New exact solutions for the Kaup-Kupershmidt equation

Mustafa Inc; Mamun Miah; Akher Chowdhury; Shahadat Ali; Hadi Rezazadeh; Mehmet Ali Akinlar; Yu-Ming Chu; Mustafa Inc; Mamun Miah; Akher Chowdhury; Shahadat Ali; Hadi Rezazadeh; Mehmet Ali Akinlar; Yu-Ming Chu

doi:10.3934/math.2020432

AIMS Mathematics

2020, Volume 5, Issue 6: 6726-6738. doi: 10.3934/math.2020432

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Research article

New exact solutions for the Kaup-Kupershmidt equation

1.
Department of Mathematics, Firat University, Elazig, Turkey
2.
Department of Medical Research, China Medical University Hospital, China Medical University, Taichung, Taiwan
3.
Department of Mathematics, Khulna University of Engineering and Technology, Bangladesh
4.
Department of Mathematics, Bangladesh Army University of Engineering and Technology, Bangladesh
5.
Department of Applied Mathematics, Noakhali Science and Technology University, Bangladesh
6.
Faculty of Engineering Technology, Amol University of Special Modern Technology, Amol, Iran
7.
Department of Mathematical Engineering, Yildiz Technical University, Istanbul, Turkey
8.
Department of Mathematics, Huzhou University, Huzhou 313000, China
9.
Hunan Provincial Key Laboratory of Mathematical Modeling and Analysis in Engineering, Changsha University of Science & Technology, Changsha 410114, China

Received: 23 June 2020 Accepted: 23 August 2020 Published: 28 August 2020
MSC : 35A09, 35E05

We present new exact solutions for the (1+1)-dimensional Kaup-Kupershmidt (KK) equation by employing method of double

$(G'/G, 1/G)$ -expansion. We express solutions by hyperbolic, trigonometric and rational functions explicitly. Computational results indicate the efficiency and applicability potential of the method.

Keywords:

Kaup-Kupershmidt equation,
the method of double $(G'/G,1/G)$ -expansion,
exact solutions

Citation: Mustafa Inc, Mamun Miah, Akher Chowdhury, Shahadat Ali, Hadi Rezazadeh, Mehmet Ali Akinlar, Yu-Ming Chu. New exact solutions for the Kaup-Kupershmidt equation[J]. AIMS Mathematics, 2020, 5(6): 6726-6738. doi: 10.3934/math.2020432

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Abstract

We present new exact solutions for the (1+1)-dimensional Kaup-Kupershmidt (KK) equation by employing method of double

$(G'/G, 1/G)$ -expansion. We express solutions by hyperbolic, trigonometric and rational functions explicitly. Computational results indicate the efficiency and applicability potential of the method.

1. Introduction

Nonlinear evolution equations (NLEEs) model many complex phenomena in physics including plasma, solid state, chemical and optical fibers, nonlinear optics, fluid mechanics, etc. Exploring exact traveling wave solutions plays a significant role in nonlinear physics. For this purpose, a number of techniques were developed including method of modified Khater ^[1,2], first integral ^[3,4], functional variable ^[5], expansions ^[6,7] of new generalized $(G'/G)$ ^[8,9,10], new $\Phi6$ -model ^[11], Jacobi elliptic function ^[12,13], sine-Gordon ^[14], bifurcation ^[15,16], exp-function ^[17,18], new auxiliary equation ^[19], exp(- $\phi (\xi)$ )-expansion ^[20,21], fan sub-equation ^[22,23], inverse scattering ^[24], generalized Kudryshov ^[25,26,27], Hirota's bilinear ^[28,29], extended direct algebraic ^[30], Lie group ^[31].

Consider the (2+1)-dimensional KK equations ^[32]

$\begin{equation} \begin{array}{l} 9{u_t} + {u_{5x}} + 15u{u_{xxx}} + \frac{{75}}{2}{u_x}{u_{xx}} + 45{u^2}{u_x} + 5\sigma {u_{xxy}} - 5\sigma \partial _x^{ - 1}{u_{yy}} + 15\sigma u{u_y}\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + 15\sigma {u_x}\partial _x^{ - 1}{u_y} = 0. \end{array} \end{equation}$

(1.1)

where $\sigma^2 = 1, \, \, \partial_x^{-1} = \int dx.$ This equation has been widely applied in many branches of physics like plasma physics, fluid dynamics, nonlinear optics, and so forth. If we take $u(x, y, t) = u(x, t),$ Eq (1.1) becomes the (1+1)-dimensional KK equation ^[32]

$\begin{equation} 9 u_{t} +u_{5x} +15 u u_{xxx} +\frac{75}{2} u_{x} u_{xx} +45 u^{2} u_{x} = 0, \end{equation}$

(1.2)

In ^[33], method of exp-function was applied to Eq (1.2). In ^[32], symmetric method was applied to the nonlinear (2+1)-KK equation.

The method of the present paper, a candid, succinct and efficient technique, considered as a generalization of ( $G'$ / $G$ )-expansion technique ^{[34,35,36,37]} was developed in ^{[38,39,40,41,42,43,44,45]}. Main purpose of this paper is to investigate the applicability of the method to (1+1)-dimensional KK equation which was not considered in the history of research so far.

2. Double $(G'/G, 1/G)$ -expansion technique

We shortly overview the method in such a fashion that maintains four remarks and five basic postulates:

Remark I. If we set up

$\begin{equation} \phi = G'/G, \psi = 1/G, \end{equation}$

(2.1)

$\begin{equation} G''(\xi )+\lambda G(\xi ) = \beta , \end{equation}$

(2.2)

then we must have the relations

$\begin{equation} \phi ' = -\phi ^{2} +\beta \psi -\lambda , \;\;\;\;\; \varphi ' = -\varphi \psi , \end{equation}$

(2.3)

wherein $\lambda$ and $\beta$ are parameters.

Remark II. If $\lambda$ is negative, general solution of (2.2) is:

$\begin{equation} G\left(\xi \right) = D_{1} \; \sinh (\xi \sqrt{-\lambda } )+D_{2} \; \cosh (\xi \sqrt{-\lambda } )+\frac{\beta }{\lambda }. \end{equation}$

(2.4)

and we receive the following relation

$\begin{equation} \psi ^{2} = \frac{-\lambda }{\lambda ^{2} \alpha _{1} +\beta ^{2} } \left(\varphi ^{2} -2\beta \psi +\lambda \right), \end{equation}$

(2.5)

wherein $D_{1}$ and $D_{2}$ are arbitrary constants and $\alpha _{1} = D_{1} ^{2} -D_{2} ^{2}$ .

Remark III. If $\lambda$ is positive, general solution of (2.2) is:

$\begin{equation} G\left(\xi \right) = D_{1} \; \sin (\xi \sqrt{\lambda } )+D_{2} \; \cos (\xi \sqrt{\lambda } )+\frac{\beta }{\lambda }, \end{equation}$

(2.6)

consequently, we obtain

$\begin{equation} \psi ^{2} = \frac{\lambda }{\lambda ^{2} \alpha_{2} -\beta ^{2} } (\varphi ^{2} -2\beta \psi +\lambda ), \end{equation}$

(2.7)

wherein $\alpha _{2} = D_{1} ^{2}+D_{2} ^{2}$ .

Remark IV. If $\lambda = 0$ , the general solution of (2.2),

$\begin{equation} G(\xi ) = \frac{\beta }{2} \xi ^{2}+D_{1} \xi +D_{2}, \end{equation}$

(2.8)

and therefore we get,

$\begin{equation} \psi ^{2} = \frac{\varphi ^{2} -2\beta \psi }{D_{1} ^{2} -2 \beta D_{2}}. \end{equation}$

(2.9)

Now let us consider:

$\begin{equation} R (u, u_{t}, u_{x}, u_{y}, u_{tt}, u_{xx}, u_{yy}, u_{xt}, \cdots ) = 0, \end{equation}$

(2.10)

wherein $R$ is a polynomial function in $u$ and $u_{t} = \frac{\partial u}{\partial t}$ , $u_{x} = \frac{\partial u}{\partial x}$ , $u_{y} = \frac{\partial u}{\partial y}$ , $u_{xx} = \frac{\partial ^{2} u}{\partial x^{2} }$ , $u_{yy} = \frac{\partial ^{2} u}{\partial y^{2} }$ , $u_{xy} = \frac{\partial ^{2} u}{\partial x \partial y}$ and so on.

Postulate 1. Consider:

$\begin{equation} u (x, y, t) = u (\xi ), \;\;\;\text{and}\;\;\;\; \xi = \eta x+ \omega y+c t, \end{equation}$

(2.11)

wherein $\eta$ , $\omega$ and $c$ are parameters. By traveling wave transformations (2.11), the Eq.(2.10) can be reduced to:

$\begin{equation} T (u, cu', \eta u', \omega u', c^{2} u'', \eta ^{2} u'', \omega ^{2} u'', \eta \omega u'', c \eta u'', \cdots ) = 0, \end{equation}$

(2.12)

wherein $T$ is a polynomial.

Postulate 2. Let us assume that the following relation is the general solution expressed by a polynomial:

$\begin{equation} u (\xi ) = a_{0}+\sum _{i = 1}^{N}\left(a_{i} \varphi ^{i} (\xi )+b_{i} \varphi ^{i-1} (\xi ) \psi (\xi )\right), \end{equation}$

(2.13)

wherein $a_{0}$ , $a_{i}$ and $b_{i} (i = 1, 2, 3, ..., N)$ are the constant coefficients such that $a_{N}^{2} +b_{N}^{2} \ne 0$ .

Postulate 3. By homogeneous balance, we determine $N$ in Eq (2.13).

Postulate 4. To convert the left-hand-side of Eq (2.12) into a polynomial function in $\psi$ and $\phi$ , we write Eq (2.13) into Eq (2.12) with Eq (2.3) and Eq (2.5). By solving polynomial, we obtain the system: in $a_{0}$ , $a_{i}$ , $b_{i} (i = 1, 2, 3, ..., N)$ , $\lambda (<0)$ , $\beta$ , $\eta$ , $\omega$ , $c$ , $D_{1}$ and $D_{2}$ . We solve this system with Mathematica. Setting values of above algebraic constants in Eq (2.13), solutions by hyperbolic functions in Eq (2.12) are obtained.

Postulate 5. Similar to Postulate 4, substituting Eq (2.13) into Eq (2.12), using Eq (2.3) and Eq (2.5) (or Eq (2.3) and Eq (2.7)), we obtain the exact traveling wave solutions of Eq (2.12) demonstrated by trigonometric functions.

3. Applications

Let us consider transformation:

$\begin{equation} u(x, t) = u(\xi ), \;\;\;\;\;\; \xi = x+c t, \end{equation}$

(3.1)

wherein $c$ is a parameter, which reduces Eq (1.2) to:

$\begin{equation} 9cu'+u^{(5)} +15uu'''+\frac{75}{2}u'u''+45u^{2}u' = 0. \end{equation}$

(3.2)

According to postulate 2, the positive number $N = 2$ is obtained by balancing between $u^{(5)}$ and $u^{2}u'$ , thus general solutions of Eq (3.2) is:

$\begin{equation} u (\xi) = a_{0} +a_{1} \varphi (\xi)+a_{2} \varphi^{2} (\xi )+b_{1} \psi (\xi )+b_{2} \varphi (\xi)\psi (\xi ), \end{equation}$

(3.3)

wherein $a_{0}$ , $a_{i}$ and $b_{i} (i = 1, 2)$ are constant coefficients such that $a_{N} ^{2}+b_{N} ^{2} \ne 0 (N = 1, 2)$ , $\phi (\xi)$ and $\psi (\xi)$ are satisfied by the Eq (2.3). Now, there are three categories of solutions of Eq (3.2):

Category 1: When $\lambda < 0$ (solutions by hyperbolic functions):

Writing Eq (3.3) with Eq (2.3) and Eq (2.5) into Eq (3.2), Eq (3.2) forms a polynomial in $\psi (\xi)$ and $\phi (\xi)$ . Solving this polynomial, we obtain a system: $a_{0}$ , $a_{1}$ , $a_{2}$ , $b_{1}$ , $b_{2}$ , $\lambda (<0)$ , $\beta$ , $c$ and $\alpha _{1}$ . Solving this system with Mathematica, we obtain the values of $a_{0}$ $a_{1}$ , $a_{2}$ , $b_{1}$ , $b_{2}$ , $\beta$ and $c$ as:

Result 1:

$\begin{equation} a_{0} = \frac{-10\lambda }{3}, \;\; a_{1} = 0, \;\;a_{2} = -4, \;\;b_{1} = 4\beta , \;\; b_{2} = \frac{\pm 4\sqrt{\beta ^{2} +\lambda ^{2} \alpha _{1} } }{\sqrt{-\lambda } } , \;\; c = \frac{-11\lambda ^{2} }{9} , \;\; \beta = \beta . \end{equation}$

(3.4)

Writing these constants from Eq (3.4) into (3.3) and by Eq (2.1) and Eq (2.4), we obtain explicit solutions of Eq (1.2):

$\begin{equation} \begin{matrix} u(\xi) = \frac{-10\lambda}{3}+\frac{4\lambda\left\{D_1\cosh(\xi\sqrt{-\lambda})D_2\sinh(\xi\sqrt{-\lambda})\right\}^2}{\left\{D_1\sinh(\xi\sqrt{-\lambda}+D_2\cosh(\xi\sqrt{-\lambda})+\frac{\beta}{\lambda}\right\}^2}+ \frac{4\beta}{\left\{D_1\sinh(\xi\sqrt{-\lambda}+D_2\cosh(\xi\sqrt{-\lambda})+\frac{\beta}{\lambda}\right\}}\\\\ \pm\frac{4\sqrt{\beta^2+\lambda^2\alpha_1}\lambda\left\{D_1\cosh(\xi\sqrt{-\lambda})D_2\sinh(\xi\sqrt{-\lambda})\right\}^2}{\left\{D_1\sinh(\xi\sqrt{-\lambda}+D_2\cosh(\xi\sqrt{-\lambda})+\frac{\beta}{\lambda}\right\}^2} \end{matrix} \end{equation}$

(3.5)

wherein $\xi = x-\frac{11\lambda ^{2} t}{9}$ and $\alpha _{1} = D_{1} ^{2} -D_{2} ^{2}$ .

In particular, if we choose $D_{1} \ne 0$ , $D_{2} = 0$ and $\beta = 0$ in Eq (3.5), we get:

$\begin{equation} \begin{array}{l} {u (x, t) = \frac{-10 \lambda }{3} +4 \lambda \coth \left(\sqrt{-\lambda } (x-\frac{11 \lambda ^{2} t}{9} ) \right) \left\{\coth \left(\sqrt{-\lambda } (x-\frac{11 \lambda ^{2} t}{9} ) \right)\right.} \\ \\{\;\;\;\qquad \left. \pm \csc h \left(\sqrt{-\lambda } (x-\frac{11 \lambda ^{2} t}{9} ) \right)\right\}.} \end{array} \end{equation}$

(3.6)

Similarly, if we choose $D_{2} \ne 0$ , $D_{1} = 0$ and $\beta = 0$ in Eq (3.5), we get:

$\begin{equation} \begin{array}{l} {u (x, t) = \frac{-10 \lambda }{3} +4\lambda \tanh \left(\sqrt{-\lambda } (x-\frac{11 \lambda ^{2} t}{9} ) \right) \left\{\tanh\left(\sqrt{-\lambda } (x-\frac{11 \lambda ^{2} t}{9} ) \right)\right. } \\\\ {\;\;\; \qquad \left. \pm i \sec h \left(\sqrt{-\lambda}(x-\frac{11 \lambda ^{2} t}{9}) \right)\right\}, } \end{array} \end{equation}$

(3.7)

Figure 1. 3D, contour and 2D surfaces of absolute Eq (3.7) when

$\lambda = -1$ .

DownLoad: Full-Size Img PowerPoint

wherein $i = \sqrt{-1}$ .

Result 2:

$\begin{equation} a_{0} = \frac{-5\lambda }{12} , \;\; a_{1} = 0, a_{2} = \frac{-1}{2} , \;\;\ b_{1} = \frac{\beta }{2} , \;\; b_{2} = \frac{\pm \sqrt{\beta ^{2} +\lambda ^{2} \alpha _{1} } }{2\sqrt{-\lambda } } , \;\; c = \frac{-\lambda ^{2} }{144} , \;\; \beta = \beta . \end{equation}$

(3.8)

Explicit solutions of Eq (1.2) are given by:

$\begin{equation} \begin{array}{l} {u (\xi ) = \frac{-5\lambda }{12} + \frac{\lambda \left\{ D_{1} \; \cosh (\xi \sqrt{-\lambda } ) +D_{2} \sinh (\xi \sqrt{-\lambda } ) \right\}^{2} }{2 \left\{D_{1} \sinh (\xi \sqrt{-\lambda } ) +D_{2} \cosh (\xi \sqrt{-\lambda } ) +\frac{\beta }{\lambda } \right\}^{2} }} {+\frac{ \beta }{2 \left\{D_{1} \sinh (\xi \sqrt{-\lambda } ) +D_{2} \; \cosh (\xi \sqrt{-\lambda } ) +\frac{\beta }{\lambda } \right\}} } \\\\ {\qquad\; \pm \frac{\sqrt{\beta ^{2} +\lambda ^{2} \alpha _{1} } \left\{ D_{1} \cosh (\xi \sqrt{-\lambda } ) +D_{2} \sinh (\xi \sqrt{-\lambda } ) \right\}}{2 \left\{D_{1} \sinh (\xi \sqrt{-\lambda } ) +D_{2} \cosh (\xi \sqrt{-\lambda } ) +\frac{\beta }{\lambda } \right\}^{2}}, } \end{array} \end{equation}$

(3.9)

wherein $\xi = x-\frac{\lambda ^{2} t}{144}$ and $\alpha _{1} = D_{1} ^{2} -D_{2} ^{2}$ .

In particular, if we choose $D_{1} \ne 0$ , $D_{2} = 0$ and $\beta = 0$ in Eq (3.9), we get:

$\begin{equation} \begin{array}{l} {u (x, t) = \frac{-5\lambda }{12} +\frac{\lambda }{2} \coth \left(\sqrt{-\lambda } (x-\frac{\lambda ^{2} t}{144} )\right) \left\{\coth \left(\sqrt{-\lambda } (x-\frac{\lambda ^{2} t}{144} )\right)\right.} \\\\ {\left. \qquad \;\; \pm \csc h \left(\sqrt{-\lambda } (x-\frac{\lambda ^{2} t}{144})\right)\right\}.} \end{array} \end{equation}$

(3.10)

Similarly, if we choose $D_{2} \ne 0$ , $D_{1} = 0$ and $\beta = 0$ in Eq (3.9), we get:

$\begin{equation} \begin{array}{l} {u (x, t) = \frac{-5 \lambda }{12} +\frac{\lambda }{2} \tanh \left(\sqrt{-\lambda } (x-\frac{\lambda ^{2} t}{144} )\right) \left\{\tanh \left(\sqrt{-\lambda } (x-\frac{\lambda ^{2} t}{144} )\right)\right.} \\ \\{\left. \qquad \;\; \pm i \sec h \left(\sqrt{-\lambda } (x-\frac{\lambda ^{2} t}{144} )\right)\right\}, } \end{array} \end{equation}$

(3.11)

wherein $i = \sqrt{-1}$ .

Result3:

$\begin{equation} \begin{array}{l} a_{0} = -\frac{11\lambda \beta ^{2} +8\lambda ^{3} \alpha _{1} }{12(\beta ^{2} +\lambda ^{2} \alpha _{1} )} , \;\; a_{1} = 0, a_{2} = -1, \;\; b_{1} = \beta, \\ \\ b_{2} = 0, \;\; c = \frac{-\lambda ^{2} (\beta ^{4} -28\lambda ^{2} \beta ^{2} \alpha _{1} +16\lambda ^{4} \alpha _{1} ^{2} )}{144(\beta ^{2} +\lambda ^{2} \alpha _{1} )^{2} } , \;\; \beta = \beta. \end{array} \end{equation}$

(3.12)

wherein $\beta ^{2} +\lambda ^{2} \alpha _{1} \ne 0$ .

We get explicit solutions of Eq (1.2) as:

$\begin{equation} \begin{array}{l} {u (\xi ) = -\frac{11 \lambda \beta ^{2} +8 \lambda ^{3} \alpha _{1} }{12 (\beta ^{2} +\lambda ^{2} \alpha _{1} )} +\frac{\lambda \left\{ D_{1} \cosh (\xi \sqrt{-\lambda } ) +D_{2} \sinh (\xi \sqrt{-\lambda } ) \right\}^{2} }{ \left\{D_{1} \sinh (\xi \sqrt{-\lambda } ) +D_{2} \cosh (\xi \sqrt{-\lambda } ) +\frac{\beta }{\lambda } \right\}^{2} } } \\\\ {\qquad\;\; +\frac{\beta }{\left\{D_{1} \sinh (\xi \sqrt{-\lambda } ) +D_{2} \cosh (\xi \sqrt{-\lambda } ) +\frac{\beta }{\lambda } \right\}}, } \end{array} \end{equation}$

(3.13)

wherein $\xi = x-\frac{\lambda ^{2} t (\beta ^{4} -28 \lambda ^{2} \beta ^{2} \alpha _{1} +16 \lambda ^{4} \alpha _{1}^{2})}{144 (\beta ^{2} +\lambda ^{2} \alpha _{1})^{2} }$ and $\alpha _{1} = D_{1} ^{2} -D_{2} ^{2}$ .

In particular, if we choose $D_{1} \ne 0$ , $D_{2} = 0$ and $\beta = 0$ in Eq (3.13), we get:

$\begin{equation} u (x, t) = \frac{-2\lambda }{3} +\lambda \coth ^{2} \left(\sqrt{-\lambda } (x-\frac{ \lambda ^{2} t}{9})\right). \end{equation}$

(3.14)

Figure 2. 3D, contour and 2D surfaces of absolute Eq (3.14) when

$\lambda = -5.$ .

DownLoad: Full-Size Img PowerPoint

Similarly, if we choose $D_{2} \ne 0$ , $D_{1} = 0$ and $\beta = 0$ in Eq (3.14), we get:

$\begin{equation} u (x, t) = \frac{-2\lambda }{3} +\lambda \tanh ^{2} \left(\sqrt{-\lambda } (x-\frac{ \lambda ^{2} t}{9})\right). \end{equation}$

(3.15)

Category 2: For $\lambda > 0,$ (i.e. trigonometric functions),

According to Postulate 5, if we execute as the category 1, we attain the values of $a_{0}$ , $a_{1}$ , $a_{2}$ , $b_{1}$ , $b_{2}$ , $\beta$ and $c$ as the following results:

Result 1:

$\begin{equation} \begin{array}{l} a_{0} = \frac{-10\lambda }{3} , \;\; a_{1} = 0, a_{2} = -4, \;\; b_{1} = 4\beta , \;\; b_{2} = \frac{\pm 4\sqrt{-\beta ^{2} +\lambda ^{2} \alpha _{1} } }{\sqrt{\lambda}}, \\\\ c = \frac{-11\lambda ^{2} }{9} , \;\;\beta = \beta. \end{array} \end{equation}$

(3.16)

Writing constants in Eq (3.16) into Eq (3.3) and by Eq (2.1) and Eq (2.6), we get explicit solutions of Eq (1.2):

$\begin{equation} \begin{array}{l} {u (\xi ) = \frac{-10\lambda }{3} - \frac{4 \lambda \left\{ D_{1} \cos (\xi \sqrt{\lambda } )-D_{2} \sin (\xi \sqrt{\lambda } )\right\}^{2} }{\left\{D_{1} \sin (\xi \sqrt{\lambda } )+D_{2} \; \cos (\xi \sqrt{\lambda } )+\frac{\beta }{\lambda } \right\}^{2} } } {+\frac{4 \beta }{D_{1} \sin (\xi \sqrt{\lambda } )+D_{2} \cos (\xi \sqrt{\lambda } )+\frac{\beta }{\lambda } } } \\\\ { \qquad \; \pm \frac{4\sqrt{-\beta ^{2} +\lambda ^{2} \alpha _{2} } \left\{ D_{1} \cos (\xi \sqrt{\lambda } )-D_{2} \sin (\xi \sqrt{\lambda } )\right\}}{\left\{D_{1} \sin (\xi \sqrt{\lambda } )+D_{2} \cos (\xi \sqrt{\lambda } )+\frac{\beta }{\lambda } \right\}^{2}}, } \end{array} \end{equation}$

(3.17)

wherein $\xi = x-\frac{ 11 \lambda ^{2} t }{9}$ and $\alpha _{2} = D_{1} ^{2} +D_{2} ^{2}$ .

Result 2:

$\begin{equation} a_{0} = \frac{-5\lambda }{12} , \;\; a_{1} = 0, \;\; a_{2} = \frac{-1}{2} , \;\; b_{1} = \frac{\beta }{2} , \;\; b_{2} = \frac{\pm \sqrt{-\beta ^{2} +\lambda ^{2} \alpha _{1} } }{2\sqrt{\lambda } } , \;\; c = \frac{-\lambda ^{2} }{144} , \;\; \beta = \beta. \end{equation}$

(3.18)

We get explicit solutions of Eq (1.2) as:

$\begin{equation} \begin{array}{l} {u (\xi ) = \frac{-5\lambda }{12} - \frac{\lambda \left\{ D_{1} \cos (\xi \sqrt{\lambda } )-D_{2} \sin (\xi \sqrt{\lambda } )\right\}^{2} }{2 \left\{D_{1} \sin (\xi \sqrt{\lambda } )+D_{2} \cos (\xi \sqrt{\lambda } )+\frac{\beta }{\lambda } \right\}^{2} } } { +\frac{\beta }{2\left\{D_{1} \sin (\xi \sqrt{\lambda } )+D_{2} \cos (\xi \sqrt{\lambda } )+\frac{\beta }{\lambda } \right\}} } \\ \\{\qquad\; \pm \frac{\sqrt{-\beta ^{2} +\lambda ^{2} \alpha _{2} } \left\{ D_{1} \cos (\xi \sqrt{\lambda } )-D_{2} \sin (\xi \sqrt{\lambda } )\right\}}{2 \left\{D_{1} \sin (\xi \sqrt{\lambda } )+D_{2} \cos (\xi \sqrt{\lambda } )+\frac{\beta }{\lambda } \right\}^{2} }, } \end{array}. \end{equation}$

(3.19)

wherein $\xi = x-\frac{ \lambda ^{2} t }{144}$ and $\alpha _{2} = D_{1} ^{2} +D_{2} ^{2}$ .

Result 3:

$\begin{equation} \begin{array}{l} a_{0} = \frac{11\lambda \beta ^{2} -8\lambda ^{3} \alpha _{2} }{12(-\beta ^{2} +\lambda ^{2} \alpha _{2} )} , \;\; a_{1} = 0, a_{2} = -1, \;\; b_{1} = \beta , \\\\ b_{2} = 0, \;\; c = \frac{-\lambda ^{2} (\beta ^{4} +28\lambda ^{2} \beta ^{2} \alpha _{2} +16\lambda ^{4} \alpha _{2} ^{2} )}{144(-\beta ^{2} +\lambda ^{2} \alpha _{2} )^{2} } , \;\; \beta = \beta. \end{array} \end{equation}$

(3.20)

wherein $-\beta ^{2} +\lambda ^{2} \alpha _{2} \ne 0$ .

We get explicit solutions of Eq (1.2) as:

$\begin{equation} \begin{array}{l} {u (\xi ) = \frac{11 \lambda \beta ^{2} -8 \lambda ^{3} \alpha _{2} }{12 (-\beta ^{2} +\lambda ^{2} \alpha _{2} )} -\frac{\lambda \left\{ D_{1} cos (\xi \sqrt{\lambda } )-D_{2} \sin (\xi \sqrt{\lambda } )\right\}^{2} }{\left\{D_{1} \sin (\xi \sqrt{\lambda } )+D_{2} \cos (\xi \sqrt{\lambda } )+\frac{\beta }{\lambda } \right\}^{2} } } \\\\ {\qquad \; +\frac{\beta }{\left\{D_{1} \sin (\xi \sqrt{\lambda } )+D_{2} \cos (\xi \sqrt{\lambda } )+\frac{\beta }{\lambda } \right\}} , } \end{array} \end{equation}$

(3.21)

Figure 3. 3D, contour and 2D surfaces of Eq (3.21) when

$\lambda = 3, D_{1} = 0.8 , D_{2} = 0.5, \beta = 3$ .

DownLoad: Full-Size Img PowerPoint

wherein $\xi = x-\frac{\lambda ^{2} t (\beta ^{4} +28 \lambda ^{2} \beta ^{2} \alpha _{2} +16 \lambda ^{4} \alpha _{2}^{2})}{144 (-\beta ^{2} +\lambda ^{2} \alpha _{2})^{2} }$ and $\alpha _{2} = D_{1} ^{2} +D_{2} ^{2}$ .

Category 3: For $\lambda = 0,$ (i.e.rational functions),

According to Postulate 5, if we execute as the category 1, we attain the values of $a_{0}$ , $a_{1}$ , $a_{2}$ , $b_{1}$ , $b_{2}$ , $\beta$ and $c$ as the following results:

$\begin{equation} a_{0} = \frac{- \beta ^{2} }{4 (-D_{1} ^{2} +2 \beta D_{2} )} , \;\; a_{1} = 0, \;\; a_{2} = -1, \;\; b_{1} = \beta , \;\; b_{2} = 0, \;\; c = \frac{-5\beta ^{4} }{16 (-D_{1} ^{2} +2 \beta D)^{2} } , \;\; \beta = \beta. \end{equation}$

(3.22)

We get explicit solutions of Eq (1.2) as:

$\begin{equation} u(\xi ) = \frac{- \beta ^{2} }{4 (-D_{1} ^{2} +2 \beta D_{2} )} -\frac{ \left(\beta \xi +D_{1} \right)^{2} }{\left(\frac{\beta }{2} \xi ^{2} +D_{1} \xi +D_{2} \right)^{2} } +\frac{ \beta }{\left(\frac{\beta }{2} \xi ^{2} +D_{1} \xi +D_{2} \right)}, \end{equation}$

(3.23)

wherein $\xi = x-\frac{5 \beta ^{4} t}{16 (-D_{1} ^{2} +2 \beta D)^{2} }$ and $-D_{1} ^{2} +2 \beta D\ne 0$ .

If we set up the particular values of the arbitrary constants if we choose $D_{1}$ , $D_{2}$ and $\beta$ in the above Eq (3.17), Eq (3.19), Eq (3.21) and Eq (3.23), we attain abundant new explicit wave solutions of KK equation which are unexposed for minimalism of length of the paper.

4. Conclusions and outlook

We obtained new explicit solutions for the (1+1)-dimensional KK equation. We achieved solitary wave solutions for analogous traveling wave solutions of Eq (1.2). These affluent solutions including bell and anti-bell solitons, kink and anti-kink solitons, periodic and rational functions of KK equation indicate that double $(G'/G, 1/G)$ -expansion technique is more powerful than the method of $(G'/G, 1/G)$ -expansion. Comparing the solutions with the ones in ^[33], we presume that all the solutions are renewed which are un-indicted elsewhere. Our mentioned method is more powerful and also an offering method to demonstrate many higher order nonlinear PDEs. We will investigate the applicability of the method to (2+1)-dimensional KK equation in a future extension of the present work.

Acknowledgments

The work was supported by the Natural Science Foundation of China (Grant Nos. 61673169, 11301127, 11701176, 11626101, 11601485).

Conflict of interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

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