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Application of Homotopy Analysis MethodCombined with Elzaki Transform forFractional Porous Medium Equation

Djelloul ZianeUniversity of Oran Senia, Laboratory LAMAP,

Faculty of sciences , P.O. Box 1524, Oran, 31000, Algeria.Email: [email protected]

October 15, 2016

Abstract

The basic motivation of the present study is to apply the modi�edfractional homotopy analysis transform method for solving nonlinearporous medium equation with time-fractional derivative. The frac-tional derivative is described in the Caputo sense. The results showthat the MFHATM is an appropriate method for solving nonlinearfractional partial di¤erential equations.

Keywords: Elzaki transform, homotopy analysis method, fractionalporous medium equation.

1 Introduction

There is no secret to the researcher in the �eld of nonlinear partial di¤erentialequations, that the solution of this class of equations is not easy. So we �ndthat many researchers have done and are still doing great e¤orts to �nd meth-ods to solve this type of equations. These e¤orts resulted in the consolidationof this research �eld in many methods, among them we �nd the HomotopyAnalysis Method (HAM) ([1]-[3]), Adomian Decomposition Method (ADM)

1

Journal of Approximation Theory and Applied Mathematics, 2016 Vol. 6

([4]-[6]), Variational Iteration Method (VIM) ([7]-[9]) and Homotopy Per-turbation Method (HPM) ([10]-[12]), which have become known in a largenumber of researchers in this area. A new option emerged recently, includesthe composition of Laplace transform, sumudu transform or Elzaki trans-form with these methods. Among wich are the Laplace Homotopy AnalysisMethod [13], Homotopy Analysis Sumudu Transform Method [14], modi�edfractional homotopy analysis transform method [15], Adomian Decomposi-tion Method coupled with Laplace Transform Method [16], Sumudu Decom-position Method for Nonlinear Equations [17], An Elzaki Transform Decom-position Algorithm Applied to a Class of Non-Linear Di¤erential Equations[18], Variational Iteration Method coupled with Laplace Transform Method[19], Variational Iteration Sumudu Transform Method [20], Application ofthe ADM Elzaki and VIM Elzaki transform for solving the nonlinear partialdi¤erential equations [21], Homotopy Perturbation Transform Method [22],Homotopy Perturbation Sumudu Transform Method [23], Homotopy Pertur-bation Elzaki Transform Method [29].The aim of this paper is to directly apply modi�ed fractional homotopy

analysis transform method (MFHATM) described in [15] to obtain the exactand an approximate analytical solution of the nonlinear porous medium equa-tion with time-fractional derivative in the operator form :

cD�t u =

@

@x

�uk@u

@x

�; 0 < � 6 1; (1)

where cD�t =

@�

@t�is the Caputo fractional derivative.

When � = 1, this equation turns to the classical porous medium equationof the form :

@u

@t=@

@x

�uk@u

@x

�; (2)

where k is a rational number. There are number of physical applicationswhere this simple model appears in a natural way, mainly to describe processesinvolving �uid �ow, heat transfer or di¤usion. May be the best known of themis the description of the �ow of an isentropic gas through a porous medium,modeled independently by Leibenzon and Muskat around 1930. Also appli-cation is found in the study of groundwater in�ltration by Boussisnesq in1903. Another important application refers to heat radiation in plasmas,developed by Zel�dovich and coworkers around 1950 [24].The present paper has been organized as follows: In Section 2 some basic

de�nitions of fractional calculus, ELzaki transform and Elzaki transform of

2


fractional derivatives are montioned. In section 3 we present the methodused in solving the proposed equation. In section 4 the nonlinear porousmedium equation with time-fractional derivative is studied with the modi�edfractional homotopy analysis transform method (MFHATM). Finally, theconclusion follows.

2 Preliminaries

In this section, we give some basic de�nitions and properties of fractional cal-culus, Elzaki transform and Elzaki transform of fractional derivatives whichare used further in this paper.

2.1 Fractional calculus

There are several de�nitions of a fractional derivative of order � > 0 (see[25]-[27]). The most commonly used de�nitions are the Riemann�Liouvilleand Caputo. We give some basic de�nitions and properties of the fractionalcalculus theory which are used further in this paper .

De�nition 1 Let = [a; b] (�1 < a < b < +1) be a �nite interval on thereal axis R: The Riemann�Liouville fractional integrals I�0+f of order � 2 R(� > 0) is de�ned by :

(I�0+f)(t) =1

� (�)

Z t

0

f(�)d�

(t� �)1�� ; t > 0; � > 0; (3)

(I00+f)(t) = f(t):

Here �(�) is the gamma function.

Theorem 2 Let � > 0 and let n = [�] + 1: If f(t) 2 ACn [a; b] ; then theCaputo fractional derivatives (cD�

0+f)(t) exist almost evrywhere on [a; b] :If � =2 N; (cD�

0+f)(t) are represented by :

(cD�0+f)(t) =

1

� (n� �)

Z t

a

f (n)(�)d�

(t� �)��n+1 ; (4)

where D = ddxand n = [�] + 1:

3


Proof (see [26]).

Remark 3 In this paper, we consider the time-fractional derivative in theCaputo�s sense. When � 2 R+, the time-fractional derivative is de�ned as :

(cD�0+u)(x; t) =

@�u(x; t)

@t�

=

(1

�(m��)R t0(t� �)m��1 @

mu(x;�)@�m

; m� 1 < � < m;@mu(x;t)@tm

; � = m;(5)

where m 2 N�:

(1) Let � > 0 and let n = [�] + 1 for n =2 N; n = � for n 2 N:If f(t) 2 ACn [a; b] ; then

(I�0+cD�

0+f)(t) = f(t)�n�1Xk=0

f (k)(0)

k!tk:

(2) (I�0+x��1)(t) = �(�)

�(�+�)t�+��1; � > 0; � > 0:

(3) ( cD�0+x

��1)(t) = �(�)�(��)t

��1; � > 0; � > n; where n = [�] + 1for n =2 N; n = � for n 2 N:

(4) (cD�0+k)(t) = 0; where k is constant.

2.2 Basic de�nitions of Elzaki transform

A new integral transform called Elzaki transform ([28]-[30]) de�ned for func-tions of exponential order, is proclaimed. They consider functions in the setA de�ned by :

A =

�f(t)=M; k1; k2 > 0; jf(t)j < Me

jtjkj ; if t 2 (�1)j � [0; 1)

�:

De�nition 4 If f(t) is function de�ned for all t > 0, its Elzaki transform isthe integral of f(t) times � t

sfrom t = 0 to 1: It is a function of s and

is de�ned by E[f ] :

E [f(t)] = T (s) = s1R0

f(t)e�tsdt: (6)

4


Theorem 5 Elzaki transform ampli�es the coe¢ cients of the power seriesfunction

f(t) =1Pn=0

antn; (7)

on the new integral transform "Elzaki transform", given by

E [f(t)] = T (v) =1Pn=0

n!anvn+2: (8)

Theorem 6 Let f(t) be in A and Let Tn(v) denote Elzaki transform of nthderivative, f (n)(t) of f(t), then for n � 1,

Tn(v) =T (v)

vn�

n�1Xk=0

v2�n+kf (k)(0): (9)

To obtain Elzaki transform of partial derivative we use integration byparts, and then we have :

E�@f(x;t)@t

�= 1

vT (x; v)� vf(x; 0);

E�@2f(x;t)@t2

�= 1

v2T (x; v)� f(x; 0)� v @f(x;0)

@t;

(10)

Properties of Elzaki transform can be found in Refs.([28],[29]), we mentiononly the following :1. E(1) = v2; 3. E (tn) = n!vn+2;2. E(t) = v3; 4. E�1 (vn+2) = tn

n!:

2.3 Elzaki transform of fractional derivatives

To give the formula of Elzaki transform of Caputo fractional derivative,we use the Laplace transform formula for the Caputo fractional derivative[25]

Lf (cD�0+f)(t); sg = s�F (s)�

m�1Xk=0

s��k�1f (k)(0);

where m� 1 < � 6 m; m 2 N�:

5


Theorem 7 [31] Let A de�ned as above. With Laplace transform F (s), thenthe Elzaki transform T (v) of f(t) is given by :

T (v) = vF (1

v):

Theorem 8 Suppose T (v) is the Elzaki transform of the function f(t) then

Ef (cD�0+f)(t); vg =

T (v)

v��

n�1Xk=0

vk��+2f (k)(0): (11)

Proof (see [32]).

3 Modi�ed fractional homotopy analysis trans-form method (MFHATM)

Kangle Wang and Sanyang Liu [15] gives the idea of the basis of this method.They consider the following general time-fractional di¤erential equation withthe initial condition as :

cDn�t U(x; t) + LU(x; t) +RU(x; t) = g(x; t);

n� 1 < n� 6 n;U(x; 0) = h(x);

(12)

where cDn�t is the Caputo fractional derivative operator, cDn�

t = @n�

@tn�, L

is the linear operator, R is the general nonlinear operator and g(x; t) is acontinuous functions.Applying Elzaki transform on both sides of Eq.(12), we can get :

E [ cDn�t U(x; t)] + E [LU(x; t) +RU(x; t)� g(x; t)] = 0; (13)

Using the property of Elzaki transform, we have the following form:

E [ U(x; t)]�vn�n�1Xk=0

v2�n�+kU (k)(x; 0)+vn�E [LU(x; t) +RU(x; t)� g(x; t)] = 0;

(14)

6


De�ne the nonlinear operator :

N [�(x; t; p)] = E [ �(x; t; p)]� vn�n�1Xk=0

v2�n�+kh(k)(x; 0) (15)

+vn�E [L�(x; t; p) +R�(x; t; p)� g(x; t; p)]

By means of homotopy analysis method [1], we construct the so-called thezero-order deformation equation :

(1� q)E[�(x; t; p)� �(x; t; 0) = phH(x; t)N [�(x; t; p)]; (16)

where p is an embedding parameter and p 2 [0; 1], H(x; t) 6= 0 is an auxiliaryfunction, h 6= 0 is an auxiliary parameter, E is an auxiliary linear Elzaki op-erator. When p = 0 and p = 1, we have :�

�(x; t; 0) = u0(x; t);�(x; t; 1) = u(x; t):

(17)

When P increases from 0 to 1, the �(x; t; p) various from U0(x; t) toU(x; t). Expanding �(x; t; p) in Taylor series with respect to p, we have :

�(x; t; p) = U0(x; t) ++1Xm=1

Um(x; t)pm; (18)

where

Um(x; t) =1

m!

@m�(x; t; p)

@pmjp=0 : (19)

When p = 1, the (18) becomes :

U(x; t) = U0(x; t) ++1Xm=1

Um(x; t): (20)

De�ne the vectors :

�!U n = fU0(x; t); U1(x; t); U2(x; t); : : : ; Un(x; t)g: (21)

Di¤erentiating (16) m�times with respect to p, then setting p = 0 and�nally dividing them by m!; we obtain the so-called mth order deformationequation :

7


E[Um(x; t)� �mUm�1(x; t)] = hpH(x; t)<m(�!U m�1(x; t)); (22)

where

<m(�!U m�1(x; t)) =

1

(m� 1)!@m�1N(x; t; p)

@pm�1jp=0 ; (23)

and

�m =

�0; m 6 1;1; m > 1:

Applying the inverse Elzaki transform on both sides of Eq.(22), we canobtain :

Um(x; t) = �mUm�1(x; t) + E�1hhpH(x; t)<m(

�!U m�1(x; t))

i: (24)

The mth deformation equation (24) is a linear which can be easily solved.So, the solution of Eq.(12) can be written into the following form :

U(x; t) =NXm=0

Um(x; t); (25)

when N !1, we can obtain an accurate approximation solution of Eq.(12).The proof of the convergence of the modi�ed fractional homotopy analysis

transform method (MFHATM) (see [2]).

4 Application of the MFHATM Method

In this section, we apply the modi�ed fractional homotopy analysis transformmethod (MFHATM) for solving the following nonlinear porous medium equa-tion with time-fractional derivative (1) in the two cases : k = �1 and k =�4=3.

8


4.1 Example

First, we take k = �1 in equation (1)[24], we get :

cD�t u =

�u�1ux

�x; 0 < � 6 1; (26)

with the initial condition :u(x; 0) =

1

x: (27)

Applying Elzaki transform on both sides of Eq.(26), we can get :

E [u]� v2u(x; 0) = v�E��u�1ux

�x

�(28)

From (28) and the initial condition (27), we have :

E [u]� v2 1x� v�E

��u�1ux

�x

�= 0: (29)

We take the nonlinear part as :

N [�(x; t; p)] = E [�]� v2 1x� v�E

��1�x

�x

�: (30)

We construct the so-called the zero-order deformation equation with as-sumptionH(x; t) = 1, we have :

(1� q)E[�(x; t; p)� �(x; t; 0) = phN [�(x; t; p)]: (31)

When p = 0 and p = 1, we can obtain :��(x; t; 0) = u0(x; t);�(x; t; 1) = u(x; t):

Therefore, we have the mth order deformation equation :

E[um(x; t)� �mum�1(x; t)] = h<m(�!u m�1(x; t)): (32)

Operating the inverse Elzaki operator on both sides of Eq.(32), we get :

um(x; t) = �mum�1(x; t) + E�1[h<m(�!u m�1(x; t))]: (33)

9


From Eq.(33), we have :

u1(x; t) = hE�1[ <1(�!u 0(x; t))];u2(x; t) = u1 + hE

�1[ <2(�!u 1(x; t))]; (34)

u3(x; t) = u2 + hE�1[ <3(�!u 2(x; t))];

...

where

<1(�!u 0(x; t)) = E [u0]� v21

x� v�E

��u�10 u0x

�x

�;

<2(�!u 1(x; t)) = E [u1]� v�E��u�10 u1x � u�20 u1u0x

�x

�; (35)

<3(�!u 2(x; t)) = E [u2]� v�E��u�30 u

21u0x � u�20 u2u0x � u�20 u1u1x + u�10 u2x

�x

�;

...

Using the initial condition (27), the iteration formulas (34) and (35), weobtain :

u0(x; t) =1

x;

u1(x; t) = � hx2

t�

�(�+ 1);

u2(x; t) = � hx2

t�

�(�+ 1)� h

2

x2t�

�(�+ 1)+ 2

h2

x3t2�

�(2�+ 1);

u3(x; t) = � hx2

t�

�(�+ 1)� h

2

x2t�

�(�+ 1)+ 4

h2

x3t2�

�(2�+ 1)(36)

�h2

x2t�

�(�+ 1)� h

3

x2t�

�(�+ 1)+ 4

h3

x3t2�

�(2�+ 1)

+3h3

x4�(2�+ 1)

�(�+ 1)2t3�

�(3�+ 1)� 12h

3

x4t3�

�(3�+ 1)...

Thus, we use four terms in evaluating the approximate solution :

u(x; t) =3X

m=0

um(x; t):

10


When h = �1, the approximate solution of Eq.(26), is given by :

u(x; t) =3X

m=0

um(x; t)

=1

x+

1

�(�+ 1)

t�

x2+

2

�(2�+ 1)

t2�

x3� 3�(2�+ 1)

(�(�+ 1))2�(3�+ 1)

t3�

x4

+12

�(3�+ 1)

t3�

x4:

If � = 1; we obtain :

u(x; t) =1

x+t

x2+t2

x3+t3

x4+ � � �

That gives :

u(x; t) =1

x� t ;�� tx�� < 1; x 6= 0;

which is an exact solution to the porous medium equation as presented in[24].

(a) (b)

11


(c)

Fig. 1 : (a) Exact solution, (b) the approximate solution in the case� = 1, (c) The exact solution and approximate solutions to the Eq.(26) fordi¤erent values of � when x = 2:

4.2 Example

Second, we take k = �4=3 in equation (1)[24], we get :cD�

t u =�u�4=3ux

�x; 0 < � 6 1; (37)

with the initial condition :

u(x; 0) = (2x)�3=4: (38)

Applying Elzaki transform on both sides of Eq.(37), we get :

E [u]� v2u(x; 0) = v�E��u�4=3ux

�x

�(39)

From (39) and the initial condition (38), we have :

E [u]� v2(2x)�3=4 � v�E��u�4=3ux

�x

�= 0:

We take the nonlinear part as :

N [�(x; t; p)] = E [�]� v2(2x)�3=4 � v�Eh��4=3�x

�x

i:

12


We construct the so-called the zero-order deformation equation with as-sumptionH(x; t) = 1, we have :

(1� q)E[�(x; t; p)� �(x; t; 0) = phN [�(x; t; p)]:

When p = 0 and p = 1, we can obtain :��(x; t; 0) = u0(x; t);�(x; t; 1) = u(x; t):

Therefore, we have the mth order deformation equation :

E[um(x; t)� �mum�1(x; t)] = h<m(�!u m�1(x; t)): (40)

Operating the inverse Elzaki operator on both sides of Eq.(40), we get :

um(x; t) = �mum�1(x; t) + E�1[h<m(�!u m�1(x; t))]; (41)

From Eq.(41), we have :

u1(x; t) = hE�1[ <1(�!u 0(x; t))];u2(x; t) = u1 + hE

�1[ <2(�!u 1(x; t))]; (42)

u3(x; t) = u2 + hE�1[ <3(�!u 2(x; t))];

...

where

<1(�!u 0(x; t)) = E [u0]� v2(2x)�3=4 � v�Eh�u�4=30 u0x

�x

i;

<2(�!u 1(x; t)) = E [u1]� v�E��43u�7=30 u1u0x + u

�4=30 u1x

�x

�; (43)

<3(�!u 2(x; t)) = E [u2]� v�E"

149u�10=30 u21u0x � 4

3u�7=30 u2u0x

�43u�7=30 u1u1x + u

�4=30 u2x

!x

#;

...

13


Using the initial condition (38), the iteration formulas (42) and (43), weobtain :

u0(x; t) = (2x)�3=4;

u1(x; t) = �94h (2x)�7=4

t�

�(�+ 1);

u2(x; t) = �94h (2x)�7=4

t�

�(�+ 1)� 94h2 (2x)�7=4

t�

�(�+ 1)+189

16h2 (2x)�11=4

t2�

�(2�+ 1);

u3(x; t) = �94h (2x)�7=4

t�

�(�+ 1)� 94h2 (2x)�7=4

t�

�(�+ 1)+189

16h2 (2x)�11=4

t2�

�(2�+ 1)

�94h2 (2x)�7=4

t�

�(�+ 1)� 94h3 (2x)�7=4

t�

�(�+ 1)+189

16h3 (2x)�11=4

t2�

�(2�+ 1)

+1089

32h3 (2x)�15=4

�(2�+ 1)

(�(�+ 1))2t3�

�(3�+ 1)+189

16h2 (2x)�11=4

t2�

�(2�+ 1)

+189

16h3 (2x)�11=4

t2�

�(2�+ 1)� 14553

64h3 (2x)�15=4

t3�

�(3�+ 1)

+2079

16h2 (2x)�15=4

�(2�+ 1)

(�(�+ 1))2t3�

�(3�+ 1):

...

Thus, we use four terms in evaluating the approximate solution :

u(x; t) =3X

m=0

um(x; t):

When h = �1, the approximate solution of Eq.(37), is given by :

u(x; t) =

3Xm=0

um(x; t)

= (2x)�3=4 +9

4(2x)�7=4

t�

�(�+ 1)+189

16(2x)�11=4

t2�

�(2�+ 1)

+3069

32(2x)�15=4

�(2�+ 1)

(�(�+ 1))2t3�

�(3�+ 1)+14553

64(2x)�15=4

t3�

�(3�+ 1):

If � = 1; we obtain :

u(x; t) = (2x)�3=4 +9

4(2x)�7=4 t+

189

32(2x)�11=4 t2 +

8943

128(2x)�15=4 + � � � ;

14


or

u(x; t) = 2�3=4 � x�3=4 + (9� 2�15=4 � x�7=4)t+ (189� 2�31=4 � x�11=4)t2

+(8943� 2�43=4 � x�15=4)t3 + � � � ;which is the same approximate solution to the porous medium equation aspresented in [24].

(a) (b)

(c)

Fig. 2 : (a) The approximate solution when � = 1; (b) The approximatesolution when � = 0:98; (c) The approximate solutions to the Eq.(37) fordi¤erent values of � when x = 0:2:

15


Remark 9 For graph approximate solutions, we took only four terms.

5 Conclusion

The coupling of homotopy analysis method (HAM) and Elzaki transformmethod proved very e¤ective to solve nonlinear partial di¤erential equations.The modi�ed fractional homotopy analysis transform method (MFHATM),is suitable for such problems and is very user friendly. From the obtainedresults, it is clear that the MFHATM yields very accurate, exact and approx-imate solutions using only a few iterates. As a result, the conclusion thatcomes through this work is that MFHATM can be applied to other nonlinearfractional partial di¤erential equations of higher order, due to the e¢ ciencyand �exibility in the application to get the possible results.

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