### 3.1 Fourth-order nonlinear Ablowitz–Kaup–Newell–Segur water equation

In this section, we apply our method to the well-known AKNS equation [44] of the general form

$$ 4u_{xt}+u_{xxxt}+8u_{x}u_{xy}+ 4u_{xx}u_{y}-\gamma u_{xx} =0. $$

(9)

This equation includes some nonlinear evolution equations such as sine-Gordon, nonlinear Schrödinger, KdV, and other equations having numerous application in physical science.

The traveling wave transformations can be reduced to the form

$$ u(x,y,t)=U(\zeta ) , \quad \zeta =x+y+\omega t, $$

(10)

where *ω* is an arbitrary constant to be determined latter. Applying this transformation to Eq. (10) and integrating, we arrive at the ordinary differential equation

$$ (4\omega -\gamma )U^{\prime}+6U^{\prime {2}} +\omega U^{'''}=0. $$

(11)

Applying the homogeneous balance principle between \(U^{\prime 2}\) and \(U^{\prime\prime\prime} \) in Eq. (11), we get \(m=1\). We assume that solution of Eq. (11) is of the form

$$ U(\zeta )=A_{0}+A_{1} \biggl( \frac{G'}{G} \biggr) . $$

(12)

Substituting Eq. (12) along with Eq. (5) into Eq. (11), we get algebraic equations in parameters \(A_{0}\), \(A _{1}\), and *ω*. This system of equations can be solved with the help of Mathematica:

$$ A_{1}= \biggl( \frac{\gamma }{\lambda_{1} ^{2}-4\lambda_{2} +4} \biggr) , \qquad A_{0}=A_{0}, \qquad \omega = \biggl( \frac{\beta }{6+\lambda_{1} ^{2}-4 \lambda_{2} -2} \biggr) . $$

(13)

Substituting Eq. (13) into Eq. (12), we obtain the exact traveling solution

$$ U(\zeta )=A_{0}+ \biggl( \frac{\gamma }{\lambda_{1} ^{2}-4\lambda_{2} +4} \biggr) \biggl( \frac{G'}{G} \biggr) . $$

(14)

Now we discuss three different cases regarding the solution (14).

*Case* 1. When \(\lambda_{1}^{2}-4\lambda_{2}>0\), substituting Eq. (6) into Eq. (14), we get the following solution of Eq. (9):

$$\begin{aligned} &U_{1}(\zeta ) \\ &\quad =A_{0}+ \biggl(\frac{\gamma }{\lambda_{1} ^{2}-4 \lambda_{2} +4} \biggr) \\ &\quad \quad {}\times \biggl( \frac{\sqrt{\lambda_{1}^{2}-4\lambda_{2}}}{2} \biggl( \frac{B_{1}\sinh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4\lambda _{2}})\zeta +B_{2}\cosh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4\lambda _{2}})\zeta }{B_{1}\cosh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4\lambda _{2}})\zeta +B_{2}\sinh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4\lambda _{2}})\zeta } \biggr) -\frac{\lambda_{1}}{2} \biggr) . \end{aligned}$$

(15)

*Case* 2. When \(\lambda_{1}^{2}-4\lambda_{2}<0\), substituting Eq. (7) into Eq. (14), we get the following solution of Eq. (9):

$$\begin{aligned} &U_{2}(\zeta ) \\ &\quad =A_{0}+ \biggl(\frac{\gamma }{\lambda_{1} ^{2}-4 \lambda_{2} +4} \biggr) \\ &\quad \quad {}\times \biggl( \frac{\sqrt{4\lambda_{2}-\lambda_{1}^{2}}}{2} \biggl( \frac{-B_{1} \sin (\frac{1}{2}\sqrt{4\lambda_{2}-\lambda_{1}^{2}})\zeta +B_{2} \cos (\frac{1}{2}\sqrt{4\lambda_{2}-\lambda_{1}^{2}})\zeta }{B_{1} \cos (\frac{1}{2}\sqrt{4\lambda_{2}-\lambda_{1}^{2}})\zeta +B_{2} \sin (\frac{1}{2}\sqrt{4\lambda_{2}-\lambda_{1}^{2}})\zeta } \biggr) -\frac{ \lambda_{1}}{2} \biggr) . \end{aligned}$$

(16)

*Case* 3. When \(\lambda_{1}^{2}-4\lambda_{2}=0\), substituting Eq. (9) into Eq. (14), we get the following solution of Eq. (9):

$$ U_{3}(\zeta )=A_{0}+ \biggl( \frac{\gamma }{\lambda_{1} ^{2}-4 \lambda _{2} +4} \biggr) \biggl( \frac{B_{2}}{B_{1}+B_{2}\zeta }-\frac{\lambda _{1}}{2} \biggr) . $$

(17)

### 3.2
\((2+1)\)-D Boussinesq dynamical equation

The general form of the \((2+1)\)-dimensional Boussinesq dynamical equation [45] is the following;

$$ u_{tt}-u_{xx}-\beta \bigl(u^{2} \bigr)_{xx}-u_{yy}-u_{xxxx}=o,\quad \beta \neq 0. $$

(18)

The extreme merit of this equation is that it is used to describe the gravity waves propagation on the water surface and also clarifies the collision of oblique wave transformation.

The traveling wave transformations for Eq. (18) can be deduced as

$$ u(x,y,t)=U(\zeta ) ,\quad \zeta =x+y+k t. $$

(19)

Using these transformations to Eq. (18), we obtain the ordinary differential equation

$$ 2\beta U^{\prime 2}+2\beta U U^{\prime\prime}+\bigl(2-k^{2}\bigr)U^{\prime\prime}+U^{\prime\prime\prime\prime}=0. $$

(20)

Here, applying the homogeneous balance principle to Eq. (20), we get \(m=2\). We suppose that the solution of Eq. (20) is of the form

$$ U(\zeta )=A_{0}+A_{1} \biggl( \frac{G'}{G} \biggr) +A_{2} \biggl( \frac{G'}{G} \biggr) ^{2}. $$

(21)

Substituting Eq. (21) along with Eq. (5) into Eq. (20), we obtain numerous algebraic equations in parameters \(A_{0}\), \(A_{1}\), \(A _{2}\), and *k*. These equations can be solved with the help of Mathematica:

$$ A_{0}= \frac{-\lambda_{1} ^{2}-8\lambda_{2} +k ^{2}-2}{2 \beta },\qquad A_{1}= - \frac{6 \lambda_{1} }{\beta },\qquad A_{2}= -\frac{6}{\beta }. $$

(22)

Substituting Eq. (22) into Eq. (21), we obtain the exact traveling solution

$$\begin{aligned} U(\zeta )= \frac{-\lambda_{1} ^{2}-8\lambda_{2} +k ^{2}-2}{2 \beta } -\frac{6 \lambda_{1} }{\beta } \biggl( \frac{G'}{G} \biggr) -\frac{6}{ \beta } \biggl( \frac{G'}{G} \biggr) ^{2}. \end{aligned}$$

(23)

Now we discuss three different cases regarding solution (23).

*Case* 1. When \(\lambda_{1}^{2}-4\lambda_{2}>0\), substituting Eq. (6) into Eq. (23), we obtain the following solution of Eq. (18):

$$\begin{aligned} U_{4}(\zeta )&= \frac{-\lambda_{1} ^{2}-8 \lambda_{2} +k ^{2}-2}{2\beta } \\ &\quad {}- \frac{6 \lambda_{1} }{\beta } \biggl( \frac{\sqrt{\lambda_{1} ^{2}-4\lambda_{2}}}{2} \biggl( \frac{B_{1}\sinh (\frac{1}{2}\sqrt{ \lambda_{1}^{2}-4\lambda_{2}})\zeta +B_{2}\cosh (\frac{1}{2}\sqrt{ \lambda_{1}^{2}-4\lambda_{2}})\zeta }{B_{1}\cosh (\frac{1}{2}\sqrt{ \lambda_{1}^{2}-4\lambda_{2}})\zeta +B_{2}\sinh (\frac{1}{2}\sqrt{ \lambda_{1}^{2}-4\lambda_{2}})\zeta } \biggr) - \frac{\lambda_{1}}{2} \biggr) \\ &\quad {}-\frac{6}{\beta } \biggl( \frac{\sqrt{\lambda_{1}^{2}-4\lambda_{2}}}{2} \biggl( \frac{B_{1}\sinh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4\lambda _{2}})\zeta +B_{2}\cosh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4\lambda _{2}})\zeta }{B_{1}\cosh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4\lambda _{2}})\zeta +B_{2}\sinh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4\lambda _{2}})\zeta } \biggr) -\frac{\lambda_{1}}{2} \biggr) ^{2}, \\ &\quad \beta \neq 0. \end{aligned}$$

(24)

*Case* 2. When \(\lambda_{1}^{2}-4\lambda_{2}<0\), substituting Eq. (7) into Eq. (23), we obtain the following solution of Eq. (18):

$$\begin{aligned} U_{5}(\zeta )&= \frac{-\lambda_{1} ^{2}-8 \lambda_{2} +k ^{2}-2}{2\beta } \\ &\quad {}-\frac{6 \lambda_{1} }{\beta } \biggl( \frac{\sqrt{4\lambda _{2}-\lambda_{1}^{2}}}{2} \biggl( \frac{-B_{1}\sin (\frac{1}{2}\sqrt{4 \lambda_{2}-\lambda_{1}^{2}})\zeta +B_{2}\cos (\frac{1}{2}\sqrt{4 \lambda_{2}-\lambda_{1}^{2}})\zeta }{B_{1}\cos (\frac{1}{2}\sqrt{4 \lambda_{2}-\lambda_{1}^{2}})\zeta +B_{2}\sin (\frac{1}{2}\sqrt{4 \lambda_{2}-\lambda_{1}^{2}})\zeta } \biggr) - \frac{\lambda_{1}}{2} \biggr) \\ &\quad {}-\frac{6}{\beta } \biggl( \frac{\sqrt{4\lambda_{2}-\lambda_{1}^{2}}}{2} \biggl( \frac{-B_{1} \sin (\frac{1}{2}\sqrt{4\lambda_{2}-\lambda_{1}^{2}})\zeta +B_{2} \cos (\frac{1}{2}\sqrt{4\lambda_{2}-\lambda_{1}^{2}})\zeta }{B_{1} \cos (\frac{1}{2}\sqrt{4\lambda_{2}-\lambda_{1}^{2}})\zeta +B_{2} \sin (\frac{1}{2}\sqrt{4\lambda_{2}-\lambda_{1}^{2}})\zeta } \biggr) -\frac{\lambda_{1}}{2} \biggr) ^{2}, \\ &\quad {}\beta \neq 0. \end{aligned}$$

(25)

*Case 3.* When \(\lambda_{1}^{2}-4\lambda_{2}=0\), substituting Eq. (8) into Eq. (23), we obtain the following solution of Eq. (18):

$$\begin{aligned} &U_{6}(\zeta )\frac{-\lambda_{1} ^{2}-8 \lambda_{2} +k ^{2}-2}{2 \beta } -\frac{6 \lambda_{1} }{\beta }\biggl( \frac{B_{2}}{B_{1}+B_{2} \zeta }-\frac{\lambda_{1}}{2} \biggr) \\ &\quad {}-\frac{6}{\beta } \biggl( \frac{B _{2}}{B_{1}+B_{2}\zeta }-\frac{\lambda_{1}}{2} \biggr) ^{2},\quad \beta \neq 0. \end{aligned}$$

(26)

### 3.3
\((3+1)\)-Dimensional Yu–Toda–Sasa–Fukuyama equation

The general form of the \((3+1)\)-dimensional Yu–Toda–Sasa–Fukuyama equation [46] is

$$ -4u_{xt}+u_{xxxz}+4u_{x}u_{xz}+2u_{z}u_{xx}+3u_{yy}=0. $$

(27)

The advantage of this equation is that it is used for investigation of the dynamics of solutions and nonlinear waves in fluid dynamics, plasma physics, and weakly dispersive media.

The traveling wave transformations for the Yu–Toda–Sasa–Fukuyama equation can be deduced as

$$ u(x,y,z,t)=U(\zeta ) ,\quad \zeta =x+y+z-\eta t. $$

(28)

Using these transformations in Eq. (27) and integrating, we obtain the ordinary differential equation

$$ U^{\prime \prime \prime}+3 U^{\prime 2}+(4\eta +3)U^{\prime}=0. $$

(29)

We assume that the solution of Eq. (29) has the form (12). Substituting Eq. (12) along Eq. (5) into Eq. (29), we obtain a series of algebraic equations in parameters \(A _{0}\), \(A_{1}\), and *η*. This class of algebraic equations can be solved, We have:

$$ A_{1}= 2,\qquad A_{0}=A_{0}, \qquad \eta = \frac{1}{4} \bigl( -\lambda_{1} ^{2}+4 \lambda_{2} -3 \bigr) . $$

(30)

Substituting Eq. (30) into Eq. (12), we obtain the exact traveling wave solution

$$ U(\zeta )=A_{0}+ 2 \biggl( \frac{G'}{G} \biggr) . $$

(31)

Now we discuss three different cases regarding solution (31).

*Case* 1. When \(\lambda_{1}^{2}-4\lambda_{2}>0\), substituting Eq. (6) into Eq. (31), we obtain the following solution of Eq. (27):

$$\begin{aligned} &U_{7}(\zeta ) \\ &\quad =A_{0} \\ &\quad \quad {}+2 \biggl( \frac{\sqrt{\lambda_{1}^{2}-4\lambda _{2}}}{2} \biggl( \frac{B_{1}\sinh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4 \lambda_{2}})\zeta +B_{2}\cosh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4 \lambda_{2}})\zeta }{B_{1}\cosh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4 \lambda_{2}})\zeta +B_{2}\sinh (\frac{1}{2}\sqrt{\lambda_{1}^{2}-4 \lambda_{2}})\zeta } \biggr) -\frac{\lambda_{1}}{2} \biggr). \end{aligned}$$

(32)

*Case* 2. When \(\lambda_{1}^{2}-4\lambda_{2}<0\), substituting Eq. (7) into Eq. (31), we obtain the following solution of Eq. (27):

$$\begin{aligned} &U_{8}(\zeta ) \\ &\quad =A_{0} \\ &\quad \quad {}+2 \biggl( \frac{\sqrt{4\lambda_{2}-\lambda_{1} ^{2}}}{2} \biggl( \frac{-B_{1}\sin (\frac{1}{2}\sqrt{4\lambda_{2}- \lambda_{1}^{2}})\zeta +B_{2}\cos (\frac{1}{2}\sqrt{4\lambda_{2}- \lambda_{1}^{2}})\zeta }{B_{1}\cos (\frac{1}{2}\sqrt{4\lambda_{2}- \lambda_{1}^{2}})\zeta +B_{2}\sin (\frac{1}{2}\sqrt{4\lambda_{2}- \lambda_{1}^{2}})\zeta } \biggr) -\frac{\lambda_{1}}{2} \biggr) . \end{aligned}$$

(33)

*Case* 3. When \(\lambda_{1}^{2}-4\lambda_{2}=0\), substituting Eq. (8) into Eq. (31), we obtain the following solution of Eq. (27):

$$ U_{9}(\zeta )=A_{0}+2 \biggl( \frac{B_{2}}{B_{1}+B_{2}\zeta }-\frac{ \lambda_{1}}{2} \biggr) . $$

(34)