A further improved (G0 )- expansion method and the G ... · extended tanh- method for ﬁnding...

A further improved (G′

G )- expansion method and the

extended tanh- method for finding exact solutions of

nonlinear PDEs

Elsayed. M. E. ZayedMathematics Department

Faculty of ScienceZagazig University

Egypteme [email protected]

Abstract: In the present article, we construct the exact traveling wave solutions of nonlinear PDEs inmathematical physics via the (1 + 1) dimensional modified Kawahara equation by using the followingtwo methods: (i) A further improved (G′

G )- expansion method, where G = G(ξ) satisfies the auxiliaryordinary differential equation [G′(ξ)]2 = aG2(ξ) + bG4(ξ) + cG6(ξ), where ξ = x − V t while a, b, c andV are constants. (ii) The well known extended tanh- function method. We show that the exact solutionsobtained by these two methods are equivalent. Note that the first method (i) has not been used by anyonebefore.

Key–Words: (G′

G )- expansion method, auxiliary equation, extended tanh- function method, traveling wavesolutions, modified Kawahara equation.

1 Introduction

In recent years, the exact solutions of nonlin-ear PDEs have been investigated by many au-thors( see for example [1]–[49]) who are inter-ested in nonlinear physical phenomena. Manypowerful different methods have been presentedby those authors. For integrable nonlinear differ-ential equations, the inverse scattering transformmethod [2], the Hirota method [10], the truncatedPainleve expansion method [43], the Backlundtransform method [19]–[21] and the exp-functionmethod [4, 9, 36, 44, 45] are used in looking forthe exact solutions. Among non-integrable non-linear differential equations there is a wide classof the equations that referred to as the partiallyintegrable, because these equations become inte-grable for some values of their parameters. Thereare many different methods to look for the ex-act solutions of these equations. The most fa-mous algorithms are the truncated Painleve ex-pansion method [14], the Weierstrass elliptic func-

tion method [13], the tanh- function method[1, 7, 8, 32, 34, 39, 46] and the Jacobi elliptic func-tion expansion method [6, 16, 18, 30, 37, 38, 40].There are other methods which can be found in[11],[22],[23]-[29],[33].

Wang et al [30] have introduced a sim-ple method which is called the (G′

G )-expansionmethod to look for traveling wave solutions ofnonlinear PDEs, where G = G(ξ) satisfies thesecond order linear ordinary differential equa-tion G′′(ξ) + λG′(ξ) + µG(ξ) = 0, where ξ =x − V t while V, λ and µ are arbitrary con-stants. For further references see the articles[3, 5, 7, 20, 41, 42, 48, 49]. Recently El-Wakil et al[7] and Parkes [20] have shown that the extendedtanh- function method proposed by Fan [8] andthe basic (G′

G )- expansion method proposed byWang et al [30] are entirely equivalent in as muchas they deliver exactly the same set of solutions toa given evolution equation. This observation hasbeen pointed out recently by Kudryashov [15]. Inthis article, we introduce an alternative approach

WSEAS TRANSACTIONS on MATHEMATICS Elsayed M. E. Zayed

ISSN: 1109-2769 56 Issue 2, Volume 10, February 2011

which is called a further improved (G′

G )- expansionmethod to find the exact traveling wave solutionsof some nonlinear PDEs, where G = G(ξ) sat-isfies the auxiliary ordinary differential equation[G′(ξ)]2 = aG2(ξ) + bG4(ξ) + cG6(ξ), where a, b

and c are constants. This approach has not beenused by anyone before. It will play an importantrole in constructing many exact traveling wavesolutions for the nonlinear PDEs via the (1 + 1)dimensional modified Kawahara equation. Theobjective of this article is to show that the exactsolutions of these two equation obtained by usingthe further improved (G′

G )- expansion method andthe well known extended tanh- function methodare equivalent.

2 Description of a furtherimproved (G′

G )- expansionmethod

Suppose we have the following nonlinear partialdifferential equation

F (u, ut, ux, uy, utt, uxt, ...) = 0, (2.1)

where u = u(x, t) is an unknown function, F is apolynomial in u(x, t) and its partial derivatives inwhich the highest order derivatives and the non-linear terms are involved. In the following wegive the main steps of a further improved (G′

G )-expansion method:

Step 1. The traveling wave variable

u(x, t) = u(ξ), ξ = x− V t, (2.2)

where V is a constant, permits us reducing Eq.(2.1) to an ODE for u = u(ξ) in the form

P (u, u′, u′′, u′′′, ...) = 0 (2.3)

where ′ = ddξ .

Step 2. Suppose the solution of Eq.(2.3) can beexpressed by a polynomial in (G′

G ) as follows

u(ξ) =n∑

i=0

αi

(G′

G

)i

, (2.4)

where G = G(ξ) satisfies the following auxiliaryequation

[G′(ξ)]2 = aG2(ξ) + bG4(ξ) + cG6(ξ), (2.5)

where αi, a, b, c and V are arbitrary constants tobe determined provided αn 6= 0. The positiveinteger n can be determined by considering thehomogeneous balance between the highest orderderivatives and the nonlinear terms appearing inEq (2.1) or (2.3).

More precisely, we define the degree of u(ξ)as D[u(ξ)] = n which gives rise to the degree ofother expressions as follows{

D[dqudξq ] = n + q,

D[up(dqudξq )s] = np + s(q + n).

(2.6)

Therefore, we can get the value of n in (2.4).

Step 3. Substituting (2.4) into (2.3) and using Eq(2.5), we obtain polynomials in Gj(ξ), G′(ξ)Gj(ξ)(j = 0,±1,±2, · · · ). Equating each coefficient ofthe resulted polynomials to zero, yields a set ofalgebraic equations for αi, a, b, c and V whichcan be solved by Maple or Mathematica.

Step 4. The general solutions of the auxiliaryequation (2.5) have been well known (see, for ex-ample [35, 47]) which can be written in the form

No G(ξ)

1[

−a b sech2(√

aξ)b2−ac(1+ε tanh(

√aξ))2

]1/2, a > 0

2[

a b csch2(√

aξ)b2−ac(1+ε coth(

√aξ))2

] 1/2, a > 0

3[

2aε√

∆ cosh(2√

aξ)−b

]1/2, a > 0,∆ > 0.

4[

2aε√

∆ cos(2√−aξ)−b

]1/2, a < 0,∆ > 0

5[

2aε√−∆ sinh(2

√aξ)−b

]1/2, a > 0,∆ < 0

6[

2aε√

∆ sin(2√−aξ)−b

]1/2, a < 0,∆ > 0

7[

−a sech2(√

aξ)b+2ε

√ac tanh(

√aξ)

]1/2, a > 0, c > 0

8[

−a sec2(√−aξ)

b+2ε√−ac tan(

√−aξ)

]1/2, a < 0, c > 0

9[

a csch2(√

aξ)b+2ε

√ac coth(

√aξ)

]1/2, a > 0, c > 0

10[

−a csc2(√−aξ)

b+2ε√−ac cot(

√−aξ)

] 1/2, a < 0, c > 0

11[−a

b (1 + ε tanh(12

√aξ)

]1/2, a > 0,∆ = 0

12[−a

b (1 + ε coth(12

√aξ)

]1/2, a > 0,∆ = 0

13[

aae2ε√

aξ

(e2ε√

aξ−4b)2−64ac

]1/2, a > 0,

14[

εae2ε√

aξ

1−64ace4ε√

aξ

]1/2, a > 0, b = 0

where ∆ = b2 − 4ac and ε = ±1.



Step 5. Substituting αi, V and the general solu-tion of Eq (2.5) into (2.4) we have many exacttraveling wave solutions of the nonlinear partialdifferential equation (2.1).

3 Some applications

In this section, we apply the further improved(G′

G )- expansion method to construct the exacttraveling wave solutions for one dimensional mod-ified Kawahara equation, which are very impor-tant nonlinear evolution equations in the mathe-matical physics and have been paid attention bymany researchers.

3.1 Example 1. On solving the mod-ified Kawahara equation by a fur-ther improved (G′

G)- expansion

It is well known [12] that the modified Kawaharaequation has the form:

ut + αuux + βuxxx − γuxxxxx = 0, (3.1)

where α, β and γ are arbitrary constants. Thisequation has been derived by Kawahara [12] as amodel for water waves in the long- wave regimefor moderate values of surface tension. The Kawa-hara equation (3.1) gives an appropriate descrip-tion of several phenomena observed in the dynam-ics of the water-wave problem.

Let us now solve Eq. (3.1) by the proposedmethod. To this end, we see that the travel-ing wave variable (2.2) permits us converting Eq.(3.1) into the following ODE:

C − V u +12αu2 + βu′′ − γu(4) = 0, (3.2)

where C is a constant of integration. Consideringthe homogeneous balance between the highest or-der derivative and the nonlinear term in (3.2), wededuce from (2.6) that D(u(4)) = D(u2). There-fore n + 4 = 2n and hence n = 4. Thus, we get

u(ξ) = α4

(G′

G

)4

+ α3

(G′

G

)3

+ α2

(G′

G

)2

+α1

(G′

G

)+ α0. (3.3)

Substituting (3.3) into (3.2), collecting all theterms of powers of

(G′

G

)and setting each coef-

ficient to zero, we get the following system of al-gebraic equations:

−V α3a + αα3a2α2 + αα3aα0

+αα1α2a + αα3a3α4 + αα1α4a

2

−V α1 + αα1α0 = 0,

2αα24b

3c + 3αα4bc2α2 + 168βα4c

2b

−31008γα4c2ab +

32αα2

3c2b− 3072γα2c

2b

−8736γα4b3c + 6αα2

4abc2 = 0,

12αα2

4b4 − V α4c

2 − 840γα4b4

+32αα2

3b2c + 3αα4b

2α2c + αα4c2α0

+12αα2

2c2 + 3αα2

4a2c2 + 112βα4c

2a

+24βα2c2 + 3αα4ac2α2 − 1320γα2b

2c

+αα3c2α1 + 6αα2

4ab2c− 12960γα4b2ac

+108βα4b2c− 1920γα2c

2a− 7936γα4c2a2

+32αα2

3c2a = 0,

−2V α4bc− 1280γα4b3a + αα2

2bc

+2αα4bcα0 + 2αα3bα1c− 120γα2b3

+12αα2

3b3 + 6αα4abα2c− 5312γα4a

2bc

−1360γα2bac + 3αα23abc + 6αα2

4a2bc

+28βα2bc + 2αα24ab3 + αα4b

3α2

+128βα4abc + 20βα4b3 = 0,

−V α4b2 − V α2c + 6βα2b

2 − 2V α4ac

+αα4b2α0 + 3αα2

4a2b2 +

32αα2

3a2c

+2αα24a

3c + αα22ca + αα2cα0 + αα3b

2α1

+32αα2

3b2a + 3αα4a

2cα2 + 2αα4acα0

+12αα2

2b2 + 28βα4b

2a + 16βα2ca + 32βα4a2c

−120γα2b2a− 496γα4b

2a2 − 256γα2ca2

−512γα4a3c +

12αα2

1c + 3αα4ab2α2

+2αα3aα1c = 0,



4βα2ba− 32γα4a3b + 2αα4abα0 + 8βα4a

2b

+αα22ba− 2V α4ab− 16γα2ba

2 + 2αα24a

3b

+3αα4a2bα2 − V α2b + αα2bα0 +

12αα2

1b

+2αα3aα1b +32αα2

3a2b = 0,

αα4c3α3 − 5760γα3c

3 = 0,

3αα4bc2α3 − 8640γα3c

2b = 0,

2αα3bα2c + 2αα4bcα1 − 288γα1cb

−360γα3b3 + 54βα3bc + αα4b

3α3

−2808γα3abc + 6αα4abα3c = 0,

αα3cα0 + αα1α2c + 8βα1c− 264γα3b2a

+αα3b2α2 − V α3c + 3αα4ab2α3

+2αα4acα1 + 2αα3aα2c + αα4b2α1

+3αα4a2cα3 + 24βα3ac + 12βα3b

2

−128γα1ca− 24γα1b2 − 384γα3a

2c = 0,

6βα3ab− V α3b− 8γα1ba + 2βα1b

+αα1α2b + 2αα3aα2b + αα3bα0

+3αα4a2bα3 + 2αα4abα1 − 24γα3a

2b = 0,

−27120γα4b2c2 − 20480γα4c

3a +12αα2

3c3

+2αα24ac3 + 80βα4c

3 − 1920γα2c3

+αα4c3α2 + 3αα2

4b2c2 = 0,

−13440γα4c4 +

12αα2

4c4 = 0,

−32640γα4c3b + 2αα2

4bc3 = 0,

3αα4b2α3c− 384γα1c

2 + αα3c2α2

+48βα3c2 + αα4c

2α1 − 4224γα3c2a

+3αα4ac2α3 − 3600γα3b2c = 0,

−V α4a2 +

12αα2

3a3 − V α0 + C

+12αα2

0 + αα4a3α2 +

12αα2

4a4

−V α2a + αα2aα0 +12αα2

1a + αα3a2α1

+12αα2

2a2 + αα4a

2α0 = 0.

(3.4)

With the aid of Maple or Mathematica we cansolve the above system (3.4) to obtain the follow-ing sets of solutions:

The set 1.

α4 = 1680 β13αa ,

α2 = −3360 β13α ,

γ = β208a ,

V = αα0 − 1104βa13 ,

C = 1338α [−28704α0αβa + 887040β2a2

+169α20α

2],

α3 = α1 = 0.

(3.5)

The set 2.

α4 = 168D β13αa ,

α2 = −224β(D+5)13α ,

γ = Dβ2080a

V = 165D [65α0αD + 4368βaD + 89280βa],

C = 11690α [−7488α0αβaD + 14784β2a2D

−118560βaα0α− 1196160β2a2

+845α20α

2],

α3 = α1 = 0,(3.6)

where D = −312 ± 3i

2

√31.

For the set 1, we have the following solutions:

u(ξ) =1680 β

13αa

(G′

G

)4

− 3360 β

13α

(G′

G

)2

+ α0,

(3.7)where

ξ = x−[αα0 −

1104βa

13

]t. (3.8)

While for the set 2, we have the following solu-tions:

u(ξ) =168D β

13αa

(G′

G

)4

−224β(D + 5)13α

(G′

G

)2

+α0,

(3.9)where

ξ = x− t

65D[65α0αD + 4368βaD + 89280βa].

(3.10)According to the step 4 of section 2, we have

the following families of exact solutions



Family 1. If a > 0,∆ > 0, then the exactsolution for the set 1 has the form

u =1680 βa

13αtanh4(2

√aξ)

−3360 βa

13αtanh2(2

√aξ) + α0, (3.11)

Family 2. If a < 0, ∆ > 0, then the exactsolution for the set 1 has the form

u =1680βa

13αtan4(2

√−a ξ)

+3360βa

13αtan2(2

√−a ξ) + α0, (3.12)

or

u =1680βa

13αcot4(2

√−a ξ)

+3360βa

13αcot2(2

√−a ξ) + α0. (3.13)

Family 3. If a > 0,∆ < 0, then the exactsolution for the set 1 has the form

u =1680 βa

13αcoth4(2

√a ξ)

− 3360 βa

13αcoth2(2

√a ξ) + α0. (3.14)

Family 4. If a > 0, c > 0 then the exactsolution for the set 1 has the form

u = α0 +105βa

13α[tanh(

√aξ) + coth(

√a ξ)]4

− 840βa

13α[tanh(

√aξ)+coth(

√aξ)]2, (3.15)

or

u = α0 +105 βa3

13αc2csch8(

√aξ)

−840 βa2

13αccsch4(

√aξ). (3.16)

Family 5. If a < 0, c > 0, then the exactsolution for the set 1 has the form

u = α0 +105 βa

13α[tan(

√−aξ)− cot(

√−a ξ)]4

+840βa

13α[tan(

√−aξ)−cot(

√−aξ)]2. (3.17)

Family 6. If a > 0, b = 0, then the exactsolution for the set 1 has the form

u =105β

3328αac2coth4(

εξ

4√

c)

−1680β

299αccoth2(

εξ

4√

c) + α0. (3.18)

Similarly, we can find the exact solutions forthe set 2, using (3.9) and (3.10) which are omittedhere.

3.2 Example 2. On solving the modi-fied Kawahara equation by the ex-tended tanh-function method

With reference to the well known extended tanh-function method [1, 7, 8, 32, 34, 39, 46], the so-lution of the equation (3.1) can be written in theform:

u(ξ) = α4φ4(ξ) + α3φ

3(ξ) + α2φ2(ξ)

+α1φ(ξ) + α0, (3.19)

where φ(ξ) satisfies the Riccati equation

φ′(ξ) = R + φ2(ξ) (3.20)

The Riccati equation (3.20) have the following so-lutions: (i) If R < 0, then

φ(ξ) = −√−R tanh(

√−Rξ)

or

φ(ξ) = −√−R coth(

√−Rξ)

(3.21)

(ii) If R > 0, then

φ(ξ) =√

R tan(√

Rξ),

or

φ(ξ) = −√

R cot(√

Rξ)

(3.22)

(iii) If R = 0, then

φ(ξ) =−1ξ

. (3.23)

Substituting (3.19) along with (3.20) into (3.1) weget the following polynomial:

(−840γα4 +12αα2

4)φ8 + (α3αα4 − 360γα3)φ7

+ (20βα4+12αα2

3+αα2α4−2080γα4R−120γα2)φ6

+ (αα2α3−816γα3R−24γα1+12βα3+αα1α4)φ5

+ [12αα2

2 + αα1α3 + αα4α0 + 6βα2

−240γα2R + 32βα4R− 1696γα4R2 − V α4]φ4

+ [−40γα1R + αα1α2 + 18βα3R− V α3

−576γα3R2 + αα3α0 + 2βα1]φ3

+ [8βα2R + 12βα4R2 − 480γα4R

3 − 136γα2R2

−V α2 + αα2α0 +12αα2

1]φ2

+ [−V α1 + 2βα1R− 120γα3R3

−16γα1R2 + αα1α0 + 6βα3R

2]φ

− V α0 + C1 − 16γα2R3 +

12αα2

0



−24γα4R4 + 2βα2R

2 = 0 (3.24)

Equating the coefficients of this polynomial tozero and solving the algebraic equations by Mapleor Mathematica, we have the following two sets ofsolutions:

The set 3

α4 = −420β13αR ,

α2 = −840β13α ,

α3 = α1 = 0,

γ = −β52R ,

V = αα0 + 276βR13 ,

C = 1338α [7176α0αβR + 55440R2β2

+169α20α

2].

(3.25)

The set 4

α4 = 42D1 β13αR ,

α2 = 56 β(D1−5)13α ,

α3 = α1 = 0,

γ = D1β520R ,

V = 165D1

[−1092D1Rβ + 65α0αD1 + 22320βR],

C = 11690α [−1872α0αβRD1 − 924D1β

2R2

+29640βRα0α− 74760β2R2 + 845α20α

2],(3.26)

where D1 = 312 ± 3i

2

√31.

Thus, the exact solutions of the modi-fied Kawahara equation (3.1) have the followingforms:

For the set 3 we deduce for R < 0 that

u = −420 βR

13αtanh4(

√−Rξ)

+840 βR

13αtanh2(

√−Rξ) + α0, (3.27)

or

u = −420 βR13α coth4(

√−Rξ)

+840 βR13α coth2(

√−Rξ) + α0, (3.28)

while for R > 0 we deduce that

u = −420 βR13α tan4(

√Rξ)

−840 βR13α tan2(

√Rξ) + α0, (3.29)

or

u = −420 βR13α cot4(

√Rξ)

−840 βR13α cot2(

√Rξ) + α0, (3.30)

where

ξ = x− (αα0 +276βR

13)t

Similarly, we can write down the exact solu-tions for the set 4, which are omitted here. Fromthe previous results, we have the following re-marks:

Remark 1 If we put R = −4a where a > 0then the results (3.11) and (3.14) are equivalentto the results (3.27) and (3.28) respectively.

Remark 2 If we put R = −4a wherea < 0 thenthe results (3.12) and (3.13) are equivalent to theresults (3.29) and (3.30) respectively. From theseremarks we have the following observation:

The exact solutions of the modified Kawaharaequation obtained using the extended tanh- func-tion method are equivalent to its exact solutionsobtained using the further improved (G′

G )- expan-sion method.

Remark 3 These solutions have been checkedwith Maple by putting them back into the origi-nal equation.

4 Conclusions

In summary, we have found the exact solutions ofthe (1+1)-dimensional modified Kawahara equa-tion (3.1) using two methods via the further im-proved ( G′

G )-expansion method and the extendedtanh-function method. We have arrived at theobservation that these exact solutions are equiva-lent.

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