applications of separation variables approach in solving...

Research ArticleApplications of Separation Variables Approach in SolvingTime-Fractional PDEs

Yinghui He and Yunmei Zhao

Department of Mathematics Honghe University Mengzi Yunnan 661199 China

Correspondence should be addressed to Yinghui He heyinghui07163com

Received 16 December 2017 Accepted 15 March 2018 Published 22 April 2018

Academic Editor Chaudry M Khalique

Copyright copy 2018 YinghuiHe andYunmei ZhaoThis is an open access article distributed under theCreative CommonsAttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited

Based on the homogenous balanced principle and subequation method an improved separation variables function-expansionmethod is proposed to seek exact solutions of time-fractional nonlinear PDEsThis method is novel and meaningful without usingLeibniz rule and chain rule of fractional derivative which have been proved to be incorrect By using this method we studied anonlinear time-fractional PDE with diffusion term Some general solutions are obtained which contain many arbitrary parametersSolutions given in related reference are just our especial case And we also obtained some new type of solutions

1 Introduction

It is well known that fractional-order models are moreadequate than previously used integer-ordermodels due to anexact description of nonlinear phenomena Therefore non-linear fractional partial differential equations (nfPDEs) haveattracted more and more attention Most recently FPDEs areincreasingly used inmathematicalmodeling of fluidmechan-ics biological and chemical processes signal processingand control systems and they are also used in fractal anddifferential geometry and so on(see [1ndash8] and their refer-ences cited) Many natural phenomena associated with real-time problems which depend on both time instant and theprevious time history especially can be successfully modeledby time-fractional nonlinear partial differential equations Ofcourse investigating solutions of nfPDEs plays an importantrole in a large number of research fields Until now there isno general method Nevertheless many powerful methodswere used for solving fractional differential equations suchas adomian decomposition method [9] symmetry method[10ndash12] numerical method [13 14] first integral method [15]the homotopy perturbation method [16] invariant subspacemethod [17] and fractional variational iterationmethod [18]and have achieved significant progress

There are many definitions of fractional derivative Thefractional differential equations can be expressed in terms ofdifferent differential operators defined by Riemann-Liouville

Caputo Weyl and many others Among these definitionsCaputo definition is most frequently used Here let us brieflyreview this definition of fractional derivative

The Caputo fractional derivative of order 120572 gt 0 is definedby the following expression

119862120572119863120572119905 119891 (119905) = 1Γ (119899 minus 120572) int119905

119886(119905 minus 120591)119899minus120572minus1 119891(119899) (120591) 119889120591

= 119891(119899) (119886) (119905 minus 119886)119899minus120572Γ (119899 minus 120572 + 1)+ 1Γ (119899 minus 120572 + 1) int119905

119886(119905 minus 120591)119899minus120572 119891(119899+1) (120591) 119889120591

(1)

where 119899 = [120572] + 1 119899 minus 1 lt 120572 ⩽ 119899 119905 gt 119886 ⩾ 0 119899 isin 119873+ In thispaper we will adopt Caputo fractional derivative definition toinvestigate exact solution of the following type of nonlineartime-fractional partial differential equation

119862119863120572119905 119906 = 119865(119906 120597119906120597119909 12059721199061205971199092 120597119899119906120597119909119899) (2)

where 119862119863120572119905 = 1198620119863120572119905 is Caputo differential operator 119906 =119906(119909 119905) and 119905 gt 0 0 lt 120572 ⩽ 1 119899 isin 119873+ 119909 isin 119877Equation (2) denotes a series of nonlinear time-fractionalPDEs these models defined by (2) can be used to accurately

HindawiMathematical Problems in EngineeringVolume 2018 Article ID 3892691 10 pageshttpsdoiorg10115520183892691

2 Mathematical Problems in Engineering

describe nonlinear phenomena in connection with real-timeproblems which not only depend on time instant but alsodepend on the previous time history

Recently Feng [19] introduced a fractional 119863120572119866119866method for seeking traveling wave solutions of space-time-fractional partial differential equations under the followingmodified Riemann-Liouville derivative definition [20]

119863120572119905 119891 (119905)

=

1Γ (1 minus 120572) int1199050(119905 minus 120591)minus120572minus1 [119891 (120591) minus 119891 (0)] 119889120591 120572 lt 0

1Γ (1 minus 120572) 119889119889119905 int1199050(119905 minus 120591)minus120572 [119891 (120591) minus 119891 (0)] 119889120591 0 lt 120572 lt 1

119891(119899) (119905)(120572minus119899) 119899 ⩽ 120572 lt 119899 + 1

(3)

The following properties for the modified Riemann-Liouville derivative are usually used

119863120572119909119909120574 = Γ (1 + 120574)Γ (1 + 120574 minus 120572)119909120574minus120572 (4)

119863120572119909119891 [119906 (119909)] = 1198911015840 (119906)119863120572119909119906 (119909) = 119863120572119909119891 (119906) (1199061015840 (119909))120572 (5)

Let us briefly review this method Using traveling wavetransformation 120585 = 119896119909 + 119888119905 119906(119909 119905) = 119880(120585) and (4) and(5) they reduced the following fractional partial differentialequation

119875 (119906119863120572119905 1199061198631205721199091199061198632120572119905 1199061198632120572119909 119906 ) = 0 (6)

which can be converted into the following fractional ordinarydifferential equation with respect to the variable 120585

119875 (119880 119888120572119863120572120585119880 119896120572119863120572120585119880 11988821205721198632120572120585 119880 11989621205721198632120572120585 119880 ) = 0 (7)

Then they suppose that the solution of (7) can beexpressed by a polynomial in 119863120572119866119866 as follows

119880 (120585) = 119898sum119894=0

119886119894 (119863120572120585119866119866 ) (8)

where 119866 = 119866(120585) satisfies the following fractional ordinarydifferential equation

119860119866 (120585)1198632120572120585 119866 (120585) minus 119861119866 (120585)119863120572120585119866 (120585) minus 119862 (119863120572120585119866 (120585))2minus 1198641198662 (120585) = 0 (9)

In order to find exact solutions of (9) a nonlinearfractional complex transformation 120578 = 120585120572Γ(1 + 120572) is used

Then (9) can be turned into the following second ordinarydifferential equation

119860119867(120578)11986710158401015840 (120578) minus 119861119867 (120578)1198671015840 (120578) minus 119862 (1198671015840 (120578))2minus 1198641198672 (120578) = 0 (10)

The exact solutions of (10) are knownFinally substituting (8) into (7) equating each coefficient

of this polynomial on119863120572120585119866119866 to zero they can obtain a largenumber of exact solutions of space-time fractional partialdifferential equations (6)

We noticed two problems as follows At first in a shortcommunication [21] Tarasov proved that formula (4) and thechain rule (5) cannot be performed together for fractionalderivatives of noninteger orders 120572 = 1 Therefore usingtraveling wave transformation 120585 = 119896119909 + 119888119905 119906(119909 119905) = 119880(120585)space-time fractional partial differential equations (6) cannotbe converted into fractional ordinary differential equation(7) And (9) cannot convert to (10) under transformation120578 = 120585120572Γ(1 + 120572)

In addition it should be noted that the Leibniz rule in theform

119863120572120585 (119891 (120585) 119892 (120585)) = 119892 (120585)119863120572120585119891 (120585) + 119891 (120585)119863120572120585119892 (120585) (11)

cannot hold for fractional derivatives of order 120572 = 1 for setsof differentiable and nondifferentiable functions [22] It canbe proved easily by using a counterexample Let

119891 (119909) = 1199091205741 119892 (119909) = 1199091205742 (12)

where 1205741 1205742 isin 119873 Obviously we have

119863120572119909119891 (119909) = Γ (1 + 1205741)Γ (1 + 1205741 minus 120572)1199091205741minus120572119863120572119909119892 (119909) = Γ (1 + 1205742)Γ (1 + 1205742 minus 120572)1199091205742minus120572

(13)

Then one has

119892 (119909)119863120572119909119891 (119909) = Γ (1 + 1205741)Γ (1 + 1205741 minus 120572)1199091205741+1205742minus120572119891 (119909)119863120572119909119892 (119909) = Γ (1 + 1205742)Γ (1 + 1205742 minus 120572)1199091205741+1205742minus120572

119863120572119909 [119891 (119909) 119892 (119909)] = Γ (1 + 1205741 + 1205742)Γ (1 + 1205741 + 1205742 minus 120572)1199091205741+1205742minus120572(14)

As a result we have the conditionΓ (1 + 1205741 + 1205742)Γ (1 + 1205741 + 1205742 minus 120572) = Γ (1 + 1205741)Γ (1 + 1205741 minus 120572)

+ Γ (1 + 1205742)Γ (1 + 1205742 minus 120572) (15)

which must be performed if the Leibniz rule (11) holds Forexample if we make 1205741 1205742 isin 119873+ and take into account Γ(119899 +1) = 119899Γ(119899) then condition (15) can be represented in the form

Mathematical Problems in Engineering 3

(1205741 + 119896) (21205741 minus 120572 + 119896) (21205741 minus 120572 + 119896 minus 1) sdot sdot sdot (1205741 minus 120572 + 119896 minus 1) + 1205741 (21205741 minus 120572 + 119896) (21205741 minus 120572 + 119896 minus 1) sdot sdot sdot (1205741 minus 120572 + 1) = (21205741 + 119896) (16)

where 119896 = 1205742 minus 1205741 It is not difficult to find that (16) holdsonly if 120572 = 1 That is to say the Leibniz rule (11) does nothold with 0 lt 120572 lt 1 Therefore the following formula usedin references is incorrect

119863120572120585119880 (120585) = 119863120572120585 (119863120572120585119866 (120585)119866 (120585) )

= 119866 (120585)1198632120572120585 119866 (120585) minus (119863120572120585119866 (120585))21198662 (120585)

(17)

Therefore when substituting (8) into (7) the left-handside of (7) cannot be expressed by the polynomial in119863120572120585119866119866

So this fractional 119863120572119866119866 method is not reliable and theobtained results are incorrect By the way we should be morecautious when using this kind of methods based on chainrule (5) and Leibniz rule (11) like the subequation methodthe 1198661015840119866-expansion method the exp-function method thefunctional variable method the trial equation method thesimple equation method and so forth We cannot obtainthe exact solutions of compound function type of time-fractional PDEs (2) as in the references Encouraged by Ruirsquoswork [23] we shall introduce an improved method based onthe homogenous balanced principle by using this improvedmethod we shall investigate exact solutions of a series ofnonlinear time-fractional PDEs formed as (2)

The rest of this paper is organized as follows In Section 2wewill introduce the improved separation variable expansionmethod based on the homogenous balanced principle InSection 3 by using this newmethod we will investigate exactsolutions of a nonlinear time-fractional PDE with diffusionterm discussed in (2)

2 Introduction of Improved SeparationVariable Function-Expansion Method

Although the fractional chain rule (5) and Leibniz rule (11)do not hold they do not affect the investigation of the exactsolutions of nonlinear time-fractional PDE (2) since formulas(4) still hold

Remark 1 In reference [23] Rui points out that Leibniz rule(11) still holds which is incorrect Actually in our method itdoes not matter that Leibniz rule (11) does not hold Becausewe just need the following formula

119863120572119905 [119896119891 (119909)] = 119896119863120572119905 [119891 (119909)] (18)

which is easy to be proved by definition for Caputo fractionalderivative (1) and for themodified Riemann-Liouville deriva-tive (3)

In the following we introduce main steps of improvedfunction-expansion method of separation variable type asfollows

Step 1 According to the formulas (4) and (18) and charactersof nonlinear time-fractional partial differential equation (2)we suppose that (2) has the following exact solutions of theseparation variable type

119906 = 119898sum119896=0

119886119896 [120593 (119905)] [V (119909)]119896 (19)

where the function 120593(119905) can be taken as power function119905120574119896 Mittag-Leffler function 119864120572(120582119905120572) or 119905120573minus1119864120572120573(120582119905120572) 119898 is apositive integer and 119886119896 120574119896 (119896 = 0 1 119898) are constants tobe determined later The function V(119909) satisfies the followingsubequation

V1015840 (119909)2 = 119903 + 119901V (119909) + 119902V2 (119909) (20)

where 119903 119901 119902 are constants Some solutions of (20) are listedas follows

When 119901 = 119902 = 0 119903 = 0V (119909) = plusmnradic119903119909 + 119862 (21)

where 119862 is an arbitrary integral constantWhen 119902 = 0 119901 = 0

V (119909) = 11986221199012 minus 21198621199012119909 + 11990921199012 minus 41199034119901 (22)

When 119902 = 0V (119909) = 4119901radic119902 minus (1199012 minus 4119902119903) 119890radic119902(119862minus119909) minus 4119902119890minusradic119902(119862minus119909)

minus811990232

V (119909) = 4119901radic119902 minus (1199012 minus 4119902119903) 119890minusradic119902(119862minus119909) minus 4119902119890radic119902(119862minus119909)minus811990232

(23)

Taking integral constant 119862 = 0 under some conditionswe can obtain many special solutions of (20) which are listedin Table 1

Step 2 In order to determine the value119898 of the function V(119909)we balance power of V(119909) between the term of the highestorder in the right-hand side of (2) and the highest order termin the left-hand side of (2) By the way the highest order termin the left-hand side of (2) is still V119898(119909) because we only makederivation for 119905 in the left-hand side of (2) Once the value119898 has been determined the expansion expression (19) can befixed correspondingly

Step 3 By using specific expansion expression obtained inthe Step 2 we substitute it into (2) (make fractional-orderderivations for 119905 and make integer-order derivations for 119909)then we balance the power of 120593(119905) (such as 119905 119864120572(120582119905120572)) thusthe values of 120582 120574119896 (119896 = 0 1 119898) can be obtained


Table 1 Solutions of (20)

Case Conditions Solution of (20)1 119901 = 119902 = 0 119903 = 1 V1 (119909) = 1199092 119903 = 119902 = 0 119901 = 4 V2 (119909) = 11990923 119903 = 119901 = 0 119902 = 1 V3 (119909) = 1198901199094 119903 = 1 119901 = 0 119902 = minus1 V4 (119909) = plusmnsin (119909)5 119903 = 1 119901 = 2 119902 = 0 V5 (119909) = 11990922 minus 1199096 119903 = minus1 119901 = 0 119902 = 1 V6 (119909) = cosh (119909)7 119903 = 1 119901 = 0 119902 = minus1 V7 (119909) = plusmncos (119909)8 119903 = 1 119901 = 0 119902 = 1 V8 (119909) = plusmnsinh (119909)

Step 4 In two sides of equation obtained by Step 2 we letcoefficients of the same order for every term V(119909) in two sidesof the equation be equal thus we can determine values of allthe coefficients 119886119896 (119896 = 0 1 119898)Step 5 Finally substituting the values of all the parameters119898 120582 119886119896 120574119896 (119896 = 0 1 119898) obtained in above steps into (19)wewill obtain exact solutions of the nonlinear time-fractionalPDE (2)

Remark 2 In [23] V(119909) is taken as one of 119909 sin (119909) cos (119909)and 119890119909 which is just a special solution of our (20) So ourmethod is more general than Ruirsquos method [23] In additionour method can test a series of functions at one time whichsatisfy (20) The improved method is more efficient andsimple

3 Exact Solutions ofa Nonlinear Time-Fractional PDEwith Diffusion Term

In this subsection we will study a nonlinear time-fractionalPDE with diffusion term as follows

119888119863120572119905 119906 = (120597119906120597119909)2 [120597119891 (119906)120597119906 ] + 119891 (119906) 12059721199061205971199092 + 1205751199062119905 ⩾ 0 119909 isin 119877 0 lt 120572 ⩽ 1

(24)

where 119906 = 119906(119905 119909) the 119891(119906) is diffusion term When 120575 = 0and 120572 = 1 (24) can be rewritten as the following nonlineardiffusion PDE

120597120597119909 [119891 (119906) 120597119906120597119909] minus 120597119906120597119905 = 0 (25)

which have appeared in problems related to plasma and solidstate physics see [24] and references cited therein In [23] Ruistudied the situation of119891(119906) = 120573119906 Equation (25) becomes thefollowing time-fractional PDE with diffusion term

119888119863120572119905 119906 = 120573(120597119906120597119909)2 + 12057311990612059721199061205971199092 + 1205751199062119905 ⩾ 0 119909 isin 119877 0 lt 120572 ⩽ 1

(26)

And exact solutions of (26) with parameters 120575 = 2120573 and 120575 =minus2120573 are given Here we will investigate more exact solutionsby improved method introduced in Section 2

When 0 lt 120572 lt 1 taking 120595(119905) = 119905120574119896 we suppose that(26) has an exact solution formed as 119906(119909 119905) = sum119898119896=0 119886119896119905120574119896V(119909)We find that the highest order of V(119909) in the term 119888119863120572119905 119906is just 119898 In the right-hand side of (26) the highest orderof V(119909) in the nonlinear terms (120597119906120597119909)2 and 119906(12059721199061205971199092) is2119898 + 2 Considering the relation between parameters 120573 and120575 we can let 119898 be an arbitrary positive integer to test (someterms may be counteracted under certain conditions) Herefor simplicity we just discuss the situations of 119898 = 1 and119898 = 231 Situation of 119898 = 1 Taking 119898 = 1 we first suppose that(26) has an exact solution as the following form

119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) (27)

where V(119909) satisfies (20) and 1205740 1205741 1198860 1198861 = 0 are undeter-mined constants that can be determined later Substituting(27) into (26) using (20) it can be reduced to

1198860Γ (1205740 + 1) 119905minus120572+1205740Γ (minus120572 + 1205740 + 1) + 1198861Γ (1205741 + 1) 119905minus120572+1205741

Γ (minus120572 + 1205741 + 1) V (119909)= 1199052120574112057311990311988612 + 121199051205741+120574012057311990111988601198861+ 1199052120574012057511988602 (321199052120574112057311990111988612 + 1199051205741+120574012057311990211988601198861+ 21199051205741+120574012057511988601198861) V (119909) + (21199052120574112057311990211988612 + 1199052120574112057511988612)sdot V (119909)2

(28)

According to homogenous balanced principle we let

21199052120574112057311990211988612 + 1199052120574112057511988612 = 0 (29)

Solving (29) yields

120575 = minus2119902120573 (30)

Substituting (30) into (28) it can be reduced to


Γ (minus120572 + 1205741 + 1) V (119909)= minus12120573 (minus31199011198861211990521205741 + 611988611199051205741+12057401199021198860) V (119909)

minus 12120573 (41199021198860211990521205740 minus 21199031198861211990521205741 minus 119901119886011988611199051205741+1205740) (31)

311 119902 = 0 119901 = 0 119903 = 0 Equation (31) can be simplified to


Γ (minus120572 + 1205741 + 1) V (119909)= minus21198862011990212057311990521205740 minus 3119886011988611199021205731199051205740+1205741V (119909)

(32)



1205740 minus 120572 = 212057401205741 minus 120572 = 1205740 + 1205741 (33)

Solving (33) yields

1205740 = minus1205721205741 = 1205741 (34)


1198860Γ (1 minus 120572) 119905minus2120572Γ (1 minus 2120572) + 1198861Γ (1 + 1205741) 1199051205741minus120572Γ (1 + 1205741 minus 120572) V (119909)= minus211988620119902120573119905minus2120572 minus 3119886011988611199021205731199051205741minus120572V (119909)

(35)

Balancing the power of V(119909) one hasminus311988601198861119902120573 = 1198861Γ (1 + 1205741)Γ (1 + 1205741 minus 120572)

minus211988620119902120573 = 1198860Γ (1 minus 120572)Γ (1 minus 2120572) (36)

Solving (36) we obtain the following result

1198860 = minus Γ (1 minus 120572)2120573119902Γ (1 minus 2120572) 1198861 = 1198861

Γ (1 minus 120572)2Γ (1 minus 2120572) = Γ (1 + 1205741)3Γ (1 + 1205741 minus 120572) (37)

Substituting (37) into (27) using solution (23) of subequationwe can obtain exact solution of (26) with 120575 = minus2119902120573 as follows

119906 (119909 119905) = minus Γ (1 minus 120572)2120573119902Γ (1 minus 2120572) 119905minus120572 + 11988611199051205741119890plusmnradic119902(119862+119909) (38)

where 120572 isin (0 12) cup (12 1) and 1205741 is the root of Γ(1 minus120572)2Γ(1 minus 2120572) = Γ(1 + 1205741)3Γ(1 + 1205741 minus 120572)312 119901 = 0 119902 = 0 119903 = 0 Equation (31) can be simplifiedto


Γ (minus120572 + 1205741 + 1) V (119909)= 120573 (minus21199021198860211990521205740 + 1199031198861211990521205741) minus 3120573119886111990211988601199051205741+1205740V (119909)

(39)


1205740 minus 120572 = 1205741 + 12057401205741 minus 120572 = 21205740 = 21205741 (40)

Solving (40) yields

1205740 = minus1205721205741 = minus120572 (41)


1198860Γ (1 minus 120572)Γ (1 minus 2120572) + 1198861Γ (1 minus 120572)Γ (1 minus 2120572) V (119909)= 120573 (minus211990211988602 + 11990311988612) minus 312057311988611199021198860V (119909) (42)

Balancing the power of V(119909) one hasminus312057311988611199021198860 = 1198861Γ (1 minus 120572)Γ (1 minus 2120572)

120573 (minus211990211988602 + 11990311988612) = 1198860Γ (1 minus 120572)Γ (1 minus 2120572) (43)

Solving (43) we obtain the following results

1198860 = minus Γ (1 minus 120572)3Γ (minus2120572 + 1) 120573119902 1198861 = plusmn Γ (1 minus 120572)3Γ (minus2120572 + 1) 120573radicminus 1119902119903

(44)

where 120572 isin (0 12) cup (12 1) 120573 = 0 Substituting (44) into(27) using solution (23) of subequation we can obtain exactsolution of (26) with 120575 = minus2119902120573 as follows

119906 (119909 119905)= minus Γ (1 minus 120572)3Γ (minus2120572 + 1) 120573119902119905minus120572

plusmn Γ (1 minus 120572) (1198902119909radic119902 minus 1199031198902119862radic119902) 119890minusradic119902(119862+119909)6119902120573radicminus119903Γ (1 minus 2120572) 119905minus120572


plusmn Γ (1 minus 120572) (minus1199031198902119909radic119902 + 1198902119862radic119902) 119890minusradic119902(119862+119909)6119902120573radicminus119903Γ (1 minus 2120572) 119905minus120572

(45)

By using Table 1 taking parameters 119901 119902 119903 as someparticular values many specific exact solutions of (26) can begot parts of which are listed as follows

When 119901 = 0 119902 = minus120596 lt 0 119903 = 1 then 120575 = 2120596120573 V(119909) =plusmn(sin(radic120596(119862 minus 119909)))radic120596 Equation (45) can be reduced to

119906 (119909 119905)= Γ (1 minus 120572)3120573120596Γ (1 minus 2120572) 119905minus120572 [plusmn sin (radic120596 (119862 minus 119909)) + 1] (46)

where 120596 is positive and 119862 is an arbitrary constantWhen 119901 = 0 119902 = minus120596 lt 0 119903 = minus1 then 120575 = 2120596120573 V(119909) =(minus119868 cos(radic120596(119862 minus 119909)))radic120596 Equation (45) can be reduced to

119906 (119909 119905)= Γ (1 minus 120572)3120573120596Γ (1 minus 2120572) 119905minus120572 [plusmn cos (radic120596 (119862 minus 119909)) + 1] (47)

where 1198682 = minus1 120596 is positive and 119862 is an arbitrary constant


If one lets 119862 = 0 or 1205872 120596 = 1 then 120575 = 2120573 V(119909) =plusmnsin (0 minus 119909) = plusmnsin (119909) or V(119909) = plusmnsin (1205872 minus 119909) = plusmncos (119909)Therefore solutions (46) and (47) are reduced to

119906 (119909 119905) = Γ (1 minus 120572)3120573Γ (1 minus 2120572) 119905minus120572 (1 plusmn sin (119909)) 119906 (119909 119905) = Γ (1 minus 120572)3120573Γ (1 minus 2120572) 119905minus120572 (1 plusmn cos (119909))

(48)

Remark 3 Equations (48) are just the solutions (337) and(339) given in reference [23] so we can say that our solutionsare general including many unreported solutions

When 119901 = 0 119902 = 120596 gt 0 119903 = 1 then 120575 = minus2120596120573 V(119909) =plusmn(sinh (radic120596(119862 minus 119909)))radic120596 Equation (45) can be reduced to

1199061 (119909 119905)= Γ (1 minus 120572)3120573120596Γ (1 minus 2120572) 119905minus120572 [plusmn sinh (radic120596 (119862 minus 119909)) + 1] (49)

where 120596 is positive and 119862 is an arbitrary constantWhen 119901 = 0 119902 = 120596 gt 0 119903 = minus1 then 120575 = minus2120596120573 V(119909) =plusmn(cosh (radic120596(119862 minus 119909)))radic120596 Equation (45) can be reduced to

119906 (119909 119905)= 119868Γ (1 minus 120572)3120573120596Γ (1 minus 2120572) 119905minus120572 [plusmn cosh (radic120596 (119862 minus 119909)) + 1] (50)


313 119901 = 0 119902 = 0 Equation (31) can be simplified to


Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 (119886011988611199011199051205740+1205741 + 21199031198861211990521205741) + 321205731198862111990111990521205741V (119909)

(51)

Case 1 (119903 = 0) According to homogenous balanced principlewe let

21205741 = 1205741 + 1205740 = 1205740 minus 12057221205741 = 1205741 minus 120572 (52)

Solving (52) yields

1205740 = minus1205721205741 = minus120572 (53)


1198860Γ (1 minus 120572)Γ (1 minus 2120572) + 1198861Γ (1 minus 120572)Γ (1 minus 2120572) V (119909)= 1205732 (11988601198861119901 + 211990311988612) + 3212057311988621119901V (119909)

(54)

Balancing the power of V(119909) one has3212057311988621119901 = 1198861Γ (1 minus 120572)Γ (1 minus 2120572)

1205732 (11988601198861119901 + 211990311988612) = 1198860Γ (1 minus 120572)Γ (1 minus 2120572) (55)


1198860 = 2119903Γ (1 minus 120572)31205731199012Γ (1 minus 2120572) 1198861 = 2Γ (1 minus 120572)3120573119901Γ (1 minus 2120572)

(56)

Substituting (56) into (27) using solution (22) of subequa-tion we can obtain exact solution of (26) with 120575 = minus2119902120573 = 0as follows

119906 (119909 119905) = 2119903Γ (1 minus 120572)31205731199012Γ (1 minus 2120572) 119905minus120572

+ Γ (1 minus 120572) (11986221199012 minus 21198621199012 + 11990121199092 minus 4119903)61205731199012Γ (1 minus 2120572) 119905minus120572

= Γ (1 minus 120572)6120573Γ (1 minus 2120572) 119905minus120572 (119862 minus 119909)2 (57)

where 120572 isin (0 12) cup (12 1)Case 2 (119903 = 0) Equation (51) becomes


Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 119886011988611199011199051205740+1205741 + 321205731198862111990111990521205741V (119909)

(58)


1205740 + 1205741 = 1205740 minus 12057221205741 = 1205741 minus 120572 (59)

Solving (59) yields

1205740 = 12057401205741 = minus120572 (60)


1198860Γ (1 + 1205740) 1199051205740minus120572Γ (1 + 1205740 minus 120572) + 1198861Γ (1 minus 120572) 119905minus2120572Γ (1 minus 2120572) V (119909)= 1205732 119886011988611199011199051205740minus120572 + 3212057311988621119901119905minus2120572V (119909)

(61)

Balancing the power of V(119909) one has3212057311988621119901 = 1198861Γ (1 minus 120572)Γ (1 minus 2120572) 1205732 11988601198861119901 = 1198860Γ (1 + 1205740)Γ (1 + 1205740 minus 120572)

(62)



1198860 = 11988601198861 = 2Γ (1 minus 120572)3120573119901Γ (1 minus 2120572)

Γ (1 minus 120572)3Γ (1 minus 2120572) = Γ (1 + 1205740)Γ (1 + 1205740 minus 120572) (63)


119906 (119909 119905) = 11988601199051205740 + Γ (1 minus 120572)6120573Γ (1 minus 2120572) 119905minus120572 (119862 minus 119909)2 (64)

where 120572 isin (0 12) cup (12 1) and 1205740 is the root of Γ(1 minus120572)3Γ(1 minus 2120572) = Γ(1 + 1205740)Γ(1 + 1205740 minus 120572)314 119901 = 0 119902 = 0 According to (31) and homogenousbalanced principle we let

21205741 = 1205741 + 1205740 = 21205740 = 1205740 minus 12057221205741 = 1205741 + 1205740 = 1205741 minus 120572 (65)

Solving (65) yields

1205740 = 1205741 = minus120572 (66)


1198860Γ (1 minus 120572)Γ (1 minus 2120572) + 1198861Γ (1 minus 120572)Γ (1 minus 2120572) V (119909)= 1205732 (11990111988601198861 minus 411990211988602 + 211990311988612)

+ 12057311988612 (1199011198861 minus 21199021198860) V (119909) (67)

Balancing the power of V(119909) one has321205731198861 (1199011198861 minus 21199021198860) = 1198861Γ (1 minus 120572)Γ (1 minus 2120572)

1205732 (11990111988601198861 minus 411990211988602 + 211990311988612) = 1198860Γ (1 minus 120572)Γ (1 minus 2120572) (68)


1198860 = ΩΓ (1 minus 120572)Γ (1 minus 2120572) 120573 1198861 = 2Γ (1 minus 120572) (3119902Ω + 1)

3120573119901Γ (1 minus 2120572) (69)

where Ω = (minus1199012 + 4119902119903 plusmn radic1199014 minus 41199012119902119903)3119902(1199012 minus 4119902119903) 1199012 minus4119902119903 = 0 Substituting (69) into (27) using solution (23) of

subequation we can obtain exact solution of (26) with 120575 =minus2119902120573 as follows

119906 (119909 119905)

= Γ (1 minus 120572) (radic1199012Δ (119890(119862minus119909)radic119902Δ + 4119890minus(119862minus119909)radic119902119902) + 4119901radic119902Δ) 119905minus12057212Γ (1 minus 2120572) Δ12057311990111990232

119906 (119909 119905)

= Γ (1 minus 120572) (radic1199012Δ (119890minus(119862minus119909)radic119902Δ + 4119890(119862minus119909)radic119902119902) + 4119901radic119902Δ) 119905minus12057212Γ (1 minus 2120572) Δ12057311990111990232

(70)

where Δ = 1199012 minus 4119902119903When 119862 = 0 119903 = 0 119902 = 1 119901 = 4 then V(119909) =(52) cosh(119909)plusmn(32) sinh(119909)minus2 Equation (70) can be reduced

to

119906 (119909 119905)= minus Γ (1 minus 120572)12120573Γ (1 minus 2120572) 119905minus120572 (5 cosh (119909) plusmn 3 sinh (119909) + 4)

119906 (119909 119905)= Γ (1 minus 120572)12120573Γ (1 minus 2120572) 119905minus120572 (5 cosh (119909) plusmn 3 sinh (119909) minus 4)

(71)

where 120572 isin (0 12) cup (12 1)When 119862 = 0 119903 = 0 119902 = minus1 119901 = 4 then V(119909) =(32)119868 cos (119909)plusmn(52) sin (119909)+2 Equation (70) can be reduced

to

119906 (119909 119905)= minus Γ (1 minus 120572)12120573Γ (1 minus 2120572) 119905minus120572 (minus5 sin (119909) plusmn 3119868 cos (119909) minus 4)

119906 (119909 119905)= Γ (1 minus 120572)12120573Γ (1 minus 2120572) 119905minus120572 (minus5 sin (119909) plusmn 3119868 cos (119909) + 4)

(72)

where 1198682 = minus1If 120572 = 1 substituting (27) into (26) using (20) we can

obtain the following five families of exact solutions of (26)

Family 1

119906 (119909 119905) = 119905minus12119902120573 + 1198861119905minus32119890plusmnradic119902(119862+119909) (73)

where 119902 gt 0 120575 = minus2119902120573Family 2

119906 (119909 119905) = 1198860119905minus13 minus 16120573119905minus1 (119862 minus 119909)2 (74)

where 120575 = 0 119862 is an arbitrary constant


Family 3

119906 (119909 119905) = 13120573119902119905minus1 plusmn (1198902119909radic119902 minus 1199031198902119862radic119902) 119890minusradic119902(119862+119909)6119902120573radicminus119903 119905minus1

119906 (119909 119905) = 13120573119902119905minus1 plusmn (minus1199031198902119909radic119902 + 1198902119862radic119902) 119890minusradic119902(119862+119909)6119902120573radicminus119903 119905minus1

(75)

where 120575 = minus2119902120573 Taking parameters 119902 119903 as suitable values wecan get the following special exact solutions of (26)

119906 (119909 119905) = 13120573120596119905minus1 [minus1 plusmn sin (radic120596 (119862 minus 119909))] 119906 (119909 119905) = 13120573120596119905minus1 [minus1 plusmn cos (radic120596 (119862 minus 119909))]

(76)

where 120575 = 2120596120573 120596 is positive and 119862 is an arbitrary constant

119906 (119909 119905) = 13120573120596119905minus1 [minus1 plusmn sinh (radic120596 (119862 minus 119909))] 119906 (119909 119905) = 13120573120596119905minus1 [minus1 plusmn cosh (1 plusmn radic120596 (119862 minus 119909))]

(77)

where 120575 = minus2120596120573 120596 is positive and119862 is an arbitrary constant

Family 4

119906 (119909 119905) = minus 211990331205731199012 119905minus1 minus 11986221199012 minus 21198621199012 + 11990121199092 minus 411990361205731199012 119905minus1

= minus 16120573119905minus1 (119862 minus 119909)2 (78)

where 120575 = 0 and 119862 is an arbitrary constant

Family 5

119906 (119909 119905)= radic1199012Δ (119890(119862minus119909)radic119902Δ + 4119890minus(119862minus119909)radic119902119902) + 4119901radic119902Δ

12Δ12057311990111990232 119905minus1119906 (119909 119905)

= radic1199012Δ (119890minus(119862minus119909)radic119902Δ + 4119890(119862minus119909)radic119902119902) + 4119901radic119902Δ12Δ12057311990111990232 119905minus1

(79)

where Δ = 1199012 minus 4119902119903 119901 = 0 119902 = 0 120575 = minus2119902120573 and 119862 is anarbitrary constant

32 Situation of 119898 = 2 Taking 119898 = 2 we suppose that (26)has an exact solution as the following form

119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) + 11988621199051205742V2 (119909) (80)

where V(119909) satisfies (20) and 1205740 1205741 1205742 1198860 1198861 1198862 = 0 areconstants that can be determined later Substituting (80)

into (26) using (20) balancing the power on 119905 of the reducedequation we have

1205740 = 1205741 = 1205742 = minus120572120575 = minus8119902120573 (81)

Substituting (81) and (80) into (26) balancing the power ofV(119909) yields

3212057311990111988612 minus 1512057311990211988601198861 + 312057311990111988601198862 + 612057311990311988611198862= 1198861Γ (1 minus 120572)Γ (minus2120572 + 1)

minus 612057311990211988612 + 612057311990311988622 + 152 12057311990111988611198862 minus 1212057311990211988601198862= 1198862Γ (1 minus 120572)Γ (minus2120572 + 1)

minus 811990212057311988602 + 12057311990311988612 + 1212057311990111988601198861 + 212057311990311988601198862= 1198860Γ (1 minus 120572)Γ (minus2120572 + 1)

71205731198862 (1199011198862 minus 1199021198861) = 0

(82)

Solving the above algebraic equations we have the followingresults

Case 1 (119901 = 0 119902 = 0 119903 = 0 120572 isin (0 12) cup (12 1))1198860 = 312057311990311988612Γ (minus2120572 + 1)2Γ (1 minus 120572) 1198861 = 11988611198862 = Γ (1 minus 120572)6Γ (minus2120572 + 1) 120573119903

(83)


119906 (119909 119905) = 312057311990311988621Γ (minus2120572 + 1)2Γ (1 minus 120572) 119905minus120572 + 1198861 (plusmnradic119903119909 + 119862) 119905minus120572

+ Γ (1 minus 120572) (plusmnradic119903119909 + 119862)26Γ (minus2120572 + 1) 120573119903 119905minus120572

(84)

Case 2 (119902 = 0 120572 isin (0 12) cup (12 1))1198860 = 2119903Γ (1 minus 120572)3Δ120573Γ (1 minus 2120572) 1198861 = 2Γ (1 minus 120572) 1199013Δ120573Γ (1 minus 2120572) 1198862 = 2119902Γ (1 minus 120572)3Δ120573Γ (1 minus 2120572)


1198860 = minus 1199012Γ (1 minus 120572)6Δ120573119902Γ (1 minus 2120572) 1198861 = minus 2Γ (1 minus 120572) 1199013120573ΔΓ (1 minus 2120572) 1198862 = minus 2119902Γ (1 minus 120572)3Δ120573Γ (1 minus 2120572)

(85)

where Δ = 1199012 minus 4119902119903 = 0 Substituting (85) into (27) usingsolution (23) of subequation we can obtain the exact solutionof (26) with 120575 = minus8119902120573 as follows

119906 (119909 119905) = Γ (1 minus 120572)961199022Δ120573Γ (1 minus 2120572)sdot 119905minus120572 (4119902119890radic119902(119862minus119909) minus Δ119890minusradic119902(119862minus119909))2

119906 (119909 119905) = minus Γ (1 minus 120572)961199022Δ120573Γ (1 minus 2120572)sdot 119905minus120572 (4119902119890radic119902(119862minus119909) + Δ119890minusradic119902(119862minus119909))2

(86)

When Δ = 4119902 and 119902 = minus120596 lt 0 then 120575 = 8120596120573 Equation(86) can be reduced to

119906 (119909 119905) = Γ (1 minus 120572)6120573120596Γ (1 minus 2120572) 119905minus120572sin2 (radic119908 (119862 minus 119909)) 119906 (119909 119905) = Γ (1 minus 120572)6120573120596Γ (1 minus 2120572) 119905minus120572cos2 (radic119908 (119862 minus 119909))

(87)

where 120596 is positive and 119862 is an arbitrary constantWhen Δ = 4119902 and 119902 = 120596 gt 0 then 120575 = minus8120596120573 Equation

(86) can be reduced to

119906 (119909 119905) = Γ (1 minus 120572)6120573120596Γ (1 minus 2120572) 119905minus120572sinh2 (radic119908 (119862 minus 119909)) 119906 (119909 119905) = minus Γ (1 minus 120572)6120573120596Γ (1 minus 2120572) 119905minus120572cosh2 (radic119908 (119862 minus 119909))

(88)

where 120596 is positive and 119862 is an arbitrary constantIf 120572 = 1 substituting (80) into (26) using (20) we can

obtain the following exact solutions of (26)

119906 (119909 119905) = minus 1961199022Δ120573119905minus1 (4119902119890radic119902(119862minus119909) minus Δ119890minusradic119902(119862minus119909))2 119906 (119909 119905) = 1961199022Δ120573119905minus1 (4119902119890radic119902(119862minus119909) + Δ119890minusradic119902(119862minus119909))2

(89)

When Δ = 4119902 and 119902 = minus120596 lt 0 then 120575 = 8120596120573 Equation (89)can be reduced to

119906 (119909 119905) = minus Γ (1 minus 120572)6120573120596Γ (1 minus 2120572) 119905minus1sin2 (radic119908 (119862 minus 119909)) 119906 (119909 119905) = minus Γ (1 minus 120572)6120573120596Γ (1 minus 2120572) 119905minus1cos2 (radic119908 (119862 minus 119909))

(90)

where 120596 is positive and 119862 is an arbitrary constant

When Δ = 4119902 and 119902 = 120596 gt 0 then 120575 = minus8120596120573 Equation(89) can be reduced to

119906 (119909 119905) = minus 16120573120596119905minus1sinh2 (radic119908 (119862 minus 119909)) 119906 (119909 119905) = 16120573120596119905minus1cosh2 (radic119908 (119862 minus 119909))

(91)


4 Conclusions

In this work we proved that the fractional Leibniz rulethat appeared in many references does not hold underRiemann-Liouville definition and Caputo definition of frac-tional derivative Based on the homogenous balanced prin-ciple we introduced a general method for investigatingexact solution of nonlinear time-fractional PDEs By usingthis method called improved separation variable function-expansion method we studied a nonlinear time-fractionalPDE with diffusion term Some new results are obtained

Firstly compared with Ruirsquos method [23] it is easy tofind that our method is more general All solutions given inreference [23] can be obtained by taking special parametersin our results For example taking 120596 = 1 119862 = 0 oursolutions (46) and (47) become solutions (340) and (341)in [23] respectively Taking 119902 = 1 119862 = 0 our solutions(38) become solution (347) in [23] Taking 119903 = 1 119862 =0 our solutions (84) become solution (360) in [23] Othersolutions obtained in our work are new such as solutions(49) (50) (57) (64) (71) (72) (87) and (88) which are notreported in related references In addition we should adoptdifferent subequation for other time-fractional PDEs such asV1015840(119909) = radic119903 + 119901V2(119909) + 119902V4(119909) Finally our method is simpleand efficient for application without any skill

According to symmetrical characteristic this methodalso can be used to investigate exact solutions of spacefractional PDEs which are formed as follows

119862119863120572119909119906 = 119865(119906 120597119906120597119905 12059721199061205971199052 120597

119899119906120597119905119899 ) (92)

Conflicts of Interest

The authors declare that there are no conflicts of interestregarding the publication of this article

Acknowledgments

This research is supported by the Natural Science Foundationof China (nos 11461022 11361023) Science Foundation ofYunnan Province (2014FA037) and Middle-Aged AcademicBackbone of Honghe University (no 2014GG0105)

References

[1] W C Tan W X Pan and M Y Xu ldquoA note on unsteadyflows of a viscoelastic fluid with the fractional Maxwell model


between two parallel platesrdquo International Journal of Non-LinearMechanics vol 38 no 5 pp 645ndash650 2003

[2] D Tripathi S K Pandey and S Das ldquoPeristaltic flow ofviscoelastic fluid with fractional Maxwell model through achannelrdquo Applied Mathematics and Computation vol 215 no10 pp 3645ndash3654 2010

[3] S M Guo L Q Mei Y Li and Y F Sun ldquoThe improvedfractional sub-equation method and its applications to thespace-time fractional differential equations in fluid mechanicsrdquoPhysics Letters A vol 376 no 4 pp 407ndash411 2012

[4] A M A El-Sayed S Z Rida and A A M Arafa ldquoExactsolutions of fractional-order biological population modelrdquoCommunications in Theoretical Physics vol 52 no 6 pp 992ndash996 2009

[5] F Liu and K Burrage ldquoNovel techniques in parameter estima-tion for fractional dynamical models arising from biologicalsystemsrdquo Computers amp Mathematics with Applications vol 62no 3 pp 822ndash833 2011

[6] K S Miller and B Ross An introduction to the fractionalcalculus and fractional differential equationsWiley-IntersciencePublication New York NY USA 1993

[7] S C Pei and J J Ding ldquoRelations between Gabor transformsand fractional Fourier transforms and their applications forsignal processingrdquo IEEE Transactions on Signal Processing vol55 no 10 pp 4839ndash4850 2007

[8] D Baleanu J A T Machado and A C J Luo FractionalDynamics and Control Springer New York NY USA 2012

[9] V Daftardar-Gejji and H Jafari ldquoAdomian decomposition atool for solving a system of fractional differential equationsrdquoJournal of Mathematical Analysis and Applications vol 301 no2 pp 508ndash518 2005

[10] K Singla andR K Gupta ldquoGeneralized Lie symmetry approachfor fractional order systems of differential equations IIIrdquoJournal ofMathematical Physics vol 58 no 6 Article ID 0615012017

[11] A Akbulut and F Tascan ldquoLie symmetries symmetry reduc-tions and conservation laws of time fractional modifiedKortewegndashde Vries (mkdv) equationrdquo Chaos Solitons amp Frac-tals vol 100 pp 1ndash6 2017

[12] T Bakkyaraj and R Sahadevan ldquoInvariant analysis ofnonlinear fractional ordinary differential equations withRiemannndashLiouville fractional derivativerdquo Nonlinear Dynamicsvol 80 no 1-2 pp 447ndash455 2015

[13] A H Bhrawy and M A Zaky ldquoHighly accurate numericalschemes for multi-dimensional space variable-order fractionalSchrodinger equationsrdquo Computers amp Mathematics with Appli-cations vol 73 no 6 pp 1100ndash1117 2017

[14] Y Li and K Shah ldquoNumerical Solutions of Coupled Systemsof Fractional Order Partial Differential Equationsrdquo Advances inMathematical Physics vol 2017 Article ID 1535826 2017

[15] M Eslami B Fathi Vajargah M Mirzazadeh and A BiswasldquoApplication of first integral method to fractional partial differ-ential equationsrdquo Indian Journal of Physics vol 88 no 2 pp177ndash184 2014

[16] T Bakkyaraj and R Sahadevan ldquoApproximate Analytical Solu-tion of Two Coupled Time Fractional Nonlinear SchrodingerEquationsrdquo International Journal of Applied and ComputationalMathematics vol 2 no 1 pp 113ndash135 2016

[17] R Sahadevan and T Bakkyaraj ldquoInvariant subspace methodand exact solutions of certain nonlinear time fractional partialdifferential equationsrdquoFractional Calculus andAppliedAnalysisvol 18 no 1 pp 146ndash162 2015

[18] G C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[19] Q Feng ldquoA new analytical method for seeking traveling wavesolutions of spacendashtime fractional partial differential equationsarising in mathematical physicsrdquo Optik - International Journalfor Light and Electron Optics vol 130 pp 310ndash323 2017

[20] G Jumarie ldquoModified Riemann-Liouville derivative and frac-tional Taylor series of nondifferentiable functions furtherresultsrdquoComputersampMathematics with Applications vol 51 no9-10 pp 1367ndash1376 2006

[21] V E Tarasov ldquoOn chain rule for fractional derivativesrdquo Com-munications inNonlinear Science andNumerical Simulation vol30 no 1-3 pp 1ndash4 2016

[22] V E Tarasov ldquoNo violation of the Leibniz rule No fractionalderivativerdquo Communications in Nonlinear Science and Numeri-cal Simulation vol 18 no 11 pp 2945ndash2948 2013

[23] W G Rui ldquoApplications of homogenous balanced principleon investigating exact solutions to a series of time fractionalnonlinear PDEsrdquo Communications in Nonlinear Science andNumerical Simulation vol 47 pp 253ndash266 2017

[24] G Bluman and S Kumei ldquoOn the remarkable nonlineardiffusion equation (120597120597x)[a(u + b)-2(120597u120597x)] - (120597u120597t) = 0rdquoJournal of Mathematical Physics vol 21 no 5 pp 1019ndash10231979

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describe nonlinear phenomena in connection with real-timeproblems which not only depend on time instant but alsodepend on the previous time history

Recently Feng [19] introduced a fractional 119863120572119866119866method for seeking traveling wave solutions of space-time-fractional partial differential equations under the followingmodified Riemann-Liouville derivative definition [20]

119863120572119905 119891 (119905)

=

1Γ (1 minus 120572) int1199050(119905 minus 120591)minus120572minus1 [119891 (120591) minus 119891 (0)] 119889120591 120572 lt 0

1Γ (1 minus 120572) 119889119889119905 int1199050(119905 minus 120591)minus120572 [119891 (120591) minus 119891 (0)] 119889120591 0 lt 120572 lt 1

119891(119899) (119905)(120572minus119899) 119899 ⩽ 120572 lt 119899 + 1

(3)

The following properties for the modified Riemann-Liouville derivative are usually used

119863120572119909119909120574 = Γ (1 + 120574)Γ (1 + 120574 minus 120572)119909120574minus120572 (4)

119863120572119909119891 [119906 (119909)] = 1198911015840 (119906)119863120572119909119906 (119909) = 119863120572119909119891 (119906) (1199061015840 (119909))120572 (5)

Let us briefly review this method Using traveling wavetransformation 120585 = 119896119909 + 119888119905 119906(119909 119905) = 119880(120585) and (4) and(5) they reduced the following fractional partial differentialequation

119875 (119906119863120572119905 1199061198631205721199091199061198632120572119905 1199061198632120572119909 119906 ) = 0 (6)

which can be converted into the following fractional ordinarydifferential equation with respect to the variable 120585

119875 (119880 119888120572119863120572120585119880 119896120572119863120572120585119880 11988821205721198632120572120585 119880 11989621205721198632120572120585 119880 ) = 0 (7)

Then they suppose that the solution of (7) can beexpressed by a polynomial in 119863120572119866119866 as follows

119880 (120585) = 119898sum119894=0

119886119894 (119863120572120585119866119866 ) (8)

where 119866 = 119866(120585) satisfies the following fractional ordinarydifferential equation

119860119866 (120585)1198632120572120585 119866 (120585) minus 119861119866 (120585)119863120572120585119866 (120585) minus 119862 (119863120572120585119866 (120585))2minus 1198641198662 (120585) = 0 (9)

In order to find exact solutions of (9) a nonlinearfractional complex transformation 120578 = 120585120572Γ(1 + 120572) is used

Then (9) can be turned into the following second ordinarydifferential equation

119860119867(120578)11986710158401015840 (120578) minus 119861119867 (120578)1198671015840 (120578) minus 119862 (1198671015840 (120578))2minus 1198641198672 (120578) = 0 (10)

The exact solutions of (10) are knownFinally substituting (8) into (7) equating each coefficient

of this polynomial on119863120572120585119866119866 to zero they can obtain a largenumber of exact solutions of space-time fractional partialdifferential equations (6)

We noticed two problems as follows At first in a shortcommunication [21] Tarasov proved that formula (4) and thechain rule (5) cannot be performed together for fractionalderivatives of noninteger orders 120572 = 1 Therefore usingtraveling wave transformation 120585 = 119896119909 + 119888119905 119906(119909 119905) = 119880(120585)space-time fractional partial differential equations (6) cannotbe converted into fractional ordinary differential equation(7) And (9) cannot convert to (10) under transformation120578 = 120585120572Γ(1 + 120572)

In addition it should be noted that the Leibniz rule in theform

119863120572120585 (119891 (120585) 119892 (120585)) = 119892 (120585)119863120572120585119891 (120585) + 119891 (120585)119863120572120585119892 (120585) (11)

cannot hold for fractional derivatives of order 120572 = 1 for setsof differentiable and nondifferentiable functions [22] It canbe proved easily by using a counterexample Let

119891 (119909) = 1199091205741 119892 (119909) = 1199091205742 (12)

where 1205741 1205742 isin 119873 Obviously we have

119863120572119909119891 (119909) = Γ (1 + 1205741)Γ (1 + 1205741 minus 120572)1199091205741minus120572119863120572119909119892 (119909) = Γ (1 + 1205742)Γ (1 + 1205742 minus 120572)1199091205742minus120572

(13)

Then one has

119892 (119909)119863120572119909119891 (119909) = Γ (1 + 1205741)Γ (1 + 1205741 minus 120572)1199091205741+1205742minus120572119891 (119909)119863120572119909119892 (119909) = Γ (1 + 1205742)Γ (1 + 1205742 minus 120572)1199091205741+1205742minus120572

119863120572119909 [119891 (119909) 119892 (119909)] = Γ (1 + 1205741 + 1205742)Γ (1 + 1205741 + 1205742 minus 120572)1199091205741+1205742minus120572(14)

As a result we have the conditionΓ (1 + 1205741 + 1205742)Γ (1 + 1205741 + 1205742 minus 120572) = Γ (1 + 1205741)Γ (1 + 1205741 minus 120572)

+ Γ (1 + 1205742)Γ (1 + 1205742 minus 120572) (15)

which must be performed if the Leibniz rule (11) holds Forexample if we make 1205741 1205742 isin 119873+ and take into account Γ(119899 +1) = 119899Γ(119899) then condition (15) can be represented in the form




119863120572120585119880 (120585) = 119863120572120585 (119863120572120585119866 (120585)119866 (120585) )

= 119866 (120585)1198632120572120585 119866 (120585) minus (119863120572120585119866 (120585))21198662 (120585)

(17)







119863120572119905 [119896119891 (119909)] = 119896119863120572119905 [119891 (119909)] (18)




119906 = 119898sum119896=0

119886119896 [120593 (119905)] [V (119909)]119896 (19)


V1015840 (119909)2 = 119903 + 119901V (119909) + 119902V2 (119909) (20)




V (119909) = 11986221199012 minus 21198621199012119909 + 11990921199012 minus 41199034119901 (22)


minus811990232


(23)











119888119863120572119905 119906 = (120597119906120597119909)2 [120597119891 (119906)120597119906 ] + 119891 (119906) 12059721199061205971199092 + 1205751199062119905 ⩾ 0 119909 isin 119877 0 lt 120572 ⩽ 1

(24)


120597120597119909 [119891 (119906) 120597119906120597119909] minus 120597119906120597119905 = 0 (25)


119888119863120572119905 119906 = 120573(120597119906120597119909)2 + 12057311990612059721199061205971199092 + 1205751199062119905 ⩾ 0 119909 isin 119877 0 lt 120572 ⩽ 1

(26)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) (27)



Γ (minus120572 + 1205741 + 1) V (119909)= 1199052120574112057311990311988612 + 121199051205741+120574012057311990111988601198861+ 1199052120574012057511988602 (321199052120574112057311990111988612 + 1199051205741+120574012057311990211988601198861+ 21199051205741+120574012057511988601198861) V (119909) + (21199052120574112057311990211988612 + 1199052120574112057511988612)sdot V (119909)2

(28)


21199052120574112057311990211988612 + 1199052120574112057511988612 = 0 (29)

Solving (29) yields

120575 = minus2119902120573 (30)








(32)



1205740 minus 120572 = 212057401205741 minus 120572 = 1205740 + 1205741 (33)

Solving (33) yields

1205740 = minus1205721205741 = 1205741 (34)



(35)











(39)


1205740 minus 120572 = 1205741 + 12057401205741 minus 120572 = 21205740 = 21205741 (40)

Solving (40) yields

1205740 = minus1205721205741 = minus120572 (41)







(44)






(45)










(48)









Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 (119886011988611199011199051205740+1205741 + 21199031198861211990521205741) + 321205731198862111990111990521205741V (119909)

(51)


21205741 = 1205741 + 1205740 = 1205740 minus 12057221205741 = 1205741 minus 120572 (52)

Solving (52) yields

1205740 = minus1205721205741 = minus120572 (53)



(54)


1205732 (11988601198861119901 + 211990311988612) = 1198860Γ (1 minus 120572)Γ (1 minus 2120572) (55)



(56)







Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 119886011988611199011199051205740+1205741 + 321205731198862111990111990521205741V (119909)

(58)


1205740 + 1205741 = 1205740 minus 12057221205741 = 1205741 minus 120572 (59)

Solving (59) yields

1205740 = 12057401205741 = minus120572 (60)



(61)


(62)



1198860 = 11988601198861 = 2Γ (1 minus 120572)3120573119901Γ (1 minus 2120572)





21205741 = 1205741 + 1205740 = 21205740 = 1205740 minus 12057221205741 = 1205741 + 1205740 = 1205741 minus 120572 (65)

Solving (65) yields

1205740 = 1205741 = minus120572 (66)



+ 12057311988612 (1199011198861 minus 21199021198860) V (119909) (67)





3120573119901Γ (1 minus 2120572) (69)



119906 (119909 119905)


119906 (119909 119905)


(70)


to



(71)


to



(72)



Family 1






Family 3



(75)



(76)



(77)


Family 4




Family 5


12Δ12057311990111990232 119905minus1119906 (119909 119905)


(79)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) + 11988621199051205742V2 (119909) (80)



1205740 = 1205741 = 1205742 = minus120572120575 = minus8119902120573 (81)





71205731198862 (1199011198862 minus 1199021198861) = 0

(82)



(83)




(84)




(85)




(86)



(87)




(88)




(89)



(90)




(91)


4 Conclusions




119862119863120572119909119906 = 119865(119906 120597119906120597119905 12059721199061205971199052 120597

119899119906120597119905119899 ) (92)



Acknowledgments


References


































Journal of












Journal of







Volume 2018






Volume 2018











119863120572120585119880 (120585) = 119863120572120585 (119863120572120585119866 (120585)119866 (120585) )

= 119866 (120585)1198632120572120585 119866 (120585) minus (119863120572120585119866 (120585))21198662 (120585)

(17)







119863120572119905 [119896119891 (119909)] = 119896119863120572119905 [119891 (119909)] (18)




119906 = 119898sum119896=0

119886119896 [120593 (119905)] [V (119909)]119896 (19)


V1015840 (119909)2 = 119903 + 119901V (119909) + 119902V2 (119909) (20)




V (119909) = 11986221199012 minus 21198621199012119909 + 11990921199012 minus 41199034119901 (22)


minus811990232


(23)











119888119863120572119905 119906 = (120597119906120597119909)2 [120597119891 (119906)120597119906 ] + 119891 (119906) 12059721199061205971199092 + 1205751199062119905 ⩾ 0 119909 isin 119877 0 lt 120572 ⩽ 1

(24)


120597120597119909 [119891 (119906) 120597119906120597119909] minus 120597119906120597119905 = 0 (25)


119888119863120572119905 119906 = 120573(120597119906120597119909)2 + 12057311990612059721199061205971199092 + 1205751199062119905 ⩾ 0 119909 isin 119877 0 lt 120572 ⩽ 1

(26)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) (27)



Γ (minus120572 + 1205741 + 1) V (119909)= 1199052120574112057311990311988612 + 121199051205741+120574012057311990111988601198861+ 1199052120574012057511988602 (321199052120574112057311990111988612 + 1199051205741+120574012057311990211988601198861+ 21199051205741+120574012057511988601198861) V (119909) + (21199052120574112057311990211988612 + 1199052120574112057511988612)sdot V (119909)2

(28)


21199052120574112057311990211988612 + 1199052120574112057511988612 = 0 (29)

Solving (29) yields

120575 = minus2119902120573 (30)








(32)



1205740 minus 120572 = 212057401205741 minus 120572 = 1205740 + 1205741 (33)

Solving (33) yields

1205740 = minus1205721205741 = 1205741 (34)



(35)











(39)


1205740 minus 120572 = 1205741 + 12057401205741 minus 120572 = 21205740 = 21205741 (40)

Solving (40) yields

1205740 = minus1205721205741 = minus120572 (41)







(44)






(45)










(48)









Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 (119886011988611199011199051205740+1205741 + 21199031198861211990521205741) + 321205731198862111990111990521205741V (119909)

(51)


21205741 = 1205741 + 1205740 = 1205740 minus 12057221205741 = 1205741 minus 120572 (52)

Solving (52) yields

1205740 = minus1205721205741 = minus120572 (53)



(54)


1205732 (11988601198861119901 + 211990311988612) = 1198860Γ (1 minus 120572)Γ (1 minus 2120572) (55)



(56)







Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 119886011988611199011199051205740+1205741 + 321205731198862111990111990521205741V (119909)

(58)


1205740 + 1205741 = 1205740 minus 12057221205741 = 1205741 minus 120572 (59)

Solving (59) yields

1205740 = 12057401205741 = minus120572 (60)



(61)


(62)



1198860 = 11988601198861 = 2Γ (1 minus 120572)3120573119901Γ (1 minus 2120572)





21205741 = 1205741 + 1205740 = 21205740 = 1205740 minus 12057221205741 = 1205741 + 1205740 = 1205741 minus 120572 (65)

Solving (65) yields

1205740 = 1205741 = minus120572 (66)



+ 12057311988612 (1199011198861 minus 21199021198860) V (119909) (67)





3120573119901Γ (1 minus 2120572) (69)



119906 (119909 119905)


119906 (119909 119905)


(70)


to



(71)


to



(72)



Family 1






Family 3



(75)



(76)



(77)


Family 4




Family 5


12Δ12057311990111990232 119905minus1119906 (119909 119905)


(79)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) + 11988621199051205742V2 (119909) (80)



1205740 = 1205741 = 1205742 = minus120572120575 = minus8119902120573 (81)





71205731198862 (1199011198862 minus 1199021198861) = 0

(82)



(83)




(84)




(85)




(86)



(87)




(88)




(89)



(90)




(91)


4 Conclusions




119862119863120572119909119906 = 119865(119906 120597119906120597119905 12059721199061205971199052 120597

119899119906120597119905119899 ) (92)



Acknowledgments


References


































Journal of












Journal of







Volume 2018






Volume 2018















119888119863120572119905 119906 = (120597119906120597119909)2 [120597119891 (119906)120597119906 ] + 119891 (119906) 12059721199061205971199092 + 1205751199062119905 ⩾ 0 119909 isin 119877 0 lt 120572 ⩽ 1

(24)


120597120597119909 [119891 (119906) 120597119906120597119909] minus 120597119906120597119905 = 0 (25)


119888119863120572119905 119906 = 120573(120597119906120597119909)2 + 12057311990612059721199061205971199092 + 1205751199062119905 ⩾ 0 119909 isin 119877 0 lt 120572 ⩽ 1

(26)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) (27)



Γ (minus120572 + 1205741 + 1) V (119909)= 1199052120574112057311990311988612 + 121199051205741+120574012057311990111988601198861+ 1199052120574012057511988602 (321199052120574112057311990111988612 + 1199051205741+120574012057311990211988601198861+ 21199051205741+120574012057511988601198861) V (119909) + (21199052120574112057311990211988612 + 1199052120574112057511988612)sdot V (119909)2

(28)


21199052120574112057311990211988612 + 1199052120574112057511988612 = 0 (29)

Solving (29) yields

120575 = minus2119902120573 (30)








(32)



1205740 minus 120572 = 212057401205741 minus 120572 = 1205740 + 1205741 (33)

Solving (33) yields

1205740 = minus1205721205741 = 1205741 (34)



(35)











(39)


1205740 minus 120572 = 1205741 + 12057401205741 minus 120572 = 21205740 = 21205741 (40)

Solving (40) yields

1205740 = minus1205721205741 = minus120572 (41)







(44)






(45)










(48)









Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 (119886011988611199011199051205740+1205741 + 21199031198861211990521205741) + 321205731198862111990111990521205741V (119909)

(51)


21205741 = 1205741 + 1205740 = 1205740 minus 12057221205741 = 1205741 minus 120572 (52)

Solving (52) yields

1205740 = minus1205721205741 = minus120572 (53)



(54)


1205732 (11988601198861119901 + 211990311988612) = 1198860Γ (1 minus 120572)Γ (1 minus 2120572) (55)



(56)







Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 119886011988611199011199051205740+1205741 + 321205731198862111990111990521205741V (119909)

(58)


1205740 + 1205741 = 1205740 minus 12057221205741 = 1205741 minus 120572 (59)

Solving (59) yields

1205740 = 12057401205741 = minus120572 (60)



(61)


(62)



1198860 = 11988601198861 = 2Γ (1 minus 120572)3120573119901Γ (1 minus 2120572)





21205741 = 1205741 + 1205740 = 21205740 = 1205740 minus 12057221205741 = 1205741 + 1205740 = 1205741 minus 120572 (65)

Solving (65) yields

1205740 = 1205741 = minus120572 (66)



+ 12057311988612 (1199011198861 minus 21199021198860) V (119909) (67)





3120573119901Γ (1 minus 2120572) (69)



119906 (119909 119905)


119906 (119909 119905)


(70)


to



(71)


to



(72)



Family 1






Family 3



(75)



(76)



(77)


Family 4




Family 5


12Δ12057311990111990232 119905minus1119906 (119909 119905)


(79)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) + 11988621199051205742V2 (119909) (80)



1205740 = 1205741 = 1205742 = minus120572120575 = minus8119902120573 (81)





71205731198862 (1199011198862 minus 1199021198861) = 0

(82)



(83)




(84)




(85)




(86)



(87)




(88)




(89)



(90)




(91)


4 Conclusions




119862119863120572119909119906 = 119865(119906 120597119906120597119905 12059721199061205971199052 120597

119899119906120597119905119899 ) (92)



Acknowledgments


References


































Journal of












Journal of







Volume 2018






Volume 2018










1205740 minus 120572 = 212057401205741 minus 120572 = 1205740 + 1205741 (33)

Solving (33) yields

1205740 = minus1205721205741 = 1205741 (34)



(35)











(39)


1205740 minus 120572 = 1205741 + 12057401205741 minus 120572 = 21205740 = 21205741 (40)

Solving (40) yields

1205740 = minus1205721205741 = minus120572 (41)







(44)






(45)










(48)









Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 (119886011988611199011199051205740+1205741 + 21199031198861211990521205741) + 321205731198862111990111990521205741V (119909)

(51)


21205741 = 1205741 + 1205740 = 1205740 minus 12057221205741 = 1205741 minus 120572 (52)

Solving (52) yields

1205740 = minus1205721205741 = minus120572 (53)



(54)


1205732 (11988601198861119901 + 211990311988612) = 1198860Γ (1 minus 120572)Γ (1 minus 2120572) (55)



(56)







Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 119886011988611199011199051205740+1205741 + 321205731198862111990111990521205741V (119909)

(58)


1205740 + 1205741 = 1205740 minus 12057221205741 = 1205741 minus 120572 (59)

Solving (59) yields

1205740 = 12057401205741 = minus120572 (60)



(61)


(62)



1198860 = 11988601198861 = 2Γ (1 minus 120572)3120573119901Γ (1 minus 2120572)





21205741 = 1205741 + 1205740 = 21205740 = 1205740 minus 12057221205741 = 1205741 + 1205740 = 1205741 minus 120572 (65)

Solving (65) yields

1205740 = 1205741 = minus120572 (66)



+ 12057311988612 (1199011198861 minus 21199021198860) V (119909) (67)





3120573119901Γ (1 minus 2120572) (69)



119906 (119909 119905)


119906 (119909 119905)


(70)


to



(71)


to



(72)



Family 1






Family 3



(75)



(76)



(77)


Family 4




Family 5


12Δ12057311990111990232 119905minus1119906 (119909 119905)


(79)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) + 11988621199051205742V2 (119909) (80)



1205740 = 1205741 = 1205742 = minus120572120575 = minus8119902120573 (81)





71205731198862 (1199011198862 minus 1199021198861) = 0

(82)



(83)




(84)




(85)




(86)



(87)




(88)




(89)



(90)




(91)


4 Conclusions




119862119863120572119909119906 = 119865(119906 120597119906120597119905 12059721199061205971199052 120597

119899119906120597119905119899 ) (92)



Acknowledgments


References


































Journal of












Journal of







Volume 2018






Volume 2018











(48)









Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 (119886011988611199011199051205740+1205741 + 21199031198861211990521205741) + 321205731198862111990111990521205741V (119909)

(51)


21205741 = 1205741 + 1205740 = 1205740 minus 12057221205741 = 1205741 minus 120572 (52)

Solving (52) yields

1205740 = minus1205721205741 = minus120572 (53)



(54)


1205732 (11988601198861119901 + 211990311988612) = 1198860Γ (1 minus 120572)Γ (1 minus 2120572) (55)



(56)







Γ (minus120572 + 1205741 + 1) V (119909)= 1205732 119886011988611199011199051205740+1205741 + 321205731198862111990111990521205741V (119909)

(58)


1205740 + 1205741 = 1205740 minus 12057221205741 = 1205741 minus 120572 (59)

Solving (59) yields

1205740 = 12057401205741 = minus120572 (60)



(61)


(62)



1198860 = 11988601198861 = 2Γ (1 minus 120572)3120573119901Γ (1 minus 2120572)





21205741 = 1205741 + 1205740 = 21205740 = 1205740 minus 12057221205741 = 1205741 + 1205740 = 1205741 minus 120572 (65)

Solving (65) yields

1205740 = 1205741 = minus120572 (66)



+ 12057311988612 (1199011198861 minus 21199021198860) V (119909) (67)





3120573119901Γ (1 minus 2120572) (69)



119906 (119909 119905)


119906 (119909 119905)


(70)


to



(71)


to



(72)



Family 1






Family 3



(75)



(76)



(77)


Family 4




Family 5


12Δ12057311990111990232 119905minus1119906 (119909 119905)


(79)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) + 11988621199051205742V2 (119909) (80)



1205740 = 1205741 = 1205742 = minus120572120575 = minus8119902120573 (81)





71205731198862 (1199011198862 minus 1199021198861) = 0

(82)



(83)




(84)




(85)




(86)



(87)




(88)




(89)



(90)




(91)


4 Conclusions




119862119863120572119909119906 = 119865(119906 120597119906120597119905 12059721199061205971199052 120597

119899119906120597119905119899 ) (92)



Acknowledgments


References


































Journal of












Journal of







Volume 2018






Volume 2018










1198860 = 11988601198861 = 2Γ (1 minus 120572)3120573119901Γ (1 minus 2120572)





21205741 = 1205741 + 1205740 = 21205740 = 1205740 minus 12057221205741 = 1205741 + 1205740 = 1205741 minus 120572 (65)

Solving (65) yields

1205740 = 1205741 = minus120572 (66)



+ 12057311988612 (1199011198861 minus 21199021198860) V (119909) (67)





3120573119901Γ (1 minus 2120572) (69)



119906 (119909 119905)


119906 (119909 119905)


(70)


to



(71)


to



(72)



Family 1






Family 3



(75)



(76)



(77)


Family 4




Family 5


12Δ12057311990111990232 119905minus1119906 (119909 119905)


(79)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) + 11988621199051205742V2 (119909) (80)



1205740 = 1205741 = 1205742 = minus120572120575 = minus8119902120573 (81)





71205731198862 (1199011198862 minus 1199021198861) = 0

(82)



(83)




(84)




(85)




(86)



(87)




(88)




(89)



(90)




(91)


4 Conclusions




119862119863120572119909119906 = 119865(119906 120597119906120597119905 12059721199061205971199052 120597

119899119906120597119905119899 ) (92)



Acknowledgments


References


































Journal of












Journal of







Volume 2018






Volume 2018









Family 3



(75)



(76)



(77)


Family 4




Family 5


12Δ12057311990111990232 119905minus1119906 (119909 119905)


(79)



119906 (119909 119905) = 11988601199051205740 + 11988611199051205741V (119909) + 11988621199051205742V2 (119909) (80)



1205740 = 1205741 = 1205742 = minus120572120575 = minus8119902120573 (81)





71205731198862 (1199011198862 minus 1199021198861) = 0

(82)



(83)




(84)




(85)




(86)



(87)




(88)




(89)



(90)




(91)


4 Conclusions




119862119863120572119909119906 = 119865(119906 120597119906120597119905 12059721199061205971199052 120597

119899119906120597119905119899 ) (92)



Acknowledgments


References


































Journal of












Journal of







Volume 2018






Volume 2018










(85)




(86)



(87)




(88)




(89)



(90)




(91)


4 Conclusions




119862119863120572119909119906 = 119865(119906 120597119906120597119905 12059721199061205971199052 120597

119899119906120597119905119899 ) (92)



Acknowledgments


References


































Journal of












Journal of







Volume 2018






Volume 2018








































Journal of












Journal of







Volume 2018






Volume 2018








applications of separation variables approach in solving...

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