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Solitons and other solutions of (3 + 1)-dimensional space–time fractional modified KdV–Zakharov–Kuznetsov equation

K. S. Al-Ghafri
K. S. Al-Ghafri
 and 
Hadi Rezazadeh
Hadi Rezazadeh
   | Jul 01, 2019
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Published Online: Jul 01, 2019
Page range: 289 - 304
Received: Mar 29, 2019
Accepted: May 07, 2019
DOI: https://doi.org/10.2478/AMNS.2019.2.00026
Keywords
© 2019 K. S. Al-Ghafri et al., published by Sciendo
This work is licensed under the Creative Commons Attribution 4.0 Public License.
Introduction

Fractional differential equations (FDEs) have become under remarkable consideration as being the generalisation form of the differential equations of integer order and due to their roles in the modelling of several physical processes. Many nonlinear phenomena in physics can be described via FDEs such as chemical physics, optical fibers, plasma, electromagnetic waves, diffusion processes, vibrations in a nonlinear string and etc [1, 2, 3, 4, 5]. Seeking different types of solutions to FDEs for such phenomena has become the subject of interest for researchers. Thus, a lot of powerful mathematical methods have been applied to obtain exact analytic solutions of FDEs, namely, the extended tanh-function method [6, 7], the exp-function method [8, 9], the sub-equation method [10, 11], the improved tan(ϕ/2)-expansion method [12, 13], the (G′/G)-expansion method [14, 15], the modified trial equation method [16, 17], the new extended direct algebraic method [18, 19], the extended sinh-Gordon equation expansion method [20, 21], the unified method [22] and so on.

One of the important physical model is the (3+1)-dimensional space-time fractional modified KdV–Zakharov–Kuznetsov (fractional mKdV–ZK) equation [23, 24, 25] of the form

Dtαu+δu2Dxαu+Dx3αu+DxαDy2αu+DxαDz2αu=0, $$\begin{array}{} \displaystyle D_t^\alpha u + \delta u^2 D_x^\alpha u + D_x^{3\alpha} u + D_x^\alpha D_y^{2\alpha} u + D_x^\alpha D_z^{2\alpha} u =0, \end{array}$$

where 0 < α ≤ 1 and δ is an arbitrary constant. This model is developed for a plasma comprised of cool and hot electrons and a species of fluid ions [26]. Recently, Sahoo and Ray [10] have applied the improved fractional sub equation method with the aid of Jumarie’s modified Riemann–Liouville derivative for space-time fractional mKdV–ZK equation and they derived exact solutions. Then, the fractional derivatives in the sense of modified Riemann–Liouville derivative and three different solution methods have been employed by Guner et. al. [27] to obtain a variety of exact solutions to equation (1). Moreover, using the ansatz method and the functional variable method, Guner [28] examined exact analytic solutions for equation (1) and constructed singular soliton solutions. In the two latter studies, the fractional complex transform method is used to reduce equation (1) into ordinary differential equation (ODE). However, in a recent study, this transformation has been proved to be incorrect by Herzallah [29]. Subsequently, all obtained solutions by means of this method are wrong.

The present study focuses on investigating the exact analytic solutions to space-time fractional mKdV–ZK equation using the variable separated ODE method [30, 31]. The structure of this work is organised as follows. In Section 2, we introduce the definition of conformable fractional derivative [32] and its properties which will be utilised to reduce FDE into an ODE. Section 3 contains the description of variable separated ODE method and the technique of implementing it to ODEs. In Section 4, the proposed method will be applied to construct the solitons and periodic wave solutions of space-time fractional mKdV–ZK equation. Then, the behaviour of some obtained solutions is displayed graphically. Finally, our discussions and conclusions are presented in Section 5.
Conformable fractional derivative

Khalil, et. al. [32] introduced a completely new definition of fractional calculus with the limit operator which is more natural and effective on satisfying some conventional properties than the existing fractional derivatives.

The definition of conformable fractional derivative is given as follows:

Definition 1

Let f: (0, ∞) ⟶ ℝ, then the conformable fractional derivative of f of order α is defined as

Dtαf(t)=limε→0f(t+εt1−α)−f(t)ε $$\begin{array}{} \displaystyle D^\alpha_t f(t)=\lim_{\varepsilon\rightarrow 0}\frac{f(t+\varepsilon t^{1-\alpha}) - f(t)}{\varepsilon} \end{array}$$

for all t > 0, α ∈ (0, 1).

It is said that if the conformable fractional derivative of f of order α exists, then f is α-differentiable. The conformable fractional derivative satisfies the properties shown in the following theorems:

Theorem 1

Let α ∈ (0, 1] and f = f(t), g = g(t) be α-differentiable at a point t > 0, then:

Dtα(af+bg)=aDtαf+bDtαg $\begin{array}{} \displaystyle D^\alpha_t(a f + b g)=a D^\alpha_t f + b D^\alpha_t g \end{array}$, for all a, b ∈ ℝ.

Dtα(tμ)=μtμ−α $\begin{array}{} \displaystyle D^\alpha_t(t^\mu)=\mu\, t^{\mu-\alpha} \end{array}$, for all μ ∈ ℝ.

Dtα(fg)=fDtαg+gDtαf. $\begin{array}{} \displaystyle D^\alpha_t(f g)= f D^\alpha_t g + g D^\alpha_t f. \end{array}$

Dtαfg=gDtαf−fDtαgg2. $\begin{array}{} \displaystyle D^\alpha_t\left(\frac{f}{g}\right)=\frac{g D^\alpha_t f - f D^\alpha_t g}{g^2}. \end{array}$

Moreover, if f is differentiable, then Dtα(f(t))=t1−αdfdt $\begin{array}{} \displaystyle D^\alpha_t(f(t))=t^{1-\alpha} \frac{\mathrm{d} f}{\mathrm{d} t} \end{array}$.

Theorem 2

Let f: (0, ∞) ⟶ ℝ be a function such that f is differentiable and also α-differentiable. Let g be a function defined in the range of f and also differentiable; then, one has the following rule:

Dtα(fg)(t)=t1−αg′(t)f′(g(t)), $$\begin{array}{} \displaystyle D^\alpha_t(fg)(t)=t^{1-\alpha} g'(t) f'(g(t)), \end{array}$$

where prime denotes the classical derivatives with respect to t.
Remark 1

We may use the notation ∂α∂tαf  for  Dtα(f(t)) $\begin{array}{} \displaystyle \frac{\partial^\alpha}{\partial t^\alpha}f~~ \text{for}~~ D^\alpha_t (f(t)) \end{array}$ to denote the conformable fractional derivatives of f with respect to the variable t of order α.
The variable separated ODE method

In this section we shall describe the technique of implementing the proposed method to FDEs in order to extract exact analytic solutions. Suppose that a nonlinear conformable fractional partial differential equation, say, in four independent variables x, y, z and t is given by

P(u,∂αu∂tα,∂αu∂xα,∂αu∂yα,∂αu∂zα,∂2αu∂t2α,∂2αu∂x2α,∂2αu∂y2α,∂2αu∂z2α,…)=0, $$\begin{array}{} \displaystyle P(u, \frac{\partial^\alpha u}{\partial t^\alpha}, \frac{\partial^\alpha u}{\partial x^\alpha}, \frac{\partial^\alpha u}{\partial y^\alpha}, \frac{\partial^\alpha u}{\partial z^\alpha}, \frac{\partial^{2\alpha} u}{\partial t^{2\alpha}}, \frac{\partial^{2\alpha} u}{\partial x^{2\alpha}}, \frac{\partial^{2\alpha} u}{\partial y^{2\alpha}}, \frac{\partial^{2\alpha} u}{\partial z^{2\alpha}}, \dots)=0, \end{array}$$

where u(x, y, z, t) is an unknown function, P is a polynomial in u and its partial derivatives, in which the highest order derivatives and nonlinear terms are involved. This method consists of the following steps:

Using the wave transformation

u(x,y,z,t)=ϕ(ξ),ξ=k1xαα+k2yαα+k3zαα+νtαα, $$\begin{array}{} \displaystyle u(x,y,z,t)=\phi(\xi), \qquad \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} + \nu \frac{t^\alpha}{\alpha}, \end{array}$$

where k1, k2, k3 and ν are constants to be determined later, one can find

∂αu∂tα=νdϕdξ,∂αu∂xα=k1dϕdξ,∂αu∂yα=k2dϕdξ,∂αu∂zα=k3dϕdξ,∂2αu∂t2α=ν2d2ϕdξ2,… $$\begin{array}{} \displaystyle \frac{\partial ^\alpha u}{\partial t^\alpha}=\nu\frac{\mathrm{d} \phi}{\mathrm{d} \xi}, \quad \frac{\partial^\alpha u}{\partial x^\alpha}=k_1\frac{\mathrm{d} \phi}{\mathrm{d} \xi}, \quad \frac{\partial^\alpha u}{\partial y^\alpha}=k_2\frac{\mathrm{d} \phi}{\mathrm{d} \xi},\quad \frac{\partial^\alpha u}{\partial z^\alpha}=k_3\frac{\mathrm{d} \phi}{\mathrm{d} \xi},\quad \frac{\partial^{2\alpha} u}{\partial t^{2\alpha}}=\nu^2 \frac{\mathrm{d}^2 \phi}{\mathrm{d} \xi^2},\quad \dots \end{array}$$

Employing (5) and (6), equation (4) is reduced into the following ODE

P(ϕ,νdϕdξ,k1dϕdξ,k2dϕdξ,k3dϕdξ,ν2d2ϕdξ2,k12d2ϕdξ2,k22d2ϕdξ2,k32d2ϕdξ2,…)=0. $$\begin{array}{} \displaystyle P(\phi, \nu\frac{\mathrm{d} \phi}{\mathrm{d} \xi}, k_1\frac{\mathrm{d} \phi}{\mathrm{d} \xi}, k_2\frac{\mathrm{d} \phi}{\mathrm{d} \xi}, k_3\frac{\mathrm{d} \phi}{\mathrm{d} \xi}, \nu^2 \frac{\mathrm{d}^2 \phi}{\mathrm{d} \xi^2}, k_1^2 \frac{\mathrm{d}^2 \phi}{\mathrm{d} \xi^2}, k_2^2 \frac{\mathrm{d}^2 \phi}{\mathrm{d} \xi^2}, k_3^2 \frac{\mathrm{d}^2 \phi}{\mathrm{d} \xi^2}, \dots)=0. \end{array}$$

Assuming that equation (7) has the solution of the form

ϕ(ξ)=∑j=−NNaj(d+F(ξ))j, $$\begin{array}{} \displaystyle \phi(\xi)=\sum_{j=-N}^N a_j (d+F(\xi))^j, \end{array}$$

where a−N or aN might be zero, but both of them could not be zero simultaneously. The coefficients aj (j = 0, ±1, ±2, …, ± N) and d are constants to be determined, whereas F(ξ) satisfies the general elliptic equation of the form

F′2(ξ)=c0+c1F(ξ)+c2F2(ξ)+c3F3(ξ)+c4F4(ξ), $$\begin{array}{} \displaystyle F'^2(\xi)=c_0 + c_1 F(\xi) + c_2 F^2(\xi) + c_3 F^3(\xi) + c_4 F^4(\xi), \end{array}$$

where ci (i = 0, 1, 2, 3, 4) are constants. Recently, Sirendaoreji [31] presented five classifications of solutions for equation (9) in accordance with the presence of coefficients ci (i = 0, 1, 2, 3, 4) and they are given as: (1) c0 = c1 = 0, (2) c3 = c4 = 0, (3) c1 = c3 = 0, (4) c2 = c4 = 0, and (5) c0 = 0. Herein, we only concentrate on two cases which are Case I c0 = c1 = 0 and Case II c1 = c3 = 0, for convenience. Solution structures of both cases are exhibited as follows.

c0 = c1 = 0:

F1(ξ)=2c2εΔcosh⁡(c2ξ)−c3,Δ>0,c2>0, $$\begin{array}{} \displaystyle F_1(\xi)=\frac{2 c_2}{\varepsilon\sqrt{\Delta} \cosh(\sqrt{c_2}\xi) - c_3}, \; \Delta \gt 0, \; c_2 \gt 0, \end{array}$$

F2(ξ)=2c2ε−Δsinh⁡(c2ξ)−c3,Δ<0,c2>0, $$\begin{array}{} \displaystyle F_2(\xi)=\frac{2 c_2}{\varepsilon\sqrt{-\Delta} \sinh(\sqrt{c_2}\xi) - c_3}, \; \Delta \lt 0, \; c_2 \gt 0, \end{array}$$

F3(ξ)=2c2εΔcos⁡(−c2ξ)−c3,Δ>0,c2<0, $$\begin{array}{} \displaystyle F_3(\xi)=\frac{2 c_2}{\varepsilon\sqrt{\Delta} \cos(\sqrt{-c_2}\xi) - c_3}, \; \Delta \gt 0, \; c_2 \lt 0, \end{array}$$

F4(ξ)=2c2εΔsin⁡(−c2ξ)−c3,Δ>0,c2<0, $$\begin{array}{} \displaystyle F_4(\xi)=\frac{2 c_2}{\varepsilon\sqrt{\Delta} \sin(\sqrt{-c_2}\xi) - c_3}, \; \Delta \gt 0, \; c_2 \lt 0, \end{array}$$

F5(ξ)=−c2c31+εtanhc22ξ,Δ=0,c2>0, $$\begin{array}{} \displaystyle F_5(\xi)=-\frac{c_2}{c_3}\left[1+\varepsilon\tanh\left(\frac{\sqrt{c_2}}{2}\xi\right)\right], \; \Delta=0, \, c_2 \gt 0, \end{array}$$

F6(ξ)=−c2c31+εcothc22ξ,Δ=0,c2>0, $$\begin{array}{} \displaystyle F_6(\xi)=-\frac{c_2}{c_3}\left[1+\varepsilon\coth\left(\frac{\sqrt{c_2}}{2}\xi\right)\right], \; \Delta=0, \, c_2 \gt 0, \end{array}$$

F7(ξ)=εc4ξ,c2=c3=0,c4>0, $$\begin{array}{} \displaystyle F_7(\xi)=\frac{\varepsilon}{\sqrt{c_4}\xi}, \; c_2=c_3=0, \, c_4 \gt 0, \end{array}$$

F8(ξ)=4c3c32ξ2−4c4,c2=0, $$\begin{array}{} \displaystyle F_8(\xi)=\frac{4 c_3}{c_3^2\xi^2 - 4c_4}, \; c_2=0, \end{array}$$

where Δ=c32−4c2c4 $\begin{array}{} \displaystyle \Delta=c_3^2-4c_2 c_4 \end{array}$, ε = ±1.

c1 = c3 = 0:

F9(ξ)=ε−c22c4tanh−c22ξ,Δ1=0,c2<0,c4>0, $$\begin{array}{} \displaystyle F_9(\xi)=\varepsilon\sqrt{-\frac{c_2}{2c_4}}\tanh\left(\sqrt{-\frac{c_2}{2}}\xi\right), \; \Delta_1=0, \, c_2 \lt 0,\; c_4 \gt 0, \end{array}$$

F10(ξ)=ε−c22c4coth−c22ξ,Δ1=0,c2<0,c4>0, $$\begin{array}{} \displaystyle F_{10}(\xi)=\varepsilon\sqrt{-\frac{c_2}{2c_4}}\coth\left(\sqrt{-\frac{c_2}{2}}\xi\right),\; \Delta_1=0, \, c_2 \lt 0,\; c_4 \gt 0, \end{array}$$

F11(ξ)=εc22c4tanc22ξ,Δ1=0,c2>0,c4>0, $$\begin{array}{} \displaystyle F_{11}(\xi)=\varepsilon\sqrt{\frac{c_2}{2c_4}}\tan\left(\sqrt{\frac{c_2}{2}}\xi\right), \; \Delta_1=0, \, c_2 \gt 0,\; c_4 \gt 0, \end{array}$$

F12(ξ)=εc22c4cotc22ξ,Δ1=0,c2>0,c4>0, $$\begin{array}{} \displaystyle F_{12}(\xi)=\varepsilon\sqrt{\frac{c_2}{2c_4}}\cot\left(\sqrt{\frac{c_2}{2}}\xi\right), \; \Delta_1=0, \, c_2 \gt 0,\; c_4 \gt 0, \end{array}$$

F13(ξ)=−c2m2c4(m2+1)sn−c2m2+1ξ,c0=c22m2c4(m2+1)2,c2<0,c4>0, $$\begin{array}{} \displaystyle F_{13}(\xi)=\sqrt{\frac{-c_2 m^2}{c_4(m^2+1)}}\text{sn}\left(\sqrt{\frac{-c_2}{m^2+1}}\xi\right), \; c_0=\frac{c_2^2 m^2}{c_4(m^2+1)^2}, \, c_2 \lt 0, \, c_4 \gt 0, \end{array}$$

F14(ξ)=−c2m2c4(2m2−1)cnc22m2−1ξ,c0=c22m2(m2−1)c4(2m2−1)2,c2>0,c4<0, $$\begin{array}{} \displaystyle F_{14}(\xi)=\sqrt{\frac{-c_2 m^2}{c_4(2m^2-1)}}\text{cn}\left(\sqrt{\frac{c_2}{2m^2-1}}\xi\right), \; c_0=\frac{c_2^2 m^2(m^2-1)}{c_4(2m^2-1)^2}, \, c_2 \gt 0, \, c_4 \lt 0, \end{array}$$

F15(ξ)=−c2c4(2−m2)dnc22−m2ξ,c0=c22(1−m2)c4(2−m2)2,c2>0,c4<0, $$\begin{array}{} \displaystyle F_{15}(\xi)=\sqrt{\frac{-c_2}{c_4(2-m^2)}}\text{dn}\left(\sqrt{\frac{c_2}{2-m^2}}\xi\right), \; c_0=\frac{c_2^2(1-m^2)}{c_4(2-m^2)^2}, \, c_2 \gt 0, \, c_4 \lt 0, \end{array}$$

F16(ξ)=m22−2c2c4(2−m2)sn−2c22−m2ξ1±dn−2c22−m2ξ,c0=c22m44c4(2−m2)2,c2<0,c4>0, $$\begin{array}{} \displaystyle F_{16}(\xi)=\frac{m^2}{2}\sqrt{\frac{-2c_2}{c_4(2-m^2)}} \frac{\text{sn}\left(\sqrt{\frac{-2c_2}{2-m^2}}\xi\right)}{1 \pm \text{dn}\left(\sqrt{\frac{-2c_2}{2-m^2}}\xi\right)}, \; c_0=\frac{c_2^2m^4}{4c_4(2-m^2)^2}, \, c_2 \lt 0, \, c_4 \gt 0, \end{array}$$

F17(ξ)=122c2c4(1−2m2)sn2c21−2m2ξ1±cn2c21−2m2ξ,c0=c224c4(1−2m2)2,c2>0,c4>0, $$\begin{array}{} \displaystyle F_{17}(\xi)=\frac{1}{2}\sqrt{\frac{2c_2}{c_4(1-2m^2)}} \frac{\text{sn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi\right)}{1 \pm \text{cn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi\right)}, \; c_0=\frac{c_2^2}{4c_4(1-2m^2)^2}, \, c_2 \gt 0, \, c_4 \gt 0, \end{array}$$

F18(ξ)=ε−4c0c414ds(−4c0c4)14ξ,22,c2=0,c0c4<0, $$\begin{array}{} \displaystyle F_{18}(\xi)=\varepsilon\left(-\frac{4c_0}{c_4}\right)^{\frac{1}{4}}\text{ds}\left((-4c_0c_4)^{\frac{1}{4}} \xi, \frac{\sqrt{2}}{2} \right), \; c_2=0, \, c_0c_4 \lt 0, \end{array}$$

F19(ξ)=εc0c414ns2(c0c4)14ξ,22+cs2(c0c4)14ξ,22,c2=0,c0c4>0, $$\begin{array}{} \displaystyle F_{19}(\xi)=\varepsilon\left(\frac{c_0}{c_4}\right)^{\frac{1}{4}}\left[\text{ns}\left(2(c_0c_4)^{\frac{1}{4}} \xi, \frac{\sqrt{2}}{2} \right) + \text{cs}\left(2(c_0c_4)^{\frac{1}{4}} \xi, \frac{\sqrt{2}}{2} \right)\right], \; c_2=0, \, c_0c_4 \gt 0, \end{array}$$

where Δ1=c22−4c0c4 $\begin{array}{} \displaystyle \Delta_1=c_2^2-4c_0 c_4 \end{array}$, ε = ±1. Here, sn(ξ) = sn(ξ, m), cn(ξ) = cn(ξ, m), dn(ξ) = dn(ξ, m) are called the Jacobian elliptic sine function, the Jacobian elliptic cosine function and the Jacobian elliptic function of the third kind and 0 < m < 1 is the modulus of Jacobian elliptic function. The other Jacobian functions are generated by these three kinds of functions as follows:

ns(ξ)=1/sn(ξ),nc(ξ)=1/cn(ξ),nd(ξ)=1/dn(ξ),sc(ξ)=sn(ξ)/cn(ξ),sd(ξ)=sn(ξ)/dn(ξ),cd(ξ)=cn(ξ)/dn(ξ),cs(ξ)=cn(ξ)/sn(ξ),ds(ξ)=dn(ξ)/sn(ξ),dc(ξ)=dn(ξ)/cn(ξ). $$\begin{array}{} \displaystyle \text{ns}(\xi)=1/\text{sn}(\xi), \; \text{nc}(\xi)=1/\text{cn}(\xi), \; \text{nd}(\xi)=1/\text{dn}(\xi), \\ \displaystyle\rm{sc}(\xi)=\text{sn}(\xi)/\text{cn}(\xi), \; \text{sd}(\xi)=\text{sn}(\xi)/\text{dn}(\xi), \; \text{cd}(\xi)=\text{cn}(\xi)/\text{dn}(\xi), \\ \displaystyle\text{cs}(\xi)=\text{cn}(\xi)/\text{sn}(\xi), \; \text{ds}(\xi)=\text{dn}(\xi)/\text{sn}(\xi), \; \text{dc}(\xi)=\text{dn}(\xi)/\text{cn}(\xi). \end{array}$$

The value of the positive integer N can be determined by balancing the highest order linear terms with the nonlinear terms of the highest order emerging in equation (7).

Substituting (8) and (9) into equation (7), we collect all terms with the same power of (d + F(ξ)). Equating each coefficient of the resulting polynomial to zero, yields a set of algebraic equations for aj (j = 0, ±1, ±2, …, ± N), ci (i = 0, 1, 2, 3, 4), kl (l = 1, 2, 3), d and ν.

Substituting the values of the constants obtained by solving the algebraic equations extracted in Step 4 together with the solutions of equation (9) into (8), we arrive at different types of solutions for equation (4).

Application of the method to the (3 + 1)-dimensional space–time fractional mKdV–ZK equation

Now, we aim to solve the fractional mKdV–ZK equation (1) by applying the variable separated ODE method described above. Thus, to deal with the complex form of equation (1) we will reduce it to an ODE using the transformation (5). The substitution of the transformation (5) into equation (1) leads to

νϕ′+δk1ϕ2ϕ′+k1(k12+k22+k32)ϕ‴=0, $$\begin{array}{} \displaystyle \nu \phi' + \delta k_1 \phi^2 \phi' + k_1(k_1^2+k_2^2+k_3^2)\phi'''=0, \end{array}$$

where prime denotes the derivative with respect to ξ. Integrating equation (31) once with respect to ξ, we obtain

νϕ+Aϕ3+Bϕ″=0, $$\begin{array}{} \displaystyle \nu \phi + A \phi^3 + B \phi''=0, \end{array}$$

where A = δk1/3, B=k1(k12+k22+k32) $\begin{array}{} \displaystyle B=k_1(k_1^2+k_2^2+k_3^2) \end{array}$ and the integration constant is taken to be zero. Now, we assume that equation (32) has a solution in the form of equation (8). The homogeneous balance between the term ϕ″ and the term ϕ3 in equation (32) gives rise to N = 1. Hence, the solution to equation (32) is given in the form

ϕ(ξ)=a−1(d+F(ξ))+a0+a1(d+F(ξ)), $$\begin{array}{} \displaystyle \phi(\xi)= \frac{a_{-1}}{(d+F(\xi))} + a_0 + a_1 (d+F(\xi)), \end{array}$$

where a−1, a0, a1 are unknown constants to be computed. Substituting equation (33) into equation (32) and using equation (9), we obtain polynomials in (d + F(ξ))j and (d + F(ξ))−j, (j = 0, 1, 2, 3). Collecting all coefficients of identical power of the resulting polynomials and equating each coefficient to zero, yield the following set of algebraic equations

B(2a−1c4d4−2a−1c3d3+2a−1c2d2−2a−1c1d+2a−1c0)+Aa−13=0, $$\begin{array}{} \displaystyle B(2 a_{-1} c_4 d^4 - 2 a_{-1} c_3 d^3 + 2 a_{-1} c_2 d^2 - 2 a_{-1} c_1 d + 2 a_{-1} c_0) + A a_{-1}^3=0, \end{array}$$

B(−6a−1c4d3+92a−1c3d2−3a−1c2d+32a−1c1)+3Aa−12a0=0, $$\begin{array}{} \displaystyle B(-6 a_{-1} c_4 d^3 + \frac{9}{2} a_{-1} c_3 d^2 - 3 a_{-1} c_2 d + \frac{3}{2} a_{-1} c_1) + 3 A a_{-1}^2 a_0=0, \end{array}$$

B(6a−1c4d2−3a−1c3d+a−1c2)+A(3a−12a1+3a−1a02)+νa−1=0, $$\begin{array}{} \displaystyle B(6 a_{-1} c_4 d^2 - 3 a_{-1} c_3 d + a_{-1} c_2) + A (3 a_{-1}^2 a_1 + 3 a_{-1} a_0^2) +\nu a_{-1}=0, \end{array}$$

B(−2a−1c4d+12a−1c3−a1c2d+32a1c3d2−2a1c4d3+12a1c1)+A(6a−1a0a1+a03)+νa0=0, $$\begin{array}{} \displaystyle B(-2 a_{-1} c_4 d + \frac{1}{2} a_{-1} c_3 - a_1 c_2 d + \frac{3}{2} a_{1} c_3 d^2 - 2 a_1 c_4 d^3 + \frac{1}{2} a_1 c_1) + A(6 a_{-1} a_0 a_1 + a_0^3) + \nu a_0 =0, \end{array}$$

B(−3a1c3d+6a1c4d2+a1c2)+A(3a−1a12+3a02a1)+νa1=0, $$\begin{array}{} \displaystyle B(-3 a_1 c_3 d + 6 a_1 c_4 d^2 + a_1 c_2) + A(3 a_{-1} a_1^2 + 3 a_0^2 a_1) + \nu a_1=0, \end{array}$$

B(−6a1c4d+32a1c3)+3Aa0a12=0, $$\begin{array}{} \displaystyle B(-6 a_1 c_4 d + \frac{3}{2} a_1 c_3) + 3 A a_0 a_1^2=0, \end{array}$$

2Ba1c4+Aa13=0. $$\begin{array}{} \displaystyle 2 B a_1 c_4 + A a_1^3=0. \end{array}$$

Solving equations (34)–(40) gives the following cases of solutions.

In this case, c0 = c1 = 0. This case has seven different sets of coefficients for the solution of equation (33) displayed as follows.

a0=−d−2c4BA,a1=−2c4BA $\begin{array}{} \displaystyle a_0=-d \sqrt{-\frac{2 c_4 B}{A}},\; a_1= \sqrt{-\frac{2 c_4 B}{A}} \end{array}$, a−1 = 0, ν = −c2 B, c3 = 0

u(x,y,z,t)=ε6c2(k12+k22+k32)δsechc2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \text{sech}\left(\sqrt{c_2}\xi \right), \end{array}$$

where c2 > 0 and δ > 0.

u(x,y,z,t)=ε−6c2(k12+k22+k32)δcschc2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \text{csch}\left(\sqrt{c_2}\xi \right), \end{array}$$

where c2 > 0 and δ < 0.

u(x,y,z,t)=ε6c2(k12+k22+k32)δsec−c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \sec\left(\sqrt{-c_2}\xi \right), \end{array}$$

u(x,y,z,t)=ε6c2(k12+k22+k32)δcsc−c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \text{cs}c\left(\sqrt{-c_2}\xi \right), \end{array}$$

where c2 < 0 and δ < 0. In solutions (41)–(44), ξ=k1xαα+k2yαα+k3zαα−c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

a0=(c3−4c4d)4c4−2c4BA,a1=−2c4BA,a−1=0,ν=c2B2,c4=c324c2 $\begin{array}{} \displaystyle a_0=\frac{(c_3-4 c_4 d)}{4 c_4}\sqrt{-\frac{2 c_4 B}{A}},\; a_1= \sqrt{-\frac{2 c_4 B}{A}}, a_{-1}=0,\; \nu=\frac{c_2 B}{2},\; c_4=\frac{c_3^2}{4 c_2} \end{array}$

u(x,y,z,t)=12ε−6c2(k12+k22+k32)δtanh12c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= \frac{1}{2}\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \tanh\left(\frac{1}{2}\sqrt{c_2}\xi \right), \end{array}$$

u(x,y,z,t)=12ε−6c2(k12+k22+k32)δcoth12c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= \frac{1}{2}\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \coth\left(\frac{1}{2}\sqrt{c_2}\xi \right), \end{array}$$

where c2 > 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα+c2k12(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +\frac{c_2 k_1}{2}(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

a0=−(c3−4c4d)4c4−2c4BA,a1=0,a−1=d(c3−2c4d)2c4−2c4BA,ν=c2B2,c4=c324c2 $\begin{array}{} \displaystyle a_0=-\frac{(c_3-4 c_4 d)}{4 c_4}\sqrt{-\frac{2 c_4 B}{A}},\; a_1=0,\; a_{-1}=d\frac{(c_3-2 c_4 d)}{2 c_4} \sqrt{-\frac{2 c_4 B}{A}}, \; \nu=\frac{c_2 B}{2},\; c_4=\frac{c_3^2}{4 c_2} \end{array}$

u(x,y,z,t)=12ε−6c2(k12+k22+k32)δc2+(c2−c3d)tanh12c2ξ(c2−c3d)+c2tanh12c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= \frac{1}{2}\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{c_2+(c_2 - c_3 d) \tanh\left(\frac{1}{2}\sqrt{c_2}\xi \right)}{(c_2 - c_3 d) + c_2 \tanh\left(\frac{1}{2}\sqrt{c_2}\xi \right)}, \end{array}$$

u(x,y,z,t)=12ε−6c2(k12+k22+k32)δc2+(c2−c3d)coth12c2ξ(c2−c3d)+c2coth12c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= \frac{1}{2}\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{c_2+(c_2 - c_3 d) \coth\left(\frac{1}{2}\sqrt{c_2}\xi \right)}{(c_2 - c_3 d) + c_2 \coth\left(\frac{1}{2}\sqrt{c_2}\xi \right)}, \end{array}$$

where c2 > 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα+c2k12(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +\frac{c_2 k_1}{2}(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

a0=−2BA(c2−c4d2),a1=0,a−1=d2BA(c2−c4d2),ν=−c2B,c3=2c2d $\begin{array}{} \displaystyle a_0=-\sqrt{\frac{2 B}{A}(c_2 - c_4 d^2)},\; a_1=0,\; a_{-1}=d \sqrt{\frac{2 B}{A}(c_2 - c_4 d^2)}, \; \nu=- c_2 B,\; c_3=\frac{2 c_2}{d} \end{array}$

This gives us the solutions (41)–(44).

a0=0,a1=−2c4BA,a−1=d2−2c4BA,ν=2c2B,c3=c2d,c4=c24d2 $\begin{array}{} \displaystyle a_0=0,\; a_1= \sqrt{-\frac{2 c_4 B}{A}},\; a_{-1}=d^2 \sqrt{-\frac{2 c_4 B}{A}},\; \nu=2 c_2 B,\; c_3=\frac{c_2}{d},\; c_4=\frac{c_2}{4 d^2} \end{array}$

u(x,y,z,t)=12ε−6c2(k12+k22+k32)δ1+tanh212c2ξtanh12c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= \frac{1}{2}\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1 + \tanh^2\left(\frac{1}{2}\sqrt{c_2}\xi \right)}{\tanh\left(\frac{1}{2}\sqrt{c_2}\xi \right)}, \end{array}$$

u(x,y,z,t)=12ε−6c2(k12+k22+k32)δ1+coth212c2ξcoth12c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= \frac{1}{2}\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1 + \coth^2\left(\frac{1}{2}\sqrt{c_2}\xi \right)}{\coth\left(\frac{1}{2}\sqrt{c_2}\xi \right)}, \end{array}$$

where c2 > 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα+2c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +2{c_2 k_1}(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

a0 = 0, a1=−−2c4BA,a−1=d2−2c4BA,ν=−c2B,c3=c2d,c4=c24d2 $\begin{array}{} \displaystyle a_1=- \sqrt{-\frac{2 c_4 B}{A}},\; a_{-1}=d^2 \sqrt{-\frac{2 c_4 B}{A}},\; \nu=- c_2 B,\; c_3=\frac{c_2}{d},\; c_4=\frac{c_2}{4 d^2} \end{array}$

u(x,y,z,t)=12ε−6c2(k12+k22+k32)δsech212c2ξtanh12c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= \frac{1}{2}\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{ \text{sech}^2\left(\frac{1}{2}\sqrt{c_2}\xi \right)}{\tanh\left(\frac{1}{2}\sqrt{c_2}\xi \right)}, \end{array}$$

u(x,y,z,t)=12ε−6c2(k12+k22+k32)δcsch212c2ξcoth12c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= \frac{1}{2}\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{\text{csch}^2\left(\frac{1}{2}\sqrt{c_2}\xi \right)}{\coth\left(\frac{1}{2}\sqrt{c_2}\xi \right)}, \end{array}$$

where c2 > 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα−c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - {c_2 k_1}(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

a0=−2d−2c4BA,a1=−2c4BA,a−1=d2−2c4BA,ν=−4c2B,c3=c2d,c4=−c24d2 $\begin{array}{} \displaystyle a_0=-2d \sqrt{-\frac{2 c_4 B}{A}},\; a_1= \sqrt{-\frac{2 c_4 B}{A}},\; a_{-1}=d^2 \sqrt{-\frac{2 c_4 B}{A}},\; \nu=-4 c_2 B,\; c_3=\frac{c_2}{d},\; c_4=-\frac{c_2}{4 d^2} \end{array}$

u(x,y,z,t)=2ε6c2(k12+k22+k32)δ12cosh2c2ξ−1, $$\begin{array}{} \displaystyle u(x,y,z,t)= 2\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}}\frac{1}{2\cosh^2\left(\sqrt{c_2}\xi \right) - 1}, \end{array}$$

where c2 > 0 and δ > 0.

u(x,y,z,t)=2ε6c2(k12+k22+k32)δ12cos2−c2ξ−1, $$\begin{array}{} \displaystyle u(x,y,z,t)= 2\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}}\frac{1}{2\cos^2\left(\sqrt{-c_2}\xi \right) - 1}, \end{array}$$

where c2 < 0 and δ < 0. In solutions (53) and (54), ξ=k1xαα+k2yαα+k3zαα−4c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - 4c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

In the second case, c1 = c3 = 0. This case has five different sets of coefficients for the solution of equation (33) shown as follows.

a0=−d−2c4BA,a1=−2c4BA,a−1=0,ν=−c2B $\begin{array}{} \displaystyle a_0=-d \sqrt{-\frac{2 c_4 B}{A}} ,\; a_1= \sqrt{-\frac{2 c_4 B}{A}},\; a_{-1}= 0,\; \nu=- c_2 B \end{array}$

u(x,y,z,t)=εm6c2(k12+k22+k32)δ(m2+1)sn−c2m2+1ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon m \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(m^2+1)}} \text{sn}\left(\sqrt{-\frac{c_2}{m^2+1}}\xi\right), \end{array}$$

where c0=c22m2c4(m2+1)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^2}{c_4(m^2+1)^2} \end{array}$, c2 < 0 and δ < 0. As m → 1, solution (55) reduces to

u(x,y,z,t)=ε3c2(k12+k22+k32)δtanh−c22ξ. $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \tanh\left(\sqrt{-\frac{c_2}{2}}\xi \right). \end{array}$$

u(x,y,z,t)=εm6c2(k12+k22+k32)δ(2m2−1)cnc22m2−1ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon m \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(2m^2-1)}} \text{cn}\left(\sqrt{\frac{c_2}{2m^2-1}}\xi \right), \end{array}$$

where c0=c22m2(m2−1)c4(2m2−1)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^2 (m^2-1)}{c_4(2m^2-1)^2} \end{array}$, c2 > 0 and δ > 0. As m → 1, solution (57) reduces to solution (41).

u(x,y,z,t)=ε6c2(k12+k22+k32)δ(2−m2)dnc22−m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(2-m^2)}} \text{dn}\left(\sqrt{\frac{c_2}{2-m^2}}\xi \right), \end{array}$$

where c0=c22(1−m2)c4(2−m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 (1-m^2)}{c_4(2-m^2)^2} \end{array}$, c2 > 0 and δ > 0. As m → 1, solution (58) reduces to solution (41).

u(x,y,z,t)=εm23c2(k12+k22+k32)δ(2−m2)sn−2c22−m2ξ1+εdn−2c22−m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon m^2\sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(2-m^2)}} \frac{\text{sn}\left(\sqrt{-\frac{2c_2}{2-m^2}}\xi \right)}{1+\varepsilon\text{dn}\left(\sqrt{-\frac{2c_2}{2-m^2}}\xi \right)}, \end{array}$$

where c0=c22m44c4(2−m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^4}{4c_4(2-m^2)^2} \end{array}$, c2 < 0 and δ < 0. As m → 1, solution (59) becomes

u(x,y,z,t)=ε3c2(k12+k22+k32)δtanh−2c2ξ1+εsech−2c2ξ. $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{\tanh\left(\sqrt{-2c_2}\xi \right)}{1+\varepsilon\text{sech}\left(\sqrt{-2c_2}\xi \right)}. \end{array}$$

u(x,y,z,t)=ε−3c2(k12+k22+k32)δ(1−2m2)sn2c21−2m2ξ1+εcn2c21−2m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(1-2m^2)}} \frac{\text{sn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi \right)}{1+\varepsilon\text{cn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi \right)}, \end{array}$$

where c0=c224c4(1−2m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2}{4c_4(1-2m^2)^2} \end{array}$, c2 > 0 and δ < 0. As m → 0, solution (61) degenerates to

u(x,y,z,t)=ε−3c2(k12+k22+k32)δsin2c2ξ1+εcos2c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{\sin\left(\sqrt{2c_2}\xi \right)}{1+\varepsilon\cos\left(\sqrt{2c_2}\xi \right)}, \end{array}$$

and as m → 1, solution (61) reduces to solution (60). In solutions (55)–(62), ξ=k1xαα+k2yαα+k3zαα−c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

a0 = 0, a1 = 0, a−1=−2c0BA $\begin{array}{} \displaystyle a_{-1}= \sqrt{-\frac{2 c_0 B}{A}} \end{array}$, ν = − c2 B, d = 0

u(x,y,z,t)=ε6c2(k12+k22+k32)δ(m2+1)ns−c2m2+1ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(m^2+1)}} \text{ns}\left(\sqrt{-\frac{c_2}{m^2+1}}\xi \right), \end{array}$$

where c0=c22m2c4(m2+1)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^2}{c_4(m^2+1)^2} \end{array}$, c2 < 0 and δ < 0. As m → 0, solution (63) degenerates to solution (44) while as m → 1, solution (63) reduces to

u(x,y,z,t)=ε3c2(k12+k22+k32)δcoth−c22ξ. $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \coth\left(\sqrt{-\frac{c_2}{2}}\xi \right). \end{array}$$

u(x,y,z,t)=ε6c2(m2−1)(k12+k22+k32)δ(2m2−1)ncc22m2−1ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2(m^2-1)(k_1^2+k_2^2+k_3^2)}{\delta(2m^2-1)}} \text{nc}\left(\sqrt{\frac{c_2}{2m^2-1}}\xi \right), \end{array}$$

where c0=c22m2(m2−1)c4(2m2−1)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^2 (m^2-1)}{c_4(2m^2-1)^2} \end{array}$, c2 > 0 and δ > 0. As m → 0, solution (65) degenerates to solution (43).

u(x,y,z,t)=ε6c2(1−m2)(k12+k22+k32)δ(2−m2)ndc22−m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2(1-m^2)(k_1^2+k_2^2+k_3^2)}{\delta(2-m^2)}} \text{nd}\left(\sqrt{\frac{c_2}{2-m^2}}\xi \right), \end{array}$$

where c0=c22(1−m2)c4(2−m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 (1-m^2)}{c_4(2-m^2)^2} \end{array}$, c2 > 0 and δ > 0.

u(x,y,z,t)=ε3c2(k12+k22+k32)δ(2−m2)1+ε dn−2c22−m2ξsn−2c22−m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(2-m^2)}} \frac{1+\varepsilon~\text{dn}\left(\sqrt{-\frac{2c_2}{2-m^2}}\xi \right)}{\text{sn}\left(\sqrt{-\frac{2c_2}{2-m^2}}\xi \right)}, \end{array}$$

where c0=c22m44c4(2−m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^4}{4c_4(2-m^2)^2} \end{array}$, c2 < 0 and δ < 0. As m → 0, solution (67) gives rise to solution (44) and as m → 1, solution (67) changes to

u(x,y,z,t)=ε3c2(k12+k22+k32)δ1+ε sech−2c2ξtanh−2c2ξ. $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1+\varepsilon~\text{sech}\left(\sqrt{-2c_2}\xi \right)}{\tanh\left(\sqrt{-2c_2}\xi \right)}. \end{array}$$

u(x,y,z,t)=ε−3c2(k12+k22+k32)δ(1−2m2)1+ε cn2c21−2m2ξsn2c21−2m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(1-2m^2)}} \frac{1+\varepsilon~\text{cn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi \right)}{\text{sn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi \right)}, \end{array}$$

where c0=c224c4(1−2m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2}{4c_4(1-2m^2)^2} \end{array}$, c2 > 0 and δ < 0. As m → 0, solution (69) transforms to

u(x,y,z,t)=ε−3c2(k12+k22+k32)δ1+εcos2c2ξsin2c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1+\varepsilon\cos\left(\sqrt{2c_2}\xi \right)}{\sin\left(\sqrt{2c_2}\xi \right)}, \end{array}$$

and as m → 1, solution (69) degenerates to solution (68). In solutions (63)–(70), ξ=k1xαα+k2yαα+k3zαα−c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

a0=−d−2c4BA,a1=0,a−1=(c2+2c4d2)2c4−2c4BA,ν=−c2B,c0=c224c4 $\begin{array}{} \displaystyle a_0=-d \sqrt{-\frac{2 c_4 B}{A}},\; a_1=0,\; a_{-1}=\frac{(c_2+2 c_4 d^2)}{2 c_4} \sqrt{-\frac{2 c_4 B}{A}}, \; \nu=- c_2 B,\; c_0=\frac{c_2^2}{4 c_4} \end{array}$

u(x,y,z,t)=ε6c2(k12+k22+k32)δ−c2c4+2c4dtanh−c22ξ2c4d+−2c2c4tanh−c22ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{\sqrt{-c_2 c_4}+\sqrt{2} c_4 d \tanh\left(\sqrt{-\frac{c_2}{2}}\xi \right)}{ 2 c_4 d + \sqrt{-2 c_2 c_4} \tanh\left(\sqrt{-\frac{c_2}{2}}\xi \right)}, \end{array}$$

u(x,y,z,t)=ε6c2(k12+k22+k32)δ−c2c4+2c4dcoth−c22ξ2c4d+−2c2c4coth−c22ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{\sqrt{-c_2 c_4}+\sqrt{2} c_4 d \coth\left(\sqrt{-\frac{c_2}{2}}\xi \right)}{ 2 c_4 d + \sqrt{-2 c_2 c_4} \coth\left(\sqrt{-\frac{c_2}{2}}\xi \right)}, \end{array}$$

where c2 < 0, c4 > 0 and δ < 0. Note that when d = 0 solution (71) converts to solution (64) whereas solution (72) turns to solution (56).

u(x,y,z,t)=ε−6c2(k12+k22+k32)δ2c4dtanc22ξ−c2c42c2c4tanc22ξ+2c4d, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{\sqrt{2} c_4 d \tan\left(\sqrt{\frac{c_2}{2}}\xi \right) - \sqrt{c_2 c_4}}{ \sqrt{2 c_2 c_4} \tan\left(\sqrt{\frac{c_2}{2}}\xi \right) + 2 c_4 d}, \end{array}$$

u(x,y,z,t)=ε−6c2(k12+k22+k32)δ2c4dcotc22ξ−c2c42c2c4cotc22ξ+2c4d, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{\sqrt{2} c_4 d \cot\left(\sqrt{\frac{c_2}{2}}\xi \right) - \sqrt{c_2 c_4}}{ \sqrt{2 c_2 c_4} \cot\left(\sqrt{\frac{c_2}{2}}\xi \right) + 2 c_4 d}, \end{array}$$

where c2 > 0, c4 > 0 and δ < 0. In solutions (71)–(74), ξ=k1xαα+k2yαα+k3zαα−c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

a0=−12−2BA(c2+2c4d2),a1=0,a−1=d−2BA(c2+2c4d2),ν=B2(c2−6c4d2),c0=c4d4 $\begin{array}{} \displaystyle a_0=-\frac{1}{2}\sqrt{-\frac{2 B}{A}(c_2 + 2 c_4 d^2)},\; a_1=0,\; a_{-1}=d \sqrt{-\frac{2 B}{A}(c_2 + 2 c_4 d^2)}, \; \nu=\frac{B}{2}(c_2 - 6 c_4 d^2),\; c_0=c_4 d^4 \end{array}$

u(x,y,z,t)=εm+12−6c2(k12+k22+k32)δ(m2+1)1−−m sn−c2m2+1ξ1+−m sn−c2m2+1ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \frac{m+1}{2} \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(m^2+1)}} \frac{1-\sqrt{-m} ~\text{sn}\left(\sqrt{-\frac{c_2}{m^2+1}}\xi \right)}{1+\sqrt{-m} ~\text{sn}\left(\sqrt{-\frac{c_2}{m^2+1}}\xi \right)}, \end{array}$$

where c0=c22m2c4(m2+1)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^2}{c_4(m^2+1)^2} \end{array}$, c2 < 0, δ > 0 and ξ=k1xαα+k2yαα+k3zαα+(m−1)2−4m2(m2+1)c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} + \frac{(m-1)^2-4m}{2(m^2+1)}c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$. As m → 1, solution (75) becomes

u(x,y,z,t)=ε−3c2(k12+k22+k32)δ1−itanh−c22ξ1+itanh−c22ξ. $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1-i \tanh\left(\sqrt{-\frac{c_2}{2}}\xi \right)}{1+i \tanh\left(\sqrt{-\frac{c_2}{2}}\xi \right)}. \end{array}$$

where ξ=k1xαα+k2yαα+k3zαα−c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$. Using the relation tanh(θ) = −i tan(iθ) solution (76) converts to

u(x,y,z,t)=ε−3c2(k12+k22+k32)δ1+tanc22ξ1−tanc22ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1 + \tan\left(\sqrt{\frac{c_2}{2}}\xi \right)}{1 - \tan\left(\sqrt{\frac{c_2}{2}}\xi \right)}, \end{array}$$

where c2 > 0 and δ < 0.

u(x,y,z,t)=εm2−1+m2−6c2(k12+k22+k32)δ(2m2−1)1−−mm2−1cnc22m2−1ξ1+−mm2−1cnc22m2−1ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \frac{\sqrt{m^2-1}+m}{2} \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(2m^2-1)}} \frac{1-\sqrt{-\frac{m}{\sqrt{m^2-1}}} \text{cn}\left(\sqrt{\frac{c_2}{2m^2-1}}\xi \right)}{1+\sqrt{-\frac{m}{\sqrt{m^2-1}}} \text{cn}\left(\sqrt{\frac{c_2}{2m^2-1}}\xi \right)}, \end{array}$$

where c0=c22m2(m2−1)c4(2m2−1)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^2 (m^2-1)}{c_4(2m^2-1)^2} \end{array}$, c2 > 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα+2m2−1−6mm2−12(2m2−1)c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} + \frac{2m^2-1-6m\sqrt{m^2-1}}{2(2m^2-1)}c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

u(x,y,z,t)=ε1−m2+12−6c2(k12+k22+k32)δ(2−m2)1−−11−m2dnc22−m2ξ1+−11−m2dnc22−m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \frac{\sqrt{1-m^2}+1}{2} \sqrt{-\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(2-m^2)}} \frac{1-\sqrt{-\frac{1}{\sqrt{1-m^2}}} \text{dn}\left(\sqrt{\frac{c_2}{2-m^2}}\xi \right)}{1+\sqrt{-\frac{1}{\sqrt{1-m^2}}} \text{dn}\left(\sqrt{\frac{c_2}{2-m^2}}\xi \right)}, \end{array}$$

where c0=c22(1−m2)c4(2−m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 (1-m^2)}{c_4(2-m^2)^2} \end{array}$, c2 > 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα+2−m2−61−m22(2−m2)c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} + \frac{2-m^2-6\sqrt{1-m^2}}{2(2-m^2)}c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

u(x,y,z,t)=ε−3c2(1−m2)(k12+k22+k32)δ(1−2m2)1+ε cn2c21−2m2ξ−sn2c21−2m2ξ1+ε cn2c21−2m2ξ+sn2c21−2m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{3 c_2 (1-m^2) (k_1^2+k_2^2+k_3^2)}{\delta(1-2m^2)}} \frac{1+\varepsilon~\text{cn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi\right)-\text{sn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi \right)}{1+\varepsilon~\text{cn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi \right)+\text{sn}\left(\sqrt{\frac{2c_2}{1-2m^2}}\xi \right)}, \end{array}$$

where c0=c224c4(1−2m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2}{4c_4(1-2m^2)^2} \end{array}$, c2 > 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα−(1+m2)c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - (1+m^2)c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$. As m → 0, solution (69) transforms to

u(x,y,z,t)=ε−3c2(k12+k22+k32)δ1+εcos2c2ξ−sin2c2ξ1+εcos2c2ξ+sin2c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{-\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1+\varepsilon\cos\left(\sqrt{2c_2}\xi \right)-\sin\left(\sqrt{2c_2}\xi \right)}{1+\varepsilon\cos\left(\sqrt{2c_2}\xi \right)+\sin\left(\sqrt{2c_2}\xi \right)}, \end{array}$$

where ξ=k1xαα+k2yαα+k3zαα−c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

u(x,y,z,t)=εd−3c4(k12+k22+k32)δ1−−2 dsd−2c4ξ,221+−2 dsd−2c4ξ,22, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon d \sqrt{-\frac{3 c_4 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1-\sqrt{-2} ~\text{ds}\left(d\sqrt{-2c_4}\xi, \frac{\sqrt{2}}{2}\right)}{1+\sqrt{-2} ~\text{ds}\left(d\sqrt{-2c_4}\xi, \frac{\sqrt{2}}{2} \right)}, \end{array}$$

where c2 = 0, c4 < 0 and δ > 0.

u(x,y,z,t)=εd−3c4(k12+k22+k32)δ1−ns2dc4ξ,22−cs2dc4ξ,221+ns2dc4ξ,22+cs2dc4ξ,22, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon d \sqrt{-\frac{3 c_4 (k_1^2+k_2^2+k_3^2)}{\delta}}\; \frac{1- \text{ns}\left(2d\sqrt{c_4}\xi, \frac{\sqrt{2}}{2}\right) - \text{cs}\left(2d\sqrt{c_4} \xi, \frac{\sqrt{2}}{2}\right)}{1+ \text{ns}\left(2d\sqrt{c_4} \xi, \frac{\sqrt{2}}{2} \right) + \text{cs}\left(2d\sqrt{c_4} \xi, \frac{\sqrt{2}}{2}\right)}, \end{array}$$

where c2 = 0, c4 > 0 and δ < 0. In solutions (82) and (83), ξ=k1xαα+k2yαα+k3zαα−3c4d2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} - 3 c_4 d^2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

a0=0,a1=−2c4BA,a−1=−2c0BA,ν=B(6c0c4−c2),d=0 $\begin{array}{} \displaystyle a_0=0, \;a_1= \sqrt{-\frac{2 c_4 B}{A}} ,\; a_{-1}= \sqrt{-\frac{2 c_0 B}{A}},\; \nu= B(6\sqrt{c_0 c_4}-c_2), \; d=0 \end{array}$

u(x,y,z,t)=ε6c2(k12+k22+k32)δ(m2+1)1−m sn2−c2m2+1ξsn−c2m2+1ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(m^2+1)}} \frac{1 - m ~\text{sn}^2\left(\sqrt{-\frac{c_2}{m^2+1}}\xi \right)}{\text{sn}\left(\sqrt{-\frac{c_2}{m^2+1}} \xi \right)}, \end{array}$$

where c0=c22m2c4(m2+1)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^2}{c_4(m^2+1)^2} \end{array}$, c2 < 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα+6m−m2−1m2+1c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +\frac{6m-m^2-1}{m^2+1} c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$. As m → 0, solution (84) converts to solution (44) while as m → 1, solution (84) degenerates to

u(x,y,z,t)=ε3c2(k12+k22+k32)δsech2−c22ξtanh−c22ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{ \text{sech}^2\left(\sqrt{-\frac{c_2}{2}}\xi \right)}{\tanh\left(\sqrt{-\frac{c_2}{2}}\xi \right)}, \end{array}$$

where ξ=k1xαα+k2yαα+k3zαα+2c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} + 2c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

u(x,y,z,t)=ε6c2(k12+k22+k32)δ(2m2−1)m2−1−m cn2c22m2−1ξcnc22m2−1ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(2m^2-1)}} \frac{\sqrt{m^2-1} - m ~\text{cn}^2\left(\sqrt{\frac{c_2}{2m^2-1}}\xi \right)}{\text{cn}\left(\sqrt{\frac{c_2}{2m^2-1}} \xi \right)}, \end{array}$$

where c0=c22m2(m2−1)c4(2m2−1)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^2(m^2-1)}{c_4(2m^2-1)^2} \end{array}$, c2 > 0, δ > 0 and ξ=k1xαα+k2yαα+k3zαα+6mm2−1−2m2+12m2−1c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +\frac{6m\sqrt{m^2-1}-2m^2+1}{2m^2-1} c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$. As m → 0, solution (86) converts to solution (43) while as m → 1, solution (86) degenerates to solution (41).

u(x,y,z,t)=ε6c2(k12+k22+k32)δ(2−m2)1−m2−dn2c22−m2ξdnc22−m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon \sqrt{\frac{6 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(2-m^2)}} \frac{\sqrt{1-m^2} - \text{dn}^2\left(\sqrt{\frac{c_2}{2-m^2}}\xi \right)}{\text{dn}\left(\sqrt{\frac{c_2}{2-m^2}} \xi \right)}, \end{array}$$

where c0=c22(1−m2)c4(2−m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 (1-m^2)}{c_4(2-m^2)^2} \end{array}$, c2 > 0, δ > 0 and ξ=k1xαα+k2yαα+k3zαα+61−m2+m2−22−m2c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +\frac{6\sqrt{1-m^2}+m^2-2}{2-m^2} c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$. As m → 1, solution (87) reduces to solution (41).

u(x,y,z,t)=2ε−3c2(k12+k22+k32)δ(1−2m2)ns2c21−2m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= 2\varepsilon\sqrt{-\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(1-2m^2)}} \text{ns}\left(\sqrt{\frac{2c_2}{1-2m^2}} \xi \right), \end{array}$$

where c0=c224c4(1−2m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2}{4c_4(1-2m^2)^2} \end{array}$, c2 > 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα+2(1+m2)1−2m2c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +\frac{2(1+m^2)}{1-2m^2} c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$. As m → 0, solution (88) reduces to

u(x,y,z,t)=2ε−3c2(k12+k22+k32)δcsc2c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= 2\varepsilon\sqrt{-\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \text{cs}c\left(\sqrt{2c_2} \xi \right), \end{array}$$

where ξ=k1xαα+k2yαα+k3zαα+2c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +2 c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$. As m → 1, solution (88) becomes

u(x,y,z,t)=2ε3c2(k12+k22+k32)δcoth−2c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= 2\varepsilon\sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \coth\left(\sqrt{-2c_2} \xi \right), \end{array}$$

where c2 < 0, δ < 0 and ξ=k1xαα+k2yαα+k3zαα−4c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} -4 c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

u(x,y,z,t)=2ε3c2(k12+k22+k32)δ(2−m2)ds−2c22−m2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= 2\varepsilon\sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta(2-m^2)}} \text{ds}\left(\sqrt{-\frac{2c_2}{2-m^2}} \xi \right), \end{array}$$

where c0=c22m44c4(2−m2)2 $\begin{array}{} \displaystyle c_0=\frac{c_2^2 m^4}{4c_4(2-m^2)^2} \end{array}$, c2 < 0, δ > 0 and ξ=k1xαα+k2yαα+k3zαα+4m2−22−m2c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +\frac{4m^2-2}{2-m^2} c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$. As m → 0, solution (91) turns to solution (44) while as m → 1, solution (91) reduces to

u(x,y,z,t)=2ε3c2(k12+k22+k32)δcsch−2c2ξ, $$\begin{array}{} \displaystyle u(x,y,z,t)= 2\varepsilon\sqrt{\frac{3 c_2 (k_1^2+k_2^2+k_3^2)}{\delta}} \text{csch}\left(\sqrt{-2c_2}\xi \right), \end{array}$$

where c2 < 0, ξ=k1xαα+k2yαα+k3zαα+2c2k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} +2 c_2 k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

u(x,y,z,t)=ε(−c0c4)1/43(k12+k22+k32)δ1+2ids2(−4c0c4)1/4ξ,22dn(−4c0c4)1/4ξ,22, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon (- c_0 c_4)^{1/4} \sqrt{\frac{3 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1 + 2 i \text{ds}^2\left((-4 c_0 c_4)^{1/4}\xi, \frac{\sqrt{2}}{2} \right)}{\text{dn}\left((-4 c_0 c_4)^{1/4} \xi, \frac{\sqrt{2}}{2} \right)}, \end{array}$$

where c2 = 0, c0c4 < 0 and δ > 0.

u(x,y,z,t)=ε(c0c4)1/4−6(k12+k22+k32)δ1+ns2(c0c4)1/4ξ,22+cs2(c0c4)1/4ξ,222ns2(c0c4)1/4ξ,22+cs2(c0c4)1/4ξ,22, $$\begin{array}{} \displaystyle u(x,y,z,t)=\varepsilon (c_0 c_4)^{1/4} \sqrt{-\frac{6 (k_1^2+k_2^2+k_3^2)}{\delta}} \frac{1 + \left(\text{ns}\left(2(c_0 c_4)^{1/4}\xi, \frac{\sqrt{2}}{2} \right)+\text{cs}\left(2(c_0 c_4)^{1/4}\xi, \frac{\sqrt{2}}{2} \right)\right)^2 }{\text{ns}\left(2(c_0 c_4)^{1/4}\xi, \frac{\sqrt{2}}{2} \right)+\text{cs}\left(2(c_0 c_4)^{1/4}\xi, \frac{\sqrt{2}}{2} \right)}, \end{array}$$

where c2 = 0, c0c4 > 0 and δ < 0. In solutions (93) and (94), ξ=k1xαα+k2yαα+k3zαα+6c0c4k1(k12+k22+k32)tαα $\begin{array}{} \displaystyle \xi=k_1\frac{x^\alpha}{\alpha}+k_2\frac{y^\alpha}{\alpha}+k_3\frac{z^\alpha}{\alpha} + 6 \sqrt{c_0 c_4} k_1(k_1^2+k_2^2+k_3^2)\frac{t^\alpha}{\alpha} \end{array}$.

In what follows, we depict the dynamics of solitons and periodic waves in the model of the (3+1)-dimensional space-time fractional modified KdV–Zakharov–Kuznetsov equation to show a clear understanding of the physical properties of obtained results. The numerical results of some of the derived solutions are exhibited by selecting different values for the constants ε, δ, α, c2 and kl (l = 1, 2, 3). For example, the 3D and 2D plots of the bell-shaped solitary wave solution (41) are displayed in Figure 1 with ε = 1, k1 = 1, k2 = 1.2, k3 = 2.2, c2 = 1.5, δ = 2.5 when α = 0.95. Figure 2 shows the 3D and 2D plots of the kink-shaped solitary wave solution (45) for ε = −1, k1 = 0.2, k2 = 1.5, k3 = 0.25, c2 = 0.5, δ = −1.2 when α = 0.9. In Figure 3, the 3D and 2D plots of the singular soliton solution (50) are depicted for ε = 1, k1 = 0.5, k2 = 1.5, k3 = 0.3, c2 = 1.3, δ = −1 when α = 0.95. Further to this, the 3D and 2D plots of the singular periodic solution (70) are described in Figure 4 for ε = 1, k1 = 1, k2 = 0.5, k3 = 1.5, c2 = 2, δ = -0.5 when α = 1. Figure 5 shows the 3D and 2D plots of the singular periodic solution (89) for ε = 1, k1 = −1.5, k2 = 0.75, k3 = 1, c2 = 1.1, δ = −1.2 when α = 1. Figure 6 presents the 3D and 2D plots of the singular soliton solution (92) with ε = −1, k1 = 0.5, k2 = -0.5, k3 = 0.5, c2 = −1.2, δ = -1.5$ when α = 1.

Fig. 1

The solitary wave solution (41) for k1 = 1, k2 = 1.2, k3 = 2.2, c2 = 1.5, δ = 2.5, α = 0.95.

Fig. 2

The kink wave solution (45) for k1 = 0.2, k2 = 1.5, k3 = 0.25, c2 = 0.5, δ = −1.2, α = 0.9.

Fig. 3

The singular soliton solution (50) for k1 = 0.5, k2 = 1.5, k3 = 0.3, c2 = 1.3, δ = −1, α = 0.95.

Fig. 4

The singular periodic solution (70) for k1 = 1, k2 = 0.5, k3 = 1.5, c2 = 2, δ = −0.5, α = 1.

Fig. 5

The singular periodic solution (89) for k1 = −1.5, k2 = 0.75, k3 = 1, c2 = 1.1, δ = −1.2, α = 1.

Fig. 6

The singular soliton solution (92) for k1 = 0.5, k2 = -0.5, k3 = 0.5, c2 = −1.2, δ = −1.5, α = 1.

Remark 2

All of the results are calculated by using Maple, when t = 1 and z = 2 with the interval 0 < x, y ≤ 10.
Conclusions

In this study, we have investigated the exact analytic solutions to the (3+1)-dimensional space-time fractional mKdV–ZK equation. By means of conformable fractional derivative and wave transformation, the fractional mKdV–ZK equation is changed to an ODE. Then, the resulting ODE is solved by applying the variable separated ODE method. All obtained solutions are verified by utilising symbolic computation. To shed light on the behaviour of extracted results, some of derived solutions are displayed graphically. Moreover, it is found that the implemented method is a powerful mathematical tool for solving ODE and provides more exact solutions such as solitary, kink and periodic waves. Consequently, it can be applied to different physical models to generate various types of solutions.
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