Conservation laws, analytical solutions and stability analysis for the time-fractional Schamel–Zakharov–Kuznetsov–Burgers equation

Abstract

In this paper, we consider the $(3+1)$-dimensional time-fractional Schamel–Zakharov–Kuznetsov–Burgers (SZKB) equation. With the help of the Riemann–Liouville derivatives, the Lie point symmetries of the $(3+1)$-dimensional time-fractional SZKB equation are derived. By applying the Lie point symmetry method as well as Erdélyi–Kober fractional operator, we get the similarity reductions of the time-fractional SZKB equation. Conservation laws of the time-fractional SZKB are constructed. Moreover, we obtain its power series solutions with the convergence analysis. In addition, the analytical solution is obtained by modified trial equation method. Finally, stability is analyzed graphically in different planes.

1 Introduction

Partial differential equations (PDEs) are frequently used to describe most of the phenomena that arise in engineering fields, mathematical physics, plasma physics, solid state physics, quantum mechanics, fluid mechanics, ecology, optical fibers, biology, chemical kinematics, geochemistry, meteorology, electricity and so on. Therefore, investigating analytical solutions (traveling wave solutions or soliton solutions) is very interesting. As a result, many new techniques have been successfully proposed, developed and extended by groups of researchers to find exact or analytical solutions for PDEs, such as the $(G^{\prime}/G)$-expansion method [1,2,3], the Kudryashov method [4, 5], the functional variable method [5, 6], the first integral method [7], the $\exp(-\phi(\xi))$-expansion method [8], the sine–cosine function method and Bernoulli’s equation approach [5, 9], the trial solution method [5, 10], Hirota’s bilinear method [3] etc.

Fractional calculus was developed as one of the best tools for studying various models in each discipline [11, 12]. Using fractional models to describe the special characteristic phenomenon of fractional order, it can well reveal the essence of the nature of phenomena and their behavior. Fractional calculus is a generalization of integer calculus, integer calculus is a special case of fractional calculus. It has universal significance to study the fractional calculus system. Compared with the integer order model in describing the actual physical process, the physical meaning of the fractional order model is clearer and the expression is more concise.

On the other hand, in recent years, fractional ordinary differential equations (FODEs) and fractional partial differential equations (FPDEs) have been widely used to describe many various physical effects and many complex nonlinear phenomena. This is due to their accurate description of nonlinear phenomena in fluid mechanics, viscoelasticity, electrical chemistry, quantum biology, physics, engineering mechanics [13,14,15,16,17,18,19] and other scientific fields. So the study of FODEs and FPDEs has attracted much attention. In other words, by using the theory of derivatives and integrals of fractional order, many physical phenomena can be accurately modeled.

The exact solutions of most of the FPDEs cannot be found easily, so analytical and numerical methods [20, 21] must be used. The solutions of the FPDEs are investigated by many authors using powerful analytical methods. Several numerical methods such as the homotopy perturbation method [22, 23], the Adomian decomposition method [24], the variational iteration method [25], the differential transform method [26], the fractional Riccati expansion method [27], the fractional sub-equation method [28,29,30,31,32,33,34] have been suggested for solving FDEs. However, solutions obtained through all these methods are of a local nature and it is important to explore other techniques in order to find exact analytical solutions of FDEs.

Lie point symmetry plays a very important role in various fields of science, especially in integrable systems where infinitely many symmetries exist. So Lie symmetry analysis is considered to be one of the efficient approaches for obtaining analytical solutions of nonlinear partial differential equations (NLPDEs). A large number of studies are devoted to the theory of Lie point symmetry and their applications to DEs. The symmetry analysis of FDEs and the fractional derivatives are proposed by Gazizov and his collaborators [35]. They proposed a prolongation formula for two basic fractional derivatives: Riemann–Liouville and Caputo. This method has been used to study many of the FDEs [36,37,38]. Furthermore Lie point symmetry is used to construct conservation laws which play an important role in the study of nonlinear physical phenomena [12]. Conservation laws are a mathematical formulation which statement that the total amount of a certain physical quantity stays the same during the evolution of a physical system [12]. Conservation laws are also used in the development of numerical methods to establish the existence and uniqueness of solution. There are many methods of constructing conservation laws for differential equations (DEs) [39,40,41,42,43,44,45]. The well-known Noether theorem [46] establishes a connection between Lie point symmetries and conservation laws of DEs, provided that the equations are Euler–Lagrange equations. Ibragimov [47] suggested a new conservation law theorem. In [39], conservation laws for time-fractional sub-diffusion and diffusion-wave equations were obtained based on the new conservation laws theorem [47]. Also there are researches that discussed conservation laws for time FPDEs [12, 48, 49].

In this work, we focus on the time-fractional SZKB equation of the form

$$ \frac{\partial^{\alpha}u}{\partial t^{\alpha}}+a \sqrt{u} u_{x}+b u_{xxx}+c(u_{xyy}+u_{xzz})+d u_{xx}=0, $$

(1.1)

where $\partial^{\alpha}_{t}$ is the fractional derivative of order α, with $0<\alpha<1$ and $u(x,y,z,t)$ is the potential function of space x, y, z and time t. If $\alpha=1$, Eq. (1.1) is reduced to the classical SZKB equation, which describes the nonlinear plasma-dust ion acoustic waves DIAWs in a magnetized dusty plasma and it is derived using the standard reductive perturbation technique in small amplitude. The coefficients of dispersion b, non-linearity a, mixed derivative c and dissipation d are given in [50, 51]. Note that Sahoo and Ray [52] have studied the $(3+1)$-dimensional time-fractional mKdV-ZK equation without the Schamel and Burgers terms. The main purpose of this paper is to obtain the Lie point symmetries, conservation laws and analytical solutions of the time-fractional SZKB equation.

The paper is organized as follows: The introduction is presented in Sect. 1. In Sect. 2, some definitions and description of Lie symmetry analysis for fractional partial differential equations (FPDEs) are briefly presented. Lie point symmetries and similarity reduction of the Eq. (1.1) are obtained In Sect. 3. In Sect. 4, the conservation laws of the Eq. (1.1) are obtained. In Sect. 5, which is based on the power series, the analytical solution of the Eq. (1.1) is constructed with convergence analysis. We construct the analytical solution of Eq. (1.1) by using a modified trial equation method In Sect. 6. Finally, the discussions and conclusions of this paper are presented in Sect. 7.

2 Preliminaries

Here in this section, we focus on some of the concepts that revolve around the subject of our article

Definition 1

Let $\alpha>0$. The operator $I^{\alpha}$ defined by

$$ I^{\alpha}f(t)=\frac{1}{\varGamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha -1}f(s) \,ds, $$

(2.1)

is called the Riemann–Liouville (R-L) fractional integral operator of order α, and $\varGamma(\cdot)$ denotes the gamma function.

Definition 2

Let $\alpha>0$. The operator $D^{\alpha}_{t}$ defined by

$$ D^{\alpha}_{t} f(t)= \textstyle\begin{cases} \frac{1}{\varGamma(n-\alpha)}\frac{d^{n}}{d t^{n}}\int_{0}^{t} (t-s)^{n-\alpha -1}f(s) \,ds &\text{if }n-1< \alpha< n, n\in N,\\ \frac{d^{n}f(t)}{d t^{n}} &\text{if }\alpha=n, n\in N, \end{cases} $$

(2.2)

is called the R-L fractional partial derivative [16, 53,54,55,56].

2.1 Description of Lie point symmetries

Let us consider the symmetry analysis for a FPDE of the form

$$ D^{\alpha}_{t} u(x,y,z,t)=G(x,y,z,t,u,u_{x},u_{t},u_{y},u_{z},u_{xx},u_{xy}, \ldots),\quad 0< \alpha< 1. $$

(2.3)

Now, let Eq. (2.3) be invariant under the following one-parameter Lie group of point transformations acting on both the dependent and the independent variables:

$$\begin{aligned}& \bar{x}=x+\varepsilon\xi(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \\& \bar{y}=y+\varepsilon\zeta(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \\& \bar{z}=z+\varepsilon\nu(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \\& \bar{t}=t+\varepsilon\tau(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \\& \bar{u}=u+\varepsilon\eta(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \\& D^{\alpha}_{\bar{t}}\bar{u}=D^{\alpha}_{t} u+\varepsilon\eta_{\alpha}^{0}(x,y,z,t)+O \bigl( \varepsilon^{2} \bigr), \\& \frac{\partial\bar{u}}{\partial\bar{x}}=\frac{\partial u}{\partial x}+\varepsilon\eta^{x}(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \\& \frac{\partial^{2} \bar{u}}{\partial\bar{x}^{2}}=\frac{\partial^{2} u}{\partial x^{2}}+\varepsilon\eta^{xx}(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \\& \frac{\partial^{3} \bar{u}}{\partial\bar{x}^{3}}=\frac{\partial^{3} u}{\partial x^{3}}+\varepsilon \eta^{xxx}(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \\& \frac{\partial^{3} \bar{u}}{\partial\bar{x}\,\partial\bar {y}^{2}}=\frac {\partial^{3} u}{\partial x\,\partial y^{2}}+\varepsilon\eta ^{xyy}(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \\& \frac{\partial^{3} \bar{u}}{\partial\bar{x}\,\partial\bar {z}^{2}}=\frac {\partial^{3} u}{\partial x\,\partial z^{2}}+\varepsilon\eta ^{xzz}(x,y,z,t)+O \bigl(\varepsilon^{2} \bigr), \end{aligned}$$

(2.4)

where $\varepsilon\ll1$ is the Lie group parameter and ξ, ζ, ν, τ, η are the infinitesimals of the transformations for dependent and independent variables, respectively. The explicit expressions of $\eta^{x}$, $\eta^{xx}$, $\eta^{xxx}$, $\eta^{xyy}$, $\eta^{xzz}$ are given by

$$ \begin{gathered}\eta^{x}=D_{x}( \eta)-u_{x}D_{x}( \xi)-u_{y}D_{x}( \zeta)-u_{z}D_{x}( \nu)-u_{t}D_{x}(\tau), \\ \eta^{xx}=D_{x} \bigl(\eta^{x} \bigr)-u_{xx}D_{x}(\xi)-u_{xy}D_{x}( \zeta)-u_{xz}D_{x}(\nu )-u_{xt}D_{x}( \tau), \\ \eta^{xxx}=D_{x} \bigl(\eta^{xx} \bigr)-u_{xxx}D_{x}(\xi)-u_{xxy}D_{x}(\zeta )-u_{xxz}D_{x}(\nu)-u_{xxt}D_{x}(\tau), \\ \eta^{xyy}=D_{x} \bigl(\eta^{yy} \bigr)-u_{xxy}D_{x}(\xi)-u_{xyy}D_{x}(\zeta )-u_{xyz}D_{x}(\nu)-u_{xyt}D_{x}(\tau), \\ \eta^{xzz}=D_{x} \bigl(\eta^{zz} \bigr)-u_{xxz}D_{x}(\xi)-u_{xyz}D_{x}(\zeta )-u_{xzz}D_{x}(\nu)-u_{xzt}D_{x}(\tau), \end{gathered} $$

(2.5)

where $D_{x}$, $D_{y}$, $D_{z}$, and $D_{t}$ are the total derivatives with respect to x, y, z, and t, respectively, that are defined for $x^{1}=x$, $x^{2}=y$, $x^{3}=z$ as

$$ D_{xj}=\frac{\partial}{\partial x^{j}}+u_{j}\frac{\partial}{\partial u}+u_{jk} \frac{\partial}{\partial u_{k}}+\cdots,\quad j,k=1,2,3,\ldots, $$

where $u_{j}=\frac{\partial u}{\partial x^{j}}$, $u_{jk}=\frac{\partial ^{2}u}{\partial x^{j} \partial x^{k}}$ and so on.

The corresponding Lie algebra of symmetries consists of a set of vector fields of the form

$$ V=\xi\frac{\partial}{\partial x}+\zeta\frac{\partial}{\partial y}+\nu \frac{\partial}{\partial z}+\tau \frac{\partial}{\partial t}+\eta \frac {\partial}{\partial u}. $$

The invariance condition of Eq. (2.3) under the infinitesimal transformations is given as

$$ \operatorname{Pr}^{(n)} V(\Delta)\big|_{\Delta=0}=0,\quad n=1,2,3, \ldots, $$

where

$$ \Delta:=D^{\alpha}_{t} u(x,y,z,t)-G(x,y,z,t,u,u_{x},u_{t},u_{y},u_{z},u_{xx},u_{xy}, \ldots). $$

Also, the invariance condition gives

$$ \tau(x,y,z,t,u)|_{t=0}=0. $$

(2.6)

The αth extended infinitesimal related to RL fractional time derivative with Eq. (2.6) can be represented as follows:

$$ \begin{aligned}[b] \eta_{\alpha}^{0}={}&D_{t}^{\alpha}( \eta)+\xi D_{t}^{\alpha}(u_{x})-D_{t}^{\alpha}( \xi u_{x})+\zeta D_{t}^{\alpha}(u_{y})-D_{t}^{\alpha}( \zeta u_{y}) +\nu D_{t}^{\alpha}(u_{z}) \\ &-D_{t}^{\alpha}(\nu u_{z})+D_{t}^{\alpha}\bigl(D_{t}(\tau)u \bigr)-D_{t}^{\alpha+1}(\tau u)+\tau D_{t}^{\alpha+1}(u),\end{aligned} $$

(2.7)

where $D_{t}^{\alpha}$ is the total fractional derivative operator and by using the generalized Leibnitz rule

$$ D_{t}^{\alpha}\bigl(f(t)g(t) \bigr)=\sum _{n=0}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )D_{t}^{\alpha-n}f(t)D_{t}^{n} g(t), \quad \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )= \frac{(-1)^{n-1}\alpha\varGamma(n-\alpha)}{\varGamma (1-\alpha)\varGamma(n+1)}. $$

By applying the Leibnitz rule, Eq. (2.7) becomes

$$ \begin{aligned}[b]\eta_{\alpha}^{0}={}&D_{t}^{\alpha}( \eta)-\alpha D_{t}^{\alpha}(\tau)\frac {\partial ^{\alpha}u}{\partial t^{\alpha}}-\sum _{n=1}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )D_{t}^{n} (\xi)D_{t}^{\alpha-n}u_{x} -\sum_{n=1}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )D_{t}^{n} ( \zeta)D_{t}^{\alpha-n}u_{y} \\ &-\sum_{n=1}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )D_{t}^{n} ( \nu)D_{t}^{\alpha-n}u_{z} -\sum _{n=1}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n+1 \end{array}\displaystyle } \right )D_{t}^{n+1} (\xi)D_{t}^{\alpha-n}u. \end{aligned} $$

(2.8)

Now the chain rule for the compound function is defined as follows:

$$ \frac{d^{n}\phi(h(t))}{dt^{n}}=\sum_{k=0}^{n}\sum _{r=0}^{k}\left ({ \textstyle\begin{array}{c} k \\ r \end{array}\displaystyle } \right )\frac{1}{k!} \bigl[-h(t) \bigr]^{r} \frac{d^{n}}{dt^{n}} \bigl[-h(t)^{k-r} \bigr]\times\frac{d^{k}\phi(h)}{dh^{k}}. $$

By applying this rule and the generalized Leibnitz rule with $f(t)=1$, we have

$$ D_{t}^{\alpha}(\eta)=\frac{\partial^{\alpha}\eta}{\partial t^{\alpha}}+\eta _{u} \frac{\partial^{\alpha}u}{\partial t^{\alpha}}-u\frac{\partial ^{\alpha}\eta _{u}}{\partial t^{\alpha}}+\sum_{n=1}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )\frac{\partial^{n}\eta_{u}}{\partial t^{n}}D_{t}^{\alpha-n}(u)+ \mu, $$

where

$$ \mu=\sum_{n=2}^{\infty}\sum _{m=2}^{n} \sum_{k=2}^{m} \sum_{r=0}^{k-1} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )\left ({ \textstyle\begin{array}{c} n \\ m \end{array}\displaystyle } \right )\left ({ \textstyle\begin{array}{c} k \\ r \end{array}\displaystyle } \right ) \frac{1}{k!}\times\frac{t^{n-\alpha}}{\varGamma (n+1-\alpha)} [-u ]^{r}\frac{\partial^{m}}{\partial t^{m}} \bigl[u^{k-r} \bigr]\frac {\partial ^{n-m+k}\eta}{\partial t^{n-m}\partial u^{k}}. $$

Therefore, Eq. (2.8) yields

$$\begin{aligned} \eta_{\alpha}^{0}={}&\frac{\partial^{\alpha}\eta}{\partial t^{\alpha}}+ \bigl( \eta _{u}-\alpha D_{t}^{\alpha}(\tau) \bigr) \frac{\partial^{\alpha}u}{\partial t^{\alpha}}-u\frac{\partial^{\alpha}\eta_{u}}{\partial t^{\alpha}}+\mu \\ &+\sum_{n=1}^{\infty}\biggl[\left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )\frac{\partial^{\alpha}\eta_{u}}{\partial t^{\alpha}}- \left ({ \textstyle\begin{array}{c} \alpha\\ n+1 \end{array}\displaystyle } \right )D_{t}^{n+1}( \tau) \biggr]D_{t}^{\alpha-n}(u)+\sum_{n=1}^{\infty}({ \alpha } )D_{t}^{n} ( \xi)D_{t}^{\alpha-n}u_{x} \\ & -\sum_{n=1}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )D_{t}^{n} ( \zeta)D_{t}^{\alpha-n}u_{y} -\sum _{n=1}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )D_{t}^{n} (\nu)D_{t}^{\alpha-n}u_{z}. \end{aligned}$$

(2.9)

Definition 3

The function $u = \theta(x,y,z,t)$ is an invariant solution of Eq. (2.3) associated with the vector field W, such that

• 1. $u = \theta(x,y,z,t)$ is an invariant surface of Eq. (2.3), this means

  $$V \theta=0 \quad\Leftrightarrow \quad \biggl(\xi\frac{\partial}{\partial x}+\zeta \frac{\partial}{\partial y}+\nu\frac{\partial}{\partial z}+\tau \frac {\partial}{\partial t}+\eta \frac{\partial}{\partial u} \biggr)\theta=0, $$

• 2. $u = \theta(x,y,z,t)$ satisfies Eq. (2.3).

3 Lie point symmetries and similarity reduction for Eq. (1.1)

In this section, we implemented Lie group method for Eq. (1.1) and then, used these symmetries to reduce Eq. (1.1) to be a FODE as shown in the next two subsections

3.1 Lie point symmetries for Eq. (1.1)

Let us consider Eq. (1.1) as an invariant under Eq. (2.4), we get

$$ \frac{\partial^{\alpha}\bar{u}}{\partial\bar{t}^{\alpha}}+a \sqrt {\bar {u}} \bar{u}_{x}+b \bar{u}_{xxx}+c(\bar{u}_{xyy}+\bar{u}_{xzz})+d \bar{u}_{xx}=0, $$

(3.1)

such that $u = u(x,y,z, t)$ satisfies Eq. (1.1), then using the point transformations Eq. (2.4) in Eq. (3.1), we get the invariant equation

$$ \eta^{0}_{\alpha}+a \sqrt{u} \eta^{x}+ \frac{a}{2\sqrt{u}}\eta u_{x}+b \eta ^{xxx}+c \bigl( \eta^{xyy}+\eta^{xzz} \bigr)+d \eta^{xx}=0, $$

(3.2)

By substituting Eq. (2.5) and Eq. (2.9) into Eq. (3.2), we acquire

$$\begin{aligned}& \frac{\partial^{\alpha}\eta}{\partial t^{\alpha}}+ \bigl(\eta_{u}-\alpha D_{t}^{\alpha}(\tau) \bigr)\frac{\partial^{\alpha}u}{\partial t^{\alpha}}-u \frac{\partial ^{\alpha}\eta_{u}}{\partial t^{\alpha}}+ \mu \\& \quad{}+\sum_{n=1}^{\infty}\biggl[\left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )\frac{\partial^{\alpha}\eta_{u}}{\partial t^{\alpha}}-\left ({ \textstyle\begin{array}{c} \alpha\\ n+1 \end{array}\displaystyle } \right )D_{t}^{n+1}(\tau) \biggr]D_{t}^{\alpha-n}(u)+ \sum_{n=1}^{\infty}({ \alpha } )D_{t}^{n} ( \xi)D_{t}^{\alpha-n}u_{x} \\& \quad{} -\sum_{n=1}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )D_{t}^{n} (\zeta)D_{t}^{\alpha-n}u_{y} -\sum _{n=1}^{\infty} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right )D_{t}^{n} (\nu)D_{t}^{\alpha-n}u_{z} \\& \quad{} +a \sqrt{u} \bigl(\eta_{x}+(\eta _{u}- \xi_{x})u_{x}-\xi_{u}u^{2}_{x}-( \zeta_{x}+ \zeta_{u}u_{x})u_{y}-( \nu_{x}+ \nu_{x}u_{x})u_{z}-(\tau _{x}+\tau _{u}u_{x})u_{t} \bigr) \\& \quad{}+\frac{a}{2\sqrt{u}}\eta u_{x}+b \bigl(\eta _{xxx}+( \eta _{xxu}- \xi_{xxx})u_{x}+(3\eta_{xu}-3 \xi_{xx})u_{xx}+(\eta_{u}-3\xi_{x})u_{xxx} \\& \quad{} +(3 \eta_{xuu}-3\xi_{xxu})u_{x}^{2}+ ( \eta_{uuu}-\xi_{xuu})u_{x}^{3}- \xi_{uuu}u_{x}^{4}+(3\eta_{uu}-9\xi _{xu})u_{x}u_{xx} \\& \quad{} -3\xi_{uu}u^{2}_{x}u_{xx}-3 \xi_{u}u_{xx}^{2}-4\xi_{u}u_{x}u_{xxx} -3( \zeta_{x}+\zeta_{u}u_{x})u_{xxy}-3( \nu_{x}+\nu_{x}u_{x})u_{xxz} \\& \quad{} -3(\tau _{x}+\tau _{u}u_{x})u_{xxt}-3 \bigl( \zeta_{xx}+2\zeta_{xu}u_{x}+ \zeta_{uu}u^{2}_{x}+ \zeta _{u}u_{xx} \bigr)u_{xy} \\& \quad{} -3 \bigl(\nu_{xx}+2\nu_{xu}u_{x}+ \nu_{uu}u^{2}_{x}+\nu_{u}u_{xx} \bigr)u_{xz}-3 \bigl(\tau _{xx}+2\tau_{xu}u_{x}+ \tau_{uu}u^{2}_{x}+\tau_{u}u_{xx} \bigr)u_{xt} \\& \quad{} - \bigl(\zeta_{xxx}+3\zeta_{xxu}u_{x}+3 \zeta_{xuu}u^{2}_{x}+\zeta_{uuu}u^{3}_{x}+3 \zeta_{uu}u_{x}u_{xx}+3 \zeta_{xu}u_{xx}+ \zeta_{u}u_{xxx} \bigr)u_{y} \\& \quad{} - \bigl(\nu_{xxx}+3\nu_{xxu}u_{x}+3 \nu_{xuu}u^{2}_{x}+\nu_{uuu}u^{3}_{x}+3 \nu _{uu}u_{x}u_{xx}+3\nu_{xu}u_{xx}+ \nu_{u}u_{xxx} \bigr)u_{z} \\& \quad{} - \bigl(\tau_{xxx}+3\tau_{xxu}u_{x}+3 \tau_{xuu}u^{2}_{x}+\tau_{uuu}u^{3}_{x}+3 \tau _{uu}u_{x}u_{xx}+3\tau_{xu}u_{xx}+ \tau_{u}u_{xxx} \bigr)u_{t} \bigr) \\& \quad{} + c ( \eta_{xyy}+(\eta_{uyy}- \xi_{xyy})u_{x}+(2\eta_{uy}-\zeta_{yy}-2 \xi_{xy})u_{xy} \\& \quad{} +(2 \eta_{xuy}+2\eta _{uuy}u_{x}- \zeta_{xyy}- \zeta_{uyy}u_{x})u_{y} +( \eta_{u}-2 \zeta_{y}-\xi_{x}- \zeta_{u}u_{x})u_{xyy} \\& \quad{} +2(\eta_{uu}-2\zeta_{uy})u_{y}u_{xy}+ (\eta_{xu}+\eta_{uu}u_{x}-2\zeta_{xy}-2 \zeta_{uy}u_{x})u_{yy}-3\zeta _{uu}u_{y}^{2}u_{xy} \\& \quad{} +(3 \eta_{uux}+\eta_{uuu}u_{x}-2 \zeta_{xuy}-2 \zeta _{uuy}u_{x})u^{2}_{y}-3 \zeta_{u}u_{xy}u_{yy}-( \zeta_{uux}+ \zeta_{uuu}u_{x})u^{3}_{y} \\& \quad{} -3 \zeta _{u}u_{y}u_{xyy}-3( \zeta_{xu}+ \zeta_{uu}u_{x})u_{y}u_{yy} -(2 \xi_{y}+2\xi_{u}u_{y}+\xi_{x}+ \xi_{u}u_{x})u_{xxy} \\& \quad{} -2(\xi_{uy}u_{x}+\xi_{xu}u_{y}+ \xi_{uu}u_{x}u_{y}+\xi_{u}u_{xy})u_{xy}-(2 \nu _{y}+2\nu_{u}u_{y}+\nu_{x}+ \nu_{u}u_{x})u_{xyz} \\& \quad{} -2(\nu_{uy}u_{x}+ \nu_{xu}u_{y}+ \nu_{uu}u_{x}u_{y}+ \nu_{u}u_{xy})u_{zy}(2 \tau_{y}+2\tau_{u}u_{y}+ \tau_{x}+ \tau_{u}u_{x})u_{xyt} \\& \quad{} -2( \tau_{uy}u_{x}+\tau_{xu}u_{y}+ \tau_{uu}u_{x}u_{y}+\tau _{u}u_{xy})u_{yt}- \bigl(\xi _{yy}u_{x}+2\xi_{uy}u_{y}+ \xi_{uu}u^{2}_{y}+\xi_{u}u_{yy} \bigr)u_{xx} \\& \quad{} - \bigl(\nu_{yy}+2\nu_{uy}u_{y}+ \nu_{uu}u^{2}_{y}+\nu_{u}u_{yy} \bigr)u_{xz}- \bigl(\tau _{yy}+2\tau _{uy}u_{y}+ \tau_{uu}u^{2}_{y}+\tau_{u}u_{yy} \bigr)u_{xt} \\& \quad{} - \bigl(\xi_{xyy}+\xi_{uyy}u_{x}+2 \xi_{xyu}u_{y}+2\xi_{uuy}u_{x}u_{y}+2 \xi_{uy}u_{xy}+\xi_{xuu}u_{y}^{2}+ \xi_{uuu}u_{x}u_{y}^{2}+ 2 \xi_{uu}u_{x}u_{xy} \\& \quad{} +\xi_{xu}u_{yy}+ \xi_{xu}u_{x}u_{yy}+ \xi_{u}u_{xyy} \bigr)u_{x}- \bigl( \nu_{xyy}+\nu_{uyy}u_{x}+2 \nu_{xyu}u_{y}+2 \nu_{uuy}u_{x}u_{y}+2 \nu_{uy}u_{xy} \\& \quad{} +\nu_{xuu}u_{y}^{2}+ \nu _{uuu}u_{x}u_{y}^{2}+2\nu _{uu}u_{x}u_{xy}+ \nu_{xu}u_{yy}+ \nu_{xu}u_{x}u_{yy}+\nu_{u}u_{xyy} \bigr)u_{z} \\& \quad{} - \bigl(\tau_{xyy}+\tau_{uyy}u_{x}+2 \tau_{xyu}u_{y}+2\tau_{uuy}u_{x}u_{y}+2 \tau_{uy}u_{xy}+\tau _{xuu}u_{y}^{2}+ \tau _{uuu}u_{x}u_{y}^{2}+2 \tau_{uu}u_{x}u_{xy} \\& \quad{} + \tau_{xu}u_{yy}+ \tau_{xu}u_{x}u_{yy} +\tau_{u}u_{xyy} \bigr)u_{t}+ \eta_{xzz}+(\eta_{uzz}-\xi_{xzz})u_{x}+(2 \eta_{uy}-\zeta_{yy}-2\xi _{xy})u_{xz} \\& \quad{} +(2 \eta _{xuz}+2\eta_{uuz}u_{x}- \nu_{xzz}- \nu_{uzz}u_{x})u_{z} +( \eta_{u}-2 \nu_{z}-\xi_{x}-\nu_{u}u_{x})u_{xzz} \\& \quad{} +2(\eta_{uu}-2\nu_{uz})u_{z}u_{xz}+ (\eta_{xu}+\eta_{uu}u_{x}-2\nu_{xz}-2 \nu_{uz}u_{x})u_{zz}-3\nu _{uu}u_{z}^{2}u_{xz} \\& \quad{} +(3 \eta_{uux}+\eta_{uuu}u_{x}-2 \nu_{xuz}-2 \nu_{uuz}u_{x})u^{2}_{z}-3 \nu_{u}u_{xz}u_{zz}-( \nu_{uux}+ \nu_{uuu}u_{x})u^{3}_{z}-3 \nu _{u}u_{z}u_{xzz} \\& \quad{} -3(\nu _{xu}+ \nu_{uu}u_{x})u_{z}u_{zz} -(2 \xi_{z}+2\xi_{u}u_{z}+\xi_{x}+ \xi_{u}u_{x})u_{xxz} \\& \quad{} -2(\xi_{uz}u_{x}+\xi_{xu}u_{z}+ \xi_{uu}u_{x}u_{z}+\xi_{u}u_{xz})u_{xz}-(2 \zeta _{z}+2\zeta_{u}u_{z}+ \zeta_{x}+ \zeta_{u}u_{x})u_{xyz} \\& \quad{} -2(\zeta_{uz}u_{x}+ \zeta_{xu}u_{z}+ \zeta_{uu}u_{x}u_{z}+ \zeta_{u}u_{xz})u_{zy}-(2 \tau_{z}+2\tau_{u}u_{z}+ \tau_{x}+ \tau_{u}u_{x})u_{xzt} \\& \quad{} -2( \tau_{uz}u_{x}+\tau_{xu}u_{z}+ \tau_{uu}u_{x}u_{z}+\tau _{u}u_{xz})u_{zt}- \bigl(\xi _{zz}u_{x}+2\xi_{uz}u_{z}+ \xi_{uu}u^{2}_{z}+\xi_{u}u_{zz} \bigr)u_{xx} \\& \quad{} - \bigl(\zeta_{zz}+2\zeta_{uz}u_{y}+ \zeta_{uu}u^{2}_{z}+\zeta _{u}u_{zz} \bigr)u_{xy}- \bigl(\tau _{zz}+2\tau_{uz}u_{y}+ \tau_{uu}u^{2}_{z}+\tau_{u}u_{zz} \bigr)u_{xt} \\& \quad{} - \bigl(\xi_{xzz}+\xi_{uzz}u_{x}+2 \xi_{xzu}u_{z}+2\xi_{uuz}u_{x}u_{z}+2 \xi_{uz}u_{xz}+\xi_{xuu}u_{z}^{2}+ \xi_{uuu}u_{x}u_{z}^{2}+ 2 \xi_{uu}u_{x}u_{xz} \\& \quad{} +\xi_{xu}u_{zz}+ \xi_{xu}u_{x}u_{zz}+ \xi_{u}u_{xzz} \bigr)u_{x}- \bigl( \zeta_{xzz}+\zeta_{uzz}u_{x}+2 \zeta_{xzu}u_{z}+2 \zeta_{uuz}u_{x}u_{z}+2 \zeta_{uz}u_{xz} \\& \quad{} +\zeta_{xuu}u_{z}^{2}+ \zeta _{uuu}u_{x}u_{z}^{2}+2 \zeta_{uu}u_{x}u_{xz}+ \zeta_{xu}u_{zz}+ \zeta_{xu}u_{x}u_{zz}+\zeta_{u}u_{xzz} \bigr)u_{y} \\& \quad{} - \bigl(\tau_{xzz}+\tau_{uzz}u_{x}+2 \tau_{xzu}u_{z}+2\tau_{uuz}u_{x}u_{z}+2 \tau_{uz}u_{xz}+\tau _{xuu}u_{z}^{2} \\& \quad{} + \tau _{uuu}u_{x}u_{z}^{2}+2 \tau_{uu}u_{x}u_{xz}+ \tau_{xu}u_{zz}+ \tau_{xu}u_{x}u_{zz}+\tau_{u}u_{xzz} \bigr)u_{t} ) \\& \quad{} +d \bigl(\eta_{xx}+(2\eta_{xu}- \xi_{xx})u_{x}+(\eta_{u}-2\xi _{x})u_{xx}+( \eta_{uu}-2\xi_{xu})u^{2}_{x}- \xi_{uu}u_{x}^{3}-3\xi_{u}u_{x}u_{xx} \\& \quad{} -2( \zeta_{x}+\zeta_{u}u_{x})u_{xy}-2( \nu_{x}+\nu_{u}u_{x})u_{xz}-2( \tau_{x}+\tau_{u}u_{x})u_{xt} \\& \quad{}- \bigl( \zeta_{xx}+2\zeta_{xu}u_{x}+ \zeta{uu}u_{x}^{2}+ \zeta_{u}u_{xx} \bigr)u_{y} - \bigl(\nu _{xx}+2\nu_{xu}u_{x}+ \nu{uu}u_{x}^{2}+ \nu_{u}u_{xx} \bigr) u_{z} \\& \quad{} - \bigl(\tau_{xx}+2\tau_{xu}u_{x}+ \tau{uu}u_{x}^{2}+\tau_{u}u_{xx} \bigr)u_{t} \bigr)=0, \end{aligned}$$

(3.3)

now, for Eq. (3.3) by making the whole powers of derivatives of u to zero, we get the following system of equations:

$$ \begin{gathered} \left ({ \textstyle\begin{array}{c} \alpha\\ n \end{array}\displaystyle } \right ) \frac{\partial^{\alpha}\eta_{u}}{\partial t^{\alpha}}-\left ({ \textstyle\begin{array}{c} \alpha\\ n+1 \end{array}\displaystyle } \right )D_{t}^{n+1}( \tau)=0,\quad n=1,2,3,\ldots, \\ \frac{a}{2\sqrt{u}}\eta+a \sqrt{u}(\eta_{u}- \xi_{x})+b(3\eta _{xxu}-\xi _{xxx})+c( \eta_{yyu}-\xi_{xyy}+\eta_{zzu}-\xi_{xzz})+d(2 \xi _{xu}-\xi_{xx})=0, \\ a \sqrt{u} \eta_{x}+b \eta_{xxx}+c( \eta_{xyy}+\eta_{xzz})+d \eta _{xx}+\mu=0, \\ \tau_{u}=\xi_{u}=\nu_{u}= \tau_{x}=\zeta_{u}=\nu_{u}=0, \\ \tau_{uu}=\xi_{uu}=\nu_{uu}= \zeta_{uu}=\nu_{uu}=0, \\ D_{t}^{n}(\xi)=0,\quad n=1,2,3,\ldots, \\ \eta_{u}-\alpha D_{t}(\tau)=0, \\ \eta_{u}-2\zeta_{y}-\xi_{x}=0, \\ \eta_{u}-3\xi_{x}=0, \\ \eta_{u}-2\nu_{z}-\xi_{x}=0. \end{gathered} $$

On solving the previous system of equations, we get the following infinitesimals:

$$ \xi=c_{1}x+c_{2},\qquad \tau= \frac{3}{\alpha}c_{1}t+c_{3},\qquad \zeta=c_{1}y+c_{4}, \qquad\nu =c_{1}z+c_{5},\qquad \eta=-4c_{1} u , $$

(3.4)

where $c_{i}$, $i=1,2,3,4,5$ are arbitrary constants. So, the associated vector fields are given by

$$ \begin{gathered}V_{1}=\frac{\partial}{\partial x},\qquad V_{2}= \frac{\partial}{\partial t},\qquad V_{3}=\frac{\partial}{\partial y}, \qquad V_{4}= \frac{\partial}{\partial z}, \\ V_{5}=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z \frac {\partial}{\partial z}+ \frac{3}{\alpha}t\frac{\partial}{\partial t}-4 u \frac{\partial}{\partial u},\end{gathered} $$

(3.5)

and then, all of the infinitesimal generators of Eq. (1.1) can be expressed as follows:

$$ V=c_{1}V_{1}+c_{2}V_{2}+c_{3}V_{3}+c_{4}V_{4}+c_{5}V_{5}. $$

3.2 The similarity reduction for Eq. (1.1)

In order to reduce Eq. (1.1) to become a FODE, we used the infinitesimal generator $V_{5}$ defined in Eq. (3.5) to form the following characteristic equation:

$$ \frac{dx}{ x}=\frac{dy}{ y}=\frac{dz}{z}=\frac{dt}{\frac{3}{\alpha }t}= \frac{du}{-4u}. $$

Solving the above equation yields the following similarity function:

$$ u=t^{-\frac{4\alpha}{3}}f(\xi),\quad \xi=(x+y+z)t^{-\frac{\alpha}{3}}. $$

(3.6)

By means of the similarity transformation $u=t^{-\frac{4\alpha }{3}}f(\xi )$, the time-fractional SZKB Eq. (1.1) can be reduced to a nonlinear FODE with a new independent variable $\xi=(x+y+z)t^{-\frac {\alpha}{3}}$. Consequently, one can get the following theorem.

Theorem 1

The transformation equation (3.6) reduces the time-fractional SZKB Eq. (1.1) to the following nonlinear FODE:

$$ \bigl(P^{1-\frac{7\alpha}{3},\alpha}_{\frac{3}{\alpha}}f \bigr) (\xi)+a \sqrt{f}f_{\xi}+(b+2c)f_{\xi\xi\xi}+d t^{\frac{\alpha}{3}}f_{\xi \xi}=0, $$

(3.7)

with Erdélyi–Kober (EK) fractional differential operator $(P^{\tau,\alpha}_{\beta} f )(\xi)$, which is defined as

$$ \bigl(P^{\tau,\alpha}_{\beta} f \bigr) (\xi)=\prod _{j=0}^{n-1} \biggl(\tau +j- \frac{1}{\beta}\frac{d}{d\tau} \biggr) \bigl(K^{\tau+\alpha ,n-\alpha }_{\beta} f \bigr) (\xi), \quad n= \textstyle\begin{cases} |\alpha|+1,& n\notin N,\\ \alpha, &n\in N, \end{cases} $$

(3.8)

where

$$ \bigl(K^{\tau+\alpha,n-\alpha}_{\beta} f \bigr) (\xi)= \textstyle\begin{cases} \frac{1}{\varGamma(\alpha)}\int_{1}^{\infty}(u-1)^{\alpha-1}u^{-(\tau +\alpha )}f(\xi u^{\frac{1}{\beta}}) \,du, &\alpha>0,\\ f(\xi), &\alpha=0, \end{cases} $$

(3.9)

is the EK fractional integral operator.

Proof

Let $n-1<\alpha<1$, $n=1,2,3,\ldots$ , we obtain the similarity transformation based on Riemann–Liouville fractional derivatives as follows:

$$ \frac{\partial^{\alpha}u}{\partial t^{\alpha}}=\frac{\partial^{n} }{\partial t^{n}} \biggl[\frac{1}{\varGamma(n-\alpha)} \int_{1}^{t} (t-s)^{n-\alpha -1}s^{\frac {-4}{3}\alpha}f \bigl((x+y+z) s^{-\frac{\alpha}{3}} \bigr) \,ds \biggr]. $$

(3.10)

Let $v=\frac{t}{s}$, one can get $ds=-\frac{t}{v^{2}}$, thus Eq. (3.10) can be written as

$$ \frac{\partial^{\alpha}u}{\partial t^{\alpha}}=\frac{\partial^{n} }{\partial t^{n}} \biggl[\frac{t^{n-\frac{7}{3}\alpha}}{\varGamma(n-\alpha)} \int _{1}^{\infty}(v-1)^{n-\alpha-1}v^{-(n+1-\frac{7}{3}\alpha)}f \bigl(\xi v^{\frac{\alpha }{3}} \bigr) \,dv \biggr], $$

(3.11)

by using the definition of EK fractional differential operator, then Eq. (3.11) becomes

$$ \frac{\partial^{\alpha}u}{\partial t^{\alpha}}=\frac{\partial^{n} }{\partial t^{n}} \bigl[t^{n-\frac{7}{3}\alpha} \bigl(K^{1-\frac{\alpha }{3},n-\alpha }_{\frac{3}{\alpha}} f \bigr) (\xi) \bigr], $$

(3.12)

and now it is appropriate to simplify the right hand side of Eq. (3.11). Consider $\xi=\chi t^{-\frac{\alpha}{3}}$, $\varphi\in (0,\infty )$ we get

$$ t\frac{\partial}{\partial t }\varphi(\xi)=t \chi \biggl(-\frac{\alpha }{3} \biggr)t^{-\frac{\alpha}{3}-1}\varphi^{\prime}(\xi)=-\frac{\alpha }{3}\xi \frac{d}{d \xi}\varphi(\xi). $$

From this,

$$ \begin{aligned} \frac{\partial^{n} }{\partial t^{n}} \bigl[t^{n-\frac{7}{3}\alpha} \bigl(K^{1-\frac{\alpha}{3},n-\alpha}_{\frac{3}{\alpha}} f \bigr) (\xi ) \bigr]&=\frac{\partial^{n-1} }{\partial t^{n-1}} \biggl[\frac{\partial }{\partial t} \bigl(t^{n-\frac{7}{3}\alpha} \bigl(K^{1-\frac{\alpha }{3},n-\alpha}_{\frac{3}{\alpha}} f \bigr) (\xi) \bigr) \biggr] \\ &=\frac{\partial^{n-1} }{\partial t^{n-1}} \biggl[t^{n-\frac {7}{3}\alpha -1} \biggl(n- \frac{7}{3}\alpha-\frac{\alpha}{3}\xi\frac{\partial }{\partial\xi} \biggr) \bigl(K^{1-\frac{\alpha}{3},n-\alpha}_{\frac {3}{\alpha}} f \bigr) (\xi) \biggr],\end{aligned} $$

and thus repeating we have

$$ \begin{aligned} \frac{\partial^{n} }{\partial t^{n}} \bigl[t^{n-\frac{7}{3}\alpha} \bigl(K^{1-\frac{\alpha}{3},n-\alpha}_{\frac{3}{\alpha}} f \bigr) (\xi ) \bigr]&=\frac{\partial^{n-1} }{\partial t^{n-1}} \biggl[\frac{\partial }{\partial t} \bigl(t^{n-\frac{7}{3}\alpha} \bigl(K^{1-\frac{\alpha }{3},n-\alpha}_{\frac{3}{\alpha}} f \bigr) (\xi) \bigr) \biggr] \\ &=\frac{\partial^{n-1} }{\partial t^{n-1}} \biggl[t^{n-\frac {7}{3}\alpha -1} \biggl(n- \frac{7}{3}\alpha-\frac{\alpha}{3}\xi\frac{d}{d \xi } \biggr) \bigl(K^{1-\frac{\alpha}{3},n-\alpha}_{\frac{3}{\alpha}} f \bigr) (\xi ) \biggr] \\ &{}\vdots \\ &=t^{-\frac{7}{3}\alpha}\prod_{j=0}^{n-1} \biggl[ \biggl(1-\frac {7}{3}\alpha +j -\frac{\alpha}{3}\xi \frac{d}{d \varsigma } \biggr) \bigl(K^{1-\frac {\alpha }{3},n-\alpha}_{\frac{3}{\alpha}} f \bigr) (\xi) \biggr].\end{aligned} $$

On using the definition of the EK fractional differential operator Eq. (3.9), we get

$$ \frac{\partial^{n} }{\partial t^{n}} \bigl[t^{n-\frac{7}{3}\alpha} \bigl(K^{1-\frac{\alpha}{3},n-\alpha}_{\frac{3}{\alpha}} f \bigr) (\xi ) \bigr]=t^{-\frac{7}{3}\alpha} \bigl(P^{1-\frac{7}{3}\alpha,\alpha }_{\frac {3}{\alpha}}f \bigr) (\xi). $$

(3.13)

By substituting Eq. (3.13) into Eq. (3.12), we have

$$ \frac{\partial^{\alpha}u }{\partial t^{\alpha}}=t^{-\frac{7}{3}\alpha } \bigl(P^{1-\frac{7}{3}\alpha,\alpha}_{\frac{3}{\alpha}}f \bigr) (\xi). $$

(3.14)

Thus, Eq. (3.13) can be reduced into a fractional ordinary differential equation

$$ \bigl(P^{1-\frac{7}{3}\alpha,\alpha}_{\frac{3}{\alpha}}f \bigr) (\xi)+a \sqrt{f}f_{\xi}+(b+2c)f_{\xi\xi\xi}+d t^{\frac{\alpha}{3}}f_{\xi \xi}=0. $$

(3.15)

The proof of the theorem is completed. □

4 Conservation laws for Eq. (1.1)

In this section, the conservation laws of the time-fractional SZKB equation (1.1) are derived, based on the formal lagrangian and Lie point symmetries as described in the following explanation:

Consider a vector $C=(C^{t},C^{x},C^{y},C^{z})$ admitting the following conservation equation:

$$ \bigl[D_{t} \bigl(C^{t} \bigr)+D_{x} \bigl(C^{x} \bigr)+D_{y} \bigl(C^{y} \bigr)+D_{z} \bigl(C^{z} \bigr) \bigr]_{\text{Eq. {(1.1)}}}=0, $$

(4.1)

where $C^{t}=C^{t}(x,y,z,t,u,\ldots)$, $C^{x}=C^{x}(x,y,z,t,u,\ldots)$, $C^{y}=C^{y}(x,y,z,t,u,\ldots)$, and $C^{z}=C^{z}(x,y,z,t,u,\ldots)$ are called the conserved vectors for Eq. (1.1). According to the new conservation theorem due to Ibragimov [47], the formal Lagrangian for Eq. (1.1) can be given by

$$ L=\omega(x,y,z,t) \biggl[\frac{\partial^{\alpha}u}{\partial t^{\alpha}}+a \sqrt{u} u_{x}+b u_{xxx}+c(u_{xyy}+u_{xzz})+d u_{xx} \biggr]=0, $$

(4.2)

here $\omega(x,y,z,t)$ is a new dependent variable. Depending on the definition of the Lagrangian, we get an action integral as follows:

$$ \int_{0}^{t} \int_{\varOmega_{1}} \int_{\varOmega_{2}} \int_{\varOmega _{3}}L \bigl(x,y,z,t,u,\omega,D_{t}^{\alpha},u_{x},u_{xxx},u_{xyy},u_{xzz},u_{xx} \bigr)\, dx \,dy \,dz \,dt. $$

The Euler–Lagrange operator is defined as

$$ \frac{\delta}{\delta u}=\frac{\partial}{\partial u}+ \bigl(D_{t}^{\alpha}\bigr)^{*}\frac{\partial}{\partial D_{t}^{\alpha}u}-D_{x}\frac{\partial }{\partial u_{x}}+D_{x}^{2} \frac{\partial}{\partial u_{xx}}-D_{x}^{3}\frac{\partial }{\partial u_{xxx}}-D_{x}D_{y}^{2} \frac{\partial}{\partial u_{xyy}}-D_{x}D_{z}^{2} \frac{\partial}{\partial u_{xzz}}, $$

where $(D_{t}^{\alpha})^{*}$ denotes to the adjoint operator of $D_{t}^{\alpha}$, and the adjoint equation to the nonlinear by means of the Euler–Lagrange equation given by

$$ \frac{\delta L }{\delta u}=0. $$

The adjoint operator $(D_{t}^{\alpha})^{*}$ for R-L is defined by

$$ \bigl(D_{t}^{\alpha}\bigr)^{*}=(-1)^{n}I_{T}^{n-\alpha} \bigl(D_{t}^{n} \bigr)\equiv{}_{t}^{C}D_{T}^{\alpha}, $$

where $I_{T}^{n-\alpha}$ is the right-sided operator of fractional integration of order $n-\alpha$ that is defined by

$$ I_{T}^{n-\alpha}f(t,x)=\frac{1}{\varGamma(n-\alpha)} \int_{t}^{T} (\tau -t)^{n-\alpha-1}f(\tau,x) \,d \tau. $$

Considering the case of one dependent variable $u(x,y,z,t)$ with four independent variables x, y, z, t, we get

$$ \bar{X}+D_{t}(\tau)I+D_{x}(\xi)I+D_{y}( \zeta)I+D_{z}(\nu)I=W\frac{\delta }{\delta u}+D_{t} \bigl(C^{t} \bigr)+D_{x} \bigl(C^{x} \bigr)+D_{y} \bigl(C^{y} \bigr)+D_{z} \bigl(C^{z} \bigr), $$

where X̄ is defined by

$$ \begin{aligned} \bar{X}={}&\tau\frac{\partial}{\partial t}+\xi\frac{\partial }{\partial x}+\zeta \frac{\partial}{\partial y}+\nu \frac{\partial}{\partial z}+\eta\frac{\partial}{\partial u}+ \eta^{0}_{\alpha}\frac{\partial }{\partial D^{\alpha}_{t} u} \\ &+\eta^{x}\frac{\partial}{\partial u_{x}}+\eta ^{xx} \frac{\partial}{\partial u_{xx}}+ \eta^{xxx}\frac{\partial }{\partial u_{xxx}}+\eta^{xyy} \frac{\partial}{\partial u_{xyy}}+\eta ^{xzz}\frac{\partial}{\partial u_{xzz}},\end{aligned} $$

and the Lie characteristic function is defined by

$$ \begin{gathered}W=\eta-\tau u_{t}-\xi u_{x}- \zeta u_{y}-\nu u_{z},\quad\text{and then}\\ W_{1}=-u_{x},\qquad W_{2}=-u_{t}, \qquad W_{3}=-u_{y},\qquad W_{t}=-u_{z}. \end{gathered} $$

(4.3)

For the R-L time-fractional derivative, the density component $C^{t}$ of conservation law is defined by

$$ C^{t}=\tau L+\sum_{k=0}^{n-1}(-1)^{k} _{0}D_{t}^{\alpha -1-k}(W_{m})D^{k}_{t} \frac {\partial L}{\partial_{0}D^{\alpha}_{t} u}-(-1)^{n}J \biggl(W_{m},D^{n}_{t} \frac {\partial L}{\partial_{0}D^{\alpha}_{t} u} \biggr), $$

(4.4)

where the operator $J(\cdot)$ is defined by

$$ J(f,g)=\frac{1}{\varGamma(n-\alpha)} \int_{0}^{t} \int_{t}^{T} \frac{f(\tau ,x,y,z)g(\mu,x,y,z)}{(\mu-\tau)^{\alpha+1-n}} \,d\mu\,d\tau, $$

and the other (flux) components are defined as

$$ \begin{aligned}[b] C^{i}={}&\xi^{i} L+W_{m} \biggl[ \frac{\partial L}{\partial u_{i}^{m}}-D_{j} \biggl( \frac {\partial L}{\partial u_{ij}^{m}} \biggr)+D_{j}D_{k} \biggl( \frac{\partial L}{\partial u_{ijk}^{m}}-\cdots \biggr) \biggr] \\ &+D_{j}(W_{m}) \biggl[\frac {\partial L}{\partial u_{ij}^{m}}-D_{k} \biggl(\frac{\partial L}{\partial u_{ijk}^{m}} \biggr)+ \cdots \biggr] +D_{j}D_{k}(W_{m}) \biggl( \frac{\partial L}{\partial u_{ijk}^{m}}-\cdots \biggr)+\cdots,\end{aligned} $$

(4.5)

where $\xi^{1}=\xi$, $\xi^{2}=\zeta$, $\xi^{3}=\nu$ and $m=1,2,\ldots,5$.

Now by using Eq. (4.3) with the help of Eqs. (4.4) and (4.5), we obtain the components of the conservation laws for the time-fractional SZKB equation as follows.

Case 1: $W_{1}=-u_{x}$ where $\xi^{x}=1$, $\xi^{t}=0$, $\xi ^{y}=0$, $\xi^{z}=0$ and $\eta=0$ we get the following conserved vectors:

$$ \begin{gathered} C^{t}_{1}=\omega_{0}D_{t}^{\alpha-1}(-u_{x}) \frac{\partial L}{\partial _{0}D^{\alpha}_{t} u}-J \biggl(-u_{x},D^{n}_{t} \frac{\partial L}{\partial _{0}D^{\alpha}_{t} u} \biggr)=-\omega D_{t}^{\alpha}(u_{x})-u_{x}D_{t}^{\alpha}( \omega), \\ C^{x}_{1}=\omega \biggl[D_{t}^{\alpha}u+a \sqrt{u} u_{x}+\frac {c}{3}(u_{yxy}+u_{yyx}+u_{zxz}+u_{zzx}) \biggr] \\ \phantom{C^{x}_{1}=}{}-u_{x} \biggl[-d \omega_{x}+b \omega_{xx}+\frac{c}{3}(\omega_{yy}+ \omega_{zz}) \biggr]+b \omega_{xx}\omega _{x}+ \frac{c}{3}(u_{xy}\omega_{y}+u_{xz} \omega_{z}), \\ C^{y}_{1}=-\frac{c}{3}u_{x}[ \omega_{xy}+\omega_{yx}]+\frac{c}{3} (u_{xx}\omega _{y}+u_{xy}\omega_{x})- \frac{c}{3}\omega(u_{xyx}+u_{yxx}), \\ C^{z}_{1}=-\frac{c}{3}u_{x}[ \omega_{xz}+\omega_{zx}]+\frac{c}{3} (u_{xx}\omega _{z}+u_{xz}\omega_{x})- \frac{c}{3}\omega(u_{xzx}+u_{zxx}). \end{gathered}$$

Case 2: $W_{2}=-u_{t}$ where $\xi^{x}=0$, $\xi^{t}=1$, $\xi ^{y}=0$, $\xi^{z}=0$ and $\eta=0$ we acquire the following conserved vectors:

$$ \begin{gathered} C^{t}_{2}=\omega L\omega_{0}D_{t}^{\alpha-1}(-u_{t}) \frac{\partial L}{\partial _{0}D^{\alpha}_{t} u}-J \biggl(-u_{t},D^{n}_{t} \frac{\partial L}{\partial _{0}D^{\alpha}_{t} u} \biggr) \\ \phantom{C^{t}_{2}}=\omega \biggl[D_{t}^{\alpha}u+a \sqrt{u} u_{x}+\frac {c}{3}(u_{yxy}+u_{yyx}+u_{zxz}+u_{zzx})+du_{xx} \biggr]-\omega D_{t}^{\alpha}(u_{t})-u_{t}D_{t}^{\alpha}( \omega), \\ C^{x}_{2}=-u_{t} \biggl[a\omega \sqrt{u}-d \omega_{x}+b \omega_{xx}+\frac {c}{3}( \omega_{yy}+\omega_{zz}) \biggr] \\ \phantom{C^{x}_{2}=}{}- u_{xt}[d \omega_{x}-b\omega _{x}]+\frac {c}{3}(u_{yt} \omega_{y}+u_{zt} \omega_{z})-b\omega u_{xxt}-\frac{c}{3} \omega(u_{yyt}+u_{zzt}), \\ C^{y}_{2}=-\frac{c}{3}u_{t}[ \omega_{xy}+\omega_{yx}]+\frac {c}{3}(u_{xt} \omega _{y}+u_{yt}\omega_{x})-\frac{c}{3} \omega(u_{xyt}+u_{yxt}), \\ C^{z}_{2}=-\frac{c}{3}u_{t}[ \omega_{xz}+\omega_{zx}]+\frac {c}{3}(u_{xt} \omega _{z}+u_{zt}\omega_{x})-\frac{c}{3} \omega(u_{xzt}+u_{zxt}).\end{gathered} $$

Case 3: $W_{3}=-u_{y}$ where $\xi^{x}=0$, $\xi^{t}=0$, $\xi ^{y}=1$, $\xi^{z}=0$ and $\eta=0$ we obtain the following conserved vectors:

$$\begin{aligned}& C^{t}_{3}=\omega_{0}D_{t}^{\alpha-1}(-u_{y}) \frac{\partial L}{\partial _{0}D^{\alpha}_{t} u}-J \biggl(-u_{y},D^{n}_{t} \frac{\partial L}{\partial _{0}D^{\alpha}_{t} u} \biggr)=-\omega D_{t}^{\alpha}(u_{y})-u_{y}D_{t}^{\alpha}( \omega), \\& C^{x}_{3}=-u_{y} \biggl[a\omega \sqrt{u}-d \omega_{x}+b \omega_{xx}+\frac {c}{3}( \omega_{yy}+\omega_{zz} )\biggr] \\& \phantom{C^{x}_{3}=}{}- u_{xy}[d \omega_{x}-b\omega _{x}]+\frac {c}{3}(u_{yy} \omega_{y}+u_{yz} \omega_{z})-b\omega u_{xxy}-\frac{c}{3} \omega(u_{yyy}+u_{zzy}), \\& C^{y}_{3}=\omega \biggl[D_{t}^{\alpha}u+a \sqrt{u} u_{x}+\frac {c}{3}(u_{yxy}+u_{yyx}+u_{zxz}+u_{zzx})+du_{xx} \biggr] \\& \phantom{C^{y}_{3}=}{}-\frac {c}{3}u_{y}[ \omega_{xy}+ \omega_{yx}]+\frac{c}{3} (u_{xy}\omega_{y}+u_{yy} \omega_{x})- \frac{c}{3}\omega (u_{xyy}+u_{yxy}), \\& C^{z}_{3}=-\frac{c}{3}u_{y}[ \omega_{xz}+\omega_{zx}]+\frac {c}{3}(u_{xy} \omega _{z}+u_{yz}\omega_{x})-\frac{c}{3} \omega(u_{xzy}+u_{zxy}). \end{aligned}$$

Case 4: $W_{4}=-u_{z}$ where $\xi^{x}=0$, $\xi^{t}=0$, $\xi ^{y}=0$, $\xi^{z}=0$ and $\eta=0$ we obtain the following conserved vectors:

$$ \begin{gathered} C^{t}_{4}=\omega_{0}D_{t}^{\alpha-1}(-u_{z}) \frac{\partial L}{\partial _{0}D^{\alpha}_{t} u}-J \biggl(-u_{z},D^{n}_{t} \frac{\partial L}{\partial _{0}D^{\alpha}_{t} u} \biggr)=-\omega D_{t}^{\alpha}(u_{z})-u_{z}D_{t}^{\alpha}( \omega), \\ C^{x}_{4}=-u_{z} \biggl[a\omega \sqrt{u}-d \omega_{x}+b \omega_{xx}+\frac {c}{3}( \omega_{yy}+\omega_{zz} )\biggr]- u_{xy}[d \omega_{x}-b\omega _{x}]\\ \phantom{C^{x}_{4}=}{}+\frac {c}{3}(u_{yy} \omega_{y}+u_{yz}\omega_{z})- b\omega u_{xxy}-\frac{c}{3}\omega(u_{yyy}+u_{zzy}), \\ C^{y}_{4}=\omega \biggl[D_{t}^{\alpha}u+a \sqrt{u} u_{x}+\frac {c}{3}(u_{yxy}+u_{yyx}+u_{zxz}+u_{zzx})+du_{xx} \biggr] \\ \phantom{C^{y}_{4}=}{}-\frac {c}{3}u_{y}[ \omega_{xy}+ \omega_{yx}]+\frac{c}{3} (u_{xy}\omega_{y}+u_{yy} \omega_{x})- \frac{c}{3}\omega (u_{xyy}+u_{yxy}), \\ C^{z}_{4}=-\frac{c}{3}u_{y}[ \omega_{xz}+\omega_{zx}]+\frac {c}{3}(u_{xy} \omega _{z}+u_{yz}\omega_{x})-\frac{c}{3} \omega(u_{xzy}+u_{zxy}).\end{gathered} $$

Remark

We have verified that all the solutions satisfy the original equation.

5 Explicit power series and convergence analysis for Eq. (1.1)

In this section, we derived the analytic solution for Eq. (1.1) via the power series method [57, 58] and proved the convergence of the power series solution as demonstrated in the next two subsections.

5.1 Power series and analytical solutions for Eq. (1.1)

Considering Eq. (1.1) and by using $u(x,y,z,t)=\psi ^{2}(x,y,z,t)$, Eq. (1.1) becomes

$$ \frac{\partial^{\alpha}\psi^{2}}{\partial t^{\alpha}}+a \psi \bigl(\psi^{2} \bigr)_{x}+b \bigl(\psi^{2} \bigr)_{xxx}+c \bigl( \bigl( \psi^{2} \bigr)_{xyy}+ \bigl(\psi^{2} \bigr)_{xzz} \bigr)+d \bigl(\psi^{2} \bigr)_{xx}=0. $$

(5.1)

Let us introduce the important transformation

$$ \psi(x,y,z,t)=\psi(\xi), \quad\xi=px+qy+rz-\frac{k t^{\alpha}}{\varGamma (1+\alpha)}, $$

(5.2)

where p, q, r and k are constants to be determined later. The substitution of (5.2) into (5.1) leads to the following nonlinear ODE:

$$ -\frac{k}{2} \bigl(\psi^{2} \bigr)^{\prime}+ \frac{a}{3} \bigl(\psi^{3} \bigr)^{\prime}+ \frac {1}{2}p \bigl(bp^{2}+c \bigl(q^{2}+r^{2} \bigr) \bigr) \bigl(\psi^{2} \bigr)^{\prime\prime\prime}+dp^{2} \bigl( \psi ^{2} \bigr)^{\prime\prime}=0. $$

(5.3)

By integrating Eq. (5.3) with respect to ξ, we get

$$ -\frac{k}{2} \bigl(\psi^{2} \bigr)+ \frac{a}{3} \bigl(\psi^{3} \bigr)+\frac {1}{2}p \bigl(bp^{2}+c \bigl(q^{2}+r^{2} \bigr) \bigr) \bigl( \psi^{2} \bigr)^{\prime\prime}+dp^{2} \bigl(\psi ^{2} \bigr)^{\prime }+\lambda=0, $$

(5.4)

where λ is the constant of integration. Let us assume that the solution of Eq. (5.4) has the following form:

$$ \psi(\xi)=\sum_{n=0}^{\infty}a_{n} \xi^{n}, $$

(5.5)

where $a_{n}$ are constants to be determined later. On substituting (5.5) into (5.4), we have the following relation:

$$ \begin{aligned}[b] &{-}\frac{k}{2}\sum _{n=0}^{\infty} \sum_{m=0}^{n} (a_{m}a_{n-m} ) \xi^{n}+\frac{a}{3}\sum _{n=0}^{\infty} \sum _{m=0}^{n}\sum_{l=0}^{n-m} (a_{l}a_{m-l}a_{n-m} )\xi^{n}+p \bigl(bp^{2}+c \bigl(q^{2}+r^{2} \bigr) \bigr) \\ &\quad\times \Biggl(\sum_{n=0}^{\infty} \sum_{m=0}^{n} (n-m+1) (n-m+2)a_{m}a_{n-m+2} \\ &\quad+ \sum_{n=0}^{\infty}\sum _{m=0}^{n}(n-m+1) (m+1)a_{m+1}a_{n-m+1} \Biggr)\xi^{n} \\ &\quad+2dp^{2}\sum_{n=0}^{\infty} \sum_{m=0}^{n} \bigl((n-m+1)a_{m}a_{n-m+1} \bigr)\xi^{n} +\lambda=0.\end{aligned} $$

(5.6)

From Eq. (5.6), by comparing the coefficients for $n =0$ one can get

$$ a_{2}=-\frac{2aa_{0}^{3}-3k a_{0}^{2}+12dp^{2}a_{0}a_{1}+6 (\lambda +bp^{2}+c(q^{2}+r^{2}) )a_{1}^{2}}{12p (bp^{2}+c(q^{2}+r^{2}) )a_{0}}, $$

(5.7)

where $a_{0}$ and $a_{1}$ are arbitrary constants where p and $a_{0}\neq 0$. Generally, for $n \geq1$ we can get

$$ \begin{aligned}[b]a_{n+2}={}&\frac{1}{6\rho(n+1)(n+2)a_{0}} \Biggl(3k\sum_{m=0}^{n} (a_{m}a_{n-m} ) -2a\sum_{m=0}^{n} \sum_{l=0}^{n-m} (a_{l}a_{m-l}a_{n-m} )-6\rho \\ & \times \Biggl(\sum_{m=0}^{n} (n-m+1) (n-m+2)a_{m}a_{n-m+2}+\sum _{m=0}^{n} (n+1) (n-m+1)a_{m+1}a_{n-m+1} \Biggr) \hspace{-24pt}\\ &-12dp^{2}\sum_{m=0}^{n} \bigl((n-m+1)a_{m}a_{n-m+1} \bigr) -6\lambda \Biggr), \end{aligned} $$

(5.8)

where $\rho=p(bp^{2}+c(q^{2}+r^{2}))$. Now, from (5.8) and (5.7) we can get all the coefficients $a_{n+2}$, $n\geq1$ of the power series (5.5) in which $a_{0}$, p, q, r, and k are the arbitrary constants. Thus, the power series solution of Eq. (5.4) could be written as follows:

$$\begin{aligned} u(\xi)={}& \Bigl(a_{0}+a_{1} \xi+a_{2} \xi^{2}+{\sum}_{n=1}^{\infty}a_{n+2} \xi ^{n+2} \Bigr)^{2} \\ ={}& (a_{0}+a_{1}\xi+ \biggl(-\frac{6\lambda-3k a_{0}^{2}+2aa_{0}^{3}+6p (2dp+bp^{2}+c(q^{2}+r^{2}) )a_{0}a_{1}}{12p (bp^{2}+c(q^{2}+r^{2}) )a_{0}} \biggr)\xi^{2} \\ &+\sum_{n=1}^{\infty} \Biggl( \frac{1}{6\rho(n+1)(n+2)a_{0}} \Biggl(3k\sum_{m=0}^{n} (a_{n}a_{n-m} ) -2a\sum_{m=0}^{n} \sum_{l=0}^{n-m} (a_{l}a_{m-l}a_{n-m} )-6\rho \\ &\times \Biggl(\sum_{m=0}^{n} (n-m+1) (n-m+2)a_{m}a_{n-m+2}+\sum _{m=0}^{n} (n+1) (n-m+1)a_{m+1}a_{n-m+1} \Biggr) \\ &-12dp^{2}\sum_{m=0}^{n} \bigl((n-m+1)a_{m}a_{n-m+1} \bigr) -6\lambda \Biggr) \xi^{n+2} \Biggr)^{2}. \end{aligned}$$

(5.9)

Thus, we obtain the exact power series solution for Eq. (3.15) as follows:

$$ \begin{aligned}[b] u(x,y,z,t)= {}&\Biggl(a_{0}+a_{1} \biggl(px+qy+rz-\frac{k t^{\alpha}}{\varGamma (1+\alpha } \biggr)\\ &+ \biggl(-\frac{6\lambda-3k a_{0}^{2}+2aa_{0}^{3}+6p (2dp+bp^{2}+c(q^{2}+r^{2}) )a_{0}a_{1}}{12p (bp^{2}+c(q^{2}+r^{2}) )a_{0}} \biggr) \\ &\times \biggl(px+qy+rz-\frac{k t^{\alpha}}{\varGamma(1+\alpha} \biggr)^{2}\\ &+ \sum _{n=1}^{\infty} \Biggl(\frac{1}{6\rho(n+1)(n+2)a_{0}} \Biggl(3k \sum_{m=0}^{n} (a_{m}a_{n-m} ) -2a\sum_{m=0}^{n} \sum_{l=0}^{n-m} (a_{l}a_{m-l}a_{n-m} ) \hspace{-24pt}\\ &-6\rho\Biggl(\sum_{m=0}^{n} (n-m+1) (n-m+2)a_{m}a_{n-m+2}\\ &+\sum_{m=0}^{n} (n+1) (n-m+1)a_{m+1}a_{n-m+1} \Biggr) \\ &-12dp^{2}\sum_{m=0}^{n} \bigl((n-m+1)a_{m}a_{n-m+1} \bigr) -6\lambda \Biggr) \Biggr)\\ &\times \biggl(px+qy+rz-\frac{k t^{\alpha}}{\varGamma(1+\alpha} \biggr)^{n+2} \Biggr)^{2}.\end{aligned} $$

(5.10)

We represented the solution defined by Eq. (5.10) using the 2D plot, see Fig. 1b, and the 3D plots, see Fig. 1a and Fig. 2. The result in Eq. (5.10) shows that there is an analytical power series solution for Eq. (5.4). It is important to find new solutions, because either new exact solutions or numerical approximate solutions and analytical solutions may deepen our understanding of the physical phenomena.

Figure 1

(a) The effect of α on the 3D solitary wave solution of equation (5.10) with a fixed time t and the parameters $k=0.3$, $p=0.2$, $q=0.7$, $r=0.3$, $a_{0}=6$, $a_{1}=1.7$. (b) The corresponding 2D plot of (a), which shows that the envelope of the wave decays far away from the peak quickly when the value of α increases meaning that the solitary wave solution behaves as a dynamical system

Figure 2

The effect of α on the 3D solitary wave solution of equation (5.10) at different values of $\alpha=0.1,0.3$ with the suitable parameters $k=0.3$, $p=0.2$, $q=0.7$, $r=0.3$, $a_{0}=6$, $a_{1}=1.7$

5.2 Convergence analysis

In this subsection, to complete the analysis of Eq. (1.1), we study the convergence of the power series solution Eq. (5.5) with the coefficients denoted by Eq. (5.7) and (5.8), where Eq. (5.8) can be enlarged to the following form:

$$ \begin{aligned}[b] \vert a_{n+2} \vert \leq {}&M \Biggl[\sum _{m=0}^{n} \vert a_{m} \vert \vert a_{n-m} \vert -\sum_{m=0}^{n} \sum_{l=0}^{n-m} \vert a_{l} \vert \vert a_{m-l} \vert \vert a_{n-m} \vert -\sum _{m=0}^{n} \vert a_{m} \vert \vert a_{n-m+2} \vert \\ &-\sum_{m=0}^{n} \vert a_{m+1} \vert \vert a_{n-m+1} \vert -\sum _{m=0}^{n} \vert a_{m} \vert \vert a_{n-m+1} \vert \Biggr],\end{aligned} $$

(5.11)

where $M=\max\{\frac{3k}{\rho}, \frac{2a}{\rho}, 12dp^{2}\}$, $\rho =bp^{2}+c(q^{2}+r^{2})$. Now, consider another power series,

$$ S(\xi)=\sum_{n=0}^{\infty}s_{n} \xi^{n}, $$

(5.12)

with $s_{i}= \vert a_{i} \vert$, $i=0,1$. Then we can get

$$ \begin{aligned}[b] s_{n+2}={}& M \Biggl[\sum_{m=0}^{n} \vert s_{m} \vert \vert s_{n-m} \vert \\& - \sum _{m=0}^{n}\sum_{l=0}^{n-m} \vert s_{l} \vert \vert s_{m-l} \vert \vert s_{n-m} \vert - \vert s_{m} \vert \vert s_{n-m+2} \vert - \vert s_{m+1} \vert \vert s_{n-m+1} \vert - \vert s_{m} \vert \vert s_{n-m+1} \vert \Biggr],\end{aligned} $$

(5.13)

where $n =0,1,2,\ldots$ thus, it is easily noted that

$$ \vert a_{n+2} \vert \leq s_{n+2} \quad\text{this leads to } \vert a_{n} \vert \leq s_{n}, $$

depending on this result, we could say that the series defined by Eq. (5.12) is a majorant series of Eq. (5.5). Next, we have shown that the series $S=S(\xi)$ has positive radius of convergence. Then, we write this series in the following form:

$$\begin{aligned} S(\xi)={}&s_{0}+s_{1}\xi+\sum_{n=2}^{\infty}s_{n} \xi^{n} \\ ={}&s_{0}+s_{1}\xi+M \Biggl[\sum_{m=0}^{n} \vert s_{m} \vert \vert s_{n-m} \vert \\ &- \sum _{m=0}^{n}\sum_{l=0}^{n-m} \vert s_{l} \vert \vert s_{m-l} \vert \vert s_{n-m} \vert - \vert s_{m} \vert \vert s_{n-m+2} \vert - \vert s_{m+1} \vert \vert s_{n-m+1} \vert - \vert s_{m} \vert \vert s_{n-m+1} \vert \Biggr] \xi^{n+2} \\ ={}&s_{0}+s_{1}\xi+M \bigl[S^{2}(\xi) \xi^{2}-S^{3}(\xi)\xi ^{2}-2(S-s_{0}) (S-s_{0}) \bigr]. \end{aligned}$$

Consider an implicit functional system with respect to the independent variable ξ as follows:

$$ \vartheta(\xi,S)=S-s_{0}-s_{1}\xi-M \bigl[S^{2}(\xi)\xi^{2}-S^{3}(\xi)\xi ^{2}-2(S-s_{0}) (S-s_{0}) \bigr], $$

(5.14)

since ϑ is an analytic in a neighborhood of $(o, s_{0})$, where $\vartheta(o, s_{0}) =0$ and, $\frac{\partial\vartheta}{\partial S}(o, s_{0})\neq0$.

Theorem

([59])

Let f be a $\wp^{\prime}$-mapping of an open set $E\subset s^{n+m}$ into $S^{n}$, such that $f (a, b) = 0$ for some point $(a, b)\in E$. Assume that $A = f^{\prime}(a, b)$ and $A_{x}$ is invertible. Then the following properties hold in the open sets $U\subset s^{n+m}$ and $W\subset S^{m}$ with $(a, b)\in U$ and $b\in W$.

• (i) For each $y\in W$, there exists a unique x such that $(x,y)\in U$ and $f(x,y)=0$.

• (ii) If x is defined to be $g(y)$, then

  $$\begin{gathered} g(b)=a, \\ f \bigl(g(y),y \bigr)=0 \quad(y\in W), \\ g^{\prime}(b)=-(A_{x})^{-1}A_{y},\end{gathered} $$

  where g is a $\wp^{\prime}$-mapping of W into $S^{n}$.

• (iii) The function g is implicitly defined by (ii).

One can see that $S=S(\xi)$ is analytical in a neighborhood of the point $(o, s_{0})$ and has positive radius. It shows that the power series Eq. (5.5) is convergent in a neighborhood of the point $(o, s_{0})$.

6 The modified trial equation method for Eq. (1.1)

In this section, we gave a brief description of the modified trial method that was proposed by Liu [60, 61]; some authors [62,63,64] improved this method. Now, let us consider the time FPDEs, say in the four variables x, y, z and t. In the following we give the main steps of the modified trial method and we obtain some traveling wave solutions of Eq. (1.1):

Step 1: Consider the time FPDE defined by Eq. (2.3) and take the wave transformation as Eq. (5.2). Under this transformation, we were permitted to reduce Eq. (2.3) to ODE.

Step 2: Take the trial equation as follows:

$$ \begin{gathered}\psi^{\prime}=\frac{F(\psi)}{G(\psi)}=\frac{\sum_{i=0}^{n}A_{i}\psi ^{i}}{\sum_{j=0}^{l}B_{j}\psi^{j}}= \frac{A_{0}+a_{1}\psi+A_{2}\psi^{2}+\cdots+A_{n}\psi^{n}}{B_{0}+B_{1}\psi+B_{2}\psi ^{2}+\cdots+B_{l}\psi^{l}}, \\ \psi^{\prime\prime}=\frac{F(\psi)(F^{\prime}(\psi)G(\psi)-F(\psi )G^{\prime}(\psi))}{G^{3}(\psi)},\end{gathered} $$

(6.1)

where F and G are polynomials in ψ. On substituting (6.1) into (5.4) yields an equation of $\varOmega(\psi )$ of polynomial in ψ

$$ \varOmega(\psi)=p_{s}\psi^{s}+\cdots+p_{1} \psi+p_{0}, $$

we found a relation between n and l according to the balance principle and determined some values of them.

Step 3: Setting the coefficients of $\varOmega(\psi)$ to zero yields a system of algebraic equations:

$$ p_{i}=0,\quad i=0,1,\ldots,s. $$

By solving the obtained system of algebraic equations, we can determine the values of $A_{0},\ldots,A_{n}$ and $B_{0},\ldots, B_{l}$.

Step 4: Rewrite Eq. (5.11) by the following integral form:

$$ \xi-\xi_{0}= \int\frac{F(\psi)}{G(\psi)} \,d\psi, $$

(6.2)

where $\xi_{0}$ is the constant of integration. When we classify the roots of $F(\psi)$ using the complete discrimination system, we obtain the analytical solutions of Eq. (1.1).

Now, applying the modified trial equation method for the $(3+1)$-dimensional time-fractional SZKB equation with the wave transformation defined by Eq. (5.2), by employing (6.1), (6.2), and using the balance principle yields $n=l+2$. This resolution procedure was applied and we obtained results as follows.

Case 1: If we take $l=0$, then $n = 2$, and

$$ \begin{gathered}\psi^{\prime}=\frac{A_{0}+A_{1}\psi+A_{2}\psi^{2}}{B_{0}}, \\ \psi^{\prime\prime}=\frac{(A_{1}+2A_{2}\psi)(A_{0}+A_{1}\psi+A_{2}\psi^{2})}{B_{0}^{2}},\end{gathered} $$

(6.3)

thus, we have a system of algebraic equations from the coefficients of polynomial of u. Solving this system, we get

$$ A_{0}=0,\quad\quad A_{2}=-\frac{a B_{0}}{6 d p},\qquad r=\pm\sqrt{- \frac{b p^{2}+c q^{2}}{c}},\quad\quad k=-\frac{4 d p^{2}A_{1}}{B_{0}}. $$

Substituting from the above coefficients into (6.2) and integrating, we get the solutions to (5.4), as follows:

$$ \xi-\xi_{0}= \int\frac{B_{0}}{A_{2}\psi^{2}+a_{1}\psi} \,d\psi, $$

(6.4)

this implies

$$ \psi(\xi)=\frac{3 k B_{0}}{2p(6d P \exp(\pm\frac{k}{4dp^{2}B_{0}}(\xi -\xi _{0}))-a B_{0})}, $$

(6.5)

thus,

$$ u(x,y,z,t)= \biggl(\frac{3 k B_{0}}{2p(6d P \exp(\pm\frac {k}{4dp^{2}B_{0}}(px+qy+rz-\frac{k t^{\alpha}}{\varGamma(1+\alpha)}-\xi_{0}))-a B_{0})} \biggr)^{2}. $$

(6.6)

For simplicity, take $\xi_{0}=0$, $A_{1}=A_{2}=1$ then the solution Eq. (6.6) is reduced to the following solution:

$$ u(x,y,z,t)= \biggl(\frac{1}{\exp(\gamma(px+qy+rz-\frac{kt^{\alpha}}{\varGamma (\alpha+1)}))-1} \biggr)^{2}, $$

(6.7)

where $\gamma=\pm\frac{A_{1}}{B_{0}}$. We represent the solution defined by Eq. (6.7) using the 2D plot, see Fig. 3b, and the 3D plots, see Fig. 3a and Fig. 4.

Figure 3

(a) The effect of α on the 3D double-layer solution of equation (6.7) with a fixed time t and the parameters $k=0.3$, $p=0.2$, $q=0.2$. (b) The corresponding 2D plot of (a) with the same parameters

Figure 4

The 3D double-layer solution of equation (6.7) at different values of $\alpha=0.5,0.9$ with the suitable parameters $k=0.3$, $p=0.2$, $q=0.2$ which shows that when the value of α decreases the singularity becomes clearer

Case 2: If we take $l=1$, then $n = 3$, and

$$ \begin{gathered}\psi^{\prime}=\frac{A_{0}+A_{1}\psi+A_{2}\psi^{2}+A_{3}\psi^{3}}{B_{0}+B_{1}\psi}, \\ \psi^{\prime\prime}=\Bigl(\bigl(A_{0}+A_{1}\psi+A_{2}\psi^{2}+A_{3}\psi ^{3}\bigr)\bigl(\bigl(B_{0}+B_{1}\psi \bigr)\bigl(A_{1}+2A_{2}\psi+3A_{3}\psi^{2}\bigr)\bigr)\\ \phantom{\psi^{\prime\prime}=}-B_{1}\bigl(A_{0}+A_{1}\psi+A_{2}\psi^{2}+ A_{3}\psi^{3}\bigr)\Bigr)\\ \phantom{\psi^{\prime\prime}=}/(B_{0}+B_{1}\psi)^{3},\end{gathered} $$

(6.8)

thus, we have a system of algebraic equations from the coefficients of polynomial of u. Solving this system, we get

$$\begin{gathered} A_{0}=0,\qquad A_{2}=\frac{6 d p A_{1}B_{1}-aB_{0}^{2}}{6 d p B_{0}},\qquad A_{3}=- \frac{a B_{1}}{6 d p},\\ k=\frac{4 d p^{2}A_{1}}{B_{0}},\qquad r=\pm\sqrt{-\frac{b p^{2}+c q^{2}}{c}}.\end{gathered} $$

Substituting from the above coefficients into Eq. (6.8) and integrating, we get the solutions to Eq. (5.4), as follows:

$$ \xi-\xi_{0}= \int\frac{B_{0}+B_{1}u}{A_{3}\psi^{3}+A_{2}\psi^{2}+a_{1}\psi} \, d\psi, $$

(6.9)

this implies

$$ \begin{aligned}[b]\xi-\xi_{0}={}&{-}\frac{1}{2A_{1}} \biggl(B_{0} \ln \bigl(A_{1}+\psi(A_{2}+A_{3}\psi ) \bigr)-2B_{0}\ln (\psi)\\ &+\frac{1}{\sigma} \biggl(2(A_{2}B_{0}-2A_{1}B_{1}) tan^{-1} \biggl(\frac {A_{2}+2A_{3}\psi}{\sigma} \biggr) \biggr) \biggr),\end{aligned} $$

(6.10)

where $\sigma=\sqrt{4A_{1}A_{3}-2A_{2}^{2}}$. Substituting the values of $A_{i}$, $i=1,2,3$, thus

$$ \psi(\xi)= \biggl(\frac{3k}{2 a p+\exp(\frac{k(\xi_{0}-\xi)}{4 d p^{2}})} \biggr), $$

(6.11)

where $\xi=px+qy+rz-\frac{kt^{\alpha}}{\varGamma(\alpha+1)}$.

For simplicity, take $\xi_{0}=0$ then the solution Eq. (6.11) is reduced to the following solution:

$$ u(x,y,z,t)= \biggl(\frac{3k}{2 a p+\exp (\frac{-k (px+qy+rz-\frac {kt^{\alpha}}{\varGamma(\alpha+1)} )}{4 d p^{2}} )} \biggr)^{2}. $$

(6.12)

We represent the solution defined by Eq. (6.12) using the 2D plot, see Fig. 5b, and the 3D plots, see Fig. 5a and Fig. 6.

Figure 5

(a) The effect of α on the 3D double-layer solution of equation (6.12) with a fixed time t and the parameters $k=0.3$, $p=0.2$, $q=0.2$. (b) The corresponding 2D plot of (a) which shows that when the value of α increases the interval of the stability of the solution also increases

Figure 6

The 3D double-layer solution of equation (6.12) at different values of $\alpha=0.5,0.9$ with the suitable parameters $k=0.3$, $p=0.2$, $q=0.2$

7 Conclusion

In this paper, we considered a $(3+1)$-dimensional time-fraction Schamel–Zakharov–Kuznetsov–Burgers equation, which described the nonlinear plasma-dust ion acoustic waves (DIAWs) in a magnetized dusty plasma. The Lie point symmetries were applied successfully to the study of the $(3+1)$-dimensional time-fraction Schamel–Zakharov–Kuznetsov–Burgers equation. Based on the Riemann–Liouville derivatives, we deduced the corresponding vector fields, which helped us to construct the symmetry reductions of the time-fractional SZKB equation. Furthermore, we obtained four kinds of conservation laws with independent variables laying the foundation of Lie point symmetries. Also, we constructed a new set of analytical solutions via two powerful methods, which are the explicit power series method and the modified trial equation method. Moreover, to introduce a better understanding of the dynamics of these solutions, we provided their graphic analysis of Eq. (1.1). As shown in Fig. 3, It was noticed that the solution $u(x,y,z,t)$ defined by Eq. (6.7) remained stable for $-4.35< x<2.75$ and has a singularity outside this interval. It can easily be observed in Figs. 4 and 6 that the solutions are in line with the values of α, it means that the singularity becomes clearer when the value of α decreases. Thus, we expect that the obtained results might serve as the explanation of the physical meaning of the time-fractional SZKB equation with more accuracy. However, we can extend the symmetry analysis to the time-space fractional $(3+1)$-dimensional Schamel–Zakharov–Kuznetsov–Burgers equation where that work may be presented later elsewhere.