ELLIPTIC FUNCTION SOLUTIONS FOR SOME NONLINEAR PDES IN

Journal of Applied Analysis and Computation Volume 7, Number 1, February 2017, 372–391

Website:http://jaac-online.com/ DOI:10.11948/2017024

ELLIPTIC FUNCTION SOLUTIONS FOR SOME NONLINEAR PDES IN MATHEMATICAL PHYSICS Yusuf Gurefe1 , Yusuf Pandir2,† , Tolga Akturk3 and Hasan Bulut3 Abstract In this work, we have constructed various types of soliton solutions of the generalized regularized long wave and generalized nonlinear KleinGordon equations by the using of the extended trial equation method. Some of the obtained exact traveling wave solutions to these nonlinear problems are the rational function, 1-soliton, singular, the elliptic integral functions F, E, Π and the Jacobi elliptic function sn solutions. Also, all of the solutions are compared with the exact solutions in literature, and it is seen that some of the solutions computed in this paper are new wave solutions. Keywords Extended trial equation method, generalized regularized long wave, generalized Klein-Gordon equation, soliton solutions, elliptic solutions. MSC(2010) 35C07, 35C08, 37K40.

1. Introduction The nonlinear physical phenomena have very importance in various fields of physics and applied mathematics. Recently, a lot of methods have been defined for getting new wave solutions to nonlinear partial differential equations that model to express these phenomena. Some of them are Hirota method, Exp-function method, (G/G)expansion method, ansatz method, mapping method, Kudryashovs method, multiple and double exp-function methods, three wave method, superposition method, sine-cosine method, asymptotic method [1, 3, 5, 8, 10, 13–15, 19–21, 25–31]. Moreover, the trial equation method is proposed to solve nonlinear wave equations by Liu [16–18]. After this method, Du has defined a new version of the trial equation method called as irrational trial equation method. Also, the trial equation method is modified by Gurefe et al. [12] and given two important applications of this method. Then, the extended trial method, which is a more general form of the trial equation methods, is constructed by Pandir et al. [4, 11, 22–24]. The elliptic integral functions and Jacobi elliptic functions solutions are determined by the applying of this method to some nonlinear partial differential equations. These new solutions appear in various fields of physics. In Section 2, we explain the extended trial equation method as detail. In Section 3, as applications, we obtain some exact † the

corresponding author. Email address:[email protected](Y. Pandir) of Econometrics, Faculty of Economics and Administrative Sciences, Usak University, 64200 Usak, Turkey 2 Department of Mathematics, Faculty of Science and Arts, Bozok University, 66100 Yozgat, Turkey 3 Department of Mathematics, Faculty of Science, Firat University, 23100 Elazig, Turkey 1 Department

Elliptic function solutions

373

solutions to the generalized regularized long wave given by [2, 9] ut + ux + δuq ux − γuxxt = 0,

q ∈ Z+ ,

(1.1)

where δ and γ are positive constants that describe the behavior of the undular bore, and generalized nonlinear Klein-Gordon equation [7] utt − a2 uxx + αu − βuγ = 0,

γ ∈ Z+ ,

(1.2)

where a is positive constant and α, β 6= 0.

2. The extended trial equation method Step 1. For a given nonlinear partial differential equation P (u, ut , ux , uxx , . . . ) = 0,

(2.1)

take the wave transformation   N X η = λ xj − ct ,

u(x1 , x2 , . . . , xN , t) = u(η),

(2.2)

j=1

where λ 6= 0 and c 6= 0. Substituting Eq. (2.2) into Eq. (2.1) yields a nonlinear ordinary differential equation, N (u, u0 , u00 , ...) = 0.

(2.3)

Step 2. Take transformation and trial equation as follows: u=

δ X

τi Γi ,

(2.4)

i=0

where (Γ0 )2 = Λ(Γ) =

ξθ Γθ + ... + ξ1 Γ + ξ0 Φ(Γ) = . Ψ(Γ) ζ Γ + ... + ζ1 Γ + ζ0

(2.5)

Using the relations 2.4 and 2.5, we can find δ X

Φ(Γ) (u ) = Ψ(Γ) 0 2

Φ0 (Γ)Ψ(Γ) − Φ(Γ)Ψ0 (Γ) u = 2Ψ2 (Γ) 00

δ X i=0

!2 i−1

iτi Γ

,

(2.6)

i=0

! i−1

iτi Γ

Φ(Γ) + Ψ(Γ)

δ X

! i(i − 1)τi Γ

i−2

, (2.7)

i=0

where Φ(Γ) and Ψ(Γ) are polynomials. Substituting these terms into Eq. (2.3) yields an equation of polynomial Ω(Γ) of Γ : Ω(Γ) = %s Γs + ... + %1 Γ + %0 = 0.

(2.8)

According to the balance principle we can determine a relation of θ, , and δ. We can take some values of θ, , and δ.

374

H. Bulut, Y. Pandir, Y. Gurefe & T. Akturk

Step 3. Let the coefficients of Ω(Γ) all be zero will yield an algebraic equations system: %i = 0, i = 0, ..., s. (2.9) Solving this equations system (2.9), we will determine the values of ξ0 , ..., ξθ ; ζ0 , ..., ζ and τ0 , ..., τδ . Step 4. Reduce Eq. (2.5) to the elementary integral form, Z ± (η − η0 ) =

dΓ p = Λ(Γ)

Z s

Ψ(Γ) dΓ. Φ(Γ)

(2.10)

Using a complete discrimination system for polynomial to classify the roots of Φ(Γ), we solve the infinite integral (2.10) and with the help of MATHEMATICA and classify the exact approximate solutions to Eq (2.1) respectively.

3. Application to the extended trial equation method In this section, we apply the method developed in Section 2 to the generalized regularized long wave (GRLW ) and the generalized nonlinear Klein-Gordon equation, respectively.

3.1. Application to the GRLW equation In order to look for travelling wave solutions of eq. (1.1), we make the transformation u(x, t) = u(η), η = kx+wt, where w is an arbitrary constant. Then, integrating the resulting equation with respect to η and setting the integration constant to zero, we obtain δk q+1 u − γwk 2 u00 = 0. (3.1) wu + ku + q+1 Using the transformation 1

u = vq,

(3.2)

Eq. (3.1) turns into the equation q 2 (q + 1)(w + k)v 2 + δkq 2 v 3 − γwk 2 (1 − q)(1 + q)(v 0 )2 − γwk 2 q(q + 1)vv 00 = 0. (3.3) Substituting Eqs. (2.6) and (2.7) into Eq. (3.3), and using the balance principle, we find θ = + δ + 2. After this solution procedure, we obtain the results as follows: Case 1: If we take = 0, δ = 1 and θ = 3, then τ12 (ξ3 Γ3 + ξ2 Γ2 + ξ1 Γ + ξ0 ) , ζ0 τ1 (3ξ3 Γ2 + 2ξ2 Γ + ξ1 ) v 00 = , 2ζ0 (v 0 )2 =

(3.4) (3.5)


375

where ξ3 6= 0, ζ0 6= 0. Respectively, solving the algebraic equation system (2.9) yields ξ3 q2 ξ12 q(q − 4) , ξ3 = − 2 1 , τ0 = τ0 , 2 4ξ0 (q − 1) 4ξ0 (q − 1)3 k(1 + q + δτ0 ) qξ1 τ0 k 2 (q + 2)γξ12 (1 + q + δτ0 ) , w=− τ1 = , ζ0 = . 2ξ0 (q − 1) 4qδξ0 τ0 (q − 1)2 q+1

ξ0 = ξ0 , ξ1 = ξ1 , ξ2 =

Substituting these results into Eqs. (2.5) and (2.10), we get √ Z dΓ ± (η − η0 ) = A q 4ξ 2 (q−1)3 ξ (q−1)(q−4) Γ3 − 0 ξ1 q Γ2 − 0ξ2 q2 Γ − 1

4ξ03 (q−1)3 ξ13 q 2

,

(3.6) (3.7)

(3.8)

2 0 (1+q+δτ0 ) . − k (q−1)(q+2)γξ δq 3 ξ1 τ0

Integrating eq. (3.8), we obtain the solutions where A = to the eq. (1.1) as follows: √ 1 ± (η − η0 ) = −2 A √ , (3.9) Γ − α1 r r A Γ − α2 arctan , α2 > α1 , (3.10) ± (η − η0 ) = 2 α2 − α1 α2 − α1 r √ √ Γ − α2 − α1 − α2 A , α1 > α2 , (3.11) ln √ ± (η − η0 ) = √ α1 − α2 Γ − α2 + α1 − α2 r A ± (η − η0 ) = 2 F (ϕ1 , l1 ), α1 > α2 > α3 , (3.12) α1 − α3 where

Z F (ϕ1 , l1 ) = 0

and

ϕ1

dψ q

,

(3.13)

1 − l12 sin2 ψ

r

Γ − α3 α2 − α3 , l12 = . α2 − α3 α1 − α3 Also α1 , α2 and α3 are the roots of the polynomial equation ϕ1 = arcsin

Γ3 +

ξ0 ξ2 2 ξ1 Γ + Γ+ = 0. ξ3 ξ3 ξ3

(3.14)

(3.15)

Substituting the solutions (3.9-3.12) into (2.4) and (3.2), we can find the following exact traveling wave solutions such as rational function solution, hyperbolic function solutions and Jacobi elliptic function solutions to Eq. (1.1), respectively:   q1 4τ1 A   u(x, t) = τ0 + τ1 α1 + (3.16) 2  , 0) kx − k(1+q+δτ t − η 0 q+1 q1 1 + q + δτ0 η0 2 u(x, t) = τ0 + τ1 α1 + τ1 (α2 − α1 ) tanh B x − t− , q+1 k (3.17) q1 1 + q + δτ0 η0 u(x, t) = τ0 + τ1 α1 + τ1 (α1 − α2 )cosech2 B x − t− , q+1 k (3.18)

376


and

Figure 1. The solution (3.20) is shown at τ0 = τ1 = α1 = ξ0 = ξ1 = δ = γ = k = 1, η0 = 0, and the second graph represents the exact approximate solution of Eq. (3.20) for t = 1.

u(x, t) = where B =

q=2

q1 η0 k(1 + q + δτ0 ) t− , τ0 + τ1 α3 + τ1 (α2 − α3 )sn Bi kx − , l12 q+1 k (3.19) q q 2

k 2

α1 −α2 A

and Bi =

(−1)i k 2

α1 −α3 A , (i

= 1, 2). If we take τ0 = −τ1 α1

Figure 2. The solution (3.21) is shown at τ0 = τ1 = α1 = ξ0 = ξ1 = δ = γ = k = 1, η0 = 0, α2 = q = 2 and the second graph represents the exact approximate solution of Eq. (3.21) for t = 1.

and η0 = 0 for simplicity, then the solutions (3.16)-(3.18) can reduce to rational function solution p ! q2 2 A˜ u(x, t) = , (3.20) k(x − vt) 1-soliton solution u(x, t) =

A1 2 q

cosh [B(x − vt)]

,

(3.21)

,

(3.22)

singular soliton solution u(x, t) =

A2 2 q

sinh [B(x − vt)]

1 1 1 α1 where A˜ = τ1 A, A1 = (τ1 (α2 − α1 )) q , A2 = (τ1 (α1 − α2 )) q and v = 1+q−δτ . q+1 Here, A1 and A2 are the amplitudes of the solitons, while v is the velocity and B is the inverse width of the solitons. Thus, we can say that the solitons exist for


377

Figure 3. The solution (3.23) is shown at τ0 = τ1 = α1 = ξ0 = ξ1 = δ = γ = k = 1, η0 = 0, α3 = q = 2, α1 = 3 and the second graph represents the exact approximate solution of Eq. (3.23) for t = 1.

τ1 > 0. On the other hand, if we take τ0 = −τ1 α3 and η0 = 0, the Jacobi elliptic function solution (3.19) can be written in the form α2 − α3 ui (x, t) = A3 sn Bi (x − vt), , α1 − α3 2 q

1

where A3 = (τ1 (α2 − α3 )) q and Bi =

(−1)i k 2

q

α1 −α3 A ,

(3.23)

(i = 1, 2).

Remark 3.1. Substituting the exact solutions (3.20)-(3.23), that the extended trial equation method gives us, into Eq. (1.1), it is seen that these solutions provide Eq. (1.1). Also, rational function and Jacobi elliptic function solutions are not found in the previous literature. Remark 3.2. When the modulus l1 → 1, the solution (3.23) can be converted into dark soliton solutions of the generalized regularized long wave equation 2

ui (x, t) = A3 tanh q [Bi (x − vt)] , (i = 1, 2),

(3.24)

where α1 = α2 , and v represents the velocity of the dark soliton. Case 2 If we take = 0, δ = 2 and θ = 4, then (τ1 + 2τ2 Γ)2 (ξ4 Γ4 + ξ3 Γ3 + ξ2 Γ2 + ξ1 Γ + ξ0 ) , ζ0 (τ1 + 2τ1 Γ)(4ξ4 Γ3 + 3ξ3 Γ2 + 2ξ2 Γ + ξ1 ) v 00 = 2ζ0 4 3 2τ2 (ξ4 Γ + ξ3 Γ + ξ2 Γ2 + ξ1 Γ + ξ0 ) + , ζ0

(v 0 )2 =

(3.25) (3.26) (3.27)

where ξ4 6= 0, ζ0 6= 0. Respectively, solving the algebraic equation system (2.9)

378


yields as follows:

ξ0 ξ2 ξ3 ζ0

q 2 δζ0 τ14 + 4k 2 γξ1 τ1 τ2 δτ12 + 2τ2 (q + 1)(q + 2) τ12 , ξ = ξ , τ = = , 1 1 0 32k 2 γτ22 (δτ12 + τ2 (q + 1)(q + 2)) 4τ2 2k 2 γξ1 τ1 τ2 3δτ12 + τ2 (q + 1)(q + 2) − q 2 δζ0 τ13 , τ1 = τ1 , τ2 = τ2 , = 2k 2 γτ1 (δτ12 + τ2 (q + 1)(q + 2)) δτ22 4k 2 δξ1 τ2 − q 2 ζ0 τ1 δτ2 4k 2 δξ1 τ2 − q 2 ζ0 τ1 , ξ4 = 2 , = 2 k γ (δτ12 + τ2 (q + 1)(q + 2)) 2k γτ1 (δτ12 + τ2 (q + 1)(q + 2)) kq 2 ζ0 τ1 δτ12 + τ2 (q + 1)(q + 2) = ζ0 , w = − . (3.28) τ2 (q + 1)(q + 2) (q 2 ζ0 τ1 − 4k 2 δξ1 τ2 )

Substituting these results into Eqs. (2.5) and (2.10), we get Z ± (η − η0 ) = C

r where C =

dΓ q

Γ4

+

2k2 γτ1 ζ0 (δτ12 +τ2 (q+1)(q+2)) . δτ22 (4k2 δξ1 τ2 −q 2 ζ0 τ1 )

ξ3 3 ξ4 Γ

+

ξ2 2 ξ4 Γ

+

ξ1 ξ4 Γ

+

ξ0 ξ4

,

(3.29)

Integrating Eq. (3.29), we obtain the

solutions to the eq. (1.1) as follows: C , Γ − α1 r Γ − α2 2C , α1 > α2 , α1 − α2 Γ − α1 Γ − α1 C , ln α1 − α2 Γ − α2 Y1 (Γ) − G1 (Γ) 2C , α1 > α2 > α3 , p ln Y1 (Γ) − G1 (Γ) (α1 − α2 )(α1 − α3 ) 2C p F (ϕ2 , l2 ), α1 > α2 > α3 > α4 , (α1 − α3 )(α2 − α4 )

± (η − η0 ) = −

(3.30)

± (η − η0 ) =

(3.31)

± (η − η0 ) = ± (η − η0 ) = ± (η − η0 ) =

(3.32) (3.33) (3.34)

where Y1 (Γ) =

p

(Γ − α2 )(α1 − α3 ), G1 (Γ) =

p

(Γ − α3 )(α1 − α2 )

and s ϕ2 = arcsin

(Γ − α1 )(α2 − α4 ) (α2 − α3 )(α1 − α4 ) , l22 = . (Γ − α2 )(α1 − α4 ) (α1 − α3 )(α2 − α4 )

(3.35)

Also α1 , α2 , α3 and α4 are the roots of the polynomial equation Γ4 +

ξ3 3 ξ2 2 ξ1 ξ0 Γ + Γ + Γ+ = 0. ξ4 ξ4 ξ4 ξ4

(3.36)


379

Figure 4. The solution (3.45) is shown at τ0 = τ2 = α1 = ξ0 = ζ0 = δ = γ = q = k = 1, η0 = 0, ξ1 = τ1 = −1, α2 = 2 and the second graph represents the exact approximate solution of Eq. (3.45) for t = 1.

Substituting the solutions (3.30)-(3.34) into (2.4) and (3.2), we have 2 # q1 τ1 C C + τ2 α1 ± , (3.37) u(x, t) = τ0 + τ1 α1 ± kx + wt − η0 kx + wt − η0 ( 4C 2 (α2 − α1 )τ1 (3.38) u(x, t) = τ0 + τ1 α1 + 2 4C 2 − [(α1 − α2 )(kx + wt − η0 )] !2 ) q1 4C 2 (α2 − α1 ) + τ2 α1 + , 2 4C 2 − [(α1 − α2 )(kx + wt − η0 )] ( (α2 − α1 )τ1 u(x, t) = τ0 + τ1 α2 + (3.39) exp B3 x − vt − ηk0 − 1 !2 ) q1 α2 − α1 , + τ2 α2 + exp B3 x − vt − ηk0 − 1 ( (α1 − α2 )τ1 u(x, t) = τ0 + τ1 α1 + (3.40) exp B3 x − vt − ηk0 − 1 !2 ) q1 α1 − α2 + τ2 α1 + , exp B3 x − vt − ηk0 − 1 ( 2(α1 − α2 )(α1 − α3 )τ1 (3.41) u(x, t) = τ0 + τ1 α1 − 2α1 − α2 − α3 + (α3 − α2 ) cosh D x − vt − ηk0 !2 ) q1 2(α1 − α2 )(α1 − α3 ) + τ2 α1 − , 2α1 − α2 − α3 + (α3 − α2 ) cosh D x − vt − ηk0 ( τ1 (α1 − α2 )(α4 − α2 ) u(x, t) = τ0 + τ1 α2 + (3.42) α4 − α2 + (α1 − α4 )sn2 E x − vt − ηk0 , l22 !2 ) q1 τ1 (α1 − α2 )(α4 − α2 ) + τ2 α2 + , α4 − α2 + (α1 − α4 )sn2 E x − vt − ηk0 , l22 "

380


where B3 =

k(α1 −α2 ) , C

D =

k

√

(α1 −α2 )(α1 −α3 ) , C

E =

k

√

(α1 −α3 )(α2 −α4 ) 2C

and v =

Figure 5. The solution (3.46) is shown at τ0 = τ2 = α1 = ξ0 = ζ0 = δ = γ = q = k = 1, η0 = 0, ξ1 = τ1 = −1, α2 = 2, α3 = 3 and the second graph represents the exact approximate solution of Eq. (3.46) for t = 1.

Figure 6. The solution (3.47) is shown at τ2 = α1 = ξ0 = ζ0 = δ = γ = q = k = 1, η0 = 0, ξ1 = τ1 = −1, τ0 = α2 = 12 , α3 = 3, α4 = 2 and the second graph represents the exact approximate solution of Eq. (3.47) for t = 1. q 2 ζ0 τ1 (δτ12 +τ2 (q+1)(q+2)) τ2 (q+1)(q+2)(q 2 ζ0 τ1 −4k2 δξ1 τ2 ) .

For simplicity, if we take η0 = 0, then we can write the solutions (3.37)-(3.42) as follows: " u(x, t) =

2 X

τi

i=0

 2 X u(x, t) =  τi

α1 +

i=0

" u(x, t) =

2 X

τi

i=0

" u(x, t) =

2 X i=0

τi

C α1 ± k(x − vt)

i # q1 ,

(3.43) !i  q1

4C 2 (α1 − α2 ) 2

4C 2 − [(α1 − α2 )k(x − vt)]

α2 − α1 α2 + exp [B3 (x − vt)] − 1 α1 − α2 α1 + exp [B3 (x − vt)] − 1

 ,

(3.44)

i # q1 ,

(3.45)

,

(3.46)

i # q1


381

i # q1 2(α1 − α2 )(α1 − α3 ) u(x, t) = , (3.47) τi α1 − 2α1 − α2 − α3 + (α3 − α2 ) cosh [D (x − vt)] i=0 " 2 i # q1 X (α1 − α2 )(α4 − α2 ) u(x, t) = . (3.48) τi α2 + α4 − α2 + (α1 − α4 )sn2 [E(x − vt), l2 ] i=0 "

2 X

Here, C is the amplitude of the soliton, while v is the velocity and D and E are the inverse width of the solitons. Remark 3.3. To our knowledge, the traveling wave solutions (3.43)-(3.48), that we find in this paper, are not shown in the previous literature. These are new exact solutions of Eq. (1.1).

3.2. Application to the generalized nonlinear Klein-Gordon equation In order to look for traveling wave solutions of Eq. (1.2), we make the transformation u(x, t) = u(η), η = k(x−ct), where k and c are an arbitrary constant. Consequently, Eq. (1.2) is reduced to the ODE k 2 (c2 − a2 )u00 (η) + αu − βuγ = 0.

(3.49)

Making use of the transformation: 2

u = v γ−1 ,

(3.50)

Eq. (3.49) converts to the nonlinear ODE: 2k 2 (c2 −a2 )(γ−1)vv 00 +2k 2 (c2 −a2 )(3−γ)(v 0 )2 +α(γ−1)2 v 2 −β(γ−1)2 v 4 = 0. (3.51) Substituting Eqs. (2.6) and (2.7) into Eq. (3.51), and using balance principle, we obtain θ = + 2δ + 2. After this solution procedure, we obtain the results as follows: Case 1: If we take = 0, δ = 1 and θ = 4, then τ12 (ξ4 Γ4 + ξ3 Γ3 + ξ2 Γ2 + ξ1 Γ + ξ0 ) , ζ0 τ1 (4ξ4 Γ3 + 3ξ3 Γ2 + 2ξ2 Γ + ξ1 ) , v 00 = 2ζ0

(v 0 )2 =

(3.52) (3.53)

where ξ4 6= 0, ζ0 6= 0. Respectively, solving the algebraic equation system (2.9) yields ξ4 τ02 α + γα − 2βτ02 ξ4 τ0 4βτ02 − α(γ + 1) 4ξ4 τ0 ξ0 = − , ξ1 = , ξ3 = , (3.54) 2βτ12 βτ13 τ1 s ξ4 α + γα − 12βτ02 2a2 k 2 (γ + 1)ξ4 + β(γ − 1)2 ζ0 τ12 ξ2 = − , c = ± , (3.55) 2βτ12 2k 2 (γ + 1)ξ4

382


where ξ4 , ζ0 , τ0 and τ1 are free parameters. Substituting these results into Eqs. (2.5) and (2.10), we can write Z dΓ q , (3.56) ± (η − η0 ) = H ξ Γ4 + ξ34 Γ3 + ξξ42 Γ2 + ξξ14 Γ + ξξ40 q τ0 (4βτ02 −α(γ+1)) α+γα−12βτ02 ξ3 ξ2 ζ0 ξ1 4τ0 where H = = , , ξξ40 = 2 ξ4 , ξ4 = τ 1 , ξ4 = − ξ 2βτ1 βτ13 4 τ 2 (α+γα−2βτ 2 ) − 0 2βτ 2 0 . Integrating Eq. (3.56), we obtain the solutions to the Eq. (1.2) 1 as follows: H , Γ − α1 r 2H Γ − α2 , α2 > α1 , α1 − α2 Γ − α1 Γ − α1 H , ln α1 − α2 Γ − α2 Y2 (Γ) − G2 (Γ) H , α1 > α2 > α3 , p ln Y2 (Γ) + G2 (Γ) (α1 − α2 )(α1 − α3 ) 2H p F (ϕ3 , l3 ), α1 > α2 > α3 > α4 , (α1 − α3 )(α2 − α4 )

± (η − η0 ) = −

(3.57)

± (η − η0 ) =

(3.58)

± (η − η0 ) = ± (η − η0 ) = ± (η − η0 ) =

(3.59) (3.60) (3.61)

where Y2 (Γ) =

p

(Γ − α2 )(α1 − α3 ),

G2 (Γ) =

p (Γ − α3 )(α1 − α2 )

and s ϕ3 = arcsin

(Γ − α1 )(α2 − α4 ) , (Γ − α2 )(α1 − α4 )

l32 =

(α2 − α3 )(α1 − α4 ) . (α1 − α3 )(α2 − α4 )

(3.62)

Also α1 , α2 , α3 and α4 are the roots of the polynomial equation Γ4 +

ξ3 3 ξ2 2 ξ1 ξ0 Γ + Γ + Γ+ = 0. ξ4 ξ4 ξ4 ξ4

(3.63) 2

Substituting the solutions (3.57)-(3.61) into (2.4) and u = v γ−1 , we obtain the following traveling wave solutions of Eq. (1.2), respectively: 2 γ−1 τ1 H u(x, t) = τ0 + τ1 α1 ± , k (x − ct) − η0 2 ( ) γ−1 4H 2 (α2 − α1 )τ1 u(x, t) = τ0 + τ1 α1 + , 2 4H 2 − [(α1 − α2 )k (x − ct − η0 )] 2   γ−1   (α2 − α1 )τ1 h i u(x, t) = τ0 + τ1 α2 + , k(α1 −α2 )  exp (x − ct − η ) − 1 

u(x, t) =



(α1 − α2 )τ1

τ0 + τ1 α1 + exp

h

(3.65)

(3.66)

0

H

 

(3.64)

k(α1 −α2 ) H

2  γ−1 

i (x − ct − η0 ) − 1 

,

(3.67)


τ0 + τ1 α1 −

u(x, t) =

383

2(α1 − α2 )(α1 − α3 )τ1 2α1 − α2 − α3 + (α3 − α2 ) cosh [A4 (x − ct)]

2 γ−1

, (3.68)

2 γ−1 τ1 (α1 − α2 )(α4 − α2 ) , (3.69) α4 − α2 + (α1 − α4 )sn2 [A5 (x − ct − η0 ) , l32 ] √ √ k (α1 −α2 )(α1 −α3 ) k (α1 −α3 )(α2 −α4 ) where A4 = and A = ∓ . If we take τ0 = 5 H 2H

u(x, t) =

τ0 + τ1 α2 +

Figure 7. The solution Eq.(3.70) is shown at τ0 = τ1 = α1 = ξ4 = ζ0 = β = δ = a = k = 1, η0 = 0, α2 = 2, γ = 3 and the second graph represents the exact approximate solution of Eq. (3.70) for t = 1.

−τ1 α1 and η0 = 0, then the solutions (3.64)-(3.68) can reduce to rational function solutions 2 γ−1 Hτ1 u(x, t) = ± , (3.70) k(x − ct) 2 ) γ−1 ( 4H 2 (α2 − α1 )τ1 u(x, t) = , (3.71) 2 4H 2 − [(α1 − α2 )k (x − ct)] traveling wave solutions 2 γ−1 k(α1 − α2 ) (α2 − α1 )τ1 1 ∓ coth (x − ct) u(x, t) = , 2 2H soliton solution u(x, t) =

A6 , 2 γ−1 D1 + cosh A4 (x − ct)

(3.72)

(3.73)

2 γ−1 1 −α3 )τ1 2 −α3 where A6 = 2(α1 −αα23)(α and D1 = 2α1α−α . Here, A6 is the ampli−α2 3 −α2 tude of the soliton, while c is the velocity and A4 is the inverse width of the soliton. Thus, we can say that the solitons exist for τ1 < 0. On the other hand, if we take τ0 = τ1 α2 and η0 = 0, the Jacobi elliptic function solution (3.69) can be written in the form A6 u(x, t) = , (3.74) 2 2 (D2 + sn [A5 (x − ct), l32 ]) γ−1 2 γ−1 2 )(α4 −α2 ) 4 −α2 where A6 = τ1 (α1 −α and D2 = α α1 −α4 α1 −α4 .

384


Figure 8. The solution (3.71) is shown at τ0 = τ1 = α1 = ξ4 = ζ0 = β = δ = a = k = 1, η0 = 0, α2 = 2, γ = 3 and the second graph represents the exact approximate solution of Eq. (3.71) for t = 1.

Remark 3.4. The traveling wave solutions (3.70)-(3.74) of Eq. (1.2) are new solutions that are not computed by any methods in the previous papers. Remark 3.5. When the modulus l3 → 1, then the solution (3.74) can be reduced to the solitary wave solution u(x, t) =

A6 , 2 γ−1 D2 + tanh2 A5 (x − vt)

(3.75)

where α3 = α4 . Remark 3.6. When the modulus l3 → 0, then the solution (3.74) can be converted into the compacton solution u(x, t) =

A6 2 γ−1 D2 + sin A5 (x − vt)

,

(3.76)

2

where α2 = α3 . Case 2: If we take = 1, δ = 1 and θ = 5, then τ12 (ξ5 Γ5 + ξ4 Γ4 + ξ3 Γ3 + ξ2 Γ2 + ξ1 Γ + ξ0 ) , ζ0 + ζ1 Γ τ1 (ζ0 + ζ1 Γ) 5ξ5 Γ4 + 4ξ4 Γ3 + 3ξ3 Γ2 + 2ξ2 Γ + ξ1 − ζ1 Φ5 (Γ)

(v 0 )2 = v 00 =

2

2 (ζ0 + ζ1 Γ)

(3.77) ,

(3.78)

where ξ5 6= 0, ζ1 6= 0 and Φ5 (Γ) = ξ5 Γ5 +ξ4 Γ4 +ξ3 Γ3 +ξ2 Γ2 +ξ1 Γ+ξ0 . Respectively, solving the algebraic equation system (2.9) yields as follows: ξ5 ζ1 α + γα − 12βτ02 − 8βζ0 τ0 τ1 ζ0 ξ5 τ02 α + γα − 2βτ02 ξ0 = − , ξ3 = − , 2βζ1 τ14 2βζ1 τ12 ξ5 τ0 ζ1 τ0 2βτ02 − α(γ + 1) − 2ζ0 τ1 α + γα − 4βτ02 ξ1 = , 2βζ1 τ14 ξ5 2ζ1 τ0 α + γα − 4βτ02 + ζ0 τ1 α + γα − 12βτ02 , ξ2 = − 2βζ1 τ13


ξ4 =

385

s

ξ5 (ζ0 τ1 + 4ζ1 τ0 ) , ζ1 τ1

c=±

2a2 k 2 (γ + 1)ξ5 + β(γ − 1)2 ζ1 τ12 , 2k 2 (γ + 1)ξ5

(3.79)

where ξ5 = ξ5 , ζ0 = ζ0 , ζ1 = ζ1 , τ0 = τ0 , τ1 = τ1 . Substituting these results into Eqs. (2.5) and (2.10), we get s ± (η − η0 ) =

v Z u u t

ζ1 ξ5

ζ0 ζ1 ξ3 3 ξ2 2 Γ + ξ5 ξ5 Γ

Γ+

Γ5 +

ξ4 4 ξ5 Γ

+

+

ξ1 ξ5 Γ

ξ0 ξ5

+

dΓ.

(3.80)

Integrating Eq. (3.80), we obtain the following exact approximate solutions to the Eq. (1.2). When Φ(Γ) = (Γ − α1 )5 , we have 2H1 ± (η − η0 ) = − √ 3 ζ1 (ζ0 + ζ1 α1 )

ζ0 + ζ1 Γ Γ − α1

32 .

(3.81)

If we take Φ(Γ) = (Γ − α1 )4 (Γ − α2 ) and α1 > α2 , then we get "

−H1 ± (η − η0 ) = α1 − α2

#

(ζ0 + ζ1 α2 )

2

p ln |Y3 (Γ)| + G3 (Γ) , ζ1 (α1 − α2 )(ζ0 + ζ1 α1 )

(3.82)

where Γ − α1 , K(Γ) + ζ0 (α1 − 2α2 ) − ζ1 α2 α1 + L(Γ) s (ζ0 + ζ1 Γ)(Γ − α2 ) 1 G3 (Γ) = Γ − α1 ζ1

Y3 (Γ) =

and K(Γ) = (ζ0 + 2ζ1 α1 − ζ1 α2 ) Γ, L(Γ) = 2

p

(ζ0 + ζ1 Γ)(ζ0 + ζ1 α1 )(Γ − α2 )(α1 − α2 ).

When Φ(Γ) = (Γ − α1 )3 (Γ − α2 )2 and α1 > α2 , we obtain s " # −2H1 ζ0 + ζ1 α2 ± (η − η0 ) = Y4 (Γ) + arctan (G4 (Γ)) , α1 − α2 ζ1 (α1 − α2 )

(3.83)

where s Y4 (Γ) =

ζ0 + ζ1 Γ , G4 (Γ) = ζ1 (Γ − α1 )

s

(Γ − α1 )(ζ0 + ζ1 α2 ) . (α1 − α2 )(ζ0 + ζ1 Γ)

If we take Φ(Γ) = (Γ − α1 )2 (Γ − α2 )2 (Γ − α3 ) and α1 > α2 > α3 , then we get ± (η − η0 ) =

−H1 √ [Y5 (Γ) + G5 (Γ)] , (α1 − α3 ) ζ1

(3.84)

386


where α2 − Γ , Y5 (Γ) = Y ln P (Γ) + ζ0 (α2 − 2α3 ) − ζ1 α2 α3 + Q(Γ) R(Γ) + ζ0 (α1 − 2α3 ) − ζ1 α1 α3 + S(Γ) , G5 (Γ) = Z ln Γ − α2 r r ζ0 + ζ1 α2 ζ0 + ζ1 α1 Y = , Z= , α2 − α3 α1 − α3 p P (Γ) = 2 (ζ0 + ζ1 Γ)(ζ0 + ζ1 α2 )(Γ − α3 )(α2 − α3 ), Q(Γ) = (ζ0 + 2ζ1 α2 − ζ1 α3 )Γ, (3.85) p R(Γ) = 2 (ζ0 + ζ1 Γ)(ζ0 + ζ1 α1 )(Γ − α3 )(α1 − α3 ), S(Γ) = (ζ0 + 2ζ1 α1 − ζ1 α3 )Γ. (3.86) When Φ(Γ) = (Γ − α1 )3 (Γ − α2 )(Γ − α3 ) and α1 > α2 > α3 , then we obtain s −2H1 ζ0 + ζ1 α 3 ± (η − η0 ) = E(ϕ4 , l4 ), (3.87) α1 − α3 ζ1 (α1 − α2 ) where ϕ4

Z E(ϕ4 , l4 ) =

q

1 − l42 sin2 ψdψ,

0

s ϕ4 = arcsin

(Γ − α3 )(α2 − α1 ) 2 (α3 − α2 )(ζ0 + ζ1 α1 ) , l4 = . (Γ − α1 )(α2 − α3 ) (α1 − α2 )(ζ0 + ζ1 α3 )

(3.88)

If we take Φ(Γ) = (Γ − α1 )2 (Γ − α2 )(Γ − α3 )(Γ − α4 ) and α1 > α2 > α3 > α3 , then we get ζ0 + ζ1 Γ ζ0 + ζ1 α 2 ± (η − η0 ) = H2 π(ϕ5 , n, l5 ) − F (ϕ5 , l5 ) , (3.89) α1 − α4 α2 − α4 where s H1 =

ζ1 , ξ5

2H (α − α4 ) p 1 2 , (α1 − α2 ) ζ1 (α2 − α3 )(ζ0 + ζ1 α4 ) dψ (α1 − α2 )(α3 − α4 ) q (3.90) , n=− (α2 − α3 )(α1 − α4 ) 2 2 2 (1 + n sin ψ) 1 − l sin ψ

H2 = Z

ϕ5

π(ϕ5 , n, l5 ) = 0

5

and s ϕ5 = arcsin

(Γ − α4 )(α3 − α2 ) , (Γ − α2 )(α3 − α4 )

l52 =

(α4 − α3 )(ζ0 + ζ1 α2 ) . (α2 − α3 )(ζ0 + ζ1 α4 )

(3.91)

Case 3: If we take = 0, δ = 2 and θ = 6, then (τ1 + 2τ2 Γ)2 (ξ6 Γ6 + ξ5 Γ5 + ξ4 Γ4 + ξ3 Γ3 + ξ2 Γ2 + ξ1 Γ + ξ0 ) , (3.92) ζ0 4τ2 (Φ6 (Γ)) + (τ1 + 2τ1 Γ) 6ξ6 Γ5 + 5ξ5 Γ4 + 4ξ4 Γ3 + 3ξ3 Γ2 + 2ξ2 Γ + ξ1 Γ 00 v = , 2ζ0 (3.93) (v 0 )2 =


387

where ξ6 6= 0, ζ0 6= 0 and Φ6 (Γ) = ξ6 Γ6 + ξ5 Γ5 + ξ4 Γ4 + ξ3 Γ3 + ξ2 Γ2 + ξ1 Γ + ξ0 . Respectively, solving the algebraic equation system (2.9) yields as follows: ξ6 τ1 3βτ14 − 8α(γ + 1)τ22 ξ6 τ12 βτ14 − 8α(γ + 1)τ22 , ξ1 = , ξ0 = 64βτ26 16βτ25 ξ6 15βτ14 − 8α(γ + 1)τ22 5ξ6 τ13 15ξ6 τ12 3ξ6 τ1 , ξ = , ξ = , ξ5 = ξ2 = , 3 4 4 2 2 16βτ2 2τ2 4τ2 τ2 s τ2 8a2 k 2 (γ + 1)ξ6 + β(γ − 1)2 ζ0 τ22 τ0 = 1 , c = ± , (3.94) 4τ2 8k 2 (γ + 1)ξ6 where ξ6 , τ1 , τ2 and ζ0 are free parameters. Substituting these results into Eqs. (2.5) and (2.10), we get Z dΓ q , (3.95) ± (η − η0 ) = H3 ξ ξ Γ6 + ξ65 Γ5 + ξ46 Γ4 + ξξ63 Γ3 + ξξ26 Γ2 + ξξ61 Γ + ξξ16 where H3 = as follows:

q

ζ0 ξ6 .

± (η − η0 ) = −

Integrating Eq. (3.95), we obtain the solutions to the Eq. (1.2)

H3 , 2(Γ − α1 )2

(3.96) s

2H3 (2Γ − 3α1 + α2 ) Γ − α2 , α1 > α2 , 2 3(α1 − α2 ) (Γ − α1 )3 Γ−α1 Γ−α2 + 1 − Γ ln H3 α2 − α1 ln Γ−α Γ−α2 1 ± (η − η0 ) = , 2 (Γ − α1 )(α1 − α2 ) −2H3 (2Γ − α1 − α2 ) p ± (η − η0 ) = , (α1 − α2 )2 (Γ − α1 )(Γ − α2 ) Γ−α3 Γ−α1 2 −H3 α1 ln Γ−α + α ln + α ln 2 3 Γ−α3 Γ−α1 Γ−α2 ± (η − η0 ) = , α1 > α2 (α1 − α2 )(α2 − α3 )(α1 − α3 ) ± (η − η0 ) =

± (η − η0 ) = H3

(3.97)

(3.98) (3.99)

> α3 ,

(3.100) ! p (Γ − α1 )(Γ − α3 ) (α1 − 2α2 + α3 )i log (V (Γ)) + , 3 3 (Γ − α2 )(α1 − α2 )(α2 − α3 ) 2(α1 − α2 ) 2 (α1 − α2 ) 2 (3.101)

where p −4(α1 − α2 )(α2 − α3 ) (Γ − α1 )(Γ − α3 ) (Γ − α2 )(α1 − 2α2 + α3 ) p 2i (α1 − α2 )(α2 − α3 )(α1 (Γ + α2 − 2α3 ) − 2Γα2 + α3 (Γ + α2 )) + , (Γ − α2 )(α1 − 2α2 + α3 ) q h i   1 )(α3 −α2 ) 1 2 arctan (Γ−α 2 Γ−α (Γ−α2 )(α1 −α3 ) Γ−α2 , p ± (η − η0 ) = H3  − (α1 − α2 )(α2 − α3 ) (α2 − α3 ) (α1 − α3 )(α3 − α2 )

V (Γ) =

(3.102)

388


± (η − η0 ) = Y6 (G6 (Γ) + M F (ϕ5 , l5 ) + N [(α1 − α4 )F (ϕ5 , l5 ) − (α2 − α4 )E(ϕ5 , l5 )]) , (3.103) where 2H3 , Y6 = (α2 − α3 )(α3 − α4 )

s G6 (Γ) =

(Γ − α2 )(Γ − α4 ) , (Γ − α1 )(Γ − α3 )

1 α1 − α2 + α3 − α4 p , M=p , N= (α1 − α4 )(α2 − α3 ) (α1 − α3 ) (α2 − α4 ) s (Γ − α2 )(α1 − α4 ) (α1 − α3 )(α2 − α4 ) ϕ5 = arcsin , l52 = , (Γ − α1 )(α2 − α4 ) (α2 − α3 )(α1 − α4 ) −H3 (Γ − α3 ) ± (η − η0 ) = Γ − α4

i log (G7 (Γ))

log (Y7 (Γ)) p

(α1 − α4 )(α2 − α4 )

(3.104) !

+p (α1 − α3 )(α3 − α2 ) (3.105)

−Γ+α4 , (Γ−α1 )(Γ−α2 )(α1 −α4 )(α2 −α4 )+(2α4 −α2 −α1 )Γ−α2 α4 +2α1 α2 −α1 α4 √ −(α3 −α4 ) 2 (Γ−α1 )(Γ−α2 )(α1 −α3 )(α3 −α2 )−i(−2Γα3 +α2 (Γ+α3 )+α1 (Γ−2α2 +α3 ))

where Y7 (Γ) = √ 2

G7 (Γ) =

√

± (η − η0 ) = H4

(α1 −α3 )(α3 −α2 )(−Γ+α4 )

F (ϕ6 , l6 ) π(ϕ6 , n, l6 ) + (α2 − α4 ) (α1 − α2 )

,

,

α1 > α2 > α3 > α4 > α5 , (3.106)

where H4 = p

2H3

s ,

ϕ6 = arcsin

(Γ − α2 )(α1 − α5 ) , (Γ − α1 )(α2 − α5 )

(α2 − α3 )(α1 − α5 ) (α (α1 − α4 )(α2 − α5 ) 1 − α3 )(α2 − α5 ) , n= . l62 = (α2 − α3 )(α1 − α5 ) (α2 − α4 )(α1 − α5 )

(3.107)

Also α1 , α2 , α3 , α4 , α5 , and α6 are the roots of the polynomial equation Γ6 +

ξ5 5 ξ4 4 ξ3 3 ξ2 2 ξ1 ξ0 Γ + Γ + Γ + Γ + Γ+ = 0. ξ6 ξ6 ξ6 ξ6 ξ6 ξ6

(3.108)

Substituting the solutions (3.96)-(3.103) and (3.105)-(3.106) into (2.4) and (3.2), we have s s 2 2 γ−1 H3 H3 + τ2 α1 ± ± u(x, t) = τ0 + τ1 α1 ± τ1 ± , 2k (x ± ct − η0 ) 2k (x ± ct − η0 ) (3.109) 2 h i γ−1 2 , (3.110) u(x, t) = τ0 + τ1 (α1 − W − 2T ) + τ2 (α1 − W − 2T ) 2 2 γ−1 √ √ u(x, t) = τ0 + τ1 α1 + W + (1 − i 3)T + τ2 α1 + W + (1 − i 3)T , (3.111) 2 γ−1 2 √ √ u(x, t) = τ0 + τ1 α1 − W1 + (1 + i 3)T + τ2 α1 − W1 + (1 + i 3)T , (3.112)


where W = T =

389

√ 1 4 3 (1 + i 3)H32 (α1 − α1 )2 H32 (α1

−

α1 )3 H52

+

3H32 (η

− η0 )(α1 − α2

)5

H32 (α1 − α1 )3 H52 + 3H32 (η − η0 )(α1 − α2 )5

p

p

H53

H53

31 ,

31

H5

,

2 √ W, H5 = 16H32 − 9(α1 − α2 )4 (η − η0 )2 , 1+i 3 " √ (α1 + α2 )H6 ± (α1 − α3 )3 (η − η0 ) −H6 u(x, t) = τ0 + τ1 2H6 2 √ 2 # γ−1 (α1 + α2 )H6 ± (α1 − α3 )3 (η − η0 ) −H6 , + τ2 2H6

W1 =

(3.113)

(3.114)

where H6 = 16H32 − (α1 − α2 )4 (η − η0 )2 . For simplicity, we can write the solutions (3.109) and (3.114) as follows:   v 2 u X   u τi α1 ± t± u(x, t) =  i=0

2 i  γ−1

H3 2k(x ±

q

8a2 k2 (γ+1)ξ6 +β(γ−1)2 ζ0 τ22 t 8k2 (γ+1)ξ6

− η0

 

, (3.115)

" u(x, t) =

2 X i=0

τi

√ (α1 + α2 )H6 ± (α1 − α3 )3 (η − η0 ) −H6 2H6

2 i # γ−1

.

(3.116)

Remark 3.7. The hyperbolic function solutions of the generalized Klein-Gordon equation was also found by the exp-function method [2]. In this study, we obtain some new classifications of the exact solutions to Eq. (1.2), such as rational function solutions, Jacobi elliptic function solutions, elliptic integral F, E and Π function solutions.

4. Conclusion In this study, we studied the extended trial equation method to establish traveling wave solutions to generalized nonlinear evolution equations arising in applied physics and engineering. Some new exact solutions for the GRLW equation and generalized Klein-Gordon equation have been successfully found. Besides, the extended trial equation method is based on the elliptic differential equation. Also, we discussed a new trial equation method. We think that the idea introduced in this paper can be applied to other nonlinear partial differential equations. Competing Interests The authors declare that they have no competing interests. Authors’ Contributions All authors contributed equally to the manuscript and read and approved of the final draft.

390


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