0.22136428007912 �7.4997E�0110.00000000000000 �0.00000000001945 1.9454E�011

analytical solution � numerical solution.

Table 4Numeri

x

0.00.10.20.30.40.50.60.70.80.91.0

*Error =

S.-U. Islam et al. / Commun Nonlinear Sci Numer Simulat 14 (2009) 1132–1138 1137

ðaÞuðkÞð0Þ ¼ ð�1Þk; uðkÞð1Þ ¼ ð�1Þke�1; k ¼ 0ð1Þ5; ð21ÞðbÞuð2kÞð0Þ ¼ 1; uð2kÞð1Þ ¼ e�1; k ¼ 0ð1Þ5: ð22Þ

Applying differential transform to Eq. (20), we get:

Uðr þ 12Þ ¼ r!

ðr þ 12Þ! 2Xr

l¼0

Xl

m¼0

1m!

Uðl�mÞUðr � lÞ þY3

i¼1

ðr þ iÞUðr þ 3Þ" #

: ð23Þ

The transformed set of boundary conditions (a) at x ¼ 0 and x ¼ 1 are given by

UðkÞ ¼ ð�1Þk

k!; k ¼ 0ð1Þ5; ð24Þ

XN

r¼0

UðrÞ ¼ e�1;XN

r¼0

Yk

i¼0

ðr � iÞUðrÞ" #

¼ ð�1Þkþ1e�1; k ¼ 0ð1Þ4: ð25Þ

Using Eqs. (23)–(25), we get the following truncated series solution of order Oðx14Þ.

uðxÞ ¼ 1� xþ x2

2!� x3

3!þ x4

4!� x5

5!þ 1:00000540321350

6!x6 � 1:00022073774403

7!x7 þ 1:00424092040339

8!x8

� 1:047754340368599!

x9 þ 1:3188097896927510!

x10 � 2:0545873600058311!

x11 þ 1:0002207377440312!

x12

þ 7:00002161285413!

x13 þ Oðx14Þ: ð26Þ

Numerical results obtained form the above equation are compared with exact solution in Table 4.Now, we consider the same differential Eq. (20) but with the set of boundary conditions (b) i.e.,

uð2iÞð0Þ ¼ 1; uð2iÞð1Þ ¼ e�1; i ¼ 0ð1Þ5

The transformed set of boundary conditions (b) at x ¼ 0 and x ¼ 1 are given by

UðkÞ ¼ 1k!; k ¼ 0ð2Þ10; ð27Þ

XN

r¼0

UðrÞ ¼ e�1;XN

r¼0

Yk

i¼0

ðr � iÞUðrÞ" #

¼ e�1; k ¼ 1ð2Þ9: ð28Þ

Using Eqs. (23), (27) and (28), we find UðiÞ; i ¼ 0;1;2; . . . for N ¼ 12. These values along with (3) give rise to the seriessolution of Oðx13Þ of the problem Eqs. (20) and (22)in the following form:

uðxÞ ¼ 1� 0:999998xþ x2

2!� 0:166669x3 þ x4

4!� 0:00833201x5 þ x6

6!� 0:000198722x7 þ x8

8!� 2:71467� 10�6x9

þ x10

10!� 2:83618� 10�8x11 þ 2:08764� 10�9x12 þ Oðx13Þ: ð29Þ

The numerical results are given in Table 5 which are exactly the same as obtained by decomposition method [1].Tables 2–5 show excellent performance of the DTM technique. It is clear from the Tables 4 and 5 that the numerical

results corresponding to problem 3 with first set of boundary conditions are superior than those of second set of boundaryconditions. The reason for improved performance in the first case is that the function values as well as its derivative are takenat the consecutive points whereas in the second case alternative points are chosen by skipping nonzero even nodal points.

cal results for Problem 3(a)

Analytic solution Solution by DTM Error*

1.00000000000000 1.00000000000000 0.00000.90483741803596 0.90483741803596 �4.1078E�0150.81873075307798 0.81873075307811 �1.3023E�0130.74081822068172 0.74081822068239 �6.7535E�0130.67032004603564 0.67032004603717 �1.5278E�0120.60653065971263 0.60653065971462 �1.9817E�0120.54881163609403 0.54881163609560 �1.5745E�0120.49658530379141 0.49658530379213 �7.1704E�0130.44932896411722 0.44932896411736 �1.4222E�0130.40656965974060 0.40656965974060 �4.1633E�0150.36787944117144 0.36787944117144 1.2212E�015

analytical solution � numerical solution.

Table 5Comparison of numerical results Problem 3(b)

x Errors* (Decomposition method [1]) Differential transform technique

0.0 0.000 0.0000.1 �1:61� 10�7 �1:61� 10�7

0.2 �3:07� 10�7 �3:07� 10�7

0.3 �4:22� 10�7 �4:22� 10�7

0.4 �4:97� 10�7 �4:97� 10�7

0.5 �5:22� 10�7 �5:22� 10�7

0.6 �4:97� 10�7 �4:96� 10�7

0.7 �4:22� 10�7 �4:22� 10�7

0.8 �3:07� 10�7 �3:07� 10�7

0.9 �1:61� 10�7 �1:61� 10�7

1.0 2:00� 10�10 1:11� 10�16

*Errror = analytical solution � numerical solution.

1138 S.-U. Islam et al. / Commun Nonlinear Sci Numer Simulat 14 (2009) 1132–1138

5. Closure

In this study, we have introduced DTM to solve special 12th-order linear and nonlinear boundary value problems. Thisapproach is simple in applicability as it does not require linearization, discretization or perturbation like other numericaland approximate methods. Comparison with the existing technique shows superiority and less computational efforts ofthe DTM.

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