An investigation of two integro-differential KP hierarchy equations to find out closed form solitons in mathematical physics

Abstract

Nonlinear partial differential equations (NLPDEs) are widely utilized in engineering and physical research to represent many physical processes of naturalistic occurrences. In this paper, we investigate two well-known NLPDEs, namely, the (2 + 1)-dimensional first integro-differential KP hierarchy equation and the (2 + 1)-dimensional second integro-differential KP hierarchy equation, through a well-stable algorithm known as the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion approach for the first time. This algorithm is generally based on the expansion of function method and has the advantage of easy implementation and can provide a reliable solution to any NLPDEs. Employing the algorithm, we have been able to perceive the closed form solitons of the two chosen NLPDEs that physically represent the solitary wave solutions like, singular, singular periodic, bell, and anti-bell-shaped types of solitons. Furthermore, we explore the graphical manifestations of the obtained solutions, which are of the mentioned soliton types. From the findings of our in-depth study, we can state that the acquired solutions for the selected two equations may greatly aid to extracting the associated natural phenomena in mathematical physics such as fluid dynamics and ocean engineering.

Keywords:

• $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion approach
• (2 + 1)-dimensional first integro-differential KP hierarchy equation
• (2 + 1)-dimensional second integro-differential KP hierarchy equation
• Appropriate wave solutions
• Closed form soliton

1. Introduction

The extraction and investigation of the appropriate progressive wave solutions to the NLPDEs are vital to exploring the intricate phenomena in nature. The NLPDEs are present in various physical, technological, engineering, and scientific puzzles. For instance, the fields related to physics, such as particle physics, optical physics, plasma physics, solid-state physics, mechanics, and relativity, are essentially dependent on the study of NLPDEs. The implications of NLPDEs are not only limited to the mentioned field but also have many applications in biology and the medical sciences. Because of this, to explore the mysterious incidents of nature, we need appropriate solutions and analysis for the NLPDEs. Many experts have developed numerous competent and tactual methods to detect the appropriate solutions for NLPDEs. While some researchers relied on the Riccati equation method (Malwe, Betchewe, Doka, & Kofane, 2016; Tverdyi & Parovik, 2022), others used common expansion-type methods in their work (Abdou, 2007; Jiang, Sheng, & Qing, 2005). Many researchers in (Akinyemi et al., 2023; Alanazi, Ouahid, Al Shahrani, Abdou, & Kumar, 2023; Djennadi et al., 2021; El-Ganaini & Kumar, 2023; Kumar, Hamid, & Abdou, 2023; Kumar, Hamid, et al., 2023; Kumar & Rani, 2021, 2022; Miah, Iqbal, & Osman, 2023; Miah, Seadawy, Ali, & Akbar, 2019; Miah et al., 2023; Niwas & Kumar, 2023; Oad, Arshad, Shoaib, Lu, & Li, 2021; Rani, Kumar, & Mann, 2023; Redi, Obsie, & Shiferaw, 2018; Redi et al., 2018; Seadawy, Arshad, & Lu, 2020; Seadawy, Iqbal, & Lu, 2020; Seadawy, Rizvi, Ahmad, Younis, & Baleanu, 2021; Seadawy & Cheemaa, 2020; Shehzad, Seadawy, Wang, & Arshad, 2023; Tala-Tebue, Seadawy, Kamdoum-Tamo, & Lu, 2018) implemented their techniques independently to explore the solution of NLPDEs. In addition to this, a group of researchers also implemented another expansion-type method called $\left(\frac{G^{\prime }}{G},\mathrm{ }\frac{1}{G}\right)$-expansion method to their selected NLPDEs (Ali, Miah, & Akbar, 2018; Inc et al., 2020; Li, Li, & Wang, 2010; Miah, Ali, Akbar, & Wazwaz, 2017; Qureshi et al., 2023; Zayed et al. 2014). Nowadays, an excellently workable approach named the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-approach have been exercised to find the appropriate solutions to the NLPDEs, which is also an expansion of function dependent method. Very few researchers implemented $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-approach to solve their problems in NLPDEs (Iqbal et al., 2023; Khaliq et al., 2022; Mia, Miah, & Osman, 2023; Tripathy & Sahoo, 2021).

Very recently, Mia et al. (2023), used this mention approach to investigate the (2 + 1)-dimensional KPBBM equation and ascertained appropriate solutions which are essential for. It is very difficult to solve the integro-differential type equations by the conventional and the above-cited methods. With our selected novel method, most NLPDEs, including integro-differential equations, can be handled rapidly and simply. According to our consciousness, the above-listed two equations have not been analyzed with the implementation of our selected approach. As a result, in this research work, we devoted ourselves to getting the appropriate solutions of the above quoted two integro-differential equations using the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion approach. This technique provides a systematic way to arrive at precise solutions and insights into the underlying dynamics of the systems, and they have been proven to be useful tools in the study of soliton equations and nonlinear mathematical physics. Insights into the behavior of nonlinear dynamics and wave propagation throughout these models can be gained from studying and comprehending soliton solutions. In domains involving nonlinear processes, mathematical physics, and information transmission, these solutions have real world applications. The study of periodic solutions, as an example of the obtained solutions, provides insight into the stability, oscillatory properties, and long-term behavior of the dynamical process. The analysis of periodic events in areas including wave propagation, oscillations, and the development of patterned structures has applications for these solutions.

This research work is placed in the following way: A necessary review and motivation of this work is given in Section 1. In Section 2, our proposed approach the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion approach, is discussed in detail. In Section 3, the implementation of the chosen approach for selected two integro-differential equations is given. In Section 4, the settings for the graph of the obtained results and their interpretation are discussed. Finally, this paper is concluded in part 5.

2. Execution of the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion approach

This section has been expanded to provide a concise description of our chosen strategy for locating the most suitable solutions to the NLPDEs. As a particular instance, we are going to utilize one of the following NLPDEs: (1) $$F\left(u,\mathrm{ }{\mathrm{ }u}_x,\mathrm{ }{\mathrm{ }u}_y,\mathrm{ }{u}_t,\mathrm{ }{u}_{xx},\mathrm{ }{u}_{xt},\mathrm{ }{u}_{\mathit{xyt}}\dots \right)=0.$$(1)

Here, $F$ is a polynomial function of $u,\mathrm{ }{u}_x,\mathrm{ }{u}_{xx},\dots ,$ etc. and u is the function of three independent variables, $x,$ $y,$ and $t.$ Also, the terms $u_x,\mathrm{ }{u}_{xx},\dots ,$ etc. stand for partial derivative. Now, the procedures of this approach are described below:

Procedure I:

Initially, first we need to turn the attached partial differential equation into the form of an ordinary differential equation (ODE). To accomplish this, we define a new variable $\psi$ as the linear culmination of x, y, z, t, etc.

Consequently, we can adopt the transformation principles shown below. (2) $$u(x,\mathrm{ }y,\mathrm{ }t)=u\left(\psi \right);\psi =x+y-\nu t.$$(2)

Differentiating Equation (2)(2) $$u(x,\mathrm{ }y,\mathrm{ }t)=u\left(\psi \right);\psi =x+y-\nu t.$$(2) as many times as needed and then putting the respective terms into Equation (1)(1) $$F\left(u,\mathrm{ }{\mathrm{ }u}_x,\mathrm{ }{\mathrm{ }u}_y,\mathrm{ }{u}_t,\mathrm{ }{u}_{xx},\mathrm{ }{u}_{xt},\mathrm{ }{u}_{\mathit{xyt}}\dots \right)=0.$$(1) yields the ODE, (3) $$\mathrm{Z}(u,\mathrm{ }{(2-\nu )u}^{\prime },{\mathrm{ }(1-\nu )u}^{\prime \prime },\mathrm{ }{-\nu \mathrm{ }u}^{\prime \prime \prime }, \dots )=0,$$(3) where the primes (') stands for derivative with $\psi .$

Procedure II:

Let the following expresion represents a compact formation of $u$: (4) $$u\left(\psi \right)=\sum _{f=0}^T{a}_f{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^f.$$(4)

Here, $G$ is a function of $\psi$ associated with the given following ODE, (5) $$\frac{d^{2}G}{d{\psi }^2}+E\frac{dG}{d\psi }+FG+AF=0,$$(5) where $A,$ $E,$ $F$ and $a_f\mathrm{ }(f=0,\mathrm{ }1,\mathrm{ }2,\mathrm{ }\dots T)$ are constants to be calculated.

From Equation (3)(3) $$\mathrm{Z}(u,\mathrm{ }{(2-\nu )u}^{\prime },{\mathrm{ }(1-\nu )u}^{\prime \prime },\mathrm{ }{-\nu \mathrm{ }u}^{\prime \prime \prime }, \dots )=0,$$(3) , we get the value of $T$ by applying the homogenous balance rule and then inserting the acquired number in Equation (4)(4) $$u\left(\psi \right)=\sum _{f=0}^T{a}_f{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^f.$$(4) . After placing the balance number into Equation (4)(4) $$u\left(\psi \right)=\sum _{f=0}^T{a}_f{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^f.$$(4) , we take the derivative as many times as needed then putting into Equation (3)(3) $$\mathrm{Z}(u,\mathrm{ }{(2-\nu )u}^{\prime },{\mathrm{ }(1-\nu )u}^{\prime \prime },\mathrm{ }{-\nu \mathrm{ }u}^{\prime \prime \prime }, \dots )=0,$$(3) , we will have a system of algebraic equations. By solving the obtained system of equations, we get the values of $a_f\mathrm{ }(f=0,\mathrm{ }1,\mathrm{ }2,\mathrm{ }\dots T)$ then we can find the function $G\left(\psi \right)$ by treating the constants $A,$ $\mathrm{ }E$ and $F$ in Equation (5)(5) $$\frac{d^{2}G}{d{\psi }^2}+E\frac{dG}{d\psi }+FG+AF=0,$$(5) .

Procedure III:

Lastly, from Equation (3)(3) $$\mathrm{Z}(u,\mathrm{ }{(2-\nu )u}^{\prime },{\mathrm{ }(1-\nu )u}^{\prime \prime },\mathrm{ }{-\nu \mathrm{ }u}^{\prime \prime \prime }, \dots )=0,$$(3) and Equation (4)(4) $$u\left(\psi \right)=\sum _{f=0}^T{a}_f{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^f.$$(4) , we get a polynomial of $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right).$ By assimilating the coefficients of same indices of $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$ from left and right sides, we can consider a set of equations in terms of $\nu ,$ $a_f,$ $A,$ $E$ and $F.$

3. Applications of the proposed method

In this segment, our selected two integro-differential equations have been investigated by the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion approach. The first one is the (2 + 1)-dimensional first integro-differential (FID) KP hierarchy equation which is given below, (6) $${u_t=\frac{1}{2}u}_{\mathit{xxy}}+{\frac{1}{2}{\partial }_x^{-2}\left[u_{\mathit{yyy}}\right]+2u}_x{\partial }_x^{-1}\left[u_y\right]+4u{u}_y,$$(6) where $u=u(x,y,t)$ is a scalar function, $x$ and $y$ are respectively the longitudinal and transverse spatial coordinates, subscripts $x,\mathrm{ }y,\mathrm{ }t$ denote partial derivatives, and ${\partial }_x^{-1}\mathrm{\blacksquare }=\int \mathrm{\blacksquare }dx.$ This model describes the evolution of nonlinear long waves of small amplitude with slow dependence on the transverse coordinate.

The second one is the (2 + 1)-dimensional second integro-differential (SID) KP hierarchy equation, which is given below, (7) $${u_t=\frac{1}{16}u}_{\mathit{xxxxx}}+\frac{5}{6}{\partial }_x^{-1}\left[{uu}_{yy}\right]+{\frac{5}{6}\partial }_x^{-1}\left[u_y^2\right]+\frac{5}{16}{\partial }_x^{-3}\left[u_{\mathit{yyyy}}\right]+\frac{5}{4}{u}_x{\partial }_x^{-2}\left[u_{yy}\right]+\frac{5}{2}u{\partial }_x^{-1}\left[u_{yy}\right]+\frac{5}{2}{u}_y{\partial }_x^{-1}\left[u_y\right]+\frac{15}{2}{u}^2{u}_x+\frac{5}{2}{u}_x{u}_{xx}+\frac{5}{4}u{u}_{\mathit{xxx}}+\frac{5}{8}{u}_{\mathit{xyy}},$$(7) where $u=u(x,y,t)$ is a scalar function, $x$ and $y$ are respectively the longitudinal and transverse spatial coordinates, subscripts $x,\mathrm{ }y,\mathrm{ }t$ denote partial derivatives, and ${\partial }_x^{-1}\mathrm{\blacksquare }=\int \mathrm{\blacksquare }dx.$ It describes the processes of interaction of exponentially localized structures.

By envisaging the connection $u\left(x,\mathrm{ }y,\mathrm{ }t\right)=V_{xx}(x,\mathrm{ }y,\mathrm{ }t),$ we convert Equation (6)(6) $${u_t=\frac{1}{2}u}_{\mathit{xxy}}+{\frac{1}{2}{\partial }_x^{-2}\left[u_{\mathit{yyy}}\right]+2u}_x{\partial }_x^{-1}\left[u_y\right]+4u{u}_y,$$(6) into the given following pattern, (8) $$V_{\mathit{xxt}}=\frac{1}{2}{V}_{\mathit{xxxxy}}+\frac{1}{2}{V}_{\mathit{yyy}}+2V_{\mathit{xxx}}\mathrm{ }{V}_{xy}+4V_{xx}{V}_{\mathit{xxy}}=0.$$(8)

Through the transformation $V\left(x,\mathrm{ }y,\mathrm{ }t\right)=R\left(\psi \right);$ $\psi =x+y-\nu t,$ we convert Equation (8)(8) $$V_{\mathit{xxt}}=\frac{1}{2}{V}_{\mathit{xxxxy}}+\frac{1}{2}{V}_{\mathit{yyy}}+2V_{\mathit{xxx}}\mathrm{ }{V}_{xy}+4V_{xx}{V}_{\mathit{xxy}}=0.$$(8) into the following ODE form, (9) $$(\nu +\frac{1}{2})R^{\prime \prime }+\frac{1}{2}{R}^{\left(iv\right)}+3{\left(R^{\prime \prime }\right)}^2+d_1=0.$$(9)

For simplicity putting $R^{″}=H$ in Equation (9)(9) $$(\nu +\frac{1}{2})R^{\prime \prime }+\frac{1}{2}{R}^{\left(iv\right)}+3{\left(R^{\prime \prime }\right)}^2+d_1=0.$$(9) , (10) $$\left(\nu +\frac{1}{2}\right)H+\frac{1}{2}{H}^{″}+3H^2+d_1=0,$$(10) where $d_1$ is the constant of integration. Similarly, exercising the setting $u\left(x,\mathrm{ }y,\mathrm{ }t\right)=V_{\mathit{xxx}}(x,\mathrm{ }y,\mathrm{ }t),$ we can turn Equation (7)(7) $${u_t=\frac{1}{16}u}_{\mathit{xxxxx}}+\frac{5}{6}{\partial }_x^{-1}\left[{uu}_{yy}\right]+{\frac{5}{6}\partial }_x^{-1}\left[u_y^2\right]+\frac{5}{16}{\partial }_x^{-3}\left[u_{\mathit{yyyy}}\right]+\frac{5}{4}{u}_x{\partial }_x^{-2}\left[u_{yy}\right]+\frac{5}{2}u{\partial }_x^{-1}\left[u_{yy}\right]+\frac{5}{2}{u}_y{\partial }_x^{-1}\left[u_y\right]+\frac{15}{2}{u}^2{u}_x+\frac{5}{2}{u}_x{u}_{xx}+\frac{5}{4}u{u}_{\mathit{xxx}}+\frac{5}{8}{u}_{\mathit{xyy}},$$(7) into the given following pattern, (11) $${V_{\mathit{xxxt}}=\frac{1}{16}V}_{\mathit{xxxxxxxx}}+\frac{15}{4}{V}_{xx}{V}_{\mathit{xxyy}}+\frac{15}{4}{\left(V_{\mathit{xxy}}\right)}^2+\frac{5}{16}{V}_{\mathit{yyyy}}+\frac{15}{4}{V}_{\mathit{xxx}}{V}_{\mathit{xyy}}+\frac{5}{4}{V}_{\mathit{xxxx}}{V}_{yy}+\frac{5}{2}{V}_{\mathit{xxxy}}{V}_{xy}+\frac{15}{2}{\left({\left[V_{xx}\right]}^2\mathrm{ }{V}_{\mathit{xxx}}\right)}_x+\frac{5}{2}{\left(V_{\mathit{xxx}}{V}_{\mathit{xxxx}}\right)}_x+\frac{5}{4}{\left(V_{xx}{V}_{\mathit{xxxxx}}\right)}_x+\frac{5}{8}{V}_{\mathit{xxxxyy}}.$$(11)

Again, through the transformation $V\left(\mathrm{x},\mathrm{ }y,\mathrm{ }t\right)=R\left(\psi \right);$ $\psi =(x+y-\nu t),$ we convert Eq. (11)(11) $${V_{\mathit{xxxt}}=\frac{1}{16}V}_{\mathit{xxxxxxxx}}+\frac{15}{4}{V}_{xx}{V}_{\mathit{xxyy}}+\frac{15}{4}{\left(V_{\mathit{xxy}}\right)}^2+\frac{5}{16}{V}_{\mathit{yyyy}}+\frac{15}{4}{V}_{\mathit{xxx}}{V}_{\mathit{xyy}}+\frac{5}{4}{V}_{\mathit{xxxx}}{V}_{yy}+\frac{5}{2}{V}_{\mathit{xxxy}}{V}_{xy}+\frac{15}{2}{\left({\left[V_{xx}\right]}^2\mathrm{ }{V}_{\mathit{xxx}}\right)}_x+\frac{5}{2}{\left(V_{\mathit{xxx}}{V}_{\mathit{xxxx}}\right)}_x+\frac{5}{4}{\left(V_{xx}{V}_{\mathit{xxxxx}}\right)}_x+\frac{5}{8}{V}_{\mathit{xxxxyy}}.$$(11) into the following ODE form, (12) $$(\nu +\frac{5}{16})R+\frac{1}{16}{R}^{\left(iv\right)}+\frac{15}{4}{R}^2+\frac{5}{2}{R}^3+\frac{5}{8}{\left(R^{\prime }\right)}^2+\frac{5}{4}R{R}^{\prime \prime }+\frac{5}{8}{R}^{\prime \prime }+d_2=0,$$(12) where $d_2$ is the constant of integration.

3.1. Implementation of the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion method to the (2 + 1)-dimensional FID KP hierarchy equation

In Equation (10)(10) $$\left(\nu +\frac{1}{2}\right)H+\frac{1}{2}{H}^{″}+3H^2+d_1=0,$$(10) , we use the homogenous balance method, and we find the balance number $T=2.$

Putting the value $T=2$ in Equation (4)(4) $$u\left(\psi \right)=\sum _{f=0}^T{a}_f{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^f.$$(4) , we get (13) $$H\left(\psi \right)=a_0+a_1\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)+a_2{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^2.$$(13)

Here, the constants $a_0,\mathrm{ }{a}_1,$ and $a_2$ must be determined.

Equations (13)(13) $$H\left(\psi \right)=a_0+a_1\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)+a_2{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^2.$$(13) and (10)(10) $$\left(\nu +\frac{1}{2}\right)H+\frac{1}{2}{H}^{″}+3H^2+d_1=0,$$(10) together with computer programs like Maple are used to create a system of equations. Consequently, the values of constants $a_0,$ $a_1,$ $a_2,$ $\nu$ and $d_1$ are given below: $$a_0=a_0, a_1=-3EF+2F^2+E^2+2F-E$$ $$a_2=-{\left(E-F-1\right)}^2,\mathrm{ }\nu =-\frac{1}{2}-\frac{1}{2}{E}^2-4F+6EF-6F^2-6a_0$$ $$d_1=\frac{7}{2}{E}^2{F}^2-6E{F}^3-4E{F}^2+3F^4+4F^3+F^2+\frac{1}{2}{E}^2{a}_0+4E{a}_0-6EF{a}_0+6F^2{a}_0+3a_0^2-\frac{1}{2}{E}^{3}F+\frac{1}{2}{E}^{2}F.$$

Fixing the values of the constants $a_0,$ $a_1$ and $a_2$ into Equation (13)(13) $$H\left(\psi \right)=a_0+a_1\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)+a_2{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^2.$$(13) , we get the solution of Equation (10)(10) $$\left(\nu +\frac{1}{2}\right)H+\frac{1}{2}{H}^{″}+3H^2+d_1=0,$$(10) . To get the appropriate solutions of Equation (6)(6) $${u_t=\frac{1}{2}u}_{\mathit{xxy}}+{\frac{1}{2}{\partial }_x^{-2}\left[u_{\mathit{yyy}}\right]+2u}_x{\partial }_x^{-1}\left[u_y\right]+4u{u}_y,$$(6) there are two occurrences here.

Occurrence 1.

While the discriminant, $D=E^2-4F>0:$ (14) $$H\left(\psi \right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]-{\left(E-F-1\right)}^2\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]}^2.$$(14)

Employing, $\psi =x+y-\nu t\mathrm{ }$ in Equation (14)(14) $$H\left(\psi \right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]-{\left(E-F-1\right)}^2\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]}^2.$$(14) , we attain, (15) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-{\left(E-F-1\right)}^2\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(15) where $\nu =-\frac{1}{2}-\frac{1}{2}{E}^2-4F+6EF-6F^2-6a_0$ for Eq. (6)(6) $${u_t=\frac{1}{2}u}_{\mathit{xxy}}+{\frac{1}{2}{\partial }_x^{-2}\left[u_{\mathit{yyy}}\right]+2u}_x{\partial }_x^{-1}\left[u_y\right]+4u{u}_y,$$(6) .

Occurrence 2.

While the discriminant, D $=E^2-4F<0:$ (16) $$H\left(\psi \right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]-{\left(E-F-1\right)}^2{\times \mathrm{ }\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]}^2.$$(16)

Employing $\psi =x+y-\nu t\mathrm{ }$ in Equation (16)(16) $$H\left(\psi \right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]-{\left(E-F-1\right)}^2{\times \mathrm{ }\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]}^2.$$(16) , we attain the respective wave solution for Equation (6)(6) $${u_t=\frac{1}{2}u}_{\mathit{xxy}}+{\frac{1}{2}{\partial }_x^{-2}\left[u_{\mathit{yyy}}\right]+2u}_x{\partial }_x^{-1}\left[u_y\right]+4u{u}_y,$$(6) as follows, (17) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]-{\left(E-F-1\right)}^2{\times \mathrm{ }\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(17) where $\nu =-\frac{1}{2}-\frac{1}{2}{E}^2-4F+6EF-6F^2-6a_0$.

3.2. Implementation of the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion method to the (2 + 1)-dimensional SID KP hierarchy equation

From Equation (12)(12) $$(\nu +\frac{5}{16})R+\frac{1}{16}{R}^{\left(iv\right)}+\frac{15}{4}{R}^2+\frac{5}{2}{R}^3+\frac{5}{8}{\left(R^{\prime }\right)}^2+\frac{5}{4}R{R}^{\prime \prime }+\frac{5}{8}{R}^{\prime \prime }+d_2=0,$$(12) , we get the homogenous balance number $T=2.$ Now, setting the value $T=2$ in Equation (4)(4) $$u\left(\psi \right)=\sum _{f=0}^T{a}_f{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^f.$$(4) , we get the corresponding equation in terms of $R$ which is given bellow, (18) $$R\left(\psi \right)=a_0+a_1\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)+a_2{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^2.$$(18)

Here the constants $a_0$ and $a_1$ have to be determined.

By combining Equation (18)(18) $$R\left(\psi \right)=a_0+a_1\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)+a_2{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^2.$$(18) with Equation (12)(12) $$(\nu +\frac{5}{16})R+\frac{1}{16}{R}^{\left(iv\right)}+\frac{15}{4}{R}^2+\frac{5}{2}{R}^3+\frac{5}{8}{\left(R^{\prime }\right)}^2+\frac{5}{4}R{R}^{\prime \prime }+\frac{5}{8}{R}^{\prime \prime }+d_2=0,$$(12) , we get a system of equations. By solving this system,two sets of values of constants $a_0,$ $a_1,$ $a_2,$ $\nu$ and $d_2$ will be obtained:

Set 1: $$a_0=-\left(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2}\right),\mathrm{ }{a}_1=3E^2-3E-9EF+6F+6F^2$$ $$a_2=-\left(3E^2-6E-6EF+6F+3F^2+3\right),\mathrm{ }\nu =-(\frac{7}{32}{E}^4-\frac{7}{4}{E}^{2}F+\frac{7}{2}{F}^2-\frac{25}{16})$$ $$d_2=-(\frac{1}{64}{E}^6+\frac{7}{64}{E}^4-\frac{3}{16}{E}^{4}F-\frac{7}{8}{E}^{2}F+\frac{3}{4}{E}^2{F}^2+\frac{7}{4}{F}^2-F^3-\frac{5}{16}).$$

Set 2: $$a_0=a_0,\mathrm{ }{a}_1=\left(E^2-E-3EF+2F+2F^2\right),\mathrm{ }{a}_2=-\left(E^2-2E-2EF+F^2+2F+1\right),$$ $$\nu =-(\frac{1}{16}{E}^4-\frac{5}{4}{E}^{3}F+\frac{3}{4}{E}^{2}F+\frac{5}{4}{a}_0{E}^2+\frac{5}{8}{E}^2+\frac{35}{4}{\mathrm{E}}^2{F}^2-10E{F}^2-15a_{0}EF-\frac{15}{2}EF-15E{F}^3+10F^3+5F+15a_0{F}^2+10a_{0}F+11F^2+\frac{15}{2}{F}^4+\frac{15}{2}{a}_0^2+\frac{15}{2}{a}_0+\frac{5}{16}),$$ $$d_2=-(\frac{5}{8}{E}^{3}F-\frac{41}{8}{E}^2{F}^2-\frac{5}{8}{E}^{2}F+5E{F}^2+\frac{27}{2}E{F}^3-\frac{5}{8}{a}_0{E}^2-\frac{27}{2}{a}_0{F}^2-5a_{0}F-\frac{5}{4}{a}_0^2{E}^2-10a_0^{2}F-15a_0^2{F}^2-20a_0{F}^3-15a_0{F}^4-\frac{1}{16}{a}_0{E}^4-12E^2{F}^3+20E{F}^4-\frac{1}{16}{E}^{4}F+2E^3{F}^2-\frac{65}{4}{E}^2{F}^4+15E{F}^5+\frac{1}{16}{E}^{4}F-\frac{21}{16}{E}^4{F}^2+\frac{15}{2}{e}^3{F}^3+\frac{15}{2}{a}_{0}EF+15a_0^{2}EF-2a_0{E}^{2}F+\frac{5}{2}{a}_0{E}^{3}F+20a_{0}E{F}^2-\frac{35}{2}{a}_0{E}^2{F}^2+30a_{0}E{F}^3-6F^3-10F^5-5F^6-\frac{5}{4}{F}^2-\frac{39}{4}{F}^4-\frac{15}{4}{a}_0^2-5a_0^3).$$

Now, inserting the values of the constants $a_0,$ $a_1$ and $a_2$ from set 1 into Equation (18)(18) $$R\left(\psi \right)=a_0+a_1\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)+a_2{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^2.$$(18) , we shall get the solution of Equation (12)(12) $$(\nu +\frac{5}{16})R+\frac{1}{16}{R}^{\left(iv\right)}+\frac{15}{4}{R}^2+\frac{5}{2}{R}^3+\frac{5}{8}{\left(R^{\prime }\right)}^2+\frac{5}{4}R{R}^{\prime \prime }+\frac{5}{8}{R}^{\prime \prime }+d_2=0,$$(12) . Here also two occurrences will exist to get the appropriate solutions of Equation (7)(7) $${u_t=\frac{1}{16}u}_{\mathit{xxxxx}}+\frac{5}{6}{\partial }_x^{-1}\left[{uu}_{yy}\right]+{\frac{5}{6}\partial }_x^{-1}\left[u_y^2\right]+\frac{5}{16}{\partial }_x^{-3}\left[u_{\mathit{yyyy}}\right]+\frac{5}{4}{u}_x{\partial }_x^{-2}\left[u_{yy}\right]+\frac{5}{2}u{\partial }_x^{-1}\left[u_{yy}\right]+\frac{5}{2}{u}_y{\partial }_x^{-1}\left[u_y\right]+\frac{15}{2}{u}^2{u}_x+\frac{5}{2}{u}_x{u}_{xx}+\frac{5}{4}u{u}_{\mathit{xxx}}+\frac{5}{8}{u}_{\mathit{xyy}},$$(7) .

Occurrence 1.

When the discriminant, $D=E^2-4F>0:$ (19) $$R\left(\psi \right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+{\mathrm{k}}_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]-(3E^2-6E-6EF+6F+3F^2+3)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]}^2.$$(19)

Setting $\psi =x+y-\nu t\mathrm{ }$ in Equation (19)(19) $$R\left(\psi \right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+{\mathrm{k}}_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]-(3E^2-6E-6EF+6F+3F^2+3)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]}^2.$$(19) , we have the following for Equation (7)(7) $${u_t=\frac{1}{16}u}_{\mathit{xxxxx}}+\frac{5}{6}{\partial }_x^{-1}\left[{uu}_{yy}\right]+{\frac{5}{6}\partial }_x^{-1}\left[u_y^2\right]+\frac{5}{16}{\partial }_x^{-3}\left[u_{\mathit{yyyy}}\right]+\frac{5}{4}{u}_x{\partial }_x^{-2}\left[u_{yy}\right]+\frac{5}{2}u{\partial }_x^{-1}\left[u_{yy}\right]+\frac{5}{2}{u}_y{\partial }_x^{-1}\left[u_y\right]+\frac{15}{2}{u}^2{u}_x+\frac{5}{2}{u}_x{u}_{xx}+\frac{5}{4}u{u}_{\mathit{xxx}}+\frac{5}{8}{u}_{\mathit{xyy}},$$(7) , (20) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-(3E^2-6E-6EF+6F+3F^2+3)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(20) where $\nu =-(\frac{7}{32}{E}^4-\frac{7}{4}{E}^{2}F+\frac{7}{2}{F}^2-\frac{25}{16}).$

Occurrence 2.

For the discriminant, $D=E^2-4F<0:$ (21) $$R\left(\psi \right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+((E-2)k_1-k_2\sqrt{-D})\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]-(3E^2-6E-6EF+6F+3F^2+3)\times {\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]}^2.$$(21)

Installing $\psi =x+y-\nu t\mathrm{ }$ in Equation (21)(21) $$R\left(\psi \right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+((E-2)k_1-k_2\sqrt{-D})\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]-(3E^2-6E-6EF+6F+3F^2+3)\times {\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]}^2.$$(21) , we get the following result for Equation (7)(7) $${u_t=\frac{1}{16}u}_{\mathit{xxxxx}}+\frac{5}{6}{\partial }_x^{-1}\left[{uu}_{yy}\right]+{\frac{5}{6}\partial }_x^{-1}\left[u_y^2\right]+\frac{5}{16}{\partial }_x^{-3}\left[u_{\mathit{yyyy}}\right]+\frac{5}{4}{u}_x{\partial }_x^{-2}\left[u_{yy}\right]+\frac{5}{2}u{\partial }_x^{-1}\left[u_{yy}\right]+\frac{5}{2}{u}_y{\partial }_x^{-1}\left[u_y\right]+\frac{15}{2}{u}^2{u}_x+\frac{5}{2}{u}_x{u}_{xx}+\frac{5}{4}u{u}_{\mathit{xxx}}+\frac{5}{8}{u}_{\mathit{xyy}},$$(7) , (22) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}{\left(\left(E-2\right)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}\right]-(3E^2-6E-6EF+\mathrm{ }6F+3F^2+3)\mathrm{ }{\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(22) where $\nu =-(\frac{7}{32}{E}^4-\frac{7}{4}{E}^{2}F+\frac{7}{2}{F}^2-\frac{25}{16}).$

Again, inserting the values of the constants $a_0,$ $a_1$ and $a_2$ from set 2 into Equation (18)(18) $$R\left(\psi \right)=a_0+a_1\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)+a_2{\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)}^2.$$(18) , we have the solution of Equation (12)(12) $$(\nu +\frac{5}{16})R+\frac{1}{16}{R}^{\left(iv\right)}+\frac{15}{4}{R}^2+\frac{5}{2}{R}^3+\frac{5}{8}{\left(R^{\prime }\right)}^2+\frac{5}{4}R{R}^{\prime \prime }+\frac{5}{8}{R}^{\prime \prime }+d_2=0,$$(12) . Here also two occurrences will exist to get the appropriate solutions of Equation (7)(7) $${u_t=\frac{1}{16}u}_{\mathit{xxxxx}}+\frac{5}{6}{\partial }_x^{-1}\left[{uu}_{yy}\right]+{\frac{5}{6}\partial }_x^{-1}\left[u_y^2\right]+\frac{5}{16}{\partial }_x^{-3}\left[u_{\mathit{yyyy}}\right]+\frac{5}{4}{u}_x{\partial }_x^{-2}\left[u_{yy}\right]+\frac{5}{2}u{\partial }_x^{-1}\left[u_{yy}\right]+\frac{5}{2}{u}_y{\partial }_x^{-1}\left[u_y\right]+\frac{15}{2}{u}^2{u}_x+\frac{5}{2}{u}_x{u}_{xx}+\frac{5}{4}u{u}_{\mathit{xxx}}+\frac{5}{8}{u}_{\mathit{xyy}},$$(7) .

Occurrence 1.

Again for the discriminant, $D=E^2-4F>0:$ (23) $$R\left(\psi \right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{\mathrm{D}}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]-(E^2-2E-2EF+F^2+2F+1)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]}^2.$$(23)

Putting $\psi =x+y-\nu t\mathrm{ }$ in Equation (23)(23) $$R\left(\psi \right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{\mathrm{D}}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]-(E^2-2E-2EF+F^2+2F+1)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}\psi }}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}\psi }}\right]}^2.$$(23) , we will have the following result for Equation (7)(7) $${u_t=\frac{1}{16}u}_{\mathit{xxxxx}}+\frac{5}{6}{\partial }_x^{-1}\left[{uu}_{yy}\right]+{\frac{5}{6}\partial }_x^{-1}\left[u_y^2\right]+\frac{5}{16}{\partial }_x^{-3}\left[u_{\mathit{yyyy}}\right]+\frac{5}{4}{u}_x{\partial }_x^{-2}\left[u_{yy}\right]+\frac{5}{2}u{\partial }_x^{-1}\left[u_{yy}\right]+\frac{5}{2}{u}_y{\partial }_x^{-1}\left[u_y\right]+\frac{15}{2}{u}^2{u}_x+\frac{5}{2}{u}_x{u}_{xx}+\frac{5}{4}u{u}_{\mathit{xxx}}+\frac{5}{8}{u}_{\mathit{xyy}},$$(7) , (24) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]\text{-}\mathrm{ }(E^2-2E-2EF+F^2+2F+1)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]}^2,$$(24) where $\nu =-(\frac{1}{16}{E}^4-\frac{5}{4}{E}^{3}F+\frac{3}{4}{E}^{2}F+\frac{5}{4}{a}_0{E}^2+\frac{5}{8}{E}^2+\frac{35}{4}{E}^2{F}^2-10E{F}^2-15a_{0}EF-\frac{15}{2}EF-15E{F}^3+10F^3+5F+15a_0{F}^2+10a_{0}F+11F^2+\frac{15}{2}{F}^4+\frac{15}{2}{a}_0^2+\frac{15}{2}{a}_0+\frac{5}{16}).$

Occurrence 2.

If, $\left(D\right)=E^2-4F<0,$ (25) $$R\left(\psi \right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]-(E^2-2E-2EF+F^2+2F+1)\times {\mathrm{ }\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]}^2.$$(25)

Installing $\psi =x+y-\nu t\mathrm{ }$ in Equation (25)(25) $$R\left(\psi \right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]-(E^2-2E-2EF+F^2+2F+1)\times {\mathrm{ }\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\psi \right)}\right]}^2.$$(25) , we find the following result for Equation (7)(7) $${u_t=\frac{1}{16}u}_{\mathit{xxxxx}}+\frac{5}{6}{\partial }_x^{-1}\left[{uu}_{yy}\right]+{\frac{5}{6}\partial }_x^{-1}\left[u_y^2\right]+\frac{5}{16}{\partial }_x^{-3}\left[u_{\mathit{yyyy}}\right]+\frac{5}{4}{u}_x{\partial }_x^{-2}\left[u_{yy}\right]+\frac{5}{2}u{\partial }_x^{-1}\left[u_{yy}\right]+\frac{5}{2}{u}_y{\partial }_x^{-1}\left[u_y\right]+\frac{15}{2}{u}^2{u}_x+\frac{5}{2}{u}_x{u}_{xx}+\frac{5}{4}u{u}_{\mathit{xxx}}+\frac{5}{8}{u}_{\mathit{xyy}},$$(7) , (26) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]-\left(E^2-2E-2EF+F^2+2F+1\right)\times {\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}{\left(\left(E-2\right)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}\right]}^2,$$(26) where $\nu =-(\frac{1}{16}{E}^4-\frac{5}{4}{E}^{3}F+\frac{3}{4}{E}^{2}F+\frac{5}{4}{a}_0{E}^2+\frac{5}{8}{E}^2+\frac{35}{4}{E}^2{F}^2-10E{F}^2-15a_{0}EF-\frac{15}{2}EF-15E{F}^3+10F^3+5F+15a_0{\mathrm{F}}^2+10a_{0}F+11F^2+\frac{15}{2}{F}^4+\frac{15}{2}{a}_0^2+\frac{15}{2}{a}_0+\frac{5}{16}).$

4. Graphical representation and discussion

The physical embellishment of the obtained results to the integro-differential equations is shown here in 3D, contour, and 2D formats. The 3D graphs are of singular, singular periodic, bell and anti-bell-shaped types of solitons. First of all, we diagrammatize the solution of Equation (15)(15) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-{\left(E-F-1\right)}^2\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(15) in three orientations, one is 3D (bell shape soliton) within the scale $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10]\mathrm{ }\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=1,$ $\mathit{D}=0.2,$ ${\mathit{a}}_0=2,$ $\mathit{\nu }=-12.84$ and $\mathit{y}=$2, contour figure for the same interval and the same parametric values, 2D figure for the scale $\mathit{x}\in [-40,\mathrm{ }40]$ for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=1,$ $\mathit{D}=0.2,$ ${\mathit{a}}_0=2,$ $\mathit{\nu }=-12.84,$ $\mathit{y}=$2 and $\mathit{t}=1,$ which are given in Figure 1(a), Figure 1(b) and Figure 1(c), respectively. Again, we diagrammatize the solution of Equation (17)(17) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]-{\left(E-F-1\right)}^2{\times \mathrm{ }\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(17) in three orientations, one is 3D (periodic soliton) within the scale $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10]$ for the values $\mathit{E}=1,$ $\mathit{F}=2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=2,$ $\mathit{D}=-7,$ ${\mathit{a}}_0=2,$ $\mathrm{ }\mathit{\nu }=-33$ and $\mathit{y}=$2, contour figure for the same interval and the same parametric values, 2D figure for the scale $\mathit{x}\in [-10,\mathrm{ }10]$ for the values $\mathit{E}=1,$ $\mathit{F}=2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=2,$ $\mathit{D}=-7,$ ${\mathit{a}}_0=2,$ $\mathit{\nu }=-33,$ $\mathit{y}=$2 and $\mathit{t}=2,$ which are given in Figure 2(a), Figure 2(b) and Figure 2(c), respectively. Now, we diagrammatize the solution of Equation (20)(20) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-(3E^2-6E-6EF+6F+3F^2+3)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(20) in three orientations, one is 3D (bell shape soliton) within the scale $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10]$ for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=1,$ $\mathit{\nu }=1.55375,$ $\mathit{D}=0.2$ and $\mathit{y}=$1, contour figure for same interval and same parametric values, 2D figure for the scale $\mathit{x}\in [-20,\mathrm{ }20]$ for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=1,$ $\mathit{D}=0.2,$ $\mathit{\nu }=1.55375,$ $\mathrm{ }\mathit{y}=$1 and $\mathit{t}=1,$ which are given in Figure 3(a), Figure 3(b) and Figure 3(c), respectively. Again, we diagrammatize the solution of Equation (22)(22) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}{\left(\left(E-2\right)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}\right]-(3E^2-6E-6EF+\mathrm{ }6F+3F^2+3)\mathrm{ }{\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(22) in three orientations, one is 3D (singular periodic soliton) within the scale $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10]$ for the values $\mathit{E}=1,$ $\mathit{F}=0.3,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=2,$ $\mathit{\nu }=1.55375,$ $\mathit{D}=-0.2\mathrm{ }$ and $\mathit{y}=$1, contour figure for same interval and same parametric values, 2D figure for the scale $\mathit{x}\in [-10,\mathrm{ }10]$ for the values $\mathit{E}=1,$ $\mathit{F}=0.3,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=2,$ $\mathit{\nu }=1.55375,$ $\mathit{D}=-0.2,$ $\mathit{y}=$1 and $\mathit{t}=1,$ which are given in Figure 4(a), Figure 4(b) and Figure 4(c), respectively. Now, we diagrammatize the solution of Equation (24)(24) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]\text{-}\mathrm{ }(E^2-2E-2EF+F^2+2F+1)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]}^2,$$(24) in three orientations, one is 3D (anti-bell shape soliton) within the scale $\mathit{x}\in [-5,\mathrm{ }5]$ and $\mathit{t}\in [-5,\mathrm{ }5]$ for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=1,$ ${\mathit{a}}_0=0.2,$ $\mathit{\nu }=-2.732,$ $\mathit{D}=0.2\mathrm{ }$ and $\mathit{y}=$1, contour figure for same interval and same parametric values, 2D figure for the scale $\mathit{x}\in [-20,\mathrm{ }20]$ for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=1,$ ${\mathit{a}}_0=0.2,$ $\mathit{\nu }=-2.732,$ $\mathit{D}=0.2,$ $\mathit{y}=$1 and $\mathit{t}=1,$ which are given in Figure 5(a), Figure 5(b) and Figure 5(c), respectively.

Figure 1. (a). the 3D singular bell shape soliton in Equation (15)(15) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-{\left(E-F-1\right)}^2\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(15) for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=1,$ $\mathit{D}=0.2,$ ${\mathit{a}}_0=2,$ $\mathit{\nu }=-12.84$ and $\mathit{y}=$2 for the range $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10].$ (b). The contour figure in Eq. (15)(15) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-{\left(E-F-1\right)}^2\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(15) for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=1,$ $\mathit{D}=0.2,$ ${\mathit{a}}_0=2,$ $\mathit{\nu }=-12.84$ and $\mathit{y}=$2 for the range $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10].$ (c). The 2D shape in Eq. (15)(15) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-{\left(E-F-1\right)}^2\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(15) for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=1,$ $\mathit{D}=0.2,$ ${\mathit{a}}_0=2,$ $\mathit{\nu }=-12.84,\mathrm{ }\mathit{y}=$2 and $\mathit{t}=1$ for the range $\mathit{x}\in [-40,\mathrm{ }40]$

Figure 2. (a). the 3D periodic soliton in Equation (17)(17) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]-{\left(E-F-1\right)}^2{\times \mathrm{ }\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(17) for the values $\mathit{E}=1,$ $\mathit{F}=2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=2,$ $\mathit{D}=-7,$ ${\mathit{a}}_0=2,$ $\mathit{\nu }=-33$ and $\mathit{y}=$2 for the range $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10].$ (b). The contour shape in Eq. (17)(17) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]-{\left(E-F-1\right)}^2{\times \mathrm{ }\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(17) for the values $\mathit{E}=1,$ $\mathit{F}=2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=2,$ $\mathit{D}=-7,$ ${\mathit{a}}_0=2,$ $\mathrm{ }\mathit{\nu }=-33$ and $\mathit{y}=$2 for the range $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10].$ (c). The 2D shape in Eq. (17)(17) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(-3EF+2F^2+E^2+2F-E\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]-{\left(E-F-1\right)}^2{\times \mathrm{ }\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(17) for the values $\mathit{E}=1,$ $\mathit{F}=2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=2,$ $\mathit{D}=-7,$ ${\mathit{a}}_0=2,$ $\mathit{\nu }=-33,$ $\mathit{y}=$2 and $\mathit{t}=2$ for the range $\mathit{x}\in [-10,\mathrm{ }10].$

Figure 3. (a). the 3D bell-shape soliton in Equation (20)(20) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-(3E^2-6E-6EF+6F+3F^2+3)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(20) for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=1,$ $\mathit{\nu }=1.55375,$ $\mathit{D}=0.2\mathrm{ }$ and $\mathit{y}=$1 for the range $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10].$ (b). The contour in Eq. (20)(20) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-(3E^2-6E-6EF+6F+3F^2+3)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(20) for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=1,$ $\mathit{\nu }=1.55375,$ $\mathit{D}=0.2\mathrm{ }$ and $\mathit{y}=$1 for the range $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10].$ (c). The 2D shape in Eq. (20)(20) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]-(3E^2-6E-6EF+6F+3F^2+3)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)}}\right]}^2,$$(20) for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1,$ ${\mathit{k}}_2=1,$ $\mathit{D}=0.2,$ $\mathit{\nu }=1.55375,$ $\mathrm{ }\mathit{y}=$1 and $\mathit{t}=1$ for the range $\mathit{x}\in [-20,\mathrm{ }20].$

Figure 4. (a). the 3D singular periodic soliton in Equation (22)(22) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}{\left(\left(E-2\right)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}\right]-(3E^2-6E-6EF+\mathrm{ }6F+3F^2+3)\mathrm{ }{\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(22) for the values $\mathit{E}=1,$ $\mathit{F}=0.3,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=2,$ $\mathit{\nu }=1.55375,$ $\mathit{D}=-0.2\mathrm{ }$ and $\mathit{y}=$ 1 for the range $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10].$ (b). the contour shape in Equation (22)(22) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}{\left(\left(E-2\right)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}\right]-(3E^2-6E-6EF+\mathrm{ }6F+3F^2+3)\mathrm{ }{\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(22) for the values $\mathit{E}=1,$ $\mathit{F}=0.3,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=2,$ $\mathit{\nu }=1.55375,$ $\mathit{D}=-0.2\mathrm{ }$ and $\mathit{y}=$1 for the range $\mathit{x}\in [-10,\mathrm{ }10]$ and $\mathit{t}\in [-10,\mathrm{ }10].$ (c). The 2D shape in Eq. (22)(22) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=-(\frac{1}{4}{E}^2-3EF+2F+3F^2+\frac{1}{2})+\left(3E^2-3E-9EF+6F+6F^2\right)\times \left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}{\left(\left(E-2\right)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}\left[x+y-\nu t\right]\right)}\right]-(3E^2-6E-6EF+\mathrm{ }6F+3F^2+3)\mathrm{ }{\left[\frac{\left(E{k}_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(E{k}_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}{\left((E-2)k_2+k_1\sqrt{-D}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)+\left(\left(E-2\right)k_1-k_2\sqrt{-D}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{\sqrt{-D}}{2}[x+y-\nu t]\right)}\right]}^2,$$(22) for the values $\mathit{E}=1,$ $\mathit{F}=0.3,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=2,$ $\mathit{\nu }=1.55375,$ $\mathit{D}=-0.2,\mathrm{ }\mathit{y}=$1 and $\mathit{t}=1$ for the range $\mathit{x}\in [-10,\mathrm{ }10].$

Figure 5. (a). The 3D anti-bell shape soliton in Eq. (24)(24) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]\text{-}\mathrm{ }(E^2-2E-2EF+F^2+2F+1)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]}^2,$$(24) for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=1,$ ${\mathit{a}}_0=0.2,$ $\mathit{\nu }=-2.732,$ $\mathit{D}=0.2\mathrm{ }$ and $\mathit{y}=$1 for the range $\mathit{x}\in [-5,\mathrm{ }5]$ and $\mathit{t}\in [-5,\mathrm{ }5].$ (b). The contour shape in Eq. (24)(24) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]\text{-}\mathrm{ }(E^2-2E-2EF+F^2+2F+1)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]}^2,$$(24) for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=1,$ ${\mathit{a}}_0=0.2,$ $\mathit{\nu }=-2.732,$ $\mathit{D}=0.2\mathrm{ }$ and $\mathit{y}=$1 for the range $\mathit{x}\in [-5,\mathrm{ }5]$ and $\mathit{t}\in [-5,\mathrm{ }5].$ (c). The 2D shape of Eq. (24)(24) $$u\left(x,\mathrm{ }y,\mathrm{ }t\right)=a_0+\left(E^2-E-3EF+2F+2F^2\right)\times \left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]\text{-}\mathrm{ }(E^2-2E-2EF+F^2+2F+1)\times {\left[\frac{k_1\left(E+\sqrt{D}\right)+k_2\left(E-\sqrt{D}\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}{k_1\left(E+\sqrt{D}-2\right)+k_2\left(E-\sqrt{D}-2\right)e^{\sqrt{D}(x+y-\nu t)\mathrm{ }}}\right]}^2,$$(24) for the values $\mathit{E}=1,$ $\mathit{F}=0.2,$ ${\mathit{k}}_1=1$ and ${\mathit{k}}_2=1,$ ${\mathit{a}}_0=0.2,$ $\mathrm{\nu }=-2.732,$ $\mathit{D}=0.2,$ $\mathit{y}=$1and $\mathit{t}=1$ for the range $\mathit{x}\in [-20,\mathrm{ }20].$

It is important to mention that when the solitary wave’s center is imaginary, we note that singular solitons may be connected to them. As a result, discussing solitary solitons is important. Spikes are present in this form of solution, which may suggest a description for the emergence of rogue waves.

5. Conclusion

We have found many progressive wave soliton solutions of the two integral-differential KP hierarchy equations in the present study by using the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion approach. The results of our research indicate that the solutions that were discovered physically imply solitary wave solutions. These solitary wave solutions incorporate like singular, singular periodic, bell and anti-bell-shaped types of solitons. When tackling the natural model equations that may be utilized to solve the PDE, particularly the nonlinear evolution equations, the $\left(\frac{G^{\prime }}{G^{\prime }+G+A}\right)$-expansion approach that we have provided is easy to understand, practically applicable, and satisfying in its results. This approach enables us to quickly find the formative solutions of our proposed equation, which may be used to explain unforeseen occurrences, notably in the branches of modern engineering and mathematical physics. These results have implications for the knowledge of nonlinear PDEs and their applications in physics, engineering, optics, and mathematics. The study provides insights into the dynamics of the two integro-differential KP hierarchy equations and demonstrates the efficiency of mathematical modeling techniques. Overall, this research enhances the field of nonlinear dynamics and provides avenues for further investigation of similar models and phenomena.

Disclosure statement

No potential conflict of interest was reported by the author(s)