Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 705313 8 pageshttpdxdoiorg1011552013705313

Research ArticleA Numerical Method for Partial Differential Algebraic EquationsBased on Differential Transform Method

Murat Osmanoglu and Mustafa Bayram

Department of Mathematical Engineering Chemical and Metallurgical Faculty Yildiz Technical UniversityEsenler 34210 Istanbul Turkey

Correspondence should be addressed to Mustafa Bayram msbayramyildizedutr

Received 14 January 2013 Accepted 28 February 2013

Academic Editor Adem Kilicman

Copyright copy 2013 M Osmanoglu and M Bayram This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We have considered linear partial differential algebraic equations (LPDAEs) of the form 119860119906119905(119905 119909) + 119861119906

119909119909(119905 119909) + 119862119906(119905 119909) = 119891(119905 119909)

which has at least one singular matrix of 119860 119861 isin R119899times119899 We have first introduced a uniform differential time index and a differentialspace index The initial conditions and boundary conditions of the given system cannot be prescribed for all components of thesolution vector 119906 here To overcome this we introduced these indexes Furthermore differential transformmethod has been givento solve LPDAEs We have applied this method to a test problem and numerical solution of the problem has been compared withanalytical solution

1 Introduction

The partial differential algebraic equation was first studiedby Marszalek He also studied the analysis of the par-tial differential algebraic equations [1] Lucht et al [2ndash4] studied the numerical solution and indexes of thelinear partial differential equations with constant coeffi-cients A study about characteristics analysis and differ-ential index of the partial differential algebraic equationswas given by Martinson and Barton [5 6] Debrabantand Strehmel investigated the convergence of Runge-Kuttamethod for linear partial differential algebraic equations[7]

There are numerous LPDAEs applications in scientificareas given for instance in the field of Navier-Stokes equa-tions in chemical engineering in magnetohydrodynamicsand in the theory of elastic multibody systems [4 8ndash12]

On the other hand the differential transform methodwas used by Zhou [13] to solve linear and nonlinear ini-tial value problems in electric circuit analysis Analysisof nonlinear circuits by using differential Taylor trans-form was given by Koksal and Herdem [14] Using one-dimensional differential transform Abdel-Halim Hassan

[15] proposed a method to solve eigenvalue problemsThe two-dimensional differential transform methods havebeen applied to the partial differential equations [16ndash19] The differential transform method extended to solvedifferential-difference equations by Arikoglu and Ozkol[20] Jang et al have used differential transform methodto solve initial value problems [21] The numerical solu-tion of the differential-algebraic equation systems hasbeen studied by using differential transform method [2223]

In this paper we have considered linear partialdifferential equations with constant coefficients of theform

119860119906119905(119905 119909) + 119861119906

119909119909(119905 119909) + 119862119906 (119905 119909) = 119891 (119905 119909)

(119905 119909) isin 119869 times Ω

(1)

where 119869 = [0infin) Ω = [minus119897 119897] 119897 gt 0 and 119860 119861 119862 isin R119899times119899In (1) at least one of the matrices 119860 119861 isin R119899times119899 should besingular If 119860 = 0 or 119861 = 0 then (1) becomes ordinarydifferential equation or differential algebraic equation sowe assume that none of the matrices 119860 or 119861 is the zeromatrix

2 Abstract and Applied Analysis

2 Indexes of Partial DifferentialAlgebraic Equation

Let us consider (1) with initial values and boundary condi-tions given as follows

119906119895(119905 plusmn119897) = 0 for 119905 isin 119869

119906119894(0 119909) = 119892 (119909) for 119909 isin Ω

(2)

where 119895 isin M119861119862

sube 1 2 119899 M119861119862

is the set of indicesof components of 119906 for which boundary conditions can beprescribed arbitrarily and 119894 isin M

119868119862sube 1 2 119899 M

119868119862

is the set of indices of components of 119906 for which initialconditions can be prescribed arbitrarilyThe initial boundaryvalue problem (IBVP) (1) has only one solution where afunction 119906 is a solution of the problem if it is sufficientlysmooth uniquely determined by its initial values (IVs) andboundary values (BVs) and if it solves the LPDAE point wise

Definition of the indexes can be given using the followingassumptions

(i) Each component of the vectors 119906 119906119905 and 119891 satisfy the

following condition1003816100381610038161003816119910 (119905 119909)

1003816100381610038161003816 le 119872119890120572119905 120572 ge 0 119905 ge 0 (3)

where119872 and 120572 are independent of 119905 and 119909(ii) (119861 120585119860 + 119862)Re(120585) gt 120572 called as the matrix pencil is

regular(iii) (119860 120583

119896119861 + 119862) is regular for all 119896 where 120583

119896is an eigen-

value of the operator 12059721205971199092 together with prescribedBCs

(iv) The vector 119891(119905 119909) and the initial vector 119892(119909) aresufficiently smooth

If we use Laplace transform from assumption (ii) (1) can betransformed into11986111990610158401015840

120585(119909) + (120585119860 + 119862) 119906

120585(119909) = 119891

120585(119909) + 119860119892 (119909) Re (120585) gt 120572

(4)

if 119861 is a singular matrix then (4) is a DAE depending on theparameter 120585 To characterizeM

119861119862 we introduce 119895 isin M

(120585)

119861119862sube

1 2 119899 as the set of indices of components of 119906120585for which

boundary conditions can be prescribed arbitrarilyIn order to define a spatial index we need the Kronecker

normal form of theDAE (4) Assumption (iii) guarantees thatthere are nonsingular matrices 119875

119871120585 119876119871120585isin C119899times119899 such that

119875119871120585119861119876119871120585= (

1198681198981

0

0 119873119871120585

)

119875119871120585(120585119860 + 119862)119876

119871120585= (

119877119871120585

0

0 1198681198982

)

(5)

where 119877119871120585isin C1198981times1198982 and119873

119871120585isin R1198982times1198982 is a nilpotent Jordan

chain matrix with1198981+1198982= 119899 119868

119896is the unit matrix of order

119896 The Riesz index (or nilpotency) of 119873119871120585

is denoted by ]119871120585

(ie119873]119871120585

119871120585= 0119873

]119871120585minus1

119871120585= 0)

Here we will assume that there is a real number 120572lowast ge 120572

such that the index set M(120585)119861119862

is independent of the Laplaceparameter 120585 provided Re(120585) ge 120572lowast

Definition 1 Let 120572lowast isin R+ be a number with 120572lowast ge 120572 such thatfor all 120585 isin C with Re(120585) ge 120572lowast

(1) the matrix pencil (119861 120585119860 + 119862) is regular

(2) M(120585)119861119862

is independent of 120585 ieM(120585)119861119862

=M119861119862

(3) the nilpotency of119873119871120585

is ]119871ge 1

Then ]119889119909

= 2]119871minus 1 is called the ldquodifferential spatial indexrdquo

of the LPDAE If ]119871= 0 then the differential spatial index of

LPDAE is defined to be zero

If we use Fourier transform (1) can be transformed into

1198601015840

119896(119905) + (120583

119896119861 + 119862)

119896(119905) =

119896(119905) + 119861120588

119896(119905) (6)

with 120588119896(119905) = (120588

1198961(119905) 120588

119896119899(119905))119879 and

120588119896119894(119905) = 0 for 119894 isinM

119861119862

120588119896119895(119905) =

1

119897[1206011015840

119896(119909) 119906119895(119905 119909) minus 120601

119896(119909) 119906119909119895(119905 119909)]

119909=119897

119909=minus119897

(7)

for 119895 notin M119861119862 which results from partial integration of the

term int119897

minus119897119906119909119909(119905 119909)120601

119896(119909)119889119909

If 119860 is a singular matrix then (6) is a DAE depending onthe parameter 120583

119896which can be solved uniquely with suitable

ICs under the assumptions (iv) and (v) Analogous to the caseof the Laplace transform the above assumption (iv) impliesthat there exist regular matrices 119875

119865119896 119876119865119896

such that

119875119865119896119860119876119865119896

= (1198681198991

0

0 119873119865119896

)

119875119865119896(120583119896119861 + 119862)119876

119865119896= (

119877119865119896

0

0 1198681198992

)

(8)

With 119877119865119896

isin R1198991times1198991 119873119865119896

isin R1198992times1198992 is again a nilpotent Jordanchain matrix with Riesz index ]

119865119896 where 119899

1+ 1198992= 119899

To characterize M119868119862 we introduce M(119896)

119868119862sube 1 2 119899

as the set of indices of components of 119896for which initial

conditions can be prescribed arbitrarilyTherefore we alwaysassume in the context of a Fourier analysis of 119906 that M(119896)

119868119862is

independent of 119896 isin N+ ieM(119896)

119868119862=M119868119862

Definition 2 Assume for 119896 = 1 2 that

(1) the matrix pencil (119860 120583119896119861 + 119862) is regular

(2) M(119896)119868119862

is independent of 119896 ieM(119896)119868119862=M119868119862

(3) the nilpotency of119873119865119896

is ]119865119896

= ]119865

Then the PDAE (1) is said to have uniform differential timeindex ]

119889119905= ]119865

The differential spatial and time indexes are used todecide which initial and boundary values can be taken tosolve the problem

Abstract and Applied Analysis 3

3 Two-Dimensional DifferentialTransform Method

The two-dimensional differential transform of function119908(119909 119910) is defined as

119882(119896 ℎ) =1

119896ℎ[120597119896+ℎ119908 (119909 119910)

120597119909119896120597119910ℎ]119909=0

119910=0

(9)

where it is noted that upper case symbol 119882(119896 ℎ) is usedto denote the two-dimensional differential transform of afunction represented by a corresponding lower case symbol119908(119909 119910) The differential inverse transform of 119882(119896 ℎ) isdefined as

119908 (119909 119910) =

infin

sum

119896=0

infin

sum

ℎ=0

119882(119896 ℎ) 119909119896119910ℎ (10)

From (9) and (10) we obtain

119908 (119909 119910) =

infin

sum

119896=0

infin

sum

ℎ=0

119909119896119910ℎ

119896ℎ[120597119896+ℎ119908 (119909 119910)

120597119909119896120597119910ℎ]119909=0

119910=0

(11)

The concept of two-dimensional differential transform isderived from two-dimensional Taylor series expansion butthe method doesnrsquot evaluate the derivatives symbolically

Theorem 3 Differential transform of the function 119908(119909 119910) =119906(119909 119910) plusmn V(119909 119910) is

119882(119896 ℎ) = 119880 (119896 ℎ) plusmn 119881 (119896 ℎ) (12)

see [17]

Theorem 4 Differential transform of the function 119908(119909 119910) =120582119906(119909 119910) is

119882(119896 ℎ) = 120582119880 (119896 ℎ) (13)

see [17]

Theorem 5 Differential transform of the function 119908(119909 119910) =120597119906(119909 119910)120597119909 is

119882(119896 ℎ) = (119896 + 1)119880 (119896 + 1 ℎ) (14)

see [17]

Theorem 6 Differential transform of the function 119908(119909 119910) =120597119906(119909 119910)120597119910 is

119882(119896 ℎ) = (ℎ + 1)119880 (119896 ℎ + 1) (15)

see [17]

Theorem 7 Differential transform of the function 119908(119909 119910) =120597119903+119904119906(119909 119910)120597119909

119903120597119910119904 is

119882(119896 ℎ) = (119896 + 1) (119896 + 2) sdot sdot sdot (119896 + 119903) (ℎ + 1)

times (ℎ + 2) sdot sdot sdot (ℎ + 119904)119880 (119896 + 119903 ℎ + 119904)

(16)

see [17]

Theorem 8 Differential transform of the function 119908(119909 119910) =119906(119909 119910) sdot V(119909 119910) is

119882(119896 ℎ) =

119896

sum

119903=0

ℎ

sum

119904=0

119880 (119903 ℎ minus 119904) 119881 (119896 minus 119903 119904) (17)

see [17]

Theorem 9 Differential transform of the function 119908(119909 119910) =119909119898119910119899 is

119882(119896 ℎ) = 120575 (119896 minus 119898 ℎ minus 119899) = 120575 (119896 minus 119898) 120575 (ℎ minus 119899) (18)

see [17] where

120575 (119896 minus 119898) = 1 119896 = 119898

0 119896 =119898

120575 (ℎ minus 119899) = 1 ℎ = 119899

0 ℎ = 119899

(19)

Theorem 10 Differential transform of the function 119908(119909 119910) =119892(119909 + 119886 119910) is

119882(119896 ℎ) =

119873

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) (20)

Proof From Definition 1 we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119866 (119896 ℎ) (119909 + 119886)119896119910ℎ

= 119866 (0 0) + 119866 (0 1) 119910 + 119886119866 (1 0) + 119866 (1 0) 119909

+ 119866 (0 2) 1199102+ 119866 (1 1) 119909119910 + 119886119866 (1 1) 119910

+ 1198862119866 (2 0) + 2119886119866 (2 0) 119909 + 119866 (2 0) 119909

2

+ 119866 (0 3) 1199103+ 119866 (1 2) 119909119910

2+ 119886119866 (1 2) 119910

2

+ 119866 (2 1) 1199092119910 + 2119886119866 (2 1) 119909119910 + 119886

2119866 (2 1) 119910

+ 1198863119866 (3 0) + 3119886

2119866 (3 0) 119909 + 3119886119866 (3 0) 119909

2

+ 119866 (3 0) 1199093+ sdot sdot sdot

119908 (119909 119910) = [119866 (0 0) + 119886119866 (1 0)

+1198862119866 (2 0) + 119886

3119866 (3 0) + sdot sdot sdot]

+ 119909 [119866 (1 0) + 2119886119866 (2 0) + 31198862119866 (3 0) + sdot sdot sdot]

+ 119910 [119866 (0 1) + 119886119866 (1 1) + 1198862119866 (2 1) + sdot sdot sdot]

+ 1199092[119866 (2 0) + 3119886119866 (3 0) + sdot sdot sdot]

+ 119909119910 [119866 (1 1) + 2119886119866 (2 1) + sdot sdot sdot]

+ 1199102[119866 (0 2) + 119886119866 (1 2) + sdot sdot sdot] + sdot sdot sdot

4 Abstract and Applied Analysis

=

infin

sum

119901=0

119886119901119866 (119901 0) + 119909

infin

sum

119901=1

119901119886119901minus1

119866 (119901 0)

+ 119910

infin

sum

119901=0

119886119901119866 (119901 1) + 119909

2

infin

sum

119901=2

119901

(119901 minus 2)2

times 119886119901minus2

119866 (119901 0) + 119909119910

infin

sum

119901=1

119901119886119901minus1

119866 (119901 1) + sdot sdot sdot

(21)

where

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

infin

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) 119909119896119910ℎ (22)

hence

119882(119896 ℎ) =

119873

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) (23)

Theorem 11 Differential transform of the function 119908(119909 119910) =119892(119909 + 119886 119910 + 119887) is

119882(119896 ℎ) =

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896) 119886119901minus119896

119887119902minusℎ119866 (119901 119902) (24)

Proof From Definition 2 we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119866 (119896 ℎ) (119909 + 119886)119896(119910 + 119887)

ℎ

= 119866 (0 0) + 119866 (1 0) 119909 + 119886119866 (1 0)

+ 119866 (0 1) 119910 + 119887119866 (0 1) + 119866 (2 0) 1199092

+ 2119886119866 (2 0) 119909 + 1198862119866 (2 0) + 119866 (1 1) 119909119910

+ 119887119866 (1 1) 119909 + 119886119866 (1 1) 119910 + 119886119887119866 (1 1)

+ 119866 (0 2) 1199102+ 2119887119866 (0 2) 119910 + 119887

2119866 (0 2)

+ 119866 (3 0) 1199093+ 3119886119866 (3 0) 119909

2+ 31198862119866 (3 0) 119909

+ 1198863119866 (3 0) + 119866 (2 1) 119909

2119910 + sdot sdot sdot

119908 (119909 119910) = [119866 (0 0) + 119886119866 (1 0) + 119887119866 (0 1)

+1198862119866 (2 0) + 119886119887119866 (1 1) + sdot sdot sdot]

+ [119866 (1 0) + 2119886119866 (2 0) + 119887119866 (1 1)

+31198862119866 (3 0) + 2119886119887119866 (2 1) + sdot sdot sdot] 119909

+ [119866 (0 1) + 119886119866 (1 1) + 2119887119866 (0 2)

+1198862119866 (2 1) + sdot sdot sdot] 119910 + sdot sdot sdot

=

119873

sum

119901=0

119873

sum

119902=0

119886119901119887119902119866 (119901 119902) + 119909

119873

sum

119901=1

119873

sum

119902=0

119901119886119901minus1

119887119902119866 (119901 119902)

+ 119910

119873

sum

119901=0

119873

sum

119902=1

119902119886119901119887119902minus1119866 (119901 119902)

+ 1199092

119873

sum

119901=2

119873

sum

119902=0

119901

(119901 minus 2)2119886119901minus2

119887119902119866 (119901 119902)

+ 119909119910

119873

sum

119901=1

119873

sum

119902=1

119901119902119886119901minus1

119887119902minus1119866 (119901 119902) + sdot sdot sdot

(25)

Hence we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896)

times 119886119901minus119896

119887119902minusℎ119866 (119901 119902) 119909

119896119910ℎ

(26)

Using Definition 2 we obtain

119882(119896 ℎ) =

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896) 119886119901minus119896

119887119902minusℎ119866 (119901 119902) (27)

Theorem 12 Differential transform of the function 119908(119909 119910) =120597119903+119904119892(119909 + 119886 119910 + 119887)120597119909

119903120597119910119904 is

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ

times

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(28)

Proof Let 119862(119896 ℎ) be differential transform of the function119892(119909+119886 119910+119887) FromTheorem 7 we can write that differentialtransform of the function 119908(119909 119910) is

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ119862 (119896 + 119903 ℎ + 119904) (29)

fromTheorem 4 we can write

119862 (119896 + 119903 ℎ + 119904) =

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(30)

Abstract and Applied Analysis 5

If we substitute (30) into (29) we find

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ

times

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(31)

4 Application

We have considered the following PDAE as a test problem

(1 1

0 0) 119906119905+ (

minus1 0

minus1 0) 119906119909119909+ (

1 1

minus1 0) 119906 = 119891

119905 isin [0infin) 119909 isin [minus1 1]

(32)

with initial values and boundary values

1199061(0 119909) = 119909

3minus 119909 119906

2(0 119909) = 119909

4minus 1

1199061(119905 1) = 119906

1(119905 minus1) = 0 119906

2(119905 1) = 119906

2(119905 minus1) = 0

(33)

The right hand side function 119891 is

119891 = ((1199094minus 1) (cos 119905 minus sin 119905) minus 6119909119890minus119905 minus119890minus119905 (1199093 + 5119909))

119879

(34)

and the exact solutions are

1199061(119905 119909) = (119909

3minus 119909) 119890

minus119905 119906

2(119905 119909) = (119909

4minus 1) cos 119905 (35)

If nonsingular matrices 119875119865119896 119876119865119896 119875119871120585 and 119876

119871120585are chosen

such as

119875119865119896

= (

1minus120583119896

120583119896minus 1

01

120583119896minus 1

) 119876119865119896

= (0 minus1

1 minus1)

119875119871120585= (

01

120585 + 11

120585 + 1minus

1

120585 + 1

) 119876119871120585= (

minus120585 minus 1 0

120585 1)

(36)

matrices 119875119865119896119860119876119865119896

and 119875119871120585119861119876119871120585

are found as

119875119865119896119860119876119865119896

= (1 0

0 0) 119875

119871120585119861119876119871120585= (

1 0

0 0) (37)

From (38) we have 119873119871120585

= 0 and 119873119865119896

= 0 Then the PDAE(32) has differential spatial index 1 and differential time index1 So it is enough to takeM(120585)

119861119862= 1 andM

(119896)

119868119862= 2 to solve

the problemTaking differential transformation of (32) we obtain

(119896 + 1)1198801(119896 + 1 ℎ) + (119896 + 1)119880

2(119896 + 1 ℎ)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) + 119880

1(119896 ℎ)

+ 1198802(119896 ℎ) = 119865

1(119896 ℎ)

(38)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) minus 119880

1(119896 ℎ) = 119865

2(119896 ℎ) (39)

Table 1 The numerical and exact solution of the test problem(32)where 119906

1(119905 119909) is the exact solution and 119906

lowast

1(119905 119909) is the numerical

solution for 119909 = 01

119905 1199061(119905 119909) 119906

lowast

1(119905 119909) |119906

1(119905 119909) minus 119906

lowast

1(119905 119909)|

01 minus00895789043 minus00895789043 002 minus00810543445 minus00810543422 0000000002303 minus00733410038 minus00733409887 0000000015104 minus00663616845 minus00663616355 0000000049005 minus00600465353 minus00600464409 0000000094406 minus00543323519 minus00543322800 0000000071907 minus00491619450 minus00491621943 0000000249308 minus00444835674 minus00444849422 0000001374809 minus00402503963 minus00402546487 0000004252410 minus00364200646 minus00364305555 00000104909

Table 2 The numerical and exact solution of the test problem(32)where 119906

2(119905 119909) is exact solution and 119906lowast

2(119905 119909) is numerical solution

for 119909 = 01

119905 1199062(119905 119909) 119906

lowast

2(119905 119909) |119906

2(119905 119909) minus 119906

lowast

2(119905 119909)|

01 minus09949046649 minus09949046653 0000000000402 minus09799685711 minus09799685778 0000000006703 minus09552409555 minus09552409875 0000000032004 minus09209688879 minus09209689778 0000000089905 minus08774948036 minus08774949653 0000000163706 minus08252530813 minus08252532000 0000000118707 minus07647657031 minus07647652653 0000000437808 minus06966370386 minus06966345778 0000002460809 minus06215478073 minus06215398875 0000007919810 minus05402482757 minus05402277778 00000204979

The Taylor series of functions 1198911and 119891

2about 119909 = 0 119905 = 0

are

1198911(119905 119909) = minus 1 + 119905 minus 6119909 +

1

21199052+ 6119909119905

minus1

61199053minus 31199091199052minus1

241199054+ 1199094+ 1199091199053

minus 1199094119905 +

1

1201199055minus1

41199091199054+

1

7201199056

minus1

211990941199052+1

201199091199055+1

611990941199053

minus1

50401199057minus

1

1201199091199056+ sdot sdot sdot

(40)

1198912(119905 119909) = minus 5119909 + 5119909119905 minus

5

21199091199052minus 1199093

+5

61199091199053+ 1199093119905 minus

1

211990931199052minus5

241199091199054

+1

611990931199053+1

241199091199055minus

1

1441199091199056

minus1

2411990931199054+ sdot sdot sdot

(41)

6 Abstract and Applied Analysis

04

02

0

minus02

minus040

051

152

1

119905

minus05

minus1

11990905

0

Figure 1 The graphic of the function 1199061(119905 119909) in the test problem

(32)

04

02

0

minus02

minus04

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 2 The graphic of the function 119906lowast1(119905 119909) in the test problem

(32)

The values 1198651(119896 ℎ) and 119865

2(119896 ℎ) in (39) and (40) are coeffi-

cients of polynomials (41) and (42) If we use Theorem 3 forboundary values we obtain

7

sum

119894=0

1198801(119895 119894) = 0 119895 = 0 1 7 (42)

7

sum

119894=0

(minus1)1198941198801(119895 119894) = 0 119895 = 0 1 7 (43)

0406

020

minus02

minus04

minus06

minus08

minus1

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 3 The graphic of the function 1199062(119905 119909) in the test problem

(32)

0406

020

minus02

minus04

minus06

minus08

minus1

0

05

1

152

1

119905

minus05

minus1

119909

05

0

Figure 4 The graphic of the function 119906lowast2(119905 119909) in the test problem

(32)

In order to write 119896 = 0 and ℎ = 0 1 2 3 4 5 in (40) we have

21198801(0 2) + 119880

1(0 0) = 0 20119880

1(0 5) + 119880

1(0 3) = 1

61198801(0 3) + 119880

1(0 1) = 5 30119880

1(0 6) + 119880

1(0 4) = 0

121198801(0 4) + 119880

1(0 2) = 0 42119880

1(0 7) + 119880

1(0 5) = 0

(44)

If we take 119895 = 0 in (43) and (44) we obtain

1198801(0 0) + 119880

1(0 1) + 119880

1(0 2) + 119880

1(0 3)

+ 1198801(0 4) + 119880

1(0 5) + 119880

1(0 6) + 119880

1(0 7) = 0

1198801(0 0) minus 119880

1(0 1) + 119880

1(0 2) minus 119880

1(0 3)

+ 1198801(0 4) minus 119880

1(0 5) + 119880

1(0 6) minus 119880

1(0 7) = 0

(45)

Abstract and Applied Analysis 7

From (45) and (46) we find

1198801(0 0) = 0 119880

1(0 1) = minus1 119880

1(0 2) = 0

1198801(0 3) = 1 119880

1(0 4) = 0 119880

1(0 5) = 0

1198801(0 6) = 0 119880

1(0 7) = 0

(46)

In this manner from (40) (44) and (45) the coefficients ofthe 1199061are obtained as follows

1198801(1 0) = 0 119880

1(1 1) = 1 119880

1(1 2) = 0

1198801(1 3) = minus1 119880

1(1 4) = 0 119880

1(1 5) = 0

1198801(1 6) = 0 119880

1(2 0) = 0 119880

1(2 1) = minus

1

2

1198801(2 2) = 0 119880

1(2 3) =

1

2 119880

1(2 4) = 0

1198801(2 5) = 0 119880

1(3 0) = 0 119880

1(3 1) =

1

6

1198801(3 2) = 0 119880

1(3 3) = minus

1

6 119880

1(3 4) = 0

1198801(4 0) = 0 119880

1(4 1) = minus

1

24 119880

1(4 2) = 0

1198801(4 3) =

1

24 119880

1(5 0) = 0 119880

1(5 1) =

1

120

1198801(5 2) = 0 119880

1(6 0) = 0 119880

1(6 1) = minus

1

720

1198801(7 0) = 0

(47)

Using the initial values for the second component we obtainthe following coefficients

1198802(0 0) = minus1 119880

2(0 1) = 0 119880

2(0 2) = 0

1198802(0 3) = 0 119880

2(0 4) = 1 119880

2(0 5) = 0

1198802(0 6) = 0 119880

2(0 7) = 0

(48)

The coefficients of the 1199062can be found using (47) (48) (49)

and taking 119896 = 0 1 2 and ℎ = 0 1 2 in (39) as follows

1198802(1 2) = 0 119880

2(1 3) = 0 119880

2(1 4) = 0

1198802(1 5) = 0 119880

2(1 6) = 0 119880

2(2 1) = 0

1198802(2 2) = 0 119880

2(2 3) = 0 119880

2(2 4) = minus

1

2

1198802(2 5) = 0 119880

2(3 0) = 0 119880

2(3 1) = 0

1198802(3 2) = 0 119880

2(3 3) = 0 119880

2(3 4) = 0

1198802(4 1) = 0 119880

2(4 2) = 0 119880

2(4 3) = 0

1198802(4 0) = minus

1

24 119880

2(5 0) = 0 119880

2(5 1) = 0

1198802(5 2) = 0 119880

2(6 0) =

1

720 119880

2(6 1) = 0

1198802(7 0) = 0

(49)

If we write the above values in (39) and (40) then we have

119906lowast

1(119905 119909) = minus 119909 + 119909119905 minus

1

21199091199052+ 1199093+1

61199091199053

minus 1199093119905 +

1

211990931199052minus1

241199091199054minus1

611990931199053

+1

1201199091199055minus

1

7201199091199056+1

2411990931199054+ sdot sdot sdot

(50)

119906lowast

2(119905 119909) = minus 1 +

1

21199052minus1

241199054

+ 1199094+

1

7201199056minus1

211990941199052+ sdot sdot sdot

(51)

Numerical and exact solution of the given problem has beencompared in Tables 1 and 2 and simulations of solutions havebeen depicted in Figures 1 2 3 and 4 respectively

5 Conclusion

The computations associated with the example discussedabove were performed by using Computer Algebra Tech-niques [24] We show the results in Tables 1 and 2 for thesolution of (32) by numerical method The numerical valueson Tables 1 and 2 obtained above are in full agreement withthe exact solutions of (32) This study has shown that thedifferential transform method often shows superior perfor-mance over series approximants providing a promising toolfor using in applied fields

References

[1] W Marszalek Analysis of partial differential algebraic equations[PhD thesis] North Carolina State University Raleigh NCUSA 1997

[2] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear par-tial differential algebraic equations Part I indexes consistentboundaryinitial conditionsrdquo Report 17 Fachbereich Mathe-matik und Informatik Martin-Luther-Universitat Halle 1997

[3] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear partialdifferential algebraic equations Part II numerical solutionrdquoReport 18 Fachbereich Mathematik und Informatik Martin-Luther-Universitat Halle 1997

[4] W Lucht K Strehmel and C Eichler-Liebenow ldquoIndexes andspecial discretization methods for linear partial differentialalgebraic equationsrdquo BIT Numerical Mathematics vol 39 no3 pp 484ndash512 1999

[5] W S Martinson and P I Barton ldquoA differentiation indexfor partial differential-algebraic equationsrdquo SIAM Journal onScientific Computing vol 21 no 6 pp 2295ndash2315 2000

8 Abstract and Applied Analysis

[6] W S Martinson and P I Barton ldquoIndex and characteristicanalysis of linear PDAE systemsrdquo SIAM Journal on ScientificComputing vol 24 no 3 pp 905ndash923 2002

[7] K Debrabant and K Strehmel ldquoConvergence of Runge-Kuttamethods applied to linear partial differential-algebraic equa-tionsrdquoApplied Numerical Mathematics vol 53 no 2-4 pp 213ndash229 2005

[8] N Guzel and M Bayram ldquoOn the numerical solution of stiffsystemsrdquo Applied Mathematics and Computation vol 170 no 1pp 230ndash236 2005

[9] E Celik and M Bayram ldquoThe numerical solution of physicalproblems modeled as a system of differential-algebraic equa-tions (DAEs)rdquo Journal of the Franklin Institute vol 342 no 1pp 1ndash6 2005

[10] M Kurulay and M Bayram ldquoApproximate analytical solutionfor the fractional modified KdV by differential transformmethodrdquo Communications in Nonlinear Science and NumericalSimulation vol 15 no 7 pp 1777ndash1782 2010

[11] M Bayram ldquoAutomatic analysis of the control of metabolicnetworksrdquo Computers in Biology and Medicine vol 26 no 5pp 401ndash408 1996

[12] N Guzel and M Bayram ldquoNumerical solution of differential-algebraic equations with index-2rdquo Applied Mathematics andComputation vol 174 no 2 pp 1279ndash1289 2006

[13] J K Zhou Differential Transformation and Its Application forElectrical CircuIts Huazhong University Press Wuhan China1986

[14] M Koksal and S Herdem ldquoAnalysis of nonlinear circuits byusing differential Taylor transformrdquo Computers and ElectricalEngineering vol 28 no 6 pp 513ndash525 2002

[15] I H Abdel-Halim Hassan ldquoOn solving some eigenvalue prob-lems by using a differential transformationrdquoAppliedMathemat-ics and Computation vol 127 no 1 pp 1ndash22 2002

[16] F Ayaz ldquoOn the two-dimensional differential transformmethodrdquo Applied Mathematics and Computation vol 143 no2-3 pp 361ndash374 2003

[17] C K Chen and S H Ho ldquoSolving partial differential equationsby two-dimensional differential transform methodrdquo AppliedMathematics and Computation vol 106 no 2-3 pp 171ndash1791999

[18] M-J Jang C-L Chen and Y-C Liu ldquoTwo-dimensional dif-ferential transform for partial differential equationsrdquo AppliedMathematics and Computation vol 121 no 2-3 pp 261ndash2702001

[19] X Yang Y Liu and S Bai ldquoA numerical solution of second-order linear partial differential equations by differential trans-formrdquo Applied Mathematics and Computation vol 173 no 2pp 792ndash802 2006

[20] A Arikoglu and I Ozkol ldquoSolution of differential-differenceequations by using differential transform methodrdquo AppliedMathematics and Computation vol 181 no 1 pp 153ndash162 2006

[21] M-J Jang C-L Chen and Y-C Liy ldquoOn solving the initial-value problems using the differential transformation methodrdquoAppliedMathematics andComputation vol 115 no 2-3 pp 145ndash160 2000

[22] F Ayaz ldquoApplications of differential transform method todifferential-algebraic equationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 649ndash657 2004

[23] H Liu and Y Song ldquoDifferential transform method applied tohigh index differential-algebraic equationsrdquoAppliedMathemat-ics and Computation vol 184 no 2 pp 748ndash753 2007

[24] G FrankMAPLE V CRC Press Boca Raton Fla USA 1996

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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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

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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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

2 Indexes of Partial DifferentialAlgebraic Equation

Let us consider (1) with initial values and boundary condi-tions given as follows

119906119895(119905 plusmn119897) = 0 for 119905 isin 119869

119906119894(0 119909) = 119892 (119909) for 119909 isin Ω

(2)

where 119895 isin M119861119862

sube 1 2 119899 M119861119862

is the set of indicesof components of 119906 for which boundary conditions can beprescribed arbitrarily and 119894 isin M

119868119862sube 1 2 119899 M

119868119862

is the set of indices of components of 119906 for which initialconditions can be prescribed arbitrarilyThe initial boundaryvalue problem (IBVP) (1) has only one solution where afunction 119906 is a solution of the problem if it is sufficientlysmooth uniquely determined by its initial values (IVs) andboundary values (BVs) and if it solves the LPDAE point wise

Definition of the indexes can be given using the followingassumptions

(i) Each component of the vectors 119906 119906119905 and 119891 satisfy the

following condition1003816100381610038161003816119910 (119905 119909)

1003816100381610038161003816 le 119872119890120572119905 120572 ge 0 119905 ge 0 (3)

where119872 and 120572 are independent of 119905 and 119909(ii) (119861 120585119860 + 119862)Re(120585) gt 120572 called as the matrix pencil is

regular(iii) (119860 120583

119896119861 + 119862) is regular for all 119896 where 120583

119896is an eigen-

value of the operator 12059721205971199092 together with prescribedBCs

(iv) The vector 119891(119905 119909) and the initial vector 119892(119909) aresufficiently smooth

If we use Laplace transform from assumption (ii) (1) can betransformed into11986111990610158401015840

120585(119909) + (120585119860 + 119862) 119906

120585(119909) = 119891

120585(119909) + 119860119892 (119909) Re (120585) gt 120572

(4)

if 119861 is a singular matrix then (4) is a DAE depending on theparameter 120585 To characterizeM

119861119862 we introduce 119895 isin M

(120585)

119861119862sube

1 2 119899 as the set of indices of components of 119906120585for which

boundary conditions can be prescribed arbitrarilyIn order to define a spatial index we need the Kronecker

normal form of theDAE (4) Assumption (iii) guarantees thatthere are nonsingular matrices 119875

119871120585 119876119871120585isin C119899times119899 such that

119875119871120585119861119876119871120585= (

1198681198981

0

0 119873119871120585

)

119875119871120585(120585119860 + 119862)119876

119871120585= (

119877119871120585

0

0 1198681198982

)

(5)

where 119877119871120585isin C1198981times1198982 and119873

119871120585isin R1198982times1198982 is a nilpotent Jordan

chain matrix with1198981+1198982= 119899 119868

119896is the unit matrix of order

119896 The Riesz index (or nilpotency) of 119873119871120585

is denoted by ]119871120585

(ie119873]119871120585

119871120585= 0119873

]119871120585minus1

119871120585= 0)

Here we will assume that there is a real number 120572lowast ge 120572

such that the index set M(120585)119861119862

is independent of the Laplaceparameter 120585 provided Re(120585) ge 120572lowast

Definition 1 Let 120572lowast isin R+ be a number with 120572lowast ge 120572 such thatfor all 120585 isin C with Re(120585) ge 120572lowast

(1) the matrix pencil (119861 120585119860 + 119862) is regular

(2) M(120585)119861119862

is independent of 120585 ieM(120585)119861119862

=M119861119862

(3) the nilpotency of119873119871120585

is ]119871ge 1

Then ]119889119909

= 2]119871minus 1 is called the ldquodifferential spatial indexrdquo

of the LPDAE If ]119871= 0 then the differential spatial index of

LPDAE is defined to be zero

If we use Fourier transform (1) can be transformed into

1198601015840

119896(119905) + (120583

119896119861 + 119862)

119896(119905) =

119896(119905) + 119861120588

119896(119905) (6)

with 120588119896(119905) = (120588

1198961(119905) 120588

119896119899(119905))119879 and

120588119896119894(119905) = 0 for 119894 isinM

119861119862

120588119896119895(119905) =

1

119897[1206011015840

119896(119909) 119906119895(119905 119909) minus 120601

119896(119909) 119906119909119895(119905 119909)]

119909=119897

119909=minus119897

(7)

for 119895 notin M119861119862 which results from partial integration of the

term int119897

minus119897119906119909119909(119905 119909)120601

119896(119909)119889119909

If 119860 is a singular matrix then (6) is a DAE depending onthe parameter 120583

119896which can be solved uniquely with suitable

ICs under the assumptions (iv) and (v) Analogous to the caseof the Laplace transform the above assumption (iv) impliesthat there exist regular matrices 119875

119865119896 119876119865119896

such that

119875119865119896119860119876119865119896

= (1198681198991

0

0 119873119865119896

)

119875119865119896(120583119896119861 + 119862)119876

119865119896= (

119877119865119896

0

0 1198681198992

)

(8)

With 119877119865119896

isin R1198991times1198991 119873119865119896

isin R1198992times1198992 is again a nilpotent Jordanchain matrix with Riesz index ]

119865119896 where 119899

1+ 1198992= 119899

To characterize M119868119862 we introduce M(119896)

119868119862sube 1 2 119899

as the set of indices of components of 119896for which initial

conditions can be prescribed arbitrarilyTherefore we alwaysassume in the context of a Fourier analysis of 119906 that M(119896)

119868119862is

independent of 119896 isin N+ ieM(119896)

119868119862=M119868119862

Definition 2 Assume for 119896 = 1 2 that

(1) the matrix pencil (119860 120583119896119861 + 119862) is regular

(2) M(119896)119868119862

is independent of 119896 ieM(119896)119868119862=M119868119862

(3) the nilpotency of119873119865119896

is ]119865119896

= ]119865

Then the PDAE (1) is said to have uniform differential timeindex ]

119889119905= ]119865

The differential spatial and time indexes are used todecide which initial and boundary values can be taken tosolve the problem

Abstract and Applied Analysis 3

3 Two-Dimensional DifferentialTransform Method

The two-dimensional differential transform of function119908(119909 119910) is defined as

119882(119896 ℎ) =1

119896ℎ[120597119896+ℎ119908 (119909 119910)

120597119909119896120597119910ℎ]119909=0

119910=0

(9)

where it is noted that upper case symbol 119882(119896 ℎ) is usedto denote the two-dimensional differential transform of afunction represented by a corresponding lower case symbol119908(119909 119910) The differential inverse transform of 119882(119896 ℎ) isdefined as

119908 (119909 119910) =

infin

sum

119896=0

infin

sum

ℎ=0

119882(119896 ℎ) 119909119896119910ℎ (10)

From (9) and (10) we obtain

119908 (119909 119910) =

infin

sum

119896=0

infin

sum

ℎ=0

119909119896119910ℎ

119896ℎ[120597119896+ℎ119908 (119909 119910)

120597119909119896120597119910ℎ]119909=0

119910=0

(11)

The concept of two-dimensional differential transform isderived from two-dimensional Taylor series expansion butthe method doesnrsquot evaluate the derivatives symbolically

Theorem 3 Differential transform of the function 119908(119909 119910) =119906(119909 119910) plusmn V(119909 119910) is

119882(119896 ℎ) = 119880 (119896 ℎ) plusmn 119881 (119896 ℎ) (12)

see [17]

Theorem 4 Differential transform of the function 119908(119909 119910) =120582119906(119909 119910) is

119882(119896 ℎ) = 120582119880 (119896 ℎ) (13)

see [17]

Theorem 5 Differential transform of the function 119908(119909 119910) =120597119906(119909 119910)120597119909 is

119882(119896 ℎ) = (119896 + 1)119880 (119896 + 1 ℎ) (14)

see [17]

Theorem 6 Differential transform of the function 119908(119909 119910) =120597119906(119909 119910)120597119910 is

119882(119896 ℎ) = (ℎ + 1)119880 (119896 ℎ + 1) (15)

see [17]

Theorem 7 Differential transform of the function 119908(119909 119910) =120597119903+119904119906(119909 119910)120597119909

119903120597119910119904 is

119882(119896 ℎ) = (119896 + 1) (119896 + 2) sdot sdot sdot (119896 + 119903) (ℎ + 1)

times (ℎ + 2) sdot sdot sdot (ℎ + 119904)119880 (119896 + 119903 ℎ + 119904)

(16)

see [17]

Theorem 8 Differential transform of the function 119908(119909 119910) =119906(119909 119910) sdot V(119909 119910) is

119882(119896 ℎ) =

119896

sum

119903=0

ℎ

sum

119904=0

119880 (119903 ℎ minus 119904) 119881 (119896 minus 119903 119904) (17)

see [17]

Theorem 9 Differential transform of the function 119908(119909 119910) =119909119898119910119899 is

119882(119896 ℎ) = 120575 (119896 minus 119898 ℎ minus 119899) = 120575 (119896 minus 119898) 120575 (ℎ minus 119899) (18)

see [17] where

120575 (119896 minus 119898) = 1 119896 = 119898

0 119896 =119898

120575 (ℎ minus 119899) = 1 ℎ = 119899

0 ℎ = 119899

(19)

Theorem 10 Differential transform of the function 119908(119909 119910) =119892(119909 + 119886 119910) is

119882(119896 ℎ) =

119873

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) (20)

Proof From Definition 1 we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119866 (119896 ℎ) (119909 + 119886)119896119910ℎ

= 119866 (0 0) + 119866 (0 1) 119910 + 119886119866 (1 0) + 119866 (1 0) 119909

+ 119866 (0 2) 1199102+ 119866 (1 1) 119909119910 + 119886119866 (1 1) 119910

+ 1198862119866 (2 0) + 2119886119866 (2 0) 119909 + 119866 (2 0) 119909

2

+ 119866 (0 3) 1199103+ 119866 (1 2) 119909119910

2+ 119886119866 (1 2) 119910

2

+ 119866 (2 1) 1199092119910 + 2119886119866 (2 1) 119909119910 + 119886

2119866 (2 1) 119910

+ 1198863119866 (3 0) + 3119886

2119866 (3 0) 119909 + 3119886119866 (3 0) 119909

2

+ 119866 (3 0) 1199093+ sdot sdot sdot

119908 (119909 119910) = [119866 (0 0) + 119886119866 (1 0)

+1198862119866 (2 0) + 119886

3119866 (3 0) + sdot sdot sdot]

+ 119909 [119866 (1 0) + 2119886119866 (2 0) + 31198862119866 (3 0) + sdot sdot sdot]

+ 119910 [119866 (0 1) + 119886119866 (1 1) + 1198862119866 (2 1) + sdot sdot sdot]

+ 1199092[119866 (2 0) + 3119886119866 (3 0) + sdot sdot sdot]

+ 119909119910 [119866 (1 1) + 2119886119866 (2 1) + sdot sdot sdot]

+ 1199102[119866 (0 2) + 119886119866 (1 2) + sdot sdot sdot] + sdot sdot sdot

4 Abstract and Applied Analysis

=

infin

sum

119901=0

119886119901119866 (119901 0) + 119909

infin

sum

119901=1

119901119886119901minus1

119866 (119901 0)

+ 119910

infin

sum

119901=0

119886119901119866 (119901 1) + 119909

2

infin

sum

119901=2

119901

(119901 minus 2)2

times 119886119901minus2

119866 (119901 0) + 119909119910

infin

sum

119901=1

119901119886119901minus1

119866 (119901 1) + sdot sdot sdot

(21)

where

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

infin

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) 119909119896119910ℎ (22)

hence

119882(119896 ℎ) =

119873

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) (23)

Theorem 11 Differential transform of the function 119908(119909 119910) =119892(119909 + 119886 119910 + 119887) is

119882(119896 ℎ) =

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896) 119886119901minus119896

119887119902minusℎ119866 (119901 119902) (24)

Proof From Definition 2 we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119866 (119896 ℎ) (119909 + 119886)119896(119910 + 119887)

ℎ

= 119866 (0 0) + 119866 (1 0) 119909 + 119886119866 (1 0)

+ 119866 (0 1) 119910 + 119887119866 (0 1) + 119866 (2 0) 1199092

+ 2119886119866 (2 0) 119909 + 1198862119866 (2 0) + 119866 (1 1) 119909119910

+ 119887119866 (1 1) 119909 + 119886119866 (1 1) 119910 + 119886119887119866 (1 1)

+ 119866 (0 2) 1199102+ 2119887119866 (0 2) 119910 + 119887

2119866 (0 2)

+ 119866 (3 0) 1199093+ 3119886119866 (3 0) 119909

2+ 31198862119866 (3 0) 119909

+ 1198863119866 (3 0) + 119866 (2 1) 119909

2119910 + sdot sdot sdot

119908 (119909 119910) = [119866 (0 0) + 119886119866 (1 0) + 119887119866 (0 1)

+1198862119866 (2 0) + 119886119887119866 (1 1) + sdot sdot sdot]

+ [119866 (1 0) + 2119886119866 (2 0) + 119887119866 (1 1)

+31198862119866 (3 0) + 2119886119887119866 (2 1) + sdot sdot sdot] 119909

+ [119866 (0 1) + 119886119866 (1 1) + 2119887119866 (0 2)

+1198862119866 (2 1) + sdot sdot sdot] 119910 + sdot sdot sdot

=

119873

sum

119901=0

119873

sum

119902=0

119886119901119887119902119866 (119901 119902) + 119909

119873

sum

119901=1

119873

sum

119902=0

119901119886119901minus1

119887119902119866 (119901 119902)

+ 119910

119873

sum

119901=0

119873

sum

119902=1

119902119886119901119887119902minus1119866 (119901 119902)

+ 1199092

119873

sum

119901=2

119873

sum

119902=0

119901

(119901 minus 2)2119886119901minus2

119887119902119866 (119901 119902)

+ 119909119910

119873

sum

119901=1

119873

sum

119902=1

119901119902119886119901minus1

119887119902minus1119866 (119901 119902) + sdot sdot sdot

(25)

Hence we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896)

times 119886119901minus119896

119887119902minusℎ119866 (119901 119902) 119909

119896119910ℎ

(26)

Using Definition 2 we obtain

119882(119896 ℎ) =

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896) 119886119901minus119896

119887119902minusℎ119866 (119901 119902) (27)

Theorem 12 Differential transform of the function 119908(119909 119910) =120597119903+119904119892(119909 + 119886 119910 + 119887)120597119909

119903120597119910119904 is

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ

times

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(28)

Proof Let 119862(119896 ℎ) be differential transform of the function119892(119909+119886 119910+119887) FromTheorem 7 we can write that differentialtransform of the function 119908(119909 119910) is

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ119862 (119896 + 119903 ℎ + 119904) (29)

fromTheorem 4 we can write

119862 (119896 + 119903 ℎ + 119904) =

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(30)

Abstract and Applied Analysis 5

If we substitute (30) into (29) we find

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ

times

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(31)

4 Application

We have considered the following PDAE as a test problem

(1 1

0 0) 119906119905+ (

minus1 0

minus1 0) 119906119909119909+ (

1 1

minus1 0) 119906 = 119891

119905 isin [0infin) 119909 isin [minus1 1]

(32)

with initial values and boundary values

1199061(0 119909) = 119909

3minus 119909 119906

2(0 119909) = 119909

4minus 1

1199061(119905 1) = 119906

1(119905 minus1) = 0 119906

2(119905 1) = 119906

2(119905 minus1) = 0

(33)

The right hand side function 119891 is

119891 = ((1199094minus 1) (cos 119905 minus sin 119905) minus 6119909119890minus119905 minus119890minus119905 (1199093 + 5119909))

119879

(34)

and the exact solutions are

1199061(119905 119909) = (119909

3minus 119909) 119890

minus119905 119906

2(119905 119909) = (119909

4minus 1) cos 119905 (35)

If nonsingular matrices 119875119865119896 119876119865119896 119875119871120585 and 119876

119871120585are chosen

such as

119875119865119896

= (

1minus120583119896

120583119896minus 1

01

120583119896minus 1

) 119876119865119896

= (0 minus1

1 minus1)

119875119871120585= (

01

120585 + 11

120585 + 1minus

1

120585 + 1

) 119876119871120585= (

minus120585 minus 1 0

120585 1)

(36)

matrices 119875119865119896119860119876119865119896

and 119875119871120585119861119876119871120585

are found as

119875119865119896119860119876119865119896

= (1 0

0 0) 119875

119871120585119861119876119871120585= (

1 0

0 0) (37)

From (38) we have 119873119871120585

= 0 and 119873119865119896

= 0 Then the PDAE(32) has differential spatial index 1 and differential time index1 So it is enough to takeM(120585)

119861119862= 1 andM

(119896)

119868119862= 2 to solve

the problemTaking differential transformation of (32) we obtain

(119896 + 1)1198801(119896 + 1 ℎ) + (119896 + 1)119880

2(119896 + 1 ℎ)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) + 119880

1(119896 ℎ)

+ 1198802(119896 ℎ) = 119865

1(119896 ℎ)

(38)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) minus 119880

1(119896 ℎ) = 119865

2(119896 ℎ) (39)

Table 1 The numerical and exact solution of the test problem(32)where 119906

1(119905 119909) is the exact solution and 119906

lowast

1(119905 119909) is the numerical

solution for 119909 = 01

119905 1199061(119905 119909) 119906

lowast

1(119905 119909) |119906

1(119905 119909) minus 119906

lowast

1(119905 119909)|

01 minus00895789043 minus00895789043 002 minus00810543445 minus00810543422 0000000002303 minus00733410038 minus00733409887 0000000015104 minus00663616845 minus00663616355 0000000049005 minus00600465353 minus00600464409 0000000094406 minus00543323519 minus00543322800 0000000071907 minus00491619450 minus00491621943 0000000249308 minus00444835674 minus00444849422 0000001374809 minus00402503963 minus00402546487 0000004252410 minus00364200646 minus00364305555 00000104909

Table 2 The numerical and exact solution of the test problem(32)where 119906

2(119905 119909) is exact solution and 119906lowast

2(119905 119909) is numerical solution

for 119909 = 01

119905 1199062(119905 119909) 119906

lowast

2(119905 119909) |119906

2(119905 119909) minus 119906

lowast

2(119905 119909)|

01 minus09949046649 minus09949046653 0000000000402 minus09799685711 minus09799685778 0000000006703 minus09552409555 minus09552409875 0000000032004 minus09209688879 minus09209689778 0000000089905 minus08774948036 minus08774949653 0000000163706 minus08252530813 minus08252532000 0000000118707 minus07647657031 minus07647652653 0000000437808 minus06966370386 minus06966345778 0000002460809 minus06215478073 minus06215398875 0000007919810 minus05402482757 minus05402277778 00000204979

The Taylor series of functions 1198911and 119891

2about 119909 = 0 119905 = 0

are

1198911(119905 119909) = minus 1 + 119905 minus 6119909 +

1

21199052+ 6119909119905

minus1

61199053minus 31199091199052minus1

241199054+ 1199094+ 1199091199053

minus 1199094119905 +

1

1201199055minus1

41199091199054+

1

7201199056

minus1

211990941199052+1

201199091199055+1

611990941199053

minus1

50401199057minus

1

1201199091199056+ sdot sdot sdot

(40)

1198912(119905 119909) = minus 5119909 + 5119909119905 minus

5

21199091199052minus 1199093

+5

61199091199053+ 1199093119905 minus

1

211990931199052minus5

241199091199054

+1

611990931199053+1

241199091199055minus

1

1441199091199056

minus1

2411990931199054+ sdot sdot sdot

(41)

6 Abstract and Applied Analysis

04

02

0

minus02

minus040

051

152

1

119905

minus05

minus1

11990905

0

Figure 1 The graphic of the function 1199061(119905 119909) in the test problem

(32)

04

02

0

minus02

minus04

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 2 The graphic of the function 119906lowast1(119905 119909) in the test problem

(32)

The values 1198651(119896 ℎ) and 119865

2(119896 ℎ) in (39) and (40) are coeffi-

cients of polynomials (41) and (42) If we use Theorem 3 forboundary values we obtain

7

sum

119894=0

1198801(119895 119894) = 0 119895 = 0 1 7 (42)

7

sum

119894=0

(minus1)1198941198801(119895 119894) = 0 119895 = 0 1 7 (43)

0406

020

minus02

minus04

minus06

minus08

minus1

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 3 The graphic of the function 1199062(119905 119909) in the test problem

(32)

0406

020

minus02

minus04

minus06

minus08

minus1

0

05

1

152

1

119905

minus05

minus1

119909

05

0

Figure 4 The graphic of the function 119906lowast2(119905 119909) in the test problem

(32)

In order to write 119896 = 0 and ℎ = 0 1 2 3 4 5 in (40) we have

21198801(0 2) + 119880

1(0 0) = 0 20119880

1(0 5) + 119880

1(0 3) = 1

61198801(0 3) + 119880

1(0 1) = 5 30119880

1(0 6) + 119880

1(0 4) = 0

121198801(0 4) + 119880

1(0 2) = 0 42119880

1(0 7) + 119880

1(0 5) = 0

(44)

If we take 119895 = 0 in (43) and (44) we obtain

1198801(0 0) + 119880

1(0 1) + 119880

1(0 2) + 119880

1(0 3)

+ 1198801(0 4) + 119880

1(0 5) + 119880

1(0 6) + 119880

1(0 7) = 0

1198801(0 0) minus 119880

1(0 1) + 119880

1(0 2) minus 119880

1(0 3)

+ 1198801(0 4) minus 119880

1(0 5) + 119880

1(0 6) minus 119880

1(0 7) = 0

(45)

Abstract and Applied Analysis 7

From (45) and (46) we find

1198801(0 0) = 0 119880

1(0 1) = minus1 119880

1(0 2) = 0

1198801(0 3) = 1 119880

1(0 4) = 0 119880

1(0 5) = 0

1198801(0 6) = 0 119880

1(0 7) = 0

(46)

In this manner from (40) (44) and (45) the coefficients ofthe 1199061are obtained as follows

1198801(1 0) = 0 119880

1(1 1) = 1 119880

1(1 2) = 0

1198801(1 3) = minus1 119880

1(1 4) = 0 119880

1(1 5) = 0

1198801(1 6) = 0 119880

1(2 0) = 0 119880

1(2 1) = minus

1

2

1198801(2 2) = 0 119880

1(2 3) =

1

2 119880

1(2 4) = 0

1198801(2 5) = 0 119880

1(3 0) = 0 119880

1(3 1) =

1

6

1198801(3 2) = 0 119880

1(3 3) = minus

1

6 119880

1(3 4) = 0

1198801(4 0) = 0 119880

1(4 1) = minus

1

24 119880

1(4 2) = 0

1198801(4 3) =

1

24 119880

1(5 0) = 0 119880

1(5 1) =

1

120

1198801(5 2) = 0 119880

1(6 0) = 0 119880

1(6 1) = minus

1

720

1198801(7 0) = 0

(47)

Using the initial values for the second component we obtainthe following coefficients

1198802(0 0) = minus1 119880

2(0 1) = 0 119880

2(0 2) = 0

1198802(0 3) = 0 119880

2(0 4) = 1 119880

2(0 5) = 0

1198802(0 6) = 0 119880

2(0 7) = 0

(48)

The coefficients of the 1199062can be found using (47) (48) (49)

and taking 119896 = 0 1 2 and ℎ = 0 1 2 in (39) as follows

1198802(1 2) = 0 119880

2(1 3) = 0 119880

2(1 4) = 0

1198802(1 5) = 0 119880

2(1 6) = 0 119880

2(2 1) = 0

1198802(2 2) = 0 119880

2(2 3) = 0 119880

2(2 4) = minus

1

2

1198802(2 5) = 0 119880

2(3 0) = 0 119880

2(3 1) = 0

1198802(3 2) = 0 119880

2(3 3) = 0 119880

2(3 4) = 0

1198802(4 1) = 0 119880

2(4 2) = 0 119880

2(4 3) = 0

1198802(4 0) = minus

1

24 119880

2(5 0) = 0 119880

2(5 1) = 0

1198802(5 2) = 0 119880

2(6 0) =

1

720 119880

2(6 1) = 0

1198802(7 0) = 0

(49)

If we write the above values in (39) and (40) then we have

119906lowast

1(119905 119909) = minus 119909 + 119909119905 minus

1

21199091199052+ 1199093+1

61199091199053

minus 1199093119905 +

1

211990931199052minus1

241199091199054minus1

611990931199053

+1

1201199091199055minus

1

7201199091199056+1

2411990931199054+ sdot sdot sdot

(50)

119906lowast

2(119905 119909) = minus 1 +

1

21199052minus1

241199054

+ 1199094+

1

7201199056minus1

211990941199052+ sdot sdot sdot

(51)

Numerical and exact solution of the given problem has beencompared in Tables 1 and 2 and simulations of solutions havebeen depicted in Figures 1 2 3 and 4 respectively

5 Conclusion

The computations associated with the example discussedabove were performed by using Computer Algebra Tech-niques [24] We show the results in Tables 1 and 2 for thesolution of (32) by numerical method The numerical valueson Tables 1 and 2 obtained above are in full agreement withthe exact solutions of (32) This study has shown that thedifferential transform method often shows superior perfor-mance over series approximants providing a promising toolfor using in applied fields

References

[1] W Marszalek Analysis of partial differential algebraic equations[PhD thesis] North Carolina State University Raleigh NCUSA 1997

[2] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear par-tial differential algebraic equations Part I indexes consistentboundaryinitial conditionsrdquo Report 17 Fachbereich Mathe-matik und Informatik Martin-Luther-Universitat Halle 1997

[3] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear partialdifferential algebraic equations Part II numerical solutionrdquoReport 18 Fachbereich Mathematik und Informatik Martin-Luther-Universitat Halle 1997

[4] W Lucht K Strehmel and C Eichler-Liebenow ldquoIndexes andspecial discretization methods for linear partial differentialalgebraic equationsrdquo BIT Numerical Mathematics vol 39 no3 pp 484ndash512 1999

[5] W S Martinson and P I Barton ldquoA differentiation indexfor partial differential-algebraic equationsrdquo SIAM Journal onScientific Computing vol 21 no 6 pp 2295ndash2315 2000

8 Abstract and Applied Analysis

[6] W S Martinson and P I Barton ldquoIndex and characteristicanalysis of linear PDAE systemsrdquo SIAM Journal on ScientificComputing vol 24 no 3 pp 905ndash923 2002

[7] K Debrabant and K Strehmel ldquoConvergence of Runge-Kuttamethods applied to linear partial differential-algebraic equa-tionsrdquoApplied Numerical Mathematics vol 53 no 2-4 pp 213ndash229 2005

[8] N Guzel and M Bayram ldquoOn the numerical solution of stiffsystemsrdquo Applied Mathematics and Computation vol 170 no 1pp 230ndash236 2005

[9] E Celik and M Bayram ldquoThe numerical solution of physicalproblems modeled as a system of differential-algebraic equa-tions (DAEs)rdquo Journal of the Franklin Institute vol 342 no 1pp 1ndash6 2005

[10] M Kurulay and M Bayram ldquoApproximate analytical solutionfor the fractional modified KdV by differential transformmethodrdquo Communications in Nonlinear Science and NumericalSimulation vol 15 no 7 pp 1777ndash1782 2010

[11] M Bayram ldquoAutomatic analysis of the control of metabolicnetworksrdquo Computers in Biology and Medicine vol 26 no 5pp 401ndash408 1996

[12] N Guzel and M Bayram ldquoNumerical solution of differential-algebraic equations with index-2rdquo Applied Mathematics andComputation vol 174 no 2 pp 1279ndash1289 2006

[13] J K Zhou Differential Transformation and Its Application forElectrical CircuIts Huazhong University Press Wuhan China1986

[14] M Koksal and S Herdem ldquoAnalysis of nonlinear circuits byusing differential Taylor transformrdquo Computers and ElectricalEngineering vol 28 no 6 pp 513ndash525 2002

[15] I H Abdel-Halim Hassan ldquoOn solving some eigenvalue prob-lems by using a differential transformationrdquoAppliedMathemat-ics and Computation vol 127 no 1 pp 1ndash22 2002

[16] F Ayaz ldquoOn the two-dimensional differential transformmethodrdquo Applied Mathematics and Computation vol 143 no2-3 pp 361ndash374 2003

[17] C K Chen and S H Ho ldquoSolving partial differential equationsby two-dimensional differential transform methodrdquo AppliedMathematics and Computation vol 106 no 2-3 pp 171ndash1791999

[18] M-J Jang C-L Chen and Y-C Liu ldquoTwo-dimensional dif-ferential transform for partial differential equationsrdquo AppliedMathematics and Computation vol 121 no 2-3 pp 261ndash2702001

[19] X Yang Y Liu and S Bai ldquoA numerical solution of second-order linear partial differential equations by differential trans-formrdquo Applied Mathematics and Computation vol 173 no 2pp 792ndash802 2006

[20] A Arikoglu and I Ozkol ldquoSolution of differential-differenceequations by using differential transform methodrdquo AppliedMathematics and Computation vol 181 no 1 pp 153ndash162 2006

[21] M-J Jang C-L Chen and Y-C Liy ldquoOn solving the initial-value problems using the differential transformation methodrdquoAppliedMathematics andComputation vol 115 no 2-3 pp 145ndash160 2000

[22] F Ayaz ldquoApplications of differential transform method todifferential-algebraic equationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 649ndash657 2004

[23] H Liu and Y Song ldquoDifferential transform method applied tohigh index differential-algebraic equationsrdquoAppliedMathemat-ics and Computation vol 184 no 2 pp 748ndash753 2007

[24] G FrankMAPLE V CRC Press Boca Raton Fla USA 1996

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

3 Two-Dimensional DifferentialTransform Method

The two-dimensional differential transform of function119908(119909 119910) is defined as

119882(119896 ℎ) =1

119896ℎ[120597119896+ℎ119908 (119909 119910)

120597119909119896120597119910ℎ]119909=0

119910=0

(9)

where it is noted that upper case symbol 119882(119896 ℎ) is usedto denote the two-dimensional differential transform of afunction represented by a corresponding lower case symbol119908(119909 119910) The differential inverse transform of 119882(119896 ℎ) isdefined as

119908 (119909 119910) =

infin

sum

119896=0

infin

sum

ℎ=0

119882(119896 ℎ) 119909119896119910ℎ (10)

From (9) and (10) we obtain

119908 (119909 119910) =

infin

sum

119896=0

infin

sum

ℎ=0

119909119896119910ℎ

119896ℎ[120597119896+ℎ119908 (119909 119910)

120597119909119896120597119910ℎ]119909=0

119910=0

(11)

The concept of two-dimensional differential transform isderived from two-dimensional Taylor series expansion butthe method doesnrsquot evaluate the derivatives symbolically

Theorem 3 Differential transform of the function 119908(119909 119910) =119906(119909 119910) plusmn V(119909 119910) is

119882(119896 ℎ) = 119880 (119896 ℎ) plusmn 119881 (119896 ℎ) (12)

see [17]

Theorem 4 Differential transform of the function 119908(119909 119910) =120582119906(119909 119910) is

119882(119896 ℎ) = 120582119880 (119896 ℎ) (13)

see [17]

Theorem 5 Differential transform of the function 119908(119909 119910) =120597119906(119909 119910)120597119909 is

119882(119896 ℎ) = (119896 + 1)119880 (119896 + 1 ℎ) (14)

see [17]

Theorem 6 Differential transform of the function 119908(119909 119910) =120597119906(119909 119910)120597119910 is

119882(119896 ℎ) = (ℎ + 1)119880 (119896 ℎ + 1) (15)

see [17]

Theorem 7 Differential transform of the function 119908(119909 119910) =120597119903+119904119906(119909 119910)120597119909

119903120597119910119904 is

119882(119896 ℎ) = (119896 + 1) (119896 + 2) sdot sdot sdot (119896 + 119903) (ℎ + 1)

times (ℎ + 2) sdot sdot sdot (ℎ + 119904)119880 (119896 + 119903 ℎ + 119904)

(16)

see [17]

Theorem 8 Differential transform of the function 119908(119909 119910) =119906(119909 119910) sdot V(119909 119910) is

119882(119896 ℎ) =

119896

sum

119903=0

ℎ

sum

119904=0

119880 (119903 ℎ minus 119904) 119881 (119896 minus 119903 119904) (17)

see [17]

Theorem 9 Differential transform of the function 119908(119909 119910) =119909119898119910119899 is

119882(119896 ℎ) = 120575 (119896 minus 119898 ℎ minus 119899) = 120575 (119896 minus 119898) 120575 (ℎ minus 119899) (18)

see [17] where

120575 (119896 minus 119898) = 1 119896 = 119898

0 119896 =119898

120575 (ℎ minus 119899) = 1 ℎ = 119899

0 ℎ = 119899

(19)

Theorem 10 Differential transform of the function 119908(119909 119910) =119892(119909 + 119886 119910) is

119882(119896 ℎ) =

119873

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) (20)

Proof From Definition 1 we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119866 (119896 ℎ) (119909 + 119886)119896119910ℎ

= 119866 (0 0) + 119866 (0 1) 119910 + 119886119866 (1 0) + 119866 (1 0) 119909

+ 119866 (0 2) 1199102+ 119866 (1 1) 119909119910 + 119886119866 (1 1) 119910

+ 1198862119866 (2 0) + 2119886119866 (2 0) 119909 + 119866 (2 0) 119909

2

+ 119866 (0 3) 1199103+ 119866 (1 2) 119909119910

2+ 119886119866 (1 2) 119910

2

+ 119866 (2 1) 1199092119910 + 2119886119866 (2 1) 119909119910 + 119886

2119866 (2 1) 119910

+ 1198863119866 (3 0) + 3119886

2119866 (3 0) 119909 + 3119886119866 (3 0) 119909

2

+ 119866 (3 0) 1199093+ sdot sdot sdot

119908 (119909 119910) = [119866 (0 0) + 119886119866 (1 0)

+1198862119866 (2 0) + 119886

3119866 (3 0) + sdot sdot sdot]

+ 119909 [119866 (1 0) + 2119886119866 (2 0) + 31198862119866 (3 0) + sdot sdot sdot]

+ 119910 [119866 (0 1) + 119886119866 (1 1) + 1198862119866 (2 1) + sdot sdot sdot]

+ 1199092[119866 (2 0) + 3119886119866 (3 0) + sdot sdot sdot]

+ 119909119910 [119866 (1 1) + 2119886119866 (2 1) + sdot sdot sdot]

+ 1199102[119866 (0 2) + 119886119866 (1 2) + sdot sdot sdot] + sdot sdot sdot

4 Abstract and Applied Analysis

=

infin

sum

119901=0

119886119901119866 (119901 0) + 119909

infin

sum

119901=1

119901119886119901minus1

119866 (119901 0)

+ 119910

infin

sum

119901=0

119886119901119866 (119901 1) + 119909

2

infin

sum

119901=2

119901

(119901 minus 2)2

times 119886119901minus2

119866 (119901 0) + 119909119910

infin

sum

119901=1

119901119886119901minus1

119866 (119901 1) + sdot sdot sdot

(21)

where

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

infin

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) 119909119896119910ℎ (22)

hence

119882(119896 ℎ) =

119873

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) (23)

Theorem 11 Differential transform of the function 119908(119909 119910) =119892(119909 + 119886 119910 + 119887) is

119882(119896 ℎ) =

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896) 119886119901minus119896

119887119902minusℎ119866 (119901 119902) (24)

Proof From Definition 2 we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119866 (119896 ℎ) (119909 + 119886)119896(119910 + 119887)

ℎ

= 119866 (0 0) + 119866 (1 0) 119909 + 119886119866 (1 0)

+ 119866 (0 1) 119910 + 119887119866 (0 1) + 119866 (2 0) 1199092

+ 2119886119866 (2 0) 119909 + 1198862119866 (2 0) + 119866 (1 1) 119909119910

+ 119887119866 (1 1) 119909 + 119886119866 (1 1) 119910 + 119886119887119866 (1 1)

+ 119866 (0 2) 1199102+ 2119887119866 (0 2) 119910 + 119887

2119866 (0 2)

+ 119866 (3 0) 1199093+ 3119886119866 (3 0) 119909

2+ 31198862119866 (3 0) 119909

+ 1198863119866 (3 0) + 119866 (2 1) 119909

2119910 + sdot sdot sdot

119908 (119909 119910) = [119866 (0 0) + 119886119866 (1 0) + 119887119866 (0 1)

+1198862119866 (2 0) + 119886119887119866 (1 1) + sdot sdot sdot]

+ [119866 (1 0) + 2119886119866 (2 0) + 119887119866 (1 1)

+31198862119866 (3 0) + 2119886119887119866 (2 1) + sdot sdot sdot] 119909

+ [119866 (0 1) + 119886119866 (1 1) + 2119887119866 (0 2)

+1198862119866 (2 1) + sdot sdot sdot] 119910 + sdot sdot sdot

=

119873

sum

119901=0

119873

sum

119902=0

119886119901119887119902119866 (119901 119902) + 119909

119873

sum

119901=1

119873

sum

119902=0

119901119886119901minus1

119887119902119866 (119901 119902)

+ 119910

119873

sum

119901=0

119873

sum

119902=1

119902119886119901119887119902minus1119866 (119901 119902)

+ 1199092

119873

sum

119901=2

119873

sum

119902=0

119901

(119901 minus 2)2119886119901minus2

119887119902119866 (119901 119902)

+ 119909119910

119873

sum

119901=1

119873

sum

119902=1

119901119902119886119901minus1

119887119902minus1119866 (119901 119902) + sdot sdot sdot

(25)

Hence we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896)

times 119886119901minus119896

119887119902minusℎ119866 (119901 119902) 119909

119896119910ℎ

(26)

Using Definition 2 we obtain

119882(119896 ℎ) =

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896) 119886119901minus119896

119887119902minusℎ119866 (119901 119902) (27)

Theorem 12 Differential transform of the function 119908(119909 119910) =120597119903+119904119892(119909 + 119886 119910 + 119887)120597119909

119903120597119910119904 is

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ

times

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(28)

Proof Let 119862(119896 ℎ) be differential transform of the function119892(119909+119886 119910+119887) FromTheorem 7 we can write that differentialtransform of the function 119908(119909 119910) is

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ119862 (119896 + 119903 ℎ + 119904) (29)

fromTheorem 4 we can write

119862 (119896 + 119903 ℎ + 119904) =

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(30)

Abstract and Applied Analysis 5

If we substitute (30) into (29) we find

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ

times

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(31)

4 Application

We have considered the following PDAE as a test problem

(1 1

0 0) 119906119905+ (

minus1 0

minus1 0) 119906119909119909+ (

1 1

minus1 0) 119906 = 119891

119905 isin [0infin) 119909 isin [minus1 1]

(32)

with initial values and boundary values

1199061(0 119909) = 119909

3minus 119909 119906

2(0 119909) = 119909

4minus 1

1199061(119905 1) = 119906

1(119905 minus1) = 0 119906

2(119905 1) = 119906

2(119905 minus1) = 0

(33)

The right hand side function 119891 is

119891 = ((1199094minus 1) (cos 119905 minus sin 119905) minus 6119909119890minus119905 minus119890minus119905 (1199093 + 5119909))

119879

(34)

and the exact solutions are

1199061(119905 119909) = (119909

3minus 119909) 119890

minus119905 119906

2(119905 119909) = (119909

4minus 1) cos 119905 (35)

If nonsingular matrices 119875119865119896 119876119865119896 119875119871120585 and 119876

119871120585are chosen

such as

119875119865119896

= (

1minus120583119896

120583119896minus 1

01

120583119896minus 1

) 119876119865119896

= (0 minus1

1 minus1)

119875119871120585= (

01

120585 + 11

120585 + 1minus

1

120585 + 1

) 119876119871120585= (

minus120585 minus 1 0

120585 1)

(36)

matrices 119875119865119896119860119876119865119896

and 119875119871120585119861119876119871120585

are found as

119875119865119896119860119876119865119896

= (1 0

0 0) 119875

119871120585119861119876119871120585= (

1 0

0 0) (37)

From (38) we have 119873119871120585

= 0 and 119873119865119896

= 0 Then the PDAE(32) has differential spatial index 1 and differential time index1 So it is enough to takeM(120585)

119861119862= 1 andM

(119896)

119868119862= 2 to solve

the problemTaking differential transformation of (32) we obtain

(119896 + 1)1198801(119896 + 1 ℎ) + (119896 + 1)119880

2(119896 + 1 ℎ)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) + 119880

1(119896 ℎ)

+ 1198802(119896 ℎ) = 119865

1(119896 ℎ)

(38)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) minus 119880

1(119896 ℎ) = 119865

2(119896 ℎ) (39)

Table 1 The numerical and exact solution of the test problem(32)where 119906

1(119905 119909) is the exact solution and 119906

lowast

1(119905 119909) is the numerical

solution for 119909 = 01

119905 1199061(119905 119909) 119906

lowast

1(119905 119909) |119906

1(119905 119909) minus 119906

lowast

1(119905 119909)|

01 minus00895789043 minus00895789043 002 minus00810543445 minus00810543422 0000000002303 minus00733410038 minus00733409887 0000000015104 minus00663616845 minus00663616355 0000000049005 minus00600465353 minus00600464409 0000000094406 minus00543323519 minus00543322800 0000000071907 minus00491619450 minus00491621943 0000000249308 minus00444835674 minus00444849422 0000001374809 minus00402503963 minus00402546487 0000004252410 minus00364200646 minus00364305555 00000104909

Table 2 The numerical and exact solution of the test problem(32)where 119906

2(119905 119909) is exact solution and 119906lowast

2(119905 119909) is numerical solution

for 119909 = 01

119905 1199062(119905 119909) 119906

lowast

2(119905 119909) |119906

2(119905 119909) minus 119906

lowast

2(119905 119909)|

01 minus09949046649 minus09949046653 0000000000402 minus09799685711 minus09799685778 0000000006703 minus09552409555 minus09552409875 0000000032004 minus09209688879 minus09209689778 0000000089905 minus08774948036 minus08774949653 0000000163706 minus08252530813 minus08252532000 0000000118707 minus07647657031 minus07647652653 0000000437808 minus06966370386 minus06966345778 0000002460809 minus06215478073 minus06215398875 0000007919810 minus05402482757 minus05402277778 00000204979

The Taylor series of functions 1198911and 119891

2about 119909 = 0 119905 = 0

are

1198911(119905 119909) = minus 1 + 119905 minus 6119909 +

1

21199052+ 6119909119905

minus1

61199053minus 31199091199052minus1

241199054+ 1199094+ 1199091199053

minus 1199094119905 +

1

1201199055minus1

41199091199054+

1

7201199056

minus1

211990941199052+1

201199091199055+1

611990941199053

minus1

50401199057minus

1

1201199091199056+ sdot sdot sdot

(40)

1198912(119905 119909) = minus 5119909 + 5119909119905 minus

5

21199091199052minus 1199093

+5

61199091199053+ 1199093119905 minus

1

211990931199052minus5

241199091199054

+1

611990931199053+1

241199091199055minus

1

1441199091199056

minus1

2411990931199054+ sdot sdot sdot

(41)

6 Abstract and Applied Analysis

04

02

0

minus02

minus040

051

152

1

119905

minus05

minus1

11990905

0

Figure 1 The graphic of the function 1199061(119905 119909) in the test problem

(32)

04

02

0

minus02

minus04

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 2 The graphic of the function 119906lowast1(119905 119909) in the test problem

(32)

The values 1198651(119896 ℎ) and 119865

2(119896 ℎ) in (39) and (40) are coeffi-

cients of polynomials (41) and (42) If we use Theorem 3 forboundary values we obtain

7

sum

119894=0

1198801(119895 119894) = 0 119895 = 0 1 7 (42)

7

sum

119894=0

(minus1)1198941198801(119895 119894) = 0 119895 = 0 1 7 (43)

0406

020

minus02

minus04

minus06

minus08

minus1

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 3 The graphic of the function 1199062(119905 119909) in the test problem

(32)

0406

020

minus02

minus04

minus06

minus08

minus1

0

05

1

152

1

119905

minus05

minus1

119909

05

0

Figure 4 The graphic of the function 119906lowast2(119905 119909) in the test problem

(32)

In order to write 119896 = 0 and ℎ = 0 1 2 3 4 5 in (40) we have

21198801(0 2) + 119880

1(0 0) = 0 20119880

1(0 5) + 119880

1(0 3) = 1

61198801(0 3) + 119880

1(0 1) = 5 30119880

1(0 6) + 119880

1(0 4) = 0

121198801(0 4) + 119880

1(0 2) = 0 42119880

1(0 7) + 119880

1(0 5) = 0

(44)

If we take 119895 = 0 in (43) and (44) we obtain

1198801(0 0) + 119880

1(0 1) + 119880

1(0 2) + 119880

1(0 3)

+ 1198801(0 4) + 119880

1(0 5) + 119880

1(0 6) + 119880

1(0 7) = 0

1198801(0 0) minus 119880

1(0 1) + 119880

1(0 2) minus 119880

1(0 3)

+ 1198801(0 4) minus 119880

1(0 5) + 119880

1(0 6) minus 119880

1(0 7) = 0

(45)

Abstract and Applied Analysis 7

From (45) and (46) we find

1198801(0 0) = 0 119880

1(0 1) = minus1 119880

1(0 2) = 0

1198801(0 3) = 1 119880

1(0 4) = 0 119880

1(0 5) = 0

1198801(0 6) = 0 119880

1(0 7) = 0

(46)

In this manner from (40) (44) and (45) the coefficients ofthe 1199061are obtained as follows

1198801(1 0) = 0 119880

1(1 1) = 1 119880

1(1 2) = 0

1198801(1 3) = minus1 119880

1(1 4) = 0 119880

1(1 5) = 0

1198801(1 6) = 0 119880

1(2 0) = 0 119880

1(2 1) = minus

1

2

1198801(2 2) = 0 119880

1(2 3) =

1

2 119880

1(2 4) = 0

1198801(2 5) = 0 119880

1(3 0) = 0 119880

1(3 1) =

1

6

1198801(3 2) = 0 119880

1(3 3) = minus

1

6 119880

1(3 4) = 0

1198801(4 0) = 0 119880

1(4 1) = minus

1

24 119880

1(4 2) = 0

1198801(4 3) =

1

24 119880

1(5 0) = 0 119880

1(5 1) =

1

120

1198801(5 2) = 0 119880

1(6 0) = 0 119880

1(6 1) = minus

1

720

1198801(7 0) = 0

(47)

Using the initial values for the second component we obtainthe following coefficients

1198802(0 0) = minus1 119880

2(0 1) = 0 119880

2(0 2) = 0

1198802(0 3) = 0 119880

2(0 4) = 1 119880

2(0 5) = 0

1198802(0 6) = 0 119880

2(0 7) = 0

(48)

The coefficients of the 1199062can be found using (47) (48) (49)

and taking 119896 = 0 1 2 and ℎ = 0 1 2 in (39) as follows

1198802(1 2) = 0 119880

2(1 3) = 0 119880

2(1 4) = 0

1198802(1 5) = 0 119880

2(1 6) = 0 119880

2(2 1) = 0

1198802(2 2) = 0 119880

2(2 3) = 0 119880

2(2 4) = minus

1

2

1198802(2 5) = 0 119880

2(3 0) = 0 119880

2(3 1) = 0

1198802(3 2) = 0 119880

2(3 3) = 0 119880

2(3 4) = 0

1198802(4 1) = 0 119880

2(4 2) = 0 119880

2(4 3) = 0

1198802(4 0) = minus

1

24 119880

2(5 0) = 0 119880

2(5 1) = 0

1198802(5 2) = 0 119880

2(6 0) =

1

720 119880

2(6 1) = 0

1198802(7 0) = 0

(49)

If we write the above values in (39) and (40) then we have

119906lowast

1(119905 119909) = minus 119909 + 119909119905 minus

1

21199091199052+ 1199093+1

61199091199053

minus 1199093119905 +

1

211990931199052minus1

241199091199054minus1

611990931199053

+1

1201199091199055minus

1

7201199091199056+1

2411990931199054+ sdot sdot sdot

(50)

119906lowast

2(119905 119909) = minus 1 +

1

21199052minus1

241199054

+ 1199094+

1

7201199056minus1

211990941199052+ sdot sdot sdot

(51)

Numerical and exact solution of the given problem has beencompared in Tables 1 and 2 and simulations of solutions havebeen depicted in Figures 1 2 3 and 4 respectively

5 Conclusion

The computations associated with the example discussedabove were performed by using Computer Algebra Tech-niques [24] We show the results in Tables 1 and 2 for thesolution of (32) by numerical method The numerical valueson Tables 1 and 2 obtained above are in full agreement withthe exact solutions of (32) This study has shown that thedifferential transform method often shows superior perfor-mance over series approximants providing a promising toolfor using in applied fields

References

[1] W Marszalek Analysis of partial differential algebraic equations[PhD thesis] North Carolina State University Raleigh NCUSA 1997

[2] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear par-tial differential algebraic equations Part I indexes consistentboundaryinitial conditionsrdquo Report 17 Fachbereich Mathe-matik und Informatik Martin-Luther-Universitat Halle 1997

[3] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear partialdifferential algebraic equations Part II numerical solutionrdquoReport 18 Fachbereich Mathematik und Informatik Martin-Luther-Universitat Halle 1997

[4] W Lucht K Strehmel and C Eichler-Liebenow ldquoIndexes andspecial discretization methods for linear partial differentialalgebraic equationsrdquo BIT Numerical Mathematics vol 39 no3 pp 484ndash512 1999

[5] W S Martinson and P I Barton ldquoA differentiation indexfor partial differential-algebraic equationsrdquo SIAM Journal onScientific Computing vol 21 no 6 pp 2295ndash2315 2000

8 Abstract and Applied Analysis

[6] W S Martinson and P I Barton ldquoIndex and characteristicanalysis of linear PDAE systemsrdquo SIAM Journal on ScientificComputing vol 24 no 3 pp 905ndash923 2002

[7] K Debrabant and K Strehmel ldquoConvergence of Runge-Kuttamethods applied to linear partial differential-algebraic equa-tionsrdquoApplied Numerical Mathematics vol 53 no 2-4 pp 213ndash229 2005

[8] N Guzel and M Bayram ldquoOn the numerical solution of stiffsystemsrdquo Applied Mathematics and Computation vol 170 no 1pp 230ndash236 2005

[9] E Celik and M Bayram ldquoThe numerical solution of physicalproblems modeled as a system of differential-algebraic equa-tions (DAEs)rdquo Journal of the Franklin Institute vol 342 no 1pp 1ndash6 2005

[10] M Kurulay and M Bayram ldquoApproximate analytical solutionfor the fractional modified KdV by differential transformmethodrdquo Communications in Nonlinear Science and NumericalSimulation vol 15 no 7 pp 1777ndash1782 2010

[11] M Bayram ldquoAutomatic analysis of the control of metabolicnetworksrdquo Computers in Biology and Medicine vol 26 no 5pp 401ndash408 1996

[12] N Guzel and M Bayram ldquoNumerical solution of differential-algebraic equations with index-2rdquo Applied Mathematics andComputation vol 174 no 2 pp 1279ndash1289 2006

[13] J K Zhou Differential Transformation and Its Application forElectrical CircuIts Huazhong University Press Wuhan China1986

[14] M Koksal and S Herdem ldquoAnalysis of nonlinear circuits byusing differential Taylor transformrdquo Computers and ElectricalEngineering vol 28 no 6 pp 513ndash525 2002

[15] I H Abdel-Halim Hassan ldquoOn solving some eigenvalue prob-lems by using a differential transformationrdquoAppliedMathemat-ics and Computation vol 127 no 1 pp 1ndash22 2002

[16] F Ayaz ldquoOn the two-dimensional differential transformmethodrdquo Applied Mathematics and Computation vol 143 no2-3 pp 361ndash374 2003

[17] C K Chen and S H Ho ldquoSolving partial differential equationsby two-dimensional differential transform methodrdquo AppliedMathematics and Computation vol 106 no 2-3 pp 171ndash1791999

[18] M-J Jang C-L Chen and Y-C Liu ldquoTwo-dimensional dif-ferential transform for partial differential equationsrdquo AppliedMathematics and Computation vol 121 no 2-3 pp 261ndash2702001

[19] X Yang Y Liu and S Bai ldquoA numerical solution of second-order linear partial differential equations by differential trans-formrdquo Applied Mathematics and Computation vol 173 no 2pp 792ndash802 2006

[20] A Arikoglu and I Ozkol ldquoSolution of differential-differenceequations by using differential transform methodrdquo AppliedMathematics and Computation vol 181 no 1 pp 153ndash162 2006

[21] M-J Jang C-L Chen and Y-C Liy ldquoOn solving the initial-value problems using the differential transformation methodrdquoAppliedMathematics andComputation vol 115 no 2-3 pp 145ndash160 2000

[22] F Ayaz ldquoApplications of differential transform method todifferential-algebraic equationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 649ndash657 2004

[23] H Liu and Y Song ldquoDifferential transform method applied tohigh index differential-algebraic equationsrdquoAppliedMathemat-ics and Computation vol 184 no 2 pp 748ndash753 2007

[24] G FrankMAPLE V CRC Press Boca Raton Fla USA 1996

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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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

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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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

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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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

=

infin

sum

119901=0

119886119901119866 (119901 0) + 119909

infin

sum

119901=1

119901119886119901minus1

119866 (119901 0)

+ 119910

infin

sum

119901=0

119886119901119866 (119901 1) + 119909

2

infin

sum

119901=2

119901

(119901 minus 2)2

times 119886119901minus2

119866 (119901 0) + 119909119910

infin

sum

119901=1

119901119886119901minus1

119866 (119901 1) + sdot sdot sdot

(21)

where

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

infin

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) 119909119896119910ℎ (22)

hence

119882(119896 ℎ) =

119873

sum

119901=119896

(119901

119896) 119886119901minus119896

119866 (119901 ℎ) (23)

Theorem 11 Differential transform of the function 119908(119909 119910) =119892(119909 + 119886 119910 + 119887) is

119882(119896 ℎ) =

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896) 119886119901minus119896

119887119902minusℎ119866 (119901 119902) (24)

Proof From Definition 2 we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119866 (119896 ℎ) (119909 + 119886)119896(119910 + 119887)

ℎ

= 119866 (0 0) + 119866 (1 0) 119909 + 119886119866 (1 0)

+ 119866 (0 1) 119910 + 119887119866 (0 1) + 119866 (2 0) 1199092

+ 2119886119866 (2 0) 119909 + 1198862119866 (2 0) + 119866 (1 1) 119909119910

+ 119887119866 (1 1) 119909 + 119886119866 (1 1) 119910 + 119886119887119866 (1 1)

+ 119866 (0 2) 1199102+ 2119887119866 (0 2) 119910 + 119887

2119866 (0 2)

+ 119866 (3 0) 1199093+ 3119886119866 (3 0) 119909

2+ 31198862119866 (3 0) 119909

+ 1198863119866 (3 0) + 119866 (2 1) 119909

2119910 + sdot sdot sdot

119908 (119909 119910) = [119866 (0 0) + 119886119866 (1 0) + 119887119866 (0 1)

+1198862119866 (2 0) + 119886119887119866 (1 1) + sdot sdot sdot]

+ [119866 (1 0) + 2119886119866 (2 0) + 119887119866 (1 1)

+31198862119866 (3 0) + 2119886119887119866 (2 1) + sdot sdot sdot] 119909

+ [119866 (0 1) + 119886119866 (1 1) + 2119887119866 (0 2)

+1198862119866 (2 1) + sdot sdot sdot] 119910 + sdot sdot sdot

=

119873

sum

119901=0

119873

sum

119902=0

119886119901119887119902119866 (119901 119902) + 119909

119873

sum

119901=1

119873

sum

119902=0

119901119886119901minus1

119887119902119866 (119901 119902)

+ 119910

119873

sum

119901=0

119873

sum

119902=1

119902119886119901119887119902minus1119866 (119901 119902)

+ 1199092

119873

sum

119901=2

119873

sum

119902=0

119901

(119901 minus 2)2119886119901minus2

119887119902119866 (119901 119902)

+ 119909119910

119873

sum

119901=1

119873

sum

119902=1

119901119902119886119901minus1

119887119902minus1119866 (119901 119902) + sdot sdot sdot

(25)

Hence we can write

119908 (119909 119910) =

infin

sum

ℎ=0

infin

sum

119896=0

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896)

times 119886119901minus119896

119887119902minusℎ119866 (119901 119902) 119909

119896119910ℎ

(26)

Using Definition 2 we obtain

119882(119896 ℎ) =

119873

sum

119901=119896

119873

sum

119902=ℎ

(119902

ℎ)(

119901

119896) 119886119901minus119896

119887119902minusℎ119866 (119901 119902) (27)

Theorem 12 Differential transform of the function 119908(119909 119910) =120597119903+119904119892(119909 + 119886 119910 + 119887)120597119909

119903120597119910119904 is

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ

times

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(28)

Proof Let 119862(119896 ℎ) be differential transform of the function119892(119909+119886 119910+119887) FromTheorem 7 we can write that differentialtransform of the function 119908(119909 119910) is

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ119862 (119896 + 119903 ℎ + 119904) (29)

fromTheorem 4 we can write

119862 (119896 + 119903 ℎ + 119904) =

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(30)

Abstract and Applied Analysis 5

If we substitute (30) into (29) we find

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ

times

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(31)

4 Application

We have considered the following PDAE as a test problem

(1 1

0 0) 119906119905+ (

minus1 0

minus1 0) 119906119909119909+ (

1 1

minus1 0) 119906 = 119891

119905 isin [0infin) 119909 isin [minus1 1]

(32)

with initial values and boundary values

1199061(0 119909) = 119909

3minus 119909 119906

2(0 119909) = 119909

4minus 1

1199061(119905 1) = 119906

1(119905 minus1) = 0 119906

2(119905 1) = 119906

2(119905 minus1) = 0

(33)

The right hand side function 119891 is

119891 = ((1199094minus 1) (cos 119905 minus sin 119905) minus 6119909119890minus119905 minus119890minus119905 (1199093 + 5119909))

119879

(34)

and the exact solutions are

1199061(119905 119909) = (119909

3minus 119909) 119890

minus119905 119906

2(119905 119909) = (119909

4minus 1) cos 119905 (35)

If nonsingular matrices 119875119865119896 119876119865119896 119875119871120585 and 119876

119871120585are chosen

such as

119875119865119896

= (

1minus120583119896

120583119896minus 1

01

120583119896minus 1

) 119876119865119896

= (0 minus1

1 minus1)

119875119871120585= (

01

120585 + 11

120585 + 1minus

1

120585 + 1

) 119876119871120585= (

minus120585 minus 1 0

120585 1)

(36)

matrices 119875119865119896119860119876119865119896

and 119875119871120585119861119876119871120585

are found as

119875119865119896119860119876119865119896

= (1 0

0 0) 119875

119871120585119861119876119871120585= (

1 0

0 0) (37)

From (38) we have 119873119871120585

= 0 and 119873119865119896

= 0 Then the PDAE(32) has differential spatial index 1 and differential time index1 So it is enough to takeM(120585)

119861119862= 1 andM

(119896)

119868119862= 2 to solve

the problemTaking differential transformation of (32) we obtain

(119896 + 1)1198801(119896 + 1 ℎ) + (119896 + 1)119880

2(119896 + 1 ℎ)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) + 119880

1(119896 ℎ)

+ 1198802(119896 ℎ) = 119865

1(119896 ℎ)

(38)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) minus 119880

1(119896 ℎ) = 119865

2(119896 ℎ) (39)

Table 1 The numerical and exact solution of the test problem(32)where 119906

1(119905 119909) is the exact solution and 119906

lowast

1(119905 119909) is the numerical

solution for 119909 = 01

119905 1199061(119905 119909) 119906

lowast

1(119905 119909) |119906

1(119905 119909) minus 119906

lowast

1(119905 119909)|

01 minus00895789043 minus00895789043 002 minus00810543445 minus00810543422 0000000002303 minus00733410038 minus00733409887 0000000015104 minus00663616845 minus00663616355 0000000049005 minus00600465353 minus00600464409 0000000094406 minus00543323519 minus00543322800 0000000071907 minus00491619450 minus00491621943 0000000249308 minus00444835674 minus00444849422 0000001374809 minus00402503963 minus00402546487 0000004252410 minus00364200646 minus00364305555 00000104909

Table 2 The numerical and exact solution of the test problem(32)where 119906

2(119905 119909) is exact solution and 119906lowast

2(119905 119909) is numerical solution

for 119909 = 01

119905 1199062(119905 119909) 119906

lowast

2(119905 119909) |119906

2(119905 119909) minus 119906

lowast

2(119905 119909)|

01 minus09949046649 minus09949046653 0000000000402 minus09799685711 minus09799685778 0000000006703 minus09552409555 minus09552409875 0000000032004 minus09209688879 minus09209689778 0000000089905 minus08774948036 minus08774949653 0000000163706 minus08252530813 minus08252532000 0000000118707 minus07647657031 minus07647652653 0000000437808 minus06966370386 minus06966345778 0000002460809 minus06215478073 minus06215398875 0000007919810 minus05402482757 minus05402277778 00000204979

The Taylor series of functions 1198911and 119891

2about 119909 = 0 119905 = 0

are

1198911(119905 119909) = minus 1 + 119905 minus 6119909 +

1

21199052+ 6119909119905

minus1

61199053minus 31199091199052minus1

241199054+ 1199094+ 1199091199053

minus 1199094119905 +

1

1201199055minus1

41199091199054+

1

7201199056

minus1

211990941199052+1

201199091199055+1

611990941199053

minus1

50401199057minus

1

1201199091199056+ sdot sdot sdot

(40)

1198912(119905 119909) = minus 5119909 + 5119909119905 minus

5

21199091199052minus 1199093

+5

61199091199053+ 1199093119905 minus

1

211990931199052minus5

241199091199054

+1

611990931199053+1

241199091199055minus

1

1441199091199056

minus1

2411990931199054+ sdot sdot sdot

(41)

6 Abstract and Applied Analysis

04

02

0

minus02

minus040

051

152

1

119905

minus05

minus1

11990905

0

Figure 1 The graphic of the function 1199061(119905 119909) in the test problem

(32)

04

02

0

minus02

minus04

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 2 The graphic of the function 119906lowast1(119905 119909) in the test problem

(32)

The values 1198651(119896 ℎ) and 119865

2(119896 ℎ) in (39) and (40) are coeffi-

cients of polynomials (41) and (42) If we use Theorem 3 forboundary values we obtain

7

sum

119894=0

1198801(119895 119894) = 0 119895 = 0 1 7 (42)

7

sum

119894=0

(minus1)1198941198801(119895 119894) = 0 119895 = 0 1 7 (43)

0406

020

minus02

minus04

minus06

minus08

minus1

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 3 The graphic of the function 1199062(119905 119909) in the test problem

(32)

0406

020

minus02

minus04

minus06

minus08

minus1

0

05

1

152

1

119905

minus05

minus1

119909

05

0

Figure 4 The graphic of the function 119906lowast2(119905 119909) in the test problem

(32)

In order to write 119896 = 0 and ℎ = 0 1 2 3 4 5 in (40) we have

21198801(0 2) + 119880

1(0 0) = 0 20119880

1(0 5) + 119880

1(0 3) = 1

61198801(0 3) + 119880

1(0 1) = 5 30119880

1(0 6) + 119880

1(0 4) = 0

121198801(0 4) + 119880

1(0 2) = 0 42119880

1(0 7) + 119880

1(0 5) = 0

(44)

If we take 119895 = 0 in (43) and (44) we obtain

1198801(0 0) + 119880

1(0 1) + 119880

1(0 2) + 119880

1(0 3)

+ 1198801(0 4) + 119880

1(0 5) + 119880

1(0 6) + 119880

1(0 7) = 0

1198801(0 0) minus 119880

1(0 1) + 119880

1(0 2) minus 119880

1(0 3)

+ 1198801(0 4) minus 119880

1(0 5) + 119880

1(0 6) minus 119880

1(0 7) = 0

(45)

Abstract and Applied Analysis 7

From (45) and (46) we find

1198801(0 0) = 0 119880

1(0 1) = minus1 119880

1(0 2) = 0

1198801(0 3) = 1 119880

1(0 4) = 0 119880

1(0 5) = 0

1198801(0 6) = 0 119880

1(0 7) = 0

(46)

In this manner from (40) (44) and (45) the coefficients ofthe 1199061are obtained as follows

1198801(1 0) = 0 119880

1(1 1) = 1 119880

1(1 2) = 0

1198801(1 3) = minus1 119880

1(1 4) = 0 119880

1(1 5) = 0

1198801(1 6) = 0 119880

1(2 0) = 0 119880

1(2 1) = minus

1

2

1198801(2 2) = 0 119880

1(2 3) =

1

2 119880

1(2 4) = 0

1198801(2 5) = 0 119880

1(3 0) = 0 119880

1(3 1) =

1

6

1198801(3 2) = 0 119880

1(3 3) = minus

1

6 119880

1(3 4) = 0

1198801(4 0) = 0 119880

1(4 1) = minus

1

24 119880

1(4 2) = 0

1198801(4 3) =

1

24 119880

1(5 0) = 0 119880

1(5 1) =

1

120

1198801(5 2) = 0 119880

1(6 0) = 0 119880

1(6 1) = minus

1

720

1198801(7 0) = 0

(47)

Using the initial values for the second component we obtainthe following coefficients

1198802(0 0) = minus1 119880

2(0 1) = 0 119880

2(0 2) = 0

1198802(0 3) = 0 119880

2(0 4) = 1 119880

2(0 5) = 0

1198802(0 6) = 0 119880

2(0 7) = 0

(48)

The coefficients of the 1199062can be found using (47) (48) (49)

and taking 119896 = 0 1 2 and ℎ = 0 1 2 in (39) as follows

1198802(1 2) = 0 119880

2(1 3) = 0 119880

2(1 4) = 0

1198802(1 5) = 0 119880

2(1 6) = 0 119880

2(2 1) = 0

1198802(2 2) = 0 119880

2(2 3) = 0 119880

2(2 4) = minus

1

2

1198802(2 5) = 0 119880

2(3 0) = 0 119880

2(3 1) = 0

1198802(3 2) = 0 119880

2(3 3) = 0 119880

2(3 4) = 0

1198802(4 1) = 0 119880

2(4 2) = 0 119880

2(4 3) = 0

1198802(4 0) = minus

1

24 119880

2(5 0) = 0 119880

2(5 1) = 0

1198802(5 2) = 0 119880

2(6 0) =

1

720 119880

2(6 1) = 0

1198802(7 0) = 0

(49)

If we write the above values in (39) and (40) then we have

119906lowast

1(119905 119909) = minus 119909 + 119909119905 minus

1

21199091199052+ 1199093+1

61199091199053

minus 1199093119905 +

1

211990931199052minus1

241199091199054minus1

611990931199053

+1

1201199091199055minus

1

7201199091199056+1

2411990931199054+ sdot sdot sdot

(50)

119906lowast

2(119905 119909) = minus 1 +

1

21199052minus1

241199054

+ 1199094+

1

7201199056minus1

211990941199052+ sdot sdot sdot

(51)

Numerical and exact solution of the given problem has beencompared in Tables 1 and 2 and simulations of solutions havebeen depicted in Figures 1 2 3 and 4 respectively

5 Conclusion

The computations associated with the example discussedabove were performed by using Computer Algebra Tech-niques [24] We show the results in Tables 1 and 2 for thesolution of (32) by numerical method The numerical valueson Tables 1 and 2 obtained above are in full agreement withthe exact solutions of (32) This study has shown that thedifferential transform method often shows superior perfor-mance over series approximants providing a promising toolfor using in applied fields

References

[1] W Marszalek Analysis of partial differential algebraic equations[PhD thesis] North Carolina State University Raleigh NCUSA 1997

[2] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear par-tial differential algebraic equations Part I indexes consistentboundaryinitial conditionsrdquo Report 17 Fachbereich Mathe-matik und Informatik Martin-Luther-Universitat Halle 1997

[3] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear partialdifferential algebraic equations Part II numerical solutionrdquoReport 18 Fachbereich Mathematik und Informatik Martin-Luther-Universitat Halle 1997

[4] W Lucht K Strehmel and C Eichler-Liebenow ldquoIndexes andspecial discretization methods for linear partial differentialalgebraic equationsrdquo BIT Numerical Mathematics vol 39 no3 pp 484ndash512 1999

[5] W S Martinson and P I Barton ldquoA differentiation indexfor partial differential-algebraic equationsrdquo SIAM Journal onScientific Computing vol 21 no 6 pp 2295ndash2315 2000

8 Abstract and Applied Analysis

[6] W S Martinson and P I Barton ldquoIndex and characteristicanalysis of linear PDAE systemsrdquo SIAM Journal on ScientificComputing vol 24 no 3 pp 905ndash923 2002

[7] K Debrabant and K Strehmel ldquoConvergence of Runge-Kuttamethods applied to linear partial differential-algebraic equa-tionsrdquoApplied Numerical Mathematics vol 53 no 2-4 pp 213ndash229 2005

[8] N Guzel and M Bayram ldquoOn the numerical solution of stiffsystemsrdquo Applied Mathematics and Computation vol 170 no 1pp 230ndash236 2005

[9] E Celik and M Bayram ldquoThe numerical solution of physicalproblems modeled as a system of differential-algebraic equa-tions (DAEs)rdquo Journal of the Franklin Institute vol 342 no 1pp 1ndash6 2005

[10] M Kurulay and M Bayram ldquoApproximate analytical solutionfor the fractional modified KdV by differential transformmethodrdquo Communications in Nonlinear Science and NumericalSimulation vol 15 no 7 pp 1777ndash1782 2010

[11] M Bayram ldquoAutomatic analysis of the control of metabolicnetworksrdquo Computers in Biology and Medicine vol 26 no 5pp 401ndash408 1996

[12] N Guzel and M Bayram ldquoNumerical solution of differential-algebraic equations with index-2rdquo Applied Mathematics andComputation vol 174 no 2 pp 1279ndash1289 2006

[13] J K Zhou Differential Transformation and Its Application forElectrical CircuIts Huazhong University Press Wuhan China1986

[14] M Koksal and S Herdem ldquoAnalysis of nonlinear circuits byusing differential Taylor transformrdquo Computers and ElectricalEngineering vol 28 no 6 pp 513ndash525 2002

[15] I H Abdel-Halim Hassan ldquoOn solving some eigenvalue prob-lems by using a differential transformationrdquoAppliedMathemat-ics and Computation vol 127 no 1 pp 1ndash22 2002

[16] F Ayaz ldquoOn the two-dimensional differential transformmethodrdquo Applied Mathematics and Computation vol 143 no2-3 pp 361ndash374 2003

[17] C K Chen and S H Ho ldquoSolving partial differential equationsby two-dimensional differential transform methodrdquo AppliedMathematics and Computation vol 106 no 2-3 pp 171ndash1791999

[18] M-J Jang C-L Chen and Y-C Liu ldquoTwo-dimensional dif-ferential transform for partial differential equationsrdquo AppliedMathematics and Computation vol 121 no 2-3 pp 261ndash2702001

[19] X Yang Y Liu and S Bai ldquoA numerical solution of second-order linear partial differential equations by differential trans-formrdquo Applied Mathematics and Computation vol 173 no 2pp 792ndash802 2006

[20] A Arikoglu and I Ozkol ldquoSolution of differential-differenceequations by using differential transform methodrdquo AppliedMathematics and Computation vol 181 no 1 pp 153ndash162 2006

[21] M-J Jang C-L Chen and Y-C Liy ldquoOn solving the initial-value problems using the differential transformation methodrdquoAppliedMathematics andComputation vol 115 no 2-3 pp 145ndash160 2000

[22] F Ayaz ldquoApplications of differential transform method todifferential-algebraic equationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 649ndash657 2004

[23] H Liu and Y Song ldquoDifferential transform method applied tohigh index differential-algebraic equationsrdquoAppliedMathemat-ics and Computation vol 184 no 2 pp 748ndash753 2007

[24] G FrankMAPLE V CRC Press Boca Raton Fla USA 1996

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

If we substitute (30) into (29) we find

119882(119896 ℎ) =(119896 + 119903)

119896

(ℎ + 119904)

ℎ

times

119873

sum

119901=119896+119903

119873

sum

119902=ℎ+119904

(119902

ℎ + 119904)(

119901

119896 + 119903)

times 119886119901minus119896minus119903

119887119902minusℎminus119904

119866 (119901 119902)

(31)

4 Application

We have considered the following PDAE as a test problem

(1 1

0 0) 119906119905+ (

minus1 0

minus1 0) 119906119909119909+ (

1 1

minus1 0) 119906 = 119891

119905 isin [0infin) 119909 isin [minus1 1]

(32)

with initial values and boundary values

1199061(0 119909) = 119909

3minus 119909 119906

2(0 119909) = 119909

4minus 1

1199061(119905 1) = 119906

1(119905 minus1) = 0 119906

2(119905 1) = 119906

2(119905 minus1) = 0

(33)

The right hand side function 119891 is

119891 = ((1199094minus 1) (cos 119905 minus sin 119905) minus 6119909119890minus119905 minus119890minus119905 (1199093 + 5119909))

119879

(34)

and the exact solutions are

1199061(119905 119909) = (119909

3minus 119909) 119890

minus119905 119906

2(119905 119909) = (119909

4minus 1) cos 119905 (35)

If nonsingular matrices 119875119865119896 119876119865119896 119875119871120585 and 119876

119871120585are chosen

such as

119875119865119896

= (

1minus120583119896

120583119896minus 1

01

120583119896minus 1

) 119876119865119896

= (0 minus1

1 minus1)

119875119871120585= (

01

120585 + 11

120585 + 1minus

1

120585 + 1

) 119876119871120585= (

minus120585 minus 1 0

120585 1)

(36)

matrices 119875119865119896119860119876119865119896

and 119875119871120585119861119876119871120585

are found as

119875119865119896119860119876119865119896

= (1 0

0 0) 119875

119871120585119861119876119871120585= (

1 0

0 0) (37)

From (38) we have 119873119871120585

= 0 and 119873119865119896

= 0 Then the PDAE(32) has differential spatial index 1 and differential time index1 So it is enough to takeM(120585)

119861119862= 1 andM

(119896)

119868119862= 2 to solve

the problemTaking differential transformation of (32) we obtain

(119896 + 1)1198801(119896 + 1 ℎ) + (119896 + 1)119880

2(119896 + 1 ℎ)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) + 119880

1(119896 ℎ)

+ 1198802(119896 ℎ) = 119865

1(119896 ℎ)

(38)

minus (ℎ + 1) (ℎ + 2)1198801(119896 ℎ + 2) minus 119880

1(119896 ℎ) = 119865

2(119896 ℎ) (39)

Table 1 The numerical and exact solution of the test problem(32)where 119906

1(119905 119909) is the exact solution and 119906

lowast

1(119905 119909) is the numerical

solution for 119909 = 01

119905 1199061(119905 119909) 119906

lowast

1(119905 119909) |119906

1(119905 119909) minus 119906

lowast

1(119905 119909)|

01 minus00895789043 minus00895789043 002 minus00810543445 minus00810543422 0000000002303 minus00733410038 minus00733409887 0000000015104 minus00663616845 minus00663616355 0000000049005 minus00600465353 minus00600464409 0000000094406 minus00543323519 minus00543322800 0000000071907 minus00491619450 minus00491621943 0000000249308 minus00444835674 minus00444849422 0000001374809 minus00402503963 minus00402546487 0000004252410 minus00364200646 minus00364305555 00000104909

Table 2 The numerical and exact solution of the test problem(32)where 119906

2(119905 119909) is exact solution and 119906lowast

2(119905 119909) is numerical solution

for 119909 = 01

119905 1199062(119905 119909) 119906

lowast

2(119905 119909) |119906

2(119905 119909) minus 119906

lowast

2(119905 119909)|

01 minus09949046649 minus09949046653 0000000000402 minus09799685711 minus09799685778 0000000006703 minus09552409555 minus09552409875 0000000032004 minus09209688879 minus09209689778 0000000089905 minus08774948036 minus08774949653 0000000163706 minus08252530813 minus08252532000 0000000118707 minus07647657031 minus07647652653 0000000437808 minus06966370386 minus06966345778 0000002460809 minus06215478073 minus06215398875 0000007919810 minus05402482757 minus05402277778 00000204979

The Taylor series of functions 1198911and 119891

2about 119909 = 0 119905 = 0

are

1198911(119905 119909) = minus 1 + 119905 minus 6119909 +

1

21199052+ 6119909119905

minus1

61199053minus 31199091199052minus1

241199054+ 1199094+ 1199091199053

minus 1199094119905 +

1

1201199055minus1

41199091199054+

1

7201199056

minus1

211990941199052+1

201199091199055+1

611990941199053

minus1

50401199057minus

1

1201199091199056+ sdot sdot sdot

(40)

1198912(119905 119909) = minus 5119909 + 5119909119905 minus

5

21199091199052minus 1199093

+5

61199091199053+ 1199093119905 minus

1

211990931199052minus5

241199091199054

+1

611990931199053+1

241199091199055minus

1

1441199091199056

minus1

2411990931199054+ sdot sdot sdot

(41)

6 Abstract and Applied Analysis

04

02

0

minus02

minus040

051

152

1

119905

minus05

minus1

11990905

0

Figure 1 The graphic of the function 1199061(119905 119909) in the test problem

(32)

04

02

0

minus02

minus04

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 2 The graphic of the function 119906lowast1(119905 119909) in the test problem

(32)

The values 1198651(119896 ℎ) and 119865

2(119896 ℎ) in (39) and (40) are coeffi-

cients of polynomials (41) and (42) If we use Theorem 3 forboundary values we obtain

7

sum

119894=0

1198801(119895 119894) = 0 119895 = 0 1 7 (42)

7

sum

119894=0

(minus1)1198941198801(119895 119894) = 0 119895 = 0 1 7 (43)

0406

020

minus02

minus04

minus06

minus08

minus1

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 3 The graphic of the function 1199062(119905 119909) in the test problem

(32)

0406

020

minus02

minus04

minus06

minus08

minus1

0

05

1

152

1

119905

minus05

minus1

119909

05

0

Figure 4 The graphic of the function 119906lowast2(119905 119909) in the test problem

(32)

In order to write 119896 = 0 and ℎ = 0 1 2 3 4 5 in (40) we have

21198801(0 2) + 119880

1(0 0) = 0 20119880

1(0 5) + 119880

1(0 3) = 1

61198801(0 3) + 119880

1(0 1) = 5 30119880

1(0 6) + 119880

1(0 4) = 0

121198801(0 4) + 119880

1(0 2) = 0 42119880

1(0 7) + 119880

1(0 5) = 0

(44)

If we take 119895 = 0 in (43) and (44) we obtain

1198801(0 0) + 119880

1(0 1) + 119880

1(0 2) + 119880

1(0 3)

+ 1198801(0 4) + 119880

1(0 5) + 119880

1(0 6) + 119880

1(0 7) = 0

1198801(0 0) minus 119880

1(0 1) + 119880

1(0 2) minus 119880

1(0 3)

+ 1198801(0 4) minus 119880

1(0 5) + 119880

1(0 6) minus 119880

1(0 7) = 0

(45)

Abstract and Applied Analysis 7

From (45) and (46) we find

1198801(0 0) = 0 119880

1(0 1) = minus1 119880

1(0 2) = 0

1198801(0 3) = 1 119880

1(0 4) = 0 119880

1(0 5) = 0

1198801(0 6) = 0 119880

1(0 7) = 0

(46)

In this manner from (40) (44) and (45) the coefficients ofthe 1199061are obtained as follows

1198801(1 0) = 0 119880

1(1 1) = 1 119880

1(1 2) = 0

1198801(1 3) = minus1 119880

1(1 4) = 0 119880

1(1 5) = 0

1198801(1 6) = 0 119880

1(2 0) = 0 119880

1(2 1) = minus

1

2

1198801(2 2) = 0 119880

1(2 3) =

1

2 119880

1(2 4) = 0

1198801(2 5) = 0 119880

1(3 0) = 0 119880

1(3 1) =

1

6

1198801(3 2) = 0 119880

1(3 3) = minus

1

6 119880

1(3 4) = 0

1198801(4 0) = 0 119880

1(4 1) = minus

1

24 119880

1(4 2) = 0

1198801(4 3) =

1

24 119880

1(5 0) = 0 119880

1(5 1) =

1

120

1198801(5 2) = 0 119880

1(6 0) = 0 119880

1(6 1) = minus

1

720

1198801(7 0) = 0

(47)

Using the initial values for the second component we obtainthe following coefficients

1198802(0 0) = minus1 119880

2(0 1) = 0 119880

2(0 2) = 0

1198802(0 3) = 0 119880

2(0 4) = 1 119880

2(0 5) = 0

1198802(0 6) = 0 119880

2(0 7) = 0

(48)

The coefficients of the 1199062can be found using (47) (48) (49)

and taking 119896 = 0 1 2 and ℎ = 0 1 2 in (39) as follows

1198802(1 2) = 0 119880

2(1 3) = 0 119880

2(1 4) = 0

1198802(1 5) = 0 119880

2(1 6) = 0 119880

2(2 1) = 0

1198802(2 2) = 0 119880

2(2 3) = 0 119880

2(2 4) = minus

1

2

1198802(2 5) = 0 119880

2(3 0) = 0 119880

2(3 1) = 0

1198802(3 2) = 0 119880

2(3 3) = 0 119880

2(3 4) = 0

1198802(4 1) = 0 119880

2(4 2) = 0 119880

2(4 3) = 0

1198802(4 0) = minus

1

24 119880

2(5 0) = 0 119880

2(5 1) = 0

1198802(5 2) = 0 119880

2(6 0) =

1

720 119880

2(6 1) = 0

1198802(7 0) = 0

(49)

If we write the above values in (39) and (40) then we have

119906lowast

1(119905 119909) = minus 119909 + 119909119905 minus

1

21199091199052+ 1199093+1

61199091199053

minus 1199093119905 +

1

211990931199052minus1

241199091199054minus1

611990931199053

+1

1201199091199055minus

1

7201199091199056+1

2411990931199054+ sdot sdot sdot

(50)

119906lowast

2(119905 119909) = minus 1 +

1

21199052minus1

241199054

+ 1199094+

1

7201199056minus1

211990941199052+ sdot sdot sdot

(51)

Numerical and exact solution of the given problem has beencompared in Tables 1 and 2 and simulations of solutions havebeen depicted in Figures 1 2 3 and 4 respectively

5 Conclusion

The computations associated with the example discussedabove were performed by using Computer Algebra Tech-niques [24] We show the results in Tables 1 and 2 for thesolution of (32) by numerical method The numerical valueson Tables 1 and 2 obtained above are in full agreement withthe exact solutions of (32) This study has shown that thedifferential transform method often shows superior perfor-mance over series approximants providing a promising toolfor using in applied fields

References

[1] W Marszalek Analysis of partial differential algebraic equations[PhD thesis] North Carolina State University Raleigh NCUSA 1997

[2] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear par-tial differential algebraic equations Part I indexes consistentboundaryinitial conditionsrdquo Report 17 Fachbereich Mathe-matik und Informatik Martin-Luther-Universitat Halle 1997

[3] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear partialdifferential algebraic equations Part II numerical solutionrdquoReport 18 Fachbereich Mathematik und Informatik Martin-Luther-Universitat Halle 1997

[4] W Lucht K Strehmel and C Eichler-Liebenow ldquoIndexes andspecial discretization methods for linear partial differentialalgebraic equationsrdquo BIT Numerical Mathematics vol 39 no3 pp 484ndash512 1999

[5] W S Martinson and P I Barton ldquoA differentiation indexfor partial differential-algebraic equationsrdquo SIAM Journal onScientific Computing vol 21 no 6 pp 2295ndash2315 2000

8 Abstract and Applied Analysis

[6] W S Martinson and P I Barton ldquoIndex and characteristicanalysis of linear PDAE systemsrdquo SIAM Journal on ScientificComputing vol 24 no 3 pp 905ndash923 2002

[7] K Debrabant and K Strehmel ldquoConvergence of Runge-Kuttamethods applied to linear partial differential-algebraic equa-tionsrdquoApplied Numerical Mathematics vol 53 no 2-4 pp 213ndash229 2005

[8] N Guzel and M Bayram ldquoOn the numerical solution of stiffsystemsrdquo Applied Mathematics and Computation vol 170 no 1pp 230ndash236 2005

[9] E Celik and M Bayram ldquoThe numerical solution of physicalproblems modeled as a system of differential-algebraic equa-tions (DAEs)rdquo Journal of the Franklin Institute vol 342 no 1pp 1ndash6 2005

[10] M Kurulay and M Bayram ldquoApproximate analytical solutionfor the fractional modified KdV by differential transformmethodrdquo Communications in Nonlinear Science and NumericalSimulation vol 15 no 7 pp 1777ndash1782 2010

[11] M Bayram ldquoAutomatic analysis of the control of metabolicnetworksrdquo Computers in Biology and Medicine vol 26 no 5pp 401ndash408 1996

[12] N Guzel and M Bayram ldquoNumerical solution of differential-algebraic equations with index-2rdquo Applied Mathematics andComputation vol 174 no 2 pp 1279ndash1289 2006

[13] J K Zhou Differential Transformation and Its Application forElectrical CircuIts Huazhong University Press Wuhan China1986

[14] M Koksal and S Herdem ldquoAnalysis of nonlinear circuits byusing differential Taylor transformrdquo Computers and ElectricalEngineering vol 28 no 6 pp 513ndash525 2002

[15] I H Abdel-Halim Hassan ldquoOn solving some eigenvalue prob-lems by using a differential transformationrdquoAppliedMathemat-ics and Computation vol 127 no 1 pp 1ndash22 2002

[16] F Ayaz ldquoOn the two-dimensional differential transformmethodrdquo Applied Mathematics and Computation vol 143 no2-3 pp 361ndash374 2003

[17] C K Chen and S H Ho ldquoSolving partial differential equationsby two-dimensional differential transform methodrdquo AppliedMathematics and Computation vol 106 no 2-3 pp 171ndash1791999

[18] M-J Jang C-L Chen and Y-C Liu ldquoTwo-dimensional dif-ferential transform for partial differential equationsrdquo AppliedMathematics and Computation vol 121 no 2-3 pp 261ndash2702001

[19] X Yang Y Liu and S Bai ldquoA numerical solution of second-order linear partial differential equations by differential trans-formrdquo Applied Mathematics and Computation vol 173 no 2pp 792ndash802 2006

[20] A Arikoglu and I Ozkol ldquoSolution of differential-differenceequations by using differential transform methodrdquo AppliedMathematics and Computation vol 181 no 1 pp 153ndash162 2006

[21] M-J Jang C-L Chen and Y-C Liy ldquoOn solving the initial-value problems using the differential transformation methodrdquoAppliedMathematics andComputation vol 115 no 2-3 pp 145ndash160 2000

[22] F Ayaz ldquoApplications of differential transform method todifferential-algebraic equationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 649ndash657 2004

[23] H Liu and Y Song ldquoDifferential transform method applied tohigh index differential-algebraic equationsrdquoAppliedMathemat-ics and Computation vol 184 no 2 pp 748ndash753 2007

[24] G FrankMAPLE V CRC Press Boca Raton Fla USA 1996

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

04

02

0

minus02

minus040

051

152

1

119905

minus05

minus1

11990905

0

Figure 1 The graphic of the function 1199061(119905 119909) in the test problem

(32)

04

02

0

minus02

minus04

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 2 The graphic of the function 119906lowast1(119905 119909) in the test problem

(32)

The values 1198651(119896 ℎ) and 119865

2(119896 ℎ) in (39) and (40) are coeffi-

cients of polynomials (41) and (42) If we use Theorem 3 forboundary values we obtain

7

sum

119894=0

1198801(119895 119894) = 0 119895 = 0 1 7 (42)

7

sum

119894=0

(minus1)1198941198801(119895 119894) = 0 119895 = 0 1 7 (43)

0406

020

minus02

minus04

minus06

minus08

minus1

0

051

152

1

119905

minus05

minus1

119909

05

0

Figure 3 The graphic of the function 1199062(119905 119909) in the test problem

(32)

0406

020

minus02

minus04

minus06

minus08

minus1

0

05

1

152

1

119905

minus05

minus1

119909

05

0

Figure 4 The graphic of the function 119906lowast2(119905 119909) in the test problem

(32)

In order to write 119896 = 0 and ℎ = 0 1 2 3 4 5 in (40) we have

21198801(0 2) + 119880

1(0 0) = 0 20119880

1(0 5) + 119880

1(0 3) = 1

61198801(0 3) + 119880

1(0 1) = 5 30119880

1(0 6) + 119880

1(0 4) = 0

121198801(0 4) + 119880

1(0 2) = 0 42119880

1(0 7) + 119880

1(0 5) = 0

(44)

If we take 119895 = 0 in (43) and (44) we obtain

1198801(0 0) + 119880

1(0 1) + 119880

1(0 2) + 119880

1(0 3)

+ 1198801(0 4) + 119880

1(0 5) + 119880

1(0 6) + 119880

1(0 7) = 0

1198801(0 0) minus 119880

1(0 1) + 119880

1(0 2) minus 119880

1(0 3)

+ 1198801(0 4) minus 119880

1(0 5) + 119880

1(0 6) minus 119880

1(0 7) = 0

(45)

Abstract and Applied Analysis 7

From (45) and (46) we find

1198801(0 0) = 0 119880

1(0 1) = minus1 119880

1(0 2) = 0

1198801(0 3) = 1 119880

1(0 4) = 0 119880

1(0 5) = 0

1198801(0 6) = 0 119880

1(0 7) = 0

(46)

In this manner from (40) (44) and (45) the coefficients ofthe 1199061are obtained as follows

1198801(1 0) = 0 119880

1(1 1) = 1 119880

1(1 2) = 0

1198801(1 3) = minus1 119880

1(1 4) = 0 119880

1(1 5) = 0

1198801(1 6) = 0 119880

1(2 0) = 0 119880

1(2 1) = minus

1

2

1198801(2 2) = 0 119880

1(2 3) =

1

2 119880

1(2 4) = 0

1198801(2 5) = 0 119880

1(3 0) = 0 119880

1(3 1) =

1

6

1198801(3 2) = 0 119880

1(3 3) = minus

1

6 119880

1(3 4) = 0

1198801(4 0) = 0 119880

1(4 1) = minus

1

24 119880

1(4 2) = 0

1198801(4 3) =

1

24 119880

1(5 0) = 0 119880

1(5 1) =

1

120

1198801(5 2) = 0 119880

1(6 0) = 0 119880

1(6 1) = minus

1

720

1198801(7 0) = 0

(47)

Using the initial values for the second component we obtainthe following coefficients

1198802(0 0) = minus1 119880

2(0 1) = 0 119880

2(0 2) = 0

1198802(0 3) = 0 119880

2(0 4) = 1 119880

2(0 5) = 0

1198802(0 6) = 0 119880

2(0 7) = 0

(48)

The coefficients of the 1199062can be found using (47) (48) (49)

and taking 119896 = 0 1 2 and ℎ = 0 1 2 in (39) as follows

1198802(1 2) = 0 119880

2(1 3) = 0 119880

2(1 4) = 0

1198802(1 5) = 0 119880

2(1 6) = 0 119880

2(2 1) = 0

1198802(2 2) = 0 119880

2(2 3) = 0 119880

2(2 4) = minus

1

2

1198802(2 5) = 0 119880

2(3 0) = 0 119880

2(3 1) = 0

1198802(3 2) = 0 119880

2(3 3) = 0 119880

2(3 4) = 0

1198802(4 1) = 0 119880

2(4 2) = 0 119880

2(4 3) = 0

1198802(4 0) = minus

1

24 119880

2(5 0) = 0 119880

2(5 1) = 0

1198802(5 2) = 0 119880

2(6 0) =

1

720 119880

2(6 1) = 0

1198802(7 0) = 0

(49)

If we write the above values in (39) and (40) then we have

119906lowast

1(119905 119909) = minus 119909 + 119909119905 minus

1

21199091199052+ 1199093+1

61199091199053

minus 1199093119905 +

1

211990931199052minus1

241199091199054minus1

611990931199053

+1

1201199091199055minus

1

7201199091199056+1

2411990931199054+ sdot sdot sdot

(50)

119906lowast

2(119905 119909) = minus 1 +

1

21199052minus1

241199054

+ 1199094+

1

7201199056minus1

211990941199052+ sdot sdot sdot

(51)

Numerical and exact solution of the given problem has beencompared in Tables 1 and 2 and simulations of solutions havebeen depicted in Figures 1 2 3 and 4 respectively

5 Conclusion

The computations associated with the example discussedabove were performed by using Computer Algebra Tech-niques [24] We show the results in Tables 1 and 2 for thesolution of (32) by numerical method The numerical valueson Tables 1 and 2 obtained above are in full agreement withthe exact solutions of (32) This study has shown that thedifferential transform method often shows superior perfor-mance over series approximants providing a promising toolfor using in applied fields

References

[1] W Marszalek Analysis of partial differential algebraic equations[PhD thesis] North Carolina State University Raleigh NCUSA 1997

[2] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear par-tial differential algebraic equations Part I indexes consistentboundaryinitial conditionsrdquo Report 17 Fachbereich Mathe-matik und Informatik Martin-Luther-Universitat Halle 1997

[3] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear partialdifferential algebraic equations Part II numerical solutionrdquoReport 18 Fachbereich Mathematik und Informatik Martin-Luther-Universitat Halle 1997

[4] W Lucht K Strehmel and C Eichler-Liebenow ldquoIndexes andspecial discretization methods for linear partial differentialalgebraic equationsrdquo BIT Numerical Mathematics vol 39 no3 pp 484ndash512 1999

[5] W S Martinson and P I Barton ldquoA differentiation indexfor partial differential-algebraic equationsrdquo SIAM Journal onScientific Computing vol 21 no 6 pp 2295ndash2315 2000

8 Abstract and Applied Analysis

[6] W S Martinson and P I Barton ldquoIndex and characteristicanalysis of linear PDAE systemsrdquo SIAM Journal on ScientificComputing vol 24 no 3 pp 905ndash923 2002

[7] K Debrabant and K Strehmel ldquoConvergence of Runge-Kuttamethods applied to linear partial differential-algebraic equa-tionsrdquoApplied Numerical Mathematics vol 53 no 2-4 pp 213ndash229 2005

[8] N Guzel and M Bayram ldquoOn the numerical solution of stiffsystemsrdquo Applied Mathematics and Computation vol 170 no 1pp 230ndash236 2005

[9] E Celik and M Bayram ldquoThe numerical solution of physicalproblems modeled as a system of differential-algebraic equa-tions (DAEs)rdquo Journal of the Franklin Institute vol 342 no 1pp 1ndash6 2005

[10] M Kurulay and M Bayram ldquoApproximate analytical solutionfor the fractional modified KdV by differential transformmethodrdquo Communications in Nonlinear Science and NumericalSimulation vol 15 no 7 pp 1777ndash1782 2010

[11] M Bayram ldquoAutomatic analysis of the control of metabolicnetworksrdquo Computers in Biology and Medicine vol 26 no 5pp 401ndash408 1996

[12] N Guzel and M Bayram ldquoNumerical solution of differential-algebraic equations with index-2rdquo Applied Mathematics andComputation vol 174 no 2 pp 1279ndash1289 2006

[13] J K Zhou Differential Transformation and Its Application forElectrical CircuIts Huazhong University Press Wuhan China1986

[14] M Koksal and S Herdem ldquoAnalysis of nonlinear circuits byusing differential Taylor transformrdquo Computers and ElectricalEngineering vol 28 no 6 pp 513ndash525 2002

[15] I H Abdel-Halim Hassan ldquoOn solving some eigenvalue prob-lems by using a differential transformationrdquoAppliedMathemat-ics and Computation vol 127 no 1 pp 1ndash22 2002

[16] F Ayaz ldquoOn the two-dimensional differential transformmethodrdquo Applied Mathematics and Computation vol 143 no2-3 pp 361ndash374 2003

[17] C K Chen and S H Ho ldquoSolving partial differential equationsby two-dimensional differential transform methodrdquo AppliedMathematics and Computation vol 106 no 2-3 pp 171ndash1791999

[18] M-J Jang C-L Chen and Y-C Liu ldquoTwo-dimensional dif-ferential transform for partial differential equationsrdquo AppliedMathematics and Computation vol 121 no 2-3 pp 261ndash2702001

[19] X Yang Y Liu and S Bai ldquoA numerical solution of second-order linear partial differential equations by differential trans-formrdquo Applied Mathematics and Computation vol 173 no 2pp 792ndash802 2006

[20] A Arikoglu and I Ozkol ldquoSolution of differential-differenceequations by using differential transform methodrdquo AppliedMathematics and Computation vol 181 no 1 pp 153ndash162 2006

[21] M-J Jang C-L Chen and Y-C Liy ldquoOn solving the initial-value problems using the differential transformation methodrdquoAppliedMathematics andComputation vol 115 no 2-3 pp 145ndash160 2000

[22] F Ayaz ldquoApplications of differential transform method todifferential-algebraic equationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 649ndash657 2004

[23] H Liu and Y Song ldquoDifferential transform method applied tohigh index differential-algebraic equationsrdquoAppliedMathemat-ics and Computation vol 184 no 2 pp 748ndash753 2007

[24] G FrankMAPLE V CRC Press Boca Raton Fla USA 1996

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

From (45) and (46) we find

1198801(0 0) = 0 119880

1(0 1) = minus1 119880

1(0 2) = 0

1198801(0 3) = 1 119880

1(0 4) = 0 119880

1(0 5) = 0

1198801(0 6) = 0 119880

1(0 7) = 0

(46)

In this manner from (40) (44) and (45) the coefficients ofthe 1199061are obtained as follows

1198801(1 0) = 0 119880

1(1 1) = 1 119880

1(1 2) = 0

1198801(1 3) = minus1 119880

1(1 4) = 0 119880

1(1 5) = 0

1198801(1 6) = 0 119880

1(2 0) = 0 119880

1(2 1) = minus

1

2

1198801(2 2) = 0 119880

1(2 3) =

1

2 119880

1(2 4) = 0

1198801(2 5) = 0 119880

1(3 0) = 0 119880

1(3 1) =

1

6

1198801(3 2) = 0 119880

1(3 3) = minus

1

6 119880

1(3 4) = 0

1198801(4 0) = 0 119880

1(4 1) = minus

1

24 119880

1(4 2) = 0

1198801(4 3) =

1

24 119880

1(5 0) = 0 119880

1(5 1) =

1

120

1198801(5 2) = 0 119880

1(6 0) = 0 119880

1(6 1) = minus

1

720

1198801(7 0) = 0

(47)

Using the initial values for the second component we obtainthe following coefficients

1198802(0 0) = minus1 119880

2(0 1) = 0 119880

2(0 2) = 0

1198802(0 3) = 0 119880

2(0 4) = 1 119880

2(0 5) = 0

1198802(0 6) = 0 119880

2(0 7) = 0

(48)

The coefficients of the 1199062can be found using (47) (48) (49)

and taking 119896 = 0 1 2 and ℎ = 0 1 2 in (39) as follows

1198802(1 2) = 0 119880

2(1 3) = 0 119880

2(1 4) = 0

1198802(1 5) = 0 119880

2(1 6) = 0 119880

2(2 1) = 0

1198802(2 2) = 0 119880

2(2 3) = 0 119880

2(2 4) = minus

1

2

1198802(2 5) = 0 119880

2(3 0) = 0 119880

2(3 1) = 0

1198802(3 2) = 0 119880

2(3 3) = 0 119880

2(3 4) = 0

1198802(4 1) = 0 119880

2(4 2) = 0 119880

2(4 3) = 0

1198802(4 0) = minus

1

24 119880

2(5 0) = 0 119880

2(5 1) = 0

1198802(5 2) = 0 119880

2(6 0) =

1

720 119880

2(6 1) = 0

1198802(7 0) = 0

(49)

If we write the above values in (39) and (40) then we have

119906lowast

1(119905 119909) = minus 119909 + 119909119905 minus

1

21199091199052+ 1199093+1

61199091199053

minus 1199093119905 +

1

211990931199052minus1

241199091199054minus1

611990931199053

+1

1201199091199055minus

1

7201199091199056+1

2411990931199054+ sdot sdot sdot

(50)

119906lowast

2(119905 119909) = minus 1 +

1

21199052minus1

241199054

+ 1199094+

1

7201199056minus1

211990941199052+ sdot sdot sdot

(51)

Numerical and exact solution of the given problem has beencompared in Tables 1 and 2 and simulations of solutions havebeen depicted in Figures 1 2 3 and 4 respectively

5 Conclusion

The computations associated with the example discussedabove were performed by using Computer Algebra Tech-niques [24] We show the results in Tables 1 and 2 for thesolution of (32) by numerical method The numerical valueson Tables 1 and 2 obtained above are in full agreement withthe exact solutions of (32) This study has shown that thedifferential transform method often shows superior perfor-mance over series approximants providing a promising toolfor using in applied fields

References

[1] W Marszalek Analysis of partial differential algebraic equations[PhD thesis] North Carolina State University Raleigh NCUSA 1997

[2] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear par-tial differential algebraic equations Part I indexes consistentboundaryinitial conditionsrdquo Report 17 Fachbereich Mathe-matik und Informatik Martin-Luther-Universitat Halle 1997

[3] W Lucht K Strehmel and C Eichler-Liebenow ldquoLinear partialdifferential algebraic equations Part II numerical solutionrdquoReport 18 Fachbereich Mathematik und Informatik Martin-Luther-Universitat Halle 1997

[4] W Lucht K Strehmel and C Eichler-Liebenow ldquoIndexes andspecial discretization methods for linear partial differentialalgebraic equationsrdquo BIT Numerical Mathematics vol 39 no3 pp 484ndash512 1999

[5] W S Martinson and P I Barton ldquoA differentiation indexfor partial differential-algebraic equationsrdquo SIAM Journal onScientific Computing vol 21 no 6 pp 2295ndash2315 2000

8 Abstract and Applied Analysis

[6] W S Martinson and P I Barton ldquoIndex and characteristicanalysis of linear PDAE systemsrdquo SIAM Journal on ScientificComputing vol 24 no 3 pp 905ndash923 2002

[7] K Debrabant and K Strehmel ldquoConvergence of Runge-Kuttamethods applied to linear partial differential-algebraic equa-tionsrdquoApplied Numerical Mathematics vol 53 no 2-4 pp 213ndash229 2005

[8] N Guzel and M Bayram ldquoOn the numerical solution of stiffsystemsrdquo Applied Mathematics and Computation vol 170 no 1pp 230ndash236 2005

[9] E Celik and M Bayram ldquoThe numerical solution of physicalproblems modeled as a system of differential-algebraic equa-tions (DAEs)rdquo Journal of the Franklin Institute vol 342 no 1pp 1ndash6 2005

[10] M Kurulay and M Bayram ldquoApproximate analytical solutionfor the fractional modified KdV by differential transformmethodrdquo Communications in Nonlinear Science and NumericalSimulation vol 15 no 7 pp 1777ndash1782 2010

[11] M Bayram ldquoAutomatic analysis of the control of metabolicnetworksrdquo Computers in Biology and Medicine vol 26 no 5pp 401ndash408 1996

[12] N Guzel and M Bayram ldquoNumerical solution of differential-algebraic equations with index-2rdquo Applied Mathematics andComputation vol 174 no 2 pp 1279ndash1289 2006

[13] J K Zhou Differential Transformation and Its Application forElectrical CircuIts Huazhong University Press Wuhan China1986

[14] M Koksal and S Herdem ldquoAnalysis of nonlinear circuits byusing differential Taylor transformrdquo Computers and ElectricalEngineering vol 28 no 6 pp 513ndash525 2002

[15] I H Abdel-Halim Hassan ldquoOn solving some eigenvalue prob-lems by using a differential transformationrdquoAppliedMathemat-ics and Computation vol 127 no 1 pp 1ndash22 2002

[16] F Ayaz ldquoOn the two-dimensional differential transformmethodrdquo Applied Mathematics and Computation vol 143 no2-3 pp 361ndash374 2003

[17] C K Chen and S H Ho ldquoSolving partial differential equationsby two-dimensional differential transform methodrdquo AppliedMathematics and Computation vol 106 no 2-3 pp 171ndash1791999

[18] M-J Jang C-L Chen and Y-C Liu ldquoTwo-dimensional dif-ferential transform for partial differential equationsrdquo AppliedMathematics and Computation vol 121 no 2-3 pp 261ndash2702001

[19] X Yang Y Liu and S Bai ldquoA numerical solution of second-order linear partial differential equations by differential trans-formrdquo Applied Mathematics and Computation vol 173 no 2pp 792ndash802 2006

[20] A Arikoglu and I Ozkol ldquoSolution of differential-differenceequations by using differential transform methodrdquo AppliedMathematics and Computation vol 181 no 1 pp 153ndash162 2006

[21] M-J Jang C-L Chen and Y-C Liy ldquoOn solving the initial-value problems using the differential transformation methodrdquoAppliedMathematics andComputation vol 115 no 2-3 pp 145ndash160 2000

[22] F Ayaz ldquoApplications of differential transform method todifferential-algebraic equationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 649ndash657 2004

[23] H Liu and Y Song ldquoDifferential transform method applied tohigh index differential-algebraic equationsrdquoAppliedMathemat-ics and Computation vol 184 no 2 pp 748ndash753 2007

[24] G FrankMAPLE V CRC Press Boca Raton Fla USA 1996

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

[6] W S Martinson and P I Barton ldquoIndex and characteristicanalysis of linear PDAE systemsrdquo SIAM Journal on ScientificComputing vol 24 no 3 pp 905ndash923 2002

[7] K Debrabant and K Strehmel ldquoConvergence of Runge-Kuttamethods applied to linear partial differential-algebraic equa-tionsrdquoApplied Numerical Mathematics vol 53 no 2-4 pp 213ndash229 2005

[8] N Guzel and M Bayram ldquoOn the numerical solution of stiffsystemsrdquo Applied Mathematics and Computation vol 170 no 1pp 230ndash236 2005

[9] E Celik and M Bayram ldquoThe numerical solution of physicalproblems modeled as a system of differential-algebraic equa-tions (DAEs)rdquo Journal of the Franklin Institute vol 342 no 1pp 1ndash6 2005

[10] M Kurulay and M Bayram ldquoApproximate analytical solutionfor the fractional modified KdV by differential transformmethodrdquo Communications in Nonlinear Science and NumericalSimulation vol 15 no 7 pp 1777ndash1782 2010

[11] M Bayram ldquoAutomatic analysis of the control of metabolicnetworksrdquo Computers in Biology and Medicine vol 26 no 5pp 401ndash408 1996

[12] N Guzel and M Bayram ldquoNumerical solution of differential-algebraic equations with index-2rdquo Applied Mathematics andComputation vol 174 no 2 pp 1279ndash1289 2006

[13] J K Zhou Differential Transformation and Its Application forElectrical CircuIts Huazhong University Press Wuhan China1986

[14] M Koksal and S Herdem ldquoAnalysis of nonlinear circuits byusing differential Taylor transformrdquo Computers and ElectricalEngineering vol 28 no 6 pp 513ndash525 2002

[15] I H Abdel-Halim Hassan ldquoOn solving some eigenvalue prob-lems by using a differential transformationrdquoAppliedMathemat-ics and Computation vol 127 no 1 pp 1ndash22 2002

[16] F Ayaz ldquoOn the two-dimensional differential transformmethodrdquo Applied Mathematics and Computation vol 143 no2-3 pp 361ndash374 2003

[17] C K Chen and S H Ho ldquoSolving partial differential equationsby two-dimensional differential transform methodrdquo AppliedMathematics and Computation vol 106 no 2-3 pp 171ndash1791999

[18] M-J Jang C-L Chen and Y-C Liu ldquoTwo-dimensional dif-ferential transform for partial differential equationsrdquo AppliedMathematics and Computation vol 121 no 2-3 pp 261ndash2702001

[19] X Yang Y Liu and S Bai ldquoA numerical solution of second-order linear partial differential equations by differential trans-formrdquo Applied Mathematics and Computation vol 173 no 2pp 792ndash802 2006

[20] A Arikoglu and I Ozkol ldquoSolution of differential-differenceequations by using differential transform methodrdquo AppliedMathematics and Computation vol 181 no 1 pp 153ndash162 2006

[21] M-J Jang C-L Chen and Y-C Liy ldquoOn solving the initial-value problems using the differential transformation methodrdquoAppliedMathematics andComputation vol 115 no 2-3 pp 145ndash160 2000

[22] F Ayaz ldquoApplications of differential transform method todifferential-algebraic equationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 649ndash657 2004

[23] H Liu and Y Song ldquoDifferential transform method applied tohigh index differential-algebraic equationsrdquoAppliedMathemat-ics and Computation vol 184 no 2 pp 748ndash753 2007

[24] G FrankMAPLE V CRC Press Boca Raton Fla USA 1996

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