New solitary wave structures to the (2 + 1)-dimensional KD and KP equations with spatio-temporal dispersion

1 Introduction

Non-linear evolution equations perform a critical task to models of many applied mathematics, mathematical physics and engineering, for example, optical fibres, fluid mechanics, plasma waves, complex scalar nucleon field, non-linear optics, quantum field theory, fibre optics, solid-state physics, chemical physics, plasma physics, ocean engineering etc. The exact wave answers of non-linear wave models are executed to describe the physical device in real life. For this goal, the distinct information of the exact answer of non-linear evolution equations is hot subjects in advanced research areas. There are numerous reliable, and robust procedures have been improved to examine exact wave solutions of non-linear evolution equations, for example, improved $\mathit{exp}(-\varphi (\xi ))$ -expansion method (Chen et al., 2019), tanh-coth expansion method (Alquran et al., 2018), the exponential rational function method (Tebue et al., 2016), $\mathit{exp}(-\varphi (\xi ))$ -expansion method (Alam and Belgacem, 2016; Alam and Tunc, 2016), the Power Index Method (Shrauner, 2019), Bernoulli sub-equation method (Syam, 2019), Bcklund transformation method (Liu et al., 2019), Weierstrass elliptic function method (Krishnan and Peng, 2005), Lie symmetry approach (Ren et al., 2019), Lie point symmetries (Khalique and Moleleki, 2019), Darboux transformation method (Chen et al., 2018), singular manifold method (Peng and Krishnan, 2005), N-fold Darboux transformation (Ha et al., 2019), the simplified Hirotas method (Wazwaz and El-Tantawy, 2017), the novel $(G^{\prime }/G)$ -expansion technique (Alam et al., 2014a; Alam and Akbar, 2014; Akbar et al., 2016 ), the extended modified $\mathit{exp}(-\varphi (\xi ))$ function method (Karaagac et al., 2019),the homotopy perturbation method (Shqair, 2019), the sine-Gordon expansion method (Bulut et al., 2017; Bulut et al., 2018), the improved Bernoulli sub-equation function method (Baskonus and Bulut, 2016), rational sine-cosine method (Alqurana et al., 2019) and so on.

Firstly, we are consider the $(2+1)$ -dimensional KD equation (Konopelchenko and Dubrovsky, 1984; Song and Zhang, 2006; Wazwaz, 2007a; Wazwaz, 2007b; Taghizadeh and Mirzazadeh, 2011 ) as

Finally, we are consider the $(2+1)$ -dimensional KP equation (Kadomtsev and Petviashvili, 1970) of the form

2 Methodology

This section instantly highlights the significant ideas of the novel generalized $(\frac{G^{\prime }}{G})$ -expansion approach:

• Step 1: We assume that a nonlinear evolution equation including $x,y$ and t as follows:

  (4)
  $$\mathrm{P}(u,u_t,u_x,u_y,u_{\mathit{yy}},u_{\mathit{tt}},u_{\mathit{xx}},u_{\mathit{xt}},\dots )=0,$$ where $u(x,y,t)$ is the result of Eq. (4).

• Step 2: Take the solutions of Eq. (4) as follows:

  (5)
  $$\mathrm{u}(x,t)=u(\xi ),\xi =x+y+\mathit{Vt},$$ where V is the transformation variable and under the transformation Eq. (5), Eq. (4) converts a nonlinear ODE:
  (6)
  $$R(u,\frac{\partial u}{\partial \xi },\frac{{\partial }^{2}u}{\partial {\xi }^2},\dots )=0\text{.}$$

• Step 3: Calculate N by using balance rule in Eq. (6).

• Step 4: Consider that Eq. (6) has the solution can be represent as

  (7)
  $$u(\xi )=a_0+a_1(d+H)+\frac{b_1}{d+H}+a_2{(d+H)}^2+\frac{b_2}{{(d+H)}^2}+{\displaystyle \sum _{i=3}^N}(a_i{(d+H)}^i+b_i(d+H^{-i}),$$ where $a_0,a_1,a_2,b_1,b_2$ and d are constants. $H(\xi )=(\frac{G^{\prime }}{G})$ and $G=G(\xi )$ represent the equation
  (8)
  $$\mathrm{A}{\mathit{GG}}^{″}-{\mathit{BGG}}^{\prime }-{\mathit{EG}}^2-C{(G^{\prime })}^2=0,$$ where $A,B,C$ and E are real unknown free parameters. Eq. (8) has three solutions:
  • –

    When $B\ne 0,\mathrm{\Psi }=A-C$ and $\mathrm{\Omega }=B^2+4E(A-C)>0$ , the solution of $H(\xi )=(\frac{G^{\prime }}{G})$ is

    (9)

    $$H(\xi )=\frac{G^{\prime }}{G}=\frac{B}{2\mathrm{\Psi }}+\frac{\sqrt{\mathrm{\Omega }}}{2\mathrm{\Psi }}\frac{C_1\mathit{sinh}(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})+C_2\mathit{cosh}(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})}{C_1\mathit{cosh}(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})+C_2\mathit{sinh}(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})}\text{;}$$

  • –

    When $B\ne 0,\mathrm{\Psi }=A-C$ and $\mathrm{\Omega }=B^2+4E(A-C)<0$ , the solution of $H(\xi )=(\frac{G^{\prime }}{G})$ is

    (10)

    $$H(\xi )=\frac{G^{\prime }}{G}=\frac{B}{2\mathrm{\Psi }}+\frac{\sqrt{-\mathrm{\Omega }}}{2\mathrm{\Psi }}\frac{-C_1\mathit{sin}(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})+C_2\mathit{cos}(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})}{C_1\mathit{cos}(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})+C_2\mathit{sin}(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})}\text{;}$$

  • –

    When $B\ne 0,\mathrm{\Psi }=A-C$ and $\mathrm{\Omega }=B^2+4E(A-C)=0$ , the solution of $H(\xi )=(\frac{G^{\prime }}{G})$ is

    (11)

    $$H(\xi )=\frac{G^{\prime }}{G}=\frac{B}{2\mathrm{\Psi }}+\frac{C_2}{C_1+C_2\xi }\text{.}$$

• Step 5: Calculate the coefficients of ${(d+H)}^N$ and ${(d+H)}^{-N}$ and receive a set of algebraic equations for $a_i,b_i,d$ and V by inserting each coefficient to zero. Receive the exact wave solutions of Eq. (4) through setting the received values in Eqs. (7) and (6) including the amount of N.

3 The (2 + 1)-dimensional Konopelchenko-Dubrovsky (KD) equation

In this section, the method is utilised to obtain soliton for the (2 + 1)-dimensional KD equation. We consider that the (2 + 1)-dimensional KD equation:

Using the wave variable $\xi =x+y-\mathit{ct}$ carries the KD Eqs. (12) and (13) into a system of ODE:

Integrating on the Eq. (15), we have

From Eqs. (16) and (14), we obtain

If $a=2b$ , then from Eq. (17), we obtain

According to the novel generalized $(\frac{G^{\prime }}{G})$ -expansion scheme (Alam et al., 2014a; Alam, 2015; Alam and Li, 2019 ), applying homogeneous balance rule between $u^{″}$ with $u^3$ gives $N=1$ . Therefore, the Eq. (18) has the following solution:

• The first set: $$c=-\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2},a_0=\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}},\mathrm{\Phi }=A-C,a_1=-\frac{2\mathrm{\Phi }}{\mathit{Aa}},b_1=0\text{.}$$

• The second set: $$c=-\frac{12A^2-4E\mathrm{\Phi }-B^2}{4A^2},d=-\frac{B}{2\mathrm{\Phi }},a_0=0,a_1=0,b_1=\frac{\pm \sqrt{0.5}(4E\mathrm{\Phi }+B^2)}{2\mathrm{\Phi }\mathit{aA}}\text{.}$$

• The third set: $$c=-\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2},a_0=-\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}},a_1=\frac{2\mathrm{\Phi }}{\mathit{Aa}},b_1=0\text{.}$$

Using the values of the first set into Eq. (19) into Eq. (18), we have the exact solutions as follows: $$u_1(x,y,t)=\left\{\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}}-\frac{2\mathrm{\Phi }}{\mathit{Aa}}(d+\frac{B}{2\mathrm{\Phi }}+\frac{\sqrt{\mathrm{\Omega }}}{2\mathrm{\Phi }}\mathit{tanh}(\frac{\sqrt{\mathrm{\Omega }}}{2A})(x+y+\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2}t))\right\}\text{.}$$ $$u_2(x,y,t)=\left\{\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}}-\frac{2\mathrm{\Phi }}{\mathit{Aa}}(d+\frac{B}{2\mathrm{\Phi }}+\frac{\sqrt{\mathrm{\Omega }}}{2\mathrm{\Phi }}\mathit{coth}(\frac{\sqrt{\mathrm{\Omega }}}{2A})(x+y+\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2}t))\right\}\text{.}$$ $$u_3(x,y,t)=\left\{\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}}-\frac{2\mathrm{\Phi }}{\mathit{Aa}}(d+\frac{B}{2\mathrm{\Phi }}-\frac{\sqrt{-\mathrm{\Omega }}}{2\mathrm{\Phi }}\mathit{tan}(\frac{\sqrt{-\mathrm{\Omega }}}{2A})(x+y+\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2}t))\right\}\text{.}$$ $$u_4(x,y,t)=\left\{\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}}-\frac{2\mathrm{\Phi }}{\mathit{Aa}}(d+\frac{B}{2\mathrm{\Phi }}+\frac{\sqrt{-\mathrm{\Omega }}}{2\mathrm{\Phi }}\mathit{cot}(\frac{\sqrt{-\mathrm{\Omega }}}{2A})(x+y+\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2}t))\right\}\text{.}$$ $$u_5(x,y,t)=\left\{\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}}-\frac{2\mathrm{\Phi }}{\mathit{Aa}}(d+\frac{B}{2\mathrm{\Phi }}+\frac{C_2}{C_1+C_2(x+y+\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2}t)}\right\}\text{.}$$ $$u_6(x,y,t)=\left\{\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}}-\frac{2\mathrm{\Phi }}{\mathit{Aa}}(d+\frac{\sqrt{\mathrm{\Delta }}}{\mathrm{\Phi }}\mathit{tanh}(\frac{\sqrt{\mathrm{\Delta }}}{A})(x+y+\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2}t))\right\}\text{.}$$ $$u_7(x,y,t)=\left\{\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}}-\frac{2\mathrm{\Phi }}{\mathit{Aa}}(d+\frac{\sqrt{\mathrm{\Delta }}}{\mathrm{\Phi }}\mathit{coth}(\frac{\sqrt{\mathrm{\Delta }}}{A})(x+y+\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2}t))\right\}\text{.}$$ $$u_8(x,y,t)=\left\{\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}}-\frac{2\mathrm{\Phi }}{\mathit{Aa}}(d-\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Phi }}\mathit{tan}(\frac{\sqrt{-\mathrm{\Delta }}}{A})(x+y+\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2}t))\right\}\text{.}$$ $$u_9(x,y,t)=\left\{\frac{2d\mathrm{\Phi }+B}{\mathit{Aa}}-\frac{2\mathrm{\Phi }}{\mathit{Aa}}(d+\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Phi }}\mathit{cot}(\frac{\sqrt{-\mathrm{\Delta }}}{A})(x+y+\frac{6A^2-4E\mathrm{\Phi }-B^2}{2A^2}t))\right\}\text{.}$$

Figs. 1–5 manifest two of the answers under $2D,3D$ and the contour plot surfaces by applying Maple.

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Fig. 1

Graphical representation of $u_1(x,y,t)$ for $y=0,A=4,B=1,C=1,d=1,a=0.5$ and $E=1$ within the interval $-1⩽x,t⩽1$ and $t=0.01$ for two dimensional.

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Fig. 2

Graphical representation of $u_2(x,y,t)$ for $y=0,A=4,B=1,C=1,d=1,a=0.5$ and $E=1$ within the interval $-1⩽x,t⩽1$ and $t=0.01$ for two dimensional.

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Fig. 3

Graphical representation of $u_3(x,y,t)$ for $y=0,A=2,B=1,C=4,d=1,a=0.5$ and $E=1$ within the interval $-1⩽x,t⩽1$ and $t=0.01$ for two dimensional.

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Fig. 4

Graphical representation of $u_5(x,y,t)$ for $y=0,A=1,B=1,C=2,d=1,a=0.5$ and $E=1$ within the interval $-1⩽x,t⩽1$ and $t=0.01$ for two dimensional.

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Fig. 5

Graphical representation of $u_7(x,y,t)$ for $y=0,A=4,B=0,C=1,d=1,a=0.5$ and $E=5$ within the interval $-1⩽x,t⩽1$ and $t=0.01$ for two dimensional.

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4 The (2 + 1)-dimensional Kadomtsev-Petviashvili equation

In this section, the method is utilised to obtain compaction, cuspon, kink, periodic, soliton traveling wave solution, singular soliton traveling wave solution and various kinds of solutions for the $(2+1)$ -dimensional KP equation which are fundamental nonlinear evolution equations in the field of nonlinear dynamics. We consider the $(2+1)$ -dimensional KP equation:

Applying $u(\xi )=v(x,y,t),\xi =x+y-\mathit{Vt}$ provides the Eq. (20) into a nonlinear ODE:

It is taken upon integrating twice and putting coefficients of integration to zeros. Making the homogeneous rule in Eq. (21), we obtain $N=2$ and the solution as follows:

• The first set: $a_0=\frac{B^2+4E\mathrm{\Psi }}{A^2},a_1=0,a_2=-\frac{2{\mathrm{\Psi }}^2}{A^2},d=-\frac{B}{2\mathrm{\Psi }},b_1=0,b_2=-\frac{m_1}{8A^2{\mathrm{\Psi }}^2},V=\frac{\lambda A^2+16E\mathrm{\Psi }+4B^2}{A^2}$ , where $\mathrm{\Psi }=A-C,m_1=B^4+8{\mathit{EB}}^2\mathrm{\Psi }+16E^2{\mathrm{\Psi }}^2,A,B,C$ and E are free constants.

• The second set: $a_0=-\frac{2(-E\mathrm{\Psi }+\mathit{Bd}\mathrm{\Psi }+d^2{\mathrm{\Psi }}^2)}{A^2},a_1=\frac{2(B\mathrm{\Psi }+2d{\mathrm{\Psi }}^2)}{A^2},a_2=-\frac{2{\mathrm{\Psi }}^2}{A^2},d=d,b_1=0,b_2=0,V=\frac{\lambda A^2+4E\mathrm{\Psi }+4B^2}{A^2}$ , where $\mathrm{\Psi }=A-C,A,B,C$ and E are free constants.

• The third set: $a_0=-\frac{B^2-2E\mathrm{\Psi }+6\mathit{Bd}\mathrm{\Psi }+6d^2{\mathrm{\Psi }}^2}{3A^2},a_1=\frac{2(B\mathrm{\Psi }+2d{\mathrm{\Psi }}^2)}{A^2},a_2=-\frac{2{\mathrm{\Psi }}^2}{A^2},d=d,b_1=0,b_2=0,V=\frac{\lambda A^2-4E\mathrm{\Psi }+4B^2}{A^2}$ , where $\mathrm{\Psi }=A-C,A,B,C$ and E are free constants.

• The fourth set: $a_0=\frac{B^2+4E\mathrm{\Psi }}{A^2},a_1=0,a_2=0,d=-\frac{B}{2\mathrm{\Psi }},b_1=0,b_2=-\frac{m_2}{8A^2{\mathrm{\Psi }}^2},V=\frac{\lambda A^2+4E\mathrm{\Psi }+4B^2}{A^2}$ , where $\mathrm{\Psi }=A-C,m_2={(B^2+E\mathrm{\Psi })}^2,A,B,C$ and E are free constants.

• The fifth set: $a_0=\frac{B^2+4E\mathrm{\Psi }}{6A^2},a_1=0,a_2=0,d=-\frac{B}{2\mathrm{\Psi }},b_1=0,b_2=-\frac{m_3}{8A^2{\mathrm{\Psi }}^2},V=\frac{\lambda A^2-4E\mathrm{\Psi }-B^2}{A^2}$ , where $\mathrm{\Psi }=A-C,m_3={(B^2+E\mathrm{\Psi })}^2,A,B,C$ and E are free constants.

• The sixth set: $a_0=\frac{2m_4}{A^2},a_1=0,a_2=0,d=d,b_1=-\frac{2m_5}{A^2},b_2=-\frac{2m_6}{A^2},V=\frac{\lambda A^2+4E\mathrm{\Psi }+B^2}{A^2}$ , where $\mathrm{\Psi }=A-C,m_4=-(-E\mathrm{\Psi }+8d\mathrm{\Psi }+d^2{\mathrm{\Psi }}^2),m_5=(2d^3{\mathrm{\Psi }}^2-2\mathit{Ed}\mathrm{\Psi }-\mathit{EB}+B^{2}d+3{\mathit{Bd}}^2\mathrm{\Psi })$ , $m_6=-(d^4{\mathrm{\Psi }}^2-2{\mathit{Ed}}^2\mathrm{\Psi }+2{\mathit{Bd}}^3\mathrm{\Psi }+B^2{d}^2-2\mathit{BdE}+E^2),A,B,C$ and E are free constants.

• The seventh set: $a_0=\frac{2m_7}{3A^2},a_1=0,a_2=0,d=d,b_1=-\frac{2m_8}{A^2},b_2=-\frac{2m_9}{A^2},V=\frac{\lambda A^2-4E\mathrm{\Psi }-B^2}{A^2}$ , where $\mathrm{\Psi }=A-C,m_7=-(-B^2-2E\mathrm{\Psi }+6\mathit{Bd}\mathrm{\Psi }+6d^2{\mathrm{\Psi }}^2),m_8=(2d^3{\mathrm{\Psi }}^2-2\mathit{Ed}\mathrm{\Psi }-\mathit{EB}+B^{2}d+3{\mathit{Bd}}^2\mathrm{\Psi })$ , $m_9=-(d^4{\mathrm{\Psi }}^2-2{\mathit{Ed}}^2\mathrm{\Psi }+2{\mathit{Bd}}^3\mathrm{\Psi }+B^2{d}^2-2\mathit{BdE}+E^2),A,B,C$ and E are free constants.

By putting the values of the first set including Eq. (8) into Eq. (22) and simplifying, leads to the following exact wave solutions: $$u_{11}(\xi )=\frac{1}{2A^2}(2(B^2+4E\mathrm{\Psi })-\mathrm{\Omega }{\mathit{coth}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})-\frac{m_1}{\mathrm{\Omega }}{\mathit{tanh}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{12}(\xi )=\frac{1}{2A^2}(2(B^2+4E\mathrm{\Psi })-\mathrm{\Omega }{\mathit{tanh}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})-\frac{m_1}{\mathrm{\Omega }}{\mathit{coth}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{13}(\xi )=\frac{1}{2A^2}(2(B^2+4E\mathrm{\Psi })+\mathrm{\Omega }{\mathit{cot}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})+\frac{m_1}{\mathrm{\Omega }}{\mathit{tan}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{14}(\xi )=\frac{1}{2A^2}(2(B^2+4E\mathrm{\Psi })+\mathrm{\Omega }{\mathit{tan}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})+\frac{m_1}{\mathrm{\Omega }}{\mathit{cot}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{15}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{2{\mathrm{\Psi }}^2}{A^2}{(\frac{C_2}{C_1+C_2\xi })}^2-\frac{m_1}{8A^2{\mathrm{\Psi }}^2}{(\frac{C_2}{C_1+C_2\xi })}^{-2}),$$ $$u_{16}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{2{\mathrm{\Psi }}^2}{A^2}{(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{coth}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))}^2-\frac{m_1}{8A^2{\mathrm{\Psi }}^2}{(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{coth}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))}^{-2}),$$ $$u_{17}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{2{\mathrm{\Psi }}^2}{A^2}{(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{tanh}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))}^2-\frac{m_1}{8A^2{\mathrm{\Psi }}^2}{(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{tanh}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))}^{-2}),$$ $$u_{18}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{2{\mathrm{\Psi }}^2}{A^2}{(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{cot}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))}^2-\frac{m_1}{8A^2{\mathrm{\Psi }}^2}{(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{cot}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))}^{-2}),$$ $$u_{19}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{2{\mathrm{\Psi }}^2}{A^2}{(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{tan}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))}^2-\frac{m_1}{8A^2{\mathrm{\Psi }}^2}{(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{tan}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))}^{-2}),$$ where $\xi =x+y-(\frac{\lambda A^2+16E\mathrm{\Psi }+4B^2}{A^2})t$ .

Similarly for the second set, we have: $$u_{21}(\xi )=\frac{1}{2A^2}((B^2+4E\mathrm{\Psi })-\mathrm{\Omega }{\mathit{coth}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{22}(\xi )=\frac{1}{2A^2}((B^2+4E\mathrm{\Psi })-\mathrm{\Omega }{\mathit{tanh}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{23}(\xi )=\frac{1}{2A^2}((B^2+4E\mathrm{\Psi })+\mathrm{\Omega }{\mathit{cot}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{24}(\xi )=\frac{1}{2A^2}((B^2+4E\mathrm{\Psi })+\mathrm{\Omega }{\mathit{tan}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{25}(\xi )=\frac{1}{2A^2}((B^2+4E\mathrm{\Psi })-{(\frac{2\mathrm{\Psi }{C}_2}{C_1+C_2\xi })}^2),$$ $$u_{26}(\xi )=\frac{2}{A^2}(E\mathrm{\Psi }+\sqrt{\mathrm{\Delta }}(\mathit{Bcoth}(\frac{\sqrt{\mathrm{\Delta }}\xi }{A})-\sqrt{\mathrm{\Delta }}{\mathit{coth}}^2(\frac{\sqrt{\mathrm{\Delta }}\xi }{A}))),$$ $$u_{27}(\xi )=\frac{2}{A^2}(E\mathrm{\Psi }+\sqrt{\mathrm{\Delta }}(\mathit{Btanh}(\frac{\sqrt{\mathrm{\Delta }}\xi }{A})-\sqrt{\mathrm{\Delta }}{\mathit{tanh}}^2(\frac{\sqrt{\mathrm{\Delta }}\xi }{A}))),$$ $$u_{28}(\xi )=\frac{2}{A^2}(E\mathrm{\Psi }+\sqrt{\mathrm{\Delta }}(\mathit{IBcot}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A})-\sqrt{-\mathrm{\Delta }}{\mathit{cot}}^2(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))),$$ $$u_{29}(\xi )=\frac{2}{A^2}(E\mathrm{\Psi }-\sqrt{\mathrm{\Delta }}(\mathit{IBtan}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A})-\sqrt{-\mathrm{\Delta }}{\mathit{tan}}^2(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))),$$ where $\xi =x+y-(\frac{\lambda A^2+4E\mathrm{\Psi }+4B^2}{A^2})t$ .

Similarly again for the third set, we find: $$u_{31}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })-3\mathrm{\Omega }{\mathit{coth}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{32}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })-3\mathrm{\Omega }{\mathit{tanh}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{33}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })+3\mathrm{\Omega }{\mathit{cot}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{34}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })+3\mathrm{\Omega }{\mathit{tan}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{35}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })-3{(\frac{2\mathrm{\Psi }{C}_2}{C_1+C_2\xi })}^2),$$ $$u_{36}(\xi )=\frac{2}{3A^2}((-B^2+2E\mathrm{\Psi })+6\sqrt{\mathrm{\Delta }}(\mathit{Bcoth}(\frac{\sqrt{\mathrm{\Delta }}\xi }{A}-\sqrt{\mathrm{\Delta }}{\mathit{coth}}^2(\frac{\sqrt{\mathrm{\Delta }}\xi }{A}))),$$ $$u_{37}(\xi )=\frac{2}{3A^2}((-B^2+2E\mathrm{\Psi })+6\sqrt{\mathrm{\Delta }}(\mathit{Btanh}(\frac{\sqrt{\mathrm{\Delta }}\xi }{A}-\sqrt{\mathrm{\Delta }}{\mathit{tanh}}^2(\frac{\sqrt{\mathrm{\Delta }}\xi }{A}))),$$ $$u_{38}(\xi )=\frac{1}{3A^2}((-B^2+2E\mathrm{\Psi })+6\sqrt{\mathrm{\Delta }}(\mathit{IBcot}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}-\sqrt{\mathrm{\Delta }}{\mathit{cot}}^2(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))),$$ $$u_{39}(\xi )=\frac{2}{3A^2}((-B^2+2E\mathrm{\Psi })+6\sqrt{\mathrm{\Delta }}(\mathit{Btan}(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}+\sqrt{\mathrm{\Delta }}{\mathit{tan}}^2(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A}))),$$ where $\xi =x+y-(\frac{\lambda A^2-4E\mathrm{\Psi }+4B^2}{A^2})t$ .

Similarly for the fourth set, we get: $$u_{41}(\xi )=\frac{1}{2A^2}((B^2+4E\mathrm{\Psi })-\frac{m_2}{\mathrm{\Omega }}{\mathit{coth}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{42}(\xi )=\frac{1}{2A^2}((B^2+4E\mathrm{\Psi })-\frac{m_2}{\mathrm{\Omega }}{\mathit{tanh}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{43}(\xi )=\frac{1}{2A^2}((B^2+4E\mathrm{\Psi })+\frac{m_2}{\mathrm{\Omega }}{\mathit{cot}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{44}(\xi )=\frac{1}{2A^2}((B^2+4E\mathrm{\Psi })+\frac{m_2}{\mathrm{\Omega }}{\mathit{tan}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{45}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{m_2}{4{\mathrm{\Psi }}^2}{(\frac{C_2}{C_1+C_2\xi })}^{-2}),$$ $$u_{46}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{m_2}{4{\mathrm{\Psi }}^2}(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{coth}{(\frac{\sqrt{\mathrm{\Delta }}\xi }{A})}^{-2}),$$ $$u_{47}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{m_2}{4{\mathrm{\Psi }}^2}(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{tanh}{(\frac{\sqrt{\mathrm{\Delta }}\xi }{A})}^{-2}),$$ $$u_{48}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{m_2}{4{\mathrm{\Psi }}^2}(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{cot}{(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A})}^{-2}),$$ $$u_{49}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{m_2}{4{\mathrm{\Psi }}^2}(-\frac{-B}{2\mathrm{\Psi }}-\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{tan}{(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A})}^{-2}),$$ where $\xi =x+y-(\frac{\lambda A^2+4E\mathrm{\Psi }+4B^2}{A^2})t$ .

Similarly for the fifth set, we obtain: $$u_{51}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })-\frac{3m_3}{\mathrm{\Omega }}{\mathit{coth}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{52}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })-\frac{3m_3}{\mathrm{\Omega }}{\mathit{tanh}}^2(\frac{\sqrt{\mathrm{\Omega }}\xi }{2A})),$$ $$u_{53}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })+\frac{3m_3}{\mathrm{\Omega }}{\mathit{cot}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{54}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })+\frac{3m_3}{\mathrm{\Omega }}{\mathit{tan}}^2(\frac{\sqrt{-\mathrm{\Omega }}\xi }{2A})),$$ $$u_{55}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })-\frac{3m_3}{4{\mathrm{\Psi }}^2}{(\frac{C_2}{C_1+C_2\xi })}^{-2}),$$ $$u_{56}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })-\frac{3m_3}{4{\mathrm{\Psi }}^2}(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{coth}{(\frac{\sqrt{\mathrm{\Delta }}\xi }{A})}^{-2}),$$ $$u_{57}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })-\frac{3m_3}{4{\mathrm{\Psi }}^2}(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{tanh}{(\frac{\sqrt{\mathrm{\Delta }}\xi }{A})}^{-2}),$$ $$u_{58}(\xi )=\frac{1}{6A^2}((B^2+4E\mathrm{\Psi })-\frac{3m_3}{4{\mathrm{\Psi }}^2}(\frac{-B}{2\mathrm{\Psi }}+\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{cot}{(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A})}^{-2}),$$ $$u_{59}(\xi )=\frac{1}{A^2}((B^2+4E\mathrm{\Psi })-\frac{3m_3}{4{\mathrm{\Psi }}^2}(-\frac{-B}{2\mathrm{\Psi }}-\frac{\sqrt{-\mathrm{\Delta }}}{\mathrm{\Psi }}\mathit{tan}{(\frac{\sqrt{-\mathrm{\Delta }}\xi }{A})}^{-2}),$$ where $\xi =x+y-(\frac{\lambda A^2-4E\mathrm{\Psi }-B^2}{A^2})t$ .