Research Article Approximate Solutions of Nonlinear Partial Differential Equations...

10
Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 569674, 9 pages http://dx.doi.org/10.1155/2013/569674 Research Article Approximate Solutions of Nonlinear Partial Differential Equations by Modified -Homotopy Analysis Method Shaheed N. Huseen 1 and Said R. Grace 2 1 Mathematics Department, Faculty of Computer Science and Mathematics, i-Qar University, i-Qar, Iraq 2 Engineering Mathematics Department, Faculty of Engineering, Cairo University, Giza, Egypt Correspondence should be addressed to Shaheed N. Huseen; shn [email protected] Received 27 May 2013; Accepted 23 September 2013 Academic Editor: Alexander Timokha Copyright © 2013 S. N. Huseen and S. R. Grace. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. A modified -homotopy analysis method (m-HAM) was proposed for solving th-order nonlinear differential equations. is method improves the convergence of the series solution in the HAM which was proposed in (see Hassan and El-Tawil 2011, 2012). e proposed method provides an approximate solution by rewriting the th-order nonlinear differential equation in the form of first-order differential equations. e solution of these differential equations is obtained as a power series solution. is scheme is tested on two nonlinear exactly solvable differential equations. e results demonstrate the reliability and efficiency of the algorithm developed. 1. Introduction Homotopy analysis method (HAM) initially proposed by Liao in his Ph.D. thesis [1] is a powerful method to solve nonlinear problems. In recent years, this method has been successfully employed to solve many types of nonlinear problems in science and engineering [2–17]. HAM contains a certain auxiliary parameter ℎ, which provides us with a simple way to adjust and control the convergence region and rate of convergence of the series solution. Moreover, by means of the so-called ℎ-curve, a valid region of ℎ can be studied to gain a convergent series solution. More recently, a powerful modification of HAM was proposed [18–20]. Hassan and El- Tawil [21, 22] presented a new technique of using homotopy analysis method for solving nonlinear initial value problems (HAM). El-Tawil and Huseen [23, 24] established a method, namely, -homotopy analysis method, (-HAM) which is a more general method of HAM, e -HAM contains an auxiliary parameter as well as ℎ such that the case of = 1 (-HAM; = 1) and the standard homotopy analysis method (HAM) can be reached. In this paper, we present the modification of -homotopy analysis method (m-HAM) for solving nonlinear problems by transforming the th-order nonlinear differential equation to a system of first-order equations. we note that the HAM is a special case of m- HAM (m-HAM; =1). 2. Analysis of the -Homotopy Analysis Method (-HAM) Consider the following nonlinear partial differential equa- tion: [ (, )] = 0, (1) where is a nonlinear operator, (, ) denotes indepen- dent variables, and (, ) is an unknown function. Let us construct the so-called zero-order deformation equation as follows: (1 − ) [0 (, ; ) − 0 (, )] = ℎ (, ) [0 (, ; )] , (2) where ≥1, ∈ [0, 1/] denotes the so-called embedded parameter, is an auxiliary linear operator with the property [] = 0 when =0, ℎ ̸ =0 is an auxiliary parameter, and

Transcript of Research Article Approximate Solutions of Nonlinear Partial Differential Equations...

Page 1: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 569674 9 pageshttpdxdoiorg1011552013569674

Research ArticleApproximate Solutions of Nonlinear Partial DifferentialEquations by Modified 119902-Homotopy Analysis Method

Shaheed N Huseen1 and Said R Grace2

1 Mathematics Department Faculty of Computer Science and Mathematics Thi-Qar University Thi-Qar Iraq2 Engineering Mathematics Department Faculty of Engineering Cairo University Giza Egypt

Correspondence should be addressed to Shaheed N Huseen shn n2002yahoocom

Received 27 May 2013 Accepted 23 September 2013

Academic Editor Alexander Timokha

Copyright copy 2013 S N Huseen and S R GraceThis is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited

A modified 119902-homotopy analysis method (m119902-HAM) was proposed for solving 119899th-order nonlinear differential equations Thismethod improves the convergence of the series solution in the 119899HAMwhich was proposed in (see Hassan and El-Tawil 2011 2012)The proposed method provides an approximate solution by rewriting the 119899th-order nonlinear differential equation in the form of 119899first-order differential equationsThe solution of these 119899 differential equations is obtained as a power series solutionThis scheme istested on two nonlinear exactly solvable differential equationsThe results demonstrate the reliability and efficiency of the algorithmdeveloped

1 Introduction

Homotopy analysis method (HAM) initially proposed byLiao in his PhD thesis [1] is a powerful method to solvenonlinear problems In recent years this method has beensuccessfully employed to solve many types of nonlinearproblems in science and engineering [2ndash17] HAM containsa certain auxiliary parameter ℎ which provides us with asimple way to adjust and control the convergence region andrate of convergence of the series solutionMoreover bymeansof the so-called ℎ-curve a valid region of ℎ can be studied togain a convergent series solution More recently a powerfulmodification of HAM was proposed [18ndash20] Hassan and El-Tawil [21 22] presented a new technique of using homotopyanalysis method for solving nonlinear initial value problems(119899HAM) El-Tawil and Huseen [23 24] established a methodnamely 119902-homotopy analysis method (119902-HAM) which isa more general method of HAM The 119902-HAM contains anauxiliary parameter 119899 as well as ℎ such that the case of 119899 =1 (119902-HAM 119899 = 1) and the standard homotopy analysismethod (HAM) can be reached In this paper we present themodification of 119902-homotopy analysis method (m119902-HAM) forsolving nonlinear problems by transforming the 119899th-ordernonlinear differential equation to a system of 119899 first-order

equations we note that the 119899HAM is a special case of m119902-HAM (m119902-HAM 119899 = 1)

2 Analysis of the 119902-Homotopy AnalysisMethod (119902-HAM)

Consider the following nonlinear partial differential equa-tion

119873[119906 (119909 119905)] = 0 (1)

where 119873 is a nonlinear operator (119909 119905) denotes indepen-dent variables and 119906(119909 119905) is an unknown function Let usconstruct the so-called zero-order deformation equation asfollows

(1 minus 119899119902) 119871 [0 (119909 119905 119902) minus 1199060(119909 119905)] = 119902ℎ119867 (119909 119905)119873 [0 (119909 119905 119902)]

(2)

where 119899 ge 1 119902 isin [0 1119899] denotes the so-called embeddedparameter 119871 is an auxiliary linear operator with the property119871[119891] = 0 when 119891 = 0 ℎ = 0 is an auxiliary parameter and

2 Journal of Applied Mathematics

119867(119909 119905) denotes a non-zero auxiliary function It is obviousthat when 119902 = 0 and 119902 = 1119899 (2) becomes

0 (119909 119905 0) = 1199060(119909 119905) 0 (119909 119905

1

119899) = 119906 (119909 119905) (3)

respectively Thus as 119902 increases from 0 to 1119899 the solution0(119909 119905 119902) varies from the initial guess 119906

0(119909 119905) to the solution

119906(119909 119905) We may choose 1199060(119909 119905) 119871 ℎ and119867(119909 119905) and assume

that all of them can be properly chosen so that the solution0(119909 119905 119902) of (2) exists for 119902 isin [0 1119899]

Now by expanding 0(119909 119905 119902) in Taylor series we have

0 (119909 119905 119902) = 1199060(119909 119905) +

+infin

sum

119898=1

119906119898(119909 119905) 119902

119898 (4)

where

119906119898 (119909 119905) =

1

119898

1205971198980(119909 119905 119902)

120597119902119898

10038161003816100381610038161003816100381610038161003816119902=0

(5)

Next we assume that ℎ119867(119909 119905) 1199060(119909 119905) and 119871 are properly

chosen such that the series (4) converges at 119902 = 1119899 and that

119906 (119909 119905) = 0 (119909 1199051

119899) = 1199060(119909 119905) +

+infin

sum

119898=1

119906119898(119909 119905) (

1

119899)

119898

(6)

Let

119906119903(119909 119905) = 119906

0(119909 119905) 119906

1(119909 119905) 119906

2(119909 119905) 119906

119903(119909 119905) (7)

Differentiating equation (2) for119898 times with respect to 119902 andthen setting 119902 = 0 and dividing the resulting equation by119898 we have the so-called119898th order deformation equation asfollows

119871 [119906119898(119909 119905) minus 119896

119898119906119898minus1(119909 119905)] = ℎ119867 (119909 119905) 119877

119898(997888997888997888119906119898minus1(119909 119905))

(8)

where

119877119898(997888997888997888119906119898minus1 (119909 119905))

=1

(119898 minus 1)

120597119898minus1(119873 [0 (119909 119905 119902)] minus 119891 (119909 119905))

120597119902119898minus1

100381610038161003816100381610038161003816100381610038161003816119902=0

119896119898= 0 119898 le 1

119899 otherwise

(9)

It should be emphasized that 119906119898(119909 119905) for 119898 ge 1 is governed

by the linear equation (8) with linear boundary conditionsthat come from the original problem Due to the existence ofthe factor (1119899)119898 more chances for convergence may occuror even much faster convergence can be obtained better thanthe standard HAM It should be noted that the case of 119899 = 1in (2) standard HAM can be reached

The 119902-homotopy analysis method (119902-HAM) can be refor-matted as follows

We rewrite the nonlinear partial differential equation (1)in the following form

119871119906 (119909 119905) + 119860119906 (119909 119905) + 119861119906 (119909 119905) = 0

119906 (119909 0) = 1198910 (119909)

120597119906(119909 119905)

120597119905

10038161003816100381610038161003816100381610038161003816119905=0

= 1198911(119909)

120597119911minus1119906(119909 119905)

120597119911minus1

100381610038161003816100381610038161003816100381610038161003816119905=0

= 119891119911minus1 (119909)

(10)

where 119871 = 120597119911120597119905119911 119911 = 1 2 is the highest partial derivativewith respect to 119905 119860 is a linear term and 119861 is a nonlinear termThe so-called zero-order deformation equation (2) becomes

(1 minus 119899119902) 119871 [0 (119909 119905 119902) minus 1199060 (119909 119905)]

= 119902ℎ119867 (119909 119905) (119871119906 (119909 119905) + 119860119906 (119909 119905) + 119861119906 (119909 119905))

(11)

we have the following119898th order deformation equation

119871 [119906119898(119909 119905) minus 119896

119898119906119898minus1(119909 119905)]

= ℎ119867 (119909 119905) (119871119906119898minus1(119909 119905) + 119860119906

119898minus1(119909 119905)+119861 (

997888997888997888119906119898minus1(119909 119905)))

(12)Hence119906119898 (119909 119905) = 119896119898119906119898minus1 (119909 119905)

+ ℎ119871minus1[119867 (119909 119905) (119871119906119898minus1 (119909 119905) + 119860119906119898minus1 (119909 119905)

+119861 (997888997888997888119906119898minus1(119909 119905)))]

(13)

Now the inverse operator 119871minus1 is an integral operator which isgiven by

119871minus1(sdot) = intint sdot sdot sdot int (sdot) 119889119905 119889119905 sdot sdot sdot 119889119905⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times+ 1198881119905119911minus1+ 1198882119905119911minus2+ sdot sdot sdot + 119888

119911

(14)

where 1198881 1198882 119888

119911are integral constants

To solve (10) by means of 119902-HAM we choose thefollowing initial approximation

1199060 (119909 119905) = 1198910 (119909) + 1198911 (119909) 119905

+ 1198912 (119909)

1199052

2+ sdot sdot sdot + 119891

119911minus1 (119909)119905119911minus1

(119911 minus 1)

(15)

Let119867(119909 119905) = 1 by means of (14) and (15) then (13) becomes

119906119898(119909 119905)

= 119896119898119906119898minus1(119909 119905)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(120597119911119906119898minus1(119909 120591)

120597120591119911+ 119860119906119898minus1(119909 120591)

+ 119861 (997888997888997888119906119898minus1(119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(16)

Journal of Applied Mathematics 3

Now from int1199050int119905

0sdot sdot sdot int119905

0(120597119911119906119898minus1(119909 120591)120597120591

119911)119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times we ob-

serve that there are repeated computations in each step whichcaused more consuming time To cancel this we use thefollowing modification to (16)

119906119898(119909 119905)

= 119896119898119906119898minus1 (119909 119905)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

120597119911119906119898minus1 (119909 120591)

120597120591119911119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(119860119906119898minus1 (119909 120591)

+ 119861 (997888997888997888119906119898minus1(119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

= 119896119898119906119898minus1(119909 119905) + ℎ119906

119898minus1(119909 119905)

minus ℎ (119906119898minus1(119909 0) + 119905

120597119906119898minus1(119909 0)

120597119905

+ sdot sdot sdot +119905119911minus1

(119911 minus 1)

120597119911minus1119906119898minus1 (119909 0)

120597119905119911minus1)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(119860119906119898minus1(119909 120591)

+119861 (997888997888997888119906119898minus1(119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(17)

Now for119898 = 1 119896119898= 0 and

1199060 (119909 0) + 119905

1205971199060 (119909 0)

120597119905+1199052

2

12059721199060 (119909 0)

1205971199052

+ sdot sdot sdot +119905119911minus1

(119911 minus 1)

120597119911minus11199060(119909 0)

120597119905119911minus1

= 1198910(119909) + 119891

1(119909) 119905 + 119891

2(119909)1199052

2

+ sdot sdot sdot + 119891119911minus1(119909)

119905119911minus1

(119911 minus 1)

= 1199060(119909 119905)

(18)

Substituting this equality into (17) we obtain

1199061 (119909 119905)

= ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(1198601199060 (119909 120591) + 119861 (1199060 (119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(19)

For119898 gt 1 119896119898= 119899 and

119906119898(119909 0) = 0

120597119906119898(119909 0)

120597119905= 0

1205972119906119898(119909 0)

1205971199052= 0

120597119911minus1119906119898(119909 0)

120597119905119911minus1= 0

(20)

Substituting this equality into (17) we obtain

119906119898(119909 119905)

= (119899 + ℎ) 119906119898minus1(119909 119905)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(119860119906119898minus1(119909 120591)

+119861 (997888997888997888119906119898minus1 (119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(21)

The standard 119902-HAM is powerful when 119911 = 1 and the seriessolution expression by 119902-HAMcan bewritten in the followingform

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(22)

But when 119911 ge 2 there are too many additional termswhere harder and more time consuming computations areperformed So the closed form solution needsmore numbersof iteration

3 The Proposed Modified 119902-HomotopyAnalysis Method (m119902-HAM)

When 119911 ge 2 we rewrite (1) as in the following system of first-order differential equations

119906119905= 1199061

1199061119905= 1199062

119906 119911 minus 1119905= minus 119860119906 (119909 119905) minus 119861119906 (119909 119905)

(23)

Set the initial approximation

1199060(119909 119905) = 119891

0(119909)

11990610 (119909 119905) = 1198911 (119909)

119906 119911 minus 10(119909 119905) = 119891

119911minus1(119909)

(24)

Using the iteration formulas (19) and (21) as follows

1199061 (119909 119905) = ℎint

119905

0

(minus11990610 (119909 120591)) 119889120591

11990611(119909 119905) = ℎint

119905

0

(minus11990620(119909 120591)) 119889120591

119906119911 minus 11 (119909 119905) = ℎint

119905

0

(1198601199060 (119909 120591) + 119861 (1199060 (119909 120591))) 119889120591

(25)

4 Journal of Applied Mathematics

For119898 gt 1 119896119898= 119899 and

119906119898 (119909 0) = 0 1199061

119898 (119909 0) = 0

1199062119898(119909 0) = 0 119906 119911 minus 1

119898(119909 0) = 0

(26)

Substituting in (17) we obtain

119906119898 (119909 119905) = (119899 + ℎ) 119906119898minus1 (119909 119905) + ℎint

119905

0

(minus1199061119898minus1 (119909 120591)) 119889120591

1199061119898(119909 119905) = (119899 + ℎ) 1199061

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1199062119898minus1(119909 120591)) 119889120591

119906 119911 minus 1119898(119909 119905)

= (119899 + ℎ) 119906119911 minus 1119898minus1 (119909 119905)

+ ℎint

119905

0

(119860119906119898minus1 (119909 120591) + 119861 (119906119898minus1 (119909 120591))) 119889120591

(27)

It should be noted that the case of 119899 = 1 in (27) the 119899HAMcan be reached

To illustrate the effectiveness of the proposed m119902-HAMcomparison betweenm119902-HAMand the 119899HAMare illustratedby the following examples

4 Illustrative Examples

Example 1 Consider the following nonlinear sine-Gordonequation

119906119905119905minus 119906119909119909+ sin 119906 = 0 (28)

subject to the following initial conditions

119906 (119909 0) = 0 119906119905(119909 0) = 4 sech 119909 (29)

The exact solution is

119906 (119909 119905) = 4 tanminus1 (119905 sech119909) (30)

In order to prevent suffering from the strongly nonlinearterm sin 119906 in the frame of 119902-HAM we can use Taylor seriesexpansion of sin 119906 as follows

sin 119906 = 119906 minus 1199063

6+1199065

120 (31)

Then (28) becomes

119906119905119905minus 119906119909119909+ 119906 minus

1199063

6+1199065

120= 0 (32)

In order to solve (28) by m119902-HAM we construct system ofdifferential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus 119906 +

1199063

6minus1199065

120

(33)

with the following initial approximations

1199060(119909 119905) = 0 V

0(119909 119905) = 4 sech119909 (34)

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+ 119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1 (119909 119905) = minus

1

6

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

+1

120

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119895minus119894

119894

sum

119896=0

119906119894minus119896

119896

sum

119897=0

119906119897119906119896minus119897

(35)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1(119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+ 1199060minus1199063

0

6+1199065

0

120)119889120591

(36)

Now for119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898 (119909 119905) = (119899 + ℎ) V119898minus1 (119909 119905)

+ ℎint

119905

0

(119860119906119898minus1 (119909 120591) + 119861 (119906119898minus1 (119909 120591))) 119889120591

(37)

And the following results are obtained

1199061 (119909 119905) = minus4ℎ119905 sech 119909

V1(119909 119905) = 0

1199062 (119909 119905) = minus4ℎ (ℎ + 119899) 119905 sech 119909

V2(119909 119905) = minus4ℎ

21199052sech3119909

1199063(119909 119905) = minus4ℎ(ℎ + 119899)

2119905 sech 119909 + 4

3ℎ31199053sech3119909

(38)

119906119898(119909 119905) (119898 = 4 5 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872 (119909 119905 119899 ℎ) =119872

sum

119894=0

119906119894 (119909 119905 119899 ℎ) (

1

119899)

119894

(39)

Equation (39) is a family of approximation solutions to theproblem (28) in terms of the convergence parameters ℎ and119899 To find the valid region of ℎ the ℎ-curves given by the6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

Journal of Applied Mathematics 5

02

04

06

08

10

minus20 minus15 minus10 minus05 00

U6n

HA

M (x

t)

U6nHAM (5 2)U6nHAM (3 1)U6nHAM (25 15)

h

Figure 1 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (28) at different values of 119909 and 119905

02

04

06

08

10

minus25 minus20 minus15 minus10 minus5

U6mq

-HA

M (x

tn

)

h

U6mq-HAM (5 2 13)U6mq-HAM (3 1 13)U6mq-HAM (25 15 13)

Figure 2 ℎ-curve for the (m119902-HAM 119899 = 13) approximationsolution 119880

6(119909 119905 13) of problem (28) at different values of 119909 and 119905

the 6th-order m119902-HAM (119899 = 13) approximation at differentvalues of 119909 119905 are drawn in Figures 1 and 2 respectively andthese figures show the interval of ℎ in which the value of1198806is constant at certain 119909 119905 and 119899 we chose the horizontal

line parallel to 119909-axis (ℎ) as a valid region which providesus with a simple way to adjust and control the convergenceregion Figure 3 shows the comparison between119880

6of 119899HAM

and 1198806of m119902-HAM using different values of 119899 with the

solution (30) The absolute errors of the 6th-order solutions119899HAM approximate and the 6th-order solutions m119902-HAMapproximate using different values of 119899 are shown in Figure 4The results obtained by m119902-HAM indicate that the speed ofconvergence for m119902-HAMwith 119899 gt 1 is faster in comparisonto 119899 = 1 (119899HAM) The results show that the convergenceregion of series solutions obtained by m119902-HAM is increasingas 119902 is decreased as shown in Figures 3 and 4

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient and

00 05 10 15 20 25

2

4

6

8

Solu

tions

Exact

t

U6nHAMU6mq-HAM (n = 55)

U6mq-HAM (n = 13)U6mq-HAM (n = 30)U6mq-HAM (n = 75)

Figure 3 Comparison between1198806of 119899HAM (m119902-HAM 119899 = 1) and

1198806m119902-HAM (119899 = 55 13 30 75)with exact solution of (28) at 119909 = 1

with (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

05 10 15 20 25

05

10

15

Abso

lute

erro

r

AE U6nHAMAE U6mq-HAM (n = 55)AE U6mq-HAM (n = 13)

AE U6mq-HAM (n = 30)AE U6mq-HAM (n = 75)

t

Figure 4 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806m119902-HAM (119899 = 55 13 30 75) for problem (28) at 119909 = 1

using (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

the interval of t (convergent region) increases as shown inFigures 5 6 7 and 8

Example 2 Consider the following Klein-Gordon equation

119906119905119905minus 119906119909119909+3

4119906 minus3

21199063= 0 (40)

subject to the following initial conditions

119906 (119909 0) = minussech119909

119906119905(119909 0) =

1

2sech 119909 tanh119909

(41)

6 Journal of Applied Mathematics

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3nHAM

U6nHAMU10nHAM

t

Figure 5 The comparison between the 1198803 1198806 and 119880

10of 119899HAM

(m119902-HAM 119899 = 1) and the exact solution of (28) at ℎ = minus1 and119909 = 1

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3mq-HAM (n = 75)

U6mq-HAM (n = 75)U10mq-HAM (n = 75)

t

Figure 6The comparison between the1198803 1198806 and119880

10of m119902-HAM

(119899 = 75) and the exact solution of (28) at ℎ = minus4925 and 119909 = 1

The exact solution is

119906 (119909 119905) = minussech(119909 + 1199052) (42)

In order to solve (40) by m119902-HAM we construct systemof differential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus3

4119906 +3

21199063

(43)

with the following initial approximations

1199060(119909 119905) = minussech119909 V

0(119909 119905) =

1

2sech 119909 tanh119909 (44)

02 04 06 08 10 12 14

005

010

015

Abso

lute

erro

r

AE U3nHAMAE U6nHAMAE U10nHAM

t

Figure 7The comparison between the absolute error of 11988031198806 and

11988010

of 119899HAM (m119902-HAM 119899 = 1) of (28) at ℎ = minus1 119909 = 1 and0 le 119905 le 15

02 04 06 08 10 12 14

002

004

006

008

010

Abso

lute

erro

r

AE U3mq-HAM (n = 75)AE U6mq-HAM (n = 75)AE U10mq-HAM (n = 75)

t

Figure 8The comparison between the absolute error of 1198803 1198806 and

11988010of m119902-HAM (119899 = 75) of (28) at ℎ = minus4925 119909 = 1 and 0 le 119905 le

15

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+3

4119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1(119909 119905) = minus

3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

(45)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1 (119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+3

41199060minus3

21199063

0)119889120591

(46)

Journal of Applied Mathematics 7

minus02

minus04

minus06

minus08

U6n

HA

M (x

t)

00minus05minus10minus15minus20

h

U6nHAM (01 06)U6nHAM (03 08)U6nHAM (06 1)

Figure 9 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (40) at different values of 119909 and 119905

For119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898(119909 119905) = (119899 + ℎ) V

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1205972119906119898minus1 (119909 119905)

1205971199092+3

4119906119898minus1(119909 119905)

minus3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894)119889120591

(47)

The following results are obtained

1199061 (119909 119905) = minus

1

2ℎ119905 sech119909 tanh119909

V1 (119909 119905) = ℎ119905 (minus

3 sech1199094

+sech31199092

+ sech119909tanh2119909)

1199062(119909 119905) = ℎ (

3

16ℎ1199052sech3119909 minus 1

16ℎ1199052 cosh (2119909) sech3119909)

minus1

2ℎ (ℎ + 119899) 119905 sech 119909 tanh119909

(48)

119906119898(119909 119905) (119898 = 3 4 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(49)

Equation (49) is a family of approximation solutions tothe problem (40) in terms of the convergence parameters ℎand 119899 To find the valid region of ℎ the ℎ-curves given bythe 6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

minus04

minus05

minus06

minus07

minus08

minus09

U6mq

-HA

M (x

tn

)

0minus50minus100minus150

U6mq-HAM (01 06 100)U6mq-HAM (03 08 100)U6mq-HAM (06 1 100)

h

Figure 10 ℎ-curve for the m119902-HAM (119899 = 100) approximationsolution 119880

6(119909 119905 100) of problem (40) at different values of 119909 and

119905

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

t

ExactU6nHAMU6mq-HAM (n = 5)

U6mq-HAM (n = 20)U6mq-HAM (n = 50)U6mq-HAM (n = 100)

Figure 11 Comparison between 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) with exact solution of (40)

at 119909 = 1 with (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ =minus795) respectively

the 6th-order m119902-HAM (119899 = 100) approximation at differentvalues of 119909 119905 are drawn in Figures 9 and 10 these figures showthe interval of ℎ in which the value of119880

6is constant at certain

119909 119905 and 119899 we chose the horizontal line parallel to 119909-axis (ℎ)as a valid regionwhich provides uswith a simpleway to adjustand control the convergence region Figure 11 shows thecomparison between119880

6of 119899HAM and119880

6of m119902-HAM using

different values of 119899with the solution (42)The absolute errorsof the 6th-order solutions 119899HAM approximate and the 6th-order solutions m119902-HAM approximate using different valuesof 119899 are shown in Figure 12 The results obtained by m119902-HAM indicate that the speed of convergence for m119902-HAMwith 119899 gt 1 is faster in comparison to 119899 = 1 (119899HAM) Theresults show that the convergence region of series solutions

8 Journal of Applied Mathematics

05 10 15 20 25 30 35

002

004

006

008

Abso

lute

erro

r

t

AE U6nHAMAE U6mq-HAM (n = 5)AE U6mq-HAM (n = 20)

AE U6mq-HAM (n = 50)AE U6mq-HAM (n = 100)

Figure 12 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) for problem (40) at 119909 = 1

using (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ = minus795)respectively

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

ExactU3nHAMU6nHAM

t

Figure 13 The comparison between the 1198803 1198806of 119899HAM (m119902-

HAM 119899 = 1) and the exact solution of (40) at ℎ = minus1 and 119909 = 1

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U3mq-HAM (n = 100)U6mq-HAM (n = 100)

Figure 14 The comparison between the 1198803 1198806of m119902-HAM (119899 =

100) and the exact solution of (40) at ℎ = minus795 and 119909 = 1

05 10 15 20 25 30 35

01

02

03

04

Abso

lute

erro

r

t

AE U3nHAMAE U6nHAM

Figure 15 The comparison between the absolute error of 1198803and

1198806of 119899HAM (m119902-HAM 119899 = 1) of (40) at ℎ = minus1 119909 = 1 and

0 le 119905 le 35

05 10 15 20 25 30 35

0005

0010

0015

0020

0025Ab

solu

te er

ror

t

AE U3mq-HAM (n = 100)AE U6mq-HAM (n = 100)

Figure 16The comparison between the absolute error of 1198803and119880

6

of m119902-HAM(119899 = 100) of (40) at ℎ = minus795 119909 = 1 and 0 le 119905 le 35

obtained bym119902-HAM is increasing as 119902 is decreased as shownin Figures 11 and 12

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient andthe interval of 119905 (convergent region) increases as shown inFigures 13 14 15 and 16

Figure 17 shows that the convergence of the series solu-tions obtained by the 3rd-order m119902-HAM (119899 = 100) is fasterthan that of the series solutions obtained by the 6th order119899HAM This fact shows the importance of the convergenceparameters 119899 in the m119902-HAM

5 Conclusion

In this paper a modified 119902-homotopy analysis method wasproposed (m119902-HAM)This method provides an approximatesolution by rewriting the 119899th-order nonlinear differentialequations in the form of system of 119899 first-order differentialequations The solution of these 119899 differential equations is

Journal of Applied Mathematics 9

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U6nHAMU3mq-HAM (n = 100)

Figure 17 The comparison between the 1198803of m119902-HAM (119899 = 100)

1198806of 119899HAM (m119902-HAM 119899 = 1) and the exact solution of (40) at

(ℎ = minus795 ℎ = minus1) and 119909 = 1

obtained as a power series solution which converges to aclosed form solution The m119902-HAM contains two auxiliaryparameters 119899 and ℎ such that the case of 119899 = 1 (m119902-HAM 119899 =1) the 119899HAM which is proposed in [21 22] can be reachedIn general it was noticed from the illustrative examples thatthe convergence of m119902-HAM is faster than that of 119899HAM

References

[1] S J Liao The proposed homotopy analysis technique for thesolution of nonlinear problems [PhD thesis] Shanghai Jiao TongUniversity 1992

[2] S Abbasbandy ldquoThe application of homotopy analysis methodto nonlinear equations arising in heat transferrdquo Physics LettersA vol 360 no 1 pp 109ndash113 2006

[3] S Abbasbandy andA Shirzadi ldquoHomotopy analysismethod formultiple solutions of the fractional Sturm-Liouville problemsrdquoNumerical Algorithms vol 54 no 4 pp 521ndash532 2010

[4] S Abbasbandy M Ashtiani and E Babolian ldquoAnalytic solutionof the Sharma-Tasso-Olver equation by homotopy analysismethodrdquoZeitschrift furNaturforschungA vol 65 no 4 pp 285ndash290 2010

[5] A K Alomari M S M Noorani and R Nazar ldquoExplicitseries solutions of some linear and nonlinear Schrodingerequations via the homotopy analysis methodrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 14 no 4pp 1196ndash1207 2009

[6] M Ayub A Rasheed and T Hayat ldquoExact flow of a third gradefluid past a porous plate using homotopy analysis methodrdquoInternational Journal of Engineering Science vol 41 no 18 pp2091ndash2103 2003

[7] A Sami Bataineh M S M Noorani and I Hashim ldquoSolvingsystems of ODEs by homotopy analysis methodrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 13 no10 pp 2060ndash2070 2008

[8] A S Bataineh M S M Noorani and I Hashim ldquoHomotopyanalysis method for singular IVPs of Emden-Fowler typerdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 14 no 4 pp 1121ndash1131 2009

[9] T Hayat M Khan and S Asghar ldquoHomotopy analysis of MHDflows of an Oldroyd 8-constant fluidrdquo Acta Mechanica vol 168no 3-4 pp 213ndash232 2004

[10] S J Liao Beyond Perturbation Introduction to the HomotopyAnalysisMethod ChapmanampHall Boca Raton Fla USA 2004

[11] S J Liao ldquoAn approximate solution technique not dependingon small parameters a special examplerdquo International Journalof Non-Linear Mechanics vol 30 no 3 pp 371ndash380 1995

[12] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[13] S Liao ldquoComparison between the homotopy analysis methodand homotopy perturbationmethodrdquoAppliedMathematics andComputation vol 169 no 2 pp 1186ndash1194 2005

[14] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[15] S Nadeem and N S Akbar ldquoPeristaltic flow of Sisko fluid in auniform inclined tuberdquoActaMechanica Sinica vol 26 no 5 pp675ndash683 2010

[16] S Nadeem and N S Akbar ldquoInfluence of heat transfer on aperistaltic flow of Johnson Segalman fluid in a non uniformtuberdquo International Communications in Heat andMass Transfervol 36 no 10 pp 1050ndash1059 2009

[17] Y Tan and S Abbasbandy ldquoHomotopy analysis method forquadratic Riccati differential equationrdquo Communications inNonlinear Science and Numerical Simulation vol 13 no 3 pp539ndash546 2008

[18] A Sami Bataineh M S M Noorani and I Hashim ldquoApproxi-mate solutions of singular two-point BVPs by modified homo-topy analysis methodrdquo Physics Letters A vol 372 pp 4062ndash4066 2008

[19] A Sami Bataineh M S M Noorani and I Hashim ldquoOn a newreliablemodification of homotopy analysismethodrdquoCommuni-cations in Nonlinear Science and Numerical Simulation vol 14no 2 pp 409ndash423 2009

[20] A S Bataineh M S M Noorani and I Hashim ldquoModifiedhomotopy analysis method for solving systems of second-orderBVPsrdquo Communications in Nonlinear Science and NumericalSimulation vol 14 no 2 pp 430ndash442 2009

[21] H N Hassan and M A El-Tawil ldquoA new technique of usinghomotopy analysis method for solving high-order nonlineardifferential equationsrdquo Mathematical Methods in the AppliedSciences vol 34 no 6 pp 728ndash742 2011

[22] H N Hassan and M A El-Tawil ldquoA new technique ofusing homotopy analysis method for second order nonlineardifferential equationsrdquo Applied Mathematics and Computationvol 219 no 2 pp 708ndash728 2012

[23] M A El-Tawil and S N Huseen ldquoThe q-homotopy analysismethod (q-HAM)rdquo International Journal of Applied Mathemat-ics and Mechanics vol 8 no 15 pp 51ndash75 2012

[24] M A El-Tawil and S N Huseen ldquoOn convergence of the q-homotopy analysis methodrdquo International Journal of Contem-porary Mathematical Sciences vol 8 no 10 pp 481ndash497 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 2: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

2 Journal of Applied Mathematics

119867(119909 119905) denotes a non-zero auxiliary function It is obviousthat when 119902 = 0 and 119902 = 1119899 (2) becomes

0 (119909 119905 0) = 1199060(119909 119905) 0 (119909 119905

1

119899) = 119906 (119909 119905) (3)

respectively Thus as 119902 increases from 0 to 1119899 the solution0(119909 119905 119902) varies from the initial guess 119906

0(119909 119905) to the solution

119906(119909 119905) We may choose 1199060(119909 119905) 119871 ℎ and119867(119909 119905) and assume

that all of them can be properly chosen so that the solution0(119909 119905 119902) of (2) exists for 119902 isin [0 1119899]

Now by expanding 0(119909 119905 119902) in Taylor series we have

0 (119909 119905 119902) = 1199060(119909 119905) +

+infin

sum

119898=1

119906119898(119909 119905) 119902

119898 (4)

where

119906119898 (119909 119905) =

1

119898

1205971198980(119909 119905 119902)

120597119902119898

10038161003816100381610038161003816100381610038161003816119902=0

(5)

Next we assume that ℎ119867(119909 119905) 1199060(119909 119905) and 119871 are properly

chosen such that the series (4) converges at 119902 = 1119899 and that

119906 (119909 119905) = 0 (119909 1199051

119899) = 1199060(119909 119905) +

+infin

sum

119898=1

119906119898(119909 119905) (

1

119899)

119898

(6)

Let

119906119903(119909 119905) = 119906

0(119909 119905) 119906

1(119909 119905) 119906

2(119909 119905) 119906

119903(119909 119905) (7)

Differentiating equation (2) for119898 times with respect to 119902 andthen setting 119902 = 0 and dividing the resulting equation by119898 we have the so-called119898th order deformation equation asfollows

119871 [119906119898(119909 119905) minus 119896

119898119906119898minus1(119909 119905)] = ℎ119867 (119909 119905) 119877

119898(997888997888997888119906119898minus1(119909 119905))

(8)

where

119877119898(997888997888997888119906119898minus1 (119909 119905))

=1

(119898 minus 1)

120597119898minus1(119873 [0 (119909 119905 119902)] minus 119891 (119909 119905))

120597119902119898minus1

100381610038161003816100381610038161003816100381610038161003816119902=0

119896119898= 0 119898 le 1

119899 otherwise

(9)

It should be emphasized that 119906119898(119909 119905) for 119898 ge 1 is governed

by the linear equation (8) with linear boundary conditionsthat come from the original problem Due to the existence ofthe factor (1119899)119898 more chances for convergence may occuror even much faster convergence can be obtained better thanthe standard HAM It should be noted that the case of 119899 = 1in (2) standard HAM can be reached

The 119902-homotopy analysis method (119902-HAM) can be refor-matted as follows

We rewrite the nonlinear partial differential equation (1)in the following form

119871119906 (119909 119905) + 119860119906 (119909 119905) + 119861119906 (119909 119905) = 0

119906 (119909 0) = 1198910 (119909)

120597119906(119909 119905)

120597119905

10038161003816100381610038161003816100381610038161003816119905=0

= 1198911(119909)

120597119911minus1119906(119909 119905)

120597119911minus1

100381610038161003816100381610038161003816100381610038161003816119905=0

= 119891119911minus1 (119909)

(10)

where 119871 = 120597119911120597119905119911 119911 = 1 2 is the highest partial derivativewith respect to 119905 119860 is a linear term and 119861 is a nonlinear termThe so-called zero-order deformation equation (2) becomes

(1 minus 119899119902) 119871 [0 (119909 119905 119902) minus 1199060 (119909 119905)]

= 119902ℎ119867 (119909 119905) (119871119906 (119909 119905) + 119860119906 (119909 119905) + 119861119906 (119909 119905))

(11)

we have the following119898th order deformation equation

119871 [119906119898(119909 119905) minus 119896

119898119906119898minus1(119909 119905)]

= ℎ119867 (119909 119905) (119871119906119898minus1(119909 119905) + 119860119906

119898minus1(119909 119905)+119861 (

997888997888997888119906119898minus1(119909 119905)))

(12)Hence119906119898 (119909 119905) = 119896119898119906119898minus1 (119909 119905)

+ ℎ119871minus1[119867 (119909 119905) (119871119906119898minus1 (119909 119905) + 119860119906119898minus1 (119909 119905)

+119861 (997888997888997888119906119898minus1(119909 119905)))]

(13)

Now the inverse operator 119871minus1 is an integral operator which isgiven by

119871minus1(sdot) = intint sdot sdot sdot int (sdot) 119889119905 119889119905 sdot sdot sdot 119889119905⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times+ 1198881119905119911minus1+ 1198882119905119911minus2+ sdot sdot sdot + 119888

119911

(14)

where 1198881 1198882 119888

119911are integral constants

To solve (10) by means of 119902-HAM we choose thefollowing initial approximation

1199060 (119909 119905) = 1198910 (119909) + 1198911 (119909) 119905

+ 1198912 (119909)

1199052

2+ sdot sdot sdot + 119891

119911minus1 (119909)119905119911minus1

(119911 minus 1)

(15)

Let119867(119909 119905) = 1 by means of (14) and (15) then (13) becomes

119906119898(119909 119905)

= 119896119898119906119898minus1(119909 119905)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(120597119911119906119898minus1(119909 120591)

120597120591119911+ 119860119906119898minus1(119909 120591)

+ 119861 (997888997888997888119906119898minus1(119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(16)

Journal of Applied Mathematics 3

Now from int1199050int119905

0sdot sdot sdot int119905

0(120597119911119906119898minus1(119909 120591)120597120591

119911)119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times we ob-

serve that there are repeated computations in each step whichcaused more consuming time To cancel this we use thefollowing modification to (16)

119906119898(119909 119905)

= 119896119898119906119898minus1 (119909 119905)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

120597119911119906119898minus1 (119909 120591)

120597120591119911119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(119860119906119898minus1 (119909 120591)

+ 119861 (997888997888997888119906119898minus1(119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

= 119896119898119906119898minus1(119909 119905) + ℎ119906

119898minus1(119909 119905)

minus ℎ (119906119898minus1(119909 0) + 119905

120597119906119898minus1(119909 0)

120597119905

+ sdot sdot sdot +119905119911minus1

(119911 minus 1)

120597119911minus1119906119898minus1 (119909 0)

120597119905119911minus1)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(119860119906119898minus1(119909 120591)

+119861 (997888997888997888119906119898minus1(119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(17)

Now for119898 = 1 119896119898= 0 and

1199060 (119909 0) + 119905

1205971199060 (119909 0)

120597119905+1199052

2

12059721199060 (119909 0)

1205971199052

+ sdot sdot sdot +119905119911minus1

(119911 minus 1)

120597119911minus11199060(119909 0)

120597119905119911minus1

= 1198910(119909) + 119891

1(119909) 119905 + 119891

2(119909)1199052

2

+ sdot sdot sdot + 119891119911minus1(119909)

119905119911minus1

(119911 minus 1)

= 1199060(119909 119905)

(18)

Substituting this equality into (17) we obtain

1199061 (119909 119905)

= ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(1198601199060 (119909 120591) + 119861 (1199060 (119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(19)

For119898 gt 1 119896119898= 119899 and

119906119898(119909 0) = 0

120597119906119898(119909 0)

120597119905= 0

1205972119906119898(119909 0)

1205971199052= 0

120597119911minus1119906119898(119909 0)

120597119905119911minus1= 0

(20)

Substituting this equality into (17) we obtain

119906119898(119909 119905)

= (119899 + ℎ) 119906119898minus1(119909 119905)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(119860119906119898minus1(119909 120591)

+119861 (997888997888997888119906119898minus1 (119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(21)

The standard 119902-HAM is powerful when 119911 = 1 and the seriessolution expression by 119902-HAMcan bewritten in the followingform

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(22)

But when 119911 ge 2 there are too many additional termswhere harder and more time consuming computations areperformed So the closed form solution needsmore numbersof iteration

3 The Proposed Modified 119902-HomotopyAnalysis Method (m119902-HAM)

When 119911 ge 2 we rewrite (1) as in the following system of first-order differential equations

119906119905= 1199061

1199061119905= 1199062

119906 119911 minus 1119905= minus 119860119906 (119909 119905) minus 119861119906 (119909 119905)

(23)

Set the initial approximation

1199060(119909 119905) = 119891

0(119909)

11990610 (119909 119905) = 1198911 (119909)

119906 119911 minus 10(119909 119905) = 119891

119911minus1(119909)

(24)

Using the iteration formulas (19) and (21) as follows

1199061 (119909 119905) = ℎint

119905

0

(minus11990610 (119909 120591)) 119889120591

11990611(119909 119905) = ℎint

119905

0

(minus11990620(119909 120591)) 119889120591

119906119911 minus 11 (119909 119905) = ℎint

119905

0

(1198601199060 (119909 120591) + 119861 (1199060 (119909 120591))) 119889120591

(25)

4 Journal of Applied Mathematics

For119898 gt 1 119896119898= 119899 and

119906119898 (119909 0) = 0 1199061

119898 (119909 0) = 0

1199062119898(119909 0) = 0 119906 119911 minus 1

119898(119909 0) = 0

(26)

Substituting in (17) we obtain

119906119898 (119909 119905) = (119899 + ℎ) 119906119898minus1 (119909 119905) + ℎint

119905

0

(minus1199061119898minus1 (119909 120591)) 119889120591

1199061119898(119909 119905) = (119899 + ℎ) 1199061

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1199062119898minus1(119909 120591)) 119889120591

119906 119911 minus 1119898(119909 119905)

= (119899 + ℎ) 119906119911 minus 1119898minus1 (119909 119905)

+ ℎint

119905

0

(119860119906119898minus1 (119909 120591) + 119861 (119906119898minus1 (119909 120591))) 119889120591

(27)

It should be noted that the case of 119899 = 1 in (27) the 119899HAMcan be reached

To illustrate the effectiveness of the proposed m119902-HAMcomparison betweenm119902-HAMand the 119899HAMare illustratedby the following examples

4 Illustrative Examples

Example 1 Consider the following nonlinear sine-Gordonequation

119906119905119905minus 119906119909119909+ sin 119906 = 0 (28)

subject to the following initial conditions

119906 (119909 0) = 0 119906119905(119909 0) = 4 sech 119909 (29)

The exact solution is

119906 (119909 119905) = 4 tanminus1 (119905 sech119909) (30)

In order to prevent suffering from the strongly nonlinearterm sin 119906 in the frame of 119902-HAM we can use Taylor seriesexpansion of sin 119906 as follows

sin 119906 = 119906 minus 1199063

6+1199065

120 (31)

Then (28) becomes

119906119905119905minus 119906119909119909+ 119906 minus

1199063

6+1199065

120= 0 (32)

In order to solve (28) by m119902-HAM we construct system ofdifferential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus 119906 +

1199063

6minus1199065

120

(33)

with the following initial approximations

1199060(119909 119905) = 0 V

0(119909 119905) = 4 sech119909 (34)

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+ 119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1 (119909 119905) = minus

1

6

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

+1

120

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119895minus119894

119894

sum

119896=0

119906119894minus119896

119896

sum

119897=0

119906119897119906119896minus119897

(35)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1(119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+ 1199060minus1199063

0

6+1199065

0

120)119889120591

(36)

Now for119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898 (119909 119905) = (119899 + ℎ) V119898minus1 (119909 119905)

+ ℎint

119905

0

(119860119906119898minus1 (119909 120591) + 119861 (119906119898minus1 (119909 120591))) 119889120591

(37)

And the following results are obtained

1199061 (119909 119905) = minus4ℎ119905 sech 119909

V1(119909 119905) = 0

1199062 (119909 119905) = minus4ℎ (ℎ + 119899) 119905 sech 119909

V2(119909 119905) = minus4ℎ

21199052sech3119909

1199063(119909 119905) = minus4ℎ(ℎ + 119899)

2119905 sech 119909 + 4

3ℎ31199053sech3119909

(38)

119906119898(119909 119905) (119898 = 4 5 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872 (119909 119905 119899 ℎ) =119872

sum

119894=0

119906119894 (119909 119905 119899 ℎ) (

1

119899)

119894

(39)

Equation (39) is a family of approximation solutions to theproblem (28) in terms of the convergence parameters ℎ and119899 To find the valid region of ℎ the ℎ-curves given by the6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

Journal of Applied Mathematics 5

02

04

06

08

10

minus20 minus15 minus10 minus05 00

U6n

HA

M (x

t)

U6nHAM (5 2)U6nHAM (3 1)U6nHAM (25 15)

h

Figure 1 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (28) at different values of 119909 and 119905

02

04

06

08

10

minus25 minus20 minus15 minus10 minus5

U6mq

-HA

M (x

tn

)

h

U6mq-HAM (5 2 13)U6mq-HAM (3 1 13)U6mq-HAM (25 15 13)

Figure 2 ℎ-curve for the (m119902-HAM 119899 = 13) approximationsolution 119880

6(119909 119905 13) of problem (28) at different values of 119909 and 119905

the 6th-order m119902-HAM (119899 = 13) approximation at differentvalues of 119909 119905 are drawn in Figures 1 and 2 respectively andthese figures show the interval of ℎ in which the value of1198806is constant at certain 119909 119905 and 119899 we chose the horizontal

line parallel to 119909-axis (ℎ) as a valid region which providesus with a simple way to adjust and control the convergenceregion Figure 3 shows the comparison between119880

6of 119899HAM

and 1198806of m119902-HAM using different values of 119899 with the

solution (30) The absolute errors of the 6th-order solutions119899HAM approximate and the 6th-order solutions m119902-HAMapproximate using different values of 119899 are shown in Figure 4The results obtained by m119902-HAM indicate that the speed ofconvergence for m119902-HAMwith 119899 gt 1 is faster in comparisonto 119899 = 1 (119899HAM) The results show that the convergenceregion of series solutions obtained by m119902-HAM is increasingas 119902 is decreased as shown in Figures 3 and 4

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient and

00 05 10 15 20 25

2

4

6

8

Solu

tions

Exact

t

U6nHAMU6mq-HAM (n = 55)

U6mq-HAM (n = 13)U6mq-HAM (n = 30)U6mq-HAM (n = 75)

Figure 3 Comparison between1198806of 119899HAM (m119902-HAM 119899 = 1) and

1198806m119902-HAM (119899 = 55 13 30 75)with exact solution of (28) at 119909 = 1

with (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

05 10 15 20 25

05

10

15

Abso

lute

erro

r

AE U6nHAMAE U6mq-HAM (n = 55)AE U6mq-HAM (n = 13)

AE U6mq-HAM (n = 30)AE U6mq-HAM (n = 75)

t

Figure 4 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806m119902-HAM (119899 = 55 13 30 75) for problem (28) at 119909 = 1

using (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

the interval of t (convergent region) increases as shown inFigures 5 6 7 and 8

Example 2 Consider the following Klein-Gordon equation

119906119905119905minus 119906119909119909+3

4119906 minus3

21199063= 0 (40)

subject to the following initial conditions

119906 (119909 0) = minussech119909

119906119905(119909 0) =

1

2sech 119909 tanh119909

(41)

6 Journal of Applied Mathematics

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3nHAM

U6nHAMU10nHAM

t

Figure 5 The comparison between the 1198803 1198806 and 119880

10of 119899HAM

(m119902-HAM 119899 = 1) and the exact solution of (28) at ℎ = minus1 and119909 = 1

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3mq-HAM (n = 75)

U6mq-HAM (n = 75)U10mq-HAM (n = 75)

t

Figure 6The comparison between the1198803 1198806 and119880

10of m119902-HAM

(119899 = 75) and the exact solution of (28) at ℎ = minus4925 and 119909 = 1

The exact solution is

119906 (119909 119905) = minussech(119909 + 1199052) (42)

In order to solve (40) by m119902-HAM we construct systemof differential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus3

4119906 +3

21199063

(43)

with the following initial approximations

1199060(119909 119905) = minussech119909 V

0(119909 119905) =

1

2sech 119909 tanh119909 (44)

02 04 06 08 10 12 14

005

010

015

Abso

lute

erro

r

AE U3nHAMAE U6nHAMAE U10nHAM

t

Figure 7The comparison between the absolute error of 11988031198806 and

11988010

of 119899HAM (m119902-HAM 119899 = 1) of (28) at ℎ = minus1 119909 = 1 and0 le 119905 le 15

02 04 06 08 10 12 14

002

004

006

008

010

Abso

lute

erro

r

AE U3mq-HAM (n = 75)AE U6mq-HAM (n = 75)AE U10mq-HAM (n = 75)

t

Figure 8The comparison between the absolute error of 1198803 1198806 and

11988010of m119902-HAM (119899 = 75) of (28) at ℎ = minus4925 119909 = 1 and 0 le 119905 le

15

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+3

4119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1(119909 119905) = minus

3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

(45)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1 (119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+3

41199060minus3

21199063

0)119889120591

(46)

Journal of Applied Mathematics 7

minus02

minus04

minus06

minus08

U6n

HA

M (x

t)

00minus05minus10minus15minus20

h

U6nHAM (01 06)U6nHAM (03 08)U6nHAM (06 1)

Figure 9 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (40) at different values of 119909 and 119905

For119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898(119909 119905) = (119899 + ℎ) V

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1205972119906119898minus1 (119909 119905)

1205971199092+3

4119906119898minus1(119909 119905)

minus3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894)119889120591

(47)

The following results are obtained

1199061 (119909 119905) = minus

1

2ℎ119905 sech119909 tanh119909

V1 (119909 119905) = ℎ119905 (minus

3 sech1199094

+sech31199092

+ sech119909tanh2119909)

1199062(119909 119905) = ℎ (

3

16ℎ1199052sech3119909 minus 1

16ℎ1199052 cosh (2119909) sech3119909)

minus1

2ℎ (ℎ + 119899) 119905 sech 119909 tanh119909

(48)

119906119898(119909 119905) (119898 = 3 4 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(49)

Equation (49) is a family of approximation solutions tothe problem (40) in terms of the convergence parameters ℎand 119899 To find the valid region of ℎ the ℎ-curves given bythe 6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

minus04

minus05

minus06

minus07

minus08

minus09

U6mq

-HA

M (x

tn

)

0minus50minus100minus150

U6mq-HAM (01 06 100)U6mq-HAM (03 08 100)U6mq-HAM (06 1 100)

h

Figure 10 ℎ-curve for the m119902-HAM (119899 = 100) approximationsolution 119880

6(119909 119905 100) of problem (40) at different values of 119909 and

119905

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

t

ExactU6nHAMU6mq-HAM (n = 5)

U6mq-HAM (n = 20)U6mq-HAM (n = 50)U6mq-HAM (n = 100)

Figure 11 Comparison between 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) with exact solution of (40)

at 119909 = 1 with (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ =minus795) respectively

the 6th-order m119902-HAM (119899 = 100) approximation at differentvalues of 119909 119905 are drawn in Figures 9 and 10 these figures showthe interval of ℎ in which the value of119880

6is constant at certain

119909 119905 and 119899 we chose the horizontal line parallel to 119909-axis (ℎ)as a valid regionwhich provides uswith a simpleway to adjustand control the convergence region Figure 11 shows thecomparison between119880

6of 119899HAM and119880

6of m119902-HAM using

different values of 119899with the solution (42)The absolute errorsof the 6th-order solutions 119899HAM approximate and the 6th-order solutions m119902-HAM approximate using different valuesof 119899 are shown in Figure 12 The results obtained by m119902-HAM indicate that the speed of convergence for m119902-HAMwith 119899 gt 1 is faster in comparison to 119899 = 1 (119899HAM) Theresults show that the convergence region of series solutions

8 Journal of Applied Mathematics

05 10 15 20 25 30 35

002

004

006

008

Abso

lute

erro

r

t

AE U6nHAMAE U6mq-HAM (n = 5)AE U6mq-HAM (n = 20)

AE U6mq-HAM (n = 50)AE U6mq-HAM (n = 100)

Figure 12 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) for problem (40) at 119909 = 1

using (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ = minus795)respectively

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

ExactU3nHAMU6nHAM

t

Figure 13 The comparison between the 1198803 1198806of 119899HAM (m119902-

HAM 119899 = 1) and the exact solution of (40) at ℎ = minus1 and 119909 = 1

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U3mq-HAM (n = 100)U6mq-HAM (n = 100)

Figure 14 The comparison between the 1198803 1198806of m119902-HAM (119899 =

100) and the exact solution of (40) at ℎ = minus795 and 119909 = 1

05 10 15 20 25 30 35

01

02

03

04

Abso

lute

erro

r

t

AE U3nHAMAE U6nHAM

Figure 15 The comparison between the absolute error of 1198803and

1198806of 119899HAM (m119902-HAM 119899 = 1) of (40) at ℎ = minus1 119909 = 1 and

0 le 119905 le 35

05 10 15 20 25 30 35

0005

0010

0015

0020

0025Ab

solu

te er

ror

t

AE U3mq-HAM (n = 100)AE U6mq-HAM (n = 100)

Figure 16The comparison between the absolute error of 1198803and119880

6

of m119902-HAM(119899 = 100) of (40) at ℎ = minus795 119909 = 1 and 0 le 119905 le 35

obtained bym119902-HAM is increasing as 119902 is decreased as shownin Figures 11 and 12

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient andthe interval of 119905 (convergent region) increases as shown inFigures 13 14 15 and 16

Figure 17 shows that the convergence of the series solu-tions obtained by the 3rd-order m119902-HAM (119899 = 100) is fasterthan that of the series solutions obtained by the 6th order119899HAM This fact shows the importance of the convergenceparameters 119899 in the m119902-HAM

5 Conclusion

In this paper a modified 119902-homotopy analysis method wasproposed (m119902-HAM)This method provides an approximatesolution by rewriting the 119899th-order nonlinear differentialequations in the form of system of 119899 first-order differentialequations The solution of these 119899 differential equations is

Journal of Applied Mathematics 9

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U6nHAMU3mq-HAM (n = 100)

Figure 17 The comparison between the 1198803of m119902-HAM (119899 = 100)

1198806of 119899HAM (m119902-HAM 119899 = 1) and the exact solution of (40) at

(ℎ = minus795 ℎ = minus1) and 119909 = 1

obtained as a power series solution which converges to aclosed form solution The m119902-HAM contains two auxiliaryparameters 119899 and ℎ such that the case of 119899 = 1 (m119902-HAM 119899 =1) the 119899HAM which is proposed in [21 22] can be reachedIn general it was noticed from the illustrative examples thatthe convergence of m119902-HAM is faster than that of 119899HAM

References

[1] S J Liao The proposed homotopy analysis technique for thesolution of nonlinear problems [PhD thesis] Shanghai Jiao TongUniversity 1992

[2] S Abbasbandy ldquoThe application of homotopy analysis methodto nonlinear equations arising in heat transferrdquo Physics LettersA vol 360 no 1 pp 109ndash113 2006

[3] S Abbasbandy andA Shirzadi ldquoHomotopy analysismethod formultiple solutions of the fractional Sturm-Liouville problemsrdquoNumerical Algorithms vol 54 no 4 pp 521ndash532 2010

[4] S Abbasbandy M Ashtiani and E Babolian ldquoAnalytic solutionof the Sharma-Tasso-Olver equation by homotopy analysismethodrdquoZeitschrift furNaturforschungA vol 65 no 4 pp 285ndash290 2010

[5] A K Alomari M S M Noorani and R Nazar ldquoExplicitseries solutions of some linear and nonlinear Schrodingerequations via the homotopy analysis methodrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 14 no 4pp 1196ndash1207 2009

[6] M Ayub A Rasheed and T Hayat ldquoExact flow of a third gradefluid past a porous plate using homotopy analysis methodrdquoInternational Journal of Engineering Science vol 41 no 18 pp2091ndash2103 2003

[7] A Sami Bataineh M S M Noorani and I Hashim ldquoSolvingsystems of ODEs by homotopy analysis methodrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 13 no10 pp 2060ndash2070 2008

[8] A S Bataineh M S M Noorani and I Hashim ldquoHomotopyanalysis method for singular IVPs of Emden-Fowler typerdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 14 no 4 pp 1121ndash1131 2009

[9] T Hayat M Khan and S Asghar ldquoHomotopy analysis of MHDflows of an Oldroyd 8-constant fluidrdquo Acta Mechanica vol 168no 3-4 pp 213ndash232 2004

[10] S J Liao Beyond Perturbation Introduction to the HomotopyAnalysisMethod ChapmanampHall Boca Raton Fla USA 2004

[11] S J Liao ldquoAn approximate solution technique not dependingon small parameters a special examplerdquo International Journalof Non-Linear Mechanics vol 30 no 3 pp 371ndash380 1995

[12] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[13] S Liao ldquoComparison between the homotopy analysis methodand homotopy perturbationmethodrdquoAppliedMathematics andComputation vol 169 no 2 pp 1186ndash1194 2005

[14] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[15] S Nadeem and N S Akbar ldquoPeristaltic flow of Sisko fluid in auniform inclined tuberdquoActaMechanica Sinica vol 26 no 5 pp675ndash683 2010

[16] S Nadeem and N S Akbar ldquoInfluence of heat transfer on aperistaltic flow of Johnson Segalman fluid in a non uniformtuberdquo International Communications in Heat andMass Transfervol 36 no 10 pp 1050ndash1059 2009

[17] Y Tan and S Abbasbandy ldquoHomotopy analysis method forquadratic Riccati differential equationrdquo Communications inNonlinear Science and Numerical Simulation vol 13 no 3 pp539ndash546 2008

[18] A Sami Bataineh M S M Noorani and I Hashim ldquoApproxi-mate solutions of singular two-point BVPs by modified homo-topy analysis methodrdquo Physics Letters A vol 372 pp 4062ndash4066 2008

[19] A Sami Bataineh M S M Noorani and I Hashim ldquoOn a newreliablemodification of homotopy analysismethodrdquoCommuni-cations in Nonlinear Science and Numerical Simulation vol 14no 2 pp 409ndash423 2009

[20] A S Bataineh M S M Noorani and I Hashim ldquoModifiedhomotopy analysis method for solving systems of second-orderBVPsrdquo Communications in Nonlinear Science and NumericalSimulation vol 14 no 2 pp 430ndash442 2009

[21] H N Hassan and M A El-Tawil ldquoA new technique of usinghomotopy analysis method for solving high-order nonlineardifferential equationsrdquo Mathematical Methods in the AppliedSciences vol 34 no 6 pp 728ndash742 2011

[22] H N Hassan and M A El-Tawil ldquoA new technique ofusing homotopy analysis method for second order nonlineardifferential equationsrdquo Applied Mathematics and Computationvol 219 no 2 pp 708ndash728 2012

[23] M A El-Tawil and S N Huseen ldquoThe q-homotopy analysismethod (q-HAM)rdquo International Journal of Applied Mathemat-ics and Mechanics vol 8 no 15 pp 51ndash75 2012

[24] M A El-Tawil and S N Huseen ldquoOn convergence of the q-homotopy analysis methodrdquo International Journal of Contem-porary Mathematical Sciences vol 8 no 10 pp 481ndash497 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

Journal of Applied Mathematics 3

Now from int1199050int119905

0sdot sdot sdot int119905

0(120597119911119906119898minus1(119909 120591)120597120591

119911)119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times we ob-

serve that there are repeated computations in each step whichcaused more consuming time To cancel this we use thefollowing modification to (16)

119906119898(119909 119905)

= 119896119898119906119898minus1 (119909 119905)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

120597119911119906119898minus1 (119909 120591)

120597120591119911119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(119860119906119898minus1 (119909 120591)

+ 119861 (997888997888997888119906119898minus1(119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

= 119896119898119906119898minus1(119909 119905) + ℎ119906

119898minus1(119909 119905)

minus ℎ (119906119898minus1(119909 0) + 119905

120597119906119898minus1(119909 0)

120597119905

+ sdot sdot sdot +119905119911minus1

(119911 minus 1)

120597119911minus1119906119898minus1 (119909 0)

120597119905119911minus1)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(119860119906119898minus1(119909 120591)

+119861 (997888997888997888119906119898minus1(119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(17)

Now for119898 = 1 119896119898= 0 and

1199060 (119909 0) + 119905

1205971199060 (119909 0)

120597119905+1199052

2

12059721199060 (119909 0)

1205971199052

+ sdot sdot sdot +119905119911minus1

(119911 minus 1)

120597119911minus11199060(119909 0)

120597119905119911minus1

= 1198910(119909) + 119891

1(119909) 119905 + 119891

2(119909)1199052

2

+ sdot sdot sdot + 119891119911minus1(119909)

119905119911minus1

(119911 minus 1)

= 1199060(119909 119905)

(18)

Substituting this equality into (17) we obtain

1199061 (119909 119905)

= ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(1198601199060 (119909 120591) + 119861 (1199060 (119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(19)

For119898 gt 1 119896119898= 119899 and

119906119898(119909 0) = 0

120597119906119898(119909 0)

120597119905= 0

1205972119906119898(119909 0)

1205971199052= 0

120597119911minus1119906119898(119909 0)

120597119905119911minus1= 0

(20)

Substituting this equality into (17) we obtain

119906119898(119909 119905)

= (119899 + ℎ) 119906119898minus1(119909 119905)

+ ℎint

119905

0

int

119905

0

sdot sdot sdot int

119905

0

(119860119906119898minus1(119909 120591)

+119861 (997888997888997888119906119898minus1 (119909 120591))) 119889120591 119889120591 sdot sdot sdot 119889120591⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119911 times

(21)

The standard 119902-HAM is powerful when 119911 = 1 and the seriessolution expression by 119902-HAMcan bewritten in the followingform

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(22)

But when 119911 ge 2 there are too many additional termswhere harder and more time consuming computations areperformed So the closed form solution needsmore numbersof iteration

3 The Proposed Modified 119902-HomotopyAnalysis Method (m119902-HAM)

When 119911 ge 2 we rewrite (1) as in the following system of first-order differential equations

119906119905= 1199061

1199061119905= 1199062

119906 119911 minus 1119905= minus 119860119906 (119909 119905) minus 119861119906 (119909 119905)

(23)

Set the initial approximation

1199060(119909 119905) = 119891

0(119909)

11990610 (119909 119905) = 1198911 (119909)

119906 119911 minus 10(119909 119905) = 119891

119911minus1(119909)

(24)

Using the iteration formulas (19) and (21) as follows

1199061 (119909 119905) = ℎint

119905

0

(minus11990610 (119909 120591)) 119889120591

11990611(119909 119905) = ℎint

119905

0

(minus11990620(119909 120591)) 119889120591

119906119911 minus 11 (119909 119905) = ℎint

119905

0

(1198601199060 (119909 120591) + 119861 (1199060 (119909 120591))) 119889120591

(25)

4 Journal of Applied Mathematics

For119898 gt 1 119896119898= 119899 and

119906119898 (119909 0) = 0 1199061

119898 (119909 0) = 0

1199062119898(119909 0) = 0 119906 119911 minus 1

119898(119909 0) = 0

(26)

Substituting in (17) we obtain

119906119898 (119909 119905) = (119899 + ℎ) 119906119898minus1 (119909 119905) + ℎint

119905

0

(minus1199061119898minus1 (119909 120591)) 119889120591

1199061119898(119909 119905) = (119899 + ℎ) 1199061

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1199062119898minus1(119909 120591)) 119889120591

119906 119911 minus 1119898(119909 119905)

= (119899 + ℎ) 119906119911 minus 1119898minus1 (119909 119905)

+ ℎint

119905

0

(119860119906119898minus1 (119909 120591) + 119861 (119906119898minus1 (119909 120591))) 119889120591

(27)

It should be noted that the case of 119899 = 1 in (27) the 119899HAMcan be reached

To illustrate the effectiveness of the proposed m119902-HAMcomparison betweenm119902-HAMand the 119899HAMare illustratedby the following examples

4 Illustrative Examples

Example 1 Consider the following nonlinear sine-Gordonequation

119906119905119905minus 119906119909119909+ sin 119906 = 0 (28)

subject to the following initial conditions

119906 (119909 0) = 0 119906119905(119909 0) = 4 sech 119909 (29)

The exact solution is

119906 (119909 119905) = 4 tanminus1 (119905 sech119909) (30)

In order to prevent suffering from the strongly nonlinearterm sin 119906 in the frame of 119902-HAM we can use Taylor seriesexpansion of sin 119906 as follows

sin 119906 = 119906 minus 1199063

6+1199065

120 (31)

Then (28) becomes

119906119905119905minus 119906119909119909+ 119906 minus

1199063

6+1199065

120= 0 (32)

In order to solve (28) by m119902-HAM we construct system ofdifferential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus 119906 +

1199063

6minus1199065

120

(33)

with the following initial approximations

1199060(119909 119905) = 0 V

0(119909 119905) = 4 sech119909 (34)

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+ 119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1 (119909 119905) = minus

1

6

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

+1

120

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119895minus119894

119894

sum

119896=0

119906119894minus119896

119896

sum

119897=0

119906119897119906119896minus119897

(35)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1(119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+ 1199060minus1199063

0

6+1199065

0

120)119889120591

(36)

Now for119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898 (119909 119905) = (119899 + ℎ) V119898minus1 (119909 119905)

+ ℎint

119905

0

(119860119906119898minus1 (119909 120591) + 119861 (119906119898minus1 (119909 120591))) 119889120591

(37)

And the following results are obtained

1199061 (119909 119905) = minus4ℎ119905 sech 119909

V1(119909 119905) = 0

1199062 (119909 119905) = minus4ℎ (ℎ + 119899) 119905 sech 119909

V2(119909 119905) = minus4ℎ

21199052sech3119909

1199063(119909 119905) = minus4ℎ(ℎ + 119899)

2119905 sech 119909 + 4

3ℎ31199053sech3119909

(38)

119906119898(119909 119905) (119898 = 4 5 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872 (119909 119905 119899 ℎ) =119872

sum

119894=0

119906119894 (119909 119905 119899 ℎ) (

1

119899)

119894

(39)

Equation (39) is a family of approximation solutions to theproblem (28) in terms of the convergence parameters ℎ and119899 To find the valid region of ℎ the ℎ-curves given by the6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

Journal of Applied Mathematics 5

02

04

06

08

10

minus20 minus15 minus10 minus05 00

U6n

HA

M (x

t)

U6nHAM (5 2)U6nHAM (3 1)U6nHAM (25 15)

h

Figure 1 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (28) at different values of 119909 and 119905

02

04

06

08

10

minus25 minus20 minus15 minus10 minus5

U6mq

-HA

M (x

tn

)

h

U6mq-HAM (5 2 13)U6mq-HAM (3 1 13)U6mq-HAM (25 15 13)

Figure 2 ℎ-curve for the (m119902-HAM 119899 = 13) approximationsolution 119880

6(119909 119905 13) of problem (28) at different values of 119909 and 119905

the 6th-order m119902-HAM (119899 = 13) approximation at differentvalues of 119909 119905 are drawn in Figures 1 and 2 respectively andthese figures show the interval of ℎ in which the value of1198806is constant at certain 119909 119905 and 119899 we chose the horizontal

line parallel to 119909-axis (ℎ) as a valid region which providesus with a simple way to adjust and control the convergenceregion Figure 3 shows the comparison between119880

6of 119899HAM

and 1198806of m119902-HAM using different values of 119899 with the

solution (30) The absolute errors of the 6th-order solutions119899HAM approximate and the 6th-order solutions m119902-HAMapproximate using different values of 119899 are shown in Figure 4The results obtained by m119902-HAM indicate that the speed ofconvergence for m119902-HAMwith 119899 gt 1 is faster in comparisonto 119899 = 1 (119899HAM) The results show that the convergenceregion of series solutions obtained by m119902-HAM is increasingas 119902 is decreased as shown in Figures 3 and 4

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient and

00 05 10 15 20 25

2

4

6

8

Solu

tions

Exact

t

U6nHAMU6mq-HAM (n = 55)

U6mq-HAM (n = 13)U6mq-HAM (n = 30)U6mq-HAM (n = 75)

Figure 3 Comparison between1198806of 119899HAM (m119902-HAM 119899 = 1) and

1198806m119902-HAM (119899 = 55 13 30 75)with exact solution of (28) at 119909 = 1

with (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

05 10 15 20 25

05

10

15

Abso

lute

erro

r

AE U6nHAMAE U6mq-HAM (n = 55)AE U6mq-HAM (n = 13)

AE U6mq-HAM (n = 30)AE U6mq-HAM (n = 75)

t

Figure 4 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806m119902-HAM (119899 = 55 13 30 75) for problem (28) at 119909 = 1

using (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

the interval of t (convergent region) increases as shown inFigures 5 6 7 and 8

Example 2 Consider the following Klein-Gordon equation

119906119905119905minus 119906119909119909+3

4119906 minus3

21199063= 0 (40)

subject to the following initial conditions

119906 (119909 0) = minussech119909

119906119905(119909 0) =

1

2sech 119909 tanh119909

(41)

6 Journal of Applied Mathematics

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3nHAM

U6nHAMU10nHAM

t

Figure 5 The comparison between the 1198803 1198806 and 119880

10of 119899HAM

(m119902-HAM 119899 = 1) and the exact solution of (28) at ℎ = minus1 and119909 = 1

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3mq-HAM (n = 75)

U6mq-HAM (n = 75)U10mq-HAM (n = 75)

t

Figure 6The comparison between the1198803 1198806 and119880

10of m119902-HAM

(119899 = 75) and the exact solution of (28) at ℎ = minus4925 and 119909 = 1

The exact solution is

119906 (119909 119905) = minussech(119909 + 1199052) (42)

In order to solve (40) by m119902-HAM we construct systemof differential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus3

4119906 +3

21199063

(43)

with the following initial approximations

1199060(119909 119905) = minussech119909 V

0(119909 119905) =

1

2sech 119909 tanh119909 (44)

02 04 06 08 10 12 14

005

010

015

Abso

lute

erro

r

AE U3nHAMAE U6nHAMAE U10nHAM

t

Figure 7The comparison between the absolute error of 11988031198806 and

11988010

of 119899HAM (m119902-HAM 119899 = 1) of (28) at ℎ = minus1 119909 = 1 and0 le 119905 le 15

02 04 06 08 10 12 14

002

004

006

008

010

Abso

lute

erro

r

AE U3mq-HAM (n = 75)AE U6mq-HAM (n = 75)AE U10mq-HAM (n = 75)

t

Figure 8The comparison between the absolute error of 1198803 1198806 and

11988010of m119902-HAM (119899 = 75) of (28) at ℎ = minus4925 119909 = 1 and 0 le 119905 le

15

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+3

4119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1(119909 119905) = minus

3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

(45)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1 (119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+3

41199060minus3

21199063

0)119889120591

(46)

Journal of Applied Mathematics 7

minus02

minus04

minus06

minus08

U6n

HA

M (x

t)

00minus05minus10minus15minus20

h

U6nHAM (01 06)U6nHAM (03 08)U6nHAM (06 1)

Figure 9 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (40) at different values of 119909 and 119905

For119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898(119909 119905) = (119899 + ℎ) V

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1205972119906119898minus1 (119909 119905)

1205971199092+3

4119906119898minus1(119909 119905)

minus3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894)119889120591

(47)

The following results are obtained

1199061 (119909 119905) = minus

1

2ℎ119905 sech119909 tanh119909

V1 (119909 119905) = ℎ119905 (minus

3 sech1199094

+sech31199092

+ sech119909tanh2119909)

1199062(119909 119905) = ℎ (

3

16ℎ1199052sech3119909 minus 1

16ℎ1199052 cosh (2119909) sech3119909)

minus1

2ℎ (ℎ + 119899) 119905 sech 119909 tanh119909

(48)

119906119898(119909 119905) (119898 = 3 4 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(49)

Equation (49) is a family of approximation solutions tothe problem (40) in terms of the convergence parameters ℎand 119899 To find the valid region of ℎ the ℎ-curves given bythe 6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

minus04

minus05

minus06

minus07

minus08

minus09

U6mq

-HA

M (x

tn

)

0minus50minus100minus150

U6mq-HAM (01 06 100)U6mq-HAM (03 08 100)U6mq-HAM (06 1 100)

h

Figure 10 ℎ-curve for the m119902-HAM (119899 = 100) approximationsolution 119880

6(119909 119905 100) of problem (40) at different values of 119909 and

119905

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

t

ExactU6nHAMU6mq-HAM (n = 5)

U6mq-HAM (n = 20)U6mq-HAM (n = 50)U6mq-HAM (n = 100)

Figure 11 Comparison between 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) with exact solution of (40)

at 119909 = 1 with (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ =minus795) respectively

the 6th-order m119902-HAM (119899 = 100) approximation at differentvalues of 119909 119905 are drawn in Figures 9 and 10 these figures showthe interval of ℎ in which the value of119880

6is constant at certain

119909 119905 and 119899 we chose the horizontal line parallel to 119909-axis (ℎ)as a valid regionwhich provides uswith a simpleway to adjustand control the convergence region Figure 11 shows thecomparison between119880

6of 119899HAM and119880

6of m119902-HAM using

different values of 119899with the solution (42)The absolute errorsof the 6th-order solutions 119899HAM approximate and the 6th-order solutions m119902-HAM approximate using different valuesof 119899 are shown in Figure 12 The results obtained by m119902-HAM indicate that the speed of convergence for m119902-HAMwith 119899 gt 1 is faster in comparison to 119899 = 1 (119899HAM) Theresults show that the convergence region of series solutions

8 Journal of Applied Mathematics

05 10 15 20 25 30 35

002

004

006

008

Abso

lute

erro

r

t

AE U6nHAMAE U6mq-HAM (n = 5)AE U6mq-HAM (n = 20)

AE U6mq-HAM (n = 50)AE U6mq-HAM (n = 100)

Figure 12 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) for problem (40) at 119909 = 1

using (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ = minus795)respectively

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

ExactU3nHAMU6nHAM

t

Figure 13 The comparison between the 1198803 1198806of 119899HAM (m119902-

HAM 119899 = 1) and the exact solution of (40) at ℎ = minus1 and 119909 = 1

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U3mq-HAM (n = 100)U6mq-HAM (n = 100)

Figure 14 The comparison between the 1198803 1198806of m119902-HAM (119899 =

100) and the exact solution of (40) at ℎ = minus795 and 119909 = 1

05 10 15 20 25 30 35

01

02

03

04

Abso

lute

erro

r

t

AE U3nHAMAE U6nHAM

Figure 15 The comparison between the absolute error of 1198803and

1198806of 119899HAM (m119902-HAM 119899 = 1) of (40) at ℎ = minus1 119909 = 1 and

0 le 119905 le 35

05 10 15 20 25 30 35

0005

0010

0015

0020

0025Ab

solu

te er

ror

t

AE U3mq-HAM (n = 100)AE U6mq-HAM (n = 100)

Figure 16The comparison between the absolute error of 1198803and119880

6

of m119902-HAM(119899 = 100) of (40) at ℎ = minus795 119909 = 1 and 0 le 119905 le 35

obtained bym119902-HAM is increasing as 119902 is decreased as shownin Figures 11 and 12

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient andthe interval of 119905 (convergent region) increases as shown inFigures 13 14 15 and 16

Figure 17 shows that the convergence of the series solu-tions obtained by the 3rd-order m119902-HAM (119899 = 100) is fasterthan that of the series solutions obtained by the 6th order119899HAM This fact shows the importance of the convergenceparameters 119899 in the m119902-HAM

5 Conclusion

In this paper a modified 119902-homotopy analysis method wasproposed (m119902-HAM)This method provides an approximatesolution by rewriting the 119899th-order nonlinear differentialequations in the form of system of 119899 first-order differentialequations The solution of these 119899 differential equations is

Journal of Applied Mathematics 9

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U6nHAMU3mq-HAM (n = 100)

Figure 17 The comparison between the 1198803of m119902-HAM (119899 = 100)

1198806of 119899HAM (m119902-HAM 119899 = 1) and the exact solution of (40) at

(ℎ = minus795 ℎ = minus1) and 119909 = 1

obtained as a power series solution which converges to aclosed form solution The m119902-HAM contains two auxiliaryparameters 119899 and ℎ such that the case of 119899 = 1 (m119902-HAM 119899 =1) the 119899HAM which is proposed in [21 22] can be reachedIn general it was noticed from the illustrative examples thatthe convergence of m119902-HAM is faster than that of 119899HAM

References

[1] S J Liao The proposed homotopy analysis technique for thesolution of nonlinear problems [PhD thesis] Shanghai Jiao TongUniversity 1992

[2] S Abbasbandy ldquoThe application of homotopy analysis methodto nonlinear equations arising in heat transferrdquo Physics LettersA vol 360 no 1 pp 109ndash113 2006

[3] S Abbasbandy andA Shirzadi ldquoHomotopy analysismethod formultiple solutions of the fractional Sturm-Liouville problemsrdquoNumerical Algorithms vol 54 no 4 pp 521ndash532 2010

[4] S Abbasbandy M Ashtiani and E Babolian ldquoAnalytic solutionof the Sharma-Tasso-Olver equation by homotopy analysismethodrdquoZeitschrift furNaturforschungA vol 65 no 4 pp 285ndash290 2010

[5] A K Alomari M S M Noorani and R Nazar ldquoExplicitseries solutions of some linear and nonlinear Schrodingerequations via the homotopy analysis methodrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 14 no 4pp 1196ndash1207 2009

[6] M Ayub A Rasheed and T Hayat ldquoExact flow of a third gradefluid past a porous plate using homotopy analysis methodrdquoInternational Journal of Engineering Science vol 41 no 18 pp2091ndash2103 2003

[7] A Sami Bataineh M S M Noorani and I Hashim ldquoSolvingsystems of ODEs by homotopy analysis methodrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 13 no10 pp 2060ndash2070 2008

[8] A S Bataineh M S M Noorani and I Hashim ldquoHomotopyanalysis method for singular IVPs of Emden-Fowler typerdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 14 no 4 pp 1121ndash1131 2009

[9] T Hayat M Khan and S Asghar ldquoHomotopy analysis of MHDflows of an Oldroyd 8-constant fluidrdquo Acta Mechanica vol 168no 3-4 pp 213ndash232 2004

[10] S J Liao Beyond Perturbation Introduction to the HomotopyAnalysisMethod ChapmanampHall Boca Raton Fla USA 2004

[11] S J Liao ldquoAn approximate solution technique not dependingon small parameters a special examplerdquo International Journalof Non-Linear Mechanics vol 30 no 3 pp 371ndash380 1995

[12] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[13] S Liao ldquoComparison between the homotopy analysis methodand homotopy perturbationmethodrdquoAppliedMathematics andComputation vol 169 no 2 pp 1186ndash1194 2005

[14] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[15] S Nadeem and N S Akbar ldquoPeristaltic flow of Sisko fluid in auniform inclined tuberdquoActaMechanica Sinica vol 26 no 5 pp675ndash683 2010

[16] S Nadeem and N S Akbar ldquoInfluence of heat transfer on aperistaltic flow of Johnson Segalman fluid in a non uniformtuberdquo International Communications in Heat andMass Transfervol 36 no 10 pp 1050ndash1059 2009

[17] Y Tan and S Abbasbandy ldquoHomotopy analysis method forquadratic Riccati differential equationrdquo Communications inNonlinear Science and Numerical Simulation vol 13 no 3 pp539ndash546 2008

[18] A Sami Bataineh M S M Noorani and I Hashim ldquoApproxi-mate solutions of singular two-point BVPs by modified homo-topy analysis methodrdquo Physics Letters A vol 372 pp 4062ndash4066 2008

[19] A Sami Bataineh M S M Noorani and I Hashim ldquoOn a newreliablemodification of homotopy analysismethodrdquoCommuni-cations in Nonlinear Science and Numerical Simulation vol 14no 2 pp 409ndash423 2009

[20] A S Bataineh M S M Noorani and I Hashim ldquoModifiedhomotopy analysis method for solving systems of second-orderBVPsrdquo Communications in Nonlinear Science and NumericalSimulation vol 14 no 2 pp 430ndash442 2009

[21] H N Hassan and M A El-Tawil ldquoA new technique of usinghomotopy analysis method for solving high-order nonlineardifferential equationsrdquo Mathematical Methods in the AppliedSciences vol 34 no 6 pp 728ndash742 2011

[22] H N Hassan and M A El-Tawil ldquoA new technique ofusing homotopy analysis method for second order nonlineardifferential equationsrdquo Applied Mathematics and Computationvol 219 no 2 pp 708ndash728 2012

[23] M A El-Tawil and S N Huseen ldquoThe q-homotopy analysismethod (q-HAM)rdquo International Journal of Applied Mathemat-ics and Mechanics vol 8 no 15 pp 51ndash75 2012

[24] M A El-Tawil and S N Huseen ldquoOn convergence of the q-homotopy analysis methodrdquo International Journal of Contem-porary Mathematical Sciences vol 8 no 10 pp 481ndash497 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

4 Journal of Applied Mathematics

For119898 gt 1 119896119898= 119899 and

119906119898 (119909 0) = 0 1199061

119898 (119909 0) = 0

1199062119898(119909 0) = 0 119906 119911 minus 1

119898(119909 0) = 0

(26)

Substituting in (17) we obtain

119906119898 (119909 119905) = (119899 + ℎ) 119906119898minus1 (119909 119905) + ℎint

119905

0

(minus1199061119898minus1 (119909 120591)) 119889120591

1199061119898(119909 119905) = (119899 + ℎ) 1199061

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1199062119898minus1(119909 120591)) 119889120591

119906 119911 minus 1119898(119909 119905)

= (119899 + ℎ) 119906119911 minus 1119898minus1 (119909 119905)

+ ℎint

119905

0

(119860119906119898minus1 (119909 120591) + 119861 (119906119898minus1 (119909 120591))) 119889120591

(27)

It should be noted that the case of 119899 = 1 in (27) the 119899HAMcan be reached

To illustrate the effectiveness of the proposed m119902-HAMcomparison betweenm119902-HAMand the 119899HAMare illustratedby the following examples

4 Illustrative Examples

Example 1 Consider the following nonlinear sine-Gordonequation

119906119905119905minus 119906119909119909+ sin 119906 = 0 (28)

subject to the following initial conditions

119906 (119909 0) = 0 119906119905(119909 0) = 4 sech 119909 (29)

The exact solution is

119906 (119909 119905) = 4 tanminus1 (119905 sech119909) (30)

In order to prevent suffering from the strongly nonlinearterm sin 119906 in the frame of 119902-HAM we can use Taylor seriesexpansion of sin 119906 as follows

sin 119906 = 119906 minus 1199063

6+1199065

120 (31)

Then (28) becomes

119906119905119905minus 119906119909119909+ 119906 minus

1199063

6+1199065

120= 0 (32)

In order to solve (28) by m119902-HAM we construct system ofdifferential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus 119906 +

1199063

6minus1199065

120

(33)

with the following initial approximations

1199060(119909 119905) = 0 V

0(119909 119905) = 4 sech119909 (34)

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+ 119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1 (119909 119905) = minus

1

6

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

+1

120

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119895minus119894

119894

sum

119896=0

119906119894minus119896

119896

sum

119897=0

119906119897119906119896minus119897

(35)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1(119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+ 1199060minus1199063

0

6+1199065

0

120)119889120591

(36)

Now for119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898 (119909 119905) = (119899 + ℎ) V119898minus1 (119909 119905)

+ ℎint

119905

0

(119860119906119898minus1 (119909 120591) + 119861 (119906119898minus1 (119909 120591))) 119889120591

(37)

And the following results are obtained

1199061 (119909 119905) = minus4ℎ119905 sech 119909

V1(119909 119905) = 0

1199062 (119909 119905) = minus4ℎ (ℎ + 119899) 119905 sech 119909

V2(119909 119905) = minus4ℎ

21199052sech3119909

1199063(119909 119905) = minus4ℎ(ℎ + 119899)

2119905 sech 119909 + 4

3ℎ31199053sech3119909

(38)

119906119898(119909 119905) (119898 = 4 5 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872 (119909 119905 119899 ℎ) =119872

sum

119894=0

119906119894 (119909 119905 119899 ℎ) (

1

119899)

119894

(39)

Equation (39) is a family of approximation solutions to theproblem (28) in terms of the convergence parameters ℎ and119899 To find the valid region of ℎ the ℎ-curves given by the6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

Journal of Applied Mathematics 5

02

04

06

08

10

minus20 minus15 minus10 minus05 00

U6n

HA

M (x

t)

U6nHAM (5 2)U6nHAM (3 1)U6nHAM (25 15)

h

Figure 1 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (28) at different values of 119909 and 119905

02

04

06

08

10

minus25 minus20 minus15 minus10 minus5

U6mq

-HA

M (x

tn

)

h

U6mq-HAM (5 2 13)U6mq-HAM (3 1 13)U6mq-HAM (25 15 13)

Figure 2 ℎ-curve for the (m119902-HAM 119899 = 13) approximationsolution 119880

6(119909 119905 13) of problem (28) at different values of 119909 and 119905

the 6th-order m119902-HAM (119899 = 13) approximation at differentvalues of 119909 119905 are drawn in Figures 1 and 2 respectively andthese figures show the interval of ℎ in which the value of1198806is constant at certain 119909 119905 and 119899 we chose the horizontal

line parallel to 119909-axis (ℎ) as a valid region which providesus with a simple way to adjust and control the convergenceregion Figure 3 shows the comparison between119880

6of 119899HAM

and 1198806of m119902-HAM using different values of 119899 with the

solution (30) The absolute errors of the 6th-order solutions119899HAM approximate and the 6th-order solutions m119902-HAMapproximate using different values of 119899 are shown in Figure 4The results obtained by m119902-HAM indicate that the speed ofconvergence for m119902-HAMwith 119899 gt 1 is faster in comparisonto 119899 = 1 (119899HAM) The results show that the convergenceregion of series solutions obtained by m119902-HAM is increasingas 119902 is decreased as shown in Figures 3 and 4

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient and

00 05 10 15 20 25

2

4

6

8

Solu

tions

Exact

t

U6nHAMU6mq-HAM (n = 55)

U6mq-HAM (n = 13)U6mq-HAM (n = 30)U6mq-HAM (n = 75)

Figure 3 Comparison between1198806of 119899HAM (m119902-HAM 119899 = 1) and

1198806m119902-HAM (119899 = 55 13 30 75)with exact solution of (28) at 119909 = 1

with (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

05 10 15 20 25

05

10

15

Abso

lute

erro

r

AE U6nHAMAE U6mq-HAM (n = 55)AE U6mq-HAM (n = 13)

AE U6mq-HAM (n = 30)AE U6mq-HAM (n = 75)

t

Figure 4 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806m119902-HAM (119899 = 55 13 30 75) for problem (28) at 119909 = 1

using (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

the interval of t (convergent region) increases as shown inFigures 5 6 7 and 8

Example 2 Consider the following Klein-Gordon equation

119906119905119905minus 119906119909119909+3

4119906 minus3

21199063= 0 (40)

subject to the following initial conditions

119906 (119909 0) = minussech119909

119906119905(119909 0) =

1

2sech 119909 tanh119909

(41)

6 Journal of Applied Mathematics

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3nHAM

U6nHAMU10nHAM

t

Figure 5 The comparison between the 1198803 1198806 and 119880

10of 119899HAM

(m119902-HAM 119899 = 1) and the exact solution of (28) at ℎ = minus1 and119909 = 1

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3mq-HAM (n = 75)

U6mq-HAM (n = 75)U10mq-HAM (n = 75)

t

Figure 6The comparison between the1198803 1198806 and119880

10of m119902-HAM

(119899 = 75) and the exact solution of (28) at ℎ = minus4925 and 119909 = 1

The exact solution is

119906 (119909 119905) = minussech(119909 + 1199052) (42)

In order to solve (40) by m119902-HAM we construct systemof differential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus3

4119906 +3

21199063

(43)

with the following initial approximations

1199060(119909 119905) = minussech119909 V

0(119909 119905) =

1

2sech 119909 tanh119909 (44)

02 04 06 08 10 12 14

005

010

015

Abso

lute

erro

r

AE U3nHAMAE U6nHAMAE U10nHAM

t

Figure 7The comparison between the absolute error of 11988031198806 and

11988010

of 119899HAM (m119902-HAM 119899 = 1) of (28) at ℎ = minus1 119909 = 1 and0 le 119905 le 15

02 04 06 08 10 12 14

002

004

006

008

010

Abso

lute

erro

r

AE U3mq-HAM (n = 75)AE U6mq-HAM (n = 75)AE U10mq-HAM (n = 75)

t

Figure 8The comparison between the absolute error of 1198803 1198806 and

11988010of m119902-HAM (119899 = 75) of (28) at ℎ = minus4925 119909 = 1 and 0 le 119905 le

15

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+3

4119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1(119909 119905) = minus

3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

(45)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1 (119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+3

41199060minus3

21199063

0)119889120591

(46)

Journal of Applied Mathematics 7

minus02

minus04

minus06

minus08

U6n

HA

M (x

t)

00minus05minus10minus15minus20

h

U6nHAM (01 06)U6nHAM (03 08)U6nHAM (06 1)

Figure 9 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (40) at different values of 119909 and 119905

For119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898(119909 119905) = (119899 + ℎ) V

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1205972119906119898minus1 (119909 119905)

1205971199092+3

4119906119898minus1(119909 119905)

minus3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894)119889120591

(47)

The following results are obtained

1199061 (119909 119905) = minus

1

2ℎ119905 sech119909 tanh119909

V1 (119909 119905) = ℎ119905 (minus

3 sech1199094

+sech31199092

+ sech119909tanh2119909)

1199062(119909 119905) = ℎ (

3

16ℎ1199052sech3119909 minus 1

16ℎ1199052 cosh (2119909) sech3119909)

minus1

2ℎ (ℎ + 119899) 119905 sech 119909 tanh119909

(48)

119906119898(119909 119905) (119898 = 3 4 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(49)

Equation (49) is a family of approximation solutions tothe problem (40) in terms of the convergence parameters ℎand 119899 To find the valid region of ℎ the ℎ-curves given bythe 6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

minus04

minus05

minus06

minus07

minus08

minus09

U6mq

-HA

M (x

tn

)

0minus50minus100minus150

U6mq-HAM (01 06 100)U6mq-HAM (03 08 100)U6mq-HAM (06 1 100)

h

Figure 10 ℎ-curve for the m119902-HAM (119899 = 100) approximationsolution 119880

6(119909 119905 100) of problem (40) at different values of 119909 and

119905

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

t

ExactU6nHAMU6mq-HAM (n = 5)

U6mq-HAM (n = 20)U6mq-HAM (n = 50)U6mq-HAM (n = 100)

Figure 11 Comparison between 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) with exact solution of (40)

at 119909 = 1 with (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ =minus795) respectively

the 6th-order m119902-HAM (119899 = 100) approximation at differentvalues of 119909 119905 are drawn in Figures 9 and 10 these figures showthe interval of ℎ in which the value of119880

6is constant at certain

119909 119905 and 119899 we chose the horizontal line parallel to 119909-axis (ℎ)as a valid regionwhich provides uswith a simpleway to adjustand control the convergence region Figure 11 shows thecomparison between119880

6of 119899HAM and119880

6of m119902-HAM using

different values of 119899with the solution (42)The absolute errorsof the 6th-order solutions 119899HAM approximate and the 6th-order solutions m119902-HAM approximate using different valuesof 119899 are shown in Figure 12 The results obtained by m119902-HAM indicate that the speed of convergence for m119902-HAMwith 119899 gt 1 is faster in comparison to 119899 = 1 (119899HAM) Theresults show that the convergence region of series solutions

8 Journal of Applied Mathematics

05 10 15 20 25 30 35

002

004

006

008

Abso

lute

erro

r

t

AE U6nHAMAE U6mq-HAM (n = 5)AE U6mq-HAM (n = 20)

AE U6mq-HAM (n = 50)AE U6mq-HAM (n = 100)

Figure 12 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) for problem (40) at 119909 = 1

using (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ = minus795)respectively

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

ExactU3nHAMU6nHAM

t

Figure 13 The comparison between the 1198803 1198806of 119899HAM (m119902-

HAM 119899 = 1) and the exact solution of (40) at ℎ = minus1 and 119909 = 1

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U3mq-HAM (n = 100)U6mq-HAM (n = 100)

Figure 14 The comparison between the 1198803 1198806of m119902-HAM (119899 =

100) and the exact solution of (40) at ℎ = minus795 and 119909 = 1

05 10 15 20 25 30 35

01

02

03

04

Abso

lute

erro

r

t

AE U3nHAMAE U6nHAM

Figure 15 The comparison between the absolute error of 1198803and

1198806of 119899HAM (m119902-HAM 119899 = 1) of (40) at ℎ = minus1 119909 = 1 and

0 le 119905 le 35

05 10 15 20 25 30 35

0005

0010

0015

0020

0025Ab

solu

te er

ror

t

AE U3mq-HAM (n = 100)AE U6mq-HAM (n = 100)

Figure 16The comparison between the absolute error of 1198803and119880

6

of m119902-HAM(119899 = 100) of (40) at ℎ = minus795 119909 = 1 and 0 le 119905 le 35

obtained bym119902-HAM is increasing as 119902 is decreased as shownin Figures 11 and 12

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient andthe interval of 119905 (convergent region) increases as shown inFigures 13 14 15 and 16

Figure 17 shows that the convergence of the series solu-tions obtained by the 3rd-order m119902-HAM (119899 = 100) is fasterthan that of the series solutions obtained by the 6th order119899HAM This fact shows the importance of the convergenceparameters 119899 in the m119902-HAM

5 Conclusion

In this paper a modified 119902-homotopy analysis method wasproposed (m119902-HAM)This method provides an approximatesolution by rewriting the 119899th-order nonlinear differentialequations in the form of system of 119899 first-order differentialequations The solution of these 119899 differential equations is

Journal of Applied Mathematics 9

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U6nHAMU3mq-HAM (n = 100)

Figure 17 The comparison between the 1198803of m119902-HAM (119899 = 100)

1198806of 119899HAM (m119902-HAM 119899 = 1) and the exact solution of (40) at

(ℎ = minus795 ℎ = minus1) and 119909 = 1

obtained as a power series solution which converges to aclosed form solution The m119902-HAM contains two auxiliaryparameters 119899 and ℎ such that the case of 119899 = 1 (m119902-HAM 119899 =1) the 119899HAM which is proposed in [21 22] can be reachedIn general it was noticed from the illustrative examples thatthe convergence of m119902-HAM is faster than that of 119899HAM

References

[1] S J Liao The proposed homotopy analysis technique for thesolution of nonlinear problems [PhD thesis] Shanghai Jiao TongUniversity 1992

[2] S Abbasbandy ldquoThe application of homotopy analysis methodto nonlinear equations arising in heat transferrdquo Physics LettersA vol 360 no 1 pp 109ndash113 2006

[3] S Abbasbandy andA Shirzadi ldquoHomotopy analysismethod formultiple solutions of the fractional Sturm-Liouville problemsrdquoNumerical Algorithms vol 54 no 4 pp 521ndash532 2010

[4] S Abbasbandy M Ashtiani and E Babolian ldquoAnalytic solutionof the Sharma-Tasso-Olver equation by homotopy analysismethodrdquoZeitschrift furNaturforschungA vol 65 no 4 pp 285ndash290 2010

[5] A K Alomari M S M Noorani and R Nazar ldquoExplicitseries solutions of some linear and nonlinear Schrodingerequations via the homotopy analysis methodrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 14 no 4pp 1196ndash1207 2009

[6] M Ayub A Rasheed and T Hayat ldquoExact flow of a third gradefluid past a porous plate using homotopy analysis methodrdquoInternational Journal of Engineering Science vol 41 no 18 pp2091ndash2103 2003

[7] A Sami Bataineh M S M Noorani and I Hashim ldquoSolvingsystems of ODEs by homotopy analysis methodrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 13 no10 pp 2060ndash2070 2008

[8] A S Bataineh M S M Noorani and I Hashim ldquoHomotopyanalysis method for singular IVPs of Emden-Fowler typerdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 14 no 4 pp 1121ndash1131 2009

[9] T Hayat M Khan and S Asghar ldquoHomotopy analysis of MHDflows of an Oldroyd 8-constant fluidrdquo Acta Mechanica vol 168no 3-4 pp 213ndash232 2004

[10] S J Liao Beyond Perturbation Introduction to the HomotopyAnalysisMethod ChapmanampHall Boca Raton Fla USA 2004

[11] S J Liao ldquoAn approximate solution technique not dependingon small parameters a special examplerdquo International Journalof Non-Linear Mechanics vol 30 no 3 pp 371ndash380 1995

[12] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[13] S Liao ldquoComparison between the homotopy analysis methodand homotopy perturbationmethodrdquoAppliedMathematics andComputation vol 169 no 2 pp 1186ndash1194 2005

[14] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[15] S Nadeem and N S Akbar ldquoPeristaltic flow of Sisko fluid in auniform inclined tuberdquoActaMechanica Sinica vol 26 no 5 pp675ndash683 2010

[16] S Nadeem and N S Akbar ldquoInfluence of heat transfer on aperistaltic flow of Johnson Segalman fluid in a non uniformtuberdquo International Communications in Heat andMass Transfervol 36 no 10 pp 1050ndash1059 2009

[17] Y Tan and S Abbasbandy ldquoHomotopy analysis method forquadratic Riccati differential equationrdquo Communications inNonlinear Science and Numerical Simulation vol 13 no 3 pp539ndash546 2008

[18] A Sami Bataineh M S M Noorani and I Hashim ldquoApproxi-mate solutions of singular two-point BVPs by modified homo-topy analysis methodrdquo Physics Letters A vol 372 pp 4062ndash4066 2008

[19] A Sami Bataineh M S M Noorani and I Hashim ldquoOn a newreliablemodification of homotopy analysismethodrdquoCommuni-cations in Nonlinear Science and Numerical Simulation vol 14no 2 pp 409ndash423 2009

[20] A S Bataineh M S M Noorani and I Hashim ldquoModifiedhomotopy analysis method for solving systems of second-orderBVPsrdquo Communications in Nonlinear Science and NumericalSimulation vol 14 no 2 pp 430ndash442 2009

[21] H N Hassan and M A El-Tawil ldquoA new technique of usinghomotopy analysis method for solving high-order nonlineardifferential equationsrdquo Mathematical Methods in the AppliedSciences vol 34 no 6 pp 728ndash742 2011

[22] H N Hassan and M A El-Tawil ldquoA new technique ofusing homotopy analysis method for second order nonlineardifferential equationsrdquo Applied Mathematics and Computationvol 219 no 2 pp 708ndash728 2012

[23] M A El-Tawil and S N Huseen ldquoThe q-homotopy analysismethod (q-HAM)rdquo International Journal of Applied Mathemat-ics and Mechanics vol 8 no 15 pp 51ndash75 2012

[24] M A El-Tawil and S N Huseen ldquoOn convergence of the q-homotopy analysis methodrdquo International Journal of Contem-porary Mathematical Sciences vol 8 no 10 pp 481ndash497 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

Journal of Applied Mathematics 5

02

04

06

08

10

minus20 minus15 minus10 minus05 00

U6n

HA

M (x

t)

U6nHAM (5 2)U6nHAM (3 1)U6nHAM (25 15)

h

Figure 1 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (28) at different values of 119909 and 119905

02

04

06

08

10

minus25 minus20 minus15 minus10 minus5

U6mq

-HA

M (x

tn

)

h

U6mq-HAM (5 2 13)U6mq-HAM (3 1 13)U6mq-HAM (25 15 13)

Figure 2 ℎ-curve for the (m119902-HAM 119899 = 13) approximationsolution 119880

6(119909 119905 13) of problem (28) at different values of 119909 and 119905

the 6th-order m119902-HAM (119899 = 13) approximation at differentvalues of 119909 119905 are drawn in Figures 1 and 2 respectively andthese figures show the interval of ℎ in which the value of1198806is constant at certain 119909 119905 and 119899 we chose the horizontal

line parallel to 119909-axis (ℎ) as a valid region which providesus with a simple way to adjust and control the convergenceregion Figure 3 shows the comparison between119880

6of 119899HAM

and 1198806of m119902-HAM using different values of 119899 with the

solution (30) The absolute errors of the 6th-order solutions119899HAM approximate and the 6th-order solutions m119902-HAMapproximate using different values of 119899 are shown in Figure 4The results obtained by m119902-HAM indicate that the speed ofconvergence for m119902-HAMwith 119899 gt 1 is faster in comparisonto 119899 = 1 (119899HAM) The results show that the convergenceregion of series solutions obtained by m119902-HAM is increasingas 119902 is decreased as shown in Figures 3 and 4

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient and

00 05 10 15 20 25

2

4

6

8

Solu

tions

Exact

t

U6nHAMU6mq-HAM (n = 55)

U6mq-HAM (n = 13)U6mq-HAM (n = 30)U6mq-HAM (n = 75)

Figure 3 Comparison between1198806of 119899HAM (m119902-HAM 119899 = 1) and

1198806m119902-HAM (119899 = 55 13 30 75)with exact solution of (28) at 119909 = 1

with (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

05 10 15 20 25

05

10

15

Abso

lute

erro

r

AE U6nHAMAE U6mq-HAM (n = 55)AE U6mq-HAM (n = 13)

AE U6mq-HAM (n = 30)AE U6mq-HAM (n = 75)

t

Figure 4 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806m119902-HAM (119899 = 55 13 30 75) for problem (28) at 119909 = 1

using (ℎ = minus1 ℎ = minus49 ℎ = minus108 ℎ = minus2315 ℎ = minus4925)respectively

the interval of t (convergent region) increases as shown inFigures 5 6 7 and 8

Example 2 Consider the following Klein-Gordon equation

119906119905119905minus 119906119909119909+3

4119906 minus3

21199063= 0 (40)

subject to the following initial conditions

119906 (119909 0) = minussech119909

119906119905(119909 0) =

1

2sech 119909 tanh119909

(41)

6 Journal of Applied Mathematics

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3nHAM

U6nHAMU10nHAM

t

Figure 5 The comparison between the 1198803 1198806 and 119880

10of 119899HAM

(m119902-HAM 119899 = 1) and the exact solution of (28) at ℎ = minus1 and119909 = 1

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3mq-HAM (n = 75)

U6mq-HAM (n = 75)U10mq-HAM (n = 75)

t

Figure 6The comparison between the1198803 1198806 and119880

10of m119902-HAM

(119899 = 75) and the exact solution of (28) at ℎ = minus4925 and 119909 = 1

The exact solution is

119906 (119909 119905) = minussech(119909 + 1199052) (42)

In order to solve (40) by m119902-HAM we construct systemof differential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus3

4119906 +3

21199063

(43)

with the following initial approximations

1199060(119909 119905) = minussech119909 V

0(119909 119905) =

1

2sech 119909 tanh119909 (44)

02 04 06 08 10 12 14

005

010

015

Abso

lute

erro

r

AE U3nHAMAE U6nHAMAE U10nHAM

t

Figure 7The comparison between the absolute error of 11988031198806 and

11988010

of 119899HAM (m119902-HAM 119899 = 1) of (28) at ℎ = minus1 119909 = 1 and0 le 119905 le 15

02 04 06 08 10 12 14

002

004

006

008

010

Abso

lute

erro

r

AE U3mq-HAM (n = 75)AE U6mq-HAM (n = 75)AE U10mq-HAM (n = 75)

t

Figure 8The comparison between the absolute error of 1198803 1198806 and

11988010of m119902-HAM (119899 = 75) of (28) at ℎ = minus4925 119909 = 1 and 0 le 119905 le

15

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+3

4119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1(119909 119905) = minus

3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

(45)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1 (119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+3

41199060minus3

21199063

0)119889120591

(46)

Journal of Applied Mathematics 7

minus02

minus04

minus06

minus08

U6n

HA

M (x

t)

00minus05minus10minus15minus20

h

U6nHAM (01 06)U6nHAM (03 08)U6nHAM (06 1)

Figure 9 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (40) at different values of 119909 and 119905

For119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898(119909 119905) = (119899 + ℎ) V

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1205972119906119898minus1 (119909 119905)

1205971199092+3

4119906119898minus1(119909 119905)

minus3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894)119889120591

(47)

The following results are obtained

1199061 (119909 119905) = minus

1

2ℎ119905 sech119909 tanh119909

V1 (119909 119905) = ℎ119905 (minus

3 sech1199094

+sech31199092

+ sech119909tanh2119909)

1199062(119909 119905) = ℎ (

3

16ℎ1199052sech3119909 minus 1

16ℎ1199052 cosh (2119909) sech3119909)

minus1

2ℎ (ℎ + 119899) 119905 sech 119909 tanh119909

(48)

119906119898(119909 119905) (119898 = 3 4 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(49)

Equation (49) is a family of approximation solutions tothe problem (40) in terms of the convergence parameters ℎand 119899 To find the valid region of ℎ the ℎ-curves given bythe 6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

minus04

minus05

minus06

minus07

minus08

minus09

U6mq

-HA

M (x

tn

)

0minus50minus100minus150

U6mq-HAM (01 06 100)U6mq-HAM (03 08 100)U6mq-HAM (06 1 100)

h

Figure 10 ℎ-curve for the m119902-HAM (119899 = 100) approximationsolution 119880

6(119909 119905 100) of problem (40) at different values of 119909 and

119905

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

t

ExactU6nHAMU6mq-HAM (n = 5)

U6mq-HAM (n = 20)U6mq-HAM (n = 50)U6mq-HAM (n = 100)

Figure 11 Comparison between 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) with exact solution of (40)

at 119909 = 1 with (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ =minus795) respectively

the 6th-order m119902-HAM (119899 = 100) approximation at differentvalues of 119909 119905 are drawn in Figures 9 and 10 these figures showthe interval of ℎ in which the value of119880

6is constant at certain

119909 119905 and 119899 we chose the horizontal line parallel to 119909-axis (ℎ)as a valid regionwhich provides uswith a simpleway to adjustand control the convergence region Figure 11 shows thecomparison between119880

6of 119899HAM and119880

6of m119902-HAM using

different values of 119899with the solution (42)The absolute errorsof the 6th-order solutions 119899HAM approximate and the 6th-order solutions m119902-HAM approximate using different valuesof 119899 are shown in Figure 12 The results obtained by m119902-HAM indicate that the speed of convergence for m119902-HAMwith 119899 gt 1 is faster in comparison to 119899 = 1 (119899HAM) Theresults show that the convergence region of series solutions

8 Journal of Applied Mathematics

05 10 15 20 25 30 35

002

004

006

008

Abso

lute

erro

r

t

AE U6nHAMAE U6mq-HAM (n = 5)AE U6mq-HAM (n = 20)

AE U6mq-HAM (n = 50)AE U6mq-HAM (n = 100)

Figure 12 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) for problem (40) at 119909 = 1

using (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ = minus795)respectively

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

ExactU3nHAMU6nHAM

t

Figure 13 The comparison between the 1198803 1198806of 119899HAM (m119902-

HAM 119899 = 1) and the exact solution of (40) at ℎ = minus1 and 119909 = 1

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U3mq-HAM (n = 100)U6mq-HAM (n = 100)

Figure 14 The comparison between the 1198803 1198806of m119902-HAM (119899 =

100) and the exact solution of (40) at ℎ = minus795 and 119909 = 1

05 10 15 20 25 30 35

01

02

03

04

Abso

lute

erro

r

t

AE U3nHAMAE U6nHAM

Figure 15 The comparison between the absolute error of 1198803and

1198806of 119899HAM (m119902-HAM 119899 = 1) of (40) at ℎ = minus1 119909 = 1 and

0 le 119905 le 35

05 10 15 20 25 30 35

0005

0010

0015

0020

0025Ab

solu

te er

ror

t

AE U3mq-HAM (n = 100)AE U6mq-HAM (n = 100)

Figure 16The comparison between the absolute error of 1198803and119880

6

of m119902-HAM(119899 = 100) of (40) at ℎ = minus795 119909 = 1 and 0 le 119905 le 35

obtained bym119902-HAM is increasing as 119902 is decreased as shownin Figures 11 and 12

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient andthe interval of 119905 (convergent region) increases as shown inFigures 13 14 15 and 16

Figure 17 shows that the convergence of the series solu-tions obtained by the 3rd-order m119902-HAM (119899 = 100) is fasterthan that of the series solutions obtained by the 6th order119899HAM This fact shows the importance of the convergenceparameters 119899 in the m119902-HAM

5 Conclusion

In this paper a modified 119902-homotopy analysis method wasproposed (m119902-HAM)This method provides an approximatesolution by rewriting the 119899th-order nonlinear differentialequations in the form of system of 119899 first-order differentialequations The solution of these 119899 differential equations is

Journal of Applied Mathematics 9

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U6nHAMU3mq-HAM (n = 100)

Figure 17 The comparison between the 1198803of m119902-HAM (119899 = 100)

1198806of 119899HAM (m119902-HAM 119899 = 1) and the exact solution of (40) at

(ℎ = minus795 ℎ = minus1) and 119909 = 1

obtained as a power series solution which converges to aclosed form solution The m119902-HAM contains two auxiliaryparameters 119899 and ℎ such that the case of 119899 = 1 (m119902-HAM 119899 =1) the 119899HAM which is proposed in [21 22] can be reachedIn general it was noticed from the illustrative examples thatthe convergence of m119902-HAM is faster than that of 119899HAM

References

[1] S J Liao The proposed homotopy analysis technique for thesolution of nonlinear problems [PhD thesis] Shanghai Jiao TongUniversity 1992

[2] S Abbasbandy ldquoThe application of homotopy analysis methodto nonlinear equations arising in heat transferrdquo Physics LettersA vol 360 no 1 pp 109ndash113 2006

[3] S Abbasbandy andA Shirzadi ldquoHomotopy analysismethod formultiple solutions of the fractional Sturm-Liouville problemsrdquoNumerical Algorithms vol 54 no 4 pp 521ndash532 2010

[4] S Abbasbandy M Ashtiani and E Babolian ldquoAnalytic solutionof the Sharma-Tasso-Olver equation by homotopy analysismethodrdquoZeitschrift furNaturforschungA vol 65 no 4 pp 285ndash290 2010

[5] A K Alomari M S M Noorani and R Nazar ldquoExplicitseries solutions of some linear and nonlinear Schrodingerequations via the homotopy analysis methodrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 14 no 4pp 1196ndash1207 2009

[6] M Ayub A Rasheed and T Hayat ldquoExact flow of a third gradefluid past a porous plate using homotopy analysis methodrdquoInternational Journal of Engineering Science vol 41 no 18 pp2091ndash2103 2003

[7] A Sami Bataineh M S M Noorani and I Hashim ldquoSolvingsystems of ODEs by homotopy analysis methodrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 13 no10 pp 2060ndash2070 2008

[8] A S Bataineh M S M Noorani and I Hashim ldquoHomotopyanalysis method for singular IVPs of Emden-Fowler typerdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 14 no 4 pp 1121ndash1131 2009

[9] T Hayat M Khan and S Asghar ldquoHomotopy analysis of MHDflows of an Oldroyd 8-constant fluidrdquo Acta Mechanica vol 168no 3-4 pp 213ndash232 2004

[10] S J Liao Beyond Perturbation Introduction to the HomotopyAnalysisMethod ChapmanampHall Boca Raton Fla USA 2004

[11] S J Liao ldquoAn approximate solution technique not dependingon small parameters a special examplerdquo International Journalof Non-Linear Mechanics vol 30 no 3 pp 371ndash380 1995

[12] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[13] S Liao ldquoComparison between the homotopy analysis methodand homotopy perturbationmethodrdquoAppliedMathematics andComputation vol 169 no 2 pp 1186ndash1194 2005

[14] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[15] S Nadeem and N S Akbar ldquoPeristaltic flow of Sisko fluid in auniform inclined tuberdquoActaMechanica Sinica vol 26 no 5 pp675ndash683 2010

[16] S Nadeem and N S Akbar ldquoInfluence of heat transfer on aperistaltic flow of Johnson Segalman fluid in a non uniformtuberdquo International Communications in Heat andMass Transfervol 36 no 10 pp 1050ndash1059 2009

[17] Y Tan and S Abbasbandy ldquoHomotopy analysis method forquadratic Riccati differential equationrdquo Communications inNonlinear Science and Numerical Simulation vol 13 no 3 pp539ndash546 2008

[18] A Sami Bataineh M S M Noorani and I Hashim ldquoApproxi-mate solutions of singular two-point BVPs by modified homo-topy analysis methodrdquo Physics Letters A vol 372 pp 4062ndash4066 2008

[19] A Sami Bataineh M S M Noorani and I Hashim ldquoOn a newreliablemodification of homotopy analysismethodrdquoCommuni-cations in Nonlinear Science and Numerical Simulation vol 14no 2 pp 409ndash423 2009

[20] A S Bataineh M S M Noorani and I Hashim ldquoModifiedhomotopy analysis method for solving systems of second-orderBVPsrdquo Communications in Nonlinear Science and NumericalSimulation vol 14 no 2 pp 430ndash442 2009

[21] H N Hassan and M A El-Tawil ldquoA new technique of usinghomotopy analysis method for solving high-order nonlineardifferential equationsrdquo Mathematical Methods in the AppliedSciences vol 34 no 6 pp 728ndash742 2011

[22] H N Hassan and M A El-Tawil ldquoA new technique ofusing homotopy analysis method for second order nonlineardifferential equationsrdquo Applied Mathematics and Computationvol 219 no 2 pp 708ndash728 2012

[23] M A El-Tawil and S N Huseen ldquoThe q-homotopy analysismethod (q-HAM)rdquo International Journal of Applied Mathemat-ics and Mechanics vol 8 no 15 pp 51ndash75 2012

[24] M A El-Tawil and S N Huseen ldquoOn convergence of the q-homotopy analysis methodrdquo International Journal of Contem-porary Mathematical Sciences vol 8 no 10 pp 481ndash497 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

6 Journal of Applied Mathematics

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3nHAM

U6nHAMU10nHAM

t

Figure 5 The comparison between the 1198803 1198806 and 119880

10of 119899HAM

(m119902-HAM 119899 = 1) and the exact solution of (28) at ℎ = minus1 and119909 = 1

00 02 04 06 08 10 12 14

05

10

15

20

25

30

Solu

tions

ExactU3mq-HAM (n = 75)

U6mq-HAM (n = 75)U10mq-HAM (n = 75)

t

Figure 6The comparison between the1198803 1198806 and119880

10of m119902-HAM

(119899 = 75) and the exact solution of (28) at ℎ = minus4925 and 119909 = 1

The exact solution is

119906 (119909 119905) = minussech(119909 + 1199052) (42)

In order to solve (40) by m119902-HAM we construct systemof differential equations as follows

119906119905(119909 119905) = V (119909 119905)

V119905(119909 119905) =

1205972119906 (119909 119905)

1205971199092minus3

4119906 +3

21199063

(43)

with the following initial approximations

1199060(119909 119905) = minussech119909 V

0(119909 119905) =

1

2sech 119909 tanh119909 (44)

02 04 06 08 10 12 14

005

010

015

Abso

lute

erro

r

AE U3nHAMAE U6nHAMAE U10nHAM

t

Figure 7The comparison between the absolute error of 11988031198806 and

11988010

of 119899HAM (m119902-HAM 119899 = 1) of (28) at ℎ = minus1 119909 = 1 and0 le 119905 le 15

02 04 06 08 10 12 14

002

004

006

008

010

Abso

lute

erro

r

AE U3mq-HAM (n = 75)AE U6mq-HAM (n = 75)AE U10mq-HAM (n = 75)

t

Figure 8The comparison between the absolute error of 1198803 1198806 and

11988010of m119902-HAM (119899 = 75) of (28) at ℎ = minus4925 119909 = 1 and 0 le 119905 le

15

and the following auxiliary linear operators

119871119906 (119909 119905) =120597119906 (119909 119905)

120597119905 119871V (119909 119905) =

120597V (119909 119905)120597119905

119860119906119898minus1 (119909 119905) = minus

1205972119906119898minus1(119909 119905)

1205971199092+3

4119906119898minus1 (119909 119905)

119861997888997888997888119906119898minus1(119909 119905) = minus

3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894

(45)

From (25) and (27) we obtain

1199061(119909 119905) = ℎint

119905

0

(minusV0(119909 120591)) 119889120591

V1 (119909 119905) = ℎint

119905

0

(minus12059721199060

1205971199092+3

41199060minus3

21199063

0)119889120591

(46)

Journal of Applied Mathematics 7

minus02

minus04

minus06

minus08

U6n

HA

M (x

t)

00minus05minus10minus15minus20

h

U6nHAM (01 06)U6nHAM (03 08)U6nHAM (06 1)

Figure 9 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (40) at different values of 119909 and 119905

For119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898(119909 119905) = (119899 + ℎ) V

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1205972119906119898minus1 (119909 119905)

1205971199092+3

4119906119898minus1(119909 119905)

minus3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894)119889120591

(47)

The following results are obtained

1199061 (119909 119905) = minus

1

2ℎ119905 sech119909 tanh119909

V1 (119909 119905) = ℎ119905 (minus

3 sech1199094

+sech31199092

+ sech119909tanh2119909)

1199062(119909 119905) = ℎ (

3

16ℎ1199052sech3119909 minus 1

16ℎ1199052 cosh (2119909) sech3119909)

minus1

2ℎ (ℎ + 119899) 119905 sech 119909 tanh119909

(48)

119906119898(119909 119905) (119898 = 3 4 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(49)

Equation (49) is a family of approximation solutions tothe problem (40) in terms of the convergence parameters ℎand 119899 To find the valid region of ℎ the ℎ-curves given bythe 6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

minus04

minus05

minus06

minus07

minus08

minus09

U6mq

-HA

M (x

tn

)

0minus50minus100minus150

U6mq-HAM (01 06 100)U6mq-HAM (03 08 100)U6mq-HAM (06 1 100)

h

Figure 10 ℎ-curve for the m119902-HAM (119899 = 100) approximationsolution 119880

6(119909 119905 100) of problem (40) at different values of 119909 and

119905

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

t

ExactU6nHAMU6mq-HAM (n = 5)

U6mq-HAM (n = 20)U6mq-HAM (n = 50)U6mq-HAM (n = 100)

Figure 11 Comparison between 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) with exact solution of (40)

at 119909 = 1 with (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ =minus795) respectively

the 6th-order m119902-HAM (119899 = 100) approximation at differentvalues of 119909 119905 are drawn in Figures 9 and 10 these figures showthe interval of ℎ in which the value of119880

6is constant at certain

119909 119905 and 119899 we chose the horizontal line parallel to 119909-axis (ℎ)as a valid regionwhich provides uswith a simpleway to adjustand control the convergence region Figure 11 shows thecomparison between119880

6of 119899HAM and119880

6of m119902-HAM using

different values of 119899with the solution (42)The absolute errorsof the 6th-order solutions 119899HAM approximate and the 6th-order solutions m119902-HAM approximate using different valuesof 119899 are shown in Figure 12 The results obtained by m119902-HAM indicate that the speed of convergence for m119902-HAMwith 119899 gt 1 is faster in comparison to 119899 = 1 (119899HAM) Theresults show that the convergence region of series solutions

8 Journal of Applied Mathematics

05 10 15 20 25 30 35

002

004

006

008

Abso

lute

erro

r

t

AE U6nHAMAE U6mq-HAM (n = 5)AE U6mq-HAM (n = 20)

AE U6mq-HAM (n = 50)AE U6mq-HAM (n = 100)

Figure 12 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) for problem (40) at 119909 = 1

using (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ = minus795)respectively

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

ExactU3nHAMU6nHAM

t

Figure 13 The comparison between the 1198803 1198806of 119899HAM (m119902-

HAM 119899 = 1) and the exact solution of (40) at ℎ = minus1 and 119909 = 1

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U3mq-HAM (n = 100)U6mq-HAM (n = 100)

Figure 14 The comparison between the 1198803 1198806of m119902-HAM (119899 =

100) and the exact solution of (40) at ℎ = minus795 and 119909 = 1

05 10 15 20 25 30 35

01

02

03

04

Abso

lute

erro

r

t

AE U3nHAMAE U6nHAM

Figure 15 The comparison between the absolute error of 1198803and

1198806of 119899HAM (m119902-HAM 119899 = 1) of (40) at ℎ = minus1 119909 = 1 and

0 le 119905 le 35

05 10 15 20 25 30 35

0005

0010

0015

0020

0025Ab

solu

te er

ror

t

AE U3mq-HAM (n = 100)AE U6mq-HAM (n = 100)

Figure 16The comparison between the absolute error of 1198803and119880

6

of m119902-HAM(119899 = 100) of (40) at ℎ = minus795 119909 = 1 and 0 le 119905 le 35

obtained bym119902-HAM is increasing as 119902 is decreased as shownin Figures 11 and 12

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient andthe interval of 119905 (convergent region) increases as shown inFigures 13 14 15 and 16

Figure 17 shows that the convergence of the series solu-tions obtained by the 3rd-order m119902-HAM (119899 = 100) is fasterthan that of the series solutions obtained by the 6th order119899HAM This fact shows the importance of the convergenceparameters 119899 in the m119902-HAM

5 Conclusion

In this paper a modified 119902-homotopy analysis method wasproposed (m119902-HAM)This method provides an approximatesolution by rewriting the 119899th-order nonlinear differentialequations in the form of system of 119899 first-order differentialequations The solution of these 119899 differential equations is

Journal of Applied Mathematics 9

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U6nHAMU3mq-HAM (n = 100)

Figure 17 The comparison between the 1198803of m119902-HAM (119899 = 100)

1198806of 119899HAM (m119902-HAM 119899 = 1) and the exact solution of (40) at

(ℎ = minus795 ℎ = minus1) and 119909 = 1

obtained as a power series solution which converges to aclosed form solution The m119902-HAM contains two auxiliaryparameters 119899 and ℎ such that the case of 119899 = 1 (m119902-HAM 119899 =1) the 119899HAM which is proposed in [21 22] can be reachedIn general it was noticed from the illustrative examples thatthe convergence of m119902-HAM is faster than that of 119899HAM

References

[1] S J Liao The proposed homotopy analysis technique for thesolution of nonlinear problems [PhD thesis] Shanghai Jiao TongUniversity 1992

[2] S Abbasbandy ldquoThe application of homotopy analysis methodto nonlinear equations arising in heat transferrdquo Physics LettersA vol 360 no 1 pp 109ndash113 2006

[3] S Abbasbandy andA Shirzadi ldquoHomotopy analysismethod formultiple solutions of the fractional Sturm-Liouville problemsrdquoNumerical Algorithms vol 54 no 4 pp 521ndash532 2010

[4] S Abbasbandy M Ashtiani and E Babolian ldquoAnalytic solutionof the Sharma-Tasso-Olver equation by homotopy analysismethodrdquoZeitschrift furNaturforschungA vol 65 no 4 pp 285ndash290 2010

[5] A K Alomari M S M Noorani and R Nazar ldquoExplicitseries solutions of some linear and nonlinear Schrodingerequations via the homotopy analysis methodrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 14 no 4pp 1196ndash1207 2009

[6] M Ayub A Rasheed and T Hayat ldquoExact flow of a third gradefluid past a porous plate using homotopy analysis methodrdquoInternational Journal of Engineering Science vol 41 no 18 pp2091ndash2103 2003

[7] A Sami Bataineh M S M Noorani and I Hashim ldquoSolvingsystems of ODEs by homotopy analysis methodrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 13 no10 pp 2060ndash2070 2008

[8] A S Bataineh M S M Noorani and I Hashim ldquoHomotopyanalysis method for singular IVPs of Emden-Fowler typerdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 14 no 4 pp 1121ndash1131 2009

[9] T Hayat M Khan and S Asghar ldquoHomotopy analysis of MHDflows of an Oldroyd 8-constant fluidrdquo Acta Mechanica vol 168no 3-4 pp 213ndash232 2004

[10] S J Liao Beyond Perturbation Introduction to the HomotopyAnalysisMethod ChapmanampHall Boca Raton Fla USA 2004

[11] S J Liao ldquoAn approximate solution technique not dependingon small parameters a special examplerdquo International Journalof Non-Linear Mechanics vol 30 no 3 pp 371ndash380 1995

[12] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[13] S Liao ldquoComparison between the homotopy analysis methodand homotopy perturbationmethodrdquoAppliedMathematics andComputation vol 169 no 2 pp 1186ndash1194 2005

[14] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[15] S Nadeem and N S Akbar ldquoPeristaltic flow of Sisko fluid in auniform inclined tuberdquoActaMechanica Sinica vol 26 no 5 pp675ndash683 2010

[16] S Nadeem and N S Akbar ldquoInfluence of heat transfer on aperistaltic flow of Johnson Segalman fluid in a non uniformtuberdquo International Communications in Heat andMass Transfervol 36 no 10 pp 1050ndash1059 2009

[17] Y Tan and S Abbasbandy ldquoHomotopy analysis method forquadratic Riccati differential equationrdquo Communications inNonlinear Science and Numerical Simulation vol 13 no 3 pp539ndash546 2008

[18] A Sami Bataineh M S M Noorani and I Hashim ldquoApproxi-mate solutions of singular two-point BVPs by modified homo-topy analysis methodrdquo Physics Letters A vol 372 pp 4062ndash4066 2008

[19] A Sami Bataineh M S M Noorani and I Hashim ldquoOn a newreliablemodification of homotopy analysismethodrdquoCommuni-cations in Nonlinear Science and Numerical Simulation vol 14no 2 pp 409ndash423 2009

[20] A S Bataineh M S M Noorani and I Hashim ldquoModifiedhomotopy analysis method for solving systems of second-orderBVPsrdquo Communications in Nonlinear Science and NumericalSimulation vol 14 no 2 pp 430ndash442 2009

[21] H N Hassan and M A El-Tawil ldquoA new technique of usinghomotopy analysis method for solving high-order nonlineardifferential equationsrdquo Mathematical Methods in the AppliedSciences vol 34 no 6 pp 728ndash742 2011

[22] H N Hassan and M A El-Tawil ldquoA new technique ofusing homotopy analysis method for second order nonlineardifferential equationsrdquo Applied Mathematics and Computationvol 219 no 2 pp 708ndash728 2012

[23] M A El-Tawil and S N Huseen ldquoThe q-homotopy analysismethod (q-HAM)rdquo International Journal of Applied Mathemat-ics and Mechanics vol 8 no 15 pp 51ndash75 2012

[24] M A El-Tawil and S N Huseen ldquoOn convergence of the q-homotopy analysis methodrdquo International Journal of Contem-porary Mathematical Sciences vol 8 no 10 pp 481ndash497 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 7: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

Journal of Applied Mathematics 7

minus02

minus04

minus06

minus08

U6n

HA

M (x

t)

00minus05minus10minus15minus20

h

U6nHAM (01 06)U6nHAM (03 08)U6nHAM (06 1)

Figure 9 ℎ-curve for the 119899HAM (m119902-HAM 119899 = 1) approximationsolution 119880

6(119909 119905) of problem (40) at different values of 119909 and 119905

For119898 ge 2 we get

119906119898(119909 119905) = (119899 + ℎ) 119906

119898minus1(119909 119905) + ℎint

119905

0

(minusV119898minus1(119909 120591)) 119889120591

V119898(119909 119905) = (119899 + ℎ) V

119898minus1(119909 119905)

+ ℎint

119905

0

(minus1205972119906119898minus1 (119909 119905)

1205971199092+3

4119906119898minus1(119909 119905)

minus3

2

119898minus1

sum

119895=0

119906119898minus1minus119895

119895

sum

119894=0

119906119894119906119895minus119894)119889120591

(47)

The following results are obtained

1199061 (119909 119905) = minus

1

2ℎ119905 sech119909 tanh119909

V1 (119909 119905) = ℎ119905 (minus

3 sech1199094

+sech31199092

+ sech119909tanh2119909)

1199062(119909 119905) = ℎ (

3

16ℎ1199052sech3119909 minus 1

16ℎ1199052 cosh (2119909) sech3119909)

minus1

2ℎ (ℎ + 119899) 119905 sech 119909 tanh119909

(48)

119906119898(119909 119905) (119898 = 3 4 ) can be calculated similarly Then the

series solution expression by m119902-HAM can be written in thefollowing form

119906 (119909 119905 119899 ℎ) 2 119880119872(119909 119905 119899 ℎ) =

119872

sum

119894=0

119906119894(119909 119905 119899 ℎ) (

1

119899)

119894

(49)

Equation (49) is a family of approximation solutions tothe problem (40) in terms of the convergence parameters ℎand 119899 To find the valid region of ℎ the ℎ-curves given bythe 6th-order 119899HAM (m119902-HAM 119899 = 1) approximation and

minus04

minus05

minus06

minus07

minus08

minus09

U6mq

-HA

M (x

tn

)

0minus50minus100minus150

U6mq-HAM (01 06 100)U6mq-HAM (03 08 100)U6mq-HAM (06 1 100)

h

Figure 10 ℎ-curve for the m119902-HAM (119899 = 100) approximationsolution 119880

6(119909 119905 100) of problem (40) at different values of 119909 and

119905

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

t

ExactU6nHAMU6mq-HAM (n = 5)

U6mq-HAM (n = 20)U6mq-HAM (n = 50)U6mq-HAM (n = 100)

Figure 11 Comparison between 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) with exact solution of (40)

at 119909 = 1 with (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ =minus795) respectively

the 6th-order m119902-HAM (119899 = 100) approximation at differentvalues of 119909 119905 are drawn in Figures 9 and 10 these figures showthe interval of ℎ in which the value of119880

6is constant at certain

119909 119905 and 119899 we chose the horizontal line parallel to 119909-axis (ℎ)as a valid regionwhich provides uswith a simpleway to adjustand control the convergence region Figure 11 shows thecomparison between119880

6of 119899HAM and119880

6of m119902-HAM using

different values of 119899with the solution (42)The absolute errorsof the 6th-order solutions 119899HAM approximate and the 6th-order solutions m119902-HAM approximate using different valuesof 119899 are shown in Figure 12 The results obtained by m119902-HAM indicate that the speed of convergence for m119902-HAMwith 119899 gt 1 is faster in comparison to 119899 = 1 (119899HAM) Theresults show that the convergence region of series solutions

8 Journal of Applied Mathematics

05 10 15 20 25 30 35

002

004

006

008

Abso

lute

erro

r

t

AE U6nHAMAE U6mq-HAM (n = 5)AE U6mq-HAM (n = 20)

AE U6mq-HAM (n = 50)AE U6mq-HAM (n = 100)

Figure 12 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) for problem (40) at 119909 = 1

using (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ = minus795)respectively

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

ExactU3nHAMU6nHAM

t

Figure 13 The comparison between the 1198803 1198806of 119899HAM (m119902-

HAM 119899 = 1) and the exact solution of (40) at ℎ = minus1 and 119909 = 1

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U3mq-HAM (n = 100)U6mq-HAM (n = 100)

Figure 14 The comparison between the 1198803 1198806of m119902-HAM (119899 =

100) and the exact solution of (40) at ℎ = minus795 and 119909 = 1

05 10 15 20 25 30 35

01

02

03

04

Abso

lute

erro

r

t

AE U3nHAMAE U6nHAM

Figure 15 The comparison between the absolute error of 1198803and

1198806of 119899HAM (m119902-HAM 119899 = 1) of (40) at ℎ = minus1 119909 = 1 and

0 le 119905 le 35

05 10 15 20 25 30 35

0005

0010

0015

0020

0025Ab

solu

te er

ror

t

AE U3mq-HAM (n = 100)AE U6mq-HAM (n = 100)

Figure 16The comparison between the absolute error of 1198803and119880

6

of m119902-HAM(119899 = 100) of (40) at ℎ = minus795 119909 = 1 and 0 le 119905 le 35

obtained bym119902-HAM is increasing as 119902 is decreased as shownin Figures 11 and 12

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient andthe interval of 119905 (convergent region) increases as shown inFigures 13 14 15 and 16

Figure 17 shows that the convergence of the series solu-tions obtained by the 3rd-order m119902-HAM (119899 = 100) is fasterthan that of the series solutions obtained by the 6th order119899HAM This fact shows the importance of the convergenceparameters 119899 in the m119902-HAM

5 Conclusion

In this paper a modified 119902-homotopy analysis method wasproposed (m119902-HAM)This method provides an approximatesolution by rewriting the 119899th-order nonlinear differentialequations in the form of system of 119899 first-order differentialequations The solution of these 119899 differential equations is

Journal of Applied Mathematics 9

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U6nHAMU3mq-HAM (n = 100)

Figure 17 The comparison between the 1198803of m119902-HAM (119899 = 100)

1198806of 119899HAM (m119902-HAM 119899 = 1) and the exact solution of (40) at

(ℎ = minus795 ℎ = minus1) and 119909 = 1

obtained as a power series solution which converges to aclosed form solution The m119902-HAM contains two auxiliaryparameters 119899 and ℎ such that the case of 119899 = 1 (m119902-HAM 119899 =1) the 119899HAM which is proposed in [21 22] can be reachedIn general it was noticed from the illustrative examples thatthe convergence of m119902-HAM is faster than that of 119899HAM

References

[1] S J Liao The proposed homotopy analysis technique for thesolution of nonlinear problems [PhD thesis] Shanghai Jiao TongUniversity 1992

[2] S Abbasbandy ldquoThe application of homotopy analysis methodto nonlinear equations arising in heat transferrdquo Physics LettersA vol 360 no 1 pp 109ndash113 2006

[3] S Abbasbandy andA Shirzadi ldquoHomotopy analysismethod formultiple solutions of the fractional Sturm-Liouville problemsrdquoNumerical Algorithms vol 54 no 4 pp 521ndash532 2010

[4] S Abbasbandy M Ashtiani and E Babolian ldquoAnalytic solutionof the Sharma-Tasso-Olver equation by homotopy analysismethodrdquoZeitschrift furNaturforschungA vol 65 no 4 pp 285ndash290 2010

[5] A K Alomari M S M Noorani and R Nazar ldquoExplicitseries solutions of some linear and nonlinear Schrodingerequations via the homotopy analysis methodrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 14 no 4pp 1196ndash1207 2009

[6] M Ayub A Rasheed and T Hayat ldquoExact flow of a third gradefluid past a porous plate using homotopy analysis methodrdquoInternational Journal of Engineering Science vol 41 no 18 pp2091ndash2103 2003

[7] A Sami Bataineh M S M Noorani and I Hashim ldquoSolvingsystems of ODEs by homotopy analysis methodrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 13 no10 pp 2060ndash2070 2008

[8] A S Bataineh M S M Noorani and I Hashim ldquoHomotopyanalysis method for singular IVPs of Emden-Fowler typerdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 14 no 4 pp 1121ndash1131 2009

[9] T Hayat M Khan and S Asghar ldquoHomotopy analysis of MHDflows of an Oldroyd 8-constant fluidrdquo Acta Mechanica vol 168no 3-4 pp 213ndash232 2004

[10] S J Liao Beyond Perturbation Introduction to the HomotopyAnalysisMethod ChapmanampHall Boca Raton Fla USA 2004

[11] S J Liao ldquoAn approximate solution technique not dependingon small parameters a special examplerdquo International Journalof Non-Linear Mechanics vol 30 no 3 pp 371ndash380 1995

[12] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[13] S Liao ldquoComparison between the homotopy analysis methodand homotopy perturbationmethodrdquoAppliedMathematics andComputation vol 169 no 2 pp 1186ndash1194 2005

[14] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[15] S Nadeem and N S Akbar ldquoPeristaltic flow of Sisko fluid in auniform inclined tuberdquoActaMechanica Sinica vol 26 no 5 pp675ndash683 2010

[16] S Nadeem and N S Akbar ldquoInfluence of heat transfer on aperistaltic flow of Johnson Segalman fluid in a non uniformtuberdquo International Communications in Heat andMass Transfervol 36 no 10 pp 1050ndash1059 2009

[17] Y Tan and S Abbasbandy ldquoHomotopy analysis method forquadratic Riccati differential equationrdquo Communications inNonlinear Science and Numerical Simulation vol 13 no 3 pp539ndash546 2008

[18] A Sami Bataineh M S M Noorani and I Hashim ldquoApproxi-mate solutions of singular two-point BVPs by modified homo-topy analysis methodrdquo Physics Letters A vol 372 pp 4062ndash4066 2008

[19] A Sami Bataineh M S M Noorani and I Hashim ldquoOn a newreliablemodification of homotopy analysismethodrdquoCommuni-cations in Nonlinear Science and Numerical Simulation vol 14no 2 pp 409ndash423 2009

[20] A S Bataineh M S M Noorani and I Hashim ldquoModifiedhomotopy analysis method for solving systems of second-orderBVPsrdquo Communications in Nonlinear Science and NumericalSimulation vol 14 no 2 pp 430ndash442 2009

[21] H N Hassan and M A El-Tawil ldquoA new technique of usinghomotopy analysis method for solving high-order nonlineardifferential equationsrdquo Mathematical Methods in the AppliedSciences vol 34 no 6 pp 728ndash742 2011

[22] H N Hassan and M A El-Tawil ldquoA new technique ofusing homotopy analysis method for second order nonlineardifferential equationsrdquo Applied Mathematics and Computationvol 219 no 2 pp 708ndash728 2012

[23] M A El-Tawil and S N Huseen ldquoThe q-homotopy analysismethod (q-HAM)rdquo International Journal of Applied Mathemat-ics and Mechanics vol 8 no 15 pp 51ndash75 2012

[24] M A El-Tawil and S N Huseen ldquoOn convergence of the q-homotopy analysis methodrdquo International Journal of Contem-porary Mathematical Sciences vol 8 no 10 pp 481ndash497 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

8 Journal of Applied Mathematics

05 10 15 20 25 30 35

002

004

006

008

Abso

lute

erro

r

t

AE U6nHAMAE U6mq-HAM (n = 5)AE U6mq-HAM (n = 20)

AE U6mq-HAM (n = 50)AE U6mq-HAM (n = 100)

Figure 12 The absolute error of 1198806of 119899HAM (m119902-HAM 119899 = 1)

and 1198806of m119902-HAM (119899 = 5 20 50 100) for problem (40) at 119909 = 1

using (ℎ = minus1 ℎ = minus485 ℎ = minus1855 ℎ = minus4311 ℎ = minus795)respectively

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

ExactU3nHAMU6nHAM

t

Figure 13 The comparison between the 1198803 1198806of 119899HAM (m119902-

HAM 119899 = 1) and the exact solution of (40) at ℎ = minus1 and 119909 = 1

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U3mq-HAM (n = 100)U6mq-HAM (n = 100)

Figure 14 The comparison between the 1198803 1198806of m119902-HAM (119899 =

100) and the exact solution of (40) at ℎ = minus795 and 119909 = 1

05 10 15 20 25 30 35

01

02

03

04

Abso

lute

erro

r

t

AE U3nHAMAE U6nHAM

Figure 15 The comparison between the absolute error of 1198803and

1198806of 119899HAM (m119902-HAM 119899 = 1) of (40) at ℎ = minus1 119909 = 1 and

0 le 119905 le 35

05 10 15 20 25 30 35

0005

0010

0015

0020

0025Ab

solu

te er

ror

t

AE U3mq-HAM (n = 100)AE U6mq-HAM (n = 100)

Figure 16The comparison between the absolute error of 1198803and119880

6

of m119902-HAM(119899 = 100) of (40) at ℎ = minus795 119909 = 1 and 0 le 119905 le 35

obtained bym119902-HAM is increasing as 119902 is decreased as shownin Figures 11 and 12

By increasing the number of iterations by m119902-HAM theseries solution becomes more accurate more efficient andthe interval of 119905 (convergent region) increases as shown inFigures 13 14 15 and 16

Figure 17 shows that the convergence of the series solu-tions obtained by the 3rd-order m119902-HAM (119899 = 100) is fasterthan that of the series solutions obtained by the 6th order119899HAM This fact shows the importance of the convergenceparameters 119899 in the m119902-HAM

5 Conclusion

In this paper a modified 119902-homotopy analysis method wasproposed (m119902-HAM)This method provides an approximatesolution by rewriting the 119899th-order nonlinear differentialequations in the form of system of 119899 first-order differentialequations The solution of these 119899 differential equations is

Journal of Applied Mathematics 9

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U6nHAMU3mq-HAM (n = 100)

Figure 17 The comparison between the 1198803of m119902-HAM (119899 = 100)

1198806of 119899HAM (m119902-HAM 119899 = 1) and the exact solution of (40) at

(ℎ = minus795 ℎ = minus1) and 119909 = 1

obtained as a power series solution which converges to aclosed form solution The m119902-HAM contains two auxiliaryparameters 119899 and ℎ such that the case of 119899 = 1 (m119902-HAM 119899 =1) the 119899HAM which is proposed in [21 22] can be reachedIn general it was noticed from the illustrative examples thatthe convergence of m119902-HAM is faster than that of 119899HAM

References

[1] S J Liao The proposed homotopy analysis technique for thesolution of nonlinear problems [PhD thesis] Shanghai Jiao TongUniversity 1992

[2] S Abbasbandy ldquoThe application of homotopy analysis methodto nonlinear equations arising in heat transferrdquo Physics LettersA vol 360 no 1 pp 109ndash113 2006

[3] S Abbasbandy andA Shirzadi ldquoHomotopy analysismethod formultiple solutions of the fractional Sturm-Liouville problemsrdquoNumerical Algorithms vol 54 no 4 pp 521ndash532 2010

[4] S Abbasbandy M Ashtiani and E Babolian ldquoAnalytic solutionof the Sharma-Tasso-Olver equation by homotopy analysismethodrdquoZeitschrift furNaturforschungA vol 65 no 4 pp 285ndash290 2010

[5] A K Alomari M S M Noorani and R Nazar ldquoExplicitseries solutions of some linear and nonlinear Schrodingerequations via the homotopy analysis methodrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 14 no 4pp 1196ndash1207 2009

[6] M Ayub A Rasheed and T Hayat ldquoExact flow of a third gradefluid past a porous plate using homotopy analysis methodrdquoInternational Journal of Engineering Science vol 41 no 18 pp2091ndash2103 2003

[7] A Sami Bataineh M S M Noorani and I Hashim ldquoSolvingsystems of ODEs by homotopy analysis methodrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 13 no10 pp 2060ndash2070 2008

[8] A S Bataineh M S M Noorani and I Hashim ldquoHomotopyanalysis method for singular IVPs of Emden-Fowler typerdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 14 no 4 pp 1121ndash1131 2009

[9] T Hayat M Khan and S Asghar ldquoHomotopy analysis of MHDflows of an Oldroyd 8-constant fluidrdquo Acta Mechanica vol 168no 3-4 pp 213ndash232 2004

[10] S J Liao Beyond Perturbation Introduction to the HomotopyAnalysisMethod ChapmanampHall Boca Raton Fla USA 2004

[11] S J Liao ldquoAn approximate solution technique not dependingon small parameters a special examplerdquo International Journalof Non-Linear Mechanics vol 30 no 3 pp 371ndash380 1995

[12] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[13] S Liao ldquoComparison between the homotopy analysis methodand homotopy perturbationmethodrdquoAppliedMathematics andComputation vol 169 no 2 pp 1186ndash1194 2005

[14] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[15] S Nadeem and N S Akbar ldquoPeristaltic flow of Sisko fluid in auniform inclined tuberdquoActaMechanica Sinica vol 26 no 5 pp675ndash683 2010

[16] S Nadeem and N S Akbar ldquoInfluence of heat transfer on aperistaltic flow of Johnson Segalman fluid in a non uniformtuberdquo International Communications in Heat andMass Transfervol 36 no 10 pp 1050ndash1059 2009

[17] Y Tan and S Abbasbandy ldquoHomotopy analysis method forquadratic Riccati differential equationrdquo Communications inNonlinear Science and Numerical Simulation vol 13 no 3 pp539ndash546 2008

[18] A Sami Bataineh M S M Noorani and I Hashim ldquoApproxi-mate solutions of singular two-point BVPs by modified homo-topy analysis methodrdquo Physics Letters A vol 372 pp 4062ndash4066 2008

[19] A Sami Bataineh M S M Noorani and I Hashim ldquoOn a newreliablemodification of homotopy analysismethodrdquoCommuni-cations in Nonlinear Science and Numerical Simulation vol 14no 2 pp 409ndash423 2009

[20] A S Bataineh M S M Noorani and I Hashim ldquoModifiedhomotopy analysis method for solving systems of second-orderBVPsrdquo Communications in Nonlinear Science and NumericalSimulation vol 14 no 2 pp 430ndash442 2009

[21] H N Hassan and M A El-Tawil ldquoA new technique of usinghomotopy analysis method for solving high-order nonlineardifferential equationsrdquo Mathematical Methods in the AppliedSciences vol 34 no 6 pp 728ndash742 2011

[22] H N Hassan and M A El-Tawil ldquoA new technique ofusing homotopy analysis method for second order nonlineardifferential equationsrdquo Applied Mathematics and Computationvol 219 no 2 pp 708ndash728 2012

[23] M A El-Tawil and S N Huseen ldquoThe q-homotopy analysismethod (q-HAM)rdquo International Journal of Applied Mathemat-ics and Mechanics vol 8 no 15 pp 51ndash75 2012

[24] M A El-Tawil and S N Huseen ldquoOn convergence of the q-homotopy analysis methodrdquo International Journal of Contem-porary Mathematical Sciences vol 8 no 10 pp 481ndash497 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 9: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

Journal of Applied Mathematics 9

05 10 15 20 25 30 35

minus05

minus04

minus03

minus02

Solu

tions

Exact

t

U6nHAMU3mq-HAM (n = 100)

Figure 17 The comparison between the 1198803of m119902-HAM (119899 = 100)

1198806of 119899HAM (m119902-HAM 119899 = 1) and the exact solution of (40) at

(ℎ = minus795 ℎ = minus1) and 119909 = 1

obtained as a power series solution which converges to aclosed form solution The m119902-HAM contains two auxiliaryparameters 119899 and ℎ such that the case of 119899 = 1 (m119902-HAM 119899 =1) the 119899HAM which is proposed in [21 22] can be reachedIn general it was noticed from the illustrative examples thatthe convergence of m119902-HAM is faster than that of 119899HAM

References

[1] S J Liao The proposed homotopy analysis technique for thesolution of nonlinear problems [PhD thesis] Shanghai Jiao TongUniversity 1992

[2] S Abbasbandy ldquoThe application of homotopy analysis methodto nonlinear equations arising in heat transferrdquo Physics LettersA vol 360 no 1 pp 109ndash113 2006

[3] S Abbasbandy andA Shirzadi ldquoHomotopy analysismethod formultiple solutions of the fractional Sturm-Liouville problemsrdquoNumerical Algorithms vol 54 no 4 pp 521ndash532 2010

[4] S Abbasbandy M Ashtiani and E Babolian ldquoAnalytic solutionof the Sharma-Tasso-Olver equation by homotopy analysismethodrdquoZeitschrift furNaturforschungA vol 65 no 4 pp 285ndash290 2010

[5] A K Alomari M S M Noorani and R Nazar ldquoExplicitseries solutions of some linear and nonlinear Schrodingerequations via the homotopy analysis methodrdquo Communicationsin Nonlinear Science and Numerical Simulation vol 14 no 4pp 1196ndash1207 2009

[6] M Ayub A Rasheed and T Hayat ldquoExact flow of a third gradefluid past a porous plate using homotopy analysis methodrdquoInternational Journal of Engineering Science vol 41 no 18 pp2091ndash2103 2003

[7] A Sami Bataineh M S M Noorani and I Hashim ldquoSolvingsystems of ODEs by homotopy analysis methodrdquo Communica-tions in Nonlinear Science and Numerical Simulation vol 13 no10 pp 2060ndash2070 2008

[8] A S Bataineh M S M Noorani and I Hashim ldquoHomotopyanalysis method for singular IVPs of Emden-Fowler typerdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 14 no 4 pp 1121ndash1131 2009

[9] T Hayat M Khan and S Asghar ldquoHomotopy analysis of MHDflows of an Oldroyd 8-constant fluidrdquo Acta Mechanica vol 168no 3-4 pp 213ndash232 2004

[10] S J Liao Beyond Perturbation Introduction to the HomotopyAnalysisMethod ChapmanampHall Boca Raton Fla USA 2004

[11] S J Liao ldquoAn approximate solution technique not dependingon small parameters a special examplerdquo International Journalof Non-Linear Mechanics vol 30 no 3 pp 371ndash380 1995

[12] S Liao ldquoOn the homotopy analysis method for nonlinearproblemsrdquo Applied Mathematics and Computation vol 147 no2 pp 499ndash513 2004

[13] S Liao ldquoComparison between the homotopy analysis methodand homotopy perturbationmethodrdquoAppliedMathematics andComputation vol 169 no 2 pp 1186ndash1194 2005

[14] A Molabahrami and F Khani ldquoThe homotopy analysis methodto solve the Burgers-Huxley equationrdquoNonlinear Analysis RealWorld Applications vol 10 no 2 pp 589ndash600 2009

[15] S Nadeem and N S Akbar ldquoPeristaltic flow of Sisko fluid in auniform inclined tuberdquoActaMechanica Sinica vol 26 no 5 pp675ndash683 2010

[16] S Nadeem and N S Akbar ldquoInfluence of heat transfer on aperistaltic flow of Johnson Segalman fluid in a non uniformtuberdquo International Communications in Heat andMass Transfervol 36 no 10 pp 1050ndash1059 2009

[17] Y Tan and S Abbasbandy ldquoHomotopy analysis method forquadratic Riccati differential equationrdquo Communications inNonlinear Science and Numerical Simulation vol 13 no 3 pp539ndash546 2008

[18] A Sami Bataineh M S M Noorani and I Hashim ldquoApproxi-mate solutions of singular two-point BVPs by modified homo-topy analysis methodrdquo Physics Letters A vol 372 pp 4062ndash4066 2008

[19] A Sami Bataineh M S M Noorani and I Hashim ldquoOn a newreliablemodification of homotopy analysismethodrdquoCommuni-cations in Nonlinear Science and Numerical Simulation vol 14no 2 pp 409ndash423 2009

[20] A S Bataineh M S M Noorani and I Hashim ldquoModifiedhomotopy analysis method for solving systems of second-orderBVPsrdquo Communications in Nonlinear Science and NumericalSimulation vol 14 no 2 pp 430ndash442 2009

[21] H N Hassan and M A El-Tawil ldquoA new technique of usinghomotopy analysis method for solving high-order nonlineardifferential equationsrdquo Mathematical Methods in the AppliedSciences vol 34 no 6 pp 728ndash742 2011

[22] H N Hassan and M A El-Tawil ldquoA new technique ofusing homotopy analysis method for second order nonlineardifferential equationsrdquo Applied Mathematics and Computationvol 219 no 2 pp 708ndash728 2012

[23] M A El-Tawil and S N Huseen ldquoThe q-homotopy analysismethod (q-HAM)rdquo International Journal of Applied Mathemat-ics and Mechanics vol 8 no 15 pp 51ndash75 2012

[24] M A El-Tawil and S N Huseen ldquoOn convergence of the q-homotopy analysis methodrdquo International Journal of Contem-porary Mathematical Sciences vol 8 no 10 pp 481ndash497 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article Approximate Solutions of Nonlinear Partial Differential Equations …downloads.hindawi.com/journals/jam/2013/569674.pdf · 2019-07-31 · Approximate Solutions of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Some Approximate Solutions for Shallow Tunnels

Some Approximate Solutions for Shallow Tunnels

Approximate Ad Hoc Parametric Solutions for Nonlinear ...downloads.hindawi.com/journals/mpe/2010/874540.pdf · Approximate Ad Hoc Parametric Solutions for ... We also mention Charpit’s

Approximate Ad Hoc Parametric Solutions for Nonlinear ...downloads.hindawi.com/journals/mpe/2010/874540.pdf · Approximate Ad Hoc Parametric Solutions for ... We also mention Charpit’s

Approximate Solutions of Coupled Nonlinear Oscillations ...

Approximate Solutions of Coupled Nonlinear Oscillations ...

STAT 35000 Exam Jam Solutions Contents - Indiana … 35000 Exam Jam Solutions Contents ... Con dence Intervals ... Mathematical Sciences Department @ IUPUI STAT 35000 Exam Jam 3.Chris

STAT 35000 Exam Jam Solutions Contents - Indiana … 35000 Exam Jam Solutions Contents ... Con dence Intervals ... Mathematical Sciences Department @ IUPUI STAT 35000 Exam Jam 3.Chris

Chapter 10: Approximate Solutions of the Navier-Stokes Equation · 2011-03-14 · Fondamenti di Meccanica dei Continui 9 Chapter 10: Approximate Solutions Nondimensionalization of

Chapter 10: Approximate Solutions of the Navier-Stokes Equation · 2011-03-14 · Fondamenti di Meccanica dei Continui 9 Chapter 10: Approximate Solutions Nondimensionalization of

APPROXIMATE SOLUTIONS OF COUPLED RAMANI EQUATION BY …

APPROXIMATE SOLUTIONS OF COUPLED RAMANI EQUATION BY …

Fast Approximate Hierarchical Solutions of MDPs

Fast Approximate Hierarchical Solutions of MDPs

Chapter 10: Approximate Solutions of the Navier-Stokes Equation

Chapter 10: Approximate Solutions of the Navier-Stokes Equation

Research Article Approximate Analytic Solutions of ...

Research Article Approximate Analytic Solutions of ...

C ASCADED F ILTERS Approximate and Exact Solutions.

C ASCADED F ILTERS Approximate and Exact Solutions.

APPROXIMATE SOLUTIONS OF GENERALIZED RIEMANN ......APPROXIMATE SOLUTIONS OF GENERALIZED RIEMANN PROBLEMS 3 We restrict our analysis to systems for which every characteristic field

APPROXIMATE SOLUTIONS OF GENERALIZED RIEMANN ......APPROXIMATE SOLUTIONS OF GENERALIZED RIEMANN PROBLEMS 3 We restrict our analysis to systems for which every characteristic field

APPROXIMATE SOLUTIONS: THE INTEGRAL METHOD

APPROXIMATE SOLUTIONS: THE INTEGRAL METHOD

Possible Causes & Solutions of Traffic Jam and Their ...

Possible Causes & Solutions of Traffic Jam and Their ...

Approximate solutions to boundary layer problems … · approximate solutions to boundary layer problems in linear kinetic theory proefschrift ter verkrijging van de graad van doctor

Approximate solutions to boundary layer problems … · approximate solutions to boundary layer problems in linear kinetic theory proefschrift ter verkrijging van de graad van doctor

Approximate Solutions to Time-Fractional Schrödinger Equation via ...

Approximate Solutions to Time-Fractional Schrödinger Equation via ...

Approximate Solutions, Thermal Properties and ...

Approximate Solutions, Thermal Properties and ...

The van der Waals equation: analytical and approximate solutions

The van der Waals equation: analytical and approximate solutions

Approximate Solutions for Mixed Boundary Value Problems by ... · Approximate Solutions for Mixed Boundary Value Problems by Finite-Difference Methods By V. Thuraisamy* Abstract.

Approximate Solutions for Mixed Boundary Value Problems by ... · Approximate Solutions for Mixed Boundary Value Problems by Finite-Difference Methods By V. Thuraisamy* Abstract.

Approximate solutions of the Schrödinger equation with ...

Approximate solutions of the Schrödinger equation with ...

Probabilistic Data Structures and Approximate Solutions

Probabilistic Data Structures and Approximate Solutions