Exact solutions of the Biswas-Milovic equation, the ZK ...

Open Phys. 2016; 14:129–139

Research Article

Open Access

EL Sayed M. E. Zayed* and Abdul-Ghani Al-Nowehy

Exact solutions of the Biswas-Milovic equation, the ZK(m,n,k) equation and the K(m,n) equation using the generalized Kudryashov method DOI 10.1515/phys-2016-0013 Received October 3, 2015; accepted February 12, 2016

Abstract: In this article, we apply the generalized Kudryashov method for finding exact solutions of three nonlinear partial differential equations (PDEs), namely: the Biswas-Milovic equation with dual-power law nonlinearity; the Zakharov–Kuznetsov equation (ZK(m,n,k)); and the K(m,n) equation with the generalized evolution term. As a result, many analytical exact solutions are obtained including symmetrical Fibonacci function solutions, and hyperbolic function solutions. Physical explanations for certain solutions of the three nonlinear PDEs are obtained. Keywords: Nonlinear PDEs; Generalized Kudryashov method; Exact solutions; Symmetrical hyperbolic Fibonacci function; The Biswas-Milovic equation; The Zakharov–Kuznetsov equation; The K(m,n) equation PACS: 02.30.Jr, 02.30.Ik, 05.45.Yv

1 Introduction The research area of nonlinear partial differential equations (PDEs) has been very active for the past few decades. There are many types of nonlinear PDEs that appear in various areas of the physical and mathematical sciences. Much effort has been expended on constructing exact solutions to these nonlinear PDEs, motivated by their important role in the study of nonlinear physical phenomena. Nonlinear phenomena appears in various scientific and engineering fields, such as fluid mechanics, plasma physics, optical fibers, biology, solid state

physics, chemical kinematics, chemical physics, and geochemistry. In recent years, a number of powerful and efficient methods for finding analytic solutions to nonlinear equations have drawn a lot of interest by a diverse group of scientists. These include, for example: Hirota’s bilinear transformation method [1, 2]; the tanh-function method [3, 4]; the (G′ /G)-expansion method [5–10]; the Exp-function method [11–14]; the multiple exp-function method [15–17]; the symmetry method [18, 19]; the modified simple equation method [20–22]; the improved (G′ /G)expansion method [23]; a multiple extended trial equation method [24]; the Jacobi elliptic function expansion method [25, 26]; the Bäcklund transform method [27, 28]; the generalized Riccati equation method [29]; the modified extended Fan sub equation method [30]; the auxiliary equation method [31, 32]; the first integral method [33, 34]; the modified Kudryashov method [35–42], and the soliton ansatz method [43–70]. The objective of this paper is to apply the generalized Kudryashov method [42] to find the exact solutions of the Biswas-Milovic equation with dual-power law nonlinearity [71, 72], the ZK(m,n,k) [73, 74], and the K(m,n) equation with the generalized evolution term [75]. This paper is organized as follows: in Sec. 2, the description of the generalized Kudryashov method is given. In Sec. 3, we use this method to solve the three aforementioned nonlinear PDEs. In Sec. 4, physical explanations of certain results are presented, and conclusions are discussed in Sec. 5.

2 Description of the generalized Kudryashov method Starting with a nonlinear PDE in the following form:

*Corresponding Author: EL Sayed M. E. Zayed: Department of Mathematics, Faculty of Sciences, Zagazig University, Zagazig, Egypt, E-mail: [email protected] Abdul-Ghani Al-Nowehy: Department of Mathematics, Faculty of Education, Ain Shams University, Roxy, Hiliopolis, Cairo, Egypt, E-mail: [email protected]

F(u, u t , u x , u tt , u xt , u xx , ...) = 0,

(2.1)

where u = u(x, t) is an unknown function, F is a polynomial in u = u(x, t) and its partial derivatives, in which

© 2016 EL Sayed M. E. Zaye and Abdul-Ghani Al-Nowehy, published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 License.

Unauthenticated Download Date | 5/19/16 8:06 PM

130 | EL Sayed M. E. Zayed and Abdul-Ghani Al-Nowehy the highest order derivatives and nonlinear terms are involved. The main steps of the generalized Kudryashov method are described as follows: Step 1. First, we use the wave transformation:

Step 3. Under the terms of the given method, we suppose that the solution of Eq. (2.3) can be written in the following form:

u(x, t) = U(ζ ), ζ = kx ± λt,

To calculate the values m and n in (2.9), i.e. the pole order for the general solution of Eq. (2.3), we progress as per the classical Kudryashov method by balancing the highest order nonlinear terms and the highest order derivatives of U(ζ ) in Eq. (2.3). This allows us to derive a formula for m and n and determine their values. Step 4. We substitute (2.4) into Eq. (2.3) to get a polynomial R(Q) of Q and equate all the coefficients of Q i , (i = 0, 1, 2, ...) to zero, to yield a system of algebraic equations for a i (i = 0, 1, ..., n) and b j (j = 0, 1, ..., m). Step 5. We solve the algebraic equations obtained in Step 4 using Mathematica or Maple, to get k, λ and the coefficients of a i (i = 0, 1, ..., n) and b j (j = 0, 1, ..., m). In this way, we attain the exact solutions to Eq. (2.3). The obtained solutions depended on the symmetrical hyperbolic Fibonacci functions given in [76]. The symmetrical Fibonacci sine, cosine, tangent, and cotangent functions are, respectively, defined as:

(2.2)

where k and λ are arbitrary constants with k, λ ≠ 0, in order to reduce equation (2.1) into a nonlinear ordinary differential equation (ODE) with respect to the variable ζ of the form H(U, U ′ , U ′′ , U ′′′ , ...) = 0, (2.3) where H is a polynomial in U(ζ ) and its total derivatives U ′ , U ′′ , U ′′′ , ... such that U ′ = dU dζ , 2

U ′′ = ddζU2 and so on. Step 2. We assume that the formal solution of the ODE (2.3) can be written in the following rational form: [︀ ]︀ ∑︀n i A Q(ζ ) i=0 a i Q (ζ ) ]︀ , = [︀ U(ζ ) = ∑︀m (2.4) j B Q(ζ ) j=0 b j Q (ζ ) [︀ ]︀ ∑︀n [︀ ]︀ i 1 = where Q = 1±a ζ , A Q(ζ ) i=0 a i Q (ζ ) and B Q(ζ ) = ∑︀m j j=0 b j Q (ζ ). The function Q is the solution of the equation Q′ = Q (Q − 1) ln(a), 0 < a ≠ 1. (2.5) Taking into consideration (2.4), we obtain [︂ ′ ]︂ A B − AB′ ′ U (ζ ) = Q (Q − 1) ln(a), B2

U(ζ ) =

sFs(x) = tan Fs(x) =

a x + a−x a x − a−x √ , cFs(x) = √ , 5 5 a x − a−x a x + a−x , cot Fs(x) = x , a x + a−x a − a−x

(2.9)

(2.10)

(2.6) sFs(x) =

]︂ A′ B − AB′ U (ζ ) = Q (Q − 1) (2Q − 1) ln2 (a) B2 [︂ ]︂ ′′ ′′ ′ ′ ′ 2 2 2 B(A B − AB ) − 2A B B + 2A(B ) + Q ( Q − 1) ln2 (a), B3 ′′

a0 + a1 Q + a2 Q2 + + a n Q n . b0 + b1 Q + b2 Q2 + + b m Q m

[︂

(2.7) U ′′′ (ζ ) = Q3 (Q − 1)3 ln3 (a) [︂ ′′′ (A B − AB′′′ − 3A′′ B′ − 3A′ B′′ )B + 6B′ (A′ B′ + AB′′ ) × B3 ]︂ 6A(B′ )3 − + 3Q2 (Q − 1)2 (2Q − 1) B4 [︂ ′′ ]︂ (A B − AB′′ − 2A′ B′ )B + 2A(B′ )2 × ln3 (a) B3 [︂ ′ ]︂ A B − AB′ 2 + Q (Q − 1) (6Q − 6Q + 1) ln3 (a), (2.8) B2 and that similar solutions apply for higher order differentiation terms.

tan Fs(x) =

2 5 tanh(x. ln(a)),

√ sh(x. ln(a)),

2 cFs(x) = √ ch(x. ln(a)), 5 cot Fs(x) = coth(x. ln(a)). (2.11)

3 Applications In this section we construct the exact solutions in terms of the symmetrical hyperbolic Fibonacci functions of the following three nonlinear PDEs using the generalized Kudryashov method described in Sec. 2:

3.1 Example 1. The Biswas-Milovic equation with dual-power law nonlinearity In this subsection, we study the Biswas-Milovic equation with dual-power law nonlinearity [71, 72] (︁ )︁ i(q m )t + a(q m )xx + b |q|2n + k |q|4n q m = 0, (3.1.1)


Exact solutions of the Biswas-Milovic equation

where a, b and k are constants, while m, n are positive integers. If m = 1, then Eq. (3.1.1) becomes the nonlinear Schrödinger equation (NLSE) with dual-power law nonlinearity. In Eq. (3.1.1), the first term is the temporal evolution, while a is the coefficient of group-velocity dispersion (GVD), while b and k are the coefficients of the nonlinear terms. Let us now solve Eq. (3.1.1) using the method of Sec. 2. To this end, we use the wave transformation q(x, t) = U(ζ ) exp (iθ) , θ = λx−wt, ζ = µ(x−νt), (3.1.2) where λ, w, µ and ν are constants. Substituting (3.1.2) into (3.1.1), we obtain the nonlinear ODE: 2

m ′′

m ′

2

aµ (U ) + iµ(−ν + 2amλ)(U ) + m(w − amλ )U

m

+ bU 2n+m + bkU 4n+m = 0. (3.1.3)

and consequently, Q (Q − 1) [︀ (a1 + 2a2 Q)(b0 + b1 Q) (b0 + b1 Q)2 ]︁ −b1 (a0 + a1 Q + a2 Q2 ) ln(a),

v′ (ζ ) =

v′′ (ζ ) =

(3.1.11)

Q (Q − 1) (2Q − 1) [(a1 + 2a2 Q)(b0 + b1 Q) (b0 + b1 Q)2

Q2 (Q − 1)2 ln2 (a) − b1 (a0 + a1 Q + a2 Q2 )] ln2 (a) + (b0 + b1 Q)3 [︁ 2 × 2a2 (b0 + b1 Q) − 2b1 (a1 + 2a2 Q)(b0 + b1 Q) ]︁ +2b21 (a0 + a1 Q + a2 Q2 ) . (3.1.12) Substituting (3.1.10), (3.1.11) and (3.1.12) into (3.1.8), collecting the coefficients of each power of Q i , (i = 0, 1, ..., 8) and setting each of the coefficients to zero, we obtain the following system of algebraic equations: Q8 :am2 a22 b21 µ2 ln2 (a) + 2amna22 b21 µ2 ln2 (a)

From Eq. (3.1.3) , we deduce that ν = 2amλ,

| 131

+ 4bkn2 a42 = 0, (3.1.4)

Q7 : − 2amna22 b21 µ2 ln2 (a) + 16bkn2 a1 a32 + 4am2 a22 b0 b1 µ2 ln2 (a) + 4amna22 b0 b1 µ2 ln2 (a)

and aµ2 (U m )′′ + m(w − amλ2 )U m + bU 2n+m + bkU 4n+m = 0. (3.1.5) Balancing (U m )′′ and U 4n+m in (3.1.5), then the following relation is attained: 1 mN + 2 = (4n + m)N ⇒ N = . 2n

(3.1.6)

(3.1.7)

Substituting (3.1.7) into equation (3.1.5) we have the new equation (︁ )︁2 2amnµ2 vv′′ + amµ2 (m − 2n) v′ + 4mn2 (w − amλ2 )v2

Q6 :12bn2 a1 a22 b1 + 4amna22 b20 µ2 ln2 (a) − 4n2 m2 aλ2 a22 b21 − 2am2 a2 b21 a0 µ2 ln2 (a) − 8am2 a22 b0 b1 µ2 ln2 (a) + 24bkn2 a21 a22 + 8amna1 b0 b1 a2 µ2 ln2 (a) + 4bn2 a32 b0 − 6amna2 b21 a1 µ2 ln2 (a) + 4am2 a22 b20 µ2 ln2 (a) + am2 a22 b21 µ2 ln2 (a) + 16bkn2 a0 a32 + 4n2 mwa22 b21 + 8amna2 b21 a0 µ2 ln2 (a) = 0, Q5 :12bn2 a21 a2 b1 + 48bkn2 a0 a1 a22 + 16amna2 b0 b1 a0 µ2 ln2 (a) + 8n2 mwa22 b0 b1

+ 4bn2 v3 + 4bkn2 v4 = 0. (3.1.8) Balancing the vv′′ and v4 in (3.1.8), then the following relation is obtained: (N − M) + (N − M) + 2 = 4(N − M) ⇒ N = M + 1. (3.1.9) If we choose M = 1 and N = 2, then the formal solution of Eq. (3.1.8) has the form: a + a1 Q + a2 Q2 , v(ζ ) = 0 b0 + b1 Q

+ 4bn2 a32 b1 = 0,

− 2amna22 b0 b1 µ2 ln2 (a) + 2am2 a1 b0 a2 b1 µ2 ln2 (a)

Then we take into consideration the transformation [︀ ]︀ 1 U(ζ ) = v(ζ ) 2n .

+ 4amna2 b21 a1 µ2 ln2 (a) − 2am2 a22 b21 µ2 ln2 (a)

+ 12bn2 a0 a22 b1 + 4am2 a22 b0 b1 µ2 ln2 (a) − 8amna1 b0 b1 a2 µ2 ln2 (a) − 2amna22 b0 b1 µ2 ln2 (a) + 12bn2 a1 a22 b0 − 4am2 a1 b0 a2 b1 µ2 ln2 (a) − 16amna2 b21 a0 µ2 ln2 (a) + 4am2 a1 b20 a2 µ2 ln2 (a) − 4am2 a2 b0 b1 a0 µ2 ln2 (a) + 8amna2 b20 a1 µ2 ln2 (a) + 16bkn2 a31 a2 + 8n2 mwa1 a2 b21 + 2amna2 b21 a1 µ2 ln2 (a) + 4am2 a2 b21 a0 µ2 ln2 (a) − 8n2 m2 aλ2 a22 b0 b1 − 8n2 m2 aλ2 a1 a2 b21

(3.1.10)

− 4amna22 b20 µ2 ln2 (a) − 8am2 a22 b20 µ2 ln2 (a) = 0,


132 | EL Sayed M. E. Zayed and Abdul-Ghani Al-Nowehy Q1 : − 8n2 m2 aλ2 a0 a1 b20 − 2amnb1 a20 b0 µ2 ln2 (a) − 8n2 m2 aλ2 a20 b0 b1 + 8n2 mwa0 a1 b20

Q4 : − 8n2 m2 aλ2 a0 a2 b21 − 2am2 a2 b21 a0 + 16n2 mwa1 a2 b0 b1 + 2am2 a1 b0 a2 b1 µ2 ln2 (a) + 4bkn2 a41 + 8amna2 b21 a0 µ2 ln2 (a) − 4n2 m2 aλ2 a21 b21 − 4n2 m2 aλ2 a22 b20 + 8n2 mwa0 a2 b21 + 24bn2 a0 a1 a2 b1 + 48bkn2 a0 a21 a2 + am2 a21 b20 µ2 ln2 (a) + 4bn2 a31 b1 + 2amna21 b20 µ2 ln2 (a) + 8am2 a2 b0 b1 a0 µ2 ln2 (a) − 2amnb21 a20 µ2 ln2 (a) + 4n2 mwa21 b21 + 4n2 mwa22 b20 2

+ 12bn a0 a22 b0 − 16n2 m2 aλ2 a1 a2 b0 b1 + 12bn2 a21 a2 b0 + 4am2 a22 b20 µ2 ln2 (a) − 8am2 a1 b20 a2 µ2 ln2 (a) − 28amna2 b0 b1 a0 µ2 ln2 (a) − 10amna2 b20 a1 µ2 ln2 (a) + 24bkn2 a20 a22 + am2 b21 a20 µ2 ln2 (a) + 2amna21 b0 b1 µ2 ln2 (a) + 12amna2 b20 a0 µ2 ln2 (a) − 2amnb21 a0 a1 µ2 ln2 (a) − 2am2 a1 b0 b1 a0 µ2 ln2 (a) = 0,

+ 4bn2 a30 b1 + 16bkn2 a30 a1 + 12bn2 a20 a1 b0 + 2amna1 b20 a0 µ2 ln2 (a) + 8n2 mwa20 b0 b1 = 0, Q0 : − 4n2 m2 aλ2 a20 b20 + 4bkn2 a40 + 4bn2 a30 b0 + 4n2 mwa20 b20 = 0.

Solving the system of algebraic equations (3.1.13) by Maple or Mathematica, we obtain the following sets: Set 1: √︃ 1 bn2 (2n + m) µ= − , ln(a) amk(n + m)2 b(2n + m) , 4k(n + m)2 b (2n + m) a0 =0, a1 = − 0 , 2k(n + m) b (2n + m) , a2 = − 1 2k(n + m) w =amλ2 +

Q3 : − 2am2 a21 b20 µ2 ln2 (a) − 2amna21 b20 µ2 ln2 (a)

b0 =b0 , b1 = b1 , λ = λ, m = m, n = n,

+ 2amnb21 a20 µ2 ln2 (a) − 16n2 m2 aλ2 a0 a2 b0 b1

k =k, a = a, b = b.

+ 8n2 mwa0 a1 b21 + 8n2 mwa21 b0 b1 + 8n2 mwa1 a2 b20 + 24bn2 a0 a1 a2 b0 + 48bkn2 a0 a1 a2 + 12bn2 a20 a2 b1 + 12amna2 b0 b1 a0 µ2 ln2 (a) + 12bn2 a0 a21 b1 + 16bkn2 a0 a31 + 4amna1 b20 a0 µ2 ln2 (a) − 4amnb1 a20 b0 µ2 ln2 (a) + 4am2 a1 b20 a2 µ2 ln2 (a) + 2amna2 b20 a1 µ2 ln2 (a) − 2am2 b21 a20 µ2 ln2 (a) − 8n2 m2 aλ2 a0 a1 b21 − 8n2 m2 aλ2 a21 b0 b1 − 8n2 m2 aλ2 a1 a2 b20 + 16n2 mwa0 a2 b0 b1 − 2amna21 b0 b1 µ2 ln2 (a) − 20amna2 b20 a0 µ2 ln2 (a) − 4am2 a2 b0 b1 a0 µ2 ln2 (a) + 2amnb21 a0 a1 µ2 ln2 (a) + 4am2 a1 b0 b1 a1 µ2 ln2 (a) = 0, + 4bn2 a31 b0

Q2 :12bn2 a20 a2 b0 − 4n2 m2 aλ2 a21 b20 + am2 b21 a20 µ2 ln2 (a) − 8n2 m2 aλ2 a0 a2 b20 − 4n2 m2 aλ2 a20 b21

(3.1.13)

(3.1.14)

Substituting (3.1.14) into (3.1.10), we get the following solution: (2n + m) 1 v(ζ ) = − . (3.1.15) 2k(n + m) (1 ± a ζ ) From (3.1.7) and (3.1.15) we have [︂ ]︂ 2n1 (2n + m) 1 U(ζ ) = − . 2k(n + m) (1 ± a ζ )

(3.1.16)

With the help of (2.10) and (2.11) the exact solution of Eq. (3.1.1) has the form: {︂ (2n + m) q(x, t) = − 4k(n + m) ⎡ ⎛ ⎞⎤⎫ 2n1 √︃ ⎬ 2 (2n + m) 1 bn ⎠⎦ × ⎣1 − tan Fs ⎝ − (x − 2amλt) ⎭ 2 ln(a) amk(n + m)2 × exp(iθ),

(3.1.17)

+ am2 a21 b20 µ2 ln2 (a) + 12bn2 a20 a1 b1 + 6amnb1 a20 b0 µ2 ln2 (a) + 4n2 mwa21 b20 + 16bkn2 a30 a2 + 8amna2 b20 a0 µ2 ln2 (a) + 12bn2 a0 a21 b0 + 24bkn2 a20 a21 + 4n2 mwa20 b21 − 16n2 m2 aλ2 a0 a1 b0 b1 + 8n2 mwa0 a2 b20 − 2am2 a1 b0 b1 a0 µ2 ln2 (a) + 16n2 mwa0 a1 b0 b1 − 6amna1 b20 a0 µ2 ln2 (a) = 0,

{︂ (2n + m) = − 4k(n + m) ⎡ ⎛ √︃ ⎞⎤⎫ 2n1 ⎬ 2 ⎣1 − tanh ⎝ 1 − bn (2n + m) (x − 2amλt)⎠⎦ 2 ⎭ amk(n + m)2 × exp(iθ), (3.1.18)



| 133

Set 3:

or {︂

(2n + m) q(x, t) = − 4k(n + m) ⎞⎤⎫ 2n1 ⎡ ⎛ √︃ ⎬ 2 bn (2n + m) 1 − (x − 2amλt)⎠⎦ × ⎣1 − cot Fs ⎝ 2 ⎭ 2 ln(a) amk(n + m) × exp(iθ),

(3.1.19)

{︂

(2n + m) = − 4k(n + m) ⎡ ⎛ √︃ ⎞⎤⎫ 2n1 ⎬ 2 (2n + m) bn 1 ⎠⎦ − × ⎣1 − coth ⎝ (x − 2amλt) 2 ⎭ amk(n + m)2 × exp(iθ),

(3.1.20) (︁

)︁

b(2n+m) t, provided that ab > 0 where θ = λx − amλ2 + 4k(n+m) 2 and k < 0. Set 2: √︃ bn2 (2n + m) 1 − , µ= ln(a) amk(n + m)2

1 µ= ln(a)

√︃

w = amλ2 +

−

bn2 (2n + m) , amk(n + m)2

b(2n + m) (b + b1 )(2n + m) , , a1 = − 0 4k(n + m)2 2k(n + m)

a0 = 0, a2 = 0, , b0 = b0 , b1 = b1 , λ = λ, m = m, n = n, k = k, a = a, b = b.

(3.1.26)

Consequently, we have the exact solutions of Eq. (3.1.1) in the form: [︂ ]︂ 1 (b0 + b1 )(2n + m) [1 − tanh η] 2n exp(iθ), q(x, t) = − 2k(n + m) [2b0 + b1 (1 − tanh η)] (3.1.27) or [︂ ]︂ 1 (b0 + b1 )(2n + m) [1 − coth η] 2n q(x, t) = − exp(iθ), 2k(n + m) [2b0 + b1 (1 − coth η)] (3.1.28) √︁ bn2 (2n+m) 1 θ = λx − where η = 2 − amk(n+m)2 (x − 2amλt), (︁ )︁ b(2n+m) 2 amλ + 4k(n+m)2 t, provided that ab > 0 and k < 0.

w = amλ2 +

On comparing our result (3.1.24), with the result (18) obtained in [71], we conclude that the two results are equivalent with λ = −κ, while our results (3.1.18), (3.1.20), (3.1.25), (3.1.27) and (3.1.28) are new, and not discussed elsewhere.

λ = λ, m = m, n = n, k = k, a = a, b = b. (3.1.21)

3.2 Example 2. The Zakharov–Kuznetsov equation (ZK(m,n,k))

b(2n + m) , 4k(n + m)2 b (2n + m) a0 = 0, a1 = − 1 , 2k(n + m) b (2n + m) a2 = 1 , b0 = 0, b1 = b1 , 2k(n + m)

Substituting (3.1.21) into (3.1.10), we get the following solution: (2n + m) aζ v(ζ ) = − . (3.1.22) ζ 2k(n + m) (a ± 1) From (3.1.7) and (3.1.122) we have [︂ ]︂ 2n1 (2n + m) aζ U(ζ ) = − . 2k(n + m) (a ζ ± 1)

(3.1.23)

With the help of (2.10) and (2.11) the exact solution of Eq. (3.1.1) has the form: [︂ ]︂ 2n1 (2n + m) q(x, t) = − exp(iθ), (1 + tanh η) 4k(n + m)

(3.1.24)

In this subsection, we apply the given method to solve the ZK(m,n,k) [73, 74] u t + λ0 (u m )x + λ1 (u n )xxx + λ2 (u k )xyy = 0,

(3.2.1)

where u = u(x, y, t) is a readily differentiable function, λ0 , λ1 and λ2 are arbitrary constants while m, n and k are positive integers. The function governs the behaviour of weakly nonlinear ion acoustic waves in a plasma comprising cold ions and hot isothermal electrons in the presence of a uniform magnetic field [77]. Recently, Ma et al. [73] used the auxiliary equation method to find the solutions of the ZK(2,1,1) equation with λ0 = 21 , λ1 = 13 , λ2 = 32 . This work is concerned with two cases of Eq. (3.2.1):

or [︂ ]︂ 2n1 (2n + m) q(x, t) = − exp(iθ), (3.1.25) (1 + coth η) 4k(n + m) √︁ bn2 (2n+m) where η = 12 − amk(n+m) θ = λx − 2 (x − 2amλt), (︁ )︁ b(2n+m) 2 amλ + 4k(n+m)2 t, provided that ab > 0 and k < 0.

3.2.1 Case 1: ZK(2,1,1) In this case Eq. (3.2.1) becomes the form: u t + λ0 (u2 )x + λ1 u xxx + λ2 u xyy = 0.


(3.2.2)

134 | EL Sayed M. E. Zayed and Abdul-Ghani Al-Nowehy To seek travelling wave solutions, we use the wave transformation u(x, y, t) = U(ζ ), ζ = x + βy + 𝛾 t,

(3.2.3)

where β and 𝛾 are arbitrary constants with β, 𝛾 ≠ 0, to reduce equation (3.2.2) to the ODE: 𝛾 U ′ + λ0 (U 2 )′ + λ1 U ′′′ + λ2 β2 U ′′′ = 0.

(3.2.4)

Substituting (3.2.9) into (3.1.10), we get the following solution of Eq. (3.2.6): (︀ )︀ 6 λ1 + λ2 β2 ln2 (a) aζ U(ζ ) = ± . (3.2.10) ζ λ0 (a ± 1)2 With the help of (2.10) and (2.11) the exact solution of Eq. (3.2.2) in the form:

Integrating Eq. (3.2.4) with respect to ζ , with zero constant of integration, we get 2

2

′′

𝛾 U + λ0 U + (λ1 + λ2 β )U = 0. ′′

(3.2.5)

(︀ )︀ 3 λ1 + λ2 β2 ln2 (a) 2 u(x, y, t) = sech η, 2λ0

(3.2.11)

or

2

(︀ )︀ 3 λ1 + λ2 β2 ln2 (a) 2 csch η, (3.2.12) 2λ0 [︁ ]︁ (︀ )︀ where η = 12 x + βy − λ1 + λ2 β2 ln2 (a).t ln(a). 2 3 a + a1 Q + a2 Q + a3 Q U(ζ ) = 0 , (3.2.6) Set 2: b0 + b1 Q (︀ )︀ b0 λ1 + λ2 β2 ln2 (a) and consequently, a = − , 0 λ0 Q (Q − 1) [︁ ′ 2 (︀ )︀ 2 U (ζ ) = (a1 + 2a2 Q + 3a3 Q ) (b0 + b1 Q) 2 (6b0 − b1 ) λ1 + λ2 β2 ln (a) (b0 + b1 Q) a = , (︁ )︁]︁ 1 λ0 −b1 a0 + a1 Q + a2 Q2 + a3 Q3 ln(a), (︀ )︀ 6 λ1 + λ2 β2 (b0 − b1 ) ln2 (a) a = − , 2 (3.2.7) λ0 (︀ )︀ 6b1 λ1 + λ2 β2 ln2 (a) [︁(︁ )︁ a = − , Q (Q − 1) (2Q − 1) 3 λ0 a1 + 2a2 Q + 3a3 Q2 (b0 + b1 Q) U ′′ (ζ ) = 2 (︁ )︁ (b0 + b1 Q) (︁ )︁]︁ 𝛾 = λ1 + λ2 β2 ln2 (a), b0 = b0 , b1 = b1 , −b1 a0 + a1 Q + a2 Q2 + a3 Q3 ln2 (a) β = β, λ0 = λ0 , λ1 = λ1 , λ2 = λ2 . (3.2.13) Q2 (Q − 1)2 [︁ 2 b + b Q (2a + 6a Q) + 1 ) 2 3 3 ( 0 Consequently, we have the exact solutions of Eq. (b0 + b1 Q) (︁ )︁ (3.2.2) in the form: −2b1 a1 + 2a2 Q + 3a3 Q2 (b0 + b1 Q) (︁ )︁]︁ (︀ )︀ +2b21 a0 + a1 Q + a2 Q2 + a3 Q3 ln2 (a). ]︁ λ1 + λ2 β2 ln2 (a) [︁ 1 − 3 tanh2 η , (3.2.14) u(x, y, t) = 2λ0 (3.2.8)

Balancing the U and U in (3.2.5), then we have N = M + 2. If we choose M = 1 and N = 3, then the formal solution of Eq. (3.2.5) has the form:

Substituting (3.2.6)and (3.2.8) into (3.2.5), and equating all the coefficients of Q i ,(i = 0, 1, ..., 7) to zero, we get a system of algebraic equations, which can be solved using the Maple, to get the following sets: Set 1:

)︀ 2

2

6b0 λ1 + λ2 β ln (a) , λ0 (︀ )︀ 6 (b0 − b1 ) λ1 + λ2 β2 ln2 (a) a2 = − , λ0 (︀ )︀ 6b1 λ1 + λ2 β2 ln2 (a) a3 = − , λ0 (︁ )︁ b0 = b0 , b1 = −2b0 , 𝛾 = − λ1 + λ2 β2 ln2 (a), a1 =

β = β, λ0 = λ0 , λ1 = λ1 , λ2 = λ2 .

or )︀ ]︁ λ1 + λ2 β2 ln2 (a) [︁ u(x, y, t) = 1 − 3 coth2 η , (3.2.15) 2λ0 [︁ ]︁ (︀ )︀ where η = 12 x + βy + λ1 + λ2 β2 ln2 (a).t ln(a). (︀

If we choose λ0 = 12 , λ1 = 13 , λ2 = 23 , a = e in our results (3.2.11),(3.2.12),(3.2.14) and (3.2.15) we have the wellknown results u1 , u2 of (10) and u1 , u2 of (13) obtained in [73].

a0 = 0, (︀

u(x, y, t) = −

3.2.2 Case 2: ZK(m,1,1), m > 3 In this case Eq. (3.2.1) becomes the form: (3.2.9)

u t + λ0 (u m )x + λ1 u xxx + λ2 u xyy = 0.


(3.2.16)


Using the same wave transformation (3.2.3), to reduce equation (3.2.16) to the ODE: 𝛾 U ′ + λ0 (U m )′ + λ1 U ′′′ + λ2 β2 U ′′′ = 0.

(3.2.17)

Integrating Eq. (3.2.17) with respect to ζ , with zero constant of integration, we get 𝛾 U + λ0 U

m

2

′′

+ (λ1 + λ2 β )U = 0.

By balancing U ′′ with U m we have N = Then we use the transformation [︀ ]︀ 2 U(ζ ) = v(ζ ) m−1 .

> 3.

(3.2.19)

(m − 1)2 (𝛾 v2 + λ0 v4 ) + (λ1 + λ2 β2 ) [︁ ]︁ × (6 − 2m) (v′ )2 + 2(m − 1)vv′′ = 0.

(3.2.20)

4

Balancing the vv and v in (3.2.21), then we have N = M + 1. If we choose M = 1 and N = 2, then Eq. (3.2.20) has the same formal solution (3.1.10). Substituting (3.1.10),(3.1.11) and (3.1.12) into (3.2.20), and equating all the coefficients of Q i ,(i = 0, 1, ..., 8) to zero, we get a system of algebraic equations, which can be solved using the aid of Maple or Mathematica, to get the following result: a0 = 0, √︃

(︀ )︀ 2 λ 1 + λ 2 β 2 ( m + 1) , − λ0 √︃ (︀ )︀ 2 λ 1 + λ 2 β 2 ( m + 1) 2b0 ln(a) a2 = ∓ − , λ0 ( m − 1)

2b ln(a) a1 = ± 0 ( m − 1)

)︂

4 λ +λ2 β2 ) ln2 (a) t ln(a). x + βy − ( 1 (m−1) 2

3.3 Example 3. The K(m,n) equation The K(m,n) equation with the generalized evolution term [75, 78, 79] is given by (︀ )︀ (q l )t + aq m q x + b q n xxx = 0, (3.3.1) where, the first term is the generalized evolution term, the second term represents the nonlinearity, and the third term is the dispersion. Also, a, b ∈ R and are constants, while l, m and n ∈ Z + . Eq. (3.3.1) is the generalized form of the KdV equation, where, in particular, the case l = m = n = 1 leads to the KdV equation. Eq. (3.3.1) appeared for the first time in [78] for l = 1. Thus, Eq. (3.3.1) reduces to the K(m,n) equation for l = 1. Therefore, for l = 1, K(1,1) is the KdV equation while K(2,1) is the mKdV equation. Eq. (3.3.1) has been discussed in [75] using the (G′ /G)-expansion method and its exact solutions have been found. In order to solve Eq. (3.3.1) using the method of section 2, we introduce the wave transformation q(x, t) = U(ζ ), ζ = µ (x − ct) ,

(3.3.2)

where µ and c are constants, to reduce Eq. (3.3.1) to the ODE: (︀ )︀′′′ −µc(U l )′ + µaU m U ′ + µ3 b U n = 0. (3.3.3)

b0 = b0 , b1 = −2b0 , (︀ )︀ 4 λ1 + λ2 β2 ln2 (a) 𝛾=− , 2 ( m − 1) β = β, λ0 = λ0 , λ1 = λ1 , λ2 = λ2 .

(3.2.24) (︂ where η =

Substituting (3.2.19) into equation (3.2.18) we have the new equation

′′

With the help of (2.10) and (2.11) the exact solution of Eq. (3.2.16) in the form: 2 ⎡ ⎤ m−1 √︃ (︀ )︀ 2 ( m + 1) 2 λ + λ β ln(a) 1 2 u(x, y, t) = ⎣± csch η⎦ , − λ0 ( m − 1)

(3.2.18) 2 m−1 , m

| 135

(3.2.21)

2

λ +λ β provided that ( 1 λ02 ) < 0. Substituting (3.2.21) into (3.1.10), we get the following solution of Eq. (3.2.20): √︃ (︀ )︀ 2 λ 1 + λ 2 β 2 ( m + 1) 2 ln(a) aζ v(ζ ) = ± . − λ0 ( m − 1) (a2ζ − 1) (3.2.22) From (3.2.22) and (3.2.19) we have 2 ⎡ ⎤ m−1 √︃ (︀ )︀ ζ 2 λ 1 + λ 2 β 2 ( m + 1) 2 ln(a) a ⎦ U(ζ ) = ⎣± − . λ0 ( m − 1) (a2ζ − 1)

Integrating Eq. (3..3.3) with respect to ζ , with zero constant of integration, we have (︀ )︀′′ µa −µcU l + U m+1 + µ3 b U n = 0. (3.3.4) m+1 Let l = n, balancing (U n )′′ and U m+1 in (3.3.4) yields 2 N = m+1−n , where m + 1 ≠ n. In order to obtain the closed form solutions, we use the transformation [︀ ]︀ 1 U(ζ ) = v(ζ ) m+1−n , (3.3.5) to reduce Eq. (3.3.4) into the ODE a ( m + 1 − n )2 3 v m+1 (︁ )︁2 + bnµ2 (2n − m − 1) v′ + bnµ2 (m + 1 − n)2 vv′′ = 0. − c ( m + 1 − n )2 v 2 +

(3.2.23)


(3.3.6)

136 | EL Sayed M. E. Zayed and Abdul-Ghani Al-Nowehy Balancing the vv′′ and v3 in (3.3.6), then we have N = M + 2. If we choose M = 1 and N = 3, then Eq. (3.3.6) has the same formal solution (3.2.6). Substituting (3.2.6),(3.2.7) and (3.2.8) into (3.3.6), and equating all the coefficients of Q i ,(i = 0, 1, ..., 10) to zero, we get a system of algebraic equations, which can be solved using the aid of Maple or Mathematica, to get the following result:

We now examine Figures (1–4) which illustrate a selection of the results obtained above. Specific parameter values are selected, for example: in some of the solutions (3.1.24) and (3.1.25) of the Biswas-Milovic equation with dual-power law nonlinearity with −10 < x, t < 10, solutions (3.2.11) and (3.2.12) of the ZK(2,1,1) with −10 < x, t < 10, solution (3.2.24) of the ZK(m,1,1) with −10 < x, t < 10, solutions (3.3.10) and (3.3.11) of the K(m,n) equation with the generalized evolution with −10 < x, t < 10.

2cb0 (m + 1) (m + 1 + n) , na 2c (b0 − b1 ) (m + 1) (m + 1 + n) a2 = − , na 2cb1 (m + 1) (m + 1 + n) a3 = − , na √︂ c (m + 1 − n) µ=± , b0 = b0 , b1 = b1 , c = c, b n ln(a) a0 = 0, a1 =

(3.3.7) provided that bc > 0. Substituting (3.3.7) into (3.2.6), we get the following solution of Eq. (3.3.6): v(ζ ) = ±

2c (m + 1) (m + 1 + n) a ζ . na (a ζ ± 1)

(3.3.8)

(a) 3D Plot solution (3.1.24) (|q|)

(b) 3D Plot solution (3.1.25) (|q|)

Figure 1: Symmetrical Fibonacci hyperbolic function solutions of the Biswas-Milovic equation with dual-power law nonlinearity when a = b = λ = n = m = 1, K = −1.

From (3.3.8) and (3.3.5) we have 1 [︂ ]︂ m+1−n 2c (m + 1) (m + 1 + n) a ζ . U(ζ ) = ± na (a ζ ± 1)

(3.3.9)

With the help of (2.10) and (2.11) the exact solution of Eq. (3.3.1) in the form: [︂ u(x, y, t) =

c ( m + 1) ( m + 1 + n ) 2 sech η 2na

1 ]︂ m+1−n

, (3.3.10)

or 1 ]︂ m+1−n [︂ c ( m + 1) ( m + 1 + n ) 2 csch η , (3.3.11) u(x, y, t) = − 2na √︁ where η = ± bc (m+1−n) (x − ct). 2n

(a) 3D Plot solution (3.2.11)

(b) 3D Plot solution (3.2.12)

Figure 2: Symmetrical Fibonacci hyperbolic function solutions of the ZK(2,1,1) when λ0 = λ1 = λ2 = β = 1, a = e, y = 0.

4 Physical explanations for some of our solutions In this section, we will illustrate the application of the results established above. Exact solutions of the results describe different nonlinear waves. The established exact solutions with symmetrical hyperbolic Fibonacci functions are special kinds of solitary waves.

5 Conclusions In this paper we have shown that the symmetrical hyperbolic Fibonacci function solutions can be obtained for the general Exp a - function by using generalized Kudryashov



| 137

References [1] [2] [3]

[4] (a) 3D Plot solution (3.2.24) with (b) 3D Plot solution (3.2.24) with m=3 m=4

[5]

Figure 3: Symmetrical Fibonacci hyperbolic function solutions of the ZK(m,1,1) when λ0 = −1, λ1 = λ2 = 1, β = 0, a = e.

[6]

[7]

[8]

[9]

[10] (a) 3D Plot solution (3.3.10)

(b) 3D Plot solution (3.3.11) [11]

Figure 4: Symmetrical Fibonacci hyperbolic function solutions of K(m,n) equation with the generalized evolution when a = b = c = m = n = 1.

[12] [13]

method. We have successfully extended the generalized Kudryashov method to solve three nonlinear partial differential equations. In terms of practical applications, we have obtained many new symmetrical hyperbolic Fibonacci function solutions for the Biswas-Milovic equation with dual-power law nonlinearity, the ZK(m,n,k) and the K(m,n) equation with the generalized evolution term. This demonstrates that the generalized Kudryashov method is powerful, effective and convenient for solving nonlinear PDEs. The physical explanation for certain solutions of such equations have been presented. The generalized Kudryashov method provides a powerful mathematical tool to obtain more general exact analytical solutions of many nonlinear PDEs in mathematical physics.

[14]

[15]

[16]

[17]

[18]

Acknowledgement: The authors wish to thank the referees for their comments on this paper. [19]

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