research article a simplification for exp-function method when … · 2019. 7. 31. · abstract and...

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 264549 5 pageshttpdxdoiorg1011552013264549

Research ArticleA Simplification for Exp-Function Method When the BalancedNonlinear Term Is a Certain Product

Hong-Zhun Liu

Zhijiang College Zhejiang University of Technology Hangzhou 310024 China

Correspondence should be addressed to Hong-Zhun Liu mathlhz163com

Received 24 October 2013 Revised 8 December 2013 Accepted 8 December 2013

Academic Editor Hossein Jafari

Copyright copy 2013 Hong-Zhun Liu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The Exp-function method plays an important role in searching for analytic solutions of many nonlinear differential equations Inthis paper we prove that the balancing procedure in the method is unnecessary when the balanced nonlinear term is a product ofthe dependent variable under consideration and its derivatives And in this case the ansatz of the method can be simplified to bewith less parameters so as to be easy to calculate

1 Introduction

In 2006 He and Wu firstly proposed the so-called Exp-functionmethod to search for solitary solutions and periodicsolutions of nonlinear partial differential equations (PDE) [1]Thismethod soon drew the attention ofmany researchers andwas successfully applied to many nonlinear problems [2ndash31]Among them it is worth mentioning that Zhu firstly appliedthismethod to difference-differential equations which showsthat the method is also effective in this case [8 9] AfterZhu Dai et al generalized the Exp-function method to findnew exact traveling wave solutions of nonlinear PDE andnonlinear differential-difference equations [19] Recently Maet al extended the Exp-function method to multiple Exp-function method for constructing multiple wave solutions[32 33] He elucidated how to solve fractional differentialequations with local fractional derivatives via the fractionalcomplex transformation and the Exp-function method [34]

For convenience we first introduce the Exp-functionmethod in brief

11 Outline of the Exp-Function Method Suppose that weconsider a (1+1)-dimensional nonlinear PDE in the form

119864 (119906 119906119905 119906119909 119906119909119909 119906119909119905 119906119905119905 ) = 0 (1)

Using traveling wave transformation

119906 = 119906 (120578) 120578 = 119896119909 + 120596119905 (2)

we get a nonlinear ordinary differential equation (ODE)

119875 (119906 1199061015840 11990610158401015840 ) = 0 (3)

where the prime as it is in the following denotes thederivation with respect to 120578

The Exp-function method is based on the assumptionthat the solutions of (3) can be expressed in the followingform

119906 (120578) =sum119889

119899=minus119888119886119899exp (119899120578)

sum119902

119899=minus119901119887119899exp (119899120578)

=119886minus119888exp (minus119888120578) + sdot sdot sdot + 119886

119889exp (119889120578)

119887minus119901

exp (minus119901120578) + sdot sdot sdot + 119887119902exp (119902120578)

(4)

where 119888 119889119901 and 119902 are positive integers to be determined and119886119899and 119887119899are constants to be specified

Then we can express the highest order nonlinear andlinear terms in (3) in terms of (4) In the resulting termsdetermine 119889 and 119902 through balancing the highest order Exp-function and 119888 and 119901 by balancing the lowest order one

Substituting (4) along with the determined 119888 119889 119901 and 119902into (3) we can obtain an equation for exp(120578) Setting all thecoefficients of the different powers of exp(120578) to zero leads toa set of algebraic equations for 119886

119899 119887119899 119896 and 120596 Determine

values of 119886119899 119887119899 119896 and 120596 by solving this algebraic equations

and put these values into (4) Thus we may obtain nontrivialexact traveling wave solutions of (1)

2 Abstract and Applied Analysis

12 An Open Problem Among the Exp-functionmethod thebalancing computation is laborious but prior However weobserve that the balancing procedure of the Exp-functionmethod in studied examples always leads to the same case119888 = 119901 and 119889 = 119902 [21] This fact has partly been proved in[16 22] In [16] Ali has proved it by assuming the highestorder linear and nonlinear terms as 119906(119899) and 119906119903119906(119904) (119904 lt 119899)respectively In [22] making use of the same approach as wasdone by Ebaid proved the fact for nonlinear terms in theform 119906

120574 (120574 ⩾ 2) 119906(119904)119906119896 (119904 119896 ⩾ 1) [119906(119904)]Ω (119904 ⩾ 1 Ω ⩾ 2)and [119906(119904)]Ω119906120582 (119904 Ω 120582 ⩾ 1) respectively along with linearterm 119906

(119903)(119903 ⩾ 1) Ebaid claimed in the abstract and section

ldquoConclusionsrdquo of his article that the case 119888 = 119901 and 119889 = 119902 isthe only relation that ldquocan be obtained through applying theExp-function ansatz for all possible cases of nonlinearODEsrdquo

ldquoHowever one cannot construct a general form forthe highest order nonlinear term because there are manypossibilities other than the ones consideredrdquo [21] Aslanand Marinakis concluded that these authors just took somespecial cases of the nonlinear term into account and hence theproblem still remained open In this paper wewill construct aspecial case to show that Ebaidrsquos claim is not true namely thecase 119888 = 119901 and 119889 = 119902 is not the only relation for some specialdifferential equations and hence the problem is still open

In what follows wewill discuss the relations of 119888 119889119901 and119902 in a more effective and concise way

2 Main Result

21 Some Terminology To begin with we recall some termi-nology in [35]

A monomial in a collection of variables 1199091 119909

119899is a

product

1199091205721

11199091205722

2 119909120572119899

119899 (5)

where the 120572119894are nonnegative integers

The total degree of (5) is the sum of the exponents 1205721+

sdot sdot sdot + 120572119899

A polynomial is said to be homogeneous if all themonomials appearing in it with nonzero coefficients have thesame total degree

For instance 11990621199061015840119906101584010158403 is a product of 119906 and its derivatives1199061015840 and 119906

10158401015840 and its total degree is 6 1199062 + 11990610158402+ 11990611990610158401015840 is

homogeneous

22 Two Introducing Ansatz Function For convenience weassume V(120578) is expressed in the form

V (120578) =sum1198891

119899=minus1198881

119896119899exp (119899120578)

sum1199021

119899=minus1199011

119897119899exp (119899120578)

=

119896minus1198881

exp (minus1198881120578) + sdot sdot sdot + 119896

1198891

exp (1198891120578)

119897minus1199011


1199021

exp (1199021120578)

(6)

where 1198881 1198891 1199011 and 119902

1are positive integers to be determined

and 119896119899and 119897119899are constants to be specified

The following three formulas will be used in this section

1199061015840(120578)

=

(119901 minus 119888) 119886minus119888119887minus119901

exp (minus (119888 + 119901) 120578) + sdot sdot sdot + (119889 minus 119902) 119886119889119887119902exp ((119889 + 119902) 120578)

1198872

minus119901exp (minus2119901120578) + sdot sdot sdot + 1198872

119902exp (2119902120578)

(7)

119906 (120578) sdot V (120578)

=

119886minus119888119896minus1198881

exp (minus (119888 + 1198881) 120578) + sdot sdot sdot + 119886

1198891198961198891

exp ((119889 + 1198891) 120578)

119887minus119901119897minus1199011


1199021198971199021

exp ((119902 + 1199021) 120578)

(8)

119906 (120578)

V (120578)

=

119886minus119888119897minus1199011


1198891198971199021

exp ((119889 + 1199021) 120578)

119887minus119901119896minus1198881


1199021198961198891

exp ((119902 + 1198891) 120578)

(9)

Before presenting our definitions we recall an importantfact that the constants 119886

minus119888 119886119889 119887minus119901 and 119887

119902in the ansatz of

(4) can be assumed to be nonzero during the process ofbalancing the linear term of highest order with the highestorder nonlinear term of certain ODE and so do the constants119896minus1198881

119896119889 119897minus1199011

and 119897119902in ansatz (6) Hence in this paper we

assume that all above eight constants are nonzeroTherefore for ansatz (4) we can define the ansatz func-

tions 119871(sdot) and 119877(sdot) as follows

119871 (119906) = 119871(119886minus119888exp (minus119888120578) + sdot sdot sdot + 119886

119889exp (119889120578)

119887minus119901


)

= minus119888 minus (minus119901) = 119901 minus 119888


119889exp (119889120578)

119887minus119901


) = 119889 minus 119902

(10)

In particular we define 119871(119862) = 119877(119862) = 0 for arbitrary con-stant 119862

For example we have

119871(exp (minus120578) + exp (2120578)exp (minus2120578) + exp (2120578)

) = 1


) = 0

(11)

According to the definitions we can find that in ansatz (4) 119888 =119901 equals 119871(119906) = 0 and 119889 = 119902 equals 119877(119906) = 0 Therefore theopen problem for ansatz (4) is equal to whether the relations119871(119906) = 0 and 119871(119906) = 0 hold

Abstract and Applied Analysis 3

23 Properties of the Ansatz Functions Assuming 119888 = 119901 119889 = 1199021198881= 1199011 and 119889

1= 1199021in ansatz (4) and (6) from (7) we obtain

119871 (1199061015840) = 2119901 minus (119888 + 119901) = 119901 minus 119888 = 119871 (119906)

119877 (1199061015840) = (119889 + 119902) minus 2119902 = 119889 minus 119902 = 119877 (119906)

(12)

So we have

119871 (119906) = 119871 (1199061015840) = 119871 (119906

10158401015840) = sdot sdot sdot = 119871 (119906

(119899))

119877 (119906) = 119877 (1199061015840) = sdot sdot sdot = 119877 (119906

(119899))

(13)

for arbitrary nonnegative integer 119899And from (8) and (9) we have

119871 (119906 sdot 120592) = (119901 + 1199011) minus (119888 + 119888

1)

= (119901 minus 119888) + (1199011minus 1198881)

= 119871 (119906) + 119871 (120592)

119877 (119906 sdot 120592) = (119889 + 1198891) minus (119902 + 119902

1)

= (119889 minus 119902) + (1198891minus 1199021)

= 119877 (119906) + 119877 (120592)

119871 (119906

120592) = (119901 + 119888

1) minus (119888 + 119901

1)

= (119901 minus 119888) minus (1199011minus 1198881)

= 119871 (119906) minus 119871 (120592)

119877 (119906

120592) = (119889 + 119902

1) minus (119902 + 119889

1)


= 119877 (119906) minus 119877 (120592)

(14)

Hence we have

119871 (119906120581) = 120581 sdot 119871 (119906) 119877 (119906

120581) = 120581 sdot 119877 (119906) (15)

for any integer 120581

24 Theorem and Proof In this section we assume thebalanced nonlinear term is a product of dependent variable119906 and its derivatives namely

1199061198941(1199061015840)1198942

sdot sdot sdot (119906(119898))119894119898

(16)

where 119894119895(119895 = 1 119898) are nonnegative integersThe fact that

the product (16) is a nonlinear term implies 1198941+1198942+sdot sdot sdot+119894

119898⩾ 2

In other words the total degree of (16) is at least 2

Theorem 1 Suppose that the balanced nonlinear term in (3)is a product of 119906 and its derivatives in the form of (16) and thebalancing linear term is 119906(119904) where 119904 is a nonnegative integerthen the Exp-function ansatz (4) admits 119871(119906) = 0 and 119877(119906) =0

Proof By contradiction suppose that 119871(119906) = 0 Then we have

119871 (1199061198941(1199061015840)1198942

sdot sdot sdot (119906(119898))119894119898

)

= 119871 (1199061198941) + 119871 ((119906

1015840)1198942

) + sdot sdot sdot + 119871 ((119906(119898))119894119898

)

= 1198941sdot 119871 (119906) + 119894

2sdot 119871 (1199061015840) + sdot sdot sdot + 119894

119898sdot 119871 (119906(119898))

= 1198941sdot 119871 (119906) + 119894


119898sdot 119871 (119906)

= (1198941+ 1198942+ sdot sdot sdot + 119894

119898) sdot 119871 (119906)

119871 (119906(119904)) = 119871 (119906)

(17)

Balancing linear and nonlinear terms requires

119871 (1199061198941(1199061015840)i2

(119906(119898))119894119898

) = 119871 (119906(119904)) (18)

So we obtain(1198941+ 1198942+ sdot sdot sdot + 119894

119898minus 1) sdot 119871 (119906) = 0 (19)

Since 1198941+ 1198942+ sdot sdot sdot + 119894

119898minus 1 = 0 we arrive at the result 119871(119906) = 0

This is a contradictionThe result 119877(119906) = 0 can be obtained in a similar way here

we omit the details

Remark Since our forms of the linear and nonlinear termsare in a more general setting Theorem 1 covers the resultspresented by Ali and Ebaid respectively

25 A Simplification for the Exp-FunctionMethod Accordingto Theorem 1 if the balanced nonlinear term is a productansatz (4) can be reduced to the following equivalent form

119906 =1205720+ 1205721exp (120578) + sdot sdot sdot + 120572

120591exp (120591120578)

1 + 1205731exp (120578) + sdot sdot sdot + 120573

120591exp (120591120578)

(20)

where 120591 is a free positive integer with 120591 ⩾ 2 and 120572119894and

120573119894are constants to be specified Ansatz (20) is concise and

easy to calculate and makes the Exp-function method morestraightforward

For example Naher et al applied the Exp-functionmethod to constructing the traveling wave solutions ofthe (3+1)-dimensional modified KdV-Zakharov-Kuznetsevequation in the form [25]

119906119905+ 1205721199062119906119909+ 119906119909119909119909

+ 119906119909119910119910

+ 119906119909119911119911

= 0 (21)The traveling wave transformation

119906 (119909 119910 119911 119905) = 119906 (120578) 120578 = 119909 + 119910 + 119911 minus 119881119905 (22)carries (21) into an ODE

minus1198811199061015840+ 12057211990621199061015840+ 3119906101584010158401015840= 0 (23)

Since the nonlinear term 11990621199061015840 is a product of 119906 and 1199061015840 we can

immediately assume ansatz (4) in the form

119906 =1205720+ 1205721exp (120578) + 120572

2exp (2120578)

1 + 1205731exp (120578) + 120573

2exp (2120578)

(24)

which is equivalent to the case 119888 = 119901 = 1 and 119889 = 119902 = 1 andin accord with Naherrsquos (namely (38) in [25])


3 A Counter-Example

In [22] Ebaid claimed in his abstract that the case 119888 = 119901

and 119889 = 119902 was the only relation that could be obtained byapplying this method to any nonlinear ODE In this sectionwe construct a counter-example to show that the case 119888 = 119901

and 119889 = 119902 is not the only relation In fact the claim does notapplied to each nonlinear homogeneous ODE

For example we can create the following ODE

1199062+ 11990610158402+ 11990611990610158401015840= 0 (25)

which can be rewritten as

119906 +11990610158402

119906+ 11990610158401015840= 0 (26)

We have

119871(11990610158402

119906) = 119871 (119906) 119871 (119906

10158401015840) = 119871 (119906)

119877(11990610158402

119906) = 119877 (119906) 119877 (119906

10158401015840) = 119877 (119906)

(27)

From (27) the relations 119871(11990610158402119906) = 119871(11990610158401015840) and 119877(11990610158402119906) =

119877(11990610158401015840) hold automatically Hence by balancing nonlinear term

11990610158402119906 and linear term 119906

10158401015840 we cannot determine the relationsof 119888 119889 119901 and 119902 That is to say all of them are free constantsHence it is possible that the relation 119888 = 119901 and 119889 = 119902 holds Inother words the relation 119888 = 119901 and 119889 = 119902 is not all-inclusive

4 Conclusion

In summary we present an entire novel approach to provethat the balancing procedure in Exp-function method isunnecessary when the balanced nonlinear term is a productof the dependent variable under consideration and its deriva-tives Our results cover the results presented byAli and EbaidWe believe that our work can serve as an answer to the openproblem proposed by Aslan et al

Acknowledgments

Many thanks are due to the helpful comments and sugges-tions from the anonymous referees and support from theNSFof China (11201427) and the College Foundation of ZhijiangCollege of Zhejiang University of Technology (YJJ0819)

References

[1] J-H He and X-H Wu ldquoExp-function method for nonlinearwave equationsrdquo Chaos Solitons amp Fractals vol 30 no 3 pp700ndash708 2006

[2] S Zhang ldquoApplication of Exp-function method to a KdVequation with variable coefficientsrdquo Physics Letters A vol 365no 5-6 pp 448ndash453 2007

[3] J-H He and M A Abdou ldquoNew periodic solutions fornonlinear evolution equations using Exp-function methodrdquoChaos Solitons amp Fractals vol 34 no 5 pp 1421ndash1429 2007

[4] A Bekir and A Boz ldquoExact solutions for a class of nonlinearpartial differential equations using exp-functionmethodrdquo Inter-national Journal of Nonlinear Sciences and Numerical Simula-tion vol 8 no 4 pp 505ndash512 2007

[5] M A Abdou A A Soliman and S T El-Basyony ldquoNewapplication of Exp-function method for improved Boussinesqequationrdquo Physics Letters A vol 369 no 5-6 pp 469ndash475 2007

[6] A Ebaid ldquoExact solitary wave solutions for some nonlinearevolution equations via Exp-function methodrdquo Physics LettersA vol 365 no 3 pp 213ndash219 2007

[7] S A El-Wakil M A Madkour and M A Abdou ldquoApplicationof Exp-functionmethod for nonlinear evolution equations withvariable coefficientsrdquo Physics Letters A vol 369 no 1-2 pp 62ndash69 2007

[8] S-D Zhu ldquoExp-function method for the Hybrid-Lattice sys-temrdquo International Journal of Nonlinear Sciences and NumericalSimulation vol 8 no 3 pp 461ndash464 2007

[9] S-D Zhu ldquoExp-function method for the discrete mKdV lat-ticerdquo International Journal of Nonlinear Sciences and NumericalSimulation vol 8 no 3 pp 465ndash468 2007

[10] X-H Wu and J-H He ldquoSolitary solutions periodic solutionsand compacton-like solutions using the Exp-function methodrdquoComputers ampMathematics with Applications vol 54 no 7-8 pp966ndash986 2007

[11] X-H Wu and J-H He ldquoEXP-function method and its applica-tion to nonlinear equationsrdquo Chaos Solitons amp Fractals vol 38no 3 pp 903ndash910 2008

[12] A Bekir and A Boz ldquoExact solutions for nonlinear evolutionequations using Exp-function methodrdquo Physics Letters A vol372 no 10 pp 1619ndash1625 2008

[13] J-H He ldquoAn elementary introduction to recently developedasymptotic methods and nanomechanics in textile engineer-ingrdquo International Journal of Modern Physics B vol 22 no 21pp 3487ndash3578 2008

[14] F Xu ldquoApplication of Exp-function method to symmetricregularized long wave (SRLW) equationrdquo Physics Letters A vol372 no 3 pp 252ndash257 2008

[15] S Zhang ldquoApplication of Exp-function method to Riccatiequation and new exact solutions with three arbitrary functionsof Broer-Kaup-Kupershmidt equationsrdquo Physics Letters A vol372 no 11 pp 1873ndash1880 2008

[16] A T Ali ldquoA note on the Exp-function method and its appli-cation to nonlinear equationsrdquo Physica Scripta vol 79 no 2Article ID 025006 2009

[17] L M B Assas ldquoNew exact solutions for the Kawahara equationusing Exp-function methodrdquo Journal of Computational andApplied Mathematics vol 233 no 2 pp 97ndash102 2009

[18] A Bekir and A Boz ldquoApplication of Exp-function methodfor (2 + 1)-dimensional nonlinear evolution equationsrdquo ChaosSolitons and Fractals vol 40 no 1 pp 458ndash465 2009

[19] C-Q Dai X Cen and S-S Wu ldquoExact travelling wavesolutions of the discrete sine-Gordon equation obtained via theexp-function methodrdquo Nonlinear Analysis Theory Methods ampApplications A vol 70 no 1 pp 58ndash63 2009

[20] S Zhang and H-Q Zhang ldquoExp-function method for N-soliton solutions of nonlinear differential-difference equationsrdquoZeitschrift fur Naturforschung A vol 65 no 11 pp 924ndash9342010

[21] I Aslan and V Marinakis ldquoSome remarks on exp-functionmethod and its applicationsrdquo Communications in TheoreticalPhysics vol 56 no 3 pp 397ndash403 2011


[22] A Ebaid ldquoAn improvement on the Exp-function method whenbalancing the highest order linear and nonlinear termsrdquo Journalof Mathematical Analysis and Applications vol 392 no 1 pp 1ndash5 2012

[23] J-M Kim and C Chun ldquoNew exact solutions to the KdV-Burgers-Kuramoto equation with the Exp-function methodrdquoAbstract and Applied Analysis vol 2012 Article ID 892420 10pages 2012

[24] H Jafari N Kadkhoda and C M Khalique ldquoExact solutionsof 1206014 equation using lie symmetry approach along with thesimplest equation and exp-function methodsrdquo Abstract andApplied Analysis vol 2012 Article ID 350287 7 pages 2012

[25] H Naher F A Abdullah and M A Akbar ldquoNew travelingwave solutions of the higher dimensional nonlinear partialdifferential equation by the exp-function methodrdquo Journal ofAppliedMathematics vol 2012Article ID 575387 14 pages 2012

[26] L ZhaoDHuang and S Zhou ldquoA new algorithm for automaticcomputation of solitary wave solutions to nonlinear partialdifferential equations based on the Exp-function methodrdquoAppliedMathematics and Computation vol 219 no 4 pp 1890ndash1896 2012

[27] J-H He ldquoAsymptotic methods for solitary solutions and com-pactonsrdquo Abstract and Applied Analysis vol 2012 Article ID916793 130 pages 2012

[28] A Bekir and E Aksoy ldquoExact solutions of extended shallowwater wave equations by exp-function methodrdquo InternationalJournal of Numerical Methods for Heat amp Fluid Flow vol 23 no2 pp 305ndash319 2013

[29] A H Bhrawy A Biswas M Javidi W X Ma Z Pınarand A Yıldırım ldquoNew solutions for (1 + 1)-dimensional and(2 + 1)-dimensional Kaup-Kupershmidt equationsrdquo Results inMathematics vol 63 no 1-2 pp 675ndash686 2013

[30] K Khan andM Ali Akbar ldquoTraveling wave solutions of the (2+1)-dimensional Zoomeron equation and the Burgers equationsvia theMSEmethod and the Exp-function methodrdquo Ain ShamsEngineering Journal In press

[31] G Magalakwe and C M Khalique ldquoNew exact solutionsfor a generalized double sinh-Gordon equationrdquo Abstract andApplied Analysis vol 2013 Article ID 268902 5 pages 2013

[32] W-X Ma T Huang and Y Zhang ldquoA multiple exp-functionmethod for nonlinear differential equations and its applicationrdquoPhysica Scripta vol 82 no 6 Article ID 065003 2010

[33] W-X Ma and Z Zhu ldquoSolving the (3 + 1)-dimensionalgeneralizedKP andBKP equations by themultiple exp-functionalgorithmrdquo Applied Mathematics and Computation vol 218 no24 pp 11871ndash11879 2012

[34] J-H He ldquoExp-function Method for Fractional DifferentialEquationsrdquo International Journal of Nonlinear Sciences andNumerical Simulation vol 14 no 6 pp 363ndash366 2013

[35] D A Cox J Little and D OrsquoShea Using Algebraic Geometryvol 185 Springer New York NY USA 2nd edition 2005

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Stochastic AnalysisInternational Journal of


12 An Open Problem Among the Exp-functionmethod thebalancing computation is laborious but prior However weobserve that the balancing procedure of the Exp-functionmethod in studied examples always leads to the same case119888 = 119901 and 119889 = 119902 [21] This fact has partly been proved in[16 22] In [16] Ali has proved it by assuming the highestorder linear and nonlinear terms as 119906(119899) and 119906119903119906(119904) (119904 lt 119899)respectively In [22] making use of the same approach as wasdone by Ebaid proved the fact for nonlinear terms in theform 119906

120574 (120574 ⩾ 2) 119906(119904)119906119896 (119904 119896 ⩾ 1) [119906(119904)]Ω (119904 ⩾ 1 Ω ⩾ 2)and [119906(119904)]Ω119906120582 (119904 Ω 120582 ⩾ 1) respectively along with linearterm 119906

(119903)(119903 ⩾ 1) Ebaid claimed in the abstract and section

ldquoConclusionsrdquo of his article that the case 119888 = 119901 and 119889 = 119902 isthe only relation that ldquocan be obtained through applying theExp-function ansatz for all possible cases of nonlinearODEsrdquo

ldquoHowever one cannot construct a general form forthe highest order nonlinear term because there are manypossibilities other than the ones consideredrdquo [21] Aslanand Marinakis concluded that these authors just took somespecial cases of the nonlinear term into account and hence theproblem still remained open In this paper wewill construct aspecial case to show that Ebaidrsquos claim is not true namely thecase 119888 = 119901 and 119889 = 119902 is not the only relation for some specialdifferential equations and hence the problem is still open

In what follows wewill discuss the relations of 119888 119889119901 and119902 in a more effective and concise way

2 Main Result

21 Some Terminology To begin with we recall some termi-nology in [35]

A monomial in a collection of variables 1199091 119909

119899is a

product

1199091205721

11199091205722

2 119909120572119899

119899 (5)

where the 120572119894are nonnegative integers

The total degree of (5) is the sum of the exponents 1205721+

sdot sdot sdot + 120572119899

A polynomial is said to be homogeneous if all themonomials appearing in it with nonzero coefficients have thesame total degree

For instance 11990621199061015840119906101584010158403 is a product of 119906 and its derivatives1199061015840 and 119906

10158401015840 and its total degree is 6 1199062 + 11990610158402+ 11990611990610158401015840 is

homogeneous

22 Two Introducing Ansatz Function For convenience weassume V(120578) is expressed in the form

V (120578) =sum1198891

119899=minus1198881

119896119899exp (119899120578)

sum1199021

119899=minus1199011

119897119899exp (119899120578)

=

119896minus1198881


1198891

exp (1198891120578)

119897minus1199011


1199021

exp (1199021120578)

(6)

where 1198881 1198891 1199011 and 119902

1are positive integers to be determined

and 119896119899and 119897119899are constants to be specified

The following three formulas will be used in this section

1199061015840(120578)

=

(119901 minus 119888) 119886minus119888119887minus119901

exp (minus (119888 + 119901) 120578) + sdot sdot sdot + (119889 minus 119902) 119886119889119887119902exp ((119889 + 119902) 120578)

1198872

minus119901exp (minus2119901120578) + sdot sdot sdot + 1198872

119902exp (2119902120578)

(7)

119906 (120578) sdot V (120578)

=

119886minus119888119896minus1198881


1198891198961198891

exp ((119889 + 1198891) 120578)

119887minus119901119897minus1199011


1199021198971199021

exp ((119902 + 1199021) 120578)

(8)

119906 (120578)

V (120578)

=

119886minus119888119897minus1199011


1198891198971199021

exp ((119889 + 1199021) 120578)

119887minus119901119896minus1198881


1199021198961198891

exp ((119902 + 1198891) 120578)

(9)

Before presenting our definitions we recall an importantfact that the constants 119886

minus119888 119886119889 119887minus119901 and 119887

119902in the ansatz of

(4) can be assumed to be nonzero during the process ofbalancing the linear term of highest order with the highestorder nonlinear term of certain ODE and so do the constants119896minus1198881

119896119889 119897minus1199011

and 119897119902in ansatz (6) Hence in this paper we

assume that all above eight constants are nonzeroTherefore for ansatz (4) we can define the ansatz func-

tions 119871(sdot) and 119877(sdot) as follows


119889exp (119889120578)

119887minus119901


)

= minus119888 minus (minus119901) = 119901 minus 119888


119889exp (119889120578)

119887minus119901


) = 119889 minus 119902

(10)

In particular we define 119871(119862) = 119877(119862) = 0 for arbitrary con-stant 119862

For example we have


) = 1


) = 0

(11)

According to the definitions we can find that in ansatz (4) 119888 =119901 equals 119871(119906) = 0 and 119889 = 119902 equals 119877(119906) = 0 Therefore theopen problem for ansatz (4) is equal to whether the relations119871(119906) = 0 and 119871(119906) = 0 hold




119871 (1199061015840) = 2119901 minus (119888 + 119901) = 119901 minus 119888 = 119871 (119906)

119877 (1199061015840) = (119889 + 119902) minus 2119902 = 119889 minus 119902 = 119877 (119906)

(12)

So we have

119871 (119906) = 119871 (1199061015840) = 119871 (119906

10158401015840) = sdot sdot sdot = 119871 (119906

(119899))

119877 (119906) = 119877 (1199061015840) = sdot sdot sdot = 119877 (119906

(119899))

(13)


119871 (119906 sdot 120592) = (119901 + 1199011) minus (119888 + 119888

1)

= (119901 minus 119888) + (1199011minus 1198881)

= 119871 (119906) + 119871 (120592)

119877 (119906 sdot 120592) = (119889 + 1198891) minus (119902 + 119902

1)

= (119889 minus 119902) + (1198891minus 1199021)

= 119877 (119906) + 119877 (120592)

119871 (119906

120592) = (119901 + 119888

1) minus (119888 + 119901

1)


= 119871 (119906) minus 119871 (120592)

119877 (119906

120592) = (119889 + 119902

1) minus (119902 + 119889

1)


= 119877 (119906) minus 119877 (120592)

(14)

Hence we have

119871 (119906120581) = 120581 sdot 119871 (119906) 119877 (119906

120581) = 120581 sdot 119877 (119906) (15)



1199061198941(1199061015840)1198942

sdot sdot sdot (119906(119898))119894119898

(16)



119898⩾ 2




119871 (1199061198941(1199061015840)1198942

sdot sdot sdot (119906(119898))119894119898

)

= 119871 (1199061198941) + 119871 ((119906

1015840)1198942

) + sdot sdot sdot + 119871 ((119906(119898))119894119898

)

= 1198941sdot 119871 (119906) + 119894


119898sdot 119871 (119906(119898))

= 1198941sdot 119871 (119906) + 119894


119898sdot 119871 (119906)

= (1198941+ 1198942+ sdot sdot sdot + 119894

119898) sdot 119871 (119906)

119871 (119906(119904)) = 119871 (119906)

(17)


119871 (1199061198941(1199061015840)i2

(119906(119898))119894119898

) = 119871 (119906(119904)) (18)


119898minus 1) sdot 119871 (119906) = 0 (19)




we omit the details




120591exp (120591120578)


120591exp (120591120578)

(20)





119906119905+ 1205721199062119906119909+ 119906119909119909119909

+ 119906119909119910119910

+ 119906119909119911119911



minus1198811199061015840+ 12057211990621199061015840+ 3119906101584010158401015840= 0 (23)



119906 =1205720+ 1205721exp (120578) + 120572

2exp (2120578)

1 + 1205731exp (120578) + 120573

2exp (2120578)

(24)



3 A Counter-Example





1199062+ 11990610158402+ 11990611990610158401015840= 0 (25)


119906 +11990610158402

119906+ 11990610158401015840= 0 (26)

We have

119871(11990610158402

119906) = 119871 (119906) 119871 (119906

10158401015840) = 119871 (119906)

119877(11990610158402

119906) = 119877 (119906) 119877 (119906

10158401015840) = 119877 (119906)

(27)



11990610158402119906 and linear term 119906


4 Conclusion


Acknowledgments


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119871 (1199061015840) = 2119901 minus (119888 + 119901) = 119901 minus 119888 = 119871 (119906)

119877 (1199061015840) = (119889 + 119902) minus 2119902 = 119889 minus 119902 = 119877 (119906)

(12)

So we have

119871 (119906) = 119871 (1199061015840) = 119871 (119906

10158401015840) = sdot sdot sdot = 119871 (119906

(119899))

119877 (119906) = 119877 (1199061015840) = sdot sdot sdot = 119877 (119906

(119899))

(13)


119871 (119906 sdot 120592) = (119901 + 1199011) minus (119888 + 119888

1)

= (119901 minus 119888) + (1199011minus 1198881)

= 119871 (119906) + 119871 (120592)

119877 (119906 sdot 120592) = (119889 + 1198891) minus (119902 + 119902

1)

= (119889 minus 119902) + (1198891minus 1199021)

= 119877 (119906) + 119877 (120592)

119871 (119906

120592) = (119901 + 119888

1) minus (119888 + 119901

1)


= 119871 (119906) minus 119871 (120592)

119877 (119906

120592) = (119889 + 119902

1) minus (119902 + 119889

1)


= 119877 (119906) minus 119877 (120592)

(14)

Hence we have

119871 (119906120581) = 120581 sdot 119871 (119906) 119877 (119906

120581) = 120581 sdot 119877 (119906) (15)



1199061198941(1199061015840)1198942

sdot sdot sdot (119906(119898))119894119898

(16)



119898⩾ 2




119871 (1199061198941(1199061015840)1198942

sdot sdot sdot (119906(119898))119894119898

)

= 119871 (1199061198941) + 119871 ((119906

1015840)1198942

) + sdot sdot sdot + 119871 ((119906(119898))119894119898

)

= 1198941sdot 119871 (119906) + 119894


119898sdot 119871 (119906(119898))

= 1198941sdot 119871 (119906) + 119894


119898sdot 119871 (119906)

= (1198941+ 1198942+ sdot sdot sdot + 119894

119898) sdot 119871 (119906)

119871 (119906(119904)) = 119871 (119906)

(17)


119871 (1199061198941(1199061015840)i2

(119906(119898))119894119898

) = 119871 (119906(119904)) (18)


119898minus 1) sdot 119871 (119906) = 0 (19)




we omit the details




120591exp (120591120578)


120591exp (120591120578)

(20)





119906119905+ 1205721199062119906119909+ 119906119909119909119909

+ 119906119909119910119910

+ 119906119909119911119911



minus1198811199061015840+ 12057211990621199061015840+ 3119906101584010158401015840= 0 (23)



119906 =1205720+ 1205721exp (120578) + 120572

2exp (2120578)

1 + 1205731exp (120578) + 120573

2exp (2120578)

(24)



3 A Counter-Example





1199062+ 11990610158402+ 11990611990610158401015840= 0 (25)


119906 +11990610158402

119906+ 11990610158401015840= 0 (26)

We have

119871(11990610158402

119906) = 119871 (119906) 119871 (119906

10158401015840) = 119871 (119906)

119877(11990610158402

119906) = 119877 (119906) 119877 (119906

10158401015840) = 119877 (119906)

(27)



11990610158402119906 and linear term 119906


4 Conclusion


Acknowledgments


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










3 A Counter-Example





1199062+ 11990610158402+ 11990611990610158401015840= 0 (25)


119906 +11990610158402

119906+ 11990610158401015840= 0 (26)

We have

119871(11990610158402

119906) = 119871 (119906) 119871 (119906

10158401015840) = 119871 (119906)

119877(11990610158402

119906) = 119877 (119906) 119877 (119906

10158401015840) = 119877 (119906)

(27)



11990610158402119906 and linear term 119906


4 Conclusion


Acknowledgments


References












































Volume 2014




Journal of











Journal of


Function Spaces






Algebra































Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article a simplification for exp-function method when … · 2019. 7. 31. · abstract and...

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