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SOME MODIFICATIONS OF ADOMIAN DECOMPOSITION METHODS FOR NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
M. Al-Mazmumy & H. Al-Malki
Department of Mathematics, Science Faculty for Girls, King Abdulaziz University, Saudi Arabia [email protected], [email protected]
ABSTRACT
In this paper , some modifications of Adomian decomposition method apply for solving nonlinear partial differential equations . These efficient modifications gives a simple powerful tool for obtaining the solutions without a need for large size of computations . The numerical results show the efficiency and accuracy of these methods .
Keywords: Adomian decomposition method; Modified decomposition method; new modification ; two-step Adomian decomposition method; Laplace Adomian decomposition Method; nonlinear partial differential equation.
1. INTRODUCTION
Nonlinear partial differential equations (PDE) arise in wide variety engineering and scientific application . Exact solutions of these applications are easily calculated by the decomposition method[1,2,3]. No similarity reductions are useal to reduce a nonlinear (PDE) to a system of simpler (PDE) with a lower number of independent variables , the original nonlinear equation is direct solvable preserving he actual physics and involving much less calculation.
No linearization, perturbation, or discretized method which result in intensive computation are necessary.
In this paper we shall consider the Adomian decomposition method (ADM) for solving (PDE). Adomian decomposition method has led to several modifications on the method made by various researchers in attempt to improve the convergence or accuracy of the series solution and this modifications usually involves only a slight change on the standard form . So, in this paper we will also discuss some modifications of (ADM) for solving (PDE).
This study reveals that the (ADM) is very efficient for (PDE) and it results give evidence that high accuracy can be achieved.
2. ADOMIAN DECOMPOSITION METHOD FOR SOLVING NONLINEAR PARTIAL DIFFERENTIAL EQUATION (PDE)
We first consider the nonlinear partial differential equation given in an operator form:
𝐿𝑥𝑢(𝑥, 𝑦) + 𝐿𝑦𝑢(𝑥, 𝑦) + 𝑅(𝑢(𝑥, 𝑦)) + 𝑁(𝑢(𝑥, 𝑦)) = 𝑔(𝑥, 𝑦), (1)
where:
𝐿𝑥 :is the highest order differential in 𝑥.
𝐿𝑦: is the highest order differential in 𝑦.
𝑅: contains the remaining linear terms of lower derivative.
𝑁:is an analytic nonlinear term.
𝑔: represent an inhomogeneous term (source term).
We set
𝐿𝑥𝑢(𝑥, 𝑦) = 𝑔(𝑥, 𝑦) − 𝐿𝑦𝑢(𝑥, 𝑦) − 𝑅(𝑢(𝑥, 𝑦)) − 𝑁(𝑢(𝑥, 𝑦)), (2) applying 𝐿−1𝑥 to both sides of (2) gives
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𝑢(𝑥, 𝑦) = Φ − 𝐿−1𝑥 𝑔(𝑥, 𝑦) − 𝐿−1𝑥 𝐿𝑦𝑢(𝑥, 𝑦) − 𝐿−1𝑥 𝑅(𝑢(𝑥, 𝑦)) − 𝐿−1𝑥 𝑁(𝑢(𝑥, 𝑦)), (3) Where the function Φ represents the terms arising from using the given conditions.
The Adomian decomposition method depend of decomposition the unknown function 𝑢(𝑥, 𝑦) into a sum of infinite number of components defined by the decomposition series:
𝑢(𝑥, 𝑦) = ∑∞𝑛=0𝑢𝑛(𝑥, 𝑦), (4) and the nonlinear term 𝑁(𝑢(𝑥, 𝑦)) by
𝑁(𝑢(𝑥, 𝑦)) = ∑∞𝑛=0𝐴𝑛, (5)
where 𝐴𝑛 are the Adomian polynomials that can be generated for all forms of nonlinearity term by the relation:
𝐴𝑛=
1 𝑛!
𝜕𝑛
𝜕𝜆𝑛[𝑁(∑𝑛 𝜆𝑖𝑢𝑖)]𝜆=0
𝑖=0 , 𝑛 = 0,1,2,3, … (6)
This form is more generally used based on the advantage of a convenient algorithm which is easily remembered.
Recently, F. A. Hendi et. al. [4] presented simple Mathematic program to compute Adomian polynomials.
the first few polynomials are given by
{
𝐴0= 𝑁(𝑢0) 𝐴1= 𝑢1𝑁′(𝑢0) 𝐴2= 𝑢2𝑁′(𝑢0) +1
2!𝑢12𝑁′′(𝑢0) 𝐴3= 𝑢3𝑁′(𝑢0) + 𝑢1𝑢2𝑁′′(𝑢0) +1
3!𝑢13𝑁′′′(𝑢0)
⋮
Substituting (5) and (4) in (3) gives
∑∞𝑛=0𝑢𝑛(𝑥, 𝑦)= Φ − 𝐿𝑥−1𝑔(𝑥, 𝑦) − 𝐿−1𝑥 𝐿𝑦(∑∞𝑛=0𝑢𝑛(𝑥, 𝑦)) − 𝐿−1𝑥 𝑅(∑∞𝑛=0𝑢𝑛(𝑥, 𝑦)) − 𝐿−1𝑥 𝑁(∑∞𝑛=0𝐴𝑛(𝑥, 𝑦)), (7)
the component 𝑢𝑛(𝑥, 𝑦), 𝑛 ≥ 0 of the solution 𝑢(𝑥, 𝑦) can be determined by using the recursive relation
{𝑢0= Φ − 𝐿−1𝑥 𝑔(𝑥, 𝑦),
𝑢𝑘+1= −𝐿−1𝑥 𝐿𝑦𝑢𝑘− 𝐿−1 𝑥𝑅(𝑢𝑘) − 𝐿−1𝑥 (𝐴𝑘), 𝑘 ≥ 0, (8)
having calculated the component 𝑢𝑛(𝑥, 𝑦), 𝑛 ≥ 0,the solution in a series form follows immediately .However, in many cases the exact solution in a closed form may be obtained .
The 𝑛-term approximant φ𝑛defined by φ𝑛= ∑𝑛=1𝑘=0𝑢𝑘 , 𝑢 = lim
𝑛→∞φ𝑛, (9) can be used for numerical approximations .
3. A MODIFIED DECOMPOSITION METHOD BY ADOMIAN (MADM1)
Power series solutions of linear homogeneous differential equation in initial-value problems yield simple recurrence relations for the coefficient, but they are generally not adequate for nonlinear equations, although applicable to some simple cases such as the Riccati equation.
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The Adomian series is actually (as stated by Adomian [5,6] ) a generalized Taylor series about a function rather than a point, and can reduce, in a linear case, to the well-known series. In 1992 Adomian and Rach, combining Adomian polynomials and decomposition with Maclanrin series, they could make a Maclaurin series more useful. the Adomian series resulting from the decomposition method is still superior in convergence properties.
[7,3,8]
To clarify the procedure, consider the general inhomogeneous nonlinear partial differential equation in the operator form
𝐿𝑡𝑡𝑢 + 𝐿𝑥𝑥𝑢 + 𝑅𝑢 + 𝑁𝑢 = 𝑔(𝑥, 𝑡), (10) with initial conditions𝑢(𝑥, 0) = 𝛼1(𝑥) , 𝑢𝑡(𝑥, 0) = 𝛼2(𝑥),
where 𝐿𝑡𝑡 = 𝜕2
𝜕𝑡2 , and𝐿𝑡𝑡−1= ∫ ∫ 𝑑𝑡𝑑𝑡0𝑡 0𝑡 ,applying 𝐿−1𝑡𝑡 to both side of (10)
𝑢(𝑥, 𝑡) = 𝛼1(𝑥) + 𝛼2(𝑥)𝑡 + 𝐿−1𝑡𝑡[𝑔(𝑥, 𝑡) − 𝐿𝑥𝑥𝑢 − 𝑅𝑢 − 𝑁(𝑢)], (11) we proceed the solution 𝑢(𝑥, 𝑦)in a series form
𝑢 = ∑∞𝑛=0∑∞𝑚=0𝑐𝑚,𝑛𝑥𝑚𝑡𝑛 = ∑∞𝑛=0𝑎𝑛(𝑥). 𝑡𝑛, (12)
and the inhomogeneous term by
𝑔 = ∑∞𝑛=0𝑔𝑛(𝑥)𝑡𝑛, (13) and the nonlinear term 𝑁(𝑢) by
𝑁(𝑢) = ∑∞𝑛=0𝐴𝑛𝑡𝑛, (14) substituting (12) , (13) and (14) in (11) gives
∑∞𝑛=0𝑎𝑛(𝑥)𝑡𝑛= 𝛼1(𝑥) + 𝛼2(𝑥)𝑡 + 𝐿−1𝑡𝑡[∑∞𝑛=0𝑔𝑛 (𝑥)𝑡𝑛− ∑∞𝑛=0𝑎𝑛′′(𝑥)𝑡𝑛 −𝑅 ∑∞𝑛=0𝑎𝑛𝑡𝑛−∑∞𝑛=0𝐴𝑛 𝑡𝑛], (15)
by integration and replacing 𝑛 = 𝑛 − 2 in the right hand side, and equality the index we obtain
𝑎0(𝑥) + 𝑎1(𝑥)𝑡 + ∑∞𝑛=2𝑎𝑛(𝑥)𝑡𝑛= 𝛼1(𝑥) + 𝛼2(𝑥)𝑡 +∑ [𝑔𝑛−2(𝑥) − 𝑎𝑛−2′′ (𝑥) − 𝑅𝑎𝑛−2(𝑥) − 𝐴𝑛−2(𝑥)] 𝑡𝑛
𝑛(𝑛−1)
∞𝑛=2 ,
(16) by comparing the coefficient of the same power of t in each said we have
{
𝑎0(𝑥) = 𝛼1(𝑥) 𝑎1(𝑥) = 𝛼2(𝑥) 𝑎𝑛(𝑥) = [𝑔𝑛−2(𝑥) − 𝑎𝑛−2′′ (𝑥) − 𝑅𝑎𝑛−2(𝑥) − 𝐴𝑛−2(𝑥)] 1
𝑛(𝑛−1), 𝑛 ≥ 2
(17)
Having calculated the coefficients 𝑎𝑛, 𝑛 ≥ 0 the solution 𝑛 in a series form defined by (17) follows immediately.
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4. THE MODIFIED DECOMPOSITION METHOD BY WAZWAZ
The assumptions made by Adomian were modified in (1999) by wazwaz [9] . In (2001) wazwaz considered anew modification[3].
4.1. The modified decomposition method (MADM2)
As we know the Adomian decomposition method suggest that the zeroth component 𝑢0 usually defined by function
= Φ + 𝐿−1𝑥 𝑔(𝑥, 𝑦) . but the modified form was established based on the assumption that the function 𝑓 can be divided into two parts namely 𝑓0 and 𝑓1. Under this assumption we set
𝑓 = 𝑓0+ 𝑓1 .
Based on this , we formulate the modified recursive relation as follows :
{
𝑢0= 𝑓0 𝑢1= 𝑓1− 𝐿−1(𝑅𝑢0) − 𝐿−1(𝐴0) 𝑢𝑘+2= −𝐿−1(𝑅𝑢𝑘+1) − 𝐿−1(𝐴𝑘+1), 𝑘 ≥ 0
(18) having calculated the component 𝑢𝑘(𝑥, 𝑦), 𝑛 ≥ 0,the solution in a series form follows immediately.
The choice of 𝑓0 and 𝑓1 Such that 𝑢𝑘 contains the minimal number of terms has a strong influence accelerates the convergence of the solution . The modification demonstrate a rapid convergence of the series solution if compared with standard (ADM) and it may give the exact solution for nonlinear equations by using two iterations only without using the so-called Adomain polynomials.
4.2. The new modification (MADM3)
In the new modification ,Wazwaz replace the process of dividing 𝑓 into two component by a series of infinite components . He therefore suggest that 𝑓 be expressed in Taylor series
𝑓(𝑥) = ∑∞ 𝑓𝑛
𝑛=0 , (19) moreover ,he suggest a new recursive relationship expressed in the form
{𝑢0= 𝑓0,
𝑢𝑘+1= 𝑓𝑘+1− 𝐿−1(𝑅𝑢𝑘) − 𝐿−1(𝐴𝑘), k ≥ 0 , (20) having calculated the component 𝑢𝑘(𝑥, 𝑦), 𝑛 ≥ 0,the solution in a series form follows immediately.
We can observe that algorithm (20) reduces the number of terms involved in each standard (ADM) only . Moreover this reduction of terms in Each component facilitates the construction of Adomian polynomials for nonlinear operators. The new modification overcomes the difficulty of decomposing f(x), and introduces an efficient algorithm that improves the performance of the standard (ADM.).
Note
If 𝑓 consists of one term only , then scheme (20) reduces to relation (8) .Moreover, if 𝑓 consists of two terms, then relation (20) reduces to the modified relation (18).
5. THE TWO-STEP ADOMIAN DECOMPOSITION METHOD (TSADM4)
Another modification of (ADM) , namely two-step (ADM) was introduced in ( 2005)[10,11] . This modification on (ADM) makes the calculations much simpler. The Two-step ADM over comes the difficulty of choice of the parts f1and f 2and explaine how we can choose f1and f 2 . We Present the steps for the (TSADM4):
168 Consider the partial differential equation
𝐿𝑢 + 𝑅𝑢 + 𝑁𝑢 = 𝑔, (21) the main ideas of two-step Adomian decomposition method are:
(1) Applying the inverse operator 𝐿−1 to g and using the given condition we obtain
𝜑 = Φ + 𝐿−1𝑔, (22) to achieve the objectives of this study, we set
𝜑 = 𝜑0+ 𝜑1+ ⋯ + 𝜑𝑚, (23)
where 𝜑0, 𝜑1, … , 𝜑𝑚 are the terms arising from integrating the source term 𝑔 and from using the given conditions.Based on this, we define
𝑢0= 𝜑𝑘+ ⋯ + 𝜑𝑘+𝑠, (24)
where 𝑘 = 0,1, … , 𝑚, 𝑠 = 0,1, … , 𝑚 − 𝑘 ,then we verify that 𝑢0 satisfies the original equation (21) and the given condition by substitution , once the exact solution is obtained we finish. otherwise, we go to following step two.
(2) We set 𝑢0= 𝜑and continue with the standard Adomian recursive relation
{𝑢0= Φ + 𝐿−1𝑔,
𝑢𝑘+1 = −𝐿−1(𝑅𝑢𝑘) − 𝐿−1(𝐴𝑘), 𝑘 ≥ 0, (25)
The Two-step Adomian decomposition method wants to modify the standard Adomian decomposition method and it may provide the solution by using only one iteration and reduce the size of calculation compared to the standard Adomian decomposition .
6. LAPLACE ADOMIAN DECOMPOSITION METHOD
The numerical Laplace transform algorithm which is based on the ADM is used to for solving the linear and nonlinear (PDE) [2,12]. The methods present a useful way to develop an analytic treatment for these problems . The Laplace decomposition method is a powerful tool to search for solution of various nonlinear problems.
We will represent the Laplace Adomian decomposition method and modified Laplace Adomian decomposition method with Wazwaz modification.
6.1. Laplace Adomian decomposition method (LADM5)
Consider the general form of second the partial differential equations with initial conditions given below 𝐿𝑢(𝑥, 𝑡) + 𝑅𝑢(𝑥, 𝑡) + 𝑁𝑢(𝑥, 𝑡) = 𝑔(𝑥, 𝑡), (26)
𝑢(𝑥, 0) = 𝛼1(𝑥), 𝑢𝑡(𝑥, 0) = 𝛼2(𝑥) (27) the methodology consists of applying Laplace transform first on both sides of Eq. (26) ℒ[𝐿𝑢(𝑥, 𝑡)] + ℒ[𝑅𝑢(𝑥, 𝑡) + ℒ[𝑁𝑢(𝑥, 𝑡)] = ℒ[𝑔(𝑥, 𝑡)], (28)
using the differentiation property of Laplace transform, we get ℒ[𝑢(𝑥, 𝑡)] =𝛼1(𝑥)
𝑠 +𝛼2(𝑥)
𝑠2 + 1
𝑠2ℒ[𝑔(𝑥, 𝑡)] −1
𝑠2ℒ[𝑅𝑢(𝑥, 𝑡)] − 1
𝑠2[𝑁𝑢(𝑥, 𝑡)] (29)
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the second step in Laplace decomposition method is that we represent solution as an infinite series given below
𝑢 = ∑∞𝑛=0𝑢𝑛(𝑥, 𝑡), (30) the nonlinear operator is decompose as
𝑁(𝑢) = ∑∞𝑛=0𝐴𝑛, (31) putting Eq. (31) and (30) in Eq. (29) we will get
∑ ℒ[𝑢𝑛(𝑥, 𝑡)] =𝛼1(𝑥)
𝑠
∞𝑛=0 +𝛼2(𝑥)
𝑠2 +1
𝑠2ℒ[𝑔(𝑥, 𝑡)] −1
𝑠2ℒ[𝑅𝑢(𝑥, 𝑡)] − 1
𝑠2ℒ[∑∞𝑛=0𝐴𝑛], (32) the recursive relation is given by
ℒ[𝑢𝑛+1(𝑥, 𝑡)] = −1
𝑠2ℒ[𝑅𝑢𝑛(𝑥, 𝑡)] − 1
𝑠2ℒ[𝐴𝑛], 𝑛 ≥ 1, (33) applying inverse Laplace transform to Eq. (33) so our required recursive relation is given below
{
𝑢0(𝑥, 𝑡) = ℒ−1[𝛼1(𝑥)
𝑠 +𝛼2(𝑥)
𝑠2 + 1
𝑠2ℒ[ℎ(𝑥, 𝑡)]] = ℒ−1[𝐾(𝑥, 𝑠)] = 𝐾(𝑥, 𝑡), 𝑢𝑛+1(𝑥, 𝑡) = −ℒ−1[1
𝑠2ℒ[𝑅𝑢𝑛(𝑥, 𝑡)] − 1
𝑠2ℒ[𝐴𝑛]] , 𝑛 ≥ 1,
(34)
where K(x,t) represent the term arising from source term and prescribe initial conditions.
6.2. Modified Laplace Adomian decomposition method (LADM6)
In modified Laplace transform with Wazwaz modification we assume that 𝐾(𝑥, 𝑡) in (34) can be divided into the sum of two parts namely 𝐾0(𝑥, 𝑡) and 𝐾1(𝑥, 𝑡) , therefore we get
𝐾(𝑥, 𝑡) = 𝐾0(𝑥, 𝑡) + 𝐾1(𝑥, 𝑡), (35)
under this assumption, we propose a slight variation only in the components 𝑢0, 𝑢1 and we formulate the modified recursive algorithm as follows:
{
𝑢0(𝑥, 𝑡) = 𝐾0(𝑥, 𝑡), 𝑢1(𝑥, 𝑡) = 𝐾1(𝑥, 𝑡) − ℒ−1[ 1
𝑠2ℒ [𝑅𝑢0(𝑥, 𝑡) + 1
𝑠2ℒ[𝐴0]] , 𝑢𝑛+1(𝑥, 𝑡) = −ℒ−1[ 1
𝑠2ℒ[𝑅𝑢𝑛(𝑥, 𝑡)] +1
𝑠2ℒ[𝐴𝑛], 𝑛 ≥ 1,
(36)
The solution through the modified Adomian decomposition method is highly depend upon the choice of𝐾0(𝑥, 𝑡)and 𝐾1(𝑥, 𝑡).
The laplace ADM has been successfully applied by researchers to find reliable approximate solutions to nonlinear (PDE)
7. APPLICATIONS
In this section application of Adomian decomposition method and its modifications for nonlinear (PDE) are given as in the illustrative examples , we consider the following tow problems
Example1. Consider the linear partial differential equation 𝑢𝑡𝑡+ 𝑢𝑥𝑥+ 𝑢 = 0,
with initial conditions 𝑢(𝑥, 0) = 1 + sin 𝑥 , 𝑢𝑡(𝑥, 0) = 0.
170 Solution :
In an operator form the Eq.(21) becomes 𝐿𝑡𝑡𝑢(𝑥, 𝑦) = −(𝑢 + 𝑢𝑥𝑥),
where 𝐿𝑡𝑡 = 𝜕2
𝜕𝑡2 , and 𝐿−1𝑡𝑡(. ) = ∫ ∫ (. )𝑑𝑡𝑑𝑡01 01 ,
applying 𝐿−1𝑡𝑡 to both sides of (22) and using the initial condition we obtain 𝑢(𝑥, 𝑡) = 1 + sin 𝑥 − 𝐿−1𝑡𝑡(𝑢 + 𝑢𝑥𝑥).
By using the Adomian decomposition method Based on the recurrence relation (8) we get
{𝑢0= 1 + sin 𝑥, 𝑢𝑘+1= −𝐿−1𝑡𝑡(𝑢𝑘+ 𝑢𝑘𝑥𝑥), 𝑘 ≥ 0, The first few component
𝑢0= 1 + sin 𝑥 ,
𝑢1= −𝐿−1𝑡𝑡(𝑢0+ 𝑢0𝑥𝑥) = −1
2!𝑡2, 𝑢2= −𝐿−1𝑡𝑡(𝑢1+ 𝑢1𝑥𝑥) = 1
4!𝑡4, 𝑢3= −𝐿−1𝑡𝑡(𝑢1+ 𝑢2𝑥𝑥) = −1
6!𝑡6, 𝑢4= −𝐿−1𝑡𝑡(𝑢3+ 𝑢3𝑥𝑥)= 1
8!𝑡8,
⋮
𝑢(𝑥, 𝑡) = sin 𝑥 + 1 −1
2!𝑡2+1
4!𝑡4−1
6!𝑡6+1
8!𝑡8+ ⋯ = sin 𝑥 + cos 𝑡
By using the modified decomposition method(MADM2) we divide 𝑓(𝑥) into two parts
𝑓0= 1 , 𝑎𝑛𝑑 𝑓1= sin 𝑥 ,then we have from the recursive relation {𝑢0= 1,
𝑢𝑘+1= sin 𝑥 − 𝐿−1𝑡𝑡(𝑢𝑘+ 𝑢𝑥𝑥𝑘), 𝑘 ≥ 0,
the first few component from the last recursive relation are 𝑢0= 1 ,
𝑢1= sin 𝑥 − 𝐿−1𝑡𝑡(𝑢0+ 𝑢𝑥𝑥) = sin 𝑥 −1
2!𝑡2, 𝑢2= −𝐿−1𝑡𝑡(𝑢1+ 𝑢1𝑥𝑥) = 1
4!𝑡4, 𝑢3= −𝐿−1𝑡𝑡(𝑢1+ 𝑢2𝑥𝑥) = −1
6!𝑡6, 𝑢4= −𝐿−1𝑡𝑡(𝑢3+ 𝑢3𝑥𝑥)= 1
8!𝑡8,
⋮
𝑢(𝑥, 𝑡) = sin 𝑥 + 1 −1
2!𝑡2+1
4!𝑡4−1
6!𝑡6+1
8!𝑡8+ ⋯ = sin 𝑥 + cos 𝑡
By using the New modification (MADM3) : the Taylor expansion for 𝑓(𝑥) = 1 + sin 𝑥 is : 𝑓(𝑥) = 1 + 𝑥 −1
3!𝑥3+1
5!𝑥5−1
7!𝑥7+ ⋯,
171 then we have from the recursive relation
𝑢0= 1
𝑢1= 𝑥 − 𝐿−1𝑡 (𝑢0+ 𝑢0𝑥𝑥) = 𝑥 −1
2!𝑡2 𝑢2= −1
3!𝑥3− 𝐿𝑡−1(𝑢1+ 𝑢1𝑥𝑥) =−1
3!𝑥3−1
2𝑥𝑡2+1
4!𝑡4 𝑢3= 1
5!𝑥5− 𝐿−1𝑡 (𝑢2+ 𝑢2𝑥𝑥) =1
5!𝑥5−1
6!𝑡6+ 1
24𝑥𝑡4+1
2𝑥𝑡2+ 1
12𝑥3𝑡2
⋮ 𝑢(𝑥, 𝑡) = (𝑥 −1
3!𝑥3+1
5!𝑥5− ⋯ ) + (1 −1
2!𝑡2+1
4!𝑡4+ ⋯ ) 𝑢(𝑥, 𝑡) = sin 𝑥 + cos 𝑡.
An important conclusion that can be made here is that the exact solution was accelerate by using the modification more than the standard Adomian method .
Example 2. Consider the linear equation 𝑢𝑡𝑡− 𝑢𝑥𝑥 = 2(𝑥2− 𝑡2)
With initial condition 𝑢(𝑥, 0) = sinh 𝑥 𝑢𝑡(𝑥, 0) = cosh 𝑥 Solution :
Based on the above discussed in section (3),and from recurrence relation (17) we get
{
𝑎0(𝑥) = sinh 𝑥 𝑎1(𝑥) = cosh 𝑥 𝑎𝑛(𝑥) =𝑔𝑛−2(𝑥)+𝑎𝑛−2′′ (𝑥)
𝑛(𝑛−1) , 𝑛 ≥ 2
Where, 𝑔0(𝑥) = 2𝑥2 𝑔1(𝑥) = 0 𝑔2(𝑥) = −2 𝑔𝑛(𝑥) = 0, 𝑛 ≥ 3 the first few terms 𝑎0(𝑥) = sinh 𝑥 𝑎1(𝑥) = cosh 𝑥 𝑎2(𝑥) =𝑔0(𝑥)+𝑎0′′(𝑥)
2 = 𝑥2+sinh 𝑥
2 𝑎3(𝑥) =𝑔1(𝑥)+𝑎1′′(𝑥)
6 = 1
3!cosh 𝑥 𝑎4(𝑥) =𝑔2(𝑥)+𝑎2′′(𝑥)
12 = 1
4!sinh 𝑥 𝑎5(𝑥) =𝑔3(𝑥)+𝑎3′′(𝑥)
20 = 1
5!cosh 𝑥
⋮
The exact solution is in the form
𝑢 = ∑∞𝑛=0𝑎𝑛(𝑥). 𝑡𝑛= 𝑎0(𝑥) + 𝑎1(𝑥)𝑡 + 𝑎2(𝑥)𝑡2+ 𝑎3(𝑥)𝑡3+ 𝑎4(𝑥)𝑡4+ ⋯ = 𝑠𝑖𝑛ℎ 𝑥 + 𝑡 𝑐𝑜𝑠ℎ 𝑥 + 𝑥2𝑡2+𝑡2
2!𝑠𝑖𝑛ℎ 𝑥 +𝑡3
3!𝑐𝑜𝑠ℎ 𝑥 +𝑡4
4!𝑠𝑖𝑛ℎ 𝑥 +𝑡5
5!𝑐𝑜𝑠ℎ 𝑥 +𝑡6
6!𝑠𝑖𝑛ℎ 𝑥 + ⋯
= 𝑥2𝑡2+ 𝑠𝑖𝑛ℎ 𝑥 (1 +𝑡2
2!+𝑡4
4!+𝑡6
6!… ) + 𝑐𝑜𝑠ℎ 𝑥 (𝑡 +𝑡3
3!+𝑡5
5!… ) = 𝑥2𝑡2+ 𝑠𝑖𝑛ℎ(𝑥 + 𝑡)
172 Example 3. Consider the partial differential equation
𝑢𝑥𝑥+ (1 − 2𝑥)𝑢𝑥𝑦+ (𝑥2− 𝑥 − 2)𝑢𝑦𝑦= 0 (37) With the initial condition 𝑢(𝑥, 0) = 𝑥 , 𝑢𝑦(𝑥, 0) = 1
In an operator form the equation (37) becomes 𝐿𝑦𝑦𝑢(𝑥, 𝑦) = − (1−2𝑥)
(𝑥2−𝑥−2)𝑢𝑥𝑦− 1
(𝑥2−𝑥−2)𝑢𝑥𝑥 (38) Where 𝐿𝑦𝑦= 𝜕2
𝜕𝑦2 , and 𝐿−1𝑦𝑦= ∫ ∫ (. )𝑑𝑦𝑑𝑦0𝑦 0𝑦
Applying 𝐿𝑦𝑦−1 to both side of (38) and using the initial condition we obtain
𝑢(𝑥, 𝑦) = 𝑥 + 𝑦 − 𝐿𝑦𝑦−1[ (𝑥(1−2𝑥)2
−𝑥−2)𝑢𝑥𝑦−(𝑥21
−𝑥−2)𝑢𝑥𝑥 ] (39) Using the eq. (39) gives:
𝜑 = 𝜑0+ 𝜑1= 𝑥 + 𝑦 𝜑0= 𝑥 , 𝜑1= 𝑦
It is obvious that 𝜑0and 𝜑1do not satisfy Eq.(37). By select 𝑢0= 𝑥 + 𝑦 and verify that 𝑢0 satisfies the eq.(37) and the given conditions. Then the exact solution is
𝑢(𝑥, 𝑦) = 𝑥 + 𝑦 Example4. Consider a nonlinear partial differential equation
𝜕𝑢
𝜕𝑡+ 𝑢𝑢𝑥= 𝑥 + 𝑥𝑡2,
with initial condition𝑢(𝑥, 0) = 0 Solution :
Applying the Laplace transform and using initial condition then we have ℒ[𝑢(𝑥, 𝑡)] = 𝑥
𝑠2+𝑥2!
𝑠4 −1
𝑠ℒ[𝑢𝑢𝑥],
applying inverse Laplace transform we get 𝑢(𝑥, 𝑡) = 𝑥𝑡 +𝑥𝑡3
3 − ℒ−1[1
𝑠ℒ[𝑢𝑢𝑥]],
by (LADM5) and from the recursive relation(34) we have {
𝑢0= 𝑥𝑡 +𝑥𝑡3
3 , 𝑢𝑛+1= −ℒ−1[1
𝑠ℒ[𝐴𝑛(𝑢)]] , 𝑛 ≥ 0 the first few component
𝑢1= −ℒ−1[1
𝑠ℒ[𝑢0𝑢0𝑥]]
= −ℒ−1[1
𝑠ℒ [(𝑥𝑡 +𝑥𝑡3
3 ) (𝑡 +𝑡3
3)] ] = −ℒ−1[1
𝑠ℒ[(𝑥𝑡2+𝑥𝑡4
3 +𝑥𝑡4
3 +𝑥𝑡6
9 )] = −𝑥𝑡3
3 −𝑥𝑡7
63 −2𝑥𝑡5
15 , 𝑢2= −ℒ−1[1
𝑠ℒ[𝐴1]]
= −ℒ−1[1
𝑠ℒ[(−𝑥𝑡3
3 −𝑥𝑡7
63 −2𝑥𝑡5
15 ) (𝑡 +𝑡3
3) + (𝑥𝑡 +𝑥𝑡3
3 ) (−𝑡3
3 −𝑡7
63−2𝑡5
15)]
=2𝑥𝑡5
15 +22𝑥𝑡7
315 +38𝑥𝑡9
2835 +2𝑥𝑡11
2079, the exact solution is
𝑢(𝑥, 𝑡) = ∑∞𝑛=0𝑢𝑛(𝑥, 𝑡) = 𝑥𝑡.
Another way by (MLADM6) and from the recursive relation(36) we have
173 {
𝑢0= 𝑥𝑡, 𝑢1=𝑥𝑡3
3 − ℒ−1[1
𝑠ℒ[𝐴0(𝑢)]], 𝑢𝑛+1= −ℒ−1[1
𝑠ℒ[∑∞𝑛=0𝐴0(𝑢)] ], 𝑛 ≥ 1,
the first few component are given by 𝑢0= 𝑥𝑡,
𝑢1=𝑥𝑡3
3 −𝑥𝑡3
3 = 0, 𝑢𝑛+1= 0 , 𝑛 ≥ 1, then the exact solution is 𝑢(𝑥, 𝑡) = ∑∞𝑛=0𝑢𝑛(𝑥, 𝑡) = 𝑥𝑡.
The above example has been solved by two ways, first with Laplace decomposition and secondly with modified Laplace decomposition method. The "noise terms" appear in the component of the solution series obtained by Laplace decomposition method. While solution obtained by modified Laplace decomposition method does not contain any noise terms.
This modified technique has been shown are important in the field of applied sciences. In addition, the modified Laplace decomposition method may give the exact solution for nonlinear partial differential equations or system of partial differential equations .
8. CONCLUSIONS
We applied some modification of ADM for solving nonlinear (PDE). A comparative between the modifications methods and the standard decomposition method is present from some examples to show the efficiency of each method. The computations associated with the examples discussed above were performed by using Maple 17.
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