Abundant explicit solutions for the M-fractional coupled nonlinear Schrödinger–KdV equations

Abstract

In this study, the famous fractional generalized coupled cubic nonlinear Schrödinger–KdV equations arising in many domains of physics and engineering such as depicting the propagation of long waves in dispersive media and the dynamics of short dispersive waves for narrow-bandwidth packet have been investigated. We propose two significant methods named the modified (Gʹ/G, 1/G)-expansion method and the Gʹ/(bGʹ + G + a)-expansion method. After utilizing these two efficient techniques, many types of explicit soliton pulse solutions including the well-known bell-shape bright soliton pulse, the kink and anti-kink dark solitons pulse, the mixed bright-dark soliton pulse, the W-shaped soliton pulse, the periodic waves pulse, and the blow-up soliton pattern solutions are obtained. If we select different values of the frequency, coefficients and orders, the dynamic properties and physical structures of these solutions are discussed, these important results can help us to further understand the inner characteristic of the model.

Introduction

In the last several decades, many topics in nonlinear dispersive partial differential equations especially for the fractional-order models have been characterized, such as soliton propagation for explaining the oceanic phenomena that correlate to nonlinear shallow or deep-water wave propagation,^1,2 the interactions of capillary-gravity water waves between short and long dispersive waves arising in fluid mechanics, ³ the vector rogue waves in coupled vector nonlinear Schrödinger equations, ⁴ the suppression of chaotic oscillations, ⁵ the market dynamics characteristic, ⁶ the optical fiber communication, ⁷ etc.^8–9 For better explaining the complex feature of these phenomena, searching for analytic explicit solutions including approximate solutions and exact solutions of these models plays an important and significant role. Up to now, many powerful methods for this subject such as the homotopy analysis method (HAM), ¹⁰ He’s homotopy perturbation method (HPM), ¹¹ Adomian’s decomposition method (ADM), ¹² variational iteration method (VIM), ¹³ Laplace transformation method, ¹⁴ double Laplace transform method, ¹⁵ Yang-Laplace transformation method, ¹⁶ and Shehu transformation method ¹⁷ can be used to find the approximate solutions. The Bäcklund transformation method, ¹⁸ Darboux transformation, ¹⁹ and Hirota bilinear method ²⁰ can be used to find the N-soliton solutions. The improved F-expansion method, ²¹ projective Riccati equations method, ²² Jacobi elliptic function expansion method, ²³ Gʹ/G-expansion method, ²⁴ (d + Gʹ/G)-expansion method, ²⁵ (Gʹ/G, 1/G)-expansion method, ²⁶ sine Gordon method, ²⁷ Lie symmetry method, ²⁸ new Kudryashov’s method, ²⁹ auxiliary equation method, ³⁰ exponential rational function method, ³¹ etc.^32–42 can be used to find doubly periodic solutions, solitary wave solutions, and trigonometric function solutions of these models. More research on the exact solutions, approximate solutions, and dynamic analysis of fractional systems can be found in Refs. [43-47].

As we all know, calculus was founded by Newton and Leibniz at the second half of the 17th century, and fractional order calculus has gradually become one of the new special fields in natural sciences since 1695. ⁴⁸ Till now, several types of definitions about the fractional derivative operator and integral operator are discussed and established by many researchers, the classical forms including Riemann–Liouville fractional derivatives, ⁴⁹ Caputo fractional derivatives, ⁵⁰ He’s fractional derivative, ⁵¹ Jumarie fractional derivative, ⁵² Atangana’s fractional derivative, ⁵³ conformable fractional derivative, ⁵⁴ Abu-Shady–Kaabar fractional derivative,^55–57 and the M-fractiona derivative^26,58,59 which will be utilized in this article.

In the present article, we consider the following M-fractional generalized coupled cubic nonlinear Schrödinger–KdV equations (MFCNLS-KdV) in the form Refs. [60–65].

$\{\begin{array}{l}i{u}_t^{\alpha }+{\lambda }_1{u}_x^{2\beta }+{\lambda }_2{\left|u\right|}^{2}u+{\lambda }_{3}uv=0,\hspace{1em}t>\mathrm{0,0}<\alpha ,\beta \le 1,\\ v_t^{\alpha }+{\sigma }_{1}v{v}_x^{\beta }+{\sigma }_2{v}_x^{3\beta }+{\sigma }_3{({\left|u\right|}^2)}_x^{\beta }=0.\end{array}$

(1)

where ${(\cdot )}_t^{\alpha }=D_{M,t}^{{\gamma }_2,\alpha }(\cdot ),{(\cdot )}_x^{\beta }=D_{M,x}^{{\gamma }_1,\beta }(\cdot ),{(\cdot )}_x^{2\beta }=D_{M,x}^{{\gamma }_1,\beta }(D_{M,x}^{{\gamma }_1,\beta }(\cdot )),{(\cdot )}_x^{3\beta }=D_{M,x}^{{\gamma }_1,\beta }(D_{M,x}^{{\gamma }_1,\beta }(D_{M,x}^{{\gamma }_1,\beta }(\cdot )))$ mean the M-fractional derivative. ⁵⁸ The coefficients ${\lambda }_1,{\lambda }_2,{\lambda }_3,{\sigma }_1,{\sigma }_2,{\sigma }_3$ are real constants, $u=u(x,t)$ is a complex function while $v=v(x,t)$ is a real-valued function. In fact, equation (1) occurs in fluid mechanics for explaining the phenomena of interactions between short and long dispersive waves, and $u$ represents the short wave, while $v$ represents the long wave. If we select $\alpha =\beta =1$ , the authors investigated the solition solutions of equation (1) by using the Kudryashov method, sub-equation method, and modified Sardar sub-equation method in Refs.[60,61]. If we select $\alpha =\beta =1$ , ${\lambda }_1={\lambda }_2=1,{\lambda }_3=\beta ,{\sigma }_1={\sigma }_2=1,{\sigma }_3=\frac{1}{2}\beta$ , the authors investigated the existence of bound and ground states for equation (1) in Refs. [62,63]. If we select $\alpha =\beta =1$ , ${\lambda }_1={\lambda }_3=-1,{\lambda }_2=0,{\sigma }_1=6,{\sigma }_2=1,{\sigma }_3=-1$ , three invariants for equation (1) have been found in Ref. [64], and the approximate solutions and exact solutions are obtained by utilizing the HPM and MEDAM in Refs. [65,66]. If we select $\beta =1$ , ${\lambda }_1={\lambda }_3=-1,{\lambda }_2=0,{\sigma }_1=6,{\sigma }_2=1,{\sigma }_3=-1$ , the existence and uniqueness to the solution and the approximate solution of equation (1) are established by using the fixed-point theorem and the modified Laplace decomposition method under the Atangana Baleanu fractional derivative in Ref. [67]. Here, let us review some basic definitions and properties about the M-fractional derivative which will be further used in this paper.

Definition

For a function $f(t):[0,\infty )\to R.$ We defined the M-fractional derivative operator of $f(t)$ of order $\alpha$ as ⁵⁸

$D_{M,t}^{\gamma ,\alpha }f(t)=\underset{\epsilon \to 0}{\lim }\frac{f(t{E}_{\gamma }(\epsilon t^{1-\alpha }))-f(t)}{\epsilon },\hspace{0.17em}\gamma >\mathrm{0,0}<\alpha \le 1.$

where $E_{\gamma }(t)={\displaystyle \sum }_{k=0}^{\infty }\frac{t^k}{\mathrm{\Gamma }(\gamma k+1)}$ is a truncated Mittag-Leffler function of one parameter.

Also, we have the following important properties^26,58,59

$\begin{array}{l}(1)D_{M,t}^{\gamma ,\alpha }f(t)=\frac{t^{1-\alpha }}{\mathrm{\Gamma }(\gamma +1)}\frac{df(t)}{dt}.\\ (2)D_{M,t}^{\gamma ,\alpha }(af(t)+bg(t))=a{D}_{M,t}^{\gamma ,\alpha }f(t)+b{D}_{M,t}^{\gamma ,\alpha }g(t),\forall a,b\in R.\\ (3)D_{M,t}^{\gamma ,\alpha }(f(t)g(t))=f(t)D_{M,t}^{\gamma ,\alpha }g(t)+g(t)D_{M,t}^{\gamma ,\alpha }f(t).\\ (4)D_{M,t}^{\gamma ,\alpha }(f(t)/g(t))=\left[g(t)D_{M,t}^{\gamma ,\alpha }f(t)-f(t)D_{M,t}^{\gamma ,\alpha }g(t)\right]/g^2(t).\\ (5)D_{M,t}^{\gamma ,\alpha }(f\mathrm{o}g)(t)=f\text{′}(g(t))D_{M,t}^{\gamma ,\alpha }g(t)=\frac{t^{1-\alpha }}{\mathrm{\Gamma }(\gamma +1)}f\text{′}(g(t))\frac{dg(t)}{dt}.\end{array}$

In this article, we will propose two efficient and significant methods for finding some new types of exact solutions of equation (1) and many important results are obtained. The paper is organized as follows: In the Description of the two methods section, we introduce the modified (Gʹ/G, 1/G)-expansion method and the Gʹ/(bGʹ + G + a)-expansion method. While in the Exact solutions to the MFCNLS-KdV section, some new types of soliton pulse solutions of the MFCNLS-KdV are found and discussed by utilizing the proposed method. Finally, the conclusion is presented in the Conclusion section.

Description of the two methods

The modified (Gʹ/G, 1/G)-expansion method

The (Gʹ/G, 1/G)-expansion method has been proposed by many authors recently.^68,69 We give some formal modification about this method in order to simplify the solving procedure for equation (1), a brief description of the technique is presented as follows:

Step 1

Consider the following nonlinear M-fractional partial differential equation

$E(u,u_t^{\alpha },u_x^{\beta },u{u}_x^{\beta },\cdots )=0.$

(2)

Step 2

Using a wave transformation^70,71

$u(x,t)=u(\xi ),\xi =\frac{\mathrm{\Gamma }({\gamma }_1+1)}{\beta }k{x}^{\beta }+\frac{\mathrm{\Gamma }({\gamma }_2+1)}{\alpha }\rho t^{\alpha },{\gamma }_{\mathrm{1,2}}>0.$

(3)

where constants $k$ and $\rho$ are to be determined latter. Equation (2) is converted into a nonlinear ordinary differential equation (ODE).

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

(4)

Step 3

Assume that equation (4) has the following solution

$u={\displaystyle \sum }_{i=0}^N{a}_i{\psi }^i+{\displaystyle \sum }_{i=1}^N{b}_i{\psi }^{i-1}\phi .$

(5)

where $N$ is a balance number, $\phi =\phi (\xi )=\frac{G^{\prime }}{G},\psi =\psi (\xi )=\frac{1}{G}$ , $a_i,b_i$ and variable function $\xi =\xi (x,t)$ are determined later. And $G=G(\xi )$ is a solution of the following auxiliary ODE.

$G^{″}=\epsilon G-\epsilon \mu ,$

(6)

where $\epsilon =\pm 1,\mu$ is an arbitrary real number. We can find the following constrained condition

$\{\begin{array}{l}{\phi }^{\prime }=\epsilon -\epsilon \mu \psi -{\phi }^2,\\ {\psi }^{\prime }=-\phi \psi ,\\ {\phi }^2=\epsilon -2\epsilon \mu \psi -\epsilon (b^2-\epsilon c^2-{\mu }^2){\psi }^2.\end{array}$

(7)

where arbitrary constants $\mu ,b$ and $c$ satisfied the relation $a^2+b^2+{\mu }^2\ne 0.$ equation (6) admits the following solutions.

Case 1

When $\epsilon =1$ , we have $G=b\cosh \xi +c\sinh \xi +\mu$ , thus

$\phi =\frac{G^{\prime }}{G}=\frac{b\sinh \xi +c\cosh \xi }{b\cosh \xi +c\sinh \xi +\mu },\hspace{0.17em}\psi =\frac{1}{G}=\frac{1}{b\cosh \xi +c\sinh \xi +\mu }$

(8)

Case 2

When $\epsilon =-1$ , we have $G=b\cos \xi +c\sin \xi +\mu$ , thus

$\phi =\frac{G^{\prime }}{G}=\frac{-b\sin \xi +c\cos \xi }{b\cos \xi +c\sin \xi +\mu },\hspace{0.17em}\psi =\frac{1}{G}=\frac{1}{b\cos \xi +c\sin \xi +\mu }$

(9)

Step 4

Substituting equations (7) and (5) into equation (4), and setting the coefficients of ${\psi }^i$ $(i=\mathrm{0,1,2},\cdots ,N)$ and ${\psi }^{i-1}\phi (i=\mathrm{1,2},\cdots ,N)$ to zero yield a set of algebraic equations (AEs) for $a_i,b_i,b,c,\mu ,k$ and $\rho$ . After solving the AEs and substituting each of the solutions $\phi (\xi ),\psi (\xi )$ from equations (8) and (9) along with (3) into equation (2), we can get the solutions of equation (2).

Remark 1

Notice that the Refs. [68,69] were focused on collecting the coefficients of ${\phi }^i$ and $\phi {\psi }^{i-1}$ to zero, but we are collecting the coefficients of ${\psi }^i$ and ${\psi }^{i-1}\phi$ to zero, each of these two ways is correct, the collected direction is decided by the solutions’ forms of these target equations.

The Gʹ/(bGʹ + G + a)-expansion method

With the similar procedure about the technique 2.1, we give the main steps about this method.

Step 1

Assume that equation (2) has the following solution

$u={\displaystyle \sum }_{i=0}^N{c}_i{F}^i.$

(10)

where $F=F(\xi )=\frac{G^{\prime }}{b{G}^{\prime }+G+a}$ , $c_i$ and variable function $\xi =\xi (x,t)$ are determined later. $b\ne 0,a$ are arbitrary constants, $G=G(\xi )$ is a solution of the following auxiliary ODE.

$G^{″}=-\frac{\lambda }{b}{G}^{\prime }-\frac{\mu }{b^2}G-\frac{\mu }{b^2}a.$

(11)

where $\lambda ,\mu$ are two arbitrary real numbers. We can find the following constrained condition

$F^{\prime }=\left(\lambda -\mu -1\right)F^2+\frac{1}{b}\left(2\mu -\lambda \right)F-\frac{1}{b^2}\mu .$

(12)

Equation (12) admits the following solutions.

Case 1

When $\mathrm{\Delta }={\lambda }^2-4\mu >0$ , we have $G=-a+p_1{e}^{\frac{1}{2b}(-\lambda -\sqrt{\mathrm{\Delta }})\xi }+p_2{e}^{\frac{1}{2b}(-\lambda +\sqrt{\mathrm{\Delta }})\xi }$ , $a,p_1,p_2$ are arbitrary constants satisfy $a^2+p_1^2+p_2^2\ne 0,$ so does in case 2, thus

$\begin{array}{l}{F}_1=\frac{p_1\hspace{0.17em}\left(\lambda +\sqrt{\mathrm{\Delta }}\right)+p_2\left(\lambda -\sqrt{\mathrm{\Delta }}\right)e^{\frac{\sqrt{\mathrm{\Delta }}}{b}\xi }}{b{p}_1\left(\lambda -2+\sqrt{\mathrm{\Delta }}\right)+b{p}_2\left(\lambda -2-\sqrt{\mathrm{\Delta }}\right)e^{\frac{\sqrt{\mathrm{\Delta }}}{b}\xi }}\\ =\frac{\left[\lambda \left(p_2-p_1\right)-\sqrt{\mathrm{\Delta }}\left(p_2+p_1\right)\right]\sinh \left(\frac{\sqrt{\mathrm{\Delta }}}{2b}\xi \right)+\left[\lambda \left(p_2+p_1\right)-\sqrt{\mathrm{\Delta }}\left(p_2-p_1\right)\right]\cosh \left(\frac{\sqrt{\mathrm{\Delta }}}{2b}\xi \right)}{b\left[\left(\lambda -2\right)\left(p_2-p_1\right)-\sqrt{\mathrm{\Delta }}\left(p_2+p_1\right)\right]\sinh \left(\frac{\sqrt{\mathrm{\Delta }}}{2b}\xi \right)+b\left[\left(\lambda -2\right)\left(p_2+p_1\right)-\sqrt{\mathrm{\Delta }}\left(p_2-p_1\right)\right]\cosh \left(\frac{\sqrt{\mathrm{\Delta }}}{2b}\xi \right)},\end{array}{F}_1=\{\begin{array}{l}{F}_{1.1}=\frac{\lambda -2\mu }{2b\left(\lambda -\mu -1\right)}-\frac{\sqrt{\mathrm{\Delta }}}{2b\left(\lambda -\mu -1\right)}\tanh \left(\frac{\sqrt{\mathrm{\Delta }}}{2b}\xi \right),\left(\lambda -2\right)\left(p_2-p_1\right)-\sqrt{\mathrm{\Delta }}\left(p_2+p_1\right)=0,\\ F_{1.2}=\frac{\lambda -2\mu }{2b\left(\lambda -\mu -1\right)}-\frac{\sqrt{\mathrm{\Delta }}}{2b\left(\lambda -\mu -1\right)}\coth \left(\frac{\sqrt{\mathrm{\Delta }}}{2b}\xi \right),\left(\lambda -2\right)\left(p_2+p_1\right)-\sqrt{\mathrm{\Delta }}\left(p_2-p_1\right)=0.\end{array}$

Case 2

When $\mathrm{\Delta }={\lambda }^2-4\mu <0$ , we have

$G=e^{-\frac{\lambda }{2b}\xi }\left(p_1\cos \left(\frac{\sqrt{-\mathrm{\Delta }}}{2b}\xi \right)+p_2\sin \left(\frac{\sqrt{-\mathrm{\Delta }}}{2b}\xi \right)\right)-a,$

$F_2=\frac{\left(\lambda p_1-\sqrt{-\mathrm{\Delta }}{p}_2\right)\cos \left(\frac{\sqrt{-\mathrm{\Delta }}}{2b}\xi \right)+\left(\lambda p_2+\sqrt{-\mathrm{\Delta }}{p}_1\right)\sin \left(\frac{\sqrt{-\mathrm{\Delta }}}{2b}\xi \right)}{b\left(\left(\lambda -2\right)p_1-\sqrt{-\mathrm{\Delta }}{p}_2\right)\cos \left(\frac{\sqrt{-\mathrm{\Delta }}}{2b}\xi \right)+b\left(\left(\lambda -2\right)p_2+\sqrt{-\mathrm{\Delta }}{p}_1\right)\sin \left(\frac{\sqrt{-\mathrm{\Delta }}}{2b}\xi \right)},F_2=\{\begin{array}{l}{F}_{2.1}=\frac{\lambda -2\mu }{2b\left(\lambda -\mu -1\right)}+\frac{\sqrt{-\mathrm{\Delta }}}{2b\left(\lambda -\mu -1\right)}\tan \left(\frac{\sqrt{-\mathrm{\Delta }}}{2b}\xi \right),\left(\lambda -2\right)p_2+\sqrt{-\mathrm{\Delta }}{p}_1=0,\\ F_{2.2}=\frac{\lambda -2\mu }{2b\left(\lambda -\mu -1\right)}-\frac{\sqrt{-\mathrm{\Delta }}}{2b\left(\lambda -\mu -1\right)}\cot \left(\frac{\sqrt{-\mathrm{\Delta }}}{2b}\xi \right),\left(\lambda -2\right)p_1-\sqrt{-\mathrm{\Delta }}{p}_2=0.\end{array}$

Step 2

Substituting equations (10) and (12) into equation (4), and setting the coefficients of $F^i$ zero yield a set of AEs for $c_i,b,\lambda ,\mu ,k$ and $\rho$ . After solving the AEs and substituting each of the solutions $F_1,F_2$ along with equations (10) and (3) into equation (2), we can get the solutions of equation (2).

In the following, we will use these two methods to solve the MFCNLS-KdV.

Exact solutions to the MFCNLS-KdV

Exact solutions

We can give the following function and traveling wave transformation

$u=P(\xi )e^{i\eta },v=Q(\xi ).$

(13)

$\xi =\frac{\mathrm{\Gamma }({\gamma }_1+1)}{\beta }{k}_1{x}^{\beta }+\frac{\mathrm{\Gamma }({\gamma }_2+1)}{\alpha }{\rho }_1{t}^{\alpha },\eta =\frac{\mathrm{\Gamma }({\gamma }_1+1)}{\beta }{k}_2{x}^{\beta }+\frac{\mathrm{\Gamma }({\gamma }_2+1)}{\alpha }{\rho }_2{t}^{\alpha },{\gamma }_1,{\gamma }_2>0.$

(14)

where constants $k_1,k_2$ and ${\rho }_1,{\rho }_2$ are to be determined latter.

Substituting equations (13) and (14) into equation (1), separate the real part and the imaginary part, we obtain

$\{\begin{array}{l}{\lambda }_1{k}_1^2{P}^{″}-({\rho }_2+k_2^2{\lambda }_1)P+{\lambda }_2{P}^3+{\lambda }_{3}PQ=0,\hspace{1em}(15.1)\\ {\sigma }_2{k}_1^3{Q}^{‴}+{\sigma }_1{k}_{1}QQ\prime {\rho }_{1}Q\text{′}+2{\sigma }_3{k}_{1}PP\text{′}=0,\hspace{1em}\hspace{1em}(15.2)\\ ({\rho }_1+2k_1{k}_2{\lambda }_1)P^{\prime }=0.\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{1em}\hspace{1em}\hspace{0.17em}(15.3)\end{array}$

(15)

Substituting equation (15.3) into equation (15.2), then integrate once, we obtain

$\{\begin{array}{l}{\lambda }_1{k}_1^2{P}^{″}-({\rho }_2+k_2^2{\lambda }_1)P+{\lambda }_2{P}^3+{\lambda }_{3}PQ=0,(16.1)\\ {\sigma }_2{k}_1^2{Q}^{″}+\frac{1}{2}{\sigma }_1{Q}^2-2k_2{\lambda }_{1}Q+{\sigma }_3{P}^2=H.\hspace{1em}(16.2)\end{array}$

(16)

where $H$ is an integration constant.

We assume that $H=0$ and equation (16) have the following solutions

$\begin{array}{l}P=a_0+a_1\psi +a_2\phi ,\\ Q=b_0+b_1\psi +b_2{\psi }^2+b_3\phi +b_4\phi \psi .\end{array}$

(17)

Substituting equations (17) and (7) into equation (16), and setting the coefficients of ${\psi }^i$ and ${\psi }^{i-1}\phi$ to zero yield a set of algebraic equations (AEs) for the unknowns $a_i,b_i,b,c,\mu ,k_1,k_2,{\rho }_1$ and ${\rho }_2$ .

$\begin{array}{l}{\psi }^0:a_0^3{\lambda }_2+\epsilon a_2{b}_3{\lambda }_3=a_0\left(k_2^2{\lambda }_1-3\epsilon a_2^2{\lambda }_2-b_0{\lambda }_3+{\rho }_2\right),\\ \phi :\epsilon a_2^3{\lambda }_2+a_0{b}_3{\lambda }_3=a_2\left(k_2^2{\lambda }_1-3a_0^2{\lambda }_2-b_0{\lambda }_3+{\rho }_2\right),\\ \psi :\epsilon a_2\left(-2\mu b_3+b_4\right){\lambda }_3+a_0\left(-6\epsilon \mu a_2^2{\lambda }_2+b_1{\lambda }_3\right)+a_1\left(\epsilon k_1^2{\lambda }_1-k_2^2{\lambda }_1+3a_0^2{\lambda }_2+3\epsilon a_2^2{\lambda }_2+b_0{\lambda }_3-{\rho }_2\right)=0,\\ \phi \psi :\left(a_1{b}_3+a_0{b}_4\right){\lambda }_3+a_2\left(-\epsilon \mu k_1^2{\lambda }_1+6a_0{a}_1{\lambda }_2+b_1{\lambda }_3\right)=2\epsilon \mu a_2^3{\lambda }_2,\\ {\psi }^2:3a_0{a}_1^2{\lambda }_2+\epsilon a_2\left(\left(-b^2+c^2\epsilon +{\mu }^2\right)b_3-2\mu b_4\right){\lambda }_3+a_1\left(-3\epsilon \mu k_1^2{\lambda }_1-6\epsilon \mu a_2^2{\lambda }_2+b_1{\lambda }_3\right)\\ +a_0\left(3\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)a_2^2{\lambda }_2+b_2{\lambda }_3\right)=0,\\ \phi {\psi }^2:\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)a_2^3{\lambda }_2+a_1{b}_4{\lambda }_3+a_2\left(2\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)k_1^2{\lambda }_1+3a_1^2{\lambda }_2+b_2{\lambda }_3\right)=0,\\ {\psi }^3:a_1^3{\lambda }_2+\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)a_2{b}_4{\lambda }_3+a_1\left(2\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)k_1^2{\lambda }_1+3\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)a_2^2{\lambda }_2+b_2{\lambda }_3\right)=0,\end{array}\begin{array}{l}{\psi }^0:a_0^3{\lambda }_2+\epsilon a_2{b}_3{\lambda }_3=a_0\left(k_2^2{\lambda }_1-3\epsilon a_2^2{\lambda }_2-b_0{\lambda }_3+{\rho }_2\right),\\ \phi :b_3\left(2k_2{\lambda }_1-b_0{\sigma }_1\right)=2a_0{a}_2{\sigma }_3,\\ \psi :\epsilon b_3{b}_4{\sigma }_1+b_1\left(-2k_2{\lambda }_1+b_0{\sigma }_1+\epsilon k_1^2{\sigma }_2\right)+2\left(a_0{a}_1-\epsilon \mu a_2^2\right){\sigma }_3=\epsilon \mu b_3^2{\sigma }_1,\\ \phi \psi :b_1{b}_3{\sigma }_1+b_4\left(-2k_2{\lambda }_1+b_0{\sigma }_1+\epsilon k_1^2{\sigma }_2\right)+2a_1{a}_2{\sigma }_3=\epsilon \mu b_3{k}_1^2{\sigma }_2,\\ {\psi }^2:b_1^2{\sigma }_1+c^2{\epsilon }^2{b}_3^2{\sigma }_1+\epsilon {\mu }^2{b}_3^2{\sigma }_1+\epsilon b_4^2{\sigma }_1+b_2\left(-4k_2{\lambda }_1+2b_0{\sigma }_1+8\epsilon k_1^2{\sigma }_2\right)+2a_1^2{\sigma }_3+2c^2{\epsilon }^2{a}_2^2{\sigma }_3\\ +2\epsilon {\mu }^2{a}_2^2{\sigma }_3=\epsilon \left(b^2{b}_3^2{\sigma }_1+4\mu b_3{b}_4{\sigma }_1+6\mu b_1{k}_1^2{\sigma }_2+2b^2{a}_2^2{\sigma }_3\right),\\ \phi {\psi }^2:b_2{b}_3{\sigma }_1+b_1{b}_4{\sigma }_1+2\epsilon \left(\left(-b^2+c^2\epsilon +{\mu }^2\right)b_3-3\mu b_4\right)k_1^2{\sigma }_2=0,\\ {\psi }^3:b_1\left(b_2{\sigma }_1+2\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)k_1^2{\sigma }_2\right)+\epsilon \left(\left(-b^2+c^2\epsilon +{\mu }^2\right)b_3{b}_4{\sigma }_1-\mu \left(b_4^2{\sigma }_1+10b_2{k}_1^2{\sigma }_2\right)\right)=0,\\ \phi {\psi }^3:b_4\left(b_2{\sigma }_1+6\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)k_1^2{\sigma }_2\right)=0,\\ {\psi }^4:b_2^2{\sigma }_1+\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)b_4^2{\sigma }_1+12\epsilon \left(-b^2+c^2\epsilon +{\mu }^2\right)b_2{k}_1^2{\sigma }_2=0\end{array}$

After solving the AEs along with equations (13) and (14), we could determine the following solutions

 Family 1 $\epsilon =1$

In this situation, we produce the following solutions of the above AEs

Case 1

$\begin{array}{l}b=\sqrt{c^2+{\mu }^2},{\rho }_1=-2k_1{k}_2{\lambda }_1,{\rho }_2=b_0{\lambda }_3-k_2^2{\lambda }_1+a_2^2{\lambda }_2,a_0=0,a_1=0,b_2=0,\\ b_3=0,b_4=0,a_2=\sqrt{\frac{6k_1^2{\lambda }_3{\sigma }_2-k_1^2{\lambda }_1{\sigma }_1}{2{\lambda }_2{\sigma }_1}},b_1=\frac{6\mu k_1^2{\sigma }_2}{{\sigma }_1},\\ k_1=\sqrt{\frac{b_0{\sigma }_1(3b_0{\lambda }_2{\sigma }_1{\sigma }_2+{\lambda }_1{\sigma }_1{\sigma }_3-6{\lambda }_3{\sigma }_2{\sigma }_3)}{{\sigma }_2(18{\lambda }_3{\sigma }_2{\sigma }_3-6b_0{\lambda }_2{\sigma }_1{\sigma }_2-3{\lambda }_1{\sigma }_1{\sigma }_3)}},k_2=\frac{b_0^2{\lambda }_2{\sigma }_1^2-k_1^2{\lambda }_1{\sigma }_1{\sigma }_3+6k_1^2{\lambda }_3{\sigma }_2{\sigma }_3}{4b_0{\lambda }_1{\lambda }_2{\sigma }_1}.\end{array}$

From equations (8), (13), (14), and (17), we can deduce the traveling wave solution of equation (1) as follows

$\begin{array}{l}{u}_1=\sqrt{\frac{6k_1^2{\lambda }_3{\sigma }_2-k_1^2{\lambda }_1{\sigma }_1}{2{\lambda }_2{\sigma }_1}}\frac{\sqrt{c^2+{\mu }^2}\sinh {\xi }_1+c\cosh {\xi }_1}{\sqrt{c^2+{\mu }^2}\cosh {\xi }_1+c\sinh {\xi }_1+\mu }{e}^{i{\eta }_1},\\ v_1=\left[b_0+\frac{6\mu k_1^2{\sigma }_2}{{\sigma }_1}\frac{1}{\sqrt{c^2+{\mu }^2}\cosh {\xi }_1+c\sinh {\xi }_1+\mu }\right]e^{i{\eta }_1},\end{array}$

${\xi }_1=\frac{k_1\mathrm{\Gamma }({\gamma }_1+1)}{\beta }{x}^{\beta }-\frac{2k_1{k}_2{\lambda }_1\mathrm{\Gamma }({\gamma }_2+1)}{\alpha }{t}^{\alpha },$

${\eta }_1=\frac{\mathrm{\Gamma }({\gamma }_1+1)}{\beta }{k}_2{x}^{\beta }+\frac{(b_0{\lambda }_3-k_2^2{\lambda }_1+a_2^2{\lambda }_2)\mathrm{\Gamma }({\gamma }_2+1)}{\alpha }{t}^{\alpha }.$

Selecting the following parameters in $u_1$ , we can get some graphical simulation in Figures 1–3.

$\begin{array}{c}b=\sqrt{2},\mu =c=1,{\lambda }_1={\lambda }_2={\lambda }_3={\sigma }_1={\sigma }_2={\sigma }_3=1,a_0=a_1=b_2=b_3=b_4=0,\\ b_0=-1,b_1=\frac{16}{7},a_2=2\sqrt{\frac{5}{21}},k_1=-2\sqrt{\frac{2}{21}},k_2=-\frac{61}{84},{\rho }_1=-\frac{61}{21}\sqrt{\frac{2}{21}},{\rho }_2=-\frac{4057}{7056}.\end{array}$

Figure 1.

The 3D plot, 2D plot, and contour plot of $\left|u_1\right|$ with $\alpha =1,\beta =1,{\gamma }_1={\gamma }_2=1$ .

Figure 2.

The 3D plot, 2D plot, and contour plot of $\mathrm{R}\mathrm{e}{u}_1$ with $\alpha =0.08,\beta =0.09,{\gamma }_1=2,{\gamma }_2=3$ .

Figure 3.

The 3D plot, 2D plot, and contour plot of $\mathrm{I}\mathrm{m}{u}_1$ with $\alpha =1,\beta =1,{\gamma }_1=2,{\gamma }_2=3$ .

Case 2

$\begin{array}{l}\mu =0,a_0=0,a_2=0,b_0=0,b_1=0,b_3=0,b_4=0,a_1=\frac{6{\lambda }_3{\sigma }_2-{\lambda }_1{\sigma }_1}{3{\lambda }_2{\sigma }_2}\sqrt{\frac{{\sigma }_3\left(c^2-b^2\right)}{{\sigma }_1}},\\ b_2=\frac{2{\sigma }_3\left(c^2-b^2\right)\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right)}{3{\lambda }_2{\sigma }_1{\sigma }_2},{\rho }_1=\frac{\left({\lambda }_1{\sigma }_1{\sigma }_3-6{\lambda }_3{\sigma }_2{\sigma }_3\right)\sqrt{6{\lambda }_3{\sigma }_2{\sigma }_3-{\lambda }_1{\sigma }_1{\sigma }_3}}{54\sqrt{2}{\lambda }_2^{3/2}{\sigma }_2^2},\\ {\rho }_2=\frac{\left(6{\lambda }_3{\sigma }_2-{\lambda }_1{\sigma }_1\right){\sigma }_3\left(72{\lambda }_1^2{\lambda }_2+{\lambda }_1{\sigma }_1{\sigma }_3-6{\lambda }_3{\sigma }_2{\sigma }_3\right)}{1296{\lambda }_1{\lambda }_2^2{\sigma }_2^2},k_1=\frac{\sqrt{6{\lambda }_3{\sigma }_2{\sigma }_3-{\lambda }_1{\sigma }_1{\sigma }_3}}{3{\sigma }_2\sqrt{2{\lambda }_2}},\\ k_2=\frac{6{\lambda }_3{\sigma }_2{\sigma }_3-{\lambda }_1{\sigma }_1{\sigma }_3}{36{\lambda }_1{\lambda }_2{\sigma }_2}.\end{array}$

From case 2, we can deduce the traveling wave solution of equation (1) as follows

$u_2=\frac{\frac{6{\lambda }_3{\sigma }_2-{\lambda }_1{\sigma }_1}{3{\lambda }_2{\sigma }_2}\sqrt{\frac{{\sigma }_3\left(c^2-b^2\right)}{{\sigma }_1}}}{b\cosh {\xi }_2+c\sinh {\xi }_2}{e}^{i{\eta }_2},v_2=\frac{\frac{2{\sigma }_3\left(c^2-b^2\right)\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right)}{3{\lambda }_2{\sigma }_1{\sigma }_2}}{{\left(b\cosh {\xi }_2+c\sinh {\xi }_2\right)}^2}{e}^{i{\eta }_2},$

${\xi }_2=\frac{\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }\sqrt{\frac{6{\lambda }_3{\sigma }_2{\sigma }_3-{\lambda }_1{\sigma }_1{\sigma }_3}{18{\lambda }_2{\sigma }_2^2}}{x}^{\beta }+\frac{\left({\lambda }_1{\sigma }_1{\sigma }_3-6{\lambda }_3{\sigma }_2{\sigma }_3\right)\mathrm{\Gamma }\left({\gamma }_2+1\right)}{54{\lambda }_2{\sigma }_2^2\alpha }\sqrt{\frac{6{\lambda }_3{\sigma }_2{\sigma }_3-{\lambda }_1{\sigma }_1{\sigma }_3}{2{\lambda }_2}}{t}^{\alpha },{\eta }_2=\frac{\left(6{\lambda }_3{\sigma }_2{\sigma }_3-{\lambda }_1{\sigma }_1{\sigma }_3\right)\mathrm{\Gamma }\left({\gamma }_1+1\right)}{36{\lambda }_1{\lambda }_2{\sigma }_2\beta }{x}^{\beta }+\frac{{\sigma }_3\left(6{\lambda }_3{\sigma }_2-{\lambda }_1{\sigma }_1\right)\left(72{\lambda }_1^2{\lambda }_2+{\lambda }_1{\sigma }_1{\sigma }_3-6{\lambda }_3{\sigma }_2{\sigma }_3\right)\mathrm{\Gamma }\left({\gamma }_2+1\right)}{1296{\lambda }_1{\lambda }_2^2{\sigma }_2^2\alpha }{t}^{\alpha }.$

Case 3

$\begin{array}{l}\mu =0,c=0,a_0=0,a_1=0,b_1=0,b_3=0,b_4=0,a_2=\pm \sqrt{\frac{k_2^2{\lambda }_1-b_0{\lambda }_3+{\rho }_2}{{\lambda }_2}},\\ b_2=\frac{12b^2{k}_1^2{\sigma }_2}{{\sigma }_1},{\rho }_1=-2k_1{k}_2{\lambda }_1,{\rho }_2=\frac{4b_0{k}_2{\lambda }_1{\lambda }_2-b_0^2{\lambda }_2{\sigma }_1-2k_2^2{\lambda }_1{\sigma }_3+2b_0{\lambda }_3{\sigma }_3}{2{\sigma }_3},\\ k_1=\sqrt{\frac{b_0{\sigma }_1\left(6{\lambda }_3{\sigma }_2{\sigma }_3-3b_0{\lambda }_2{\sigma }_1{\sigma }_2-{\lambda }_1{\sigma }_1{\sigma }_3\right)}{{\sigma }_2\left(24b_0{\lambda }_2{\sigma }_1{\sigma }_2+12{\lambda }_1{\sigma }_1{\sigma }_3-72{\lambda }_3{\sigma }_2{\sigma }_3\right)}},\\ k_2=\frac{6b_0^2{\lambda }_2^2{\sigma }_1^2{\sigma }_2^2+6b_0{\lambda }_2{\sigma }_1{\sigma }_2\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3+{\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right)}^2{\sigma }_3^2}{12{\lambda }_1{\lambda }_2{\sigma }_2\left(2b_0{\lambda }_2{\sigma }_1{\sigma }_2+{\lambda }_1{\sigma }_1{\sigma }_3-6{\lambda }_3{\sigma }_2{\sigma }_3\right)}.\end{array}$

In this result, we can deduce the traveling wave solution of equation (1) as follows

$\begin{array}{l}{u}_3=\pm \sqrt{\frac{k_2^2{\lambda }_1-b_0{\lambda }_3+{\rho }_2}{{\lambda }_2}}\tanh \left(\frac{k_1\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }-\frac{2k_1{k}_2{\lambda }_1\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }\right)e^{i\left(\frac{k_2\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }+\frac{{\rho }_2\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }\right)},\\ v_3=\left[b_0+\frac{12k_1^2{\sigma }_2}{{\sigma }_1}{\mathrm{sech}}^2\left(\frac{k_1\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }-\frac{2k_1{k}_2{\lambda }_1\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }\right)\right]e^{i\left(\frac{k_2\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }+\frac{{\rho }_2\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }\right)}.\end{array}$

Selecting the following parameters in $u_3,v_3$ , we obtain the famous kink and anti-kind dark solitons pulse $u_3$ and valley-type soliton pulse solution $v_3$ which is simulated in Figure 4.

$\mu =c=b_1=b_3=b_4=a_0=a_1=0,b_0=b={\lambda }_1={\lambda }_2=1,{\sigma }_2=2,{\sigma }_3=\alpha =\beta ={\gamma }_1={\gamma }_2=1,$

${\lambda }_3={\sigma }_1=-1,b_2=-\frac{5}{7},a_2=\sqrt{\frac{55}{42}},k_2=\frac{13}{168},k_1=-\frac{1}{2}\sqrt{\frac{5}{42}},{\rho }_1=\frac{13}{168}\sqrt{\frac{5}{42}},{\rho }_2=-\frac{9913}{28224}.$

Selecting the following parameters in $v_3$ , we obtain the W-shaped soliton pulse shown in Figure 5.

${\lambda }_1={\sigma }_2=2,b={\lambda }_2={\lambda }_3={\sigma }_1={\sigma }_3=1,\mu =c=a_0=a_1=b_1=b_3=b_4=0,\alpha =\beta ={\gamma }_1={\gamma }_2=1,k_2=-\frac{61}{168},k_1=\frac{1}{\sqrt{21}},{\rho }_2=-\frac{4393}{14112},{\rho }_1=\frac{61}{42\sqrt{21}},b_0=-1,b_2=\frac{8}{7},a_2=2\sqrt{\frac{5}{21}}.$

Figure 4.

The 3D plot, 2D plot, and contour plot of $\left|v_3\right|$ .

Figure 5.

The W-shaped pulse solutions’ 3D plot, 2D plot, and contour plot of $\left|v_3\right|$ .

Case 4

$\begin{array}{l}\mu =0,b=0,b_1=0,b_3=0,a_0=0,a_1=0,b_4=0,a_2=\pm \sqrt{\frac{2\left(6k_1^2{\lambda }_3{\sigma }_2-k_1^2{\lambda }_1{\sigma }_1\right)}{{\lambda }_2{\sigma }_1}},\\ b_2=-\frac{12c^2{k}_1^2{\sigma }_2}{{\sigma }_1},{\rho }_1=-2k_1{k}_2{\lambda }_1,{\rho }_2=\frac{-2k_1^2{\lambda }_1{\sigma }_1-k_2^2{\lambda }_1{\sigma }_1+b_0{\lambda }_3{\sigma }_1+12k_1^2{\lambda }_3{\sigma }_2}{{\sigma }_1},\\ k_1=\sqrt{\frac{6b_0{\lambda }_3{\sigma }_1{\sigma }_2{\sigma }_3-3b_0^2{\lambda }_2{\sigma }_1^2{\sigma }_2-b_0{\lambda }_1{\sigma }_1^2{\sigma }_3}{12\left(2b_0{\lambda }_2{\sigma }_1{\sigma }_2^2+{\lambda }_1{\sigma }_1{\sigma }_2{\sigma }_3-6{\lambda }_3{\sigma }_2^2{\sigma }_3\right)}},k_2=\frac{6b_0{\lambda }_2{\sigma }_1{\sigma }_2+24k_1^2{\lambda }_2{\sigma }_2^2+{\lambda }_1{\sigma }_1{\sigma }_3-6{\lambda }_3{\sigma }_2{\sigma }_3}{12{\lambda }_1{\lambda }_2{\sigma }_2}.\end{array}$

From case 4, we have

$u_4=\pm \sqrt{\frac{2\left(6k_1^2{\lambda }_3{\sigma }_2-k_1^2{\lambda }_1{\sigma }_1\right)}{{\lambda }_2{\sigma }_1}}\coth \left(\frac{k_1\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }-\frac{2k_1{k}_2{\lambda }_1\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }\right)e^{i\left(\frac{k_2\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }+\frac{{\rho }_2\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }\right)},v_4=\left[b_0-\frac{12c^2{k}_1^2{\sigma }_2}{{\sigma }_1}\mathrm{csch}{}^2\left(\frac{k_1\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }-\frac{2k_1{k}_2{\lambda }_1\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }\right)\right]e^{i\left(\frac{k_2\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }+\frac{{\rho }_2\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }\right)}.$

The blow-up pulse solution $v_4$ is deduced in Figure 6 after selecting the following parameters

$\begin{array}{l}{b}_0=-1,{\lambda }_1=2,{\lambda }_2={\lambda }_3={\sigma }_1={\sigma }_3=c=1,{\sigma }_2=2,\mu =b=b_1=b_3=b_4=a_0=a_1=0,\alpha =\beta =1,\\ {\gamma }_1={\gamma }_2=1,b_2=-\frac{8}{7},a_2=2\sqrt{\frac{5}{21}},k_1=\frac{1}{\sqrt{21}},k_2=-\frac{61}{168},{\rho }_1=\frac{61}{42\sqrt{21}},{\rho }_2=-\frac{4393}{14112}.\end{array}$

Figure 6.

The blow-up pulse solutions’ 3D plot, 2D plot, and contour plot of $\left|v_4\right|$ .

Remark 2

If we let $\alpha =\beta =1$ , the solutions $u_3,v_3,u_4,v_4$ turn to the solutions (32) and (33) in Refs. [60].

Family 2 $\epsilon =-1$

In this situation, we produce the following solutions of the above AEs.

Case 1

$\begin{array}{l}\mu =0,a_0=0,a_1=0,b_1=0,b_3=0,b_4=0,a_2=\pm \frac{\sqrt{-4b_0{k}_2{\lambda }_1+b_0^2{\sigma }_1}}{\sqrt{2}\sqrt{{\sigma }_3}},b_2=-\frac{12\left(b^2+c^2\right)k_1^2{\sigma }_2}{{\sigma }_1},\\ {\rho }_2=\frac{4b_0{k}_2{\lambda }_1{\lambda }_2-b_0^2{\lambda }_2{\sigma }_1-2k_2^2{\lambda }_1{\sigma }_3+2b_0{\lambda }_3{\sigma }_3}{2{\sigma }_3},k_1=\sqrt{\frac{-3b_0^2{\lambda }_2{\sigma }_1^2{\sigma }_2-b_0{\lambda }_1{\sigma }_1^2{\sigma }_3+6b_0{\lambda }_3{\sigma }_1{\sigma }_2{\sigma }_3}{72{\lambda }_3{\sigma }_2^2{\sigma }_3-24b_0{\lambda }_2{\sigma }_1{\sigma }_2^2-12{\lambda }_1{\sigma }_1{\sigma }_2{\sigma }_3}},\\ {\rho }_1=-2k_1{k}_2{\lambda }_1,k_2=\frac{6b_0^2{\lambda }_2^2{\sigma }_1^2{\sigma }_2^2+6b_0{\lambda }_2{\sigma }_1{\sigma }_2\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3+{\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right)}^2{\sigma }_3^2}{12{\lambda }_1{\lambda }_2{\sigma }_2\left(2b_0{\lambda }_2{\sigma }_1{\sigma }_2+{\lambda }_1{\sigma }_1{\sigma }_3-6{\lambda }_3{\sigma }_2{\sigma }_3\right)}.\end{array}$

From (9), (13), (14), and (17), we can deduce the traveling wave solution of equation (1) as follows

$\begin{array}{l}{u}_5=\pm \sqrt{\frac{b_0^2{\sigma }_1-4b_0{k}_2{\lambda }_1}{2{\sigma }_3}}\frac{-b\sin {\xi }_5+c\cos {\xi }_5}{b\cos {\xi }_5+c\sin {\xi }_5}{e}^{i{\eta }_5},v_5=\left[b_0-\frac{\frac{12\left(b^2+c^2\right)k_1^2{\sigma }_2}{{\sigma }_1}}{{\left(b\cos {\xi }_5+c\sin {\xi }_5\right)}^2}\right]e^{i{\eta }_5}.\\ {\xi }_5=\frac{k_1\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }-\frac{2k_1{k}_2{\lambda }_1\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha },{\eta }_5=\frac{k_2\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }+\frac{{\rho }_2\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }.\end{array}$

If we select the following parameters, the periodic pulse solutions can be simulated in Figures 7 and 8.

$a_0=a_1=b_1=b_3=b_4=\mu =0,b=b_0=1,{\lambda }_2={\lambda }_3={\sigma }_1={\sigma }_3=c=1,{\lambda }_1={\sigma }_2=2,\alpha =\beta =1,$

${\gamma }_1={\gamma }_2=1,b_2=-\frac{4}{3},a_2=\frac{\sqrt{5}}{3},{\rho }_1=\frac{1}{108},{\rho }_2=\frac{1151}{2592},k_1=\frac{1}{6},k_2=-\frac{1}{72}.$

Obviously, the solution $u_5,v_5$ contain the following solutions.

$u_{5.1}=\pm \sqrt{\frac{b_0^2{\sigma }_1-4b_0{k}_2{\lambda }_1}{2{\sigma }_3}}\tan {\xi }_5{e}^{i{\eta }_5},v_{5.1}=\left[b_0-\frac{12k_1^2{\sigma }_2}{{\sigma }_1}\sec {}^2{\xi }_5\right]e^{i{\eta }_5},c=0.$

$u_{5.2}=\pm \sqrt{\frac{b_0^2{\sigma }_1-4b_0{k}_2{\lambda }_1}{2{\sigma }_3}}\cot {\xi }_5{e}^{i{\eta }_5},v_{5.2}=\left[b_0-\frac{12k_1^2{\sigma }_2}{{\sigma }_1}\csc {}^2{\xi }_5\right]e^{i{\eta }_5},b=0.$

Figure 7.

The periodic pulse solutions’ 3D plot, 2D plot, and contour plot of $\mathrm{R}\mathrm{e}{u}_5$ .

Figure 8.

The 3D plot, 2D plot, and contour plot of $\mathrm{R}\mathrm{e}{v}_5$ with $\alpha =0.34,\beta =1$ .

Remark 3

If we let $\alpha =\beta =1$ , the solutions $u_{5.1},v_{5.1}$ and $u_{5.2},v_{5.2}$ turn to the solutions (34), (51) and (35), (52) in Refs. [60,61].

Case 2

$\begin{array}{l}\mu =0,a_0=0,a_2=0,b_1=0,b_3=0,b_4=0,{\rho }_1=-2k_1{k}_2{\lambda }_1,{\rho }_2=-k_1^2{\lambda }_1-k_2^2{\lambda }_1+b_0{\lambda }_3,\\ a_1=\sqrt{\frac{12\left(b^2+c^2\right)k_1^2{\lambda }_3{\sigma }_2-2b^2{k}_1^2{\sigma }_1^2{\lambda }_1-2c^2{k}_1^2{\sigma }_1^2{\lambda }_1}{{\sigma }_1{\lambda }_2}},b_2=-\frac{12\left(b^2+c^2\right)k_1^2{\sigma }_2}{{\sigma }_1},\\ k_1=\sqrt{\frac{3b_0{\lambda }_2{\sigma }_1{\sigma }_2+{\lambda }_1{\sigma }_1{\sigma }_3-6{\lambda }_3{\sigma }_2{\sigma }_3}{24{\lambda }_2{\sigma }_2^2}},k_2=\frac{b_0{\sigma }_1}{4{\lambda }_1}.\end{array}$

In this result, we have

$\begin{array}{l}{u}_6=\frac{\sqrt{\frac{12\left(b^2+c^2\right)k_1^2{\lambda }_3{\sigma }_2-2b^2{k}_1^2{\sigma }_1^2{\lambda }_1-2c^2{k}_1^2{\sigma }_1^2{\lambda }_1}{{\sigma }_1{\lambda }_2}}}{b\cos {\xi }_6+c\sin {\xi }_6}{e}^{i{\eta }_6},v_6=\left[b_0-\frac{\frac{12\left(b^2+c^2\right)k_1^2{\sigma }_2}{{\sigma }_1}}{{\left(b\cos {\xi }_6+c\sin {\xi }_6\right)}^2}\right]e^{i{\eta }_6},\\ {\xi }_6=\frac{k_1\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }-\frac{b_0{\sigma }_1{k}_1{\lambda }_1\mathrm{\Gamma }\left({\gamma }_2+1\right)}{2{\lambda }_1\alpha }{t}^{\alpha },{\eta }_6=\frac{k_2\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }+\left(\frac{b_0{\lambda }_3-k_1^2{\lambda }_1}{\alpha }-\frac{b_0^2{\sigma }_1^2}{16\alpha {\lambda }_1}\right)\mathrm{\Gamma }\left({\gamma }_2+1\right)t^{\alpha }.\end{array}$

Thus

$u_{6.1}=\sqrt{\frac{12k_1^2{\lambda }_3{\sigma }_2-2k_1^2{\sigma }_1^2{\lambda }_1}{{\sigma }_1{\lambda }_2}}\csc {\xi }_6{e}^{i{\eta }_6},v_{6.1}=\left[b_0-\frac{12k_1^2{\sigma }_2}{{\sigma }_1}\csc {}^2{\xi }_6\right]e^{i{\eta }_6},b=0.u_{6.2}\sqrt{\frac{12k_1^2{\lambda }_3{\sigma }_2-2k_1^2{\sigma }_1^2{\lambda }_1}{{\sigma }_1{\lambda }_2}}\sec {\xi }_6{e}^{i{\eta }_6},v_{6.2}=\left[b_0-\frac{12k_1^2{\sigma }_2}{{\sigma }_1}\sec {}^2{\xi }_6\right]e^{i{\eta }_6},c=0.$

The singular trigonometric function wave pulse can be simulated in Figure 9 by selecting

$a_0=a_2=b_1=b_3=b_4=\mu =b=0,b_0=c={\lambda }_2={\lambda }_3={\sigma }_3=1,{\lambda }_1=2,{\sigma }_1=-1,{\sigma }_2=-2,b_2=-4,$

$a_1=\sqrt{\frac{10}{3}},\alpha =\beta =1,{\gamma }_1={\gamma }_2=1,{\rho }_1=-\frac{1}{2\sqrt{6}},{\rho }_2=\frac{61}{96},k_1=-\frac{1}{\sqrt{6}},k_2=-\frac{1}{8}.$

Figure 9.

The 3D plot, 2D plot, and contour plot of the singular wave pulse $\mathrm{I}\mathrm{m}{u}_{6.1}$ .

Case 3

$\begin{array}{l}b=ic,i=\sqrt{-1},\mu =a_1=b_1=b_2=b_4=0,k_2=\frac{b_0{\lambda }_2{\sigma }_1-{\lambda }_3{\sigma }_3}{2{\lambda }_1{\lambda }_2},{\rho }_1=\frac{{\lambda }_3{\sigma }_3{k}_1-b_0{\lambda }_2{\sigma }_1{k}_1}{{\lambda }_2},\\ a_2=\sqrt{\frac{2b_0{\lambda }_3^3{\sigma }_3-b_0^2{\sigma }_1{\lambda }_2{\lambda }_3^2+2a_0^2{\sigma }_3{\lambda }_2{\lambda }_3^2}{4a_0^2{\lambda }_2^3{\sigma }_1+2{\lambda }_2{\sigma }_3{\lambda }_3^2}},b_3=-\frac{2a_0{\lambda }_2}{{\lambda }_3}\sqrt{\frac{2b_0{\lambda }_3^3{\sigma }_3-b_0^2{\sigma }_1{\lambda }_2{\lambda }_3^2+2a_0^2{\sigma }_3{\lambda }_2{\lambda }_3^2}{4a_0^2{\lambda }_2^3{\sigma }_1+2{\lambda }_2{\sigma }_3{\lambda }_3^2}},\\ {\rho }_2=a_0^2{\lambda }_2+b_0{\lambda }_3-\frac{{\left(b_0{\lambda }_2{\sigma }_1-{\lambda }_3{\sigma }_3\right)}^2}{4{\lambda }_1{\lambda }_2^2}-\frac{3{\lambda }_3^2\left(2b_0{\lambda }_3{\sigma }_3-b_0^2{\lambda }_2{\sigma }_1+2a_0^2{\lambda }_2{\sigma }_3\right)}{4a_0^2{\lambda }_2^2{\sigma }_1+2{\lambda }_3^2{\sigma }_3}\\ +a_2{\lambda }_3\sqrt{\frac{4b_0{\lambda }_2{\lambda }_3{\sigma }_3-2{\sigma }_1{b}_0^2{\lambda }_2^2+4{\sigma }_3{a}_0^2{\lambda }_2^2}{2{\sigma }_1{a}_0^2{\lambda }_2^2+{\sigma }_3{\lambda }_3^2}},k_1\in R.\end{array}$

From case 3, we obtain

$\begin{array}{l}{u}_7=\left[a_0+\sqrt{\frac{2b_0{\lambda }_3^3{\sigma }_3-b_0^2{\sigma }_1{\lambda }_2{\lambda }_3^2+2a_0^2{\sigma }_3{\lambda }_2{\lambda }_3^2}{4a_0^2{\lambda }_2^3{\sigma }_1+2{\lambda }_2{\sigma }_3{\lambda }_3^2}}\frac{\tan {\xi }_7-i}{1-i\tan {\xi }_7}\right]e^{i{\eta }_7},\\ v_7=\left[b_0-\frac{2a_0{\lambda }_2}{{\lambda }_3}\sqrt{\frac{2b_0{\lambda }_3^3{\sigma }_3-b_0^2{\sigma }_1{\lambda }_2{\lambda }_3^2+2a_0^2{\sigma }_3{\lambda }_2{\lambda }_3^2}{4a_0^2{\lambda }_2^3{\sigma }_1+2{\lambda }_2{\sigma }_3{\lambda }_3^2}}\frac{\tan {\xi }_7-i}{1-i\tan {\xi }_7}\right]e^{i{\eta }_7},\\ {\xi }_7=\frac{k_1\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }+\frac{\left({\lambda }_3{\sigma }_3{k}_1-b_0{\lambda }_2{\sigma }_1{k}_1\right)\mathrm{\Gamma }\left({\gamma }_2+1\right)}{{\lambda }_2\alpha }{t}^{\alpha },\\ {\eta }_7=\frac{k_2\mathrm{\Gamma }\left({\gamma }_1+1\right)}{\beta }{x}^{\beta }+\frac{\left(b_0{\lambda }_3-k_1^2{\lambda }_1-\frac{b_0^2{\sigma }_1^2}{16{\lambda }_1^2}{\lambda }_1\right)\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }.\end{array}$

In the following, let us consider the Gʹ/(bGʹ + G + a) method, we assume equation (16) have the following solutions

$\begin{array}{l}P=c_0+c_{1}F,\\ Q=d_0+d_{1}F+d_2{F}^2.\end{array}$

(18)

Substituting equations (18) and (12) into equation (16) and setting the coefficients of $F^i$ zero yield the following AEs

$\begin{array}{l}{F}^0:c_1(\lambda -2\mu )\mu k_1^2{\lambda }_1+c_0(-k_2^2{\lambda }_1+c_0^2{\lambda }_2+d_0{\lambda }_3-{\rho }_2)=0,\\ F^1:c_0{d}_1{\lambda }_3+c_1(({\lambda }^2+2\mu -6\lambda \mu +6{\mu }^2)k_1^2{\lambda }_1-k_2^2{\lambda }_1+3c_0^2{\lambda }_2+d_0{\lambda }_3-{\rho }_2)=0,\\ F^2:3c_0{c}_1^2{\lambda }_2+c_0{d}_2{\lambda }_3+c_1(-3({\lambda }^2+2\mu (1+\mu )-\lambda (1+3\mu ))k_1^2{\lambda }_1+d_1{\lambda }_3)=0,\\ F^3:c_1{(2(1-\lambda +\mu )}^2{k}_1^2{\lambda }_1+c_1^2{\lambda }_2+d_2{\lambda }_3)=0,\end{array}\begin{array}{l}{F}^0:d_0^2{\sigma }_1+2(-H+\lambda \mu d_1{k}_1^2{\sigma }_2-2{\mu }^2{d}_1{k}_1^2{\sigma }_2+2{\mu }^2{d}_2{k}_1^2{\sigma }_2+c_0^2{\sigma }_3)-4d_0{k}_2{\lambda }_1=0,\\ F^1:6\lambda \mu d_2{k}_1^2{\sigma }_2+d_1(-2k_2{\lambda }_1+d_0{\sigma }_1+({\lambda }^2+2\mu -6\lambda \mu +6{\mu }^2)k_1^2{\sigma }_2)+2c_0{c}_1{\sigma }_3-12{\mu }^2{d}_2{k}_1^2{\sigma }_2=0,\\ F^2:d_1^2{\sigma }_1+2d_2(-2k_2{\lambda }_1+d_0{\sigma }_1+4({\lambda }^2+2\mu -6\lambda \mu +6{\mu }^2)k_1^2{\sigma }_2)+2c_1^2{\sigma }_3\\ -6({\lambda }^2+2\mu (1+\mu )-\lambda (1+3\mu ))d_1{k}_1^2{\sigma }_2=0,\\ F^3:10({\lambda }^2+2\mu (1+\mu )-\lambda (1+3\mu ))d_2{k}_1^2{\sigma }_2-d_1{(d_2{\sigma }_1+2(1-\lambda +\mu )}^2{k}_1^2{\sigma }_2)=0,\\ F^4:d_2{(d_2{\sigma }_1+12(1-\lambda +\mu )}^2{k}_1^2{\sigma }_2)=0.\end{array}$

After solving the AEs along with equations (13) and (14), we can get the following solutions

Case 1

$\begin{array}{l}b=1,\mu =0,c_0=\sqrt{\frac{H}{{\sigma }_3}},c_1=-\frac{2H\left(\lambda -1\right)}{\lambda \sqrt{H{\sigma }_3}},d_0=0,d_1=-\frac{24H{\lambda }_2{\sigma }_2\left(\lambda -1\right)}{\lambda \left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3},d_2=\frac{24H{\lambda }_2{\sigma }_2{\left(\lambda -1\right)}^2}{{\lambda }^2\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3},\\ {\rho }_1=-\frac{{(({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2)}^2{\sigma }_3^2-12H{\lambda }_2^2{\sigma }_1{\sigma }_2^2)\sqrt{H{\sigma }_1}}{3{\sigma }_2\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3\sqrt{-2{\lambda }_2{\sigma }_3{\lambda }^2\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right)}},{\rho }_2=\frac{H{\lambda }_2}{{\sigma }_3}-\frac{{(({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2)}^2{\sigma }_3^2-12H{\lambda }_2^2{\sigma }_1{\sigma }_2^2{)}^2}{144{\lambda }_1{\lambda }_2^2{\sigma }_2^2{\sigma }_3^2{\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right)}^2},\\ k_1=\frac{\sqrt{2H{\lambda }_2{\sigma }_1}}{\sqrt{-{\lambda }^2\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3}},k_2=\frac{{\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right)}^2{\sigma }_3^2-12H{\lambda }_2^2{\sigma }_1{\sigma }_2^2}{12{\lambda }_1{\lambda }_2{\sigma }_2\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3}.\end{array}$

Case 2

$\begin{array}{l}b=1,\lambda =0,d_2=-\frac{12{\left(1+\mu \right)}^2{k}_1^2{\sigma }_2}{{\sigma }_1},d_1=\frac{24\mu \left(1+\mu \right)k_1^2{\sigma }_2}{{\sigma }_1},c_1=k_1\left(1+\mu \right)\sqrt{\frac{12{\lambda }_3{\sigma }_2-2{\lambda }_1{\sigma }_1}{{\lambda }_2{\sigma }_1}},\\ c_0=-\mu k_1\sqrt{\frac{12{\lambda }_3{\sigma }_2-2{\lambda }_1{\sigma }_1}{{\lambda }_2{\sigma }_1}},d_0=\frac{{\rho }_2{\sigma }_1-2\mu k_1^2{\lambda }_1{\sigma }_1+k_2^2{\lambda }_1{\sigma }_1-12{\mu }^2{k}_1^2{\lambda }_3{\sigma }_2}{{\lambda }_3{\sigma }_1},k_2,{\rho }_2\in R,\\ k_1=\sqrt{\frac{12k_2{\lambda }_1{\lambda }_2{\lambda }_3{\sigma }_2-6k_2^2{\lambda }_1{\lambda }_2{\sigma }_1{\sigma }_2-6{\lambda }_2{\rho }_2{\sigma }_1{\sigma }_2-{\lambda }_1{\lambda }_3{\sigma }_1{\sigma }_3+6{\lambda }_3^2{\sigma }_2{\sigma }_3}{12\mu {\lambda }_2{\sigma }_2\left(4{\lambda }_3{\sigma }_2-{\lambda }_1{\sigma }_1\right)}},{\rho }_1=-2k_1{k}_2{\lambda }_1,\end{array}\begin{array}{l}H=\frac{1}{72}(-\frac{108k_2^2{\lambda }_1^2}{{\sigma }_1}+\frac{36k_2^2{\lambda }_1^4{\sigma }_1}{{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}-\frac{144k_2^3{\lambda }_1^3{\sigma }_1{\sigma }_2}{{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}-\frac{144k_2{\lambda }_1^2{\rho }_2{\sigma }_1{\sigma }_2}{{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}+\frac{144k_2^4{\lambda }_1^2{\sigma }_1{\sigma }_2^2}{{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}+\\ \frac{288k_2^2{\lambda }_1{\rho }_2{\sigma }_1{\sigma }_2^2}{{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}+\frac{144{\rho }_2^2{\sigma }_1{\sigma }_2^2}{{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}-\frac{72k_2^2{\lambda }_1^3}{{\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2}+\frac{144k_2^3{\lambda }_1^2{\sigma }_2}{{\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2}+\frac{144k_2{\lambda }_1{\rho }_2{\sigma }_2}{{\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2}+\frac{54k_2^2{\lambda }_1{\sigma }_3}{{\lambda }_2}+\\ \frac{54{\rho }_2{\sigma }_3}{{\lambda }_2}-\frac{108k_2{\lambda }_1{\lambda }_3{\sigma }_3}{{\lambda }_2{\sigma }_1}+\frac{12k_2{\lambda }_1^3{\lambda }_3{\sigma }_1{\sigma }_3}{{\lambda }_2{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}-\frac{6k_2^2{\lambda }_1^3{\sigma }_1^2{\sigma }_3}{{\lambda }_2{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}-\frac{6{\lambda }_1^2{\rho }_2{\sigma }_1^2{\sigma }_3}{{\lambda }_2{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}-\frac{27{\lambda }_3^2{\sigma }_3^2}{{\lambda }_2^2{\sigma }_1}\\ +\frac{{\lambda }_1^2{\sigma }_1{\sigma }_3^2}{{\lambda }_2^2{\sigma }_2^2}-\frac{4{\lambda }_1{\lambda }_3{\sigma }_3^2}{{\lambda }_2^2{\sigma }_2}+\frac{{\lambda }_1^2{\lambda }_3^2{\sigma }_1{\sigma }_3^2}{{\lambda }_2^2{\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}^2}+\frac{2{\lambda }_1{\lambda }_3^2{\sigma }_3^2}{{\lambda }_2^2\left({\lambda }_1{\sigma }_1-4{\lambda }_3{\sigma }_2\right)}).\end{array}$

From case 1, we can determine the following solutions

$u_8=(\sqrt{\frac{H}{{\sigma }_3}}-\frac{2H(\lambda -1)}{\lambda \sqrt{H{\sigma }_3}}{F}_8)e^{i{\eta }_8},v_8=(-\frac{24H{\lambda }_2{\sigma }_2(\lambda -1)}{({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2)\lambda {\sigma }_3}{F}_8+\frac{24H{\lambda }_2{\sigma }_2{(\lambda -1)}^2}{({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2){\sigma }_3{\lambda }^2}{F}_8^2)e^{i{\eta }_8}.{\xi }_8=\frac{\mathrm{\Gamma }({\gamma }_1+1)\sqrt{2H{\lambda }_2{\sigma }_1}}{\beta \sqrt{{\lambda }^2(6{\lambda }_3{\sigma }_2-{\lambda }_1{\sigma }_1){\sigma }_3}}{x}^{\beta }-\frac{{(({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2)}^2{\sigma }_3^2-12H{\lambda }_2^2{\sigma }_1{\sigma }_2^2)\mathrm{\Gamma }({\gamma }_2+1)\sqrt{H{\sigma }_1}}{3{\sigma }_2({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2){\sigma }_3\alpha \sqrt{2{\lambda }_2{\sigma }_3{\lambda }^2(6{\lambda }_3{\sigma }_2-{\lambda }_1{\sigma }_1)}}{t}^{\alpha },$

$\begin{array}{l}{\eta }_8=\frac{{(({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2)}^2{\sigma }_3^2-12H{\lambda }_2^2{\sigma }_1{\sigma }_2^2)\mathrm{\Gamma }({\gamma }_1+1)}{12{\lambda }_1{\lambda }_2{\sigma }_2({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2){\sigma }_3\beta }{x}^{\beta }\\ +(\frac{H{\lambda }_2}{{\sigma }_3}-\frac{{(({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2)}^2{\sigma }_3^2-12H{\lambda }_2^2{\sigma }_1{\sigma }_2^2{)}^2}{144{\lambda }_1{\lambda }_2^2{\sigma }_2^2{\sigma }_3^2{({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2)}^2})\frac{\mathrm{\Gamma }({\gamma }_2+1)}{\alpha }{t}^{\alpha },\end{array}$

$F_8=\frac{\lambda p_1}{\left(\lambda -1\right)p_1-p_2{e}^{\lambda {\xi }_8}}=\frac{-2\lambda p_1\sinh \left(\frac{\lambda }{2}{\xi }_8\right)+2\lambda p_1\cosh \left(\frac{\lambda }{2}{\xi }_8\right)}{2\left(p_1-\lambda p_1-p_2\right)\sinh \left(\frac{\lambda }{2}{\xi }_8\right)+2\left(\lambda p_1-p_1-p_2\right)\cosh \left(\frac{\lambda }{2}{\xi }_8\right)}.$

The singular solitary wave pulse and classical bell-shape bright soliton pulse are simulated in Figures 10–13 after selecting

${\lambda }_1=2,{\lambda }_2={\lambda }_3={\sigma }_1={\sigma }_3=b=1,{\sigma }_2=2,\mu =0,\lambda =2,H=1,c_0=1,c_1=-1,\alpha =\beta =1,$

${\gamma }_{\mathrm{1,2}}=p_{\mathrm{1,2}}=1,d_0=0,d_1=\frac{12}{5},d_2=-\frac{6}{5},{\rho }_1=\frac{13}{60\sqrt{5}},{\rho }_2=\frac{7031}{7200},k_1=\frac{1}{2\sqrt{5}},k_2=-\frac{13}{120}.$

In particular, by selecting different $p_1,p_2$ , we have the following soliton pulse solutions

$\begin{array}{l}{u}_{8.1}=\left[\sqrt{\frac{H}{{\sigma }_3}}-\frac{2H\left(\lambda -1\right)}{\lambda \sqrt{H{\sigma }_3}}{F}_{8.1}\right]e^{i{\eta }_8},v_{8.1}=\left(-\frac{24H{\lambda }_2{\sigma }_2\left(\lambda -1\right)}{\lambda \left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3}{F}_{8.1}+\frac{24H{\lambda }_2{\sigma }_2{\left(\lambda -1\right)}^2}{\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3{\lambda }^2}{F}_{8.1}^2\right)e^{i{\eta }_8},\\ F_{8.1}=\frac{\lambda }{2\left(\lambda -1\right)}-\frac{\lambda }{2\left(\lambda -1\right)}\tanh \left(\frac{\lambda }{2}{\xi }_8\right).p_1=1,p_2=-1.\end{array}\begin{array}{l}{u}_{8.2}=\left[\sqrt{\frac{H}{{\sigma }_3}}-\frac{2H\left(\lambda -1\right)}{\lambda \sqrt{H{\sigma }_3}}{F}_{8.2}\right]e^{i{\eta }_8},v_{8.2}=\left(-\frac{24H{\lambda }_2{\sigma }_2\left(\lambda -1\right)}{\lambda \left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3}{F}_{8.2}+\frac{24H{\lambda }_2{\sigma }_2{\left(\lambda -1\right)}^2}{\left({\lambda }_1{\sigma }_1-6{\lambda }_3{\sigma }_2\right){\sigma }_3{\lambda }^2}{F}_{8.2}^2\right)e^{i{\eta }_8},\\ F_{8.2}=\frac{\lambda }{2\left(\lambda -1\right)}-\frac{\lambda }{2\left(\lambda -1\right)}\coth \left(\frac{\lambda }{2}{\xi }_8\right).p_1=p_2=1.\end{array}$

From case 2, we get

$\begin{array}{l}{u}_9=\left(-\mu k_1\sqrt{\frac{12{\lambda }_3{\sigma }_2-2{\lambda }_1{\sigma }_1}{{\lambda }_2{\sigma }_1}}+k_1\left(1+\mu \right)\sqrt{\frac{12{\lambda }_3{\sigma }_2-2{\lambda }_1{\sigma }_1}{{\lambda }_2{\sigma }_1}}{F}_9\right)e^{i{\eta }_9},\\ v_9=\left(\frac{{\rho }_2{\sigma }_1-2\mu k_1^2{\lambda }_1{\sigma }_1+k_2^2{\lambda }_1{\sigma }_1-12{\mu }^2{k}_1^2{\lambda }_3{\sigma }_2}{{\lambda }_3{\sigma }_1}+\frac{24\mu \left(1+\mu \right)k_1^2{\sigma }_2}{{\sigma }_1}{F}_9-\frac{12{\left(1+\mu \right)}^2{k}_1^2{\sigma }_2}{{\sigma }_1}{F}_9^2\right)e^{i{\eta }_9},\end{array}$

${\xi }_9=\frac{\mathrm{\Gamma }\left({\gamma }_1+1\right)k_1}{\beta }{x}^{\beta }-\frac{2k_1{k}_2{\lambda }_1\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha },\hspace{0.17em}{\eta }_9=\frac{\mathrm{\Gamma }\left({\gamma }_1+1\right)k_2}{\beta }{x}^{\beta }+\frac{{\rho }_2\mathrm{\Gamma }\left({\gamma }_2+1\right)}{\alpha }{t}^{\alpha }.$

When $\mu <0$ , we have

$F_{9.1}=\frac{\left[-2\sqrt{-\mu }\left(c_2+c_1\right)\right]\sinh \left(\sqrt{-\mu }{\xi }_9\right)+\left[-2\sqrt{-\mu }\left(c_2-c_1\right)\right]\cosh \left(\sqrt{-\mu }{\xi }_9\right)}{\left[-2\left(c_2-c_1\right)-2\sqrt{-\mu }\left(c_2+c_1\right)\right]\sinh \left(\sqrt{-\mu }{\xi }_9\right)+\left[-2\left(c_2+c_1\right)-2\sqrt{-\mu }\left(c_2-c_1\right)\right]\cosh \left(\sqrt{-\mu }{\xi }_9\right)}.$

If selecting

$\mu =-2,b=1,{\lambda }_1=-2,{\lambda }_2={\lambda }_3={\sigma }_1=1,{\sigma }_2=2,{\sigma }_3=1,\lambda =2,c_0=\sqrt{\frac{77}{15}},\alpha =\beta =1,{\gamma }_{\mathrm{1,2}}=p_{\mathrm{1,2}}=1,c_1=-\sqrt{\frac{77}{60}},d_0=-\frac{173}{30},d_1=\frac{22}{5},d_2=-\frac{11}{10},{\rho }_1=\sqrt{\frac{11}{15}},{\rho }_2=1,k_1=\frac{1}{4}\sqrt{\frac{11}{15}},k_2=1,H=-\frac{9611}{1800}.$

We obtain the unbounded solitary wave solution $v_{9.1}$ which is simulated in Figure 14.

When $\mu >0$ , we have

$F_{9.2}=\frac{\sqrt{\mu }{c}_2\cos \left(\sqrt{\mu }{\xi }_9\right)-\sqrt{\mu }{c}_1\sin \left(\sqrt{\mu }{\xi }_9\right)}{\left(c_1+\sqrt{\mu }{c}_2\right)\cos \left(\sqrt{\mu }{\xi }_9\right)+\left(c_2-\sqrt{\mu }{c}_1\right)\sin \left(\sqrt{\mu }{\xi }_9\right)}.$

If we select

$p_1=p_2=1,\mu =2,b=1,{\lambda }_1=-2,{\lambda }_2=1,{\lambda }_3=-1,{\sigma }_1=1,{\sigma }_2=-2,{\sigma }_3=1,\lambda =2,c_0=-\sqrt{\frac{259}{15}},c_1=\sqrt{\frac{777}{20}},d_0=\frac{437}{30},d_1=-\frac{222}{5},d_2=\frac{333}{10},{\rho }_1=\sqrt{\frac{37}{15}},{\rho }_2=1,k_1=\frac{1}{4}\sqrt{\frac{37}{15}},k_2=1,H=-\frac{18059}{1800}.$

Thus, the unbounded periodic wave solution $v_{9.2}$ can be simulated in Figure 15.

We can easily find that $F_{9.1},F_{9.2}$ contain the following solutions

$F_{\mathrm{9.1.1}}=\frac{\mu }{2}+\frac{\sqrt{-\mu }}{1+\mu }\coth \left(\sqrt{-\mu }{\xi }_9\right),p_2=\frac{\sqrt{-\mu }-1}{\sqrt{-\mu }+1}{p}_1.$

$F_{\mathrm{9.1.2}}=-\frac{1}{1+\mu }+\frac{\sqrt{-\mu }}{1+\mu }\tanh \left(\sqrt{-\mu }{\xi }_9\right),p_2=\frac{1-\sqrt{-\mu }}{1+\sqrt{-\mu }}{p}_1.$

$F_{\mathrm{9.2.1}}=\frac{\mu }{1+\mu }+\frac{\sqrt{\mu }}{1+\mu }\cot \left(\sqrt{\mu }{\xi }_9\right),p_1=-\sqrt{\mu }{p}_2.$

$F_{\mathrm{9.2.2}}=\frac{\mu }{1+\mu }-\frac{\sqrt{\mu }}{1+\mu }\tan \left(\sqrt{\mu }{\xi }_9\right),p_2=\sqrt{\mu }{p}_1.$

Figure 10.

The 3D plot, 2D plot, and contour plot of the singular wave pulse $\mathrm{R}\mathrm{e}{v}_8$ .

Figure 11.

The 3D plot, 2D plot, and contour plot of the solitary wave pulse $\mathrm{I}\mathrm{m}{v}_{8.1}$ .

Figure 12.

The 3D plot, 2D plot, and contour plot of the famous bell-shape soliton pulse $\left|v_{8.1}\right|$ .

Figure 13.

The 3D plot, 2D plot and contour plot of $\mathrm{R}\mathrm{e}{u}_{8.1}$ .

Figure 14.

The 3D plot, 2D plot, and contour plot of $\mathrm{R}\mathrm{e}{v}_{9.1}$ .

Figure 15.

The 3D plot, 2D plot, and contour plot of $\mathrm{R}\mathrm{e}{v}_{9.2}$ .

Results and discussion

After utilizing the two methods, we get many types of exact solutions of equation (1), including the bell-shape bright soliton pulse solution, the kink and anti-kink dark soliton pulse solution, the blow-up pattern solution, the W-shape soliton solution, the mixed solitary wave pattern solution, the periodic wave solution, etc. Some structures of these solutions are simulated in Figures 1–15. The visualization can help us to better understand the dynamic behavior, physical, and propagation characteristics of the coupled model. For example, by selecting $a=0.34,\beta =1$ and $a=0.1,\beta =1$ in $v_5$ , we can find the waveform of the periodic solution changes dramatically in Figure 16(a), while with the increase of the frequency $k_1$ , the W-shape modulus of $v_3$ is shrunk to a valley-type pulse shown in Figure 16(b). Figure 11 shows that Im $v_{8.1}$ decays sharply to zero with the increase of time $t$ . From Figure 13, it is easy to find that the real part of $u_{8.1}$ produced a continuous type of distortion at the origin point while an discontinuous point occurred between the region $(-\mathrm{10,0})$ for the real part of $v_{9.1}$ shown in Figure 14. The simulations of imaginary part of mixed solitary wave pattern solution $u_1$ and modulus of blow-up pattern solution $v_4$ are shown in Figures 3 and 6. Figure 8 shows that the waveforms of Re $u_5$ transformed from the lower half plane to the upper half plane through the region $(-25,-5)$ , simulations of the other pattern solutions can be seen in Figures 1, 2, 4, 5, 7, 9, 10, and 12. We believe that these simulations can help us to further understand the inner structure of this model.

Figure 16.

The change of real part for $v_5$ with the different parameters $a,\beta$ .

Remark 4

All of the nine types of solutions and parameter values obtained in this paper have been checked by mathematica software, and they are found for the first time to our knowledge.

Conclusion

In conclusion, nine types of new exact solutions for the MFCNLS-KdV have been found after utilizing the modified (Gʹ/G, 1/G) and Gʹ/(bGʹ + G + a) methods. Some propagation behavior of these solutions are discussed and simulated, the graphs of which shows that these solitary wave solutions and periodic solutions are propagated through different pattern, these patterns are important for revealing the internal structure of equation (1), the efficient and significant methods can be used to many other nonlinear models such as Hirota–Satsuma KdV equation, Ginzburg–Landau equation, Boussinesq–Burgers equation, etc. The current definition of M-fractional derivatives still has great limitations, and it is difficult to characterize the necessary connection between two real number order derivatives. On the other hand, the unified definition of fractional derivatives needs to be further explored and developed for us in the future. Finally, all simulations of these solutions obtained in the present article have been checked by the mathematica software.