Comparison differential transformation technique with Adomian decomposition method for linear and nonlinear initial value problems

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Chaos, Solitons and Fractals 36 (2008) 53–65 www.elsevier.com/locate/chaos

Comparison differential transformation technique with Adomian decomposition method for linear and nonlinear initial value problems I.H. Abdel-Halim Hassan Department of Mathematics, Faculty of Science, Zagazig University, Zagazig, Egypt Accepted 6 June 2006

Communicated by Prof. L. Marek-Crnjac

Abstract In this paper, we will compare the differential transformation method DTM and Adomian decomposition method ADM to solve partial differential equations (PDEs). The definition and operations of differential transform method was introduced by Zhou [Zhou JK. Differential transformation and its application for electrical circuits. Wuuhahn, China: Huarjung University Press; 1986 [in Chinese]]. Adomian decomposition method which is given by Adomian [Adomian G. Convergent series solution of nonlinear equation. J Comput Appl Math 1984;11:113–7; Adomian G. Solutions of nonlinear PDE. Appl Math Lett 1989;11:121–3; Adomian G. Solving Frontier problems of physics. The decomposition method, Boston, 1994] for approximate solution of linear and nonlinear differential equations and to the solutions of various scientific models see [Abulwafa EM, Abdou MA, Mahmoud AA. The solution of nonlinear coagulation problem with mass loss. Chaos, Solitons & Fractals 2006;29:313–30; El-Danaf TS, Ramadan MA, Abd Alaal FEI. The use of Adomian decomposition method for solving the regularized long-wave equation. Chaos, Solitons & Fractals 2005;26:747–57; El-Sayed SM. The decomposition method for studying the Klein–Gordon equation. Chaos, Solitions & Fractals 2003;18:1025–30; Helal MA, Mehanna MS. A comparison between two different methods for solving KdVBurgers equation. Chaos, Solitons & Fractals 2006;28:320–6, Hashim I, Noorani MSM, Ahmad R, Bakar SA, Ismail ES, Zakaria AM. Accuracy of the decomposition method applied to the Lorenz system. Chaos, Solitons & Fractals 2006;28:1149–58; Kaya D, El-Sayed SM. An application of the decomposition method for the generalized KdV and RLW equations, Chaos, Solitons & Fractals 2003;17:869–77; Lesnic D. Blow-up solutions obtained using the decomposition method. Chaos, Solitons & Fractals 2006;28:776–87; Wazwaz AM. Construction of solitary wave solutions and rational solutions for KdV equation by Adomian decomposition method. Chaos, Solitons & Fractals 2001;12:2283–93; Wazwaz AM. Construction of soliton solutions and periodic solutions of the Boussinesq equation by the modified decomposition method. Chaos, Solitons & Fractals 2001;12:1549–56]. A distinctive practical feature of the differential transformation method DTM is ability to solve linear or nonlinear differential equations. Higherorder dimensional differential transformation are applied to a few some initial value problems to show that the solutions obtained by the proposed method DTM coincide with the approximate solution ADM and the analytic solutions. Ó 2006 Published by Elsevier Ltd.

E-mail address: [email protected] 0960-0779/$ - see front matter Ó 2006 Published by Elsevier Ltd. doi:10.1016/j.chaos.2006.06.040

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1. Introduction A variety of methods, exact, approximate, and purely numerical are available for the solution of systems of differential equations. Most of these methods are computationally intensive because they are trial-and-error in nature, or need complicated symbolic computations. Integral transforms such as Laplace and Fourier transforms are commonly used to solve differential equations and usefulness of these integral transforms lies in their ability to transform differential equations into algebraic equations which allows simple and systematic solution procedures. However, using integral transform in nonlinear problems may increase its complexity. In the present work, some partial differential equations with nonhomogeneous initial conditions aimed to solve by the differential transformation method and comparison with Adomian decomposition method. The differential transformation is a numerical method for solving differential equations. The concept of differential transform was introduced by Zhou [45], who solved linear and nonlinear initial value problems in electric circuit analysis. Ref. [9,12,13,20,27,30] developed this method for PDEs and obtained closed form series solutions for linear and nonlinear initial value problems. The differential transform method an analytical solution in the form of a polynomial. It is different from the traditional high order Taylor series method, which requires symbolic computation of the necessary derivatives of the data functions. The Taylor series method is computationally taken long time for large orders. The present method reduces the size of computational domain and applicable to many problems easily. In this paper, the definitions and operations of differential transformation is introduced. Different types of partial differential equations with nonhomogeneous initial conditions are solved by the differential transformation method and Adomian decomposition method.

2. The definitions and operations of differential transform 2.1. One-dimensional differential transform As in [21,22,27–29,10], the basic definition of the differential transformation are introduced as follows: Definition 2.1. If u(t) is analytic in the domain T, then it will be differentiated continuously with respect to time t, ok uðtÞ ¼ /ðt; kÞ; otk

for all t 2 T

ð2:1Þ

for t = ti, then /(t, k) = /(ti, k), where k belongs to the set of nonnegative integers, denoted as the K-domain. Therefore, Eq. (2.1) can be rewritten as  k  o uðtÞ  U ðkÞ ¼ /ðti ; kÞ ¼ ; ð2:2Þ otk t¼ti where U(k) is called the spectrum of u(t) at t = ti. Definition 2.2. If u(t) can be expressed by Taylor’s series, then u(t) can be represented as " # 1 X ðt  ti Þk U ðkÞ: uðtÞ ¼ k! k¼0

ð2:3Þ

Eq. (2.3) is called the inverse of u(t), with the symbol D denoting the differential transformation process. Upon combining (2.2) and (2.3), we obtain " # 1 X ðt  ti Þk uðtÞ ¼ U ðkÞ  D1 U ðkÞ: ð2:4Þ k! k¼0 Using the differential transformation, a differential equation in the domain of interest can be transformed to an algebraic equation in the K-domain and the u(t) can be obtained by finite-term Taylor’s series plus a remainder, as " # n X ðt  ti Þk uðtÞ ¼ U ðkÞ þ Rnþ1 ðtÞ: ð2:5Þ k! k¼0

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55

Table 1 Operations of the one-dimensional differential transformation Original function

Transformed function

z(t) = u(t) ± v(t) z(t) = au(t) zðtÞ ¼ duðtÞ dt

Z(k) = U(k) ± V(k) Z(k) = aU(k) Z(k) = (k + 1)U(k + 1)

zðtÞ ¼ d dtuðtÞ 2

Z(k) = (k + 1)(k + 2)U(k + 2)

zðtÞ ¼ d dtuðtÞ m z(t) = u(t)v(t) z(t) = tm z(t) = exp(kt) z(t) = (1 + t)m z(t) = sin(xt + a)

Z(k) = (k P + 1)(k + 2)  (k + m)U(k + m) ZðkÞ ¼ k‘¼0 V ð‘ÞU ðk  ‘Þ  1; if k ¼ m ZðkÞ ¼ dðk  mÞ; dðk  mÞ ¼ 0; if k 6¼ m kk ZðkÞ ¼ k! mðm1Þðmkþ1Þ ZðkÞ ¼ k!  k ZðkÞ ¼ xk! sin pk 2 þa   k ZðkÞ ¼ xk! cos pk 2 þa

2

m

z(t) = cos(xt + a)

If u(t), v(t) are two uncorrelated functions with time t and U(k), V(k) are the transformed functions corresponding to u(t), v(t) then we can easily proof the fundamental mathematics (Table 1). 2.2. Two-dimensional differential transform Similarly, consider a function of two variables x(x, y), be analytic in the domain K and let (x, y) = (x0, y0) in this domain. The function x(x, y) is then represented by one power series whose center at located at (x0, y0). The differential transform of function x(x, y) is the form   1 okþh xðx; yÞ ; ð2:6Þ W ðk; hÞ ¼ k!h! oxk oy h ðx0 ;y 0 Þ where x(x, y) is the original function and W(k, h) is the transformed function. The differential inverse transform of W(k, h) is defined as xðx; yÞ ¼

1 X 1 X

W ðk; hÞðx  x0 Þk ðy  y 0 Þh ;

ð2:7Þ

k¼0 h¼0

and from (2.6) and (2.7) can be concluded   1 X 1 X 1 okþh xðx; yÞ  ðx  x0 Þk ðy  y 0 Þh : xðx; yÞ ¼  k oy h k!h! ox ðx0 ;y 0 Þ k¼0 h¼0

ð2:8Þ

In a real application, and when (x0, y0) are taken as (0, 0), then the function x(x, y) in Eq. (2.7) can be written by a finite series as n X m X W ðk; hÞxk y h : ð2:9Þ xðx; yÞ ¼ k¼0 h¼0

Based on the properties of one-dimensional transform some operations and theorems of the two-dimensional differential transformation are listed in Refs. [13,20]. 2.3. n-Dimensional differential transform [30] Consider a function of n variables x = x(x1, x2, x3, . . . , xn), and k = (k1, k2, . . . , kn) be a vector of n nonnegative integer. Based on the properties of one and two-dimensional transform, we can define n-dimensional differential transform as " # 1 ok1 þk2 þþkn W ðk 1 ; k 2 ; . . . ; k n Þ ¼ xðx1 ; x2 ; . . . ; xn Þ ; ð2:10Þ k 1 !k 2 ! . . . k n ! oxk11 oxk22 . . . oxknn ð0;0;...;0Þ where x(x) = x(x1, x2, . . . , xn) is the original function and W(k) = W(k1, k2, . . . , kn) is the transformed function.

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The differential inverse transform of W(k) is defined as xðx1 ; x2 ; . . . ; xn Þ ¼

1 X 1 X



k 1 ¼0 k 2 ¼0

1 X

W ðk 1 ; k 2 ; . . . ; k n Þ

k n ¼0

n Y

k

xj j :

ð2:11Þ

j¼1

From (2.10) and (2.11) we conclude xðx1 ; x2 ; . . . ; xn Þ ¼

1 X 1 X k 1 ¼0 k 2 ¼0



1 X k n ¼0

" # n Y 1 ok1 þk2 þþk n k  xðx1 ; x2 ; . . . ; xn Þ xj j : k1 k2 k 1 !k 2 ! . . . k n ! ox1 ox2 . . . oxknn j¼1 ð0;0;...;0Þ

ð2:12Þ

By the linearity of differential transform [9,11,22,28] the derivations of the theorems of the n-dimensional differential transformation are investigated and proofs by [30].

3. The Adomian decomposition method The topic of the Adomian decomposition method has been rapidly growing in recent years. The concept of this method was first introduced by Adomian in the beginning of [1–3]. In this method the solution of a functional equations is considered as the sum of an infinite series usually converging to the solution (see Refs. [16,23–26,31–33,42,43]). A considerable research works have been conducted recently in applying this method to a class of linear and nonlinear ordinary differential equations. Some authors have exaggerated on the applications of this method see [5,6,14–19,35– 41,44]. The solution of linear and nonlinear problem can be obtained in closed form in special cases. Adomian and Rach [4] and Wazwaz [41], have investigated the phenomena of the self-cancelling ‘‘noise’’ terms for a decomposition series solutions. With the presented decomposition method the structure of the problem will not be changed and no linearization will be made. In the simplest case the solution can be developed as a Taylor series expansion about the function not the point at which the initial condition and integrating right hand side function of the problem are determined the first term u0 of the decomposition series for n P 0. The sum of the u0, u1, u2, u3, . . . terms are simply the decomposition series uðx; tÞ ¼

1 X

un ðx; tÞ:

ð3:1Þ

n¼0

Suppose that the differential equation operator including both linear and nonlinear terms, can be formed as Lu þ Ru þ Nu ¼ /ðx; tÞ;

ð3:2Þ

with the initial conditions uð0; tÞ ¼ f ðtÞ and

ux ð0; tÞ ¼ gðtÞ;

ð3:3Þ

and L is invertible and R is the remainder of the linear operator, linear terms are written as L + R and N is the nonlinear term. To apply the decomposition method, we write Eq. (3.2) in an operator form Lu ¼ /ðx; tÞ  Ru  Nu:

ð3:4Þ

As L is reversible, equal to the above it is formed as L1 Lu ¼ L1 /ðx; tÞ  L1 Ru  L1 Nu;

ð3:5Þ 1

if these response to an initial value problem, L degree operator, L1 can be defined as Z tZ t ðÞdt dt: L1 ¼ 0

can be an integrable operator defined from t0 to t. If L is the second

ð3:6Þ

0

For the boundary value problem, Eq. (3.6) is used and the constants are found from the initial conditions (3.3). If u is solved in Eq. (3.5), then u ¼ f ðtÞ þ tgðtÞ þ L1 /ðx; tÞ  L1 Ru  L1 Nu; is found.

ð3:7Þ

I.H. Abdel-Halim Hassan / Chaos, Solitons and Fractals 36 (2008) 53–65

The nonlinear term which is of the form Nu can be written in the form Nu ¼ can be calculated for all forms of nonlinearity [3,47], An are given by !! 1  X 1 dn  k N k uk An ¼  ; n ¼ 0; 1; 2; 3; . . . ; n  n! dk k¼0

P1

n¼0 An ,

57

where these polynomials An

ð3:8Þ

k¼0

for example 8 A0 ¼ Nðu0 Þ; > > > < A1 ¼ u1 N 0 ðu0 Þ; u1 0 00 > > A2 ¼ u2 N ðu0 Þ þ 2 N ðu0 Þ; > : A3 ¼ u3 N 0 ðu0 Þ þ u1 u2 N 00 ðu0 Þ þ 16 u31 N 000 ðu0 Þ; and so on, the other polynomials can be constructed in a similar way [3,47]. When f(t) + tg(t) + L1/(x, t) is defined as u0, then 8 u0 ¼ f ðtÞ þ tgðtÞ þ L1 /ðx; tÞ; > > > > > u ¼ L1 Ru0  L1 A0 ; > > < 1 u2 ¼ L1 Ru1  L1 A1 ; > > .. > > > . > > : unþ1 ¼ L1 Run  L1 An ;

ð3:9Þ

ð3:10Þ

are obtained. Namely, An polynomials are obtained for every single nonlinear term as (3.9). Hence by calculating all components of un, the solution can be found from Eq. (3.1). The convergence of this method is proved by [7,8].

4. Numerical illustrations Let us now give the examples related with two methods for comparing the differences of convergence. Example 1. Consider the Telegrapher’s equation: 2 o2 o 2 o uðx; tÞ ¼ c uðx; tÞ þ 2a uðx; tÞ; ot ot2 ox2

ð4:1Þ

where a and c are positive constants, with the Cauchy data are given as uðx; 0Þ ¼ cos x; ut ðx; 0Þ ¼ 0:

ð4:2Þ ð4:3Þ

4.1. Differential transformation method Taking the differential transformation of (4.1), can be obtained U ðk; h þ 2Þ ¼

1 ½c2 ðk þ 1Þðk þ 2ÞU ðk þ 2; hÞ  2aðh þ 1ÞU ðk; h þ 1Þ: ðh þ 1Þðh þ 2Þ

From the initial conditions (4.2) we can write 8 1; k ¼ 0; > > > < 1 ; k ¼ 2; 6; 10; . . . U ðk; 0Þ ¼ 1k! > ; k ¼ 4; 8; 12; . . . > > : k! 0; otherwise:

ð4:4Þ

ð4:5Þ

From the initial condition (4.3) we can write U ðk; 1Þ ¼ 0:

ð4:6Þ

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Substituting from (4.5) and (4.6) into (4.4), and by recursive method the results corresponding to U(k, h) where h = 2 we have 8 1 2 > < 2k! c ; k ¼ 2; 6; 10; . . . U ðk; 2Þ ¼ 1 ð4:7Þ c2 ; k ¼ 0; 4; 8; . . . 2k! > : 0; otherwise; and so on, we can calculated U(k, h). Substituting all U(k, h) into (2.9) as a = c and m ! 1, n ! 1 we obtain 1 X 1 X

uðx; tÞ ¼

W ðk; hÞxk th

k¼0 h¼0

¼

 1 1 1 6 x þ  1  x2 þ x4  2 24 720  1 22 1 33 1 1 55 1 66 ct þ c t þ   1  ct þ c t  c t þ c4 t4  2 6 24 120 720   1 1 1 1 55 1 66 ct þ c t þ  ¼ ð1 þ ctÞ½expðctÞ cos x: þ ct 1  ct þ c2 t2  c3 t3 þ c4 t4  2 6 24 120 720

ð4:8Þ

4.2. Adomian decomposition method Now let us calculate the approximate solution of this linear differential equation by using decomposition method. If Eq. (4.1) is dealt with this method, o2 o2 o uðx; tÞ ¼ c2 2 uðx; tÞ  2a uðx; tÞ; 2 ot ot ox

ð4:9Þ

is found. In operator form, Eq. (4.9) is formed as Lt uðx; tÞ ¼ c2 Lx uðx; tÞ  2a

o uðx; tÞ; ot

ð4:10Þ

where Lt ¼

o2 ; ot2

Lx ¼

o2 : ox2

If the invertible operator L1 t ¼

ð4:11Þ Rt Rt

ðÞdt dt is applied to Eq. (4.10), then  2 1 1 o uðx; tÞ ; L1 t Lt uðx; tÞ ¼ c Lt ðLx uðx; tÞÞ  2aLt ot 0

0

ð4:12Þ

is obtained. By this 1 uðx; tÞ ¼ uðx; 0Þ þ tut ðx; 0Þ þ c2 L1 t ðLx uðx; tÞÞ  2aLt



o uðx; tÞ ; ot

ð4:13Þ

is found. Here the main point is that the solution of the decomposition method is in the form of uðx; tÞ ¼

1 X

un ðx; tÞ:

ð4:14Þ

n¼0

Substituting from Eq. (4.14) in (4.13), we find 1 X

un ðx; tÞ ¼ uðx; 0Þ þ tut ðx; 0Þ þ

c2 L1 t

n¼0

Lx

1 X

! un ðx; tÞ



n¼0

2aL1 t

! 1 o X un ðx; tÞ ; ot n¼0

ð4:15Þ

is found. Thus the approximate solution can be obtained as follows: u0 ðx; tÞ ¼ uðx; 0Þ þ tut ðx; 0Þ ¼ cos x;

ð4:16Þ

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59

and uðnþ1Þ ðx; tÞ ¼

Z 0

t

! !# Z t" 1 1 X o X 2 c Lx un ðx; tÞ  2a un ðx; tÞ dt dt: ot n¼0 0 n¼0

ð4:17Þ

On the other hand, from Eq. (4.17) we have  2 2 t cos x; u1 ¼ c 2!  4  3 t t cos x þ 2ac2 cos x; u2 ¼ c4 4! 3!  6  5  4 t t t cos x  4ac4 cos x  4a2 c2 cos x; u3 ¼ c6 6! 5! 4! .. . are found. Thus the approximate solution of the given initial value problem (4.1)–(4.3) is obtained after some calculations when c = a as  c2 t2 c3 t3 c4 t4 c5 t5  þ  þ    ¼ ð1 þ ctÞ½expðctÞ cos x: ð4:18Þ uðx; tÞ ¼ cos xð1 þ ctÞ 1  ct þ 2! 3! 4! 5! From (4.8) and (4.18) the approximate solution of the given initial value problem (4.1)–(4.3) by using differential transformation method is the same results with that obtained by the Adomian decomposition method and it is clearly appeared that the approximate solution remain closed form to the analytic solution is given by [46]. Example 2. Consider the initial value problem o2 o2 o2 uðx; yÞ  3 2 uðx; yÞ ¼ 0; uðx; yÞ þ 2 2 ox oxoy oy

ð4:19Þ

with the nonhomogeneous initial data uðx; 0Þ ¼ sin x; uy ðx; 0Þ ¼ x:

ð4:20Þ ð4:21Þ

4.3. Differential transformation method Taking the differential transformation of (4.19), can be obtained ðj þ 1Þðj þ 2ÞU ðj þ 2; iÞ þ 2ðj þ 1Þðj þ 2Þði þ 1ÞU ðj þ 1; i þ 1Þ  3ði þ 1Þði þ 2ÞU ðj; i þ 2Þ ¼ 0: From the initial condition (4.20), we have 1 1 X X uðx; 0Þ ¼ U ðj; 0Þxj ¼ sin x ¼ ð1Þj1 j¼0

j¼1

ð4:22Þ

x2j1 : ð2j  1Þ!

The corresponding spectra can be obtained as follows ( ð1Þðj1Þ=2 j!1 ; j ¼ 1; 3; 5; . . . U ðj; 0Þ ¼ 0; j ¼ 0; 2; 4; . . .

ð4:23Þ

From the initial condition (4.21), we have 1 ouðx; 0Þ X ¼ U ðj; 1Þxj ¼ x: oy j¼0 The corresponding spectra can be obtained as follows:  1; j ¼ 1; U ðj; 1Þ ¼ 0; j ¼ 0; 2; 3; 4; . . .

ð4:24Þ

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Substituting from (4.23) and (4.24) to (4.22), we have 81 ; j ¼ 0; > > > 31 > < 6 ; j ¼ 1; U ðj; 2Þ ¼ 1 1  ðjþ1Þ=2 > ð1Þ ; j ¼ 3; 5; 7; . . . ; > 6 j! > > : 0; otherwise:

  ( ðjþ2Þ=2 1 1 ; j ¼ 0; 2; 4; . . . ; ð1Þ j! 33 U ðj; 3Þ ¼ 0; otherwise;

ð4:25Þ

ð4:26Þ

and so on, we can calculated U(j, i). Substituting all values of U(j, i) into (2.9) as m ! 1, n ! 1 we obtain the series solution for the initial value problem (4.19)–(4.21) as uðx; yÞ ¼

1 X 1 X j¼0

W ðj; iÞxj y i

i¼0

(" #) 1 1 1 1 X X 1 1 x2j1 X y 2i x2i X y 2j1 ¼ xy þ y 2 þ ð1Þj1 ð1Þi ð1Þi ð1Þj1 þ 3 4 ð2j  1Þ! i¼0 ð2iÞ! i¼0 ð2iÞ! j¼1 ð2j  1Þ! j¼1 (" # )   y 2i1 X 1 1 1 1 X X 3 xj1 X y 2i x2j ð1Þj ð1Þi ð1Þi1 3 ð1Þj  þ ð2j  1Þ! i¼1 ð2iÞ! i¼1 ð2i  1Þ! j¼0 ð2jÞ! 4 j¼0



  1 1 3 y : ¼ xy þ y 2 þ sinðx þ yÞ þ sin x þ 3 4 4 3

ð4:27Þ

4.4. Adomian decomposition method Now let us solve the initial value problem (4.19)–(4.21) by the decomposition method. Since Z y Z y 1 X o2 o2 ¼ ðÞdy dy and uðx; yÞ ¼ un ðx; yÞ: Lx ¼ 2 ; Ly ¼ 2 ; L1 y ox oy 0 0 n¼0

ð4:28Þ

So Ly uðx; yÞ ¼

2 o2 1 uðx; tÞ þ Lx uðx; yÞ; 3 oxoy 3

ð4:29Þ

is found. If L1 y is applied to the both sides of (4.29), then 2 1 o2 1 uðx; tÞ þ L1 L1 Lx uðx; yÞ: y Ly uðx; yÞ ¼ Ly 3 3 y oxoy

ð4:30Þ

Using Eqs. (4.30), (4.29), (4.28), (4.21) and (4.20), then the approximate solution can be obtained as follows: u0 ðx; yÞ ¼ uðx; 0Þ þ yuy ðx; 0Þ ¼ sin x þ xy;  Z y Z y 2 o2 1 ðun ðx; yÞÞ þ ðLx un ðx; yÞÞ dy dy; uðnþ1Þ ðx; yÞ ¼ 3 oxoy 3 0 0 is calculated. From this point,  2  3 y y  sin x; u1 ¼ 3 3!   3   4 2 y 1 y cos x þ sin x; u2 ¼  9 3! 9 4!   4   5   6 4 y 4 y 1 y sin x þ cos x  sin x; u3 ¼ 27 4! 27 5! 27 6! .. .

n P 0;

ð4:31Þ

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61

P are found. If these values substituting in equation uðx; yÞ ¼ 1 n¼0 un ðx; yÞ ¼ u0 ðx; yÞ þ u1 ðx; yÞ þ u2 ðx; yÞ þ u3 ðx; yÞ þ   , thus the approximate solution which is  2 y uðx; yÞ ¼ xy þ 3 " #  2 1 1 X X y 1 ð1Þj ðyÞ2j ð1Þj1 ðyÞ2j1 ¼ xy þ þ sin x þ cos x 4 3 ð2jÞ! ð2j  1Þ! j¼0 j¼1 " #     2j 2j1 1 1 X X ð1Þj 3y ð1Þj1 3y 3  cos x þ sin x 4 ð2jÞ! ð2j  1Þ! j¼0 j¼1  2

y 1 3 y ð4:32Þ ¼ xy þ þ sinðx þ yÞ þ sin x  : 4 4 3 3 From (4.32) and (4.27), we see that the solution by using differential transformation method of the given initial value problem (4.19), (4.20) and (4.21) is the same approximate result with that obtained by the Adomian decomposition method and is too close to the analytic solution is given by [34]. Example 3. Consider nonlinear partial differential equation: ouðx; y; zÞ ouðx; y; zÞ ouðx; y; zÞ þ þ ¼ u2 ðx; y; zÞ; ox oy oz

ð4:33Þ

with the nonhomogeneous initial condition uðx; y; 0Þ ¼ x þ y:

ð4:34Þ

4.5. Differential transformation method Taking the differential transformation of (4.33) and the initial condition (4.43) respectively, we obtain ðk þ 1ÞU ðk þ 1; h; mÞ þ ðh þ 1ÞU ðk; h þ 1; mÞ þ ðm þ 1ÞUðk; h; m þ 1Þ ¼

k X h X m X

Uða; h  b; m  cÞU ðk  a; b; cÞ

ð4:35Þ

a¼0 b¼0 c¼0

and  U ðk; h; 0Þ ¼

1 0

if k ¼ 1 or h ¼ 1; otherwise:

ð4:36Þ

By applying Eq. (4.36) into Eq. (4.35) we can obtain some values of U(k, h, m) as follows: U ð0; 0; mÞ ¼ 0 if m ¼ 0; 2; 4; 6; . . . ; U ðk; h; 0Þ ¼ 0 if k ¼ h ¼ 0; 1; 2; 3; . . . ; U ðk; 0; 0Þ ¼ 0 if k ¼ 2; 3; 4; . . . ; U ðk; h; mÞ ¼ 0 if k ¼ h ¼ 1; m ¼ 2; 4; 6; . . . ; U ð0; 0; 1Þ ¼ ½U ð0; 0; 0Þ2  ½U ð1; 0; 0Þ þ U ð0; 1; 0Þ ¼ 2: And so on we can calculating another values of U(k, h, m) and by recursive method, some result are listed as follows in Table 2. Consequently substituting all U(k, h, m) into (2.12) and after some calculations, we obtain the series form solution of (4.33) and (4.34) as 1 X 1 X 1 X uðx; y; zÞ ¼ U ðk; h; mÞxk y h zm k¼0 h¼0 m¼0

¼ ½ðx þ y  2zÞ þ zðx þ y  2zÞ2 þ z2 ðx þ y  2zÞ3 þ z3 ðx þ y  2zÞ4 þ    ðx þ y  2zÞ : ¼ 1  zðx þ y  2zÞ

ð4:37Þ

62

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Table 2 Some values of U(k, h, m) of Ex. 3 U(1, 1, 1) = 2 U(2, 0, 1) = 1 U(0, 2, 1) = 1 U(1, 0, 2) = 4 U(0, 1, 2) = 4

U(3, 0, 2) = 1 U(1, 0, 4) = 12 U(0, 1, 4) = 12 U(0, 0, 7) = 16 U(2, 0, 5) = 24

U(2, 0, 3) = 6 U(1, 3, 2) = 1 U(0, 4, 3) = 1 U(3, 0, 4) = 8 U(1, 0, 6) = 32

U(2, 1, 2) = 3 U(1, 1, 3) = 12 U(0, 0, 5) = 8 U(3, 1, 3) = 4 U(1, 3, 3) = 4

U(0, 0, 3) = 4 U(0, 1, 6) = 32 (2, 1, 4) = 24 U(1, 1, 5) = 48 U(1, 2, 4) = 24

U(1, 2, 2) = 3 U(0, 2, 3) = 6 U(4, 0, 3) = 1 U(2, 2, 3) = 6 U(0, 3, 4) = 8

4.6. Adomian decomposition method Now we solve the initial value problem (4.33) and (4.34) by the decomposition method. Since Z z 1 X o o o 1 ðÞdz and uðx; y; zÞ ¼ un ðx; y; zÞ: Lx ¼ ; Ly ¼ ; Lz ¼ ; Lz ¼ ox oy oz 0 n¼0

ð4:38Þ

So Lz uðx; y; zÞ ¼ Lx uðx; y; zÞ  Ly uðx; y; zÞ þ u2 ðx; y; zÞ;

ð4:39Þ

is found. If L1 z is applied to the both sides of (4.38), then 1 1 1 2 L1 z Lz uðx; y; zÞ ¼ Lz ðLx uðx; y; zÞÞ  Lz ðLy uðx; y; zÞÞ þ Lz ðu ðx; y; zÞÞ:

ð4:40Þ

2

The decomposition method expands the nonlinear term u (x, y, z) formally in a power series, given, u2 ðx; y; zÞ ¼

1 X

An ;

n¼0

where An is the so-called Adomian polynomials corresponding to the nonlinear term u2(x, y, z), see Eqs. (3.8) and (3.9). Thus, from (4.40), (4.38) and the initial condition (4.34), we have Z z unþ1 ðx; y; zÞ ¼ u0 ðx; y; zÞ þ ðLx un ðx; y; zÞ  Ly un ðx; y; zÞ þ An Þdz; ð4:41Þ 0

is found for n P 0. Thus u0 ¼ x þ y; A0 ¼ x2 þ 2xy þ y 2 ; Z z u1 ¼ ðLx u0  Ly u0 þ A0 Þdz ¼ 2z þ zx2 þ 2xyz þ zy 2 ; 0

A1 ¼ 4zx  4zy þ 2zx3 þ 6zx2 y þ 6zxy 2 þ 2zy 3 ; Z z u2 ¼ ðLx u1  Ly u1 þ A1 Þdz ¼ 4z3 x  4z2 y þ z2 x3 þ 3z2 x2 y þ 3z2 xy 2 þ z2 y 3 ; 0

A2 ¼ 12z2 x2  24z2 xy  12z2 y 2 þ 3z2 x4 þ 12z2 x3 y þ 18z2 x2 y 2 þ 12z2 xy 3 þ 3z2 y 4 þ 4z4 ; Z z u3 ¼ ðLx u2  Ly u2 þ A2 Þdz ¼ 4z3  6z3 x2  12z3 xy  6z3 y 2 þ z3 x4 þ 4z3 x3 y þ 6z3 x2 y 2 þ 4z3 xy 3 þ z3 y 4 ; 0

.. . and so on. At the end point, we have the approximate solution as formed: uðx; y; zÞ ¼ u0 þ u1 þ u2 þ u3 þ u4 þ    ¼

x þ y  2z : ½1  zðx þ y  2zÞ

ð4:42Þ

From (4.37) and (4.42), we see that the solution by using differential transformation method of the given initial value problem (4.33) and (4.34) is the same approximate result with that obtained by the Adomian decomposition method and is too close to the analytic solution is given by [46].

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63

Example 4. Consider partial differential equation: o2 uðx; y; z; tÞ ¼ c2 r2 uðx; y; z; tÞ; ot2

ð4:43Þ

with the nonhomogeneous initial condition uðx; y; z; 0Þ ¼ sin px sin py sin pz; o uðx; y; z; 0Þ ¼ 0; ot

ð4:44Þ ð4:45Þ

and the boundary conditions u(0, y, z, t) = u(1, y, z, t) = u(x, 0, z, t) = u(x, 1, z, t) = u(x, y, 0, t) = u(x, y, 1, t) = 0 where u = u(x, y, z, t) is a function of the variables x, y, z and t. 4.7. Differential transformation method Taking the differential transformation of (4.43) we obtain U ðk; h; m; n þ 2Þ ¼

c2 ½ðk þ 1Þðk þ 2ÞU ðk þ 2; h; m; nÞ þ ðh þ 1Þðh þ 2Þ  U ðk; h þ 2; m; nÞ ðn þ 1Þðn þ 2Þ þ ðm þ 1Þðm þ 2ÞU ðk; h; m þ 2; nÞ:

ð4:46Þ

The differential transformation of the initial conditions (4.44) and (4.45), respectively are U ðk; h; m; 0Þ ¼

ð1Þ

ðkþhþm3Þ 2

pðkþhþmÞ ; k!h!m!

ð4:47Þ

U ðk; h; m; 1Þ ¼ 0:

ð4:48Þ

Then substituting (4.47) into (4.46) for n = 0 we have ðkþhþmÞ

U ðk; h; m; 2Þ ¼

1

c2 p2 ð1Þ 2 2 pðkþhþmÞ : 2! k!h!m!

ð4:49Þ

After successive substitution for n = 2, 3, 4, . . . in (4.46) we arrive the general formula ( pffiffi ðkþhþmÞ n3 þ 2 pðkþhþmÞ ð 3cÞn ð1Þ 2 ; n ¼ 2; 4; 6; . . . ; n! k!h!m! U ðk; h; m; nÞ ¼ 0; n ¼ 1; 3; 5; . . . By (2.11), the solution u(x, y, z, t) can be obtained as

pffiffiffi  uðx; y; z; tÞ ¼ sin px sin py sin pz cos 3pct :

ð4:50Þ

ð4:51Þ

4.8. Adomian decomposition method Now we solve the initial value problem (4.43)–(4.45) by the decomposition method. The equivalent canonical of partial differential equation (4.43) is uðx; y; z; tÞ ¼ L1 t ½Lx uðx; y; z; tÞ þ Ly uðx; y; z; tÞ þ Lz uðx; y; z; tÞ;

ð4:52Þ

where Lx ¼

o ; ox

Ly ¼

o ; oy

Lz ¼

o ; oz

L1 ¼ t

Z 0

t

Z

t

ðÞdt dt:

ð4:53Þ

0

Then the solution by Adomian decomposition method consists of the following schema: u0 ðx; y; z; tÞ ¼ sin px sin py sin pz; Z t Z t ½Lx un ðx; y; z; tÞ þ Ly un ðx; y; z; tÞ þ Lz un ðx; y; z; tÞdt dt; unþ1 ðx; y; z; tÞ ¼ 0

is found for n P 0.

0

ð4:54Þ

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Thus u0 ðx; y; z; tÞ ¼ sin px sin py sin pz; 3c2 p2 ðsin px sin py sin pzÞðt2 Þ; 2! 9c4 p4 ðsin px sin py sin pzÞðt4 Þ; u2 ðx; y; z; tÞ ¼ 4! 27c6 p6 ðsin px sin py sin pzÞðt6 Þ; u3 ðx; y; z; tÞ ¼ 6! .. . u1 ðx; y; z; tÞ ¼

are found. Thus the approximate solution of the given initial value problem (4.43)–(4.45) by Adomian decomposition method is obtained as uðx; y; z; tÞ ¼ u0 ðx; y; z; tÞ þ u1 ðx; y; z; tÞ þ u2 ðx; y; z; tÞ þ    " # pffiffiffi pffiffiffi pffiffiffi ð 3cptÞ2 ð 3cptÞ4 ð 3cptÞ6 þ  þ  ¼ sin px sin py sin pz 1  2! 4! 6! pffiffiffi ¼ sin px sin py sin pz cosð 3cptÞ:

ð4:55Þ

From (4.51) and (4.55), the approximate solution by Adomian decomposition method is similar to the solution given by the differential transformation method which are closed with the analytic solution [34]. 5. Conclusion This paper applied the differential transformation technique and Adomian decomposition method to solve initial value problem. Throughout the result of four examples which are found by using the two methods are compared with the analytic solutions, we show that the convergence are quite close. Too the results of four examples tell us the two methods can be alternative way for the solution of the linear and nonlinear higher-order initial value problems. Acknowledgements The author would like to thank Prof. M.S. El-Naschie and Prof. L. Marek-Crnjac for suggestions and comments of this work. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]

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