the combined of homotopy analysis method with …...the combined of homotopy analysis method with...

Malaya Journal of Matematik, Vol. 6, No. 1, 34-40, 2018

https://doi.org/10.26637/MJM0601/0005

The combined of Homotopy analysis method withnew transform for nonlinear partial differentialequationsDjelloul Ziane1*

AbstractThe idea proposed in this work is to extend the Aboodh transform method to resolve the nonlinear partialdifferential equations by combining them with the so-called homotopy analysis method (HAM). This method canbe called homotopy analysis aboodh transform method (HAATM). The results obtained by the application to theproposed examples show that this method is easy to apply and can therefore be used to solve other nonlinearpartial differential equations.

KeywordsHomotopy analysis method, Aboodh transform method, nonlinear partial differential equations.

AMS Subject Classification26A33, 44A05, 34K37, 35F61.

1Laboratory of mathematics and its applications (LAMAP), University of Oran1 Ahmed Ben Bella, Oran, 31000, Algeria.*Corresponding author: 1 [email protected];Article History: Received 24 November 2017; Accepted 19 December 2017 c©2017 MJM.

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2 Basic definitions and theorems . . . . . . . . . . . . . . . . . . . . 35

3 Homotopy analysis Aboodh transform method (HAATM)35

4 Application of the HAATM . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1. IntroductionNonlinear equations are of great importance to our contem-porary world. Nonlinear phenomena have important applica-tions in applied mathematics, physics, and issues related toengineering. Despite the importance of obtaining the exactsolution of nonlinear partial differential equations in physicsand applied mathematics, there is still the daunting problemof finding new methods to discover new exact or approximatesolutions [5]. So we find that a lot of researchers are workingto develop new methods to solve this kind of equations. Theseefforts have strengthened this area of research through manymethods, among them we find, homotopy analysis method(HAM). This method was developed in 1992 by Liao Shijun

of Shanghai Jiaotong University ([10]-[13]). Then, a newoption emerged recently, includes the composition of Laplacetransform, Sumudu transform, Natural transform or Elzakitransform with this method to solve linear and nonlinear ordi-nary partial differential equations. Among which are the ho-motopy analysis method coupled with Laplace transform ([1],[2], [8], [14], [18], [19]), homotopy analysis Sumudu trans-form method ([3], [16], [17]), homotopy Natural transformmethod ([9], [21]) and homotopy analysis Elzaki transformmethod ([20], [22]).

The objective of the present study is to combine two pow-erful methods, homotopy analysis method and Aboodh trans-form method to get a better method to solve nonlinear partialdifferential equations. The modified method is called homo-topy analysis Aboodh transform method (HAATM). Threeexamples are given to re-confirm the effectiveness of thismethod.

The present paper has been organized as follows: In Sec-tion 2 Some basic definitions and properties of the Aboodhtransform method. In section 3 We give an analysis of theproposed method. In section 4 We present three examplesexplaining how to apply the proposed method. Finally, theconclusion follows.

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2. Basic definitions and theoremsIn this section, we give some basic definitions and propertiesof Aboodh transform which are used further in this paper.A new transform called the Aboodh transform defined forfunction of exponential order, we consider functions in the setA, defined by [7]

A ={

f (t) : ∃M, k1,k2 > 0, | f (t)|< Me−vt} . (2.1)

For given function in the set A, the constant M must befinite number, k1,k2 my be finite or infinite.

Definition 2.1. The Aboodh transform denoted by the opera-tor A(·) defined by the integral equation

A [ f (t)] =K(v)=1v

∫∞

0f (t)e−vtdt, t > 0,k1 6 v6 k2. (2.2)

We will summarize here some results of simple functionsrelated to Aboodh transform in the following table [7]

f (t) A [ f (t)] f (t) A [ f (t)]1 1

v2 sinat av(v2+a2)

t 1v3 cosat 1

v2+a2

tn n!vn+2 sinhat a

v(v2−a2)

eat 1v2−av coshat 1

v2−a2

Theorem 2.2. Let K(v) is the Aboodh transform of f (t), thenone has

A[

f ′(t)]= vK(v)− f (0)

v, (2.3)

A[

f′′(t)]= v2K(v)− f ′(0)

v− f (0), (2.4)

A[

f(n)(t)]= vnK(v)−

n−1

∑k=0

f (k)(0)v2−n+k . (2.5)

Proof. (see [7]).

Theorem 2.3. Let K(x,v) is the Aboodh transform of u(x, t),to obtain Aboodh transform of partial derivativewe use inte-gration by parts, and then we have

A[

∂u(x, t)∂ t

]= vK(x,v)− u(x,0)

v, (2.6)

A[

∂ 2u(x, t)∂ t2

]= v2K(x,v)− 1

v∂u(x,0)

∂ t−u(x,0), (2.7)

Proof. (see [4]).

Theorem 2.4. Let K(x,v) is the Aboodh transform of u(x, t),then one has

A[

∂ nu(x, t)∂ tn

]= vnK(x,v)−

n−1

∑k=0

1v2−n+k

∂ ku(x,0)∂ tk . (2.8)

Proof. To demonstrate the validity of the formula (2.8), weuse mathematical induction.

If n = 1 and according to formula (2.8), we obtain

A[

∂u(x, t)∂ t

]= vK(x,v)− u(x,0)

v. (2.9)

So, according to the (2.6) we note that the formula holdswhen n = 1.

Assume inductively that the formula holds for n, so that

A[

∂ nu(x, t)∂ tn

]= vnK(x,v)−

n−1

∑k=0

1v2−n+k

∂ ku(x,0)∂ tk , (2.10)

and show that it stays true at rank n+ 1. Let ∂ nu(x,t)∂ tn =

w(x, t) and according to (2.6) and (2.10) , we have

A[

∂ n+1u(x, t)∂ tn+1

]= A

[∂w(x, t)

∂ t

]= vA(w(x, t))− w(x,0)

v

= v

[vnK(x,v)−

n−1

∑k=0

1v2−n+k

∂ ku(x,0)∂ tk

]− w(x,0)

v

= vn+1K(x,v)−n−1

∑k=0

1v1−n+k

∂ ku(x,0)∂ tk − 1

v∂ nu(x,0)

∂ tn

= vn+1K(x,v)−n

∑k=0

1v1−n+k

∂ ku(x,0)∂ tk .

Thus by the principle of mathematical induction, the for-mula (2.8) holds for all n > 1.

3. Homotopy analysis Aboodh transformmethod (HAATM)

We consider the following general nonlinear partial differen-tial equation as

∂ nX(κ,τ)∂τn +RX(κ,τ)+NX(κ,τ) = f (κ,τ), (3.1)

where n = 1,2,3, ..., with the initial conditions

∂ n−1X(κ,τ)∂τn−1

∣∣∣∣τ=0

= gn−1(κ), n = 1,2,3, ... (3.2)

35


and ∂ nX(x,t)∂τn is the partial derivative of the function X(κ,τ)

of order n (n = 1,2,3...), R is the linear partail differentialoperator, N represents the general nonlinear partail differentialoperator, and f (κ,τ) is the source term.

Applying Aboodh transform on both sides of (3.1), wecan get

A[

∂ nX(κ,τ)∂τn

]+A [RX(κ,τ)+NX(κ,τ)] = A[ f (κ,τ)] ,

(3.3)

Using the property of Aboodh transform, we have thefollowing form

A [X(κ,τ)] −n−1

∑k=0

1v2+k

∂ kX(κ,0)∂τk (3.4)

+1vn A [RX(κ,τ)+NX(κ,τ)− f (κ,τ)] = 0

Define the nonlinear operator

N[φ(κ,τ; p)] = A [ φ(κ,τ; p)]−n−1

∑k=0

1v2+k

∂ kφ(κ,0; p)∂τk (3.5)

+1vn A [Rφ(κ,τ; p)+Nφ(κ,τ; p)− f (κ,τ)]

By means of homotopy analysis method ([10]-[13]), weconstruct the so-called the zero-order deformation equation

(1−q)A[φ(κ,τ; p)−φ(κ,τ;0) = phH(κ,τ)N[φ(κ,τ; p)],

(3.6)

where p is an embedding parameter and p ∈ [0, 1],H(κ,τ) 6= 0 is an auxiliary function, h 6= 0 is an auxiliaryparameter, A is an auxiliary linear Aboodh operator. Whenp = 0 and p = 1, we have

{φ(κ,τ;0) = X0(κ,τ),φ(κ,τ;1) = X(κ,τ). (3.7)

When P increases from 0 to 1, the φ(κ,τ, p) various fromX0(κ,τ) to X(κ,τ). Expanding φ(κ,τ; p) in Taylor serieswith respect to p, we have

φ(x, t; p) = X0(κ,τ)++∞

∑m=1

Xm(κ,τ)pm, (3.8)

where

Xm(κ,τ) =1

m!∂ mφ(κ,τ; p)

∂ pm

∣∣p=0 (3.9)

When p = 1, the formula (3.8) becomes

X(κ,τ) = X0(κ,τ)++∞

∑m=1

Xm(κ,τ). (3.10)

Define the vectors

−→X = {X0(κ,τ),X1(κ,τ),X2(κ,τ), . . . ,Xm(κ,τ)}. (3.11)

Differentiating (3.6) m−times with respect to p, then set-ting p = 0 and finally dividing them by m!, we obtain theso-called mth order deformation equation

A[Xm(κ,τ)−χmXm−1(κ,τ)] = hH(κ,τ)ℜm(−→X m−1(κ,τ)),

(3.12)

where

ℜm(−→X m−1(κ,τ)) =

1(m−1)!

∂ m−1N(κ,τ; p)∂ pm−1

∣∣p=0 , (3.13)

and

χm =

{0, m 6 1,1, m > 1.

Applying the inverse Aboodh transform on both sides of(3.12), we can obtain

Xm(κ,τ)= χmXm−1(κ,τ)+hA−1[H(κ,τ)ℜm(

−→X m−1(κ,τ))

].

(3.14)

The mth deformation equation (3.14) is a linear which canbe easily solved. So, the solution of (3.1) can be written intothe following form

X(κ,τ) =N

∑m=0

Xm(κ,τ), (3.15)

when N→ ∞, we can obtain an accurate approximationsolution of (3.1).

For the proof of the convergence of the homotopy analysismethod (see [11]).

4. Application of the HAATMIn this section, we apply the homotopy analysis method (HAM)coupled with Aboodh transform method for solving sommeexemples of nonlinear partial differential equations.

36


Example 4.1. First, we consider the following nonlinear gasdynamics equation

Xτ +XXκ−X(1−X) = 0, τ > 0, (4.1)

with the initial condition

X(κ,0) = e−x. (4.2)

In view of the HAM technique and assuming H(κ,τ) = 1,we construct the the so-called the zero-order deformationequation as follows

(1− p)A[φ(κ,τ; p)−X0(κ,τ)] = phN[φ(κ,τ; p)], (4.3)

where

N[φ(κ,τ, p)]=A [φ(κ,τ; p)]− 1v2 e−x+

1v

A [XXκ−X(1−X)] .

(4.4)

The series solution of Eq.(4.1) is given by (3.10). Thus,we obtain the m-th order deformation equation

Xm(κ,τ) = χmXm−1(κ,τ)+hA−1[ℜm(−→X m−1(κ,τ))]. (4.5)

with

ℜm(−→X m−1(κ,τ)) = A [Xm−1(κ,τ)]−

1v2 (1−χm)e−x

+1v

A

[m−1

∑i=0

Xi(Xm−1−i)κ

]

+1v

[A

m−1

∑i=0

XiXm−1−i−Xm−1

],(4.6)

and

χm =

{0, m 6 1,1, m > 1. (4.7)

According to (4.5) and (4.6), the formulas of the first termsis given by

X1(x,τ) = hA−1(

1v

A[X0(X0)κ +(X0)

2−X0])

,

X2(x,τ) = (1+h)X1(κ,τ)

+hA−1(

1v

A [X0(X1)κ +X1(X0)κ +2X0X1−X1]

),

X3(x,τ) = (1+h)X2(κ,τ) (4.8)

+hA−1(

1v

A [X0(X2)κ +X1(X1)κ]

)+hA−1

(1v

A[X2(X0)κ +2X0X2 +(X1)

2−X2])

,

...

Using the initial condition (4.2) and the iteration formulas(4.8), we obtain

X0(κ,τ) = e−κ,

X1(κ,τ) = (−h)e−xτ,

X2(κ,τ) = (−h)(1+h)e−xτ +h2e−x τ2

2!, (4.9)

X3(κ,τ) = (−h)(1+h)2e−xτ +2(1+h)h2e−x τ2

2!+(−h3)e−x τ3

3!,

...

The other components of the (HAATM) can be determinedin a similar way. Finally, the approximate solution of Eq.(4.1)in a series form

X(κ,τ) = X0(κ,τ)+X1(κ,τ)+X2(κ,τ)+X3(κ,τ)+ · · ·= e−κ +

[2(−h)−h2 +(−h)(1+h)2]e−x

τ (4.10)

+[h2 +2h2(1+h)

]e−x τ2

2!+(−h3)e−x τ3

3!+ · · ·

Substiting h = −1 in (4.10), the approximate solution of(4.1), given as follows

X(κ,τ) = e−κ[

1+ τ +τ2

2!+

τ3

3!+ · · ·

]. (4.11)

And in the closed form, is given by

X(κ,τ) = e−κeτ = eτ−κ. (4.12)

This result represents the exact solution of the equation(4.1) as presented in [15].

37


Example 4.2. Second, we consider the inhomogeneous non-linear Klein Gordon equation

Xττ −Xκκ +X2 = κ2τ

2, (4.13)

with the initial conditions

X(κ,0) = 0, Xτ(κ,0) = κ. (4.14)



where

N[φ(κ,τ, p)]=A [φ(κ,τ; p)]− 1v3κ+

1v2 A

[−Xκκ +X2−κ2

τ2] .

(4.16)

The series solution of (4.13) is given by (3.10). Thus, weobtain the m-th order deformation equation

Xm(κ,τ)= χmXm−1(κ,τ)+hA−1[ℜm(−→X m−1(κ,τ))], (4.17)

with

ℜm(−→X m−1(κ,τ)) = A [Xm−1(κ,τ)]−

1v3 (1−χm)κ

+1v2 A [−(Xm−1)κκ] (4.18)

+1v2 A

[m−1

∑i=0

XiXm−1−i−κ2τ

2

],

and

χm =

{0, m 6 1,1, m > 1. (4.19)

According to (4.17) and (4.18), the formulas of the firstterms is given by

X1(x,τ) = hA−1(

1v2 A

[−(X0)κκ +X2

0 −κ2τ

2]) ,

X2(x,τ) = (1+h)X1(κ,τ)

+hA−1(

1v2 A [−(X1)κκ +2X0X1]

), (4.20)

X3(x,τ) = (1+h)X2(κ,τ)

+hA−1(

1v2 A

[−(X2)κκ +2X0X2 +X2

1])

,

...

Using the initial condition (4.14) and the iteration formu-las (4.20), we obtain

X0(κ,τ) = κτ,

X1(κ,τ) = 0,X2(κ,τ) = 0, (4.21)X3(κ,τ) = 0,

...

Finally, the approximate solution of Eq.(4.13) is given by

X(κ,τ) = X0(κ,τ)+X1(κ,τ)+X2(κ,τ)+X3(κ,τ)+ · · ·(4.22)= κτ +0+0+0+ · · ·

According to formula (3.15), we obtain

X(κ,τ) = κτ. (4.23)

This result represents the exact solution of the Klein Gor-don equation (4.13) as presented in [6].

Example 4.3. Finaly, we consider the nonlinear partial dif-ferential equation of third order

Xτττ −38[(Xκκ)

2]κ =

32

τ, (4.24)

with the initial conditions

X(κ,0) =−12κ2, Xτ(κ,0) =

13κ3, Xττ(κ,0) = 0. (4.25)



where

N[φ(κ,τ, p)] = A [φ(κ,τ; p)]+1

2v2κ2− 1

3v3κ3

+1v3 A

[−3

8[(Xκκ)

2]κ−

32

τ

].(4.27)

The series solution of (4.24) is given by (3.10). Thus, weobtain the m-th order deformation equation

Xm(κ,τ)= χmXm−1(κ,τ)+hA−1[ℜm(−→X m−1(κ,τ))], (4.28)

with

38


ℜm(−→X m−1(κ,τ)) = A [Xm−1(κ,τ)]

−(− 12v2κ

2 +1

3v3κ3)(1−χm)

+1v3 A

[−3

8

m−1

∑i=0

XiκκX(m−1−i)κκ

]

− 1v3 A

[32

τ

], (4.29)

and

χm =

{0, m 6 1,1, m > 1. (4.30)

According to (4.28) and (4.29), the formulas of the firstterms is given by

X1(x,τ) = hA−1(

1v3 A

[−3

8[(X0κκ)

2]κ−

32

τ

]),

X2(x,τ) = (1+h)X1(κ,τ)

+hA−1(

1v3 A

[−3

4(X0κκX1κκ)κ

]), (4.31)

X3(x,τ) = (1+h)X2(κ,τ)

+hA−1(

1v3 A

[−3

8(2X0κκX2κκ +(X1κκ)

2)]) ,

...

and so on.Consequently, while taking the initial conditions (4.25),

and according to the Eqs.(4.31), the first few components ofthe homotopy analysis Aboodh transform method of Eq.(4.24),are derived as follows

X0(x, t) = −12

x2 +13

x3τ,

X1(x,τ) = (−h)κτ5

20,

X2(x,τ) = (1+h)X1, (4.32)X3(x,τ) = (1+h)X2,

...

The other components of the (HAATM) can be deter-mined in a similar way. Finally, the approximate solutionof Eq.(4.24) in a series form

X(κ,τ) = X0(κ,τ)+X1(κ,τ)+X2(κ,τ)+X3(κ,τ)+ · · · (4.33)

= −12

x2 +13

x3τ +(−h)κ

τ5

20+(1+h)X1 +(1+h)X2 + · · ·

Substiting h = −1 in (4.33), we obtain

X(κ,τ) = −12

x2 +13

x3τ +κ

τ5

20+0+0+ · · ·

= −12

x2 +13

x3τ +κ

τ5

20. (4.34)

This result represents the exact solution of the equation(4.24) as presented in [21] in the case α = 3.

5. ConclusionThe coupling of homotopy analysis method (HAM) and theAboodh transform method, proved very effective to solve non-linear partial differential equations. The proposed algorithmprovides the solution in a series form that converges rapidlyto the exact solution if it exists. From the obtained results, itis clear that the HAATM yields very accurate solutions usingonly a few iterates. The results show that the homotopy anal-ysis Aboodh transform method (HAATM) is an appropriatemethod for solving nonlinear partial differential equations.

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