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Mathematical Theory and Modeling                                                                  www.iiste.org
ISSN 2224-5804 (Paper)    ISSN 2225-0522 (Online)
Vol.2, No.3, 2012


    Homotopy Perturbation and Elzaki Transform for Solving
                    Nonlinear Partial Differential Equations
                                   Tarig. M. Elzaki1* & Eman M. A. Hilal2
      1. Mathematics Department, Faculty of Sciences and Arts-Alkamil, King Abdulaziz University,
                                             Jeddah-Saudi Arabia.
   Mathematics Department, Faculty of Sciences, Sudan University of Sciences and Technology-Sudan.
           2. Mathematics Department, Faculty of Sciences for Girles King Abdulaziz University
                                              Jeddah-Saudi Arabia
                 * E-mail of the corresponding author: Tarig.alzaki@gmail.com


Abstract
In this work, we present a reliable combination of homotopy perturbation method and Elzaki transform to
investigate some nonlinear partial differential equations. The nonlinear terms can be handled by the use of
homotopy perturbation method. The proposed homotopy perturbation method is applied to the reformulated
first and second order initial value problem which leads the solution in terms of transformed variables, and
the series solution is obtained by making use of the inverse transformation. The results show the efficiency
of this method.
Keywords: Homotopy perturbation methods, Elzaki transform nonlinear partial differential equations.


 1. Introduction
Linear and nonlinear partial differential equations are of fundamental importance in science and
engineering. Some integral transform method such as Laplace and Fourier and Sumudu transforms methods
see [Kilicman and E.Gadain. (2009), (2010) lslam, Yasir Khan, Naeem Faraz and Francis Austin (2010)],
are used to solve linear partial differential equations and use fullness of these integral transform lies in their
ability to transform differential equations into algebraic equations which allows simple and systematic
solution procedures.
However, using integral transform in nonlinear differential equations may increase its complexity. In recent
years, many research workers have paid attention to find the solutions of nonlinear differential equations by
using various methods. Among these are the Adomian decomposition method [Hashim, Noorani, Ahmed.
Bakar. Ismail and Zakaria, (2006)], the tanh method, the homotopy perturbation method [ Sweilam, Khader
(2009), Sharma and Giriraj Methi (2011), Jafari, Aminataei (2010), (2011) ], the differential transform
method [(2008)], and the variational iteration method.
Elzaki transform [ Tarig and Salih, (2011), (2012)] is totally incapable of handling the nonlinear equations
because of the difficulties that are caused by the nonlinear terms. Various ways have been proposed recently
to deal with these nonlinearities, one of these combinations of homotopy perturbation method and Elzaki
transform which is studies in this paper.
The advantage of this method is its capability of combining two powerful methods for obtaining exact


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Mathematical Theory and Modeling                                                                        www.iiste.org
ISSN 2224-5804 (Paper)    ISSN 2225-0522 (Online)
Vol.2, No.3, 2012
solutions for nonlinear partial differential equations. This article considers the effectiveness of the
homotopy perturbation Elzaki transform method in solving nonlinear partial differential equations both
homogeneous and non-homogeneous.

1.1. Elzaki Transform
The basic definitions of modified of Sumudu transform or Elzaki transform is defined as follows, Elzaki

transform of the function   f (t ) is

                                            ∞               t
                                                       −
                             E [ f (t )] = v ∫ f (t ) e v dt ,              t >0                                    (1)
                                            0

Tarig M. Elzaki and Sailh M. Elzaki in [(2011), (2012)], showed the modified of Sumudu transform
[(2007), (2010)] or Elzaki transform was applied to partial differential equations, ordinary differential
equations, system of ordinary and partial differential equations and integral equations. Elzaki transform is a
powerful tool for solving some differential equations which can not solve by Sumudu transform in [(2012)].
In this paper, we combined Elzaki transform and homotopy perturbation to solve nonlinear partial
differential equations.
To obtain Elzaki transform of partial derivative we use integration by parts, and then we have:

  ∂f (x , t )  1                                                ∂ 2 f (x , t )  1                              ∂f (x , 0)
E               = T (x ,v ) −vf (x , 0)                        E                 = v 2 T (x ,v ) − f (x , 0) −v
  ∂t  v                                                         ∂t                                                 ∂t
                                                                          2
                                                                                 
   ∂f (x , t )  d                                               ∂ 2f (x , t )  d 2
 E               = [T (x ,v )]                                 E                 = 2 [T (x ,v )]
   ∂x  dx                                                       ∂x
                                                                           2
                                                                                  dx
Proof:
 To obtain ELzaki transform of partial derivatives we use integration by parts as follows:


   ∂f          ∞  ∂f −t
                                   p  −t
                                         ∂f             −t
                                                                     
                                                                        p p −t               
                                                                                             
Ε  ( x ,t ) = ∫ v    e dt = lim ∫ve
                        v             v
                                            dt = lim  ve f (x , t )  − ∫ e v f (x , t )dt 
                                                          v

   ∂t          0  ∂t        p →∞
                                   0
                                         ∂t      p →∞
                                                      
                                                                     0 0                   
                                                                                             
                                               T ( x ,v )
                                           =                    −vf ( x , 0 )
                                                   v
We assume that   f is piecewise continuous and it is of exponential order.
             ∂f   
                       ∞    −t
                               ∂f ( x , t )       ∂
                                                    ∞    −t

                    = ∫ ve                           ve v f ( x , t ) dt , using the Leibnitz rule to find:
                                                 ∂x ∫
  Now      Ε               v
                                            dt =
             ∂x    0            ∂x                0


                                           ∂f     d
                                         Ε        = dx T ( x ,v ) 
                                                                    
                                           ∂x    

                                       ∂ 2f  d 2
 By the same method we find:        Ε  2  = 2 T ( x ,v ) 
                                                           
                                       ∂x  dx

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Vol.2, No.3, 2012
                    ∂2 f       
To find:         Ε  2 ( x, t ) 
                    ∂t         
             ∂f
Let             = g , then we have:
             ∂t
                       ∂ 2f             ∂g ( x , t ) 
                    Ε  2 ( x , t ) = Ε                    g ( x , t )  −vg x , 0
                                                                         
                                                        =Ε                    ( )
                       ∂t                  ∂t                 v

                        ∂ 2f        1                                ∂f
                     Ε  2 ( x , t ) = 2 T ( x , v ) − f ( x , 0 ) −v    ( x , 0)
                        ∂t          v                                ∂t
We can easily extend this result to the nth partial derivative by using mathematical induction.


1.2. Homotopy Perturbation Method:

Let X and Y be the topological spaces. If f and g are continuous maps of the space X into Y , it is


said that f is homotopic to         g , if there is continuous map F : X × [0,1] →Y such that

F (x , 0) = f (x ) and F (x ,1) = g (x ) , for each x ∈ X , then the map is called homotopy between

f and g .
 To explain the homotopy perturbation method, we consider a general equation of the type,

                                                     L (u ) = 0                                            (2)


Where L is any differential operator, we define a convex homotopy H (u , p ) by


                                     H (u , p ) = (1 − p )F (u ) + pL (u )                                 (3)


Where F (u ) is a functional operator with known solution         v 0 which can be obtained easily. It is clear
that, for

                               H (u , p ) = 0                                                              (4)


We have:    H (u , 0) = F (u ) , H (u ,1) = L (u )

In topology this show that H (u , p ) continuously traces an implicitly defined carves from a starting point


H (v 0 , 0) to a solution function H (f ,1) . The HPM uses the embed ling parameter p as a small
parameter and write the solution as a power series

                                                       35
Mathematical Theory and Modeling                                                                         www.iiste.org
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Vol.2, No.3, 2012

                                  u = u 0 + pu 1 + p 2u 2 + p 3u 3 + .......
(5)

If p → 1 , then (5) corresponds to (3) and becomes the approximate solution of the form,

                                                            ∞
                                           f = lim u = ∑ u i                                                       (6)
                                                  p →1     i =0

We assume that (6) has a unique solution. The comparisons of like powers of p give solutions of various
orders.
2. Homotopy Perturbation Elzaki Transform Method:
Consider a general nonlinear non-homogenous partial differential equation with initial conditions of the
form:

                                Du (x , t ) + Ru (x , t ) + Nu (x , t ) = g (x , t )                               (7)


                                      u (x , 0) = h (x ) ,        u t (x , 0) = f (x )

Where D is linear differential operator of order two, R is linear differential operator of less order

than D , N is the general nonlinear differential operator and           g (x , t ) is the source term.
Taking Elzaki transform on both sides of equation (7), to get:

                             E [Du (x , t )] + E [Ru (x , t )] + E [Nu (x , t )] = E [ g (x , t )]                (8)

Using the differentiation property of Elzaki transforms and above initial conditions, we have:

          E [u (x , t )] = v 2 E [ g (x , t )] + v 2 h (x ) + v 3f (x ) −v 2 E [Ru (x , t ) + Nu (x , t )]        (9)

Appling the inverse Elzaki transform on both sides of equation (9), to find:

                                 u (x , t ) = G (x , t ) − E −1 {v 2 E [ Ru (x , t ) + Nu (x , t )]}              (10)


Where G ( x , t ) represents the term arising from the source term and the prescribed initial conditions.

Now, we apply the homotopy perturbation method.
                                                                          ∞
                                                            u ( x , t ) = ∑ p nu n ( x , t )                     (11)
                                                                         n =0

And the nonlinear term can be decomposed as
                                                                           ∞
                                                         N [u (x , t )] = ∑ p n H n (u )                          (12)
                                                                          n =0


Where     H n (u ) are given by:

                                                            36
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Vol.2, No.3, 2012
                                                           1 ∂            ∞ i 
                             H n (u 0 ,u 1 ,.....u n ) =                 N  ∑ p u i   , n = 0,1, 2,......
                                                           n ! ∂p n       i =0        p =0
(13)
Substituting equations (11) and (12) in equation (10), we get:
       ∞                                  
                                                             ∞ n                   ∞
                                                                                                     
       ∑ p nu n (x , t ) = G (x , t ) − p E −1 v 2 E
                                          
                                                               R ∑ p u n (x , t ) + ∑ p n H n (u )   
                                                               n =0                                
                                                                                                                 (14)
       n =0                                                                        n =0               
This is the coupling of the Elzaki transform and the homotopy perturbation method.
Comparing the coefficient of like powers of p , the following approximations are obtained.

p 0 : u 0 (x , t ) =G (x , t )
p 1 :u 1 (x , t ) = − E −1 {v 2 E [ Ru 0 (x , t ) + H 0 (u ) ]}
p 2 :u 2 (x , t ) = −E −1 {v 2 E [ Ru1 (x , t ) + H 1 (u )]}
p 3 :u 3 (x , t ) = −E −1 {v 2 E [ Ru 2 (x , t ) + H 2 (u )]}                                                    (15)

.                      .                                       .
.                      .                                       .
.                      .                                       .

Then the solution is       u (x , t ) = u 0 (x , t ) , +u1 (x , t ) + u 2 (x , t ) + ....

3. Applications:
In this section we apply the homotopy perturbation Elzaki transform method for solving nonlinear partial
differential equations.
Example 3.1:
Consider the following homogenous nonlinear partial differential equations

                                                 u t + uu x = 0 , u (x , 0) = − x                                (16)

Taking Elzaki transform of equation (16) subject to the initial condition, we have:

                                                    E [u (x , t ) ] = xv 2 − vE [uu x ]                          (17)


The inverse Elzaki transform implies that:

                                                     u (x , t ) = −x − E −1 {vE [uu x ]}                         (18)


    Now applying the homotopy perturbation method, we get:
                      ∞                            
                                                          ∞                  
                     ∑    p nu n (x , t ) = −x − p  E −1 vE  ∑ p n H n (u )                                 (19)
                     n =0                          
                                                           n =0              

                                                                   37
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Where   H n (u ) are He's polynomials that represents the nonlinear terms.
Or

         p [uu x ] = 0 ⇒ p (u 0 + pu 1 + p 2u 2 + ....)(u 0 x + pu 1x + p 2u 2 x + ....)
                                               u = u 0 + pu 1 + p 2u 2 + ....
The first few components of He's polynomials, are given by
                                           H 0 (u ) = u 0u 0 x
                                           H 1 (u ) = u 0u 1x + u1u 0 x
                                           H 2 (u ) = u 0u 2 x + u1u1x + u 2u 0 x
                                                                .
                                                                .
                                                                .
Comparing the coefficients of the same powers of p , we get:
                    p 0 : u 0 (x , t ) = −x            ,    H 0 (u ) = x
                    p :u 1 (x , t ) = − E
                        1                      −1
                                                    {vE [ H (u )]} = −xt
                                                            0                   , H 1 (u ) = 2xt
                    p 2 :u 2 (x , t ) = −E −1        {vE [ H (u )]} = −xt
                                                            1
                                                                            2
                                                                                , H 2 (u ) = 3xt 2
                    p 3 :u 3 (x , t ) = −E −1       {vE [ H (u )]} = −xt
                                                            2
                                                                            3


                    .                      .                                       .
                    .                      .                                       .
                    .                      .                                       .

Therefore the solution      u (x , t ) is given by:

                                                                                         x
                                    u (x , t ) = −x (1 + t + t 2 + t 3 + .....) =
                                                                                       t −1
Example 3.2:
Consider the first order nonlinear partial differential equation

                 u t + uu x = 2t + x + t 3 + xt 2 , u (x , 0) = −x                                            (20)

To find the solution by homotopy perturbation Elzaki transform method, we applying homotopy
perturbation method after taking Elzaki and inverse Elzaki transforms of equation (20), we get:
           ∞
                                                    t 4 xt 3     
                                                                        ∞                  
          ∑ p nu n (x , t ) = t 2 + xt +               +     − p  E −1 vE  ∑ p n H n (u )               (21)
          n =0                                       4   3       
                                                                         n =0              
Where




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Vol.2, No.3, 2012
                                            H 0 (u ) = u 0u 0 x
                                            H 1 (u ) = u 0u1x + u 1u 0 x
                                            H 2 (u ) = u 0u 2 x + u 1u 1x + u 2u 0 x
                                                             .
                                                             .
                                             .
Comparing the coefficients of like powers of p , we have:

                                                   xt 3 t 4
                     p : u 0 (x , t ) = t + xt +
                          0                       2
                                                        +
                                                    3      4
                                            4    3        5
                                          t   xt      2xt     7t 6 xt 7 t 8
                     p 1 :u 1 (x , t ) = − −       −        −     −     −
                                           4   3       15     72 63 98
                    .                       .                          .
                    .                             .                                    .
                    .                             .                                    .

The noise terms appear between the components                     u 0 (x , t ) ,u 1 (x , t ) , therefore, the exact solution is given

by:   u (x , t ) = t 2 + xt
Example 3.3:
Let us consider the second order nonlinear partial differential equation

                              ∂u  ∂u     ∂ 2u
                                              2

                                =     +u                        ,        u (x , 0) = x 2                                      (22)
                              ∂t  ∂x     ∂x 2
Applying Elzaki transform of equation (22), and making use of the initial condition, to find:

                               E [u (x , t ) ] = x 2v 2 + vE u x2 + uu xx 
                                                                                                                              (23)

Take the inverse Elzaki transform of equation (23), we get:

                                                         {
                              u (x , t ) = x 2 + E −1 vE u x2 + uu xx 
                                                                               }                                              (24)

Apply the homotopy perturbation method of (24), to get:
                    ∞                                    
                                                                ∞                  
                   ∑p u       n
                                  n   (x , t ) = x 2 + p  E −1 vE  ∑ p n H n (u )                                         (25)
                   n =0                                  
                                                                 n =0              

Or                      p u x2 + uu xx  = 0 , u = u 0 + pu1 + p 2u 2 + ....
                                                                                                                              (26)

Equation (26), can be written in the form;

 p (u 0 x + pu 1x + p 2u 2 x + ....) 2 + p (u 0 + pu 1 + p 2u 2 + ....)(u 0 xx + pu1xx + p 2u 2 xx + ....) = 0
The first few components of He's polynomials are given by:

                                                                      39
Mathematical Theory and Modeling                                                                        www.iiste.org
ISSN 2224-5804 (Paper)    ISSN 2225-0522 (Online)
Vol.2, No.3, 2012
                                 H 0 (u ) = u 0 x + u 0u 0 xx
                                              2


                                 H 1 (u ) = 2u 0 x u1x + u 0u1xx + u 1u 0 xx
                                                .
                                                .
                                                .
Therefore


                  p 0 : u 0 (x , t ) = x 2      ,     H 0 (u ) = 6x 2
                  p 1 :u 1 (x , t ) = E −1 {vE [ H 0 (u ) ]} = 6x 2t , H 1 (u ) = 72x 2t
                  p 2 :u 2 (x , t ) = E −1 {vE [ H 1 (u ) ]} = 36x 2t 2
                 .                      .                                      .
                 .                      .                                      .
                 .                      .                                      .

                                                                                                     x2
Then the solution of equation (22) is given by:      u (x , t ) = x 2 (1 + 6t + 36t 2 + .....) =
                                                                                                   1 − 6t
4. Conclusion
In this paper, we mixture Elzaki transform and homotopy perturbation method to solve nonlinear partial
differential equations. The solution by using Adomian decomposition method is simple, but the calculation
of Adomian polynomials is complex. The fact that the developed algorithm solves nonlinear partial
differential equations without Adomian's polynomials can be considered as a clear advantage of this
technique over the decomposition method.
References
[1] S. lslam, Yasir Khan, Naeem Faraz and Francis Austin (2010), Numerical Solution of Logistic
Differential Equations by using the Laplace Decomposition Method, World Applied Sciences Journal 8
(9):1100-1105.
[2] Nuran    Guzel and Muhammet Nurulay (2008), Solution of Shiff Systems By using Differential
Transform Method, Dunlupinar universities Fen Bilimleri Enstitusu Dergisi, ISSN 1302-3055, PP. 49-59.
[3] Shin- Hsiang Chang , I-Ling Chang (2008), A new algorithm for calculating one – dimensional
differential transform of nonlinear functions, Applied Mathematics and Computation 195 ,799-808.
[4] Tarig M. Elzaki (2011), The New Integral Transform “Elzaki Transform” Global Journal of Pure and
Applied Mathematics, ISSN 0973-1768,Number 1, pp. 57-64.
[5] Tarig M. Elzaki & Salih M. Elzaki (2011), Application of New Transform “Elzaki Transform” to Partial
Differential Equations, Global Journal of Pure and Applied Mathematics, ISSN 0973-1768,Number 1, pp.
65-70.
[6] Tarig M. Elzaki & Salih M. Elzaki (2011), On the Connections between Laplace and Elzaki transforms,
Advances in Theoretical and Applied Mathematics, ISSN 0973-4554 Volume 6, Number 1, pp. 1-11.
[7] Tarig M. Elzaki & Salih M. Elzaki (2011), On the Elzaki Transform and Ordinary Differential Equation
With Variable Coefficients, Advances in Theoretical and Applied Mathematics. ISSN 0973-4554 Volume 6,

                                                          40
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Vol.2, No.3, 2012
Number 1, pp. 13-18.
[8] Lokenath Debnath and D. Bhatta (2006). Integral transform and their Application second Edition,
Chapman & Hall /CRC
[9] A.Kilicman and H.E.Gadain. (2009), An application of double Laplace transform and Sumudu
transform, Lobachevskii J. Math.30 (3) pp.214-223.
[10] J. Zhang, (2007). A Sumudu based algorithm m for solving differential equations, Comp.           Sci. J.
Moldova 15(3), pp – 303-313.
[11] Hassan Eltayeb and Adem         kilicman, (2010),    A Note on      the   Sumudu    Transforms      and
differential   Equations, Applied   Mathematical     Sciences, VOL, 4, no.22,1089-1098
[12] Kilicman A. & H. ELtayeb. (2010), A note on Integral transform and Partial Differential Equation,
Applied Mathematical Sciences, 4(3), PP.109-118.
[13] Hassan ELtayeh and Adem kilicman (2010), on Some Applications of a new Integral Transform, Int.
Journal of Math. Analysis, Vol, 4, no.3, 123-132.
[14] N.H. Sweilam, M.M. Khader (2009). Exact Solutions of some Capled nonlinear partial differential
equations using the homotopy perturbation method. Computers and Mathematics with Applications 58,
2134-2141.
[15] P.R. Sharma and Giriraj Methi (2011). Applications of Homotopy Perturbation method to Partial
differential equations. Asian Journal of Mathematics and Statistics 4 (3): 140-150.
[16] M.A. Jafari, A. Aminataei (2010). Improved Homotopy Perturbation Method. International
Mathematical Forum, 5, no, 32, 1567-1579.
[17] Jagdev Singh, Devendra, Sushila. Homotopy Perturbation Sumudu Transform Method for Nonlinear
Equations. Adv. Theor. Appl. Mech., Vol. 4, 2011, no. 4, 165-175.
[18] Nuran Guzel and Muhammet Nurulay (2008), Solution of Shiff Systems By using Differential
Transform Method, Dunlupinar universities Fen Bilimleri Enstitusu Dergisi, ISSN 1302-3055, PP. 49-59.
[19] Shin- Hsiang Chang , I-Ling Chang (2008), A new algorithm for calculating one – dimensional
differential transform of nonlinear functions, Applied Mathematics and Computation 195, 799-808.
[20]Hashim, I, M.SM.Noorani,R.Ahmed.S.A.Bakar.E.S.I.IsmailandA.M.Zakaria, (2006). Accuracy of the
Adomian decomposition method applied to the Lorenz system chaos.08.135.
[21] Tarig M. Elzaki, Salih M. Elzaki, and Eman M. A. Hilal, (2012). Elzaki and Sumudu Transforms for Solving
Some Differential Equations, Global Journal of Pure and Applied Mathematics, ISSN 0973-1768, Volume 8,
Number 2, pp. 167-173.




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Vol.2, No.3, 2012

Table 1. Elzaki transform of some Functions

                      f (t )                                       E [ f (t )] = T (u )

                       1
                                                                            v2
                       t
                                                                          v3

                       tn                                              n ! v n +2

                      e at                                                v2
                                                                        1 − av
                     sinat
                                                                         av 3
                                                                       1 + a 2v 2
                     cosat
                                                                          v2
                                                                       1 + a 2v 2

Tarig M. Elzaki

Department of Mathematics, Faculty of Sciences and Arts-Alkamil,

King Abdulaziz University, Jeddah-Saudi Arabia
e-mail: tarig.alzaki@gmail.com and tfarah@kau.edu.sa
Eman M. A. Hilal
Department of Mathematics, Faculty of Sciences for Girls

King Abdulaziz University, Jeddah Saudi Arabia

e-mail: ehilal@kau.edu.sa




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11.homotopy perturbation and elzaki transform for solving nonlinear partial differential equations

  • 1. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012 Homotopy Perturbation and Elzaki Transform for Solving Nonlinear Partial Differential Equations Tarig. M. Elzaki1* & Eman M. A. Hilal2 1. Mathematics Department, Faculty of Sciences and Arts-Alkamil, King Abdulaziz University, Jeddah-Saudi Arabia. Mathematics Department, Faculty of Sciences, Sudan University of Sciences and Technology-Sudan. 2. Mathematics Department, Faculty of Sciences for Girles King Abdulaziz University Jeddah-Saudi Arabia * E-mail of the corresponding author: Tarig.alzaki@gmail.com Abstract In this work, we present a reliable combination of homotopy perturbation method and Elzaki transform to investigate some nonlinear partial differential equations. The nonlinear terms can be handled by the use of homotopy perturbation method. The proposed homotopy perturbation method is applied to the reformulated first and second order initial value problem which leads the solution in terms of transformed variables, and the series solution is obtained by making use of the inverse transformation. The results show the efficiency of this method. Keywords: Homotopy perturbation methods, Elzaki transform nonlinear partial differential equations. 1. Introduction Linear and nonlinear partial differential equations are of fundamental importance in science and engineering. Some integral transform method such as Laplace and Fourier and Sumudu transforms methods see [Kilicman and E.Gadain. (2009), (2010) lslam, Yasir Khan, Naeem Faraz and Francis Austin (2010)], are used to solve linear partial differential equations and use fullness of these integral transform lies in their ability to transform differential equations into algebraic equations which allows simple and systematic solution procedures. However, using integral transform in nonlinear differential equations may increase its complexity. In recent years, many research workers have paid attention to find the solutions of nonlinear differential equations by using various methods. Among these are the Adomian decomposition method [Hashim, Noorani, Ahmed. Bakar. Ismail and Zakaria, (2006)], the tanh method, the homotopy perturbation method [ Sweilam, Khader (2009), Sharma and Giriraj Methi (2011), Jafari, Aminataei (2010), (2011) ], the differential transform method [(2008)], and the variational iteration method. Elzaki transform [ Tarig and Salih, (2011), (2012)] is totally incapable of handling the nonlinear equations because of the difficulties that are caused by the nonlinear terms. Various ways have been proposed recently to deal with these nonlinearities, one of these combinations of homotopy perturbation method and Elzaki transform which is studies in this paper. The advantage of this method is its capability of combining two powerful methods for obtaining exact 33
  • 2. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012 solutions for nonlinear partial differential equations. This article considers the effectiveness of the homotopy perturbation Elzaki transform method in solving nonlinear partial differential equations both homogeneous and non-homogeneous. 1.1. Elzaki Transform The basic definitions of modified of Sumudu transform or Elzaki transform is defined as follows, Elzaki transform of the function f (t ) is ∞ t − E [ f (t )] = v ∫ f (t ) e v dt , t >0 (1) 0 Tarig M. Elzaki and Sailh M. Elzaki in [(2011), (2012)], showed the modified of Sumudu transform [(2007), (2010)] or Elzaki transform was applied to partial differential equations, ordinary differential equations, system of ordinary and partial differential equations and integral equations. Elzaki transform is a powerful tool for solving some differential equations which can not solve by Sumudu transform in [(2012)]. In this paper, we combined Elzaki transform and homotopy perturbation to solve nonlinear partial differential equations. To obtain Elzaki transform of partial derivative we use integration by parts, and then we have:  ∂f (x , t )  1  ∂ 2 f (x , t )  1 ∂f (x , 0) E = T (x ,v ) −vf (x , 0) E  = v 2 T (x ,v ) − f (x , 0) −v  ∂t  v  ∂t ∂t 2    ∂f (x , t )  d  ∂ 2f (x , t )  d 2 E = [T (x ,v )] E  = 2 [T (x ,v )]  ∂x  dx  ∂x 2   dx Proof: To obtain ELzaki transform of partial derivatives we use integration by parts as follows:  ∂f  ∞ ∂f −t p −t ∂f   −t   p p −t   Ε  ( x ,t ) = ∫ v e dt = lim ∫ve v v dt = lim  ve f (x , t )  − ∫ e v f (x , t )dt  v  ∂t  0 ∂t p →∞ 0 ∂t p →∞   0 0   T ( x ,v ) = −vf ( x , 0 ) v We assume that f is piecewise continuous and it is of exponential order.  ∂f  ∞ −t ∂f ( x , t ) ∂ ∞ −t  = ∫ ve ve v f ( x , t ) dt , using the Leibnitz rule to find: ∂x ∫ Now Ε v dt =  ∂x  0 ∂x 0  ∂f  d Ε  = dx T ( x ,v )     ∂x   ∂ 2f  d 2 By the same method we find: Ε  2  = 2 T ( x ,v )     ∂x  dx 34
  • 3. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012  ∂2 f  To find: Ε  2 ( x, t )   ∂t  ∂f Let = g , then we have: ∂t  ∂ 2f   ∂g ( x , t )  Ε  2 ( x , t ) = Ε   g ( x , t )  −vg x , 0   =Ε ( )  ∂t   ∂t  v  ∂ 2f  1 ∂f Ε  2 ( x , t ) = 2 T ( x , v ) − f ( x , 0 ) −v ( x , 0)  ∂t  v ∂t We can easily extend this result to the nth partial derivative by using mathematical induction. 1.2. Homotopy Perturbation Method: Let X and Y be the topological spaces. If f and g are continuous maps of the space X into Y , it is said that f is homotopic to g , if there is continuous map F : X × [0,1] →Y such that F (x , 0) = f (x ) and F (x ,1) = g (x ) , for each x ∈ X , then the map is called homotopy between f and g . To explain the homotopy perturbation method, we consider a general equation of the type, L (u ) = 0 (2) Where L is any differential operator, we define a convex homotopy H (u , p ) by H (u , p ) = (1 − p )F (u ) + pL (u ) (3) Where F (u ) is a functional operator with known solution v 0 which can be obtained easily. It is clear that, for H (u , p ) = 0 (4) We have: H (u , 0) = F (u ) , H (u ,1) = L (u ) In topology this show that H (u , p ) continuously traces an implicitly defined carves from a starting point H (v 0 , 0) to a solution function H (f ,1) . The HPM uses the embed ling parameter p as a small parameter and write the solution as a power series 35
  • 4. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012 u = u 0 + pu 1 + p 2u 2 + p 3u 3 + ....... (5) If p → 1 , then (5) corresponds to (3) and becomes the approximate solution of the form, ∞ f = lim u = ∑ u i (6) p →1 i =0 We assume that (6) has a unique solution. The comparisons of like powers of p give solutions of various orders. 2. Homotopy Perturbation Elzaki Transform Method: Consider a general nonlinear non-homogenous partial differential equation with initial conditions of the form: Du (x , t ) + Ru (x , t ) + Nu (x , t ) = g (x , t ) (7) u (x , 0) = h (x ) , u t (x , 0) = f (x ) Where D is linear differential operator of order two, R is linear differential operator of less order than D , N is the general nonlinear differential operator and g (x , t ) is the source term. Taking Elzaki transform on both sides of equation (7), to get: E [Du (x , t )] + E [Ru (x , t )] + E [Nu (x , t )] = E [ g (x , t )] (8) Using the differentiation property of Elzaki transforms and above initial conditions, we have: E [u (x , t )] = v 2 E [ g (x , t )] + v 2 h (x ) + v 3f (x ) −v 2 E [Ru (x , t ) + Nu (x , t )] (9) Appling the inverse Elzaki transform on both sides of equation (9), to find: u (x , t ) = G (x , t ) − E −1 {v 2 E [ Ru (x , t ) + Nu (x , t )]} (10) Where G ( x , t ) represents the term arising from the source term and the prescribed initial conditions. Now, we apply the homotopy perturbation method. ∞ u ( x , t ) = ∑ p nu n ( x , t ) (11) n =0 And the nonlinear term can be decomposed as ∞ N [u (x , t )] = ∑ p n H n (u ) (12) n =0 Where H n (u ) are given by: 36
  • 5. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012 1 ∂   ∞ i  H n (u 0 ,u 1 ,.....u n ) =  N  ∑ p u i   , n = 0,1, 2,...... n ! ∂p n   i =0   p =0 (13) Substituting equations (11) and (12) in equation (10), we get: ∞     ∞ n ∞   ∑ p nu n (x , t ) = G (x , t ) − p E −1 v 2 E   R ∑ p u n (x , t ) + ∑ p n H n (u )     n =0  (14) n =0   n =0  This is the coupling of the Elzaki transform and the homotopy perturbation method. Comparing the coefficient of like powers of p , the following approximations are obtained. p 0 : u 0 (x , t ) =G (x , t ) p 1 :u 1 (x , t ) = − E −1 {v 2 E [ Ru 0 (x , t ) + H 0 (u ) ]} p 2 :u 2 (x , t ) = −E −1 {v 2 E [ Ru1 (x , t ) + H 1 (u )]} p 3 :u 3 (x , t ) = −E −1 {v 2 E [ Ru 2 (x , t ) + H 2 (u )]} (15) . . . . . . . . . Then the solution is u (x , t ) = u 0 (x , t ) , +u1 (x , t ) + u 2 (x , t ) + .... 3. Applications: In this section we apply the homotopy perturbation Elzaki transform method for solving nonlinear partial differential equations. Example 3.1: Consider the following homogenous nonlinear partial differential equations u t + uu x = 0 , u (x , 0) = − x (16) Taking Elzaki transform of equation (16) subject to the initial condition, we have: E [u (x , t ) ] = xv 2 − vE [uu x ] (17) The inverse Elzaki transform implies that: u (x , t ) = −x − E −1 {vE [uu x ]} (18) Now applying the homotopy perturbation method, we get: ∞    ∞   ∑ p nu n (x , t ) = −x − p  E −1 vE  ∑ p n H n (u )    (19) n =0     n =0   37
  • 6. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012 Where H n (u ) are He's polynomials that represents the nonlinear terms. Or p [uu x ] = 0 ⇒ p (u 0 + pu 1 + p 2u 2 + ....)(u 0 x + pu 1x + p 2u 2 x + ....) u = u 0 + pu 1 + p 2u 2 + .... The first few components of He's polynomials, are given by H 0 (u ) = u 0u 0 x H 1 (u ) = u 0u 1x + u1u 0 x H 2 (u ) = u 0u 2 x + u1u1x + u 2u 0 x . . . Comparing the coefficients of the same powers of p , we get: p 0 : u 0 (x , t ) = −x , H 0 (u ) = x p :u 1 (x , t ) = − E 1 −1 {vE [ H (u )]} = −xt 0 , H 1 (u ) = 2xt p 2 :u 2 (x , t ) = −E −1 {vE [ H (u )]} = −xt 1 2 , H 2 (u ) = 3xt 2 p 3 :u 3 (x , t ) = −E −1 {vE [ H (u )]} = −xt 2 3 . . . . . . . . . Therefore the solution u (x , t ) is given by: x u (x , t ) = −x (1 + t + t 2 + t 3 + .....) = t −1 Example 3.2: Consider the first order nonlinear partial differential equation u t + uu x = 2t + x + t 3 + xt 2 , u (x , 0) = −x (20) To find the solution by homotopy perturbation Elzaki transform method, we applying homotopy perturbation method after taking Elzaki and inverse Elzaki transforms of equation (20), we get: ∞ t 4 xt 3    ∞   ∑ p nu n (x , t ) = t 2 + xt + + − p  E −1 vE  ∑ p n H n (u )    (21) n =0 4 3     n =0   Where 38
  • 7. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012 H 0 (u ) = u 0u 0 x H 1 (u ) = u 0u1x + u 1u 0 x H 2 (u ) = u 0u 2 x + u 1u 1x + u 2u 0 x . . . Comparing the coefficients of like powers of p , we have: xt 3 t 4 p : u 0 (x , t ) = t + xt + 0 2 + 3 4 4 3 5 t xt 2xt 7t 6 xt 7 t 8 p 1 :u 1 (x , t ) = − − − − − − 4 3 15 72 63 98 . . . . . . . . . The noise terms appear between the components u 0 (x , t ) ,u 1 (x , t ) , therefore, the exact solution is given by: u (x , t ) = t 2 + xt Example 3.3: Let us consider the second order nonlinear partial differential equation ∂u  ∂u  ∂ 2u 2 =  +u , u (x , 0) = x 2 (22) ∂t  ∂x  ∂x 2 Applying Elzaki transform of equation (22), and making use of the initial condition, to find: E [u (x , t ) ] = x 2v 2 + vE u x2 + uu xx    (23) Take the inverse Elzaki transform of equation (23), we get: { u (x , t ) = x 2 + E −1 vE u x2 + uu xx    } (24) Apply the homotopy perturbation method of (24), to get: ∞    ∞   ∑p u n n (x , t ) = x 2 + p  E −1 vE  ∑ p n H n (u )    (25) n =0     n =0   Or p u x2 + uu xx  = 0 , u = u 0 + pu1 + p 2u 2 + ....   (26) Equation (26), can be written in the form; p (u 0 x + pu 1x + p 2u 2 x + ....) 2 + p (u 0 + pu 1 + p 2u 2 + ....)(u 0 xx + pu1xx + p 2u 2 xx + ....) = 0 The first few components of He's polynomials are given by: 39
  • 8. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012 H 0 (u ) = u 0 x + u 0u 0 xx 2 H 1 (u ) = 2u 0 x u1x + u 0u1xx + u 1u 0 xx . . . Therefore p 0 : u 0 (x , t ) = x 2 , H 0 (u ) = 6x 2 p 1 :u 1 (x , t ) = E −1 {vE [ H 0 (u ) ]} = 6x 2t , H 1 (u ) = 72x 2t p 2 :u 2 (x , t ) = E −1 {vE [ H 1 (u ) ]} = 36x 2t 2 . . . . . . . . . x2 Then the solution of equation (22) is given by: u (x , t ) = x 2 (1 + 6t + 36t 2 + .....) = 1 − 6t 4. Conclusion In this paper, we mixture Elzaki transform and homotopy perturbation method to solve nonlinear partial differential equations. The solution by using Adomian decomposition method is simple, but the calculation of Adomian polynomials is complex. The fact that the developed algorithm solves nonlinear partial differential equations without Adomian's polynomials can be considered as a clear advantage of this technique over the decomposition method. References [1] S. lslam, Yasir Khan, Naeem Faraz and Francis Austin (2010), Numerical Solution of Logistic Differential Equations by using the Laplace Decomposition Method, World Applied Sciences Journal 8 (9):1100-1105. [2] Nuran Guzel and Muhammet Nurulay (2008), Solution of Shiff Systems By using Differential Transform Method, Dunlupinar universities Fen Bilimleri Enstitusu Dergisi, ISSN 1302-3055, PP. 49-59. [3] Shin- Hsiang Chang , I-Ling Chang (2008), A new algorithm for calculating one – dimensional differential transform of nonlinear functions, Applied Mathematics and Computation 195 ,799-808. [4] Tarig M. Elzaki (2011), The New Integral Transform “Elzaki Transform” Global Journal of Pure and Applied Mathematics, ISSN 0973-1768,Number 1, pp. 57-64. [5] Tarig M. Elzaki & Salih M. Elzaki (2011), Application of New Transform “Elzaki Transform” to Partial Differential Equations, Global Journal of Pure and Applied Mathematics, ISSN 0973-1768,Number 1, pp. 65-70. [6] Tarig M. Elzaki & Salih M. Elzaki (2011), On the Connections between Laplace and Elzaki transforms, Advances in Theoretical and Applied Mathematics, ISSN 0973-4554 Volume 6, Number 1, pp. 1-11. [7] Tarig M. Elzaki & Salih M. Elzaki (2011), On the Elzaki Transform and Ordinary Differential Equation With Variable Coefficients, Advances in Theoretical and Applied Mathematics. ISSN 0973-4554 Volume 6, 40
  • 9. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012 Number 1, pp. 13-18. [8] Lokenath Debnath and D. Bhatta (2006). Integral transform and their Application second Edition, Chapman & Hall /CRC [9] A.Kilicman and H.E.Gadain. (2009), An application of double Laplace transform and Sumudu transform, Lobachevskii J. Math.30 (3) pp.214-223. [10] J. Zhang, (2007). A Sumudu based algorithm m for solving differential equations, Comp. Sci. J. Moldova 15(3), pp – 303-313. [11] Hassan Eltayeb and Adem kilicman, (2010), A Note on the Sumudu Transforms and differential Equations, Applied Mathematical Sciences, VOL, 4, no.22,1089-1098 [12] Kilicman A. & H. ELtayeb. (2010), A note on Integral transform and Partial Differential Equation, Applied Mathematical Sciences, 4(3), PP.109-118. [13] Hassan ELtayeh and Adem kilicman (2010), on Some Applications of a new Integral Transform, Int. Journal of Math. Analysis, Vol, 4, no.3, 123-132. [14] N.H. Sweilam, M.M. Khader (2009). Exact Solutions of some Capled nonlinear partial differential equations using the homotopy perturbation method. Computers and Mathematics with Applications 58, 2134-2141. [15] P.R. Sharma and Giriraj Methi (2011). Applications of Homotopy Perturbation method to Partial differential equations. Asian Journal of Mathematics and Statistics 4 (3): 140-150. [16] M.A. Jafari, A. Aminataei (2010). Improved Homotopy Perturbation Method. International Mathematical Forum, 5, no, 32, 1567-1579. [17] Jagdev Singh, Devendra, Sushila. Homotopy Perturbation Sumudu Transform Method for Nonlinear Equations. Adv. Theor. Appl. Mech., Vol. 4, 2011, no. 4, 165-175. [18] Nuran Guzel and Muhammet Nurulay (2008), Solution of Shiff Systems By using Differential Transform Method, Dunlupinar universities Fen Bilimleri Enstitusu Dergisi, ISSN 1302-3055, PP. 49-59. [19] Shin- Hsiang Chang , I-Ling Chang (2008), A new algorithm for calculating one – dimensional differential transform of nonlinear functions, Applied Mathematics and Computation 195, 799-808. [20]Hashim, I, M.SM.Noorani,R.Ahmed.S.A.Bakar.E.S.I.IsmailandA.M.Zakaria, (2006). Accuracy of the Adomian decomposition method applied to the Lorenz system chaos.08.135. [21] Tarig M. Elzaki, Salih M. Elzaki, and Eman M. A. Hilal, (2012). Elzaki and Sumudu Transforms for Solving Some Differential Equations, Global Journal of Pure and Applied Mathematics, ISSN 0973-1768, Volume 8, Number 2, pp. 167-173. 41
  • 10. Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.2, No.3, 2012 Table 1. Elzaki transform of some Functions f (t ) E [ f (t )] = T (u ) 1 v2 t v3 tn n ! v n +2 e at v2 1 − av sinat av 3 1 + a 2v 2 cosat v2 1 + a 2v 2 Tarig M. Elzaki Department of Mathematics, Faculty of Sciences and Arts-Alkamil, King Abdulaziz University, Jeddah-Saudi Arabia e-mail: tarig.alzaki@gmail.com and tfarah@kau.edu.sa Eman M. A. Hilal Department of Mathematics, Faculty of Sciences for Girls King Abdulaziz University, Jeddah Saudi Arabia e-mail: ehilal@kau.edu.sa 42
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