# A Mathematical Analysis of a Model Involving an Integrable Equation for Wave Packet Envelope

Authors: Melike Kaplan; Arzu Akbulut
Journal: Journal of Mathematics
 (2022)

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## Abstract

This study investigates new optical solutions to a model with an integrable equation for wave packet envelopes. For this purpose, we have employed two reliable techniques involving the modified extended tanh function and the exponential rational function procedures. We have also given the 3D graphics of the obtained solutions.


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## Body

## 1. Introduction

Numerous complex phenomena that are encountered in relativity, fluid dynamics, mathematical physics, economics, and optical fibers are modeled via nonlinear partial differential equations (NPDEs) [1]. Recently, new methods have been recommended for finding solutions to the NPDEs. Some notable studies have been discussed. Alquran has obtained periodic solutions and bell-shaped soliton solutions to the two-component evolutionary system of homogeneous KdV equations of the second order, the Gardner equation, Benjamin–Bona–Mahony equation, and the Cassama-Holm equation via the sine-cosine method [2]. Akbulut et al. have employed the exp −φξ-expansion procedure to the ill-posed Boussinesq equation and Zakharov–Kuznetsov–Benjamin–Bona–Mahony equation to obtain trigonometric, hyperbolic, and rational functions for these equations [3]. In 2015, Kaplan et al. applied the auxiliary equation procedure to the nonlinear Zoomeron equation, equal width wave equation, and coupled Higgs equation [4]. Ali et al. discussed cubic optical solitons in a polarization-preserving fiber described via nonlinear Schrödinger equation (NLSE) via the new extended direct algebraic method [5]. Yao et al. used the improved generalized Riccati equation mapping procedure to get traveling wave solutions for the higher-order Sasa-Satsuma equation [6]. Durur et al. applied the subequation procedure to find exact solutions for the KdV6 equation [7]. In 2021, Mirzazadeh et al. have employed the generalized Riccati equation mapping method [8]. Ma et al. have employed the first integral, the exp −φξ-expansion, and sine-Gordon expansion procedures to extract soliton solutions of the Hirota–Maccari system [9]. In 2016, Osman used the generalized unified method for the Zakharov–Kuznetsov equation [10]. Inc et al. utilized the functional variable procedure [11]. Leta et al. applied the bifurcation technique to the (2 + 1)-dimensional Bogoyavlenskii coupled system [12]. Rezazadeh et al. applied the extended rational sin-cos method to the chiral nonlinear Schrödinger equation [13].In recent years, optical solitary waves, temporal and spatial solitons, have been discussed by many researchers theoretically and experimentally. For example, in 2018, Biswas et al. have founded optical solitons for the Lakshmanan–Porsezian–Daniel model via the modified simple equation procedure [14]. Manafian, in 2016, has founded optical soliton solutions for NLSE via the tan ϕξ/2-expansion procedure [15]. Kudryashov, in 2020, has proposed a procedure to find highly-dispersive optical solitons of NPDEs [16].This paper studied(1)iqt+aqxx+ibqxxx+cqxxxx+dq4q+eq2q+fq2xxq=0,where a, b, c, d, e, and f are parameters, and qx,t is a complex-valued function. It is known that this equation is the generalization of some famous equations. It is utilized to describe traveling wave solutions in optical fibers. This equation is reduced to NLSE when we choose b=c=d=f=0, and it models the slowly varying envelope dynamics of a weakly nonlinear quasimonochromatic wave packet in dispersive media [17]. Also, by setting b=c=f=0 in equation (1), the cubic-quintic NLS (cqNLS) equation is obtained which is used in the nonlinear optics [18, 19]. cqNLS is used for the nuclear hydrodynamics with Skyrme forces [20], the optical pulse propagations in dielectric media of non-Kerr type [21], and description of the boson gas with two and three-body interactions [22]. For this reason, we can consider equation (1) as a generalization of the NLSE. It could be utilized for describing nonlinear processes in optics [23].The paper consists of the following sections: we present the preliminary information in Section2. Then, we give a description of utilized techniques, respectively, and The modified extended tanh-function procedure and the exponential rational function procedure in Section 2.2. The application of the given methods is given in Section 3. Finally, the conclusion of the whole research is given in Section 4.


## 2. Preliminary Information

In this section, we firstly introduce the preliminary information for reducing NPDE to an ordinary differential equation (ODE). For this purpose, we will consider the following NPDE:(2)Pq,qt,qx,qtt,qxx,qxt,…=0,where P inherits q and its partial derivatives, and q is a complex-valued function.Wave transformation is given by(3)qx,t=uξeikx−vt,ξ=x−c0t,where v,k, and c0 are the constants. If we apply transformation (3) to equation (2) and equate the real and imaginary parts to zero, we obtain two equation systems. Then, we solve these systems to find conditions for parameters and use the values of the parameters. Therefore, we find an ODE, which will be integrated with respect to ξ possible times, as follows:(4)Qu,u′,u″,u″′,…=0,where the partial derivative is given with respect to ξ [24].The modified extended tanh function and the exponential rational function procedures are applicable to this reduced ODE.


### 2.1. The Modified Extended Tanh-Function Procedure

We will assume the exact solution of equation (4) by(5)uξ=∑i=0naiϕξi+∑j=1nbjϕξ−j,to employ the modified extended tanh-function procedure. Here, aii=0,1,…,n,bjj=1,…,n are constants, and they will be calculated; an and bn cannot be zero at the same time. In equation (5), the term n is called the balancing term. This term can be found balancing the nonlinear terms and the highest order derivatives. ϕξ is the solution of following auxiliary equation:(6)ϕ′ξ=ϕξ2+β,where β is given as a constant. The solutions of equation (6) are given in three cases as follows:(1)
Whenβ<0, the hyperbolic solution is(7)ϕξ=−−βtanh−βξorϕξ=−−βcoth−βξ.(2)
Whenβ>0, the trigonometric solution is(8)ϕξ=βtanβξorϕξ=−βcotβξ.(3)
Whenβ=0, the rational solution is(9)ϕξ=−1ξ.Surrogating equation (5) into equation (4) along with equation (6), then accepting all the coefficients of ϕiξ is zero, we get an algebraic equation system for k,v,c0,aii=0,…,n,bjj=1,…,n. If we solve an obtained determining equation system with the Maple, we find a variety of exact solutions [25].


### 2.2. The Exponential Rational Function Procedure

We will assume the solution of equation (4) as follows [26, 27]:(10)uξ=∑i=0nai1+eξn,where anan≠0 are constants. Here, n is the balancing number as the previous technique. We surrogate equation (10) into equation (4) and collect all terms with the same order of eiξi=0,1,2,… together, and we make into the left-hand side of equation (4) and another polynomial in eiξ. Then, we equate each coefficient of this polynomial to zero, and it gives a set of algebraic equations for an unknown parameters. Finally, we solve the equation system to construct a variety of exact solutions for equation (4).


## 2.1. The Modified Extended Tanh-Function Procedure

We will assume the exact solution of equation (4) by(5)uξ=∑i=0naiϕξi+∑j=1nbjϕξ−j,to employ the modified extended tanh-function procedure. Here, aii=0,1,…,n,bjj=1,…,n are constants, and they will be calculated; an and bn cannot be zero at the same time. In equation (5), the term n is called the balancing term. This term can be found balancing the nonlinear terms and the highest order derivatives. ϕξ is the solution of following auxiliary equation:(6)ϕ′ξ=ϕξ2+β,where β is given as a constant. The solutions of equation (6) are given in three cases as follows:(1)
Whenβ<0, the hyperbolic solution is(7)ϕξ=−−βtanh−βξorϕξ=−−βcoth−βξ.(2)
Whenβ>0, the trigonometric solution is(8)ϕξ=βtanβξorϕξ=−βcotβξ.(3)
Whenβ=0, the rational solution is(9)ϕξ=−1ξ.Surrogating equation (5) into equation (4) along with equation (6), then accepting all the coefficients of ϕiξ is zero, we get an algebraic equation system for k,v,c0,aii=0,…,n,bjj=1,…,n. If we solve an obtained determining equation system with the Maple, we find a variety of exact solutions [25].


## 2.2. The Exponential Rational Function Procedure

We will assume the solution of equation (4) as follows [26, 27]:(10)uξ=∑i=0nai1+eξn,where anan≠0 are constants. Here, n is the balancing number as the previous technique. We surrogate equation (10) into equation (4) and collect all terms with the same order of eiξi=0,1,2,… together, and we make into the left-hand side of equation (4) and another polynomial in eiξ. Then, we equate each coefficient of this polynomial to zero, and it gives a set of algebraic equations for an unknown parameters. Finally, we solve the equation system to construct a variety of exact solutions for equation (4).


## 3. Application of the Given Methods

In the current section, we will see new traveling wave solutions of equation (1). For this reason, firstly, we give the mathematical analysis of the considered equation.


### 3.1. Mathematical Analysis

If we substitute equation (3) into equation (1) and then equate the real and imaginary parts to zero, we find(11)b+4ckuξξξ+2ak−c0−3bk2−4ck3uξ=0,cuξξξξ+2fu2uξξ+2fuξ2u+a−3bk−6ck2uξξ+du5+eu3+v−ak2+bk3+ck4u=0.From the first equation in the above system, we obtain(12)b=−4ck,c0=2ak−3bk2−4ck3.Using (12) in the second equation in System (11), we get following reduced ODE:(13)cuξξξξ+2fu2uξξ+2fuξ2u+a+6ck2uξξ+du5+eu3+v−ak2−3ck4u=0.We will separately apply the modified extended tanh-function procedure and the exponential rational function procedure to this reduced ODE to obtain different exact solutions.


### 3.2. Modified Extended Tanh-Function Procedure

Homogeneous balance is obtained as n=1 from equation (13); thus, the solution of equation (13) is given by(14)uξ=a0+a1ϕξ+b1ϕξ.Surrogating equation (14) along with equation (6) into equation (13), then assuming the coefficients of ϕξj as zero,(15)da15+6fa13+24ca1=0,5da0a14+10fa0a12=0,10da02a13+5da14b1+8fa13β+4fa02a1+ea13+6fa12b1+12a1ck2+40ca1β+2a1a=0,10da03a12+20da0a13b1+12fa0a12β+3ea0a12+4fa0a1b1=0,5da04a1+30da02a12b1+10da13b12+2fa13β2−3a1ck4+4fa02a1β+8fa12b1β+12a1ck2β−a1ak2+3ea02a1+3ea12b1+2fa1b12+16ca1β2+2a1aβ+a1v=0,b15d+6b13β2f+24b1β4c=0,5a0b14d+10a0b12β2f=0,10a02b13d+4a02b1β2f+5a1b14d+6a1b12β2f+12b1β2ck2+8b13βf+40b1β3c2ab1β2+b13e=0.is acquired. If we solve System (15), we obtain values of the constants.


#### 3.2.1. Solution Family 1

(16)a0=0,a1=0,b1=b1,a=a,d=−6β2b12f+4cβ2b14,e=−2β6cβk2+4b12f+20cβ2+aβb12,v=3ck4−12cβk2+ak2−2b12f−16cβ2−2aβ.Whenβ<0,(17)q1x,t=−b1−βtanh−βx−8ck3+2akt×eikx−3ck4−12cβk2+ak2−2b12f−16cβ2−2aβt.Whenβ>0,(18)q2x,t=b1βtanβx−8ck3+2akt×eikx−3ck4−12cβk2+ak2−2b12f−16cβ2−2aβt.Whenβ=0,(19)q3x,t=−b1x−8ck3+2akteikx−3ck4+ak2−2b12ft.


#### 3.2.2. Solution Family 2

(20)a0=0,a1=a1,b1=0,a=a,d=−6a12f+4ca14,e=−26ck2+4a12βf+20cβ+aa12,v=−2a12β2f+3ck4−12cβk2+ak2−16cβ2−2aβ.Whenβ<0,(21)q4x,t=−a1−βtanh−βx−8ck3+2akt×eikx−−2a12β2f+3ck4−12cβk2+ak2−16cβ2−2aβt.Whenβ>0,(22)q5x,t=a1βtanβx−8ck3+2akt×eikx−−2a12β2f+3ck4−12cβk2+ak2−16cβ2−2aβt.Whenβ=0,(23)q6x,t=−a1x−8ck3+2akteikx−3ck4+ak2t.


#### 3.2.3. Solution Family 3

(24)a0=0,a1=a1,b1=a1β,a=a,d=−6a12f+4ca14,e=−26ck2−8a12βf−40cβ+aa12,v=3ck4+24cβk2+ak2−16cβ2+4aβ.Whenβ<0,(25)q7x,t=−a1−βtanh−βx−8ck3+2akt2−1tanh−βx−8ck3+2akt×eikx−3ck4+24cβk2+ak2−16cβ2+4aβt.Whenβ>0,(26)q8x,t=a1βtanβx−8ck3+2akt2+1tanhβx−8ck3+2akt×eikx−3ck4+24cβk2+ak2−16cβ2+4aβt.Whenβ=0, we obtain the same solution as the third solution in solution family 2.


#### 3.2.4. Solution Family 4

(27)a0=0,a1=a1,b1=−a1β,a=a,d=−6a12f+4ca14,e=−26ck2−16a12βf−80cβ+aa12,v=−32a12β2f+3ck4−48cβk2+ak2−256cβ2−8aβWhenβ<0,(28)q9x,t=−a1−βtanh−βx−8ck3+2akt2+1tanh−βx−8ck3+2akt×eikx−−32a12β2f+3ck4−48cβk2+ak2−256cβ2−8aβt.Whenβ>0,(29)q10x,t=a1βtanβx−8ck3+2akt2−1tanβx−8ck3+2akt×eikx−−32a12β2f+3ck4−48cβk2+ak2−256cβ2−8aβt.Whenβ=0, we obtain the same solution as the third solution in solution family 2.3.2.5. Solution Family 5(30)a0=0,a1=∓2b1−cb12f+4β2c,b1=b1,a=±6f−cb12f+4β2cb12−6ck2−20βc,d=−6β2b12f+4β2cb14,e=±4β∓2fb12−c/b12f+4β2c+15βcb12−c/b12f+4β2cv=1/b12f+4β2c±6b14f2k2±24b12β2cfk2∓44b14βf2∓176b12β3cf−cb12f+4β2c×1b12f+4β2c−3b12cfk4−12β2c2k4−20b12βcfk2−80β3c2k2−2b14f2+144b12β2cf+576β4c2Whenβ<0, we obtain(31)q11,12x,t=∓b12βtanh−βx−c0t2−c/b12f+4β2c±1−βtanh−βx−c0teikx−vt,where c0=8ck3+2k±6fb12−c/b12f+4β2c−6ck2−20βc.Whenβ>0, we get(32)q13,14x,t=∓b12βtanβx−c0t2−c/b12f+4β2c∓1βtanβx−c0teikx−vt,where c0=8ck3+2k±6fb12−c/b12f+4β2c−6ck2−20βc.Whenβ=0, we find(33)q15,16x,t=−b1x−c0t2∓2−c/b12fx−c0teikx−vt,where c0=8ck3+2k±6fb12−c/b12f−6ck2.


### 3.3. The Exponential Rational Function Procedure

Since the balancing number is 1, according to the exponential rational function procedure, the solution is(34)uξ=a0+a11+eξ.If we surrogate equation (34) into equation (13) and collect all the terms with the same order of eiξ,i=0,1,…,5 together, and we make into the left-hand side of equation (4) and another polynomial in eiξ. Then, we equate each coefficient to zero to get an algebraic equation set for aii=0,1,a1≠0 unknown parameters as follows:(35)e5ξ:da05−3ck4a0−ak2a0+ea03+va0=0,e4ξ:5da04a1+2fa02a1−ak2a1+va1+5da05+ca1−15ck4a0+6ck2a1+5ea03+3ea02a1−3ck4a1−5ak2a0+5va0+aa1=0,e3ξ:2fa02a1+12ea02a1+20da04a1+10ea03−11ca1+6ck2a1+3ea0a12+4va1−+6fa0a12+10va0+10da05+aa14ak2a1−10ak2a0−30ck4a0−12ck4a1+10da03a12=0,e2ξ:ea13−2fa02a1+11ca1+18ea02a1+10ea03+4fa13+10va0−6ck2a1+30da03a12−aa1+9ea0a12−6ak2a1+10da05−10ak2a0+2fa0a12+6va1+30da04a1+10da02a13−30ck4a0−18ck4a1=0,eξ:−2fa02a1+12ea02a1+9ea0a12−6ck2a1−4fa0a12+5da0a14−2fa13+5ea03+20da02a13−aa1+20da04a1+5va0+2ea13+30da03a12−ca1−15ck4a0+4va1+5da05−4ak2a1−5ak2a0−12ck4a1=0,e0ξ:3ea02a1+3ea0a12−ak2a1−3ck4a1+ea13+va1+da15−ak2a0−3ck4a0+5da04a1+va0+10da03a12+10da02a13+5da0a14+ea03+da05=0.Thus, we obtain the following solutions:Solution 1:(36)a=a12fk2+56a12f,a0=0,c=−a12f6,d=−2fa12,e=8f3,v=12a12fk4+56a12fk2−23a12f.By substituting these values, we find(37)q17x,t=a1eikx−1/2a12fk4+5/6a12fk2−2/3a12ft1+coshx−5/3a12fk+2/3a12fk3t+sinhx−5/3a12fk+2/3a12fk3t.Solution 2:(38)a=56a12f+a12fk2,a0=−a1,c=−a12f6,d=−2fa12,e=8f3,v=12a12fk4+56a12fk2−23a12f.By substituting these values, we find(39)q18x,t=a1eikx−1/2a12fk4+5/6a12fk2−2/3a12ft1+coshx−5a12fk/3+2a12fk3/3t+sinhx−5a12fk/3+2a12fk3/3t.Solution 3:(40)a=−ea122−5da1424−a12f4+k2da144+3a12fk22,a0=−a12,c=−da1424−a12f4,v=−ea124+k4da148+3a12fk44−da1416−k2ea122−5k2da1424−a12fk24.By substituting these values, we find(41)q19x,t=−a12+a11+coshx−a12fk3+k3da14/6−a12fk/2−5kda14/12−kea12t+sinhx−a12fk3+k3da14/6−a12fk/2−5kda14/12−kea12t×eikx−−ea12/4+k4da14/8+3a12fk4/4−da14/16−k2ea12/2−5k2da14/24−a12fk2/4t.


## 3.1. Mathematical Analysis

If we substitute equation (3) into equation (1) and then equate the real and imaginary parts to zero, we find(11)b+4ckuξξξ+2ak−c0−3bk2−4ck3uξ=0,cuξξξξ+2fu2uξξ+2fuξ2u+a−3bk−6ck2uξξ+du5+eu3+v−ak2+bk3+ck4u=0.From the first equation in the above system, we obtain(12)b=−4ck,c0=2ak−3bk2−4ck3.Using (12) in the second equation in System (11), we get following reduced ODE:(13)cuξξξξ+2fu2uξξ+2fuξ2u+a+6ck2uξξ+du5+eu3+v−ak2−3ck4u=0.We will separately apply the modified extended tanh-function procedure and the exponential rational function procedure to this reduced ODE to obtain different exact solutions.


## 3.2. Modified Extended Tanh-Function Procedure

Homogeneous balance is obtained as n=1 from equation (13); thus, the solution of equation (13) is given by(14)uξ=a0+a1ϕξ+b1ϕξ.Surrogating equation (14) along with equation (6) into equation (13), then assuming the coefficients of ϕξj as zero,(15)da15+6fa13+24ca1=0,5da0a14+10fa0a12=0,10da02a13+5da14b1+8fa13β+4fa02a1+ea13+6fa12b1+12a1ck2+40ca1β+2a1a=0,10da03a12+20da0a13b1+12fa0a12β+3ea0a12+4fa0a1b1=0,5da04a1+30da02a12b1+10da13b12+2fa13β2−3a1ck4+4fa02a1β+8fa12b1β+12a1ck2β−a1ak2+3ea02a1+3ea12b1+2fa1b12+16ca1β2+2a1aβ+a1v=0,b15d+6b13β2f+24b1β4c=0,5a0b14d+10a0b12β2f=0,10a02b13d+4a02b1β2f+5a1b14d+6a1b12β2f+12b1β2ck2+8b13βf+40b1β3c2ab1β2+b13e=0.is acquired. If we solve System (15), we obtain values of the constants.


### 3.2.1. Solution Family 1

(16)a0=0,a1=0,b1=b1,a=a,d=−6β2b12f+4cβ2b14,e=−2β6cβk2+4b12f+20cβ2+aβb12,v=3ck4−12cβk2+ak2−2b12f−16cβ2−2aβ.Whenβ<0,(17)q1x,t=−b1−βtanh−βx−8ck3+2akt×eikx−3ck4−12cβk2+ak2−2b12f−16cβ2−2aβt.Whenβ>0,(18)q2x,t=b1βtanβx−8ck3+2akt×eikx−3ck4−12cβk2+ak2−2b12f−16cβ2−2aβt.Whenβ=0,(19)q3x,t=−b1x−8ck3+2akteikx−3ck4+ak2−2b12ft.


### 3.2.2. Solution Family 2

(20)a0=0,a1=a1,b1=0,a=a,d=−6a12f+4ca14,e=−26ck2+4a12βf+20cβ+aa12,v=−2a12β2f+3ck4−12cβk2+ak2−16cβ2−2aβ.Whenβ<0,(21)q4x,t=−a1−βtanh−βx−8ck3+2akt×eikx−−2a12β2f+3ck4−12cβk2+ak2−16cβ2−2aβt.Whenβ>0,(22)q5x,t=a1βtanβx−8ck3+2akt×eikx−−2a12β2f+3ck4−12cβk2+ak2−16cβ2−2aβt.Whenβ=0,(23)q6x,t=−a1x−8ck3+2akteikx−3ck4+ak2t.


### 3.2.3. Solution Family 3

(24)a0=0,a1=a1,b1=a1β,a=a,d=−6a12f+4ca14,e=−26ck2−8a12βf−40cβ+aa12,v=3ck4+24cβk2+ak2−16cβ2+4aβ.Whenβ<0,(25)q7x,t=−a1−βtanh−βx−8ck3+2akt2−1tanh−βx−8ck3+2akt×eikx−3ck4+24cβk2+ak2−16cβ2+4aβt.Whenβ>0,(26)q8x,t=a1βtanβx−8ck3+2akt2+1tanhβx−8ck3+2akt×eikx−3ck4+24cβk2+ak2−16cβ2+4aβt.Whenβ=0, we obtain the same solution as the third solution in solution family 2.


### 3.2.4. Solution Family 4

(27)a0=0,a1=a1,b1=−a1β,a=a,d=−6a12f+4ca14,e=−26ck2−16a12βf−80cβ+aa12,v=−32a12β2f+3ck4−48cβk2+ak2−256cβ2−8aβWhenβ<0,(28)q9x,t=−a1−βtanh−βx−8ck3+2akt2+1tanh−βx−8ck3+2akt×eikx−−32a12β2f+3ck4−48cβk2+ak2−256cβ2−8aβt.Whenβ>0,(29)q10x,t=a1βtanβx−8ck3+2akt2−1tanβx−8ck3+2akt×eikx−−32a12β2f+3ck4−48cβk2+ak2−256cβ2−8aβt.Whenβ=0, we obtain the same solution as the third solution in solution family 2.3.2.5. Solution Family 5(30)a0=0,a1=∓2b1−cb12f+4β2c,b1=b1,a=±6f−cb12f+4β2cb12−6ck2−20βc,d=−6β2b12f+4β2cb14,e=±4β∓2fb12−c/b12f+4β2c+15βcb12−c/b12f+4β2cv=1/b12f+4β2c±6b14f2k2±24b12β2cfk2∓44b14βf2∓176b12β3cf−cb12f+4β2c×1b12f+4β2c−3b12cfk4−12β2c2k4−20b12βcfk2−80β3c2k2−2b14f2+144b12β2cf+576β4c2Whenβ<0, we obtain(31)q11,12x,t=∓b12βtanh−βx−c0t2−c/b12f+4β2c±1−βtanh−βx−c0teikx−vt,where c0=8ck3+2k±6fb12−c/b12f+4β2c−6ck2−20βc.Whenβ>0, we get(32)q13,14x,t=∓b12βtanβx−c0t2−c/b12f+4β2c∓1βtanβx−c0teikx−vt,where c0=8ck3+2k±6fb12−c/b12f+4β2c−6ck2−20βc.Whenβ=0, we find(33)q15,16x,t=−b1x−c0t2∓2−c/b12fx−c0teikx−vt,where c0=8ck3+2k±6fb12−c/b12f−6ck2.


## 3.2.1. Solution Family 1

(16)a0=0,a1=0,b1=b1,a=a,d=−6β2b12f+4cβ2b14,e=−2β6cβk2+4b12f+20cβ2+aβb12,v=3ck4−12cβk2+ak2−2b12f−16cβ2−2aβ.Whenβ<0,(17)q1x,t=−b1−βtanh−βx−8ck3+2akt×eikx−3ck4−12cβk2+ak2−2b12f−16cβ2−2aβt.Whenβ>0,(18)q2x,t=b1βtanβx−8ck3+2akt×eikx−3ck4−12cβk2+ak2−2b12f−16cβ2−2aβt.Whenβ=0,(19)q3x,t=−b1x−8ck3+2akteikx−3ck4+ak2−2b12ft.


## 3.2.2. Solution Family 2

(20)a0=0,a1=a1,b1=0,a=a,d=−6a12f+4ca14,e=−26ck2+4a12βf+20cβ+aa12,v=−2a12β2f+3ck4−12cβk2+ak2−16cβ2−2aβ.Whenβ<0,(21)q4x,t=−a1−βtanh−βx−8ck3+2akt×eikx−−2a12β2f+3ck4−12cβk2+ak2−16cβ2−2aβt.Whenβ>0,(22)q5x,t=a1βtanβx−8ck3+2akt×eikx−−2a12β2f+3ck4−12cβk2+ak2−16cβ2−2aβt.Whenβ=0,(23)q6x,t=−a1x−8ck3+2akteikx−3ck4+ak2t.


## 3.2.3. Solution Family 3

(24)a0=0,a1=a1,b1=a1β,a=a,d=−6a12f+4ca14,e=−26ck2−8a12βf−40cβ+aa12,v=3ck4+24cβk2+ak2−16cβ2+4aβ.Whenβ<0,(25)q7x,t=−a1−βtanh−βx−8ck3+2akt2−1tanh−βx−8ck3+2akt×eikx−3ck4+24cβk2+ak2−16cβ2+4aβt.Whenβ>0,(26)q8x,t=a1βtanβx−8ck3+2akt2+1tanhβx−8ck3+2akt×eikx−3ck4+24cβk2+ak2−16cβ2+4aβt.Whenβ=0, we obtain the same solution as the third solution in solution family 2.


## 3.2.4. Solution Family 4

(27)a0=0,a1=a1,b1=−a1β,a=a,d=−6a12f+4ca14,e=−26ck2−16a12βf−80cβ+aa12,v=−32a12β2f+3ck4−48cβk2+ak2−256cβ2−8aβWhenβ<0,(28)q9x,t=−a1−βtanh−βx−8ck3+2akt2+1tanh−βx−8ck3+2akt×eikx−−32a12β2f+3ck4−48cβk2+ak2−256cβ2−8aβt.Whenβ>0,(29)q10x,t=a1βtanβx−8ck3+2akt2−1tanβx−8ck3+2akt×eikx−−32a12β2f+3ck4−48cβk2+ak2−256cβ2−8aβt.Whenβ=0, we obtain the same solution as the third solution in solution family 2.3.2.5. Solution Family 5(30)a0=0,a1=∓2b1−cb12f+4β2c,b1=b1,a=±6f−cb12f+4β2cb12−6ck2−20βc,d=−6β2b12f+4β2cb14,e=±4β∓2fb12−c/b12f+4β2c+15βcb12−c/b12f+4β2cv=1/b12f+4β2c±6b14f2k2±24b12β2cfk2∓44b14βf2∓176b12β3cf−cb12f+4β2c×1b12f+4β2c−3b12cfk4−12β2c2k4−20b12βcfk2−80β3c2k2−2b14f2+144b12β2cf+576β4c2Whenβ<0, we obtain(31)q11,12x,t=∓b12βtanh−βx−c0t2−c/b12f+4β2c±1−βtanh−βx−c0teikx−vt,where c0=8ck3+2k±6fb12−c/b12f+4β2c−6ck2−20βc.Whenβ>0, we get(32)q13,14x,t=∓b12βtanβx−c0t2−c/b12f+4β2c∓1βtanβx−c0teikx−vt,where c0=8ck3+2k±6fb12−c/b12f+4β2c−6ck2−20βc.Whenβ=0, we find(33)q15,16x,t=−b1x−c0t2∓2−c/b12fx−c0teikx−vt,where c0=8ck3+2k±6fb12−c/b12f−6ck2.


## 3.3. The Exponential Rational Function Procedure

Since the balancing number is 1, according to the exponential rational function procedure, the solution is(34)uξ=a0+a11+eξ.If we surrogate equation (34) into equation (13) and collect all the terms with the same order of eiξ,i=0,1,…,5 together, and we make into the left-hand side of equation (4) and another polynomial in eiξ. Then, we equate each coefficient to zero to get an algebraic equation set for aii=0,1,a1≠0 unknown parameters as follows:(35)e5ξ:da05−3ck4a0−ak2a0+ea03+va0=0,e4ξ:5da04a1+2fa02a1−ak2a1+va1+5da05+ca1−15ck4a0+6ck2a1+5ea03+3ea02a1−3ck4a1−5ak2a0+5va0+aa1=0,e3ξ:2fa02a1+12ea02a1+20da04a1+10ea03−11ca1+6ck2a1+3ea0a12+4va1−+6fa0a12+10va0+10da05+aa14ak2a1−10ak2a0−30ck4a0−12ck4a1+10da03a12=0,e2ξ:ea13−2fa02a1+11ca1+18ea02a1+10ea03+4fa13+10va0−6ck2a1+30da03a12−aa1+9ea0a12−6ak2a1+10da05−10ak2a0+2fa0a12+6va1+30da04a1+10da02a13−30ck4a0−18ck4a1=0,eξ:−2fa02a1+12ea02a1+9ea0a12−6ck2a1−4fa0a12+5da0a14−2fa13+5ea03+20da02a13−aa1+20da04a1+5va0+2ea13+30da03a12−ca1−15ck4a0+4va1+5da05−4ak2a1−5ak2a0−12ck4a1=0,e0ξ:3ea02a1+3ea0a12−ak2a1−3ck4a1+ea13+va1+da15−ak2a0−3ck4a0+5da04a1+va0+10da03a12+10da02a13+5da0a14+ea03+da05=0.Thus, we obtain the following solutions:Solution 1:(36)a=a12fk2+56a12f,a0=0,c=−a12f6,d=−2fa12,e=8f3,v=12a12fk4+56a12fk2−23a12f.By substituting these values, we find(37)q17x,t=a1eikx−1/2a12fk4+5/6a12fk2−2/3a12ft1+coshx−5/3a12fk+2/3a12fk3t+sinhx−5/3a12fk+2/3a12fk3t.Solution 2:(38)a=56a12f+a12fk2,a0=−a1,c=−a12f6,d=−2fa12,e=8f3,v=12a12fk4+56a12fk2−23a12f.By substituting these values, we find(39)q18x,t=a1eikx−1/2a12fk4+5/6a12fk2−2/3a12ft1+coshx−5a12fk/3+2a12fk3/3t+sinhx−5a12fk/3+2a12fk3/3t.Solution 3:(40)a=−ea122−5da1424−a12f4+k2da144+3a12fk22,a0=−a12,c=−da1424−a12f4,v=−ea124+k4da148+3a12fk44−da1416−k2ea122−5k2da1424−a12fk24.By substituting these values, we find(41)q19x,t=−a12+a11+coshx−a12fk3+k3da14/6−a12fk/2−5kda14/12−kea12t+sinhx−a12fk3+k3da14/6−a12fk/2−5kda14/12−kea12t×eikx−−ea12/4+k4da14/8+3a12fk4/4−da14/16−k2ea12/2−5k2da14/24−a12fk2/4t.


## 4. Conclusion

NPDEs have a broad field of applications in fundamental sciences such as computer science, biology, chemistry, mathematics, physics, and engineering sciences such as chemical engineering, electricity, mechanics, and materials. The exact solutions of NPDEs have attracted considerable observation from researchers. For this reason, we have studied modified extended tanh-function and the exponential rational function procedures to determine various types of exact solutions of an integrable equation for wave packet envelope. Also, we have obtained 3D graphics of the obtained solutions (Figures1–8). The obtained exact solutions have been verified by substituting them back into the original equation using Maple 19. According to our point of view, our obtained results have the potential to be applicable in nonlinear dynamics and optics. The obtained solutions can be classified as periodic and soliton solutions. For future works, supernonlinear, quasiperiodic, chaotic wave, and solitonic solutions will be potentially obtained. Finally, despite using Maple 19 during the calculations, the authors have noted that wxMaxima can be chosen due to the fact that is free and open-source software.Figure 1
The dark soliton solution ofq4x,t when β=−0.05,c=−1,k=3,a1=0.2,f=−2.5, a=−2,within the interval−4≤x≤0,−5≤t≤5.Figure 2
The periodic solution ofq5x,t when β=0.05,c=−1,k=3,a1=0.2,f=−2.5, a=−2,within the interval−8≤x≤8,−9≤t≤9.Figure 3
The bright soliton solution ofq6x,t when β=0,c=−1,k=3,a1=0.2,f=−2.5, a=−2,within the interval−8≤x≤8,−9≤t≤9.Figure 4
The bright soliton solution ofq9x,t when β=−0.05,c=−1,k=3,a1=0.2,f=−2.5, a=−2,within the interval−2≤x≤0,−5≤t≤5.Figure 5
The periodic solution ofq10x,t when β=0.05,c=−1,k=3,a1=0.2,f=−2.5, a=−2,within the interval−8≤x≤8,−9≤t≤9.Figure 6
The antikink soliton solution ofq17x,t when a1=1,f=2,k=−1within the interval−2≤x≤0,−5≤t≤5.Figure 7
The antikink soliton solution ofq18x,t when a1=1,f=2,k=−1within the interval−2≤x≤0,−5≤t≤5.Figure 8
The antikink soliton solution ofq19x,twhena1=−4,f=0.7,k=−3.5within the interval−5≤x≤5,−5≤t≤5.



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*Source: 3486780-2022-02-09.xml*