# Conservation Laws and Exact Solutions of a Generalized Zakharov–Kuznetsov Equation

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International Institute for Symmetry Analysis and Mathematical Modelling, Department of Mathematical Sciences, North-West University, Mafikeng Campus, Private Bag X 2046, Mmabatho 2735, South Africa

Author to whom correspondence should be addressed.

Academic Editor: Sergei Odintsov

Received: 22 January 2015 / Revised: 3 April 2015 / Accepted: 21 April 2015 / Published: 3 June 2015

In this paper, we study a generalized Zakharov–Kuznetsov equation in three variables, which has applications in the nonlinear development of ion-acoustic waves in a magnetized plasma. Conservation laws for this equation are constructed for the first time by using the new conservation theorem of Ibragimov. Furthermore, new exact solutions are obtained by employing the Lie symmetry method along with the simplest equation method.

Many important phenomena and dynamic processes in physics, applied mathematics and engineering can be described by higher-dimensional extensions of the Korteweg–de Vries (KdV) equation. Zakharov and Kuznetsov successfully proposed one such model [1]. The Zakharov–Kuznetsov (ZK) equation given by:

$${u}_{t}+\alpha u{u}_{x}+\beta {\left({u}_{xx}+{u}_{yy}\right)}_{x}=0$$

This paper aims to study the generalized Zakharov–Kuznetsov (gZK) equation [4,8]:

$${u}_{t}+\alpha {u}^{n}{u}_{x}+\beta {\left({u}_{xx}+{u}_{yy}\right)}_{x}=0$$

It is of great importance to search for exact solutions of nonlinear partial differential equations (NPDEs), such as the gZK equation, because many physical phenomena are described by NPDEs. Although there is no unique method for finding exact solutions of NLPEs, a great deal of research work has been devoted to developing different methods to solve NLPEs. Some of the methods found in the literature include the inverse scattering transform method [9], Darboux transformation [10], Hirota’s bilinear method [11], Bäcklund transformation [12], the multiple exp-function method [13], the (G′/G)-expansion method [14], the sine-cosine method [15], the F -expansion method [16], the exp-function expansion method [17] and the Lie symmetry method [18,19].

There is no doubt that in the study of differential equations, conservation laws play an important role. In fact, conservation laws describe physical conserved quantities, such as mass, energy, momentum and angular momentum, as well as charge and other constants of motion [20,21]. They have been used in investigating the existence, uniqueness and stability of solutions of nonlinear partial differential equations [22–24]. Furthermore, they have been used in the development and use of numerical methods [25,26]. Recently, conservation laws were used to obtain exact solutions of some partial differential equations [27–31]. Thus, it is essential to study the conservation laws of partial differential equations.

The paper is organized as follows: In Section 2, we derive conservation laws of (2) by employing the new conservation law theorem by Ibragimov [32]. In Section 3, we obtain exact solutions of (2) using Lie symmetry analysis and the simplest equation method [33–35]. Finally, concluding remarks are presented in Section 4.

In this section, the new conservation theorem by Ibragimov [32] will be used to construct conservation laws for (2). To use the conservation theorem by Ibragimov [32], we need to know the Lie point symmetries of (2). Thus, we first compute the symmetries of (2).

The vector field:

$$X={\xi}^{1}(x,y,t,u)\frac{\partial}{\partial x}+{\xi}^{2}\left(x,y,t,u\right)\frac{\partial}{\partial y}+{\xi}^{3}\left(x,y,t,u\right)\frac{\partial}{\partial t}+\eta \left(x.y,t,u\right)\frac{\partial}{\partial u}$$

$${X}^{[3]}[{u}_{t}+\alpha {u}^{n}{u}_{x}+\beta {({u}_{xx}+{u}_{yy})}_{x}]=0$$

$$\begin{array}{l}{\xi}_{t}^{3}=0,\phantom{\rule{0.2em}{0ex}}{\xi}_{y}^{2}=0,\phantom{\rule{0.2em}{0ex}}{\xi}_{x}^{3}=0,\phantom{\rule{1em}{0ex}}{\xi}_{u}^{3}=0,\phantom{\rule{0.2em}{0ex}}{\xi}_{u}^{2}=0,\phantom{\rule{1em}{0ex}}{\xi}_{x}^{1}=0,\\ {\xi}_{y}^{1}=0,\phantom{\rule{0.2em}{0ex}}{\xi}_{u}^{1}=0,\phantom{\rule{0.2em}{0ex}}{\xi}_{xx}^{2}=0,\phantom{\rule{1em}{0ex}}{\eta}_{xu}=0,\phantom{\rule{0.2em}{0ex}}{\eta}_{uu}=0,\phantom{\rule{1em}{0ex}}{\xi}_{yy}^{3}-2{\eta}_{yu}=0,\\ \beta u{n}_{yyu}-u{\xi}_{t}^{2}+2\alpha {u}^{n+1}{\xi}_{x}^{2}+n\alpha {u}^{n}\eta =0,\\ \beta u{\eta}_{yyu}-u{\xi}_{t}^{2}+2\alpha {u}^{n+1}{\xi}_{y}^{3}+n\alpha {u}^{n}\eta =0,\\ \beta {\eta}_{xxx}+\beta {\eta}_{xyy}+\alpha {u}^{n}{\eta}_{x}+{\eta}_{t}=0,\\ \beta u{\eta}_{yyu}-u{\xi}_{t}^{2}+n\alpha {u}^{n}\eta +\alpha {u}^{n+1}{\xi}_{t}^{1}-\alpha {u}^{n+1}{\xi}_{x}^{2}=0.\end{array}$$

Solving the above system of partial differential equations, one obtains the following four Lie point symmetries:

$${X}_{1}=\frac{\partial}{\partial t},\phantom{\rule{1em}{0ex}}{X}_{2}=\frac{\partial}{\partial x},\phantom{\rule{1em}{0ex}}{X}_{3}=\frac{\partial}{\partial y},\phantom{\rule{1em}{0ex}}{X}_{4}=3nt\frac{\partial}{\partial t}+\phantom{\rule{0.2em}{0ex}}nx\frac{\partial}{\partial x}+\phantom{\rule{0.2em}{0ex}}ny\frac{\partial}{\partial y}-2u\frac{\partial}{\partial u}$$

The gZK equation together with its adjoint equation are given by:

$$F\equiv {u}_{t}+\alpha {u}^{n}{u}_{x}+\beta {u}_{xxx}+\beta {u}_{xyy}=0$$

$$F*\equiv {\upsilon}_{t}+\alpha {\upsilon}_{x}{u}^{n}+\beta {\upsilon}_{xxx}+\beta {\upsilon}_{xyy}=0$$

The third-order Lagrangian for the system of Equations (3a) and (3b) is given by:

$$L=\upsilon \left({u}_{t}+\alpha {u}^{n}{u}_{x}+\beta {u}_{xxx}+\beta {u}_{xyy}\right)$$

$$L=\upsilon \left({u}_{t}+\alpha {u}^{n}{u}_{x}\right)-\beta {\upsilon}_{x}{u}_{xx}-\beta {\upsilon}_{x}{u}_{yy}$$

We have the following four cases:

We first consider the Lie point symmetry X

_{1}= ∂_{t}of (2). Corresponding to this symmetry, the Lie characteristic functions are W^{1}= −u_{t}. and W^{2}= −v_{t}. Thus, by using the Ibragimov theorem [32], the components of the conserved vector associated with the symmetry X_{1}= ∂_{t}are given by:$$\begin{array}{l}{C}_{1}^{t}=\alpha \upsilon {u}^{n}{u}_{x}-\beta {\upsilon}_{x}\left({u}_{xx}+{u}_{yy}\right),\\ {C}_{1}^{x}=\beta {\upsilon}_{x}{u}_{tx}+\beta {\upsilon}_{t}\left({u}_{xx}+{u}_{yy}\right)-\alpha \upsilon {u}^{n}{u}_{t}-\beta {u}_{t}{\upsilon}_{xx},\\ {C}_{1}^{y}=\beta {\upsilon}_{x}{u}_{ty}-\beta {u}_{t}{\upsilon}_{xy}.\end{array}$$Likewise, the Lie point symmetry X

_{2}= ∂_{x}has the Lie characteristic functions W^{1}= −u_{x}and W^{2}= −v_{x}. Invoking Ibragimov’s theorem, we obtain the conserved vector, whose components are:$$\begin{array}{l}{C}_{2}^{t}=-\upsilon {u}_{x},\\ {C}_{2}^{x}=\beta {\upsilon}_{x}{u}_{xx}-\beta {u}_{x}{\upsilon}_{xx}+\upsilon {u}_{t},\\ {C}_{2}^{y}=\beta {\upsilon}_{x}{u}_{xy}-\beta {u}_{x}{\upsilon}_{xy}.\end{array}$$The Lie point symmetry X

_{3}= ∂_{y}has the Lie characteristic functions W^{1}= −u and W^{2}= −v_{y}, and using Ibragimov’s theorem, the components of the conserved vector are:$$\begin{array}{l}{C}_{3}^{t}=-\upsilon {u}_{y},\\ {C}_{3}^{x}=\beta {\upsilon}_{x}{u}_{xy}+\beta {\upsilon}_{y}\left({u}_{xx}+{u}_{yy}\right)-\beta {u}_{y}{\upsilon}_{xx}-\alpha \upsilon {u}_{n}{u}_{y},\\ {C}_{3}^{y}=\upsilon \left({u}_{t}+\alpha {u}^{n}{u}_{x}\right)-\beta {\upsilon}_{x}{u}_{xx}-\beta {u}_{y}{\upsilon}_{xy}.\end{array}$$Finally, the Lie point symmetry X

_{4}= 3nt∂_{t}+ nx∂_{x}+ ny∂_{y}− 2u∂_{u}gives W^{1}= −(2u + 3ntu_{t}+ nxu_{x}+ nyu_{y}) and W^{2}= (2 − 2n)v − 3ntv_{t}− nxv_{x}− nyv_{y}, and so, the associated conserved vector has components:$$\begin{array}{l}{C}_{4}^{t}=\upsilon \left(3\alpha nt{u}^{n}{u}_{x}-2u-nx{u}_{x}-ny{u}_{y}\right)-3\beta nt{\upsilon}_{x}\left({u}_{xx}+{u}_{yy}\right),\\ {C}_{4}^{x}=\upsilon \left(nx{u}_{t}-2\alpha {u}^{n+1}-3\alpha nt{u}^{n}{u}_{t}-\alpha ny{u}^{n}{u}_{y}\right)-\beta ny{\upsilon}_{y}\left({u}_{xx}+{u}_{yy}\right),\\ \phantom{\rule{2em}{0ex}}+3\beta nt{\upsilon}_{t}\left({u}_{xx}+{u}_{yy}\right)+\beta {\upsilon}_{x}\left(2{u}_{x}+3nt{u}_{tx}+n{u}_{x}+nx{u}_{xx}+ny{u}_{xy}\right)\\ \phantom{\rule{2em}{0ex}}-\beta {\upsilon}_{xx}\left(2u+3nt{u}_{t}+nx{u}_{x}+ny{u}_{y}\right)-2\beta \upsilon \left({u}_{xx}+{u}_{yy}\right)+2\beta n\upsilon \left({u}_{xx}+{u}_{yy}\right),\\ {C}_{4}^{y}=\beta {\upsilon}_{x}\left(2{u}_{y}+3nt{u}_{ty}+nx{u}_{xy}+n{u}_{y}-ny{u}_{xx}\right)-\beta {\upsilon}_{xy}\left(2u+3nt{u}_{t}+nx{u}_{x}+ny{u}_{y}\right)\\ \phantom{\rule{2em}{0ex}}+ny\upsilon \left({u}_{t}+\alpha {u}^{n}{u}_{x}\right).\end{array}$$

In this section, we obtain exact solutions of (2) using firstly its Lie point symmetries and, secondly, by employing the simplest equation method.

First of all, we utilize the linear combination of the three translation symmetries, namely X = X_{1} + νX_{2} + X_{3}, and reduce the gZK Equation (2) to a PDE in two independent variables. The associated Lagrange system is:

$$\frac{dt}{1}=\frac{dx}{\upsilon}=\frac{dy}{1}=\frac{du}{0}$$

$$f=t-y,\phantom{\rule{1em}{0ex}}g=x-\upsilon y,\phantom{\rule{1em}{0ex}}\theta -u$$

By considering θ as the new dependent variable and f and g as new independent variables, the gZK Equation (2) transforms to:

$${\theta}_{f}+\alpha {\theta}^{n}{\theta}_{g}+\beta \left({\upsilon}^{2}+1\right){\theta}_{ggg}+2\beta \upsilon {\theta}_{fgg}+\beta {\theta}_{ffg}=0$$

$${\mathrm{\Gamma}}_{1}=\frac{\partial}{\partial f},\phantom{\rule{1em}{0ex}}{\mathrm{\Gamma}}_{2}=\frac{\partial}{\partial g}$$

The combination Γ_{1} + kΓ_{2}, of the two symmetries Γ_{1} and Γ_{2}, for an arbitrary constant k, yields the two invariants:

$$z=g-kf\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}W=\theta $$

$$\beta \left(1+{\left(\nu -k\right)}^{2}\right){W}^{\u2034}+\alpha {W}^{n}{W}^{\prime}-k{W}^{\prime}=0$$

The integration of (8) yields

$$\beta \left(1+{\left(\nu -k\right)}^{2}\right){W}^{\u2033}+\frac{\alpha}{n+1}{W}^{n+1}-kW=0$$

$$\frac{\beta \left(1+{\left(\nu -k\right)}^{2}\right)}{2}{{W}^{\prime}}^{2}+\frac{\alpha}{\left(n+1\right)\left(n+2\right)}{W}^{n+2}-\frac{k}{2}{W}^{2}=0$$

$$u\left(t,x,y\right)={\left(\frac{k\left(n+1\right)\left(n+2\right)}{2\alpha}\right)}^{\frac{1}{n}}{\text{sech}}^{\frac{2}{n}}\left(R\right)$$

$$\begin{array}{l}R=\frac{n\sqrt{k}\left({C}_{1}+z\right)}{2\sqrt{\beta \left(1+{\left(\nu -k\right)}^{2}\right)}}\\ z=x-kt-\left(\nu -k\right)y\end{array}$$

Note that (11) represents a non-topological soliton solution. A sketch of the solution (11) with n = 2, α = 2, k = 5, ν = 1, β = 1, t = 0 and C_{1} = 1 is given in Figure 1.

In this subsection, we use the simplest equation method [33–35] to solve the nonlinear third-order ODE (8) for n = 1, 2. The simplest equations that we use here are the Bernoulli equation:

$${H}^{\prime}\left(z\right)=\alpha H\left(z\right)+b{H}^{2}\left(z\right)$$

$${G}^{\prime}\left(z\right)=a{G}^{2}\left(z\right)+bG\left(z\right)+c$$

$$W\left(z\right)={\displaystyle \sum _{i=0}^{M}{A}_{i}{(G(z))}^{i}}$$

The solution of Bernoulli Equation (12) we use here is given by:

$$H\left(z\right)=a\left\{\frac{\mathrm{cosh}\left[a\left(z+C\right)\right]+\mathrm{sinh}\left[a\left(z+C\right)\right]}{1-b\mathrm{cosh}\left[a\left(z+C\right)\right]-b\mathrm{sinh}\left[a\left(z+C\right)\right]}\right\}$$

$$G\left(z\right)=-\frac{b}{2a}-\frac{\theta}{2a}\mathrm{tanh}\left[\frac{1}{2}\theta \left(z+C\right)\right]$$

$$G\left(z\right)=-\frac{b}{2a}-\frac{\theta}{2a}\mathrm{tanh}\left(\frac{1}{2}\theta z\right)+\frac{\text{sech}\left(\frac{\theta z}{2}\right)}{C\mathrm{cosh}\left(\frac{\theta z}{2}\right)-\frac{2a}{\theta}\mathrm{sinh}\left(\frac{\theta z}{2}\right)}$$

**n = 1**

In this case, the balancing procedure yields M = 2 and solutions of (8) are of the form:

$$W\left(z\right)={A}_{0}+{A}_{1}G+{A}_{2}{G}^{2}$$

We insert this value of W (z) in (8). Then, using the Bernoulli Equation (12) and, thereafter, equating the coefficients of powers of G^{i} to zero, we obtain an algebraic system of five equations in terms of A_{0}, A_{1}, A_{2}, namely:

$$\begin{array}{r}24{b}^{3}\beta {k}^{2}{A}_{2}-48{b}^{3}\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{2}+24{b}^{3}\beta {\nu}^{2}{A}_{2}+24{b}^{3}\beta \phantom{\rule{0.2em}{0ex}}{A}_{2}+2\alpha b{A}_{2}{}^{2}=0,\\ {a}^{3}\beta {k}^{2}{A}_{1}-2{a}^{3}\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{1}+{a}^{3}\beta {\nu}^{2}{A}_{1}+{a}^{3}\beta \phantom{\rule{0.2em}{0ex}}{A}_{1}+a\alpha {A}_{0}{A}_{1}-ak{A}_{1}=0,\\ 54a{b}^{2}\beta {k}^{2}{A}_{2}-108\phantom{\rule{0.2em}{0ex}}a{b}^{2}\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{2}+54\phantom{\rule{0.2em}{0ex}}a{b}^{2}\beta {\nu}^{2}{A}_{2}+6{b}^{3}\beta {k}^{2}{A}_{1}-12{b}^{3}\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{1}\phantom{\rule{1.5em}{0ex}}\\ +6{b}^{3}\beta {\nu}^{2}{A}_{1}+54a{b}^{2}\beta {A}_{2}+6{b}^{3}\beta {A}_{1}+2a\alpha {A}_{2}{}^{2}+3\alpha b{A}_{1}{A}_{2}=0,\\ 38{a}^{2}b\beta {k}^{2}{A}_{2}-76{a}^{2}b\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{2}+38{a}^{2}b\beta {\nu}^{2}{A}_{2}+12a{b}^{2}\beta {k}^{2}{A}_{1}-24\phantom{\rule{0.2em}{0ex}}\alpha {b}^{2}\beta \phantom{\rule{0.2em}{0ex}}k\nu \phantom{\rule{0.2em}{0ex}}{A}_{1}\phantom{\rule{1.5em}{0ex}}\\ +12a{b}^{2}\beta {\nu}^{2}{A}_{1}+38{a}^{2}b\beta {A}_{2}+12a{b}^{2}\beta {A}_{1}+3a\alpha {A}_{1}{A}_{2}+2\alpha b{A}_{0}{A}_{2}\phantom{\rule{1.5em}{0ex}}\\ +\alpha b{A}_{1}{}^{2}-2bk{A}_{2}=0,\\ 8{a}^{3}\beta {k}^{2}{A}_{2}-16{a}^{3}\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{2}+8{a}^{3}\beta {\nu}^{2}{A}_{2}+7{a}^{2}b\beta {k}^{2}{A}_{1}-14{\alpha}^{2}b\beta \phantom{\rule{0.2em}{0ex}}k\nu \phantom{\rule{0.2em}{0ex}}{A}_{1}\phantom{\rule{1.5em}{0ex}}\\ +7{a}^{2}b\beta {\nu}^{2}{A}_{1}+8{a}^{3}\beta {A}_{2}+7{a}^{2}b\beta {A}_{1}+2a\alpha {A}_{0}{A}_{2}+a\alpha {A}_{1}{}^{2}\\ +\alpha b{A}_{0}{A}_{1}-2ak{A}_{2}-bk{A}_{1}=0.\end{array}$$

With the aid of Maple, we solve the above system and obtain:

$$\begin{array}{l}{A}_{0}=\frac{1}{\alpha}\{2{a}^{2}\beta k\nu -{a}^{2}\beta {\nu}^{2}-{a}^{2}\beta -{a}^{2}\beta {k}^{2}+k\},{A}_{1}=\frac{1}{\alpha}\{12ab\beta (2k\nu -{\nu}^{2}-{k}^{2}-1)\},\\ {A}_{2}=\frac{1}{a}\{12{b}^{2}\beta (2k\nu -{\nu}^{2}-{k}^{2}-1)\}.\end{array}$$

Therefore, the solution of (2), for n = 1 is given by:

$$\begin{array}{r}u(t,x,y)={A}_{0}+a{A}_{1}\left\{\frac{\mathrm{cosh}[a\left(z+C\right)]+\mathrm{sinh}\left[a\left(z+C\right)\right]}{1-b\mathrm{cosh}\left[a\left(z+C\right)\right]-b\mathrm{sinh}\left[a\left(z+C\right)\right]}\right\}\\ +{A}_{2}{a}^{2}{\left\{\frac{\mathrm{cosh}[a\left(z+C\right)]+\mathrm{sinh}\left[a\left(z+C\right)\right]}{1-b\mathrm{cosh}\left[a\left(z+C\right)\right]-b\mathrm{sinh}\left[a\left(z+C\right)\right]}\right\}}^{2}\end{array}$$

**n = 2**

The balancing procedure yields M = 1, so the solutions of (8) take the form:

$$W\left(z\right)={A}_{0}+{A}_{1}G$$

As before, substituting (17) into (8), we obtain the algebraic system of equations:

$$\begin{array}{r}6{b}^{3}\beta {k}^{2}{A}_{1}-12{b}^{3}\beta k\nu {A}_{1}+6{b}^{3}\beta {\nu}^{2}{A}_{1}+\alpha b{A}_{1}{}^{3}+6{b}^{3}\beta {A}_{1}=0,\\ {a}^{3}\beta {k}^{2}{A}_{1}-2{a}^{3}\beta k\nu {A}_{1}+{a}^{3}\beta {\nu}^{2}{A}_{1}+{\alpha}^{3}\beta {A}_{1}+a\alpha {A}_{0}{}^{2}{A}_{1}-ak{A}_{1}=0,\\ 12a{b}^{2}\beta {k}^{2}{A}_{1}-24a{b}^{2}\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{1}+12a{b}^{2}\beta {\nu}^{2}{A}_{1}+a\alpha {A}_{1}{}^{3}+12a{b}^{2}\beta {A}_{1}\phantom{\rule{1.5em}{0ex}}\\ +2\alpha b{A}_{0}{A}_{1}{}^{2}=0,\\ 7{a}^{2}b\beta {k}^{2}{A}_{1}-14{a}^{2}b\beta \phantom{\rule{0.2em}{0ex}}k\nu \phantom{\rule{0.2em}{0ex}}{A}_{1}+7{a}^{2}b\beta {\nu}^{2}{A}_{1}+7{a}^{2}b\beta {A}_{1}+2a\alpha {A}_{0}{A}_{1}{}^{2}\phantom{\rule{1.5em}{0ex}}\\ +\alpha b{A}_{0}{}^{2}{A}_{1}-bk{A}_{1}=0,\end{array}$$

$${A}_{0}=\pm \sqrt{\frac{3k}{\alpha},}\phantom{\rule{0.2em}{0ex}}{A}_{1}=\pm \frac{2b}{a}\sqrt{\frac{3k}{\alpha}}\beta =\frac{2k}{{a}^{2}\left(2k\nu -{\nu}^{2}-{k}^{2}-1\right)}.$$

Therefore, the solutions of (2) for n = 2 are given by:

$$u\left(t,x,y\right)={A}_{0}+a{A}_{1}\left\{\frac{\mathrm{cosh}\left[a\left(z+C\right)\right]+\mathrm{sinh}\left[a\left(z+C\right)\right]}{1-b\mathrm{cosh}\left[a\left(z+C\right)\right]-b\mathrm{sinh}\left[a\left(z+C\right)\right]}\right\}$$

**n = 1**

For this case, the balancing procedure gives M = 2, and so, (14) becomes:

$$W\left(z\right)={A}_{0}+{A}_{1}G+{A}_{2}{G}^{2}$$

The insertion of this value of W (z) into (8) and making use of the Riccati Equation (13) yields the following algebraic system of equations in terms of A_{0}, A_{1}, A_{2}:

$$\begin{array}{r}24{a}^{3}\beta {k}^{2}{A}_{2}-48{a}^{3}\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{2}+24{a}^{3}\beta {\nu}^{2}{A}_{2}+24{a}^{3}\beta {A}_{2}+2a\alpha {A}_{2}{}^{2}=0,\\ 6{a}^{3}\beta {k}^{2}{A}_{1}-12{a}^{3}\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{1}+6{a}^{3}\beta {\nu}^{2}{A}_{1}+54{a}^{2}b\beta {k}^{2}{A}_{2}-108{a}^{2}b\beta k\nu {A}_{2}\phantom{\rule{1.5em}{0ex}}\\ +54{a}^{2}b\beta {\nu}^{2}{A}_{2}+6{a}^{3}\beta {A}_{1}+54{a}^{2}b\beta {A}_{2}+3a\alpha {A}_{1}{A}_{2}+2\alpha b{A}_{2}{}^{2}=0,\\ 2a\beta {c}^{2}{k}^{2}{A}_{1}-4a\beta {c}^{2}k\nu {A}_{1}+2a\beta {c}^{2}{\nu}^{2}{A}_{1}+{b}^{2}\beta c{k}^{2}{A}_{1}-2{b}^{2}\beta ck\nu \phantom{\rule{0.2em}{0ex}}{A}_{1}\\ +{b}^{2}\beta c{\nu}^{2}{A}_{1}+6b\beta {c}^{2}{k}^{2}{A}_{2}-12b\beta {c}^{2}k\nu {A}_{2}+6b\beta {c}^{2}{\nu}^{2}{A}_{2}+2a\beta {c}^{2}{A}_{1}\\ +{b}^{2}\beta c{A}_{1}+6b\beta {c}^{2}{A}_{2}+\alpha c{A}_{0}{A}_{1}-ck{A}_{1}=0,\\ 12{a}^{2}b\beta {k}^{2}{A}_{1}-24{a}^{2}b\beta k\nu \phantom{\rule{0.2em}{0ex}}{A}_{1}+12{a}^{2}b\beta {\nu}^{2}{A}_{1}+40{a}^{2}\beta c{k}^{2}{A}_{2}-80{a}^{2}\beta ck\nu {A}_{2}\\ +40{a}^{2}\beta c{\nu}^{2}{A}_{2}+38a{b}^{2}\beta {k}^{2}{A}_{2}-76a{b}^{2}\beta k\nu {A}_{2}+38a{b}^{2}\beta {\nu}^{2}{A}_{2}+12{a}^{2}b\beta {A}_{1}\\ +40{a}^{2}\beta c{A}_{2}+38a{b}^{2}\beta {A}_{2}-2a\alpha {A}_{0}{A}_{2}+a\alpha {A}_{1}{}^{2}+3\alpha b{A}_{1}{A}_{2}\\ +2\alpha c{A}_{2}{}^{2}-2ak{A}_{2}=0,\\ 8ab\beta c{k}^{2}{A}_{1}-16ab\beta ck\nu {A}_{1}+8ab\beta c{\nu}^{2}{A}_{1}+16a\beta {c}^{2}{k}^{2}{A}_{2}-32a\beta {c}^{2}k\nu {A}_{2}\\ +16a\beta {c}^{2}{\nu}^{2}{A}_{2}+{b}^{3}\beta {k}^{2}{A}_{1}-2{b}^{3}\beta k\nu {A}_{1}+{b}^{3}\beta {\nu}^{2}{A}_{1}+14{b}^{2}\beta c{k}^{2}{A}_{2}\\ -28{b}^{2}\beta ck\nu {A}_{2}+14{b}^{2}\beta c{\nu}^{2}{A}_{2}+8ab\beta c{A}_{1}+16a\beta {c}^{2}{A}_{2}+{b}^{3}\beta {A}_{1}\\ +14{b}^{2}\beta c{A}_{2}+\alpha b{A}_{0}{A}_{1}+2\alpha c{A}_{0}{A}_{2}+\alpha c{A}_{1}{}^{2}-bk{A}_{1}-2ck{A}_{2}=0,\\ 8{a}^{2}\beta c{k}^{2}{A}_{1}-16{a}^{2}\beta ck\nu {A}_{1}+8{a}^{2}\beta c{\nu}^{2}{A}_{1}+7a{b}^{2}\beta {k}^{2}{A}_{1}-14a{b}^{2}\beta k\nu {A}_{1}\\ +7a{b}^{2}\beta {\nu}^{2}{A}_{1}+52ab\beta c{k}^{2}{A}_{2}-104ab\beta ck\nu {A}_{2}+52ab\beta c{\nu}^{2}{A}_{2}+8{b}^{3}\beta {k}^{2}{A}_{2}\\ -16{b}^{3}\beta k\nu {A}_{2}+8{b}^{3}\beta {\nu}^{2}{A}_{2}+8{a}^{2}\beta c{A}_{1}+7a{b}^{2}\beta {A}_{1}+52ab\beta c{A}_{2}+8{b}^{3}\beta {A}_{2}\\ +a\alpha {A}_{0}{A}_{1}+2\alpha b{A}_{0}{A}_{2}+\alpha b{A}_{1}{}^{2}+3\alpha c{A}_{1}{A}_{2}-ak{A}_{1}-2bk{A}_{2}=0.\end{array}$$

The solution of the above system using Maple gives:

$$\begin{array}{l}{A}_{0}=\frac{1}{\alpha}\{16a\beta ck\nu -8a\beta c{\nu}^{2}-8a\beta c-8a\beta c{k}^{2}-\beta {b}^{2}{\nu}^{2}-\beta {b}^{2}-\beta {b}^{2}{k}^{2}-2\beta {b}^{2}k\nu +k\},\\ {A}_{1}=\frac{1}{\alpha}\{12ab\beta (2k\nu -{\nu}^{2}-{k}^{2}-1)\},{A}_{2}=\frac{1}{\alpha}\{12{a}^{2}\beta (2k\nu -{\nu}^{2}-{k}^{2}-1)\}.\end{array}$$

Consequently, the solutions of (2) are:

$$\begin{array}{c}u(t,x,y)={A}_{0}+{A}_{1}\left\{-\frac{b}{2a}-\frac{\theta}{2a}\mathrm{tanh}\left(\frac{1}{2}\theta (z+C)\right)\right\}\\ +{A}_{2}{\left\{-\frac{b}{2a}-\frac{\theta}{2a}\mathrm{tanh}\left(\frac{1}{2}\theta (z+C)\right)\right\}}^{2}\end{array}$$

$$\begin{array}{c}u(t,x,y)={A}_{0}+{A}_{1}\left\{-\frac{b}{2a}-\frac{\theta}{2a}\mathrm{tanh}\left(\frac{1}{2}\theta z\right)+\frac{\text{sech}\left(\frac{\theta z}{2}\right)}{C\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(\frac{\theta z}{2}\right)-\frac{2a}{{}^{\theta}}\text{sinh}\left(\frac{\theta z}{2}\right)}\right\}\\ +{A}_{2}{\left\{-\frac{b}{2a}-\frac{\theta}{2a}\mathrm{tanh}\left(\frac{1}{2}\theta z\right)+\frac{\text{sech}\left(\frac{\theta z}{2}\right)}{C\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(\frac{\theta z}{2}\right)-\frac{2a}{{}^{\theta}}\text{sinh}\left(\frac{\theta z}{2}\right)}\right\}}^{2}\end{array}$$

**n = 2**

The balancing procedure yields M = 1, so the solutions of (8) are of the form:

$$W(z)={A}_{0}+{A}_{1}G$$

Substituting (22) into (8) and using the Riccati equation [36], we obtain the following algebraic system of equations:

$$\begin{array}{r}-6b{A}_{1}{c}^{3}\nu +3a{A}_{1}^{3}c=0,\\ 6a{A}_{1}^{2}{A}_{0}c+3a{A}_{1}^{3}d-12b{A}_{1}{c}^{2}d\nu =0,\\ 3a{A}_{1}{A}_{0}^{2}e-2b{A}_{1}c{e}^{2}\nu -{A}_{1}e\nu -{A}_{1}de\nu =0,\\ -8a{A}_{1}cde\nu -{A}_{1}d\nu -6a{A}_{1}^{2}{A}_{0}e-b{A}_{1}{d}^{3}\nu +3a{A}_{1}{A}_{0}^{2}d=0,\\ -{A}_{1}c\nu +3a{A}_{1}^{3}e-7b{A}_{1}c{d}^{2}\nu +3a{A}_{1}{A}_{0}^{2}c+6a{A}_{1}^{2}{A}_{0}d-8b{A}_{1}{c}^{2}e\nu =0.\end{array}$$

Solving the above algebraic equations, one obtains:

$$\begin{array}{l}{A}_{0}=\pm \frac{b}{2\sqrt{\alpha}}\sqrt{6\beta (2k\nu -{\nu}^{2}-{k}^{2}-1)},{A}_{1}=\pm \frac{a}{\sqrt{\alpha}}\sqrt{6\beta (2k\nu -{\nu}^{2}-{k}^{2}-1)},\\ a=\frac{{b}^{2}\beta {k}^{2}-2{b}^{2}\beta k\nu +{b}^{2}\beta {\nu}^{2}+{b}^{2}\beta +2k}{4\beta c({k}^{2}-2\nu k+{\nu}^{2}+1)}\end{array}$$

Hence, we have the following solutions of (2) for n = 2:

$$u(x,y,t)={A}_{0}+{A}_{1}\left\{-\frac{b}{2a}-\frac{\theta}{2a}\mathrm{tanh}\left(\frac{1}{2}\theta (z+C)\right)\right\}$$

$$u(t,x,y)={A}_{0}+{A}_{1}\left\{-\frac{b}{2a}-\frac{\theta}{2a}\mathrm{tanh}\left(\frac{1}{2}\theta z\right)+\frac{\text{sech}\left(\frac{\theta z}{2}\right)}{C\phantom{\rule{0.2em}{0ex}}\text{cosh}\left(\frac{\theta z}{2}\right)-\frac{2a}{{}^{\theta}}\text{sinh}\left(\frac{\theta z}{2}\right)}\right\}$$

In this paper, we studied the generalized Zakharov–Kuznetsov Equation (2). We derived the conservation laws of this equation by using the new conservation theorem by Ibragimov. Moreover, the Lie point symmetries of (2) were obtained and were used in conjunction with the simplest equation method to obtain exact solutions of the generalized Zakharov–Kuznetsov equation. The solutions obtained here are new and more general than the ones obtained before in [4] and [8]. Furthermore, the importance of the conservation laws has been emphasized in the Introduction.

D.M.M. would like to thank the FRC of Faculty of Agriculture, Science and Technology, North-West University, Mafikeng Campus, and NRF of South Africa for their financial support. D.M.M. and C.M.K. thank Tanki Motsepa for helpful discussions.

D.M.M. and C.M.K. worked together in the derivation of the mathematical results. Both authors read and approved the final manuscript.

The authors declare no conflict of interest.

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