Exact Solutions for the Generalized Atangana-Baleanu-Riemann Fractional (3 + 1)-Dimensional Kadomtsev–Petviashvili Equation

Abstract

In this article, the generalized Jacobi elliptic function expansion method with four new Jacobi elliptic functions was used to the generalized fractional (3 + 1)-dimensional Kadomtsev–Petviashvili (GFKP) equation with the Atangana-Baleanu-Riemann fractional derivative, and abundant new types of analytical solutions to the GFKP were obtained. It is well known that there is a tight connection between symmetry and travelling wave solutions. Most of the existing techniques to handle the PDEs for finding the exact solitary wave solutions are, in essence, a case of symmetry reduction, including nonclassical symmetry and Lie symmetries etc. Some 3D plots, 2D plots, and contour plots of these solutions were simulated to reveal the inner structure of the equation, which showed that the efficient method is sufficient to seek exact solutions of the nonlinear partial differential models arising in mathematical physics.

1. Introduction

In more recent years, the fractional calculus and fractional nonlinear partial differential equations (FNPDE) have played an important role in the investigation nonlinear phenomena. Many scientists in the world are absorbed by this topic, and searching for the exact solutions of these FNPDES is significant in the study of nonlinear phenomena, including the nonlinear wave observed in mechanics [1], ecological and economic systems [2], two-scale thermal science [3], optical fibers [4], chaotic oscillations [5], atmospheric science [6], telegraph transmission [7], etc. [8,9,10,11]. To date, many powerful methods have been proposed for this subject, such as the Bäcklund transformation method [12], Darboux transformation [13], Hirota bilinear method [14], improved F-expansion method [15], sine-Gordon method [16], projective Riccati equations method [17], G’/G-expansion method [18], (G’/G,1/G)-expansion method [19], improved (m + G’/G)-expansion method [20], improved G’/G²-expansion method [21], the first integral method [22], Generalized Exp-Function Method [23], Exp(−φ(ξ))-Expansion Method [24], and Lie symmetry method, which are connected to the wave transformations of equations that do not change the set of solutions [25], among others [26,27,28,29,30,31,32,33,34,35,36].

Since calculus was founded by Newton and Leibniz in the 1660s, the fractional calculus has been gradually studied by many scholars all over the world. To date, there have been many definitions of the fractional derivative. The most classic definitions are the Riemann–Liouville fractional derivative [37], Caputo fractional derivative [38], Jumaries’s fractional derivative [39], Ji-Huan He’s fractional derivative [40], Atangana’s fractional derivative [41], M-fractional derivative [42], conformable fractional derivative [43], Atangana–Baleanu–Caputo Operator [44], and the Atangana-Baleanu-Riemann derivative [45,46,47,48,49,50,51]. Each definition has its own advantages and limitations. Compared with other definitions, the ABR derivative is more convenient for integration operations due to the introduction of series and rational standardized functions.

Consider the following generalized Atangana-Baleanu-Riemann time-fractional (3 + 1)-dimensional Kadomtsev-Petviashvili (KP) equation [52,53,54,55,56,57,58]:

$${(u_t^{\gamma }+\alpha u{u}_x+\beta u_{xxx})}_x+\lambda (u_{xx}+u_{yy}+u_{zz})+\mu (u_{xy}+u_{yz}+u_{zx})=0,0<\gamma \le 1.$$

(1)

where $\alpha ,\beta ,\lambda ,\mu$ are real numbers. The wave function $u=u(x,y,z),$$u_t^{\gamma }{=}^{ABR}{D}_t^{\gamma }u$ means the Atangana-Baleanu-Riemann (ABR) fractional derivative [45,46,47,48,49,50,51]; when $\gamma =1$, the lump solitons of Equation (1) were deduced using Hirota’s bilinear transformation method in Ref. [52]. The solutions of many other deformed forms of (3 + 1)-dimensional KP equation can be found in Refs. [53,54,55,56,57,58,59]. In this paper, we extended the Jacobi elliptic function expansion method [60,61] to Equation (1) utilizing a wave transformation.

The ABR fractional derivative is defined as follows.

Definition 1.

For a function$f(t):[0,\infty )\to R.$The Atangana-Baleanu-Riemann (ABR) derivative operator of$f(t)$of order$\alpha$is defined as [45,46]:

$${}^{ ABR}D_t^{\gamma }f(t)=\{\begin{array}{l}\frac{B(\gamma )}{1-\gamma }\frac{d}{dt}{\displaystyle {\int }_{t_0}^{t}f(\tau )E_{\gamma }[\frac{-\gamma }{1-\gamma }{(t-\tau )}^{\gamma }]d\tau ,0<\gamma <1.}\\ \frac{df(t)}{dt},\gamma =1.\end{array}$$

where$E_{\gamma }$stands for the Mittag-Leffler function, given by

$$E_{\gamma }[\frac{-\gamma }{1-\gamma }{(t-\tau )}^{\gamma }]={\displaystyle \sum _{n=0}^{\infty }{(\frac{-\gamma }{1-\gamma })}^n\frac{{(t-\tau )}^{n\gamma }}{\Gamma (n\gamma +1)}}.$$

and$B(\gamma )=1-\gamma +\frac{\gamma }{\Gamma (\gamma )}$, being a normalization function, satisfies$B(0)=B(1)=1.$

Theorem 1.

The relation between the ABR fractional derivative and Riemann-Liouville integral operator is given by

$${}^{ABR}D_t^{\gamma }f(t)=\frac{B(\gamma )}{1-\gamma }{\displaystyle \sum _{n=0}^{\infty }[{(\frac{-\gamma }{1-\gamma })}^n{J}^{n\alpha }\frac{{(t-\tau )}^{n\gamma }}{\Gamma (n\gamma +1)}}f(t)].$$

Proof. 

$$\begin{array}{l}{}^{ABR}D_t^{\gamma }f(t)=\frac{B(\gamma )}{1-\gamma }\frac{d}{dt}{\displaystyle {\int }_{t_0}^{t}f(\tau )[1+{(\frac{-\gamma }{1-\gamma })}^1\frac{{(t-\tau )}^{\gamma }}{\Gamma (1\cdot \gamma +1)}+{(\frac{-\gamma }{1-\gamma })}^2\frac{{(t-\tau )}^{2\gamma }}{\Gamma (2\cdot \gamma +1)}}\\ \begin{array}{cccc}& & & \end{array}+\cdots +{(\frac{-\gamma }{1-\gamma })}^n\frac{{(t-\tau )}^{n\gamma }}{\Gamma (n\cdot \gamma +1)}+\cdots ]d\tau \\ \begin{array}{cccc}& & & \end{array}=\frac{B(\gamma )}{1-\gamma }[{(\frac{-\gamma }{1-\gamma })}^0{}^{RL}D^{-0\gamma }f(t)+{(\frac{-\gamma }{1-\gamma })}^1{}^{RL}D^{-1\gamma }f(t)\\ \begin{array}{cccc}& & & \end{array}+{(\frac{-\gamma }{1-\gamma })}^2{}^{RL}D^{-2\gamma }f(t)+\cdots +{(\frac{-\gamma }{1-\gamma })}^n{}^{RL}D^{-n\gamma }f(t)+\cdots ]\\ \begin{array}{cccc}& & & \end{array}=\frac{B(\gamma )}{1-\gamma }{\displaystyle \sum _{n=0}^{\infty }[{(\frac{-\gamma }{1-\gamma })}^n{D}^{-n\gamma }}f(t)]\\ \begin{array}{cccc}& & & \end{array}=\frac{B(\gamma )}{1-\gamma }{\displaystyle \sum _{n=0}^{\infty }[{(\frac{-\gamma }{1-\gamma })}^n{J}^{n\gamma }}f(t)].\end{array}$$

where $J$ is a Riemann–Liouville integral operator. □

The rest of the paper is organized as follows. In Section 2, we introduce the generalized Jacobi elliptic function expansion method, while in Section 3, some exact solutions of the GFKP equation are found and discussed utilizing the proposed methods. Finally, the conclusion is presented in Section 4.

2. Summary of the Generalized Jacobi Elliptic Functions Expansion Method

Consider the following (3 + 1)-dimensional ABR fractional differential equation:

$$P(u,u_t^{\gamma },u_x,u_y,u_z,u_{xx},u_{yy},u_{zz},u_{xy},u_{yz},u_{zx},u{u}_x,u{u}_y,u{u}_z,\cdots )=0$$

(2)

We use the following wave transformation:

$$u(x,y,z,t)=u(\xi ),\xi =\{\begin{array}{l}x+y+z+\frac{(1-\gamma )c}{B(\gamma ){\displaystyle \sum _{n=0}^{\infty }{(\frac{-\gamma }{1-\gamma })}^n\Gamma (1-n\gamma )}}{t}^{-n\gamma },\gamma \in (0.1).\\ x+y+z+ct,\gamma =1.\end{array}$$

(3)

where constant $c$ means the wave speed to be determined later.

For $0<\gamma <1$ we have

$${}^{ABR}D_t^{\gamma }u(x,y,z,t)=c{u}^{\prime }=c{u}^{\prime }(\xi )=c\frac{du(\xi )}{d\xi }$$

(4)

.

Thus, Equation (2) reduces to an ordinary differential equation:

$$O(u,u^{\prime },u^{″},u^{‴},u^{(4)},u{u}^{\prime },u^{\prime }{u}^{″},\cdots )=0.$$

(5)

Assume that Equation (5) has the following solution:

$$\begin{array}{l}u\left(\xi \right)={\displaystyle \sum _{i=0}^N{A}_i{F}^i(\xi )}+\\ {\displaystyle \sum _{i,j=1;i\le j\le N}^N[B_i{F}^{j-i}(\xi )E^i(\xi )+C_i{F}^{j-i}(\xi )G^i(\xi )+D_i{F}^{j-i}(\xi )H^i(\xi )].}\end{array}$$

(6)

where $A_0,A_i,B_i,C_i,D_i,(i=1,2,\cdots ,N)$ are constants to be determined later. $\xi =\xi (x,y,z,t)$ are arbitrary functions with the variables $x,y,z$ and t. The parameter $N$ can be determined by balancing the highest-order derivative terms with the nonlinear terms in Equation (2). $E(\xi ),F(\xi ),G(\xi ),H(\xi )$ is an arbitrary array of the four functions $e=e(\xi ),f=f(\xi ),g=g(\xi )$ and $h=h(\xi )$, which are expressed as follows [60,61]:

$$\{\begin{array}{l}e=\frac{1}{p+qsn\xi +rcn\xi +ldn\xi },f=\frac{sn\xi }{p+qsn\xi +rcn\xi +ldn\xi },\\ g=\frac{cn\xi }{p+qsn\xi +rcn\xi +ldn\xi },h=\frac{dn\xi }{p+qsn\xi +rcn\xi +ldn\xi }.\end{array}$$

(7)

where $p,q,r,l$ are arbitrary constants. The functions $e,f,g,h$ satisfy the restricted relations (4) and (5a–5d) mentioned in [60,61]. Using the standard steps [60,61], we can construct some new solutions of Equations (2) and (5).

3. Exact Solutions to the Fractional (3 + 1)-Dimensional KP Equation

Substituting Equation (3) into Equation (1), we obtain:

$$\beta u_{4\xi }^{(4)}+[c+3(\lambda +\mu )]u_{\xi \xi }^{″}+\alpha {(u_{\xi }^{\prime })}^2+\alpha u{u}_{\xi \xi }^{″}=0.$$

(8)

By balancing the highest-order linear term and the nonlinear term in Equation (8), we can obtain $N=2$. Thus, we assume that Equation (8) has the following solutions:

$$\begin{array}{l}u=c_0+c_{1}e+c_{2}f+c_{3}g+c_{4}h+d_1{e}^2+d_2{f}^2+d_3{g}^2+d_4{h}^2+d_{5}fg+d_{6}fh\\ \begin{array}{cc}& \end{array}+d_{7}gh+d_{8}ef+d_{9}eg+d_{10}eh.\end{array}$$

(9)

where $u=u(\xi ),e=e(\xi ),f=f(\xi ),g=g(\xi ),h=h(\xi )$.

Substituting (4) and (5a–5d) mentioned in [60,61] separately along with Equation (9) into Equation (8) and setting the coefficients of $F^i(\xi )E^{j_1}{(\xi )}^{j_2}G{(\xi )}^{j_3}H{(\xi )}^{j_4}$, $(j_{1,\cdots ,4}=0,1,j_1{j}_2{j}_3{j}_4=0)$ $(i=0,1,2,\cdots ),$ to zero yields ODEs with respect to the unknowns $c_i(i=0,\cdots ,4),$$d_j(j=1,\cdots ,10),$$\gamma ,\alpha ,\beta ,\lambda ,\mu ,$$p,q,r,l$. After solving the ODEs by mathematical software and Wu elimination, we could determine the following solutions.

Set 1. When $p=0$

Case 1

$$q=l=0,c=c_{(1)}=-c_0\alpha +4(1+m^2)\beta -3(\lambda +\mu ),d_4=-\frac{12r^2\beta }{\alpha }.$$

Case 2

$$q=r=0,c=c_{(2)}=-c_0\alpha -4(2m^2-1)\beta -3(\lambda +\mu ),d_2=-\frac{12l^2{m}^2\beta (m^2-1)}{\alpha }.$$

Case 3

$$q=0,r=\pm ml,c=c_{(3)}=-c_0\alpha -2(m^2+1)\beta -3(\lambda +\mu ),d_1=\frac{3l^2\beta (m^2-1)}{\alpha }.$$

Case 4

$$q=0,r=\mp l,c=c_{(4)}=-c_0\alpha +(m^2-5)\beta -3(\lambda +\mu ),c_3=\pm \frac{6l\beta (m^2-1)}{\alpha }.$$

Case 5

$$q=0,r=\pm l,c=c_{(5)}=-c_0\alpha -2(m^2+1)\beta -3(\lambda +\mu ),d_2=\frac{3l^2\beta (m^2-1)}{\alpha }.$$

Case 6

$$\begin{gathered} q=0,r=\mp l\sqrt{m},c=c_{(6)}=-c_0\alpha +\beta (1-18m+m^2)-3(\lambda +\mu ), \\ {d}_7=\pm \frac{12l^2\beta {(1-m)}^2\sqrt{m}}{\alpha }. \end{gathered}$$

Case 7

$$l=0,q=\pm ir,c=c_{(7)}=-c_0\alpha +2(2-m^2)\beta -3(\lambda +\mu ),d_1=\frac{3m^2{r}^2\beta }{\alpha },i=\sqrt{-1}.$$

Case 8

$$\begin{array}{l}l=0,q=\pm r\sqrt[4]{1-m^2},c=c_{(8)}=-c_0\alpha +\beta (m^2-18\sqrt{1-m^2}-2)-3(\lambda +\mu ),\\ d_5=\frac{12r^2\beta \sqrt[4]{1-m^2}(m^2-2\sqrt{1-m^2}-2)}{\alpha }.\end{array}$$

Case 9

$$l=0,q=\pm ir,c=c_{(9)}=-c_0\alpha +(4-5m^2)\beta -3(\lambda +\mu ),c_3=\frac{6m^{2}r\beta }{\alpha },i=\sqrt{-1}.$$

Case 10

$$\begin{array}{l}l=0,q=\epsilon ir,c=c_{(10)}=-c_0\alpha +(1-2m^2)\beta -3(\lambda +\mu ),c_3=\frac{3m^{2}r\beta }{\alpha },c_4=\pm \frac{3m^{2}r\beta }{\alpha },\\ i=\sqrt{-1},\epsilon =\pm 1.\end{array}$$

Case 11

$$\begin{array}{l}r=l=1,q=\pm 1,c=c_{(11)}=-c_0\alpha +\beta (4m^2-5)-3(\lambda +\mu ),\\ c_2=\mp \frac{6\beta m^2}{\alpha },c_3=-\frac{12\beta (m^2-1)}{\alpha },d_2=\frac{3\beta m^4}{\alpha },d_3=-\frac{12\beta }{\alpha }.\end{array}$$

Case 12

$$\begin{array}{l}r=l=1,q=\pm 1,c=c_{(12)}=-c_0\alpha -\beta (2m^2+5)-3(\lambda +\mu ),\\ c_3=-\frac{6\beta (m^2-2)}{\alpha },c_4=\frac{6\beta m^2}{\alpha },d_3=\frac{3\beta {(m^2-2)}^2}{\alpha (1-m^2)},d_4=\frac{3\beta m^4}{\alpha (1-m^2)}.\end{array}$$

where $c_0,c_1,c_2,c_3,c_4,d_1,\cdots ,d_{10}$ unmentioned are equal to zero, as are the following situations. According to Equations (3), (7) and (9) and case 1–case 12, we can deduce the following solutions of Equation (1):

$$u_1({\xi }_1)=c_0-\frac{12\beta }{\alpha }d{c}^2{\xi }_1.$$

$$u_2({\xi }_2)=c_0-\frac{12m^2\beta (m^2-1)}{\alpha }s{d}^2{\xi }_2.$$

$$u_3({\xi }_3)=c_0+\frac{3\beta (m^2-1)}{\alpha }\frac{1}{(\pm mcn{\xi }_3+dn{\xi }_3{)}^2}.$$

$$u_4({\xi }_4)=c_0\pm \frac{6\beta (m^2-1)}{\alpha }\frac{cn{\xi }_4}{\mp cn{\xi }_4+dn{\xi }_4}.$$

$$u_5({\xi }_5)=c_0+\frac{3\beta (m^2-1)}{\alpha }\frac{s{n}^2{\xi }_5}{(\pm cn{\xi }_5+dn{\xi }_5{)}^2}.$$

$$u_6({\xi }_6)=c_0\pm \frac{12\beta {(1-m)}^2\sqrt{m}}{\alpha }\frac{cn{\xi }_{6}dn{\xi }_6}{{(\mp \sqrt{m}cn{\xi }_6+dn{\xi }_6)}^2}.$$

$$u_7({\xi }_7)=c_0+\frac{3m^2\beta }{\alpha }\frac{1}{{(\pm isn{\xi }_7+cn{\xi }_7)}^2}.$$

$$u_8({\xi }_8)=c_0+\frac{12\beta \sqrt[4]{1-m^2}(m^2-2\sqrt{1-m^2}-2)}{\alpha }\frac{sn{\xi }_{8}cn{\xi }_8}{{(\pm \sqrt[4]{1-m^2}sn{\xi }_8+cn{\xi }_8)}^2}.$$

$$u_9({\xi }_9)=c_0+\frac{6m^2\beta }{\alpha }\frac{cn{\xi }_9}{\pm isn{\xi }_9+cn{\xi }_9}.$$

$$u_{10}({\xi }_{10})=c_0+\frac{3m^2\beta }{\alpha }\frac{cn{\xi }_{10}\pm dn{\xi }_{10}}{\epsilon isn{\xi }_{10}+cn{\xi }_{10}}.$$

$$u_{11}=c_0+\frac{\mp \frac{6\beta m^2}{\alpha }sn{\xi }_{11}-\frac{12\beta (m^2-1)}{\alpha }cn{\xi }_{11}}{\pm sn{\xi }_{11}+cn{\xi }_{11}+dn{\xi }_{11}}+\frac{\frac{3\beta m^4}{\alpha }s{n}^2{\xi }_{11}-\frac{12\beta }{\alpha }c{n}^2{\xi }_{11}}{{(\pm sn{\xi }_{11}+cn{\xi }_{11}+dn{\xi }_{11})}^2}.$$

$$u_{12}=c_0+\frac{-\frac{6\beta (m^2-2)}{\alpha }cn{\xi }_{12}+\frac{6\beta m^2}{\alpha }dn{\xi }_{12}}{\pm sn{\xi }_{12}+cn{\xi }_{12}+dn{\xi }_{12}}+\frac{\frac{3\beta {(m^2-2)}^2}{\alpha (1-m^2)}c{n}^2{\xi }_{12}+\frac{3\beta m^4}{\alpha (1-m^2)}d{n}^2{\xi }_{12}}{{(\pm sn{\xi }_{12}+cn{\xi }_{12}+dn{\xi }_{12})}^2}.$$

where

$$\{\begin{array}{l}{\xi }_k=x+y+z+\frac{(1-\gamma )}{B(\gamma ){\displaystyle \sum _{n=0}^{\infty }{(\frac{-\gamma }{1-\gamma })}^n\Gamma (1-n\gamma )}}\cdot c_{(k)}{t}^{-n\gamma },\begin{array}{cc}\gamma \in (0,1)& k=1,2,\cdots ,12.\end{array}\\ {\xi }_k=x+y+z+c_{(k)}t{,}_{}\begin{array}{cc}\gamma =1,& k=1,2,\cdots ,12.\end{array}\end{array}$$

The expression of ${\xi }_k$ has the similar forms in the following situations.

Some simulations of $u_1,u_8,u_{11},u_{12}$ are shown in Figure 1, Figure 2, Figure 3 and Figure 4. The corresponding parameter values were selected as follows:

$$\gamma =1,p=q=l=0,r=1,c_0=1,\alpha =1,\beta =1,\lambda =1,\mu =1,m=1/2,c_{(1)}=-2,d_4=-12.$$

$$\begin{array}{l}\gamma =0.5,l=0,r=1,q=\frac{\sqrt{2}}{2},p=0,c_0=1,\alpha =1,\beta =1,\lambda =1,\mu =1,m=\sqrt{3}/2,n=3,\\ d_5=-27\sqrt{2}/2,c_{(8)}=-17.25.\end{array}$$

$$\begin{array}{l}\gamma =0.5,r=l=1,q=1,p=0,c_0=1,\alpha =1,\beta =1,\lambda =1,\mu =1,m=\sqrt{2}/2,n=3,\\ c_2=-3,c_3=6,d_2=0.75,d_3=-12,c_{(11)}=-10.\end{array}$$

$$\begin{array}{l}\gamma =0.5,r=l=1,q=1,p=0,c_0=1,\alpha =1,\beta =1,\lambda =1,\mu =1,m=\sqrt{2}/2,\\ c_3=9,c_4=3,n=3,d_3=13.5,d_4=1.5,c_{(12)}=-13.\end{array}$$

With the same process, we can derive the other three families of new exact solutions of Equation (1) as follows.

Set 2. When $q=0$

Case 13

$$r=0,l=0,c=c_{(13)}=-c_0\alpha +4(1-2m^2)\beta -3(\lambda +\mu ),d_3=\frac{12m^2{p}^2\beta }{\alpha }.$$

Case 14

$$r=0,p=\pm l,c=c_{(14)}=-c_0\alpha -2(m^2-2)\beta -3(\lambda +\mu ),d_2=-\frac{3p^2{m}^4\beta }{\alpha }.$$

Case 15

$$\begin{array}{l}l=0,p=\epsilon r,c=c_{(15)}=-c_0\alpha +(m^2-2)\beta -3(\lambda +\mu ),c_4=\pm \frac{3r\beta }{\alpha },\\ c_3=\frac{3r\beta }{\alpha },\epsilon =\pm 1.\end{array}$$

Case 16

$$\begin{array}{l}l=0,r=-\epsilon \frac{mp}{\sqrt{m^2-1}},c=c_{(16)}=-c_0\alpha +\beta (m^2+1)-3(\lambda +\mu ),\\ c_2=\pm \frac{3mp\beta }{\alpha },c_3=\epsilon \frac{3mp\beta }{\alpha \sqrt{m^2-1}},\epsilon =\pm 1.\end{array}$$

Case 17

$$l=0,r=\pm p,c=c_{(17)}=-c_0\alpha +2(2m^2-1)\beta -3(\lambda +\mu ),d_2=-\frac{3p^2\beta }{\alpha }.$$

Case 18

$$\begin{array}{l}l=0,r=\mp \frac{p\sqrt{m}}{\sqrt[4]{m^2-1}},c=c_{(18)}=-c_0\alpha +(18m\sqrt{m^2-1}-2m^2+1)\beta -3(\lambda +\mu ),\\ d_9=\pm \frac{12p^2\beta (2m\sqrt{m^2-1}+1-2m^2)}{\alpha \sqrt[4]{m^2-1}}.\end{array}$$

Case 19

$$\begin{array}{l}l=0,p=\epsilon r,c=c_{(19)}=-c_0\alpha +\beta (m^2-2)-3(\lambda +\mu ),c_4=\pm \frac{3r\beta }{\alpha },\\ c_3=\frac{3r\beta }{\alpha },\epsilon =\pm 1.\end{array}$$

Case 20

$$\begin{array}{l}l=0,r=-\epsilon \frac{mp}{\sqrt{m^2-1}},c=c_{(20)}=-c_0\alpha +\beta (m^2+1)-3(\lambda +\mu ),c_2=\pm \frac{3mp\beta }{\alpha },\\ c_3=\epsilon \frac{3mp\beta }{\alpha \sqrt{m^2-1}},\epsilon =\pm 1.\end{array}$$

Case 21

$$\begin{array}{l}p=\sqrt{1-m^2},l=1,r=\pm (\sqrt{1-m^2}-\epsilon ),\epsilon =\pm 1,c_3=\pm \frac{6\beta ((m^2-1)\epsilon +\sqrt{1-m^2})}{\alpha },\\ c=c_{(21)}=-c_0\alpha +\beta (m^2-2+3\epsilon \sqrt{1-m^2})-3(\lambda +\mu ).\end{array}$$

Case 22

$$\begin{array}{l}p=\sqrt{1-m^2},l=1,r=\mp m,\epsilon =\pm 1,c_1=\frac{12\beta m^2\sqrt{1-m^2}}{\alpha },d_1=\frac{12\beta m^2(1-m^2)}{\alpha },\\ c=c_{(22)}=-c_0\alpha +\beta (m^2-2+3\epsilon \sqrt{1-m^2})-3(\lambda +\mu ).\end{array}$$

We obtain the following solutions of Equation (1):

$$u_{13}({\xi }_{13})=c_0+\frac{12m^2\beta }{\alpha }c{n}^2{\xi }_{13}.$$

$$u_{14}({\xi }_{14})=c_0-\frac{3m^4\beta }{\alpha }\frac{s{n}^2{\xi }_{14}}{{(1\pm dn{\xi }_{14})}^2}.$$

$$u_{15}({\xi }_{15})=c_0+\frac{3\beta }{\alpha }\frac{cn{\xi }_{15}\pm dn{\xi }_{15}}{\epsilon \pm cn{\xi }_{15}}.$$

$$u_{16}({\xi }_{16})=c_0\pm \frac{3m\beta }{\alpha }\frac{\sqrt{m^2-1}sn{\xi }_{16}+\epsilon cn{\xi }_{16}}{\sqrt{m^2-1}-\epsilon mcn{\xi }_{16}}.$$

$$u_{17}({\xi }_{17})=c_0-\frac{3\beta }{\alpha }\frac{s{n}^2{\xi }_{17}}{{(1\pm cn{\xi }_{17})}^2}.$$

$$u_{18}({\xi }_{18})=c_0\pm \frac{12\beta (2m\sqrt{m^2-1}+1-2m^2)}{\alpha }\frac{\sqrt[4]{1-m^2}cn{\xi }_{18}}{{(\sqrt[4]{1-m^2}\mp \sqrt{m}cn{\xi }_{18})}^2}.$$

$$u_{19}({\xi }_{19})=c_0+\frac{3\beta }{\alpha }\frac{cn{\xi }_{19}\pm dn{\xi }_{19}}{\epsilon \pm cn{\xi }_{19}}.$$

$$u_{20}({\xi }_{20})=c_0\pm \frac{3m\beta }{\alpha }\frac{\sqrt{m^2-1}sn{\xi }_{20}+\epsilon cn{\xi }_{20}}{\sqrt{m^2-1}-\epsilon mcn{\xi }_{20}}.$$

$$u_{21}=c_0\pm \frac{\frac{6\beta ((m^2-1)\epsilon +\sqrt{1-m^2})}{\alpha }cn{\xi }_{21}}{\sqrt{1-m^2}\pm (\sqrt{1-m^2}-\epsilon )cn{\xi }_{21}+dn{\xi }_{21}}.$$

$$u_{22}=c_0+\frac{\frac{12\beta m^2\sqrt{1-m^2}}{\alpha }}{\sqrt{1-m^2}\mp mcn{\xi }_{22}+dn{\xi }_{22}}+\frac{\frac{12\beta m^2(1-m^2)}{\alpha }}{{(\sqrt{1-m^2}\mp mcn{\xi }_{22}+dn{\xi }_{22})}^2}.$$

Some simulations of $u_{13},u_{21},u_{22}$ are shown in Figure 5, Figure 6, Figure 7 and Figure 8.

$$\begin{array}{l}\gamma =0.5,q=0,r=0,l=0,p=1,c_0=1,\alpha =1,\beta =1,\lambda =1,\mu =1,m=1/2,n=3,\\ d_3=3,c_{(13)}=-5.\end{array}$$

$$\begin{array}{l}\gamma =0.5,q=0,p=\sqrt{1-m^2},l=1,r=1/2,\epsilon =1,c_0=1,\alpha =1,\beta =1,\lambda =1,\mu =1,\\ m=\sqrt{3}/2,c_3=-1.5,c_{(21)}=-6.75.\end{array}$$

Set 3. When $r=0$

Case 23

$$l=0,q=\pm p,c=c_{(23)}=-c_0\alpha -2(1+m^2)\beta -3(\lambda +\mu ),d_3=\frac{3(1-m^2)p^2\beta }{\alpha }.$$

Case 24

$$l=0,q=\pm mp,c=c_{(24)}=-c_0\alpha -2(1+m^2)\beta -3(\lambda +\mu ),d_4=\frac{3(1-m^2)p^2\beta }{\alpha }.$$

Case 25

$$\begin{array}{l}l=0,q=\mp p\sqrt{m},c=c_{(25)}=-c_0\alpha +\beta (1-18m+m^2)-3(\lambda +\mu ),\\ d_8=\pm \frac{12p^2\beta {(1-m)}^2\sqrt{m}}{\alpha }.\end{array}$$

Case 26

$$l=0,q=\pm mp,c=c_{(26)}=-c_0\alpha +(1-5m^2)\beta -3(\lambda +\mu ),c_2=\mp \frac{6m(1-m^2)p\beta }{\alpha }.$$

Case 27

$$\begin{array}{l}l=0,q=\epsilon p,c=c_{(27)}=-c_0\alpha +(1-2m^2)\beta -3(\lambda +\mu ),c_1=\frac{3p\beta (m^2-1)}{\alpha },\\ c_4=\pm \frac{3p\beta }{\alpha }\sqrt{1-m^2},\epsilon =\pm 1.\end{array}$$

Case 28

$$\begin{array}{l}p=l=1,q=\pm \epsilon (1+\epsilon \sqrt{1-m^2}),c_2=\pm \frac{6\beta (\epsilon (1-m^2)+\sqrt{1-m^2})}{\alpha },\epsilon =\pm 1,\\ c=c_{(28)}=-c_0\alpha +\beta (m^2-2-3\epsilon \sqrt{1-m^2})-3(\lambda +\mu ).\end{array}$$

Case 29

$$\begin{array}{l}p=l=1,q=\pm mi,i=\sqrt{-1},c=c_{(29)}=-c_0\alpha +\beta (4+m^2)-3(\lambda +\mu ),\\ c_1=-\frac{12\beta m^2}{\alpha },d_1=\frac{12\beta m^2}{\alpha }.\end{array}$$

We obtain the following solutions of Equation (1):

$$u_{23}({\xi }_{23})=c_0+\frac{3(1-m^2)\beta }{\alpha }\frac{c{n}^2{\xi }_{23}}{{(1\pm sn{\xi }_{23})}^2}.$$

$$u_{24}({\xi }_{24})=c_0+\frac{3(1-m^2)\beta }{\alpha }\frac{d{n}^2{\xi }_{24}}{{(1\pm msn{\xi }_{24})}^2}.$$

$$u_{25}({\xi }_{25})=c_0\pm \frac{12\beta {(1-m)}^2\sqrt{m}}{\alpha }\frac{sn{\xi }_{25}}{{(1\mp \sqrt{m}sn{\xi }_{25})}^2}.$$

$$u_{26}({\xi }_{26})=c_0\mp \frac{6m(1-m^2)\beta }{\alpha }\frac{sn{\xi }_{26}}{1\pm msn{\xi }_{26}}.$$

$$u_{27}({\xi }_{27})=c_0+\frac{3\beta (m^2-1)}{\alpha }\frac{1}{1+\epsilon sn{\xi }_{27}}\pm \frac{3\beta \sqrt{1-m^2}}{\alpha }\frac{dn{\xi }_{27}}{1+\epsilon sn{\xi }_{27}}.$$

$$u_{28}=c_0+\frac{\pm \frac{6\beta (\epsilon (1-m^2)+\sqrt{1-m^2})}{\alpha }sn{\xi }_{28}}{1\pm \epsilon (1+\epsilon \sqrt{1-m^2})sn{\xi }_{28}+dn{\xi }_{28}}.$$

$$u_{29}=c_0-\frac{\frac{12\beta m^2}{\alpha }}{1\mp misn{\xi }_{29}+dn{\xi }_{29}}+\frac{\frac{12\beta m^2}{\alpha }}{{(1\mp misn{\xi }_{29}+dn{\xi }_{29})}^2}.$$

Some simulations of $u_{27},u_{28}$ are shown in Figure 9 and Figure 10.

$$\begin{array}{l}\gamma =0.5,r=0,l=0,p=1,q=1,c_0=1,\alpha =1,\beta =1,\lambda =1,\mu =1,\epsilon =-1,m=\sqrt{3}/2,\\ c_1=-0.75,c_4=1.5,c_{(27)}=-7.5.\end{array}$$

$$\begin{array}{l}\gamma =0.5,r=0,p=l=1,q=1.5,c_0=1,\alpha =1,\beta =1,\lambda =1,\mu =1,\epsilon =1,m=\sqrt{3}/2,\\ c_2=4.5,\epsilon =1,c_{(28)}=-11.25.\end{array}$$

Set 4. When $l=0$

Case 30

$$p=q=1,r=\pm i,i=\sqrt{-1},c=c_{(30)}=-c_0\alpha +\beta (1-3m+m^2)-3(\lambda +\mu ),c_3=\pm \frac{6m\beta i}{\alpha }$$

Case 31

$$p^2=1,q^2=1,r=\pm 1,c=c_{(31)}=-c_0\alpha +\beta (4m^2-5)-3(\lambda +\mu ),c_3=\pm \frac{12\beta }{\alpha },d_3=-\frac{12\beta }{\alpha }$$

We obtain the following solutions of Equation (1):

$$u_{30}=c_0\pm \frac{\frac{6m\beta i}{\alpha }cn{\xi }_{30}}{1+sn{\xi }_{30}\pm icn{\xi }_{30}}.$$

$$u_{31}=c_0\pm \frac{\frac{12\beta }{\alpha }cn{\xi }_{31}}{1\pm sn{\xi }_{31}+\epsilon cn{\xi }_{31}}-\frac{\frac{12\beta }{\alpha }c{n}^2{\xi }_{31}}{{(1\pm sn{\xi }_{31}+\epsilon cn{\xi }_{31})}^2}.$$

The simulation of $u_{31}$ for different variable space are shown in Figure 11 and Figure 12.

$$\begin{array}{l}\gamma =0.5,l=0,p=1,q=1,r=1,c_0=1,\alpha =1,\beta =1,\lambda =1,\mu =1,\epsilon =1,\\ m=1/2,c_3=12,d_3=-12,c_{(31)}=-11.\end{array}$$

In the above cases, we have:

$$\{\begin{array}{l}{\xi }_k=x+y+z+\frac{(1-\gamma )}{B(\gamma ){\displaystyle \sum _{n=0}^{\infty }{(\frac{-\gamma }{1-\gamma })}^n\Gamma (1-n\gamma )}}\cdot c_{(k)}{t}^{-n\gamma },\begin{array}{cc}\gamma \in (0,1),& k=1,2,\cdots ,31.\end{array}\\ {\xi }_k=x+y+z+c_{(k)}t{,}_{}\begin{array}{cc}\gamma =1,& k=1,2,\cdots ,31.\end{array}\end{array}$$

Notice that the Jacobian elliptic functions have the following degenerative properties. Setting the modulus $m\to 1$ yields $sn\xi \to \tanh \xi ,cn\xi \to \sec h\xi$, $dn\xi \to \sec h\xi$, and when $m\to 0$, it derives that $sn\xi \to \sin \xi ,cn\xi \to \cos \xi$ and $dn\xi \to 1$. We can obtain some solitary wave solutions and trigonometric function wave solutions.

Remark. Solutions $u_i(i=1,7,9,10,11,12,13,14,15,17,19,29,30,31)$ are degenerated to solitary solutions when the modulus $m\to 1$, and solutions $u_i(i=1,4,5,8,11,12,15,17,$$19,21,23,27,28,31)$ are degenerated to triangular functions solutions when the modulus $m\to 0$.

4. Results and Discussion

After utilizing the generalized Jacobi elliptic functions expansion method, we obtained many types of exact explicit solutions as in Equation (1); some structure of these solutions are simulated in Figure 1, Figure 2, Figure 3, Figure 4, Figure 5, Figure 6, Figure 7, Figure 8, Figure 9, Figure 10, Figure 11 and Figure 12. The visualization can help us better understanding the dynamic behavior and propagation property of these solutions. For example, the Jacobi function wave solution of Equation (1) is shown in Figure 1. From the solution, we can find that the waveform changes periodically in the upper and lower planes, while the double periodic solution $u_8$ makes cyclical changes upward, as shown in Figure 2. In Figure 3 and Figure 4, the waveform widths of $u_{11}$ and $u_{12}$ in the upper and lower directions show periodic changes in alternating size. The famous Jacobi oval cosine wave solution $u_{13}$ and bell-shape soliton solution $u_{13,m=1}$ are plotted in Figure 5 and Figure 6. The simulations of various other different forms of double periodic wave solutions are shown in Figure 7, Figure 8, Figure 9, Figure 10 and Figure 11. From Figure 10, we can find that there are two parallel asymptotes between the adjacent two cycles of the wave function. Each set of waves was suppressed in a region of equal width. When $t=x=0$, we can find that the 3D plot of $u_{31}$ on the $y,z$ dimension is different from the $x,t$ dimension, which is shown in Figure 12. If we select different values for the parameters $\gamma$ and $n$, the shape and position of Jacobi function solutions $u_1$ produces an offset, which is simulated in Figure 13. These different propagation patterns may explain the different phenomena for this model.

5. Conclusions

In this paper, we succeeded to propose the generalized Jacobi elliptic function expansion method for finding new exact solutions of nonlinear evolution equations by constructing the four new types of Jacobian elliptic functions (4). Using this method and computerized symbolic computation, we found 31 types of exact solutions for the generalized fractional (3 + 1)-dimensional KP equation, including Jacobi elliptic functions solutions, bell-shape soliton solution, solitary wave solutions, trigonometric function solutions, and periodic wave solutions in the mixed form to the GFKP. More importantly, our method is a simple and powerful technique to find new solutions to various kinds of nonlinear evolution equations. We believe that this method should play an important role in finding exact solutions in mathematical physics.