In this paper, using the extended rational sine–cosine and rational sinh–cosh methods, we find new soliton solutions for the Klein–Gordon–Zakharov equations. The extended rational sine–cosine and rational sinh–cosh methods are prospering in finding soliton solutions of the Klein–Gordon–Zakharov equations. By means of these methods, we found some young solitons of the above mentioned equation. The conclusions we receive are dark, bright, and periodic. In addition, in order to imagine the underlying dynamics of the obtained soliton solutions, 2D and 3D plots are drawn.

## I. INTRODUCTION

Research on the exact traveling wave solutions to nonlinear partial differential equations (PDEs) and fractional differential, which portray physics phenomena, has become very important in the literature.^{1,2} Because many natural phenomena can be modeled in the form of these equations, it is always important to find the exact solution to them. In particular, problems in physical phenomena, engineering, and basic sciences can be modeled in the form of the above equations. Thus, a proliferation of coalition techniques, including multiple scale method,^{3} new extended direct algebraic method,^{4} auxiliary equation method,^{5,6} sine–cosine method,^{7,8} sine-Gordon expansion method,^{9} Liu’s extended trial function method,^{10} Hirota bilinear method,^{11} the first integral method,^{12,13} similarity transformation,^{14} extended trial method,^{15} Jacobi’s elliptic function expansion method,^{16} and extended F-expansion and projective Ricatti equation methods,^{17} were involved in the study of specific solitary wave solutions of PDEs.^{18–44} Recently, the Klein–Gordon–Zakharov (KGZ) equations have engrossed much consideration in different branches, such as biology, plasma physics, and optic fibers. In this paper, our purpose is to manufacture a new solitary wave solution by the extended rational sine–cosine and rational sinh–cosh methods to the KGZ equations. The residual part of this paper is formed as follows: in Sec. II, algorithms of extended rational sinh–cosh and sine–cosine methods are presented. In Sec. III, we represent these techniques with the KGZ equations. In addition, we consider the graphical representation of some acquired solutions in Sec. IV. Indeed, conclusion is provided in Sec. V.

## II. METHOD AND ALGORITHM

Consider nonlinear PDE of the form as

where *ψ* = *ψ*(*x*, *t*) is a traveling wave solution of nonlinear PDE (1).

Using the following transformation:

where *k* is the wave speed, Eq. (1) can be converted into an ODE as follows:

### A. Extended rational sine–cosine method

- Suppose that Eq. (3) has the solution in the form ofor(4)$\psi (\xi )=a0\u2061sin(\eta \xi )a2+a1\u2061cos(\eta \xi ),cos(\eta \xi )\u2260\u2212a2a1$where(5)$\psi (\xi )=a0\u2061cos(\eta \xi )a2+a1\u2061sin(\eta \xi ),sin(\eta \xi )\u2260\u2212a2a1,$
*a*_{i}, (*i*= 0, 1, 2) are parameters to be determined and*η*is the wave number. Unknown constants can be found by substituting Eq. (4) or Eq. (5) into Eq. (3), collecting all terms with the same powers of cos(

*ηξ*)^{m}or sin(*ηξ*)^{m}and equating to zero all the coefficients of cos(*ηξ*)^{m}or sin(*ηξ*)^{m}yield a set of algebraic equations. By calculation using the Maple software, the solutions of the algebraic equations can be derived.By substituting the values of

*a*_{0},*a*_{1},*a*_{2},*c*, and*η*into Eq. (4) or Eq. (5), the solution of Eq. (3) can be found.

### B. Extended rational sinh–cosh method

- Suppose that Eq. (3) has the solution in the formor(6)$\psi (\xi )=a0\u2061sinh(\eta \xi )a2+a1\u2061cosh(\eta \xi ),cosh(\eta \xi )\u2260\u2212a2a1$where(7)$\psi (\xi )=a0\u2061cosh(\eta \xi )a2+a1\u2061sinh(\eta \xi ),sinh(\eta \xi )\u2260\u2212a2a1,$
*a*_{i}, (*i*= 0, 1, 2) are parameters to be determined and*η*is the wave number. Substituting Eq. (6) or Eq. (7) into Eq. (3), collecting all terms with the same powers of cosh(

*ηξ*)^{m}or sinh(*ηξ*)^{m}and equating to zero all the coefficients of cosh(*ηξ*)^{m}or sinh(*ηξ*)^{m}yield a set of algebraic equations. The values of the unknowns are to be determined by solving the algebraic equations.The solution of Eq. (3) can be found by substituting the values of

*a*_{0},*a*_{1},*a*_{2},*c*, and*η*into Eq. (6) or Eq. (7).

## III. IMPLEMENTATION OF THE METHOD

where *ϕ*(*x*, *t*) stands for a complex function, while *v*(*x*, *t*) is a real function, *t* shows the time, and *x* is the distance along the direction of publication. Therefore, *α* and *β* are nonzero parameters.

To construct the soliton wave solution to the KGZ equations [Eqs. (8) and (9)], we adopt the transformation in the form of

where *ψ*(*x*, *t*) = *ψ*(*ξ*) is a real function, while *k* and *ω* are constants to be determined, and *θ* stands for an arbitrary constant.

Next, to surmise *v*(*x*, *t*) = *W*(*ξ*) and insert (14) in (13), after integrating twice with a nonzero constant value, it is recovered that

where Δ = *αC* + *k*^{2} − *ω*^{2} + 1. Now, instead of solving the main equation, we solve the last equation [Eq. (16)], and with the method mentioned in the second part and using the Maple software, we obtain the equation coefficients. After solving the system of equations, the following set of solutions is obtained for numerical coefficients.^{46}

### A. Soliton wave solution by ERSCM

Suppose that Eq. (16) has solutions in the form of

Substituting Eq. (17) into Eq. (16) and collecting all terms with the same powers of cos(*ηξ*)^{m} and equating to zero all the coefficients of cos(*ηξ*)^{m}, the following algebraic equations are obtained:

After solving these algebraic equations, the following solutions are obtained:

- Taking set 1 into account, the solutions of (16) can be obtained asCombining Eqs. (10) and (18), we get(18)$\psi 1(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta tan12\u22122\Delta k2\u2212\omega 2\xi \omega .$(19)$\varphi 1(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta tan12\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+\omega t+\theta \omega ,$(20)$W1(x,t)=\Delta tan\u2212\Delta 2k2\u22122\omega 2kt+\omega x2\alpha +C.$
- Similarly, for set 2, the solutions of (16) can be obtained asCombining Eqs. (10) and (21), we get(21)$\psi 2(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta sin\u22122\Delta k2\u2212\omega 2\xi \omega cos\u22122\Delta k2\u2212\omega 2\xi \u22121.$(22)$\varphi 2(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta sin\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+t\omega +\theta \omega cos\u22122\Delta k2\u2212\omega 2kt+\omega x\u22121,$(23)$W2(x,t)=\Delta sin\u22122\Delta k2\u2212\omega 2kt+\omega x2\alpha cos\u22122\Delta k2\u2212\omega 2kt+\omega x\u221212+C.$
- Similarly, for set 2, the solutions of (16) can be obtained asCombining Eqs. (10) and (24), we get(24)$\psi 3(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta sin\u22122\Delta k2\u2212\omega 2\xi \omega cos\u22122\Delta k2\u2212\omega 2\xi +1.$(25)$\varphi 3(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta sin\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+t\omega +\theta \omega cos\u22122\Delta k2\u2212\omega 2kt+\omega x+1,$OR(26)$W3(x,t)=\Delta sin\u22122\Delta k2\u2212\omega 2kt+\omega x2\alpha cos\u22122\Delta k2\u2212\omega 2kt+\omega x+12+C.$Suppose that Eq. (16) has solutions in the form ofSubstituting Eq. (27) into Eq. (16) and collecting all terms with the same powers of sin((27)$\psi (\xi )=a0\u2061cos(\eta \xi )a2+a1\u2061sin(\eta \xi ).$
*ηξ*)^{m}and equating to zero all the coefficients of sin(*ηξ*)^{m}, the following algebraic equations are obtained:After solving these algebraic equations, the following solutions are obtained:$sin(\eta \xi )2:\u2212C\alpha k2a12+C\alpha \omega 2a12+\alpha \beta \omega 2a02\u2212k4a12+2k2\omega 2a12\u2212\omega 4a12\u2212k2a12+\omega 2a12a0,sin(\eta \xi )1:\u2212\eta 2a2a1k4+2\eta 2a2a1k2\omega 2\u2212\eta 2a2a1\omega 4\u22122C\alpha k2a2a1+2C\alpha \omega 2a2a1\u22122a1a2k4+4a1a2k2\omega 2\u22122a1a2\omega 4\u22122k2a2a1+2\omega 2a2a1a0,sin(\eta \xi )0:\u2212a0k\u2212\omega k+\omega \eta 2\u22121\omega 2+\u2212\eta 2+1k2+C\alpha +1a22+2\eta 2\omega 4a12+\u22124\eta 2k2a12+\alpha \beta a02\omega 2+2\eta 2k4a12.$$Set\u20093)\eta =\xb1\u2212\Delta 2k2\u22122\omega 2,a0=\xb1\u2212\Delta 2k2\u22122\omega 2,a1=a1,a2=0,Set\u20094)\eta =\xb1\u22122\Delta k2\u2212\omega 2,a0=\xb1a2\omega \u2212\Delta (k2\u2212\omega 2)\alpha \beta ,a1=\xb1a2,a2=a2.$ - Taking set 3 into account, the solutions of (16) can be obtained asCombining Eqs. (10) and (28), we get(28)$\psi 4(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta cot12\u22122\Delta k2\u2212\omega 2\xi \omega .$(29)$\varphi 4(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta cot12\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+\omega t+\theta \omega ,$(30)$W4(x,t)=\Delta cot\u2212\Delta 2k2\u22122\omega 2kt+\omega x2\alpha +C.$
- Similarly, for set 4, the solutions of (16) can be obtained asCombining Eqs. (10) and (31), we get(31)$\psi 5(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta cos\u22122\Delta k2\u2212\omega 2\xi \omega sin\u22122\Delta k2\u2212\omega 2\xi +1.$(32)$\varphi 5(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta cos\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+t\omega +\theta \omega sin\u22122\Delta k2\u2212\omega 2kt+\omega x+1,$(33)$W5(x,t)=\u2212\Delta cos\u22122\Delta k2\u2212\omega 2kt+\omega x2\alpha cos\u22122\Delta k2\u2212\omega 2kt+\omega x2\u22122\u2061sin\u22122\Delta k2\u2212\omega 2kt+\omega x\u22122+C.$
- Similarly, for set 4, the solutions of (16) can be obtained asCombining Eqs. (10) and (34), we get(34)$\psi 6(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta cos\u22122\Delta k2\u2212\omega 2\xi \omega sin\u22122\Delta k2\u2212\omega 2\xi +1.$(35)$\varphi 6(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta cos\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+t\omega +\theta \omega sin\u22122\Delta k2\u2212\omega 2kt+\omega x\u22121,$(36)$W6(x,t)=\u2212\Delta cos\u22122\Delta k2\u2212\omega 2kt+\omega x2\alpha cos\u22122\Delta k2\u2212\omega 2kt+\omega x2\u22122\u2061sin\u22122\Delta k2\u2212\omega 2kt+\omega x\u22122+C.$

### B. Soliton wave solution by extended rational sinh–cosh method

Suppose that Eq. (16) has solutions in the form of

Substituting Eq. (37) into Eq. (16) and collecting all terms with the same powers of cosh(*ηξ*)^{m} and equating to zero all the coefficients of cosh(*ηξ*)^{m}, the following algebraic equations are obtained:

After solving these algebraic equations, the following solutions are obtained:

- Taking set 5 into account, the solutions of (16) can be obtained asCombining Eqs. (10) and (38), we get(38)$\psi 7(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta tanh12\u22122\Delta k2\u2212\omega 2\xi \omega .$(39)$\varphi 7(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta tanh12\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+\omega t+\theta \omega ,$(40)$W7(x,t)=\Delta tanh\u2212\Delta 2k2\u22122\omega 2kt+\omega x2\alpha +C.$
- Similarly, for set 6, the solutions of (16) can be obtained asCombining Eqs. (10) and (41), we get(41)$\psi 8(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta sinh\u22122\Delta k2\u2212\omega 2\xi \omega cosh\u22122\Delta k2\u2212\omega 2\xi \u22121.$(42)$\varphi 8(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta sinh\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+t\omega +\theta \omega cosh\u22122\Delta k2\u2212\omega 2kt+\omega x\u22121,$(43)$W8(x,t)=\Delta sinh\u22122\Delta k2\u2212\omega 2kt+\omega x2\alpha cosh\u22122\Delta k2\u2212\omega 2kt+\omega x\u221212+C.$
- Similarly, for set 6, the solutions of (16) can be obtained asCombining Eqs. (10) and (44), we get(44)$\psi 9(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta sinh\u22122\Delta k2\u2212\omega 2\xi \omega cosh\u22122\Delta k2\u2212\omega 2\xi +1.$(45)$\varphi 9(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta sinh\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+t\omega +\theta \omega cosh\u22122\Delta k2\u2212\omega 2kt+\omega x+1,$OR(46)$W9(x,t)=\Delta sinh\u22122\Delta k2\u2212\omega 2kt+\omega x2\alpha cosh\u22122\Delta k2\u2212\omega 2kt+\omega x+12+C.$Suppose that Eq. (16) has solutions in the form ofSubstituting Eq. (47) into Eq. (16) and collecting all terms with the same powers of sinh((47)$\psi (\xi )=a0\u2061cosh(\eta \xi )a2+a1\u2061sinh(\eta \xi ).$
*ηξ*)^{m}and equating to zero all the coefficients of sinh(*ηξ*)^{m}, the following algebraic equations are obtained:$sinh(\eta \xi )2:C\alpha k2a12\u2212C\alpha \omega 2a12\u2212\alpha \beta \omega 2a02+k4a12\u22122k2\omega 2a12+\omega 4a12+k2a12\u2212\omega 2a12a0,$$sinh(\eta \xi )1:\u2212\eta 2a2a1k4+2\eta 2a2a1k2\omega 2\u2212\eta 2a2a1\omega 4\u22122aCk2a2a1+2aC\omega 2a2a1\u22122a2a1k4+4a2a1k2\omega 2\u22122a2a1\omega 4\u22122a2a1k2+2\omega 2a1a2a0,sinh(\eta \xi )0:\u2212a0k+\omega k\u2212\omega \eta 2\u22121\omega 2+\u2212\eta 2+1k2+\alpha C+1a22+2\eta 2\omega 4a12+\u22124\eta 2k2a12+\alpha \beta a02\omega 2+2\eta 2k4a12.$After solving these algebraic equations, the following solutions are obtained:$Set\u20097)\eta =\xb1\u2212\Delta 2k2\u22122\omega 2,a0=\xb1a1\omega \u2212\Delta (k2\u2212\omega 2)\alpha \beta ,a1=a1,a2=0,Set\u20098)\xb1\eta =\u22122\Delta k2\u2212\omega 2,a0=\xb1a2\omega \u2212\Delta (k2\u2212\omega 2)\alpha \beta ,a1=\xb1a2,a2=a2.$ - Taking set 7 into account, the solutions of (16) can be obtained asCombining Eqs. (10) and (48), we get(48)$\psi 10(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta coth12\u22122\Delta k2\u2212\omega 2\xi \omega .$(49)$\varphi 10(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta coth12\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+\omega t+\theta \omega ,$(50)$W10(x,t)=\Delta coth\u2212\Delta 2k2\u22122\omega 2kt+\omega x2\alpha +C.$
- Similarly, for set 8, the solutions of (16) can be obtained asCombining Eqs. (10) and (51), we get(51)$\psi 11(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta cosh\u22122\Delta k2\u2212\omega 2\xi \omega sinh\u22122\Delta k2\u2212\omega 2\xi +1.$(52)$\varphi 11(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta cosh\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+t\omega +\theta \omega sinh\u22122\Delta k2\u2212\omega 2kt+\omega x+1,$(53)$W11(x,t)=\Delta cosh\u22122\Delta k2\u2212\omega 2kt+\omega x2\alpha cosh\u22122\Delta k2\u2212\omega 2kt+\omega x2\u22122\u2061sinh\u22122\Delta k2\u2212\omega 2kt+\omega x+2+C.$
- Similarly, for set 8, the solutions of (16) can be obtained asCombining Eqs. (10) and (54), we get(54)$\psi 12(\xi )=\xb1\Delta k2\u2212\omega 2\alpha \beta cosh\u22122\Delta k2\u2212\omega 2\xi \omega sinh\u22122\Delta k2\u2212\omega 2\xi \u22121.$(55)$\varphi 12(x,t)=\xb1\Delta k2\u2212\omega 2\alpha \beta cosh\u22122\Delta k2\u2212\omega 2kt+\omega xeikx+t\omega +\theta \omega sinh\u22122\Delta k2\u2212\omega 2kt+\omega x\u22121,$(56)$W12(x,t)=\Delta cosh\u22122\Delta k2\u2212\omega 2kt+\omega x2\alpha cosh\u22122\Delta k2\u2212\omega 2kt+\omega x2\u22122\u2061sinh\u22122\Delta k2\u2212\omega 2kt+\omega x\u22122+C.$

## IV. PHYSICAL REPRESENTATION OF THE REPORTED RESULTS

In this section, we have provided the 2D and 3D graphs to some of the acquired conclusions based on the appropriate values of the parameters. We also checked the wave behaviors of the KGZ equations by using extended rational sine–cosine and extended rational sinh–cosh methods. The soliton wave solutions are acquired by the offered methods and graphically determined into a diversity of distinguished physical structures, such as periodic, dark, and bright soliton functions. These functions have different physical significance.

In this text, Fig. 1 illustrates 2D and 3D surfaces of the bright periodic solution of the Eq. |*ϕ*_{1}(*x*, *t*)| for the parametric values *α* = 3, *k* = 0.2, *ω* = 1.5, *β* = 1, *C* = 2, *θ* = 1, and *y* = 0.001 for the 2D graphics.

Figure 2 shows 2D and 3D bell-shaped periodic solution of Eq. (20) at *α* = 3, *k* = 0.2, *ω* = 1.5, *β* = 1, *C* = 2, *θ* = 1, and *y* = 0.001 for the 2D graphics.

Figure 3 represents 2D and 3D surfaces of the bright periodic solution of the Eq. |*ϕ*_{4}(*x*, *t*)| for the parametric values *α* = 3, *k* = 0.2, *ω* = 1.5, *β* = 1, *C* = 2, *θ* = 1, and *y* = 0.001 for the 2D graphics. Figure 4 also indicates 2D and 3D bell-shaped periodic solution of Eq. (30) at *α* = 3, *k* = 0.2, *ω* = 1.5, *β* = 1, *C* = 2, *θ* = 1, and *y* = 0.001 for the 2D graphics that are similar in shape like Fig. 2.

Figure 5 shows 2D and 3D surfaces of the dark wave soliton solution of Eq. |*ϕ*_{7}(*x*, *t*)| for the KGZ at *α* = 3, *k* = 0.2, *ω* = 1.5, *β* = 1, *C* = 2, *θ* = 1, and *y* = 0.001 for the 2D graphics.

Figure 6 also indicates 2D and 3D dark wave soliton solution of Eq. (40) at *α* = 3, *k* = 0.2, = 1.5, *β* = 1, *C* = 2, *θ* = 1, and *y* = 0.001 for the 2D graphics that are similar in shape like Fig. 5.

Figure 7 shows the cusp wave soliton solution that comes from Eq. (49) for the parametric values *α* = 3, *k* = 0.2, *ω* = 1.5, *β* = 1, *C* = 2, *θ* = 1, and *y* = 0.001 for the 2D graphics.

## V. CONCLUSION

In this paper, plentiful new soliton solutions for the KGZ equations were checked by extended rational sine–cosine (ERSC) and extended rational sinh–cosh methods. These methods are useful for finding soliton and singular periodic solutions of nonlinear PDEs and have been successfully applied to obtain some new soliton solutions to the KGZ equations to attain the purpose. First, we assume that KGZ equations have the answer as mentioned above. Then, by placing this answer in the KGZ equations, we arrive at a set of equations. Then, we find the coefficients of this system of equations using Maple or Mathematics software and then we find a new soliton wave solution by placing them in the KGZ equations. We note that comparatively with Refs. 18, 19, and 47, the received conclusions are more special. We expect that the gained one soliton solutions should be beneficial in solitary wave theory.

## ACKNOWLEDGMENTS

The authors acknowledge the National Natural Science Foundation of China (Grant No. 71601072), the Key Scientific Research Project of Higher Education Institutions in Henan Province of China (Grant No. 20B110006), and the Fundamental Research Funds for the Universities of Henan Province (Grant No. NSFRF210314).

## DATA AVAILABILITY

Data sharing is not applicable to this article as no new data were created or analyzed in this study.