# Solving Nonlinear Partial Differential Equations of Special Kinds of 3rd Order Using Balance Method and Its Models

Authors: Daba Meshesha Gusu; Wakjira Gudeta
Journal: International Journal of Differential Equations
 (2023)

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## Abstract

Most nonlinear partial differential equations have many applications in the physical world. Finding solutions to nonlinear partial differential equations is not easily solvable and hence different modified techniques are applied to get solutions to such nonlinear partial differential equations. Among them, we considered the modified Korteweg–de Vries third order using the balance method and constructing its models using certain parameters. The method is successfully implemented in solving the stated equations. We obtained kind one and two soliton solutions and their graphical models are shown using mathematical software-12. The obtained results lead to shallow wave models. A few illustrative examples were presented to demonstrate the applicability of the models. Furthermore, physical and geometrical interpretations are considered for different parameters to investigate the nature of soliton solutions to their models. Finally, the proposed method is a standard, effective, and easily computable method for solving the modified Korteweg–de Vries equations and determining its perspective models.


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## Body

## 1. Introduction

It is significantly important in nonlinear phenomena to search for exact solutions to nonlinear partial differential equations (NLPDEs). Exact solutions play a vital role in understanding various qualitative and quantitative features of nonlinear phenomena. There are diverse classes of interesting exact solutions, traveling wave solutions, and soliton solutions, but it often needs specific mathematical techniques to construct exact solutions due to the nonlinearity present in their dynamical nature [1–3]. The NLPDEs are widely used as models to depict many important complex physical phenomena in a variety of fields of science and engineering. Some nonlinear partial differential equations can be written as follows:Burger’s equation is as follows:(1)ut+uux=αuxx.The modified Korteweg–de Vries (KdV) equation is as follows:(2)ut+6uux+uxxx=0,The Kadomtsev–Petriashvili (KP) equation is as follows:(3)ut+auux+buxxxx+uyy=0.The NLPDEs are fundamentally important because lots of mathematical physics models are often described by such wave phenomena and the investigation of traveling wave solutions is becoming more and more attractive in nonlinear sciences nowadays. However, NLPDEs are very difficult to solve explicitly, specifically, or in detail. As a result, many powerful methods have been proposed and developed for finding analytical solutions to nonlinear problems.Some of the researchers used to solve NLPDEs are the simple equation method [4], the tanh-function and balance methods [5], the inverse scattering transform method [6], Backland and transformation [7], and other mathematical procedures such as combinations of the equation, transformation procedure, bilinear method, integration, and so on. Of these methods, the balance method was based on the higher order of partial derivative terms, the highest nonlinear terms, and parameters of the intended technique to solve the proposed problem of the modified KdV equation.The Kadomtsev–Petviashvili (KP) equation is a partial differential equation that describes nonlinear wave motion [8], which is usually written as(4)ut+6uux+uxxxx−δuyy=0.δ=±1 that can be applied to mathematical physics as a way to model water waves of long wavelengths. It is a two-dimensional generalization of the one-dimensional KdV (2). Like the KdV equation, the KP equation is completely integrable.KdV were among the scholars who derived the KdV equation and one of the most famous nonlinear PDEs that arise in a great number of physical situations. It was derived from fluid mechanics to describe shallow water waves in a rectangular channel.(5)ut+uux+βuxxx=0.The positive parameterβ refers to a dispersive effect.The generalized modified KdV equation is given by the following equation:(6)ut+uxxx+upux=0.x∈R, where p is a positive parameter. Formulated in the moving frame x=ξ−ct, the generalized modified KdV equation reads as follows:(7)ut−cux+uxxx+upux=0,where c denotes the wave speed.The inverse scattering transform method is a method introduced that yields a solution to the initial value problem for an NLPDE with the help of the solutions to the direct and inverse scattering problems for an associated linear ordinary differential equation. The balance method (BM) is a powerful method for finding exact or approximate solutions to given NLPDEs which was presented by Wang and other scholars in recent years and was improved by Fan with others to make it more straight forward and simple [9]. In addition to this, as El-Wakil and others [10], used the balance method and auto-Backlund transformation. We used balance method for solving nonlinear partial differential equations, the modified Korteweg–de Vries equation. It is well known the KdV equations describe the unidirectional propagation of shallow water waves and a number of generalizations.Let us consider a general nonlinear PDE, say, in two variables,(8)Pu,ux,ut,uxx,uxt,….=0,where P is a polynomial function of its arguments and the subscripts denote the partial derivatives. The balance method consists of the following steps:Step 1.
Suppose that the solution of (8) is in the form of the following equation:(9)u=apdlnωp,d+∑k,j=0,k+j≠0,p+dk=m,j=nakjlnωk,j+a00q,where u=ux,t, ω=ωx,t, lnωk,j=∂k+jlnωx,t/∂xk∂tj, and ak,jk=0,1,2,….p,j=0,1,2,…..,d balance coefficients are constants to be determined. By balancing the highest nonlinear terms and the higher order partial derivative terms in this expression; p,d, and q can be determined.Step 2.
Substituting (9) into (8) and arranging it at each order of ω yield an equation as follows:(10)∑l=oflωlwhere fll=0,1,2,…. are differential-algebraic expressions of ω and akj. Setting fl=0 and using a compatible condition ωxt=ωtx yield a set of differential-algebraic equations (DAEs).Step 3.
Solving the set of DAEs,ω and akjk=0,1,2,….p,j=0,1,2,…..,d can be determined. By substituting ω,p,d,q, and akj into (9), the exact solutions of (8) can be obtained.
Nonlinear type partial differential equations can be solved by many different methods such as the Hirota method [11, 12]. Inverse scattering transform method and similarity reductions method [13]. Among those methods, the modified Korteweg–de Vries equation was solved by a few of them with their limitations as it is. The limitation of complexities to find the solutions are one core of the problem. This study aimed to find an alternative solution to the modified KdV equation which simplifies the complexity of solving NLPDEs showing the 1 and 2− soliton solutions of the modified KdV equation, by applying the balance method.
A long wave characterizes geophysical fluid dynamics in shallow waters and deep oceans [14, 15]. From another perspective, the KdV equation appeared for the first time in 1895 as a one-dimensional evolution equation describing the waves of an along surface gravity propagation in a water shallow canal [16]. It also appeared in a numeral of diverse physical phenomena such as hydromagnetic collision-free waves, ion acoustic waves, stratified waves interior, lattice dynamics, and physics of plasma [17]. By applying the differential transform technique, the approximate result of coupled KdV has been studied in [5]. One of the most attractive and surprising wave phenomena is the creation of solitary waves or solitons. An adequate theory for solitary waves was developed, in the form of a modified wave equation known as KdV [18–21]. The modified KdV equation has defined a wide variety of physical phenomena used to model the interaction and evolution of nonlinear waves [22–24]. Korteweg and de Vries pursued the work done by Rayleigh and including the effect of surface tension leading to the now famous KdV equation given as follows: ∂η/∂τ+3/2g/h∂/∂ζ1/2η2+2/3αη+1/3δ∂2η/∂ζ2=0, where η is the surface elevation, of the wave, above the equilibrium level h, α is a constant related to the uniform motion of the liquid, with the unit of length, g is the gravitational acceleration and δ=1/3h3−Th/ρg, with T is surface tension.
The permanent profile of a soliton solution of the KdV equation results from the equilibrium between two effects; the nonlinearity which is proportional touux or ηηζ and Dispersion which is proportional to uxxx or ηζζζ. The nonlinearity tends to comprise, to constitute the wave and dispersions spread it out. The dispersive term is proportional to h3, it decreases whereas, the nonlinearity term, is proportional to 1/h, increases leading to a large wave. The balance method has been developed for the analytical solutions of nonlinear partial differential equations. Compared to other methods, the balance method gives high accuracy and has a wide range of applications in many mathematical problems and many physical phenomena. In this paper, we used the balance method in combination with the Hirota bilinear equation method to solve the modified form of the KdV equation considered using the balance method and its respective models.


## 2. Mathematical Formulation and Main Results

### 2.1. Preliminaries

#### 2.1.1. Bilinear Operators

Bilinear differential operatorD mapping a pair of functions Df.g. Unlike usual linear differential operators like ∂/∂xn, which maps a single function f into a single function ∂nf/∂xn. That is D:C∞xC∞⟶C∞:f.g↦Df.gRemark 1.
For all integersm,n≥0,(11)DtmDxnf.gx,t=∂∂t−∂∂t′m∂∂x−∂∂x′nfx,tgx′,t′x′=x,t′=t,where f is a function of x and t, whereas g is a function of x′ and t′; for which f≠g.


#### 2.1.2. Fractional Derivative

A fractional derivative is a derivative of any order, real or complex in applied mathematics or mathematical analysis.


#### 2.1.3. The ELzaki Transform

The Elzaki transform is a function of the formhx,t with respect to t(12)ψhx,t=ψx,s=s∫0∞hx,texp−tsdt,where s is the complex number, t>0.


### 2.2. Main Results

#### 2.2.1. Method of Solution

Considering the general nonlinear PDEs given in (8), let us consider the modified KdV equation that was chosen as an example to illustrate the balance method. Suppose that the solution of (5) is (9). Now considering uxxx, uux balancing it in the (5), it is required that pq+3=2pq+1 (by letting u=pq, ux=1, uxx=2, and uxxx=3, where 1, 2, and 3 represent the order of derivatives for this case or condition). Again balancing ut and uux in (9), it gave that dq+1=2dq+1, which implies dq=2dq. Choosing p=2, and then solving for the rest of the variables, we obtained that q=1 and d=0. That means from the equation pq+3=2pq+1, by simplifying it we found pq=2. Since we chosen p=2, then we have q=1. Again from the second condition dq+1=2dq+1, we have d=0.Hence (9) u=apdlnωp,d+∑k,j=0,k+j≠0,p+dk=p,j=dakjlnωk,j+a00q, …(13)u=a20lnωxx+a10lnωx+a00.Also, this equation can also be expressed as follows:(14)u=a20wxxw−wx2w2+a10wxw+a00,ux=∂∂xa20wxxw−wx2w2+a10wxw+a00=a20wxxw−wxxwxw2−2wxwxxw2−wx22wwxw4+a10wxxw−wxwxw2=a20wxxxw−wxxwxw2−2wxxwxw2+2wx3w3+a10wxxw−wx2w2=a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2,(15)ux=a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2,ut=∂∂ta20lnωxx+a10lnωx+a00=∂∂ta20wxxw−wx2w2+a10wxw+a00=a20wxxtw−wxxwtw2−2wxwxtw2−wx2wt2ww4+a10wxtw−wxwtw2=a20wxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2=a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2,(16)ut=a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2.From equation (15), we get(17)uxx=∂∂xUx=∂∂xa20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2=a20wxxxxw−wxxxwxw2−3wxxxwx+wxxwxxw2−wxxwx2wwxw4+23wx2wxx.w3−w3x3w2wxw6+a10wxxxw−wxxwxw2−2wxwxxw2−wx22w.wxw4=a20wxxxxw−wxxxwxw2−3wxxxwxw2−3w2xxw2+6wxxwx2w3+6wxxwx2w3−6w4w4+a10wxxxw−wxxwxw2−2wxxwxw2+2wx3w3=a20wxxxxw−4wxxxwx+3w2xxw2+12wxxwx2w3−6w4w4+a10wxxxw−3wxxwxw2+2wx3w3,(18)uxx=a20wxxxxw−4wxxxwx+3w2xxw2+12wxxwx2w3−6wx4w4+a10wxxxw−3wxxwxw2+2wx3w3.From (18), we get(19)uxxx=∂∂xa20wxxxxw−4wxxxwx+3w2xxw2+12wxxwx2w3−6w4w4+a10wxxxw−3wxxwxw2+2wx3w3=a20wxxxxxw−wxxxxwxw2−4wxxxxwx+wxxxwxw2−wxxxwx2w.wxw4−32wxxwxxxw2−wxx22w.wxw4+12wxxxwx2+wxx2wxwxxw3−wxxwx23w2wxw6−64wx3wxxw4−wx44w3wxw8+a10wxxxxw−wxxxwxw2−3wxxxwx+wxxwxxw2−wxxwx2w.wxw4+23wx2wxxw3−wx33w2wxw6=a20wxxxxxw−wxxxxwxw2−4wxxxxwxw2−4wxxxwxxw2+8wxxxwx2w3−6wxxxwxxw2+6wxx2wxw3+12wxxxwx2w3+24wxx2wxw3−36wxxwx3w4−24wxxwx3w4+24wx5w5+a10wxxxxw−wxxxwxw2−3wxxxwxw2−3wxx2w2+6wxxwx2w3+6wxxwx2w3−6wx4w4=a20wxxxxxw−5wxxxxwxw2−10wxxxwxxw2+20wxxxwx2w3+30wxx2wxw3−60wxxwx3w4−24wx5w5+a10wxxxxw−4wxxxwxw2−3wxx2w2+12wxxwx2w3−6wx4w4,(20)uxxx=a20wxxxxxw−5wxxxxwxw2−10wxxxwxxw2+20wxxxwx2w3+30wxx2wxw3−60wxxwx3w424wx5w5+a10wxxxxw−4wxxxwxw2−3wxx2w2+12wxxwx2w3−6wx4w4.Now substituting equations (14), (15), (16), and (20) into (5), we have;(21)a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2+a20wxxw−wx2w2+a10wxw+a00a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2+βa20wxxxxxw−5wxxxxwxw2−10wxxxwxxw2+20wxxxwx2w3+30wxx2wxw3−60wxxwx3w4−24wx5w5+a10wxxxxw−4wxxxwxw2−3wxx2w2+12wxxwx2w3−6wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a20wxxw−a20wx2w2+a10wxw+a00a20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a20wxxwa20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2−a20wx2w2a20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+a10wxwa20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+a00a20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a20wxxwa20wxxxw−a20wxxwa203wxxwxw2+a20wxxwa202wx3w3+a20wxxwa10wxxw−a20wxxwa10wx2w2−−a20wx2w2a20wxxxw−a20wx2w2a203wxxwxw2+a20wx2w2a202wx3w3+a20wx2w2a10wxxw−a20wx2w2a10wx2w2−a10wxwa20wxxxw−a10wxwa203wxxwxw2+a10wxwa202wx3w3+a10wxwa10wxxw−a10wxwa10wx2w2+a00a20wxxxw−a00a203wxxwxw2+a00a202wx3w3+a00a10wxxw−a00a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a202wxxwxxxw2−a2023wxx2wxw3+a2022wx3wxxw4+a20a10wxx2w2−a20a10wx2wxxw3−a202wx2wxxxw3+a2023wxxwx3w4−a2022wx5w5−a20a10wxxwx2w3+a20a10wx4w4+a10a20wxwxxxw2−a10a203wxxwx2w3+a10a202wx4w4+a102wxwxxw2−a102wx3w3+a00a20wxxxw−a00a203wxxwxw2+a00a202wx3w3+a00a10wxxw−a00a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a202wxxwxxxw2−a2023wxx2wxw3+a2022wx3wxxw4+a20a10wxx2w2−a20a10wx2wxxw3−a202wx2wxxxw3+a2023wxxwx3w4−a2022wx5w5−a20a10wxxwx2w3+a20a10wx4w4+a10a20wxwxxxw2−a10a203wxxwx2w3+a10a202wx4w4+a102wxwxxw2−a102wx3w3+a00a20wxxxw−a00a203wxxwxw2+a00a202wx3w3+a00a10wxxw−a00a10wx2w2+βa20wxxxxxw−βa205wxxxxwxw2−βa2010wxxxwxxw2+βa2020wxxxwx2w3+βa2030wxx2wxw3−βa2060wxxwx3w4−βa2024wx5w5+βa10wxxxxw−βa104wxxxwxw2−βa103wxx2w2+βa1012wxxwx2w3−βa106wx4w4=0Now equating the coefficients ofwx5/w5 and wx4/w4 to zero on the left-hand side, we have a20=12β and a10=0. This is to mean that; equating a20β24wx5/w5=0, a10β6wx4/w4=0, a10a202wx4/w4=0, −2a202wx5/w5=0, and a10a20wx4/w4=0. Now considering the two equations a20β24wx5/w5=0 and −2a202wx5/w5=0, it is possible to solve for a20. Since β≠0 and w≠024a20β=0 and −2a202=0, we solved for the reset a20 as 24a20β=0−2a202=0Adding these two equations together and solving fora20, as 24a20β−2a202=02a2012β−a20=0. From this, we can get a20=12βAgain considering the equations;−6βa10wx4/w4=0,2a10a20wx4/w4=0,anda10a20wx4/w4=0.And adding them together using a system of equations as−6βa10=0,2a10a20=0,a10a20=0.we have 3a20a10−6βa10=0. 3a20a10=6βa10,Simplifying it we get, a20a10=2βa10. Substituting a20=12β into this equation we get, 12βa10=2a10. Which give us 10βa10=0, β≠0. And hence it implies that a10=0Substitutinga20=12β and a10=0 into the following equation: u=a20lnωxx+a10lnωx+a00, we get u=12lnωxx+0lnωx+a00=12lnωxx+a00=12βwxx/w−wx2/w2+a00, where a00 is any arbitrary constant.(22)u=12lnωxx+a00=12βwxxw−wx2w2+a00.Putting (22) into (5), we have the following results, where, a20=12β and a10=0.(23)12βwxxtw−12βwxxwtw2−12β2wxwxtw2+12β2wx2wtw3+0wxtw−0wxwtw2+12β2wxxwxxxw2−12β23wxx2wxw3+12β22wx3wxxw4+012βwxx2w2−012βwx2wxxw3−12β2wx2wxxxw3+12β23wxxwx3w4−12β22wx5w5−012βwxxwx2w3+012βwx4w4+012βwxwxxxw2−012β3wxxwx2w3+012β2wx4w4+0wxwxxw2−0wx3w3+a0012βwxxxw−a0012β3wxxwxw2+a0012β2wx3w3+0a00wxxw−0a00wx2w2+β12βwxxxxxw−12β25wxxxxwxw2−12β210wxxxwxxw2+12β220wxxxwx2w3+12β230wxx2wxw3−12β260wxxwx3w4−12β224wx5w5+0βwxxxxw−04βwxxxwxw2−03βwxx2w2+012βwxxwx2w3−06βwx4w4=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+144β2wxxwxxxw2−144β23wxx2wxw3+144β22wx3wxxw4+144β2wx2wxxxw3+144β23wxxwx3w4−144β22wx5w5−a0012βwxxxw−a0036βwxxwxw2+a0024βwx3w3+β12βwxxxxxw−60β2wxxxxwxw2−120β2wxxxwxxw2+240β2wxxxwx2w3+360β2wxx2wxw3−720β2wxxwx3w4−288β2wx5w5=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12β12βwxxwxxxw2−12β36βwxx2wxw3+12β24βwx3wxxw4+12β12βwx2wxxxw3+12β36βwxxwx3w4−12β24βwx5w5+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12ββwxxxxxw−5βwxxxxwxw2−10βwxxxwxxw2+20βwxxxwx2w3+30βwxx2wxw3−60βwxxwx3w4−24βwx5w5=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12β12βwxxwxxxw2−36βwxx2wxw3+24βwx3wxxw4+12βwx2wxxxw3+36βwxxwx3w4−24βwx5w5+βwxxxxxw−5βwxxxxwxw2−10βwxxxwxxw2+20βwxxxwx2w3+30βwxx2wxw3−60βwxxwx3w4−24βwx5w5=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12β12βwxxwxxxw2−10βwxxxwxxw2−36βwxx2wxw3+30βwxx2wxw3+24βwx3wxxw4+36βwxxwx3w4+−60βwxxwx3w4−12βwx2wxxxw3+20βwxxxwx2w3−24βwx5w5+24βwx5w5+βwxxxxxw−5βwxxxxwxw2=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12β2βwxxwxxxw2−6βwxx2wxw3+60βwx3wxxw4−60βwxxwx3w4+8βwx2wxxxw3+βwxxxxxw−5βwxxxxwxw2=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12ββwxxxxxw+2wxxxwxx−5wxxxxwxw2+8wxxxwx2−6wxx2wxw3=012βK1+K2+K3=0,where(24)K1=wxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3,K2=a00wxxxw−a003wxxwxw2+a002wx3w3,K3=βwxxxxxw+2wxxxwxx−5wxxxxwxw2+8wxxxwx2−6wxx2wxw3,(25)12βK1+K2+K3=0.Simplifying (25) and integrating once with respect to x, we get(26)∂∂xwxtw−wxwt+βwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2=0.Equations (20) and (26) are identical with(27)wxtw−wxwt+βwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2−Ctw2=0,where Ct is an arbitrary function of t, and a00 is an arbitrary constant. Especially, taking Ct as zero in equation (27), we get the bilinear equation of (5) as follows:(28)wxtw−wxwt+βwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2=0.Equation (28) can be written concisely in terms of D-operator as(29)DxDt+βDX4+a00Dx2w,w=0,where DxmDtna.b=∂x−∂x′m∂t−∂t′nax,tbx′,t′x=x′,t=t′.Applying Hirota’s method, the bilinear equation of (5) can be written as(30)DxDt+βDX4w,w=0.Equation (30) is obtained by setting a00=0 in equation (29). Obviously, equation (30) is a special case of equation (29). Therefore, a more general bilinear equation of the modified KdV equation is obtained by using Hirota’s method.Let us consider the modified KdV equationut+6uux+uxxx=0 by using Hirota’s method.Substituting(31)u=wx,which gives the following new equation: wxt+6wxwxx+wxxxx=0. Now integrating both sides with respect to x, we have;(32)∫wxt+6wxwxx+wxxxxdx=0,∫wxtdx+∫6wxwxxdx+∫wxxxxdx=gt,wt+6wx22+wxxx=C=gt.where C=gt is constant of integration. Likely we can get, without loss of generality, the integration “constant” with respect to x, gt can be observed by a redefinition of w that does not change the KdV field, u=wxwold=wnew+∫t0tdt′gt′, dt′ is the time derivative.Using the neww, we have(33)wt+3wx2+wxxx=0.Now considering the 1-soliton solution of the modified KdV equation, let(34)u=2μ2sech2μx−x0−4μ2t.with(35)μ=v2≥0,where v is the velocity of the wave propagating.It can be written asu=wx, with(36)w=2μ2tanhμx−x0−4μ2t.In fact, we can integrate the right-hand side of equation (36) once again, using tanh y=d/dy log cosh y. Therefore, equation (34) can be written as follows:(37)u=2∂2∂x2logcoshμx−x0−4μ2t.Let(38)X=x−x0−4μ2t.Then,u=2∂2/∂X2logexp−μX1+exp2μX/2, where coshy=exp−μX1+exp2μX/2(39)=2∂2∂x2−μx−log2+log1+exp2μX=−2∂2∂x2μx−2∂2∂x2log2+2∂2∂x2log1+exp2μX=2∂2∂X2log1+exp2μX.That is,ux,t=2∂2/∂x2log1+exp2μx−x0−4μ2t=2∂2/∂x2log1+exp2μX(40)ux,t=2∂2∂x2log1+exp2μx−x0−4μ2t=2∂2∂x2log1+exp2μX,which is the 1-soliton solution of the modified KdV equation, that we use in the following:Considering equations (31) and (33), the KdV equation in a bilinear form, and rewriting it in a quadratic form, we have the following equation. Inspired by equation (34), let us substitute(41)w=2∂∂xlogf=2fxf⇔u=2∂2∂x2logf,where f=1+exp2μx−x0−4μ2t. Substituting in equation (33), we get(42)12wt=fxt−fxftf2,12wx=fxx−fx2f2,12wxx=fxxxf−3fxfxxf2+2fx3f3,12wxxx=fxxxxf−4fxfxxxf2−3fxx2f2+12fxxfx2f3−6fx4f4.(43)12wxxx=fxxxxf−4fxfxxxf2−3fxx2f2+12fxxfx2f3−6fx4f4,Equation (43) is quadratic in f.Thus, equation (33) for w becomes(44)wt+3wx2+wxxx=0,=2fxt−fxftf2+322fxxf−fx2f22+2fxxxxf−42fxfxxxf2−32fxx2f2+122fxxfx2f3−62fx4f4=2fxt−fxftf2+12fxx2f2−2fxxfx2f3+fx4f4+2fxxxxf−8fxfxxxf2−6fxx2f2+24fxxfx2f3−12fx4f4=2fxt−fxftf2+12fxx2f2−24fxxfx2f3+12fx4f4+2fxxxxf−8fxfxxxf2−6fxx2f2+24fxxfx2f3−12fx4f412=2fxt−fxftf2+2fxxxxf−8fxfxxxf2+6fxx2f2=fxt−fxftf2+fxxxxf−4fxfxxxf2+3fxx2f2.Multiplying byf2 throughout, we have(45)ffxt−fxft+ffxxxx−4fxfxxx+3fxx2=0This is the bilinear form of the modified KdV equation. Despite the fact that some nontrivial cancellations took place it looks more complicated than the original problem, however, its special form makes it possible to solve using bilinear operators.Claim 1.
∂/∂x∂/∂t1/2f2=∂/∂xfft=ffxt+fxft
To express equation (45), we use the following expressions:(46)Dtf.gx,t=∂∂t−∂∂t′∂∂x−∂∂x′fx,tgx′,t′x′=x,t′=t=ftx,tgx′,t−fx,tgt′x′,t′x′=x,t′=t=ftx,tgx,t−fx,tgtx,tDtf.gx,t=ftx,tgx,t−fx,tgtx,tDtf.fx,t=0,Dxf.fx,t=0forf=gDtDxf.g=∂∂t−∂∂t′∂∂x−∂∂x′fx,tgx′,t′x′=x,t′=t=∂∂t−∂∂t′fxx,tgx′,t′−fx,tgx′x′,t′x′=x,t′=t=fxtg−ftgx−fxgt−fgxtfxtg−ftgx−fxgt+fgxtDtDxf.f=fxtf−ftfx−fxft+ffxt=2ffxt−ftfxDtDxf.f=2ffxt−ftfx
Like the first two terms in (45).(47)Dx2f.g=∂∂x−∂∂x′2fxx,tgx′,t′x′=x,t′=t=∂∂x−∂∂x′fxx,tgx′,t′−fx,tgx′x′,t′x′=x,t′=t=fxxg−fxgx−fxgx+fgxx=fxxg−2fxgx+fgxx,Dx2f.g=fxxg−2fxgx+fgxx,Dx2f.f=fxxf−2fxfx+ffxx=2ffxx−fx2 all functions of x,t.
Again this looks a bit like the differentiation of a product. But, we haveDx2f,f=2ffxx−2fx2 even though Dxf.fx,t=0. So importantly the operators are not associative since Dx2f.f≠DxDxf.f.
In fact, the right-hand side is meaningless, becauseDxf.f is a single function, but the outer Dx needs to act on a pair of functions.(48)Dx4f.g=∂∂x−∂∂x′4fxx,tgx′,t′x′=x,t′=t=∂∂x−∂∂x′3fxx,tgx′,t′−fx,tgx′x′,t′x′=x,t′=t=∂∂x−∂∂x′2fxxg−2fxgx′+fgx′x′x′=x,t′=t=∂∂x−∂∂x′fxxxg−3fxxgx′+3fxgx′x′−fgx′x′x′x′=x,t′=t=fxxxxg−4fxxxgx′+6fxxgx′x′−4fxgx′x′x′+fgx′x′x′x′x′=x,t′=t=fxxxxg−4fxxxgx+6fxxgxx−4fxgxxx+fgxxxxDx4f.g=fxxxxg−4fxxxgx+6fxxgxx−4fxgxxx+fgxxxx.
Again it looks just like a differentiated product off and g but with significant changes on the odd terms. Dx4f.f=fxxxxf−4fxxxfx+6fxxfxx−4fxfxxx+ffxxxx=2fxxxxf−8fxxxfx+6fxx2.(49)Dx4f.f=2fxxxxf−8fxxxfx+6fxx2
Here, we have to notice that (49) is like ∂x4fg, but we have an alternative signs;(50)Dx4f.f=2fxxxxf−4fxxxfx+3fxx2.
So, the KdV equation is in a quadratic form in equation (45) and can be recast as DtDx+Dx4f.f=0 is a bilinear form of the KdV equation.(51)DtDx+Dx4f.f=0,DtDx+Dx4=DxDt+Dx3.
(1) Solutions of the Bilinear KdV Equation. For example (1) and considering u=2∂x2logf, we have two cases.Case 1.
Forf=1, then u=0, which is a vacuum.
Forf=1+exp2μx−x0−4μ2t, then u=1−soliton solution. Since equation (51) is a bilinear equation, then this suggests that multisoliton solutions might be the sums of exponents of linear functions of x and t.
(2). The 1-Soliton Solution. In Hirota form the multisoliton solutions are sums of exponential of linear expressions in x and t.
For the one-soliton solution try(52)f=e0+eθ.where =ax+bt+c, ⇒x=ax,c=−x0, and −4μ2=tTheorem 1.
For anyθ1 and θ2, we have DtmDxneθ1.eθ2 = b1−b2ma1−a2neθ1+θ2, where m,n≥0.Proof.
By induction.
Let ifθi=aix+bit+ci,i=1,2. Then,(53)DtmDxneθ1.eθ2=b1−b2ma1−a2neθ1+θ2=b1−b2ma1−a2neθ1x,t+θ2x′,t′x′=x,t′=t
Then,(54)DtmDxn+1eθ1.eθ2=b1−b2ma1−a2n∂∂x−∂∂x′eθ1x,t+θ2x′,t′x′=x,t′=t=b1−b2ma1−a2na1−a2eθ1x,t+θ2x′,t′x′=x,t′=t=b1−b2ma1−a2n+1eθ1+θ2DtmDxn+1eθ1.eθ2=b1−b2ma1−a2n+1eθ1+θ2.and similarly for Dt operator.
Finally, the equation is obviously true forn=1,m=0 and n=0,m=1 so by induction must be true for all n and m. In particular, we find(55)DtmDxneθ.eθ=0.
Unlessm=n=0(56)DtmDxneθ.1=bmaneθDtmDxn1.eθ=−1m+nbman1,eθ.
Then, the bilinear form of the KdV equation forf=1+eθ is(57)DtDx+Dx41+eθ,1+eθ=0=DtDx+Dx41.1+1.eθ+eθ.1+eθ.eθ=DtDx+Dx4eθ.1+1.eθ=2baeθ+2a4eθ2aba+a3eθ=0.
Sinceeθ≠0, we have two opportunities. These two cases or opportunities are as follows:(1)
a=0. If a=0, then f is independent of x. So u=0, (which is a boundary condition).(2)
b=−a3. If b=−a3, then f=1+eax−a3t+c and hence u=2∂x2logf, f=1+eax−a3t+c, which is u=2∂x2log1+eax−a3t+c(58)=a22sech212ax−a3t+c,u=a22sech212ax−a3t+c.
This is the one-soliton solution withv=a2.
Figure1 shows the disturbance of the water surface having kink-shaped traveling waves with amplitude and antibell soliton. The follow of this wave is to the front.
Figure2 shows the propagation of waves having kink shaped and like some periodic soliton. And follow the wave from back to front.
Figure3 shows surface waves of long wave length having kink-shaped and bright periodic soliton. And the wave is from left to right.
Figure4 surface waves of long wave length having kink shaped and bright periodic soliton. And the wave is from left to right. This is the other form of Figure 3, but it is different by the value of x, that is why it is broken from the above.
(3) Multisoliton Solutions.Figure 1
The result of one-soliton solution fora=1/2, c = 0.1.Figure 2
The result of one-soliton solution fora=0.6, c = 0.2.Figure 3
The result of one-soliton solution fora=0.7,c=0.3.Figure 4
The result of one-soliton solution fora=0.7, c=0.3.Case 2.
For the power series in an auxiliary with parameterε(59)fx,t=∑n=0∞εnfnx,t,
Withf0=1 in which the series will terminate at some power of ε to give exact results rather than an infinite series in ε. So, we can take ε finite ε = 1.
To begin with, we putfx,t=1+∑n=0∞εnfnx,t and collect powers of ε. We start with 1 as the ε0 term as it appeared in the one-soliton solution. Defining(60)B=DtDx+Dx4.
We want to find solutions such that(61)Bf.f=0.
Substituting (59) into (60), we have 0=B∑n1=0∞εn1fn1,∑n2=0∞εn2fn2, with f0=1=∑n1=0∞∑n2=0∞εn1+n2Bfn1,fn2 and gathering terms of the same degree n=n1+n2inε and also expanding as for the one soliton case we have(62)Bf.f=B1.1+εB1.f1+Bf1.1+ε2B1.f2+Bf1.f1+Bf2.1=0,=∑n=0∞εn∑m=0nBfn−m.fm.so we have to solve an infinite set of equations order by order in ε(63)∑m=0nBfn−m.fm=0,for all n = 1, 2, ….
Equation (63) can be written as(64)Bfn,1+B1,fn=−∑m=1n−1Bfn−m.fm.
(Expression involving inf1,f2,….,fn−1).
Now we can solve (63) which is equivalent to (64) recursively to determine the coefficients fn. That is we begin with f1 and solve iteratively to get f2,3,.. hoping that it terminates at some point. To do this we need the following:Theorem 2.
For any givenf, we have; DtmDxnf.1=−1n+mDtmDxn1.f, where m,n≥0Proof.
By considering to interchange primed and unprimed values as a simple label; i.e.,(65)DtmDxnf.g=∂∂t−∂∂t′m∂∂x−∂∂x′nfx,tgx′,t′x′=x,t′=t=−1m+n∂∂t′−∂∂tm∂∂x′−∂∂xngx′,t′fx,tx′=x,t′=t=−1m+n∂∂t−∂∂t′m∂∂x−∂∂x′ngx,tfx′,t′x′=x,t′=t=−1m+nDtmDxng.fDtmDxnf.1=−1n+mDtmDxn1.f=∂m∂tm∂n∂xnfDtmDxnf.1=∂m∂tm∂n∂xnf,where f is any function of x and t
Thus, (70) became(66)∂∂x∂∂t+∂3∂x3fn=−12∑m=1n−1Bfn−m.fm.
Forn=1, it reduced to(67)∂∂x∂∂t+∂3∂x3f1=0⇒∂∂t+∂3∂x3f1=0.
Beginning withf1 we may iterate(repeat) to find all the fn and note that at any order we are now solving linear partial differential equations. A simple particular solution to the last equation would be(68)f1=∑i=1Nεaix−a3it+ci=∑i=1Nεθi,f1=εaix−a3it+ci.
Using (63) for f2, we have ∂/∂x∂/∂t+∂3/∂x3f2=−1/2Bf1.f1=0, where this result follows (55). Here is the fact, with f1, the expansion of (59), terminated at order N or (the f3…∞ all vanish trivially). The series has terminated with(69)f=1+εf1.
All higher-order equations withn>N are solved by fn=0 for n>N. We can always absorb ε into the constant c and we recognize the single soliton solution. Our hope is to find other solutions where the series terminates in this way, so that the solution is exact. Consider beginning with(70)f1=∑i=1Nεaix−a3it+ci.f=1+ε1f1+ε2f2+…εNfN. We have already seen the single soliton solution. Let’s verify that we recover the two soliton solutions. Take for N=2,(71)f1=eθ1+eθ2.where θ1,2 are as above chosen to satisfy the f1 equation, and θi=aix−ai3t+ci. The f2 equation became(72)∂∂x∂∂t+∂3∂x3fn=−12Bf1.f1,=−12Beθ1+eθ2.eθ1+eθ2=−12Beθ1.eθ2−12Beθ2.eθ1=−a1−a2b1−b2+a1−a23eθ1+θ2=−a1−a2−a13−a23+a1−a23eθ1+θ2=a1−a22a12+a1a2+a22−a12+2a1a2−a22eθ1+θ2=3a1a2a1−a22eθ1+θ2∂∂x∂∂t+∂3∂x3f2=3a1a2a1−a22eθ1+θ2.
Since the equation is linear inf2 it has an obvious solution of the form f2=Aeθ1+θ2. So, we just have to determine A. Substituting into the f2 equation; that is, let(73)∂x=a1+a2,∂t=−a13+a23,∂x3=a1+a23,∂∂x∂∂t+∂3∂x3f2=3a1a2a1−a22eθ1+θ2a1+a2−a13+a23+a1+a23Aeθ1+θ2=3a1a2a1−a22eθ1+θ2a1+a2−a13+a23+a1+a23A=3a1a2a1−a223a1a2a1+a22A=3a1a2a1−a22A=a1−a22a1+a22.
The solution is then,f=1+εeθ1+εeθ2+ε2a1−a22/a1+a22eθ1+θ2, which is a two-soliton solution of the KdV equation. Setting ε=1, we have(74)f=1+eθ1+eθ2+a1−a22a1+a22eθ1+θ2.
Figure5 shows follow of the presence of a solitary wave to the front, having two kinks shaped with long amplitude, dark of high speed.
Figure6 shows the solitary wave to the front, having two kinks shaped, a valley of the wave, and dark high speed.
Figure7 shows the solitary wave from left to right, having two kink-shaped solitons, a valley-shaped wave, long amplitudes, and dark high speed.
Figure8 shows follow of the solitary wave to the front, having two kink--shaped soliton profiles, periodic, and dark and bell slow speed.
Finally, we just need to differentiate to get the originalw or u using w=2fx/f=a1eθ1+a2eθ2+a1−a22/a1+a22eθ1+θ2/1+eθ1+eθ2+a1−a22/a1+a22eθ1+θ2. It is easier in this format to study the asymptotic of the two soliton solutions. As in the BT discussion, first, we choose a coordinate system with the origin at the center of the around i=1,2 soliton; taking t⟶ti+δt with δx=x−x0−viti, where vi=ai2 and let ti⟶±∞, we have θi=vi/2δx−viδt(75)θj≠i=vj2δx−vjδt+vj2vi−vjti.
Near thei soliton, we have (by definition) δx,δt≈0, so that θi is small but θj≠i⟶±∞ as follows: (t⟶−∞, if i=1 , θ1 is fixed and θj=2⟶−∞, and if i=2, θj=1⟶−∞), (t⟶+∞, if i=1, θj=1⟶−∞, and if i=2, θj=2⟶+∞). When θj≠i⟶−∞ the f clearly reduces to the single soliton case, f=1+eθi. When θj≠i⟶+∞, we have to do a little more work. To lead order we have(76)f=eθj1+a1−a22a1+a22eθi+…..where the ellipsis indicates terms of order e−θ2. But then w=2∂xlogf(77)w=2aj+2∂xlog1+a1−a22a1+a22eθi+…..
Sincew⟶w+const gives equivalent solutions we might ignore the constant piece. The remaining solution is equivalent to an f of the for(78)f=1+a1−a22a1+a22eθi.
This is the original singlei-soliton solution but with a phase shift of(79)δx0−2vilogv1−v2v1−v2.
Precisely what we found for the 2 KdV soliton solution using the much more complicated BT method. Finally, it was just presented the general solution which might be proved by induction. Define anNxN matrix SN with elements. Considering the 2-soliton solution for equation (68), f=1+εeθ1+εeθ2+ε2a1−a22/a1+a22eθ1+θ2(80)=1+εeθ11+εeθ2−ε2eθ1+θ2+ε2a1−a2a1+a22eθ1+θ2=1+εeθ11+εeθ2−ε24a1a2a1+a22eθ1+θ2=1+εeθ1ε2a1a1+a2eθ2ε2a2a1+a2eθ11+εeθ2.f=detS,whereSNij=δij+ε2aiai+ajeθj.and i,j=1,2. Then, the N-soliton solution U=2∂x2logf, where f=detSN.Figure 5
The result of the two-soliton solution fora1=1,a2=1.5,c=0.5.Figure 6
The result of the two-soliton solution fora1=−1,a2=−0.5,c=−1.5.Figure 7
The result of the two-soliton solution fora1=−3,a2=−2.5,c=−3.5.Figure 8
The result of the two-soliton solution fora1=−5,a2=−3.5,c=−4.5.


### 2.3. Illustrative Examples

Example 1.
Consider the fractional-order nonlinear KdV system is given as follows:(81)∂ρμ∂τρ=−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,∂ρv∂τρ=−a∂3v∂ζ3−3aμ∂v∂ζ′.0<ρ<1.with the initial condition,(82)μζ,0=η2sech2α2+μζ2,vζ,0=α2η2sech2α2+μζ2.
Forρ=1, the exact results of the KdV scheme (82) are given by(83)μζ,τ=η2sech2α2+ηζ2−aη3τ2,vζ,τ=α2η2sech2α2+ηζ2−aη3τ2.
Using the ELzaki transform to (83), we obtain,(84)1sρEμζ,τ−∑k=0m−1μkζ,0s2−ρ+k=E−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,1sρEvζ,τ−∑k=0m−1vkζ,0s2−ρ+k=E−a∂3v∂ζ3−3aμ∂v∂ζ,(85)1sρEμζ,τ=μkζ,0s2−ρ+E−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,1sρEvζ,τ=vkζ,0s2−ρ+E−a∂3v∂ζ3−3aμ∂v∂ζ,Eμζ,τ=s2μζ,0+Esρ−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,Evζ,τ=s2vζ,0+Esρ−a∂3v∂ζ3−3aμ∂v∂ζ.
Applying the inverse ELzaki transform of (85), we have(86)μζ,τ=E−1s2μζ,0+E−1sρE−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,vζ,τ=E−1s2vζ,0+E−1sρE−a∂3v∂ζ3−3aμ∂v∂ζ.
Now, by using the suggested analytical method, we get(87)μ0ζ,τ=η2sech2α2+μζ2,v0ζ,τ=α2η2sech2α2+μζ2,μ1ζ,τ=E−1sρE−a∂3μ0∂ζ3−6aμ0∂μ0∂ζ+6v0∂v0∂ζ,v1ζ,τ=E−1sρE−a∂3v0∂ζ3−3aμ0∂v0∂ζ,μ1ζ,τ=η5atanhα2+ηζ2sech2α2+ηζ2τρΓρ+1,v1ζ,τ=η5a3/22tanhα2+ηζ2sech2α2+ηζ2τρΓρ+1,μ2ζ,τ=E−1sρE−a∂3μ1∂ζ3−6aμ1∂μ1∂ζ+6v1∂v1∂ζ,v2ζ,τ=E−1sρE−a∂3v1∂ζ3−3aμ1∂v1∂ζ,μ2ζ,τ=η8a222cosh2α2+ηζ2−3sech4α2+ηζ2τ2ρΓ2ρ+1,v2ζ,τ=η5a5/2242cosh2α2+ηζ2−3sech4α2+ηζ2τ2ρΓ2ρ+1,μ3ζ,τ=E−1sρE−a∂3μ2∂ζ3−6aμ2∂μ2∂ζ+6v2∂v2∂ζ,v3ζ,τ=E−1sρE−a∂3v2∂ζ3−3aμ2∂v2∂ζ,μ3ζ,τ=sinhα/2+ηζ/2τ3ρa3η42Γ3ρ+1Γρ+12cosh7α/2+ηζ/22Γρ+12cosh4α2+ηζ2−18Γρ+12cosh2α2+ηζ2+6Γ2ρ+1cosh2α2+ηζ2−18Γρ+12−9Γ2ρ+1,v3ζ,τ=2sinhα/2+ηζ/2τ3ρa7/2η114Γ3ρ+1Γρ+12cosh7α/2+ηζ/22Γρ+12cosh4α2+ηζ2−18Γρ+12cosh2α2+ηζ2+6Γ2ρ+1cosh2α2+ηζ2−18Γρ+12−9Γ2ρ+1.
Generally, thenth solution result was obtained as follows:(88)μnζ,τ=E−1sρE−a∂3μn−1∂ζ3−6aμn−1∂μn−1∂ζ+6vn−1∂vn−1∂ζ,vnζ,τ=E−1sρE−a∂3vn−1∂ζ3−3aμn−1∂vn−1∂ζ,n≥0.
The series form result is as follows:(89)μζ,τ=μ0ζ,τ+μ1ζ,τ+μ2ζ,τ+μ3ζ,τ+…..μnζ,τ,vζ,τ=v0ζ,τ+v1ζ,τ+v2ζ,τ+v3ζ,τ+…..vnζ,τ,μζ,τ=η2sech2α2+μζ2+E−1sρE−a∂3μ0∂ζ3−6aμ0∂μ0∂ζ+6v0∂v0∂ζ+E−1sρE−a∂3μ1∂ζ3−6aμ1∂μ1∂ζ+6v1∂v1∂ζ+sinhα/2+ηζ/2τ3ρa3η42Γ3ρ+1Γρ+12cosh7α/2+ηζ/22Γρ+12cosh4α2+ηζ2−18Γρ+12cosh2α2+ηζ2+6Γ2ρ+1cosh2α/2+ηζ/2−18Γρ+12−9Γ2ρ+1+….vζ,τ=α2η2sech2α2+μζ2+E−1sρE−a∂3v0∂ζ3−3aμ0∂v0∂ζ+E−1sρE−a∂3v1∂ζ3−3aμ1∂v1∂ζ+2sinhα/2+ηζ/2τ3ρa7/2η114Γ3ρ+1Γρ+12cosh7α/2+ηζ/22Γρ+12cosh4α2+ηζ2−18Γρ+12cosh2α2+ηζ2+6Γ2ρ+1cosh2α2+ηζ2−18Γρ+12−9Γ2ρ+1+….
Forρ=1, the exact results of the KdV scheme (81) are given by(90)μζ,τ=η2sech2α2+ηζ2−aη3τ2,vζ,τ=α2η2sech2α2+ηζ2−aη3τ2.Example 2.
Consider the fractional-order nonlinear dispersive long wave scheme.(91)∂ρμ∂τρ=−∂v∂ζ−12∂μ2∂ζ,∂ρμ∂τρ=−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ,0<ρ<1,and the initial condition;(92)μζ,0=atanhη2+αζ2+1,vζ,0=−1+12sech2η2+αζ2.
While using ELzaki transform of (91), we obtained(93)1sρEμζ,τ−∑k=0m−1μkζ,0s2−ρ+k=E−∂v∂ζ−12∂μ2∂ζ,1sρEvζ,τ−∑k=0m−1vkζ,0s2−ρ+k=E−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ,1sρEμζ,τ=∑k=0m−1μkζ,0s2−ρ+k+E−∂v∂ζ−12∂μ2∂ζ,1sρEvζ,τ=∑k=0m−1vkζ,0s2−ρ+k+E−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ,1sρEμζ,τ=μ0ζ,0s2−ρ+E−∂v∂ζ−12∂μ2∂ζ,1sρEvζ,τ=v0ζ,0s2−ρ+E−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ,Eμζ,τ=s2μζ,0+sρE−∂v∂ζ−12∂μ2∂ζ,Evζ,τ=s2vζ,0+sρE−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ.
Applying the inverse ELzaki transform of equation (93), we get(94)μζ,τ=E−1s2μζ,0+E−1sρE−∂v∂ζ−12∂μ2∂ζ,vζ,τ=E−1s2vζ,0+E−1sρE−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ.
Now, by using the current analytical method, we get(95)μ0ζ,0=atanhη2+αζ2+1,v0ζ,0=−1+12a2sech2η2+αζ2μ1ζ,τ=E−1sρE−∂v0∂ζ−12∂μ02∂ζ,v1ζ,τ=E−1sρE−∂3μ0∂ζ3−∂μ0∂ζ−∂μ0v0∂ζμ1ζ,τ=−a22sech2η2+αζ2τρΓρ+1v1ζ,τ=a32sinhη2+αζ2sech3η2+αζ2τρΓρ+1μ2ζ,τ=E−1sρE−∂v1∂ζ−12∂μ12∂ζ,v2ζ,τ=E−1sρE−∂3μ1∂ζ3−∂μ1∂ζ−∂μ1v1∂ζμ2ζ,τ=−a54sech2η2+αζ2τ2ρΓ2ρ+1+3a54sinh2η2+αζ2+a74sinhη2+αζ2sech4η2+αζ2τ2ρΓ2ρ+1+sech5η2+αζ2Γ2ρ+1τ3ρΓ3ρ+1Γρ+12v2ζ,τ=a642cosh2η2+αζ2−3sech4η2+αζ2τ2ρΓ2ρ+1μnζ,τ=E−1sρE−∂vn−1∂ζ−12∂μn−12∂ζ,vnζ,τ=E−1sρE−∂3μn−1∂ζ3−∂μn−1∂ζ−∂μn−1vn−1∂ζn≥0
Thus, the series form of the result is as follows:(96)μζ,τ=μ0ζ,τ+μ1ζ,τ+μ2ζ,τ+⋯+μnζ,τ,vζ,τ=v0ζ,τ+v1ζ,τ+v2ζ,τ+⋯+vnζ,τ,μζ,τ=atanhη2+αζ2+1−a22sech2η2+αζ2τρσρ+1−a54sech2η2+αζ2τ2ρΓ2ρ+1,+3a54sinh2η2+αζ2sech4η2+αζ2τ2ρΓ2ρ+1+a74sinhη2+αζ2sech5η2+αζ2Γ2ρ+1τ3ρσ3ρ+1Γρ+12+⋯,vζ,τ=−1+12a2sech2η2+αζ2+a32sinhη2+αζ2sech3η2+αζ2τρΓρ+1+a642cosh2η2+αζ2−3sech4η2+αζ2τ2ρΓ2ρ+1+⋯.
Forρ=1, the exact results of the KdV scheme (91)is as follows:(97)μζ,τ=atanhη2+αζ2−a2τ2+1,vζ,τ=−1+12sech2η2+αζ2−a2τ2.Example 3.
Considering the nonlinear fractional-order new coupled modified KdV system,we get(98)∂ρμ∂τρ=12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,∂ρv∂τρ=−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,∂ρw∂τρ=−∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ,and the initial conditions are(99)μζ,0=1+12tanhζ,vζ,0=12−14tanhζ,wζ,0=2−tanhζ.
Using the ELzaki transform to (98), we get(100)1sρEμζ,τ−∑k=0m−1μkζ,0s2−ρ+k=E12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,1sρEvζ,τ−∑k=0m−1vkζ,0s2−ρ+k=E−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,1sρEwζ,τ−∑k=0m−1wkζ,0s2−ρ+k=E∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ,1sρEμζ,τ=μ0ζ,0s2−ρ+E12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,1sρEvζ,τ=v0ζ,0s2−ρ+E−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,1sρEwζ,τ=w0ζ,0s2−ρ+E∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ,(101)Eμζ,τ=s2μζ,0+sρE12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,Evζ,τ=s2vζ,0+sρE−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,Ewζ,τ=s2wζ,0+sρE∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ.
Applying the inverse ELzaki transform of (101)(102)μζ,τ=E−1s2μζ,0+E−1sρE12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,vζ,τ=E−1s2vζ,0+E−1sρE−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,wζ,τ=E−1s2wζ,0+E−1sρE∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ.
Now, by using the suggested analytical method, we get(103)μζ,0=1+12tanhζ,vζ,0=12−14tanhζ,wζ,0=2−tanhζ,μ1ζ,τ=E−1sρE12∂3μ0∂τ3−3μ02∂μ0∂ζ+32w0∂2v0∂ζ2+3∂v0∂ζ∂w0∂ζ+32v0∂2w0∂ζ2+3v0w0∂μ0∂ζ+3μ0w0∂v0∂ζ+3μ0v0∂w0∂ζ,v1ζ,τ=E−1sρE−∂3v0∂τ3−3∂v0∂ζ∂μ0∂ζ−3v0∂2μ0∂ζ2−3v02∂w0∂ζ+6μ0v0∂μ0∂ζ+3μ02∂v0∂ζ,w1ζ,τ=E−1sρE∂3w0∂τ3−3∂w0∂ζ∂μ0∂ζ−3w0∂2μ0∂ζ2−3w02∂v0∂ζ+6μ0w0∂μ0∂ζ+3μ02∂w0∂ζ,μ1ζ,τ=112sech2ζηρΓρ+1,v1ζ,τ=−118sech2ζηρΓρ+1,w1ζ,τ=−112sech2ζηρΓρ+1,μ2ζ,τ=E−1sρE12∂3μ1∂τ3−3μ12∂μ1∂ζ+32w1∂2v1∂ζ2+3∂v1∂ζ∂w1∂ζ+32v1∂2w1∂ζ2+3v1w1∂μ1∂ζ+3μ1w1∂v1∂ζ+3μ1v1∂w1∂ζ,v2ζ,τ=E−1sρE−∂3v1∂τ3−3∂v1∂ζ∂μ1∂ζ−3v1∂2μ1∂ζ2−3v12∂w1∂ζ+6μ1v1∂μ1∂ζ+3μ12∂v1∂ζ,w2ζ,τ=E−1sρE∂3w1∂τ3−3∂w1∂ζ∂μ1∂ζ−3w1∂2μ1∂ζ2−3w12∂v1∂ζ+6μ1w1∂μ1∂ζ+3μ12∂w1∂ζ,μ2ζ,τ=−1218tanhζsech2ζη2ρΓ2ρ+1,v2ζ,τ=1218tanhζsech2ζη2ρΓ2ρ+1,w2ζ,τ=1214tanhζsech2ζη2ρΓ2ρ+1,μ3ζ,τ=133148cosh2ζ−2sech4ζη3ρΓ3ρ+1,v3ζ,τ=−133196cosh2ζ−2sech4ζη3ρΓ3ρ+1,w3ζ,τ=−133124cosh2ζ−2sech4ζη3ρΓ3ρ+1,μnζ,τ=E−1sρE12∂3μn−1∂τ3−3μn−12∂μn−1∂ζ+32wn−1∂2vn−1∂ζ2+3∂vn−1∂ζ∂wn−1∂ζ+32vn−1∂2wn−1∂ζ2+3vn−1wn−1∂μn−1∂ζ+3μn−1wn−1∂vn−1∂ζ+3μn−1vn−1∂wn−1∂ζ,vnζ,τ=E−1sρE−∂3vn−1∂τ3−3∂vn−1∂ζ∂μn−1∂ζ−3vn−1∂2μn−1∂ζ2−3vn−12∂wn−1∂ζ+6μn−1vn−1∂μn−1∂ζ+3μn−12∂vn−1∂ζ,wnζ,τ=E−1sρE∂3wn−1∂τ3−3∂wn−1∂ζ∂μn−1∂ζ−3wn−1∂2μn−1∂ζ2−3wn−12∂vn−1∂ζ+6μn−1wn−1∂μn−1∂ζ+3μn−12∂wn−1∂ζ.
The series form result is given as follows:(104)μζ,τ=μ0ζ,τ+μ1ζ,τ+μ2ζ,τ+μ3ζ,τ+⋯+μnζ,τ,vζ,τ=v0ζ,τ+v1ζ,τ+v2ζ,τ+v3ζ,τ+⋯+vnζ,τ,wζ,τ=w0ζ,τ+w1ζ,τ+w2ζ,τ+w2ζ,τ+⋯+wnζ,τ,μζ,τ=1+12tanhζ+112sech2ζηρΓρ+1−1218tanhζsech2ζη2ρΓ2ρ+1+133148cosh2ζ−2sech4ζη3ρΓ3ρ+1+⋯,vζ,τ=12−14tanhζ−118sech2ζηρΓρ+1+1218tanhζsech2ζη2ρΓ2ρ+1−133196cosh2ζ−2sech4ζη3ρΓ3ρ+1+⋯wζ,τ=2−tanhζ−112sech2ζηρΓρ+1+1214tanhζsech2ζη2ρΓ2ρ+1−133124cosh2ζ−2sech4ζη3ρΓ3ρ+1+⋯.
The exact result of (98) is as follows:(105)μζ,τ=1+12tanhζ−112η,vζ,τ=12−14tanhζ−112η,wζ,τ=2−tanhζ−112η.


## 2.1. Preliminaries

### 2.1.1. Bilinear Operators

Bilinear differential operatorD mapping a pair of functions Df.g. Unlike usual linear differential operators like ∂/∂xn, which maps a single function f into a single function ∂nf/∂xn. That is D:C∞xC∞⟶C∞:f.g↦Df.gRemark 1.
For all integersm,n≥0,(11)DtmDxnf.gx,t=∂∂t−∂∂t′m∂∂x−∂∂x′nfx,tgx′,t′x′=x,t′=t,where f is a function of x and t, whereas g is a function of x′ and t′; for which f≠g.


### 2.1.2. Fractional Derivative

A fractional derivative is a derivative of any order, real or complex in applied mathematics or mathematical analysis.


### 2.1.3. The ELzaki Transform

The Elzaki transform is a function of the formhx,t with respect to t(12)ψhx,t=ψx,s=s∫0∞hx,texp−tsdt,where s is the complex number, t>0.


## 2.1.1. Bilinear Operators

Bilinear differential operatorD mapping a pair of functions Df.g. Unlike usual linear differential operators like ∂/∂xn, which maps a single function f into a single function ∂nf/∂xn. That is D:C∞xC∞⟶C∞:f.g↦Df.gRemark 1.
For all integersm,n≥0,(11)DtmDxnf.gx,t=∂∂t−∂∂t′m∂∂x−∂∂x′nfx,tgx′,t′x′=x,t′=t,where f is a function of x and t, whereas g is a function of x′ and t′; for which f≠g.


## 2.1.2. Fractional Derivative

A fractional derivative is a derivative of any order, real or complex in applied mathematics or mathematical analysis.


## 2.1.3. The ELzaki Transform

The Elzaki transform is a function of the formhx,t with respect to t(12)ψhx,t=ψx,s=s∫0∞hx,texp−tsdt,where s is the complex number, t>0.


## 2.2. Main Results

### 2.2.1. Method of Solution

Considering the general nonlinear PDEs given in (8), let us consider the modified KdV equation that was chosen as an example to illustrate the balance method. Suppose that the solution of (5) is (9). Now considering uxxx, uux balancing it in the (5), it is required that pq+3=2pq+1 (by letting u=pq, ux=1, uxx=2, and uxxx=3, where 1, 2, and 3 represent the order of derivatives for this case or condition). Again balancing ut and uux in (9), it gave that dq+1=2dq+1, which implies dq=2dq. Choosing p=2, and then solving for the rest of the variables, we obtained that q=1 and d=0. That means from the equation pq+3=2pq+1, by simplifying it we found pq=2. Since we chosen p=2, then we have q=1. Again from the second condition dq+1=2dq+1, we have d=0.Hence (9) u=apdlnωp,d+∑k,j=0,k+j≠0,p+dk=p,j=dakjlnωk,j+a00q, …(13)u=a20lnωxx+a10lnωx+a00.Also, this equation can also be expressed as follows:(14)u=a20wxxw−wx2w2+a10wxw+a00,ux=∂∂xa20wxxw−wx2w2+a10wxw+a00=a20wxxw−wxxwxw2−2wxwxxw2−wx22wwxw4+a10wxxw−wxwxw2=a20wxxxw−wxxwxw2−2wxxwxw2+2wx3w3+a10wxxw−wx2w2=a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2,(15)ux=a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2,ut=∂∂ta20lnωxx+a10lnωx+a00=∂∂ta20wxxw−wx2w2+a10wxw+a00=a20wxxtw−wxxwtw2−2wxwxtw2−wx2wt2ww4+a10wxtw−wxwtw2=a20wxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2=a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2,(16)ut=a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2.From equation (15), we get(17)uxx=∂∂xUx=∂∂xa20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2=a20wxxxxw−wxxxwxw2−3wxxxwx+wxxwxxw2−wxxwx2wwxw4+23wx2wxx.w3−w3x3w2wxw6+a10wxxxw−wxxwxw2−2wxwxxw2−wx22w.wxw4=a20wxxxxw−wxxxwxw2−3wxxxwxw2−3w2xxw2+6wxxwx2w3+6wxxwx2w3−6w4w4+a10wxxxw−wxxwxw2−2wxxwxw2+2wx3w3=a20wxxxxw−4wxxxwx+3w2xxw2+12wxxwx2w3−6w4w4+a10wxxxw−3wxxwxw2+2wx3w3,(18)uxx=a20wxxxxw−4wxxxwx+3w2xxw2+12wxxwx2w3−6wx4w4+a10wxxxw−3wxxwxw2+2wx3w3.From (18), we get(19)uxxx=∂∂xa20wxxxxw−4wxxxwx+3w2xxw2+12wxxwx2w3−6w4w4+a10wxxxw−3wxxwxw2+2wx3w3=a20wxxxxxw−wxxxxwxw2−4wxxxxwx+wxxxwxw2−wxxxwx2w.wxw4−32wxxwxxxw2−wxx22w.wxw4+12wxxxwx2+wxx2wxwxxw3−wxxwx23w2wxw6−64wx3wxxw4−wx44w3wxw8+a10wxxxxw−wxxxwxw2−3wxxxwx+wxxwxxw2−wxxwx2w.wxw4+23wx2wxxw3−wx33w2wxw6=a20wxxxxxw−wxxxxwxw2−4wxxxxwxw2−4wxxxwxxw2+8wxxxwx2w3−6wxxxwxxw2+6wxx2wxw3+12wxxxwx2w3+24wxx2wxw3−36wxxwx3w4−24wxxwx3w4+24wx5w5+a10wxxxxw−wxxxwxw2−3wxxxwxw2−3wxx2w2+6wxxwx2w3+6wxxwx2w3−6wx4w4=a20wxxxxxw−5wxxxxwxw2−10wxxxwxxw2+20wxxxwx2w3+30wxx2wxw3−60wxxwx3w4−24wx5w5+a10wxxxxw−4wxxxwxw2−3wxx2w2+12wxxwx2w3−6wx4w4,(20)uxxx=a20wxxxxxw−5wxxxxwxw2−10wxxxwxxw2+20wxxxwx2w3+30wxx2wxw3−60wxxwx3w424wx5w5+a10wxxxxw−4wxxxwxw2−3wxx2w2+12wxxwx2w3−6wx4w4.Now substituting equations (14), (15), (16), and (20) into (5), we have;(21)a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2+a20wxxw−wx2w2+a10wxw+a00a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2+βa20wxxxxxw−5wxxxxwxw2−10wxxxwxxw2+20wxxxwx2w3+30wxx2wxw3−60wxxwx3w4−24wx5w5+a10wxxxxw−4wxxxwxw2−3wxx2w2+12wxxwx2w3−6wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a20wxxw−a20wx2w2+a10wxw+a00a20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a20wxxwa20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2−a20wx2w2a20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+a10wxwa20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+a00a20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a20wxxwa20wxxxw−a20wxxwa203wxxwxw2+a20wxxwa202wx3w3+a20wxxwa10wxxw−a20wxxwa10wx2w2−−a20wx2w2a20wxxxw−a20wx2w2a203wxxwxw2+a20wx2w2a202wx3w3+a20wx2w2a10wxxw−a20wx2w2a10wx2w2−a10wxwa20wxxxw−a10wxwa203wxxwxw2+a10wxwa202wx3w3+a10wxwa10wxxw−a10wxwa10wx2w2+a00a20wxxxw−a00a203wxxwxw2+a00a202wx3w3+a00a10wxxw−a00a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a202wxxwxxxw2−a2023wxx2wxw3+a2022wx3wxxw4+a20a10wxx2w2−a20a10wx2wxxw3−a202wx2wxxxw3+a2023wxxwx3w4−a2022wx5w5−a20a10wxxwx2w3+a20a10wx4w4+a10a20wxwxxxw2−a10a203wxxwx2w3+a10a202wx4w4+a102wxwxxw2−a102wx3w3+a00a20wxxxw−a00a203wxxwxw2+a00a202wx3w3+a00a10wxxw−a00a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a202wxxwxxxw2−a2023wxx2wxw3+a2022wx3wxxw4+a20a10wxx2w2−a20a10wx2wxxw3−a202wx2wxxxw3+a2023wxxwx3w4−a2022wx5w5−a20a10wxxwx2w3+a20a10wx4w4+a10a20wxwxxxw2−a10a203wxxwx2w3+a10a202wx4w4+a102wxwxxw2−a102wx3w3+a00a20wxxxw−a00a203wxxwxw2+a00a202wx3w3+a00a10wxxw−a00a10wx2w2+βa20wxxxxxw−βa205wxxxxwxw2−βa2010wxxxwxxw2+βa2020wxxxwx2w3+βa2030wxx2wxw3−βa2060wxxwx3w4−βa2024wx5w5+βa10wxxxxw−βa104wxxxwxw2−βa103wxx2w2+βa1012wxxwx2w3−βa106wx4w4=0Now equating the coefficients ofwx5/w5 and wx4/w4 to zero on the left-hand side, we have a20=12β and a10=0. This is to mean that; equating a20β24wx5/w5=0, a10β6wx4/w4=0, a10a202wx4/w4=0, −2a202wx5/w5=0, and a10a20wx4/w4=0. Now considering the two equations a20β24wx5/w5=0 and −2a202wx5/w5=0, it is possible to solve for a20. Since β≠0 and w≠024a20β=0 and −2a202=0, we solved for the reset a20 as 24a20β=0−2a202=0Adding these two equations together and solving fora20, as 24a20β−2a202=02a2012β−a20=0. From this, we can get a20=12βAgain considering the equations;−6βa10wx4/w4=0,2a10a20wx4/w4=0,anda10a20wx4/w4=0.And adding them together using a system of equations as−6βa10=0,2a10a20=0,a10a20=0.we have 3a20a10−6βa10=0. 3a20a10=6βa10,Simplifying it we get, a20a10=2βa10. Substituting a20=12β into this equation we get, 12βa10=2a10. Which give us 10βa10=0, β≠0. And hence it implies that a10=0Substitutinga20=12β and a10=0 into the following equation: u=a20lnωxx+a10lnωx+a00, we get u=12lnωxx+0lnωx+a00=12lnωxx+a00=12βwxx/w−wx2/w2+a00, where a00 is any arbitrary constant.(22)u=12lnωxx+a00=12βwxxw−wx2w2+a00.Putting (22) into (5), we have the following results, where, a20=12β and a10=0.(23)12βwxxtw−12βwxxwtw2−12β2wxwxtw2+12β2wx2wtw3+0wxtw−0wxwtw2+12β2wxxwxxxw2−12β23wxx2wxw3+12β22wx3wxxw4+012βwxx2w2−012βwx2wxxw3−12β2wx2wxxxw3+12β23wxxwx3w4−12β22wx5w5−012βwxxwx2w3+012βwx4w4+012βwxwxxxw2−012β3wxxwx2w3+012β2wx4w4+0wxwxxw2−0wx3w3+a0012βwxxxw−a0012β3wxxwxw2+a0012β2wx3w3+0a00wxxw−0a00wx2w2+β12βwxxxxxw−12β25wxxxxwxw2−12β210wxxxwxxw2+12β220wxxxwx2w3+12β230wxx2wxw3−12β260wxxwx3w4−12β224wx5w5+0βwxxxxw−04βwxxxwxw2−03βwxx2w2+012βwxxwx2w3−06βwx4w4=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+144β2wxxwxxxw2−144β23wxx2wxw3+144β22wx3wxxw4+144β2wx2wxxxw3+144β23wxxwx3w4−144β22wx5w5−a0012βwxxxw−a0036βwxxwxw2+a0024βwx3w3+β12βwxxxxxw−60β2wxxxxwxw2−120β2wxxxwxxw2+240β2wxxxwx2w3+360β2wxx2wxw3−720β2wxxwx3w4−288β2wx5w5=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12β12βwxxwxxxw2−12β36βwxx2wxw3+12β24βwx3wxxw4+12β12βwx2wxxxw3+12β36βwxxwx3w4−12β24βwx5w5+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12ββwxxxxxw−5βwxxxxwxw2−10βwxxxwxxw2+20βwxxxwx2w3+30βwxx2wxw3−60βwxxwx3w4−24βwx5w5=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12β12βwxxwxxxw2−36βwxx2wxw3+24βwx3wxxw4+12βwx2wxxxw3+36βwxxwx3w4−24βwx5w5+βwxxxxxw−5βwxxxxwxw2−10βwxxxwxxw2+20βwxxxwx2w3+30βwxx2wxw3−60βwxxwx3w4−24βwx5w5=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12β12βwxxwxxxw2−10βwxxxwxxw2−36βwxx2wxw3+30βwxx2wxw3+24βwx3wxxw4+36βwxxwx3w4+−60βwxxwx3w4−12βwx2wxxxw3+20βwxxxwx2w3−24βwx5w5+24βwx5w5+βwxxxxxw−5βwxxxxwxw2=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12β2βwxxwxxxw2−6βwxx2wxw3+60βwx3wxxw4−60βwxxwx3w4+8βwx2wxxxw3+βwxxxxxw−5βwxxxxwxw2=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12ββwxxxxxw+2wxxxwxx−5wxxxxwxw2+8wxxxwx2−6wxx2wxw3=012βK1+K2+K3=0,where(24)K1=wxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3,K2=a00wxxxw−a003wxxwxw2+a002wx3w3,K3=βwxxxxxw+2wxxxwxx−5wxxxxwxw2+8wxxxwx2−6wxx2wxw3,(25)12βK1+K2+K3=0.Simplifying (25) and integrating once with respect to x, we get(26)∂∂xwxtw−wxwt+βwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2=0.Equations (20) and (26) are identical with(27)wxtw−wxwt+βwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2−Ctw2=0,where Ct is an arbitrary function of t, and a00 is an arbitrary constant. Especially, taking Ct as zero in equation (27), we get the bilinear equation of (5) as follows:(28)wxtw−wxwt+βwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2=0.Equation (28) can be written concisely in terms of D-operator as(29)DxDt+βDX4+a00Dx2w,w=0,where DxmDtna.b=∂x−∂x′m∂t−∂t′nax,tbx′,t′x=x′,t=t′.Applying Hirota’s method, the bilinear equation of (5) can be written as(30)DxDt+βDX4w,w=0.Equation (30) is obtained by setting a00=0 in equation (29). Obviously, equation (30) is a special case of equation (29). Therefore, a more general bilinear equation of the modified KdV equation is obtained by using Hirota’s method.Let us consider the modified KdV equationut+6uux+uxxx=0 by using Hirota’s method.Substituting(31)u=wx,which gives the following new equation: wxt+6wxwxx+wxxxx=0. Now integrating both sides with respect to x, we have;(32)∫wxt+6wxwxx+wxxxxdx=0,∫wxtdx+∫6wxwxxdx+∫wxxxxdx=gt,wt+6wx22+wxxx=C=gt.where C=gt is constant of integration. Likely we can get, without loss of generality, the integration “constant” with respect to x, gt can be observed by a redefinition of w that does not change the KdV field, u=wxwold=wnew+∫t0tdt′gt′, dt′ is the time derivative.Using the neww, we have(33)wt+3wx2+wxxx=0.Now considering the 1-soliton solution of the modified KdV equation, let(34)u=2μ2sech2μx−x0−4μ2t.with(35)μ=v2≥0,where v is the velocity of the wave propagating.It can be written asu=wx, with(36)w=2μ2tanhμx−x0−4μ2t.In fact, we can integrate the right-hand side of equation (36) once again, using tanh y=d/dy log cosh y. Therefore, equation (34) can be written as follows:(37)u=2∂2∂x2logcoshμx−x0−4μ2t.Let(38)X=x−x0−4μ2t.Then,u=2∂2/∂X2logexp−μX1+exp2μX/2, where coshy=exp−μX1+exp2μX/2(39)=2∂2∂x2−μx−log2+log1+exp2μX=−2∂2∂x2μx−2∂2∂x2log2+2∂2∂x2log1+exp2μX=2∂2∂X2log1+exp2μX.That is,ux,t=2∂2/∂x2log1+exp2μx−x0−4μ2t=2∂2/∂x2log1+exp2μX(40)ux,t=2∂2∂x2log1+exp2μx−x0−4μ2t=2∂2∂x2log1+exp2μX,which is the 1-soliton solution of the modified KdV equation, that we use in the following:Considering equations (31) and (33), the KdV equation in a bilinear form, and rewriting it in a quadratic form, we have the following equation. Inspired by equation (34), let us substitute(41)w=2∂∂xlogf=2fxf⇔u=2∂2∂x2logf,where f=1+exp2μx−x0−4μ2t. Substituting in equation (33), we get(42)12wt=fxt−fxftf2,12wx=fxx−fx2f2,12wxx=fxxxf−3fxfxxf2+2fx3f3,12wxxx=fxxxxf−4fxfxxxf2−3fxx2f2+12fxxfx2f3−6fx4f4.(43)12wxxx=fxxxxf−4fxfxxxf2−3fxx2f2+12fxxfx2f3−6fx4f4,Equation (43) is quadratic in f.Thus, equation (33) for w becomes(44)wt+3wx2+wxxx=0,=2fxt−fxftf2+322fxxf−fx2f22+2fxxxxf−42fxfxxxf2−32fxx2f2+122fxxfx2f3−62fx4f4=2fxt−fxftf2+12fxx2f2−2fxxfx2f3+fx4f4+2fxxxxf−8fxfxxxf2−6fxx2f2+24fxxfx2f3−12fx4f4=2fxt−fxftf2+12fxx2f2−24fxxfx2f3+12fx4f4+2fxxxxf−8fxfxxxf2−6fxx2f2+24fxxfx2f3−12fx4f412=2fxt−fxftf2+2fxxxxf−8fxfxxxf2+6fxx2f2=fxt−fxftf2+fxxxxf−4fxfxxxf2+3fxx2f2.Multiplying byf2 throughout, we have(45)ffxt−fxft+ffxxxx−4fxfxxx+3fxx2=0This is the bilinear form of the modified KdV equation. Despite the fact that some nontrivial cancellations took place it looks more complicated than the original problem, however, its special form makes it possible to solve using bilinear operators.Claim 1.
∂/∂x∂/∂t1/2f2=∂/∂xfft=ffxt+fxft
To express equation (45), we use the following expressions:(46)Dtf.gx,t=∂∂t−∂∂t′∂∂x−∂∂x′fx,tgx′,t′x′=x,t′=t=ftx,tgx′,t−fx,tgt′x′,t′x′=x,t′=t=ftx,tgx,t−fx,tgtx,tDtf.gx,t=ftx,tgx,t−fx,tgtx,tDtf.fx,t=0,Dxf.fx,t=0forf=gDtDxf.g=∂∂t−∂∂t′∂∂x−∂∂x′fx,tgx′,t′x′=x,t′=t=∂∂t−∂∂t′fxx,tgx′,t′−fx,tgx′x′,t′x′=x,t′=t=fxtg−ftgx−fxgt−fgxtfxtg−ftgx−fxgt+fgxtDtDxf.f=fxtf−ftfx−fxft+ffxt=2ffxt−ftfxDtDxf.f=2ffxt−ftfx
Like the first two terms in (45).(47)Dx2f.g=∂∂x−∂∂x′2fxx,tgx′,t′x′=x,t′=t=∂∂x−∂∂x′fxx,tgx′,t′−fx,tgx′x′,t′x′=x,t′=t=fxxg−fxgx−fxgx+fgxx=fxxg−2fxgx+fgxx,Dx2f.g=fxxg−2fxgx+fgxx,Dx2f.f=fxxf−2fxfx+ffxx=2ffxx−fx2 all functions of x,t.
Again this looks a bit like the differentiation of a product. But, we haveDx2f,f=2ffxx−2fx2 even though Dxf.fx,t=0. So importantly the operators are not associative since Dx2f.f≠DxDxf.f.
In fact, the right-hand side is meaningless, becauseDxf.f is a single function, but the outer Dx needs to act on a pair of functions.(48)Dx4f.g=∂∂x−∂∂x′4fxx,tgx′,t′x′=x,t′=t=∂∂x−∂∂x′3fxx,tgx′,t′−fx,tgx′x′,t′x′=x,t′=t=∂∂x−∂∂x′2fxxg−2fxgx′+fgx′x′x′=x,t′=t=∂∂x−∂∂x′fxxxg−3fxxgx′+3fxgx′x′−fgx′x′x′x′=x,t′=t=fxxxxg−4fxxxgx′+6fxxgx′x′−4fxgx′x′x′+fgx′x′x′x′x′=x,t′=t=fxxxxg−4fxxxgx+6fxxgxx−4fxgxxx+fgxxxxDx4f.g=fxxxxg−4fxxxgx+6fxxgxx−4fxgxxx+fgxxxx.
Again it looks just like a differentiated product off and g but with significant changes on the odd terms. Dx4f.f=fxxxxf−4fxxxfx+6fxxfxx−4fxfxxx+ffxxxx=2fxxxxf−8fxxxfx+6fxx2.(49)Dx4f.f=2fxxxxf−8fxxxfx+6fxx2
Here, we have to notice that (49) is like ∂x4fg, but we have an alternative signs;(50)Dx4f.f=2fxxxxf−4fxxxfx+3fxx2.
So, the KdV equation is in a quadratic form in equation (45) and can be recast as DtDx+Dx4f.f=0 is a bilinear form of the KdV equation.(51)DtDx+Dx4f.f=0,DtDx+Dx4=DxDt+Dx3.
(1) Solutions of the Bilinear KdV Equation. For example (1) and considering u=2∂x2logf, we have two cases.Case 1.
Forf=1, then u=0, which is a vacuum.
Forf=1+exp2μx−x0−4μ2t, then u=1−soliton solution. Since equation (51) is a bilinear equation, then this suggests that multisoliton solutions might be the sums of exponents of linear functions of x and t.
(2). The 1-Soliton Solution. In Hirota form the multisoliton solutions are sums of exponential of linear expressions in x and t.
For the one-soliton solution try(52)f=e0+eθ.where =ax+bt+c, ⇒x=ax,c=−x0, and −4μ2=tTheorem 1.
For anyθ1 and θ2, we have DtmDxneθ1.eθ2 = b1−b2ma1−a2neθ1+θ2, where m,n≥0.Proof.
By induction.
Let ifθi=aix+bit+ci,i=1,2. Then,(53)DtmDxneθ1.eθ2=b1−b2ma1−a2neθ1+θ2=b1−b2ma1−a2neθ1x,t+θ2x′,t′x′=x,t′=t
Then,(54)DtmDxn+1eθ1.eθ2=b1−b2ma1−a2n∂∂x−∂∂x′eθ1x,t+θ2x′,t′x′=x,t′=t=b1−b2ma1−a2na1−a2eθ1x,t+θ2x′,t′x′=x,t′=t=b1−b2ma1−a2n+1eθ1+θ2DtmDxn+1eθ1.eθ2=b1−b2ma1−a2n+1eθ1+θ2.and similarly for Dt operator.
Finally, the equation is obviously true forn=1,m=0 and n=0,m=1 so by induction must be true for all n and m. In particular, we find(55)DtmDxneθ.eθ=0.
Unlessm=n=0(56)DtmDxneθ.1=bmaneθDtmDxn1.eθ=−1m+nbman1,eθ.
Then, the bilinear form of the KdV equation forf=1+eθ is(57)DtDx+Dx41+eθ,1+eθ=0=DtDx+Dx41.1+1.eθ+eθ.1+eθ.eθ=DtDx+Dx4eθ.1+1.eθ=2baeθ+2a4eθ2aba+a3eθ=0.
Sinceeθ≠0, we have two opportunities. These two cases or opportunities are as follows:(1)
a=0. If a=0, then f is independent of x. So u=0, (which is a boundary condition).(2)
b=−a3. If b=−a3, then f=1+eax−a3t+c and hence u=2∂x2logf, f=1+eax−a3t+c, which is u=2∂x2log1+eax−a3t+c(58)=a22sech212ax−a3t+c,u=a22sech212ax−a3t+c.
This is the one-soliton solution withv=a2.
Figure1 shows the disturbance of the water surface having kink-shaped traveling waves with amplitude and antibell soliton. The follow of this wave is to the front.
Figure2 shows the propagation of waves having kink shaped and like some periodic soliton. And follow the wave from back to front.
Figure3 shows surface waves of long wave length having kink-shaped and bright periodic soliton. And the wave is from left to right.
Figure4 surface waves of long wave length having kink shaped and bright periodic soliton. And the wave is from left to right. This is the other form of Figure 3, but it is different by the value of x, that is why it is broken from the above.
(3) Multisoliton Solutions.Figure 1
The result of one-soliton solution fora=1/2, c = 0.1.Figure 2
The result of one-soliton solution fora=0.6, c = 0.2.Figure 3
The result of one-soliton solution fora=0.7,c=0.3.Figure 4
The result of one-soliton solution fora=0.7, c=0.3.Case 2.
For the power series in an auxiliary with parameterε(59)fx,t=∑n=0∞εnfnx,t,
Withf0=1 in which the series will terminate at some power of ε to give exact results rather than an infinite series in ε. So, we can take ε finite ε = 1.
To begin with, we putfx,t=1+∑n=0∞εnfnx,t and collect powers of ε. We start with 1 as the ε0 term as it appeared in the one-soliton solution. Defining(60)B=DtDx+Dx4.
We want to find solutions such that(61)Bf.f=0.
Substituting (59) into (60), we have 0=B∑n1=0∞εn1fn1,∑n2=0∞εn2fn2, with f0=1=∑n1=0∞∑n2=0∞εn1+n2Bfn1,fn2 and gathering terms of the same degree n=n1+n2inε and also expanding as for the one soliton case we have(62)Bf.f=B1.1+εB1.f1+Bf1.1+ε2B1.f2+Bf1.f1+Bf2.1=0,=∑n=0∞εn∑m=0nBfn−m.fm.so we have to solve an infinite set of equations order by order in ε(63)∑m=0nBfn−m.fm=0,for all n = 1, 2, ….
Equation (63) can be written as(64)Bfn,1+B1,fn=−∑m=1n−1Bfn−m.fm.
(Expression involving inf1,f2,….,fn−1).
Now we can solve (63) which is equivalent to (64) recursively to determine the coefficients fn. That is we begin with f1 and solve iteratively to get f2,3,.. hoping that it terminates at some point. To do this we need the following:Theorem 2.
For any givenf, we have; DtmDxnf.1=−1n+mDtmDxn1.f, where m,n≥0Proof.
By considering to interchange primed and unprimed values as a simple label; i.e.,(65)DtmDxnf.g=∂∂t−∂∂t′m∂∂x−∂∂x′nfx,tgx′,t′x′=x,t′=t=−1m+n∂∂t′−∂∂tm∂∂x′−∂∂xngx′,t′fx,tx′=x,t′=t=−1m+n∂∂t−∂∂t′m∂∂x−∂∂x′ngx,tfx′,t′x′=x,t′=t=−1m+nDtmDxng.fDtmDxnf.1=−1n+mDtmDxn1.f=∂m∂tm∂n∂xnfDtmDxnf.1=∂m∂tm∂n∂xnf,where f is any function of x and t
Thus, (70) became(66)∂∂x∂∂t+∂3∂x3fn=−12∑m=1n−1Bfn−m.fm.
Forn=1, it reduced to(67)∂∂x∂∂t+∂3∂x3f1=0⇒∂∂t+∂3∂x3f1=0.
Beginning withf1 we may iterate(repeat) to find all the fn and note that at any order we are now solving linear partial differential equations. A simple particular solution to the last equation would be(68)f1=∑i=1Nεaix−a3it+ci=∑i=1Nεθi,f1=εaix−a3it+ci.
Using (63) for f2, we have ∂/∂x∂/∂t+∂3/∂x3f2=−1/2Bf1.f1=0, where this result follows (55). Here is the fact, with f1, the expansion of (59), terminated at order N or (the f3…∞ all vanish trivially). The series has terminated with(69)f=1+εf1.
All higher-order equations withn>N are solved by fn=0 for n>N. We can always absorb ε into the constant c and we recognize the single soliton solution. Our hope is to find other solutions where the series terminates in this way, so that the solution is exact. Consider beginning with(70)f1=∑i=1Nεaix−a3it+ci.f=1+ε1f1+ε2f2+…εNfN. We have already seen the single soliton solution. Let’s verify that we recover the two soliton solutions. Take for N=2,(71)f1=eθ1+eθ2.where θ1,2 are as above chosen to satisfy the f1 equation, and θi=aix−ai3t+ci. The f2 equation became(72)∂∂x∂∂t+∂3∂x3fn=−12Bf1.f1,=−12Beθ1+eθ2.eθ1+eθ2=−12Beθ1.eθ2−12Beθ2.eθ1=−a1−a2b1−b2+a1−a23eθ1+θ2=−a1−a2−a13−a23+a1−a23eθ1+θ2=a1−a22a12+a1a2+a22−a12+2a1a2−a22eθ1+θ2=3a1a2a1−a22eθ1+θ2∂∂x∂∂t+∂3∂x3f2=3a1a2a1−a22eθ1+θ2.
Since the equation is linear inf2 it has an obvious solution of the form f2=Aeθ1+θ2. So, we just have to determine A. Substituting into the f2 equation; that is, let(73)∂x=a1+a2,∂t=−a13+a23,∂x3=a1+a23,∂∂x∂∂t+∂3∂x3f2=3a1a2a1−a22eθ1+θ2a1+a2−a13+a23+a1+a23Aeθ1+θ2=3a1a2a1−a22eθ1+θ2a1+a2−a13+a23+a1+a23A=3a1a2a1−a223a1a2a1+a22A=3a1a2a1−a22A=a1−a22a1+a22.
The solution is then,f=1+εeθ1+εeθ2+ε2a1−a22/a1+a22eθ1+θ2, which is a two-soliton solution of the KdV equation. Setting ε=1, we have(74)f=1+eθ1+eθ2+a1−a22a1+a22eθ1+θ2.
Figure5 shows follow of the presence of a solitary wave to the front, having two kinks shaped with long amplitude, dark of high speed.
Figure6 shows the solitary wave to the front, having two kinks shaped, a valley of the wave, and dark high speed.
Figure7 shows the solitary wave from left to right, having two kink-shaped solitons, a valley-shaped wave, long amplitudes, and dark high speed.
Figure8 shows follow of the solitary wave to the front, having two kink--shaped soliton profiles, periodic, and dark and bell slow speed.
Finally, we just need to differentiate to get the originalw or u using w=2fx/f=a1eθ1+a2eθ2+a1−a22/a1+a22eθ1+θ2/1+eθ1+eθ2+a1−a22/a1+a22eθ1+θ2. It is easier in this format to study the asymptotic of the two soliton solutions. As in the BT discussion, first, we choose a coordinate system with the origin at the center of the around i=1,2 soliton; taking t⟶ti+δt with δx=x−x0−viti, where vi=ai2 and let ti⟶±∞, we have θi=vi/2δx−viδt(75)θj≠i=vj2δx−vjδt+vj2vi−vjti.
Near thei soliton, we have (by definition) δx,δt≈0, so that θi is small but θj≠i⟶±∞ as follows: (t⟶−∞, if i=1 , θ1 is fixed and θj=2⟶−∞, and if i=2, θj=1⟶−∞), (t⟶+∞, if i=1, θj=1⟶−∞, and if i=2, θj=2⟶+∞). When θj≠i⟶−∞ the f clearly reduces to the single soliton case, f=1+eθi. When θj≠i⟶+∞, we have to do a little more work. To lead order we have(76)f=eθj1+a1−a22a1+a22eθi+…..where the ellipsis indicates terms of order e−θ2. But then w=2∂xlogf(77)w=2aj+2∂xlog1+a1−a22a1+a22eθi+…..
Sincew⟶w+const gives equivalent solutions we might ignore the constant piece. The remaining solution is equivalent to an f of the for(78)f=1+a1−a22a1+a22eθi.
This is the original singlei-soliton solution but with a phase shift of(79)δx0−2vilogv1−v2v1−v2.
Precisely what we found for the 2 KdV soliton solution using the much more complicated BT method. Finally, it was just presented the general solution which might be proved by induction. Define anNxN matrix SN with elements. Considering the 2-soliton solution for equation (68), f=1+εeθ1+εeθ2+ε2a1−a22/a1+a22eθ1+θ2(80)=1+εeθ11+εeθ2−ε2eθ1+θ2+ε2a1−a2a1+a22eθ1+θ2=1+εeθ11+εeθ2−ε24a1a2a1+a22eθ1+θ2=1+εeθ1ε2a1a1+a2eθ2ε2a2a1+a2eθ11+εeθ2.f=detS,whereSNij=δij+ε2aiai+ajeθj.and i,j=1,2. Then, the N-soliton solution U=2∂x2logf, where f=detSN.Figure 5
The result of the two-soliton solution fora1=1,a2=1.5,c=0.5.Figure 6
The result of the two-soliton solution fora1=−1,a2=−0.5,c=−1.5.Figure 7
The result of the two-soliton solution fora1=−3,a2=−2.5,c=−3.5.Figure 8
The result of the two-soliton solution fora1=−5,a2=−3.5,c=−4.5.


## 2.2.1. Method of Solution

Considering the general nonlinear PDEs given in (8), let us consider the modified KdV equation that was chosen as an example to illustrate the balance method. Suppose that the solution of (5) is (9). Now considering uxxx, uux balancing it in the (5), it is required that pq+3=2pq+1 (by letting u=pq, ux=1, uxx=2, and uxxx=3, where 1, 2, and 3 represent the order of derivatives for this case or condition). Again balancing ut and uux in (9), it gave that dq+1=2dq+1, which implies dq=2dq. Choosing p=2, and then solving for the rest of the variables, we obtained that q=1 and d=0. That means from the equation pq+3=2pq+1, by simplifying it we found pq=2. Since we chosen p=2, then we have q=1. Again from the second condition dq+1=2dq+1, we have d=0.Hence (9) u=apdlnωp,d+∑k,j=0,k+j≠0,p+dk=p,j=dakjlnωk,j+a00q, …(13)u=a20lnωxx+a10lnωx+a00.Also, this equation can also be expressed as follows:(14)u=a20wxxw−wx2w2+a10wxw+a00,ux=∂∂xa20wxxw−wx2w2+a10wxw+a00=a20wxxw−wxxwxw2−2wxwxxw2−wx22wwxw4+a10wxxw−wxwxw2=a20wxxxw−wxxwxw2−2wxxwxw2+2wx3w3+a10wxxw−wx2w2=a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2,(15)ux=a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2,ut=∂∂ta20lnωxx+a10lnωx+a00=∂∂ta20wxxw−wx2w2+a10wxw+a00=a20wxxtw−wxxwtw2−2wxwxtw2−wx2wt2ww4+a10wxtw−wxwtw2=a20wxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2=a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2,(16)ut=a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2.From equation (15), we get(17)uxx=∂∂xUx=∂∂xa20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2=a20wxxxxw−wxxxwxw2−3wxxxwx+wxxwxxw2−wxxwx2wwxw4+23wx2wxx.w3−w3x3w2wxw6+a10wxxxw−wxxwxw2−2wxwxxw2−wx22w.wxw4=a20wxxxxw−wxxxwxw2−3wxxxwxw2−3w2xxw2+6wxxwx2w3+6wxxwx2w3−6w4w4+a10wxxxw−wxxwxw2−2wxxwxw2+2wx3w3=a20wxxxxw−4wxxxwx+3w2xxw2+12wxxwx2w3−6w4w4+a10wxxxw−3wxxwxw2+2wx3w3,(18)uxx=a20wxxxxw−4wxxxwx+3w2xxw2+12wxxwx2w3−6wx4w4+a10wxxxw−3wxxwxw2+2wx3w3.From (18), we get(19)uxxx=∂∂xa20wxxxxw−4wxxxwx+3w2xxw2+12wxxwx2w3−6w4w4+a10wxxxw−3wxxwxw2+2wx3w3=a20wxxxxxw−wxxxxwxw2−4wxxxxwx+wxxxwxw2−wxxxwx2w.wxw4−32wxxwxxxw2−wxx22w.wxw4+12wxxxwx2+wxx2wxwxxw3−wxxwx23w2wxw6−64wx3wxxw4−wx44w3wxw8+a10wxxxxw−wxxxwxw2−3wxxxwx+wxxwxxw2−wxxwx2w.wxw4+23wx2wxxw3−wx33w2wxw6=a20wxxxxxw−wxxxxwxw2−4wxxxxwxw2−4wxxxwxxw2+8wxxxwx2w3−6wxxxwxxw2+6wxx2wxw3+12wxxxwx2w3+24wxx2wxw3−36wxxwx3w4−24wxxwx3w4+24wx5w5+a10wxxxxw−wxxxwxw2−3wxxxwxw2−3wxx2w2+6wxxwx2w3+6wxxwx2w3−6wx4w4=a20wxxxxxw−5wxxxxwxw2−10wxxxwxxw2+20wxxxwx2w3+30wxx2wxw3−60wxxwx3w4−24wx5w5+a10wxxxxw−4wxxxwxw2−3wxx2w2+12wxxwx2w3−6wx4w4,(20)uxxx=a20wxxxxxw−5wxxxxwxw2−10wxxxwxxw2+20wxxxwx2w3+30wxx2wxw3−60wxxwx3w424wx5w5+a10wxxxxw−4wxxxwxw2−3wxx2w2+12wxxwx2w3−6wx4w4.Now substituting equations (14), (15), (16), and (20) into (5), we have;(21)a20wxxtw−wxxwt+2wxwxtw2+2wx2wtw3+a10wxtw−wxwtw2+a20wxxw−wx2w2+a10wxw+a00a20wxxxw−3wxxwxw2+2wx3w3+a10wxxw−wx2w2+βa20wxxxxxw−5wxxxxwxw2−10wxxxwxxw2+20wxxxwx2w3+30wxx2wxw3−60wxxwx3w4−24wx5w5+a10wxxxxw−4wxxxwxw2−3wxx2w2+12wxxwx2w3−6wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a20wxxw−a20wx2w2+a10wxw+a00a20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a20wxxwa20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2−a20wx2w2a20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+a10wxwa20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+a00a20wxxxw−a203wxxwxw2+a202wx3w3+a10wxxw−a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a20wxxwa20wxxxw−a20wxxwa203wxxwxw2+a20wxxwa202wx3w3+a20wxxwa10wxxw−a20wxxwa10wx2w2−−a20wx2w2a20wxxxw−a20wx2w2a203wxxwxw2+a20wx2w2a202wx3w3+a20wx2w2a10wxxw−a20wx2w2a10wx2w2−a10wxwa20wxxxw−a10wxwa203wxxwxw2+a10wxwa202wx3w3+a10wxwa10wxxw−a10wxwa10wx2w2+a00a20wxxxw−a00a203wxxwxw2+a00a202wx3w3+a00a10wxxw−a00a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a202wxxwxxxw2−a2023wxx2wxw3+a2022wx3wxxw4+a20a10wxx2w2−a20a10wx2wxxw3−a202wx2wxxxw3+a2023wxxwx3w4−a2022wx5w5−a20a10wxxwx2w3+a20a10wx4w4+a10a20wxwxxxw2−a10a203wxxwx2w3+a10a202wx4w4+a102wxwxxw2−a102wx3w3+a00a20wxxxw−a00a203wxxwxw2+a00a202wx3w3+a00a10wxxw−a00a10wx2w2+βa20wxxxxxw−a205wxxxxwxw2−a2010wxxxwxxw2+a2020wxxxwx2w3+a2030wxx2wxw3−a2060wxxwx3w4−a2024wx5w5+a10wxxxxw−a104wxxxwxw2−a103wxx2w2+a1012wxxwx2w3−a106wx4w4=0a20wxxtw−a20wxxwt+2wxwxtw2+a202wx2wtw3+a10wxtw−a10wxwtw2+a202wxxwxxxw2−a2023wxx2wxw3+a2022wx3wxxw4+a20a10wxx2w2−a20a10wx2wxxw3−a202wx2wxxxw3+a2023wxxwx3w4−a2022wx5w5−a20a10wxxwx2w3+a20a10wx4w4+a10a20wxwxxxw2−a10a203wxxwx2w3+a10a202wx4w4+a102wxwxxw2−a102wx3w3+a00a20wxxxw−a00a203wxxwxw2+a00a202wx3w3+a00a10wxxw−a00a10wx2w2+βa20wxxxxxw−βa205wxxxxwxw2−βa2010wxxxwxxw2+βa2020wxxxwx2w3+βa2030wxx2wxw3−βa2060wxxwx3w4−βa2024wx5w5+βa10wxxxxw−βa104wxxxwxw2−βa103wxx2w2+βa1012wxxwx2w3−βa106wx4w4=0Now equating the coefficients ofwx5/w5 and wx4/w4 to zero on the left-hand side, we have a20=12β and a10=0. This is to mean that; equating a20β24wx5/w5=0, a10β6wx4/w4=0, a10a202wx4/w4=0, −2a202wx5/w5=0, and a10a20wx4/w4=0. Now considering the two equations a20β24wx5/w5=0 and −2a202wx5/w5=0, it is possible to solve for a20. Since β≠0 and w≠024a20β=0 and −2a202=0, we solved for the reset a20 as 24a20β=0−2a202=0Adding these two equations together and solving fora20, as 24a20β−2a202=02a2012β−a20=0. From this, we can get a20=12βAgain considering the equations;−6βa10wx4/w4=0,2a10a20wx4/w4=0,anda10a20wx4/w4=0.And adding them together using a system of equations as−6βa10=0,2a10a20=0,a10a20=0.we have 3a20a10−6βa10=0. 3a20a10=6βa10,Simplifying it we get, a20a10=2βa10. Substituting a20=12β into this equation we get, 12βa10=2a10. Which give us 10βa10=0, β≠0. And hence it implies that a10=0Substitutinga20=12β and a10=0 into the following equation: u=a20lnωxx+a10lnωx+a00, we get u=12lnωxx+0lnωx+a00=12lnωxx+a00=12βwxx/w−wx2/w2+a00, where a00 is any arbitrary constant.(22)u=12lnωxx+a00=12βwxxw−wx2w2+a00.Putting (22) into (5), we have the following results, where, a20=12β and a10=0.(23)12βwxxtw−12βwxxwtw2−12β2wxwxtw2+12β2wx2wtw3+0wxtw−0wxwtw2+12β2wxxwxxxw2−12β23wxx2wxw3+12β22wx3wxxw4+012βwxx2w2−012βwx2wxxw3−12β2wx2wxxxw3+12β23wxxwx3w4−12β22wx5w5−012βwxxwx2w3+012βwx4w4+012βwxwxxxw2−012β3wxxwx2w3+012β2wx4w4+0wxwxxw2−0wx3w3+a0012βwxxxw−a0012β3wxxwxw2+a0012β2wx3w3+0a00wxxw−0a00wx2w2+β12βwxxxxxw−12β25wxxxxwxw2−12β210wxxxwxxw2+12β220wxxxwx2w3+12β230wxx2wxw3−12β260wxxwx3w4−12β224wx5w5+0βwxxxxw−04βwxxxwxw2−03βwxx2w2+012βwxxwx2w3−06βwx4w4=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+144β2wxxwxxxw2−144β23wxx2wxw3+144β22wx3wxxw4+144β2wx2wxxxw3+144β23wxxwx3w4−144β22wx5w5−a0012βwxxxw−a0036βwxxwxw2+a0024βwx3w3+β12βwxxxxxw−60β2wxxxxwxw2−120β2wxxxwxxw2+240β2wxxxwx2w3+360β2wxx2wxw3−720β2wxxwx3w4−288β2wx5w5=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12β12βwxxwxxxw2−12β36βwxx2wxw3+12β24βwx3wxxw4+12β12βwx2wxxxw3+12β36βwxxwx3w4−12β24βwx5w5+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12ββwxxxxxw−5βwxxxxwxw2−10βwxxxwxxw2+20βwxxxwx2w3+30βwxx2wxw3−60βwxxwx3w4−24βwx5w5=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12β12βwxxwxxxw2−36βwxx2wxw3+24βwx3wxxw4+12βwx2wxxxw3+36βwxxwx3w4−24βwx5w5+βwxxxxxw−5βwxxxxwxw2−10βwxxxwxxw2+20βwxxxwx2w3+30βwxx2wxw3−60βwxxwx3w4−24βwx5w5=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12β12βwxxwxxxw2−10βwxxxwxxw2−36βwxx2wxw3+30βwxx2wxw3+24βwx3wxxw4+36βwxxwx3w4+−60βwxxwx3w4−12βwx2wxxxw3+20βwxxxwx2w3−24βwx5w5+24βwx5w5+βwxxxxxw−5βwxxxxwxw2=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12β2βwxxwxxxw2−6βwxx2wxw3+60βwx3wxxw4−60βwxxwx3w4+8βwx2wxxxw3+βwxxxxxw−5βwxxxxwxw2=012βwxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3+12βa00wxxxw−a003wxxwxw2+a002wx3w3+12ββwxxxxxw+2wxxxwxx−5wxxxxwxw2+8wxxxwx2−6wxx2wxw3=012βK1+K2+K3=0,where(24)K1=wxxtw−wxxwtw2−2wxwxtw2+2wx2wtw3,K2=a00wxxxw−a003wxxwxw2+a002wx3w3,K3=βwxxxxxw+2wxxxwxx−5wxxxxwxw2+8wxxxwx2−6wxx2wxw3,(25)12βK1+K2+K3=0.Simplifying (25) and integrating once with respect to x, we get(26)∂∂xwxtw−wxwt+βwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2=0.Equations (20) and (26) are identical with(27)wxtw−wxwt+βwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2−Ctw2=0,where Ct is an arbitrary function of t, and a00 is an arbitrary constant. Especially, taking Ct as zero in equation (27), we get the bilinear equation of (5) as follows:(28)wxtw−wxwt+βwxxxxw−4wxwxxx+3wxx2+a00wxxw−wx2w2=0.Equation (28) can be written concisely in terms of D-operator as(29)DxDt+βDX4+a00Dx2w,w=0,where DxmDtna.b=∂x−∂x′m∂t−∂t′nax,tbx′,t′x=x′,t=t′.Applying Hirota’s method, the bilinear equation of (5) can be written as(30)DxDt+βDX4w,w=0.Equation (30) is obtained by setting a00=0 in equation (29). Obviously, equation (30) is a special case of equation (29). Therefore, a more general bilinear equation of the modified KdV equation is obtained by using Hirota’s method.Let us consider the modified KdV equationut+6uux+uxxx=0 by using Hirota’s method.Substituting(31)u=wx,which gives the following new equation: wxt+6wxwxx+wxxxx=0. Now integrating both sides with respect to x, we have;(32)∫wxt+6wxwxx+wxxxxdx=0,∫wxtdx+∫6wxwxxdx+∫wxxxxdx=gt,wt+6wx22+wxxx=C=gt.where C=gt is constant of integration. Likely we can get, without loss of generality, the integration “constant” with respect to x, gt can be observed by a redefinition of w that does not change the KdV field, u=wxwold=wnew+∫t0tdt′gt′, dt′ is the time derivative.Using the neww, we have(33)wt+3wx2+wxxx=0.Now considering the 1-soliton solution of the modified KdV equation, let(34)u=2μ2sech2μx−x0−4μ2t.with(35)μ=v2≥0,where v is the velocity of the wave propagating.It can be written asu=wx, with(36)w=2μ2tanhμx−x0−4μ2t.In fact, we can integrate the right-hand side of equation (36) once again, using tanh y=d/dy log cosh y. Therefore, equation (34) can be written as follows:(37)u=2∂2∂x2logcoshμx−x0−4μ2t.Let(38)X=x−x0−4μ2t.Then,u=2∂2/∂X2logexp−μX1+exp2μX/2, where coshy=exp−μX1+exp2μX/2(39)=2∂2∂x2−μx−log2+log1+exp2μX=−2∂2∂x2μx−2∂2∂x2log2+2∂2∂x2log1+exp2μX=2∂2∂X2log1+exp2μX.That is,ux,t=2∂2/∂x2log1+exp2μx−x0−4μ2t=2∂2/∂x2log1+exp2μX(40)ux,t=2∂2∂x2log1+exp2μx−x0−4μ2t=2∂2∂x2log1+exp2μX,which is the 1-soliton solution of the modified KdV equation, that we use in the following:Considering equations (31) and (33), the KdV equation in a bilinear form, and rewriting it in a quadratic form, we have the following equation. Inspired by equation (34), let us substitute(41)w=2∂∂xlogf=2fxf⇔u=2∂2∂x2logf,where f=1+exp2μx−x0−4μ2t. Substituting in equation (33), we get(42)12wt=fxt−fxftf2,12wx=fxx−fx2f2,12wxx=fxxxf−3fxfxxf2+2fx3f3,12wxxx=fxxxxf−4fxfxxxf2−3fxx2f2+12fxxfx2f3−6fx4f4.(43)12wxxx=fxxxxf−4fxfxxxf2−3fxx2f2+12fxxfx2f3−6fx4f4,Equation (43) is quadratic in f.Thus, equation (33) for w becomes(44)wt+3wx2+wxxx=0,=2fxt−fxftf2+322fxxf−fx2f22+2fxxxxf−42fxfxxxf2−32fxx2f2+122fxxfx2f3−62fx4f4=2fxt−fxftf2+12fxx2f2−2fxxfx2f3+fx4f4+2fxxxxf−8fxfxxxf2−6fxx2f2+24fxxfx2f3−12fx4f4=2fxt−fxftf2+12fxx2f2−24fxxfx2f3+12fx4f4+2fxxxxf−8fxfxxxf2−6fxx2f2+24fxxfx2f3−12fx4f412=2fxt−fxftf2+2fxxxxf−8fxfxxxf2+6fxx2f2=fxt−fxftf2+fxxxxf−4fxfxxxf2+3fxx2f2.Multiplying byf2 throughout, we have(45)ffxt−fxft+ffxxxx−4fxfxxx+3fxx2=0This is the bilinear form of the modified KdV equation. Despite the fact that some nontrivial cancellations took place it looks more complicated than the original problem, however, its special form makes it possible to solve using bilinear operators.Claim 1.
∂/∂x∂/∂t1/2f2=∂/∂xfft=ffxt+fxft
To express equation (45), we use the following expressions:(46)Dtf.gx,t=∂∂t−∂∂t′∂∂x−∂∂x′fx,tgx′,t′x′=x,t′=t=ftx,tgx′,t−fx,tgt′x′,t′x′=x,t′=t=ftx,tgx,t−fx,tgtx,tDtf.gx,t=ftx,tgx,t−fx,tgtx,tDtf.fx,t=0,Dxf.fx,t=0forf=gDtDxf.g=∂∂t−∂∂t′∂∂x−∂∂x′fx,tgx′,t′x′=x,t′=t=∂∂t−∂∂t′fxx,tgx′,t′−fx,tgx′x′,t′x′=x,t′=t=fxtg−ftgx−fxgt−fgxtfxtg−ftgx−fxgt+fgxtDtDxf.f=fxtf−ftfx−fxft+ffxt=2ffxt−ftfxDtDxf.f=2ffxt−ftfx
Like the first two terms in (45).(47)Dx2f.g=∂∂x−∂∂x′2fxx,tgx′,t′x′=x,t′=t=∂∂x−∂∂x′fxx,tgx′,t′−fx,tgx′x′,t′x′=x,t′=t=fxxg−fxgx−fxgx+fgxx=fxxg−2fxgx+fgxx,Dx2f.g=fxxg−2fxgx+fgxx,Dx2f.f=fxxf−2fxfx+ffxx=2ffxx−fx2 all functions of x,t.
Again this looks a bit like the differentiation of a product. But, we haveDx2f,f=2ffxx−2fx2 even though Dxf.fx,t=0. So importantly the operators are not associative since Dx2f.f≠DxDxf.f.
In fact, the right-hand side is meaningless, becauseDxf.f is a single function, but the outer Dx needs to act on a pair of functions.(48)Dx4f.g=∂∂x−∂∂x′4fxx,tgx′,t′x′=x,t′=t=∂∂x−∂∂x′3fxx,tgx′,t′−fx,tgx′x′,t′x′=x,t′=t=∂∂x−∂∂x′2fxxg−2fxgx′+fgx′x′x′=x,t′=t=∂∂x−∂∂x′fxxxg−3fxxgx′+3fxgx′x′−fgx′x′x′x′=x,t′=t=fxxxxg−4fxxxgx′+6fxxgx′x′−4fxgx′x′x′+fgx′x′x′x′x′=x,t′=t=fxxxxg−4fxxxgx+6fxxgxx−4fxgxxx+fgxxxxDx4f.g=fxxxxg−4fxxxgx+6fxxgxx−4fxgxxx+fgxxxx.
Again it looks just like a differentiated product off and g but with significant changes on the odd terms. Dx4f.f=fxxxxf−4fxxxfx+6fxxfxx−4fxfxxx+ffxxxx=2fxxxxf−8fxxxfx+6fxx2.(49)Dx4f.f=2fxxxxf−8fxxxfx+6fxx2
Here, we have to notice that (49) is like ∂x4fg, but we have an alternative signs;(50)Dx4f.f=2fxxxxf−4fxxxfx+3fxx2.
So, the KdV equation is in a quadratic form in equation (45) and can be recast as DtDx+Dx4f.f=0 is a bilinear form of the KdV equation.(51)DtDx+Dx4f.f=0,DtDx+Dx4=DxDt+Dx3.
(1) Solutions of the Bilinear KdV Equation. For example (1) and considering u=2∂x2logf, we have two cases.Case 1.
Forf=1, then u=0, which is a vacuum.
Forf=1+exp2μx−x0−4μ2t, then u=1−soliton solution. Since equation (51) is a bilinear equation, then this suggests that multisoliton solutions might be the sums of exponents of linear functions of x and t.
(2). The 1-Soliton Solution. In Hirota form the multisoliton solutions are sums of exponential of linear expressions in x and t.
For the one-soliton solution try(52)f=e0+eθ.where =ax+bt+c, ⇒x=ax,c=−x0, and −4μ2=tTheorem 1.
For anyθ1 and θ2, we have DtmDxneθ1.eθ2 = b1−b2ma1−a2neθ1+θ2, where m,n≥0.Proof.
By induction.
Let ifθi=aix+bit+ci,i=1,2. Then,(53)DtmDxneθ1.eθ2=b1−b2ma1−a2neθ1+θ2=b1−b2ma1−a2neθ1x,t+θ2x′,t′x′=x,t′=t
Then,(54)DtmDxn+1eθ1.eθ2=b1−b2ma1−a2n∂∂x−∂∂x′eθ1x,t+θ2x′,t′x′=x,t′=t=b1−b2ma1−a2na1−a2eθ1x,t+θ2x′,t′x′=x,t′=t=b1−b2ma1−a2n+1eθ1+θ2DtmDxn+1eθ1.eθ2=b1−b2ma1−a2n+1eθ1+θ2.and similarly for Dt operator.
Finally, the equation is obviously true forn=1,m=0 and n=0,m=1 so by induction must be true for all n and m. In particular, we find(55)DtmDxneθ.eθ=0.
Unlessm=n=0(56)DtmDxneθ.1=bmaneθDtmDxn1.eθ=−1m+nbman1,eθ.
Then, the bilinear form of the KdV equation forf=1+eθ is(57)DtDx+Dx41+eθ,1+eθ=0=DtDx+Dx41.1+1.eθ+eθ.1+eθ.eθ=DtDx+Dx4eθ.1+1.eθ=2baeθ+2a4eθ2aba+a3eθ=0.
Sinceeθ≠0, we have two opportunities. These two cases or opportunities are as follows:(1)
a=0. If a=0, then f is independent of x. So u=0, (which is a boundary condition).(2)
b=−a3. If b=−a3, then f=1+eax−a3t+c and hence u=2∂x2logf, f=1+eax−a3t+c, which is u=2∂x2log1+eax−a3t+c(58)=a22sech212ax−a3t+c,u=a22sech212ax−a3t+c.
This is the one-soliton solution withv=a2.
Figure1 shows the disturbance of the water surface having kink-shaped traveling waves with amplitude and antibell soliton. The follow of this wave is to the front.
Figure2 shows the propagation of waves having kink shaped and like some periodic soliton. And follow the wave from back to front.
Figure3 shows surface waves of long wave length having kink-shaped and bright periodic soliton. And the wave is from left to right.
Figure4 surface waves of long wave length having kink shaped and bright periodic soliton. And the wave is from left to right. This is the other form of Figure 3, but it is different by the value of x, that is why it is broken from the above.
(3) Multisoliton Solutions.Figure 1
The result of one-soliton solution fora=1/2, c = 0.1.Figure 2
The result of one-soliton solution fora=0.6, c = 0.2.Figure 3
The result of one-soliton solution fora=0.7,c=0.3.Figure 4
The result of one-soliton solution fora=0.7, c=0.3.Case 2.
For the power series in an auxiliary with parameterε(59)fx,t=∑n=0∞εnfnx,t,
Withf0=1 in which the series will terminate at some power of ε to give exact results rather than an infinite series in ε. So, we can take ε finite ε = 1.
To begin with, we putfx,t=1+∑n=0∞εnfnx,t and collect powers of ε. We start with 1 as the ε0 term as it appeared in the one-soliton solution. Defining(60)B=DtDx+Dx4.
We want to find solutions such that(61)Bf.f=0.
Substituting (59) into (60), we have 0=B∑n1=0∞εn1fn1,∑n2=0∞εn2fn2, with f0=1=∑n1=0∞∑n2=0∞εn1+n2Bfn1,fn2 and gathering terms of the same degree n=n1+n2inε and also expanding as for the one soliton case we have(62)Bf.f=B1.1+εB1.f1+Bf1.1+ε2B1.f2+Bf1.f1+Bf2.1=0,=∑n=0∞εn∑m=0nBfn−m.fm.so we have to solve an infinite set of equations order by order in ε(63)∑m=0nBfn−m.fm=0,for all n = 1, 2, ….
Equation (63) can be written as(64)Bfn,1+B1,fn=−∑m=1n−1Bfn−m.fm.
(Expression involving inf1,f2,….,fn−1).
Now we can solve (63) which is equivalent to (64) recursively to determine the coefficients fn. That is we begin with f1 and solve iteratively to get f2,3,.. hoping that it terminates at some point. To do this we need the following:Theorem 2.
For any givenf, we have; DtmDxnf.1=−1n+mDtmDxn1.f, where m,n≥0Proof.
By considering to interchange primed and unprimed values as a simple label; i.e.,(65)DtmDxnf.g=∂∂t−∂∂t′m∂∂x−∂∂x′nfx,tgx′,t′x′=x,t′=t=−1m+n∂∂t′−∂∂tm∂∂x′−∂∂xngx′,t′fx,tx′=x,t′=t=−1m+n∂∂t−∂∂t′m∂∂x−∂∂x′ngx,tfx′,t′x′=x,t′=t=−1m+nDtmDxng.fDtmDxnf.1=−1n+mDtmDxn1.f=∂m∂tm∂n∂xnfDtmDxnf.1=∂m∂tm∂n∂xnf,where f is any function of x and t
Thus, (70) became(66)∂∂x∂∂t+∂3∂x3fn=−12∑m=1n−1Bfn−m.fm.
Forn=1, it reduced to(67)∂∂x∂∂t+∂3∂x3f1=0⇒∂∂t+∂3∂x3f1=0.
Beginning withf1 we may iterate(repeat) to find all the fn and note that at any order we are now solving linear partial differential equations. A simple particular solution to the last equation would be(68)f1=∑i=1Nεaix−a3it+ci=∑i=1Nεθi,f1=εaix−a3it+ci.
Using (63) for f2, we have ∂/∂x∂/∂t+∂3/∂x3f2=−1/2Bf1.f1=0, where this result follows (55). Here is the fact, with f1, the expansion of (59), terminated at order N or (the f3…∞ all vanish trivially). The series has terminated with(69)f=1+εf1.
All higher-order equations withn>N are solved by fn=0 for n>N. We can always absorb ε into the constant c and we recognize the single soliton solution. Our hope is to find other solutions where the series terminates in this way, so that the solution is exact. Consider beginning with(70)f1=∑i=1Nεaix−a3it+ci.f=1+ε1f1+ε2f2+…εNfN. We have already seen the single soliton solution. Let’s verify that we recover the two soliton solutions. Take for N=2,(71)f1=eθ1+eθ2.where θ1,2 are as above chosen to satisfy the f1 equation, and θi=aix−ai3t+ci. The f2 equation became(72)∂∂x∂∂t+∂3∂x3fn=−12Bf1.f1,=−12Beθ1+eθ2.eθ1+eθ2=−12Beθ1.eθ2−12Beθ2.eθ1=−a1−a2b1−b2+a1−a23eθ1+θ2=−a1−a2−a13−a23+a1−a23eθ1+θ2=a1−a22a12+a1a2+a22−a12+2a1a2−a22eθ1+θ2=3a1a2a1−a22eθ1+θ2∂∂x∂∂t+∂3∂x3f2=3a1a2a1−a22eθ1+θ2.
Since the equation is linear inf2 it has an obvious solution of the form f2=Aeθ1+θ2. So, we just have to determine A. Substituting into the f2 equation; that is, let(73)∂x=a1+a2,∂t=−a13+a23,∂x3=a1+a23,∂∂x∂∂t+∂3∂x3f2=3a1a2a1−a22eθ1+θ2a1+a2−a13+a23+a1+a23Aeθ1+θ2=3a1a2a1−a22eθ1+θ2a1+a2−a13+a23+a1+a23A=3a1a2a1−a223a1a2a1+a22A=3a1a2a1−a22A=a1−a22a1+a22.
The solution is then,f=1+εeθ1+εeθ2+ε2a1−a22/a1+a22eθ1+θ2, which is a two-soliton solution of the KdV equation. Setting ε=1, we have(74)f=1+eθ1+eθ2+a1−a22a1+a22eθ1+θ2.
Figure5 shows follow of the presence of a solitary wave to the front, having two kinks shaped with long amplitude, dark of high speed.
Figure6 shows the solitary wave to the front, having two kinks shaped, a valley of the wave, and dark high speed.
Figure7 shows the solitary wave from left to right, having two kink-shaped solitons, a valley-shaped wave, long amplitudes, and dark high speed.
Figure8 shows follow of the solitary wave to the front, having two kink--shaped soliton profiles, periodic, and dark and bell slow speed.
Finally, we just need to differentiate to get the originalw or u using w=2fx/f=a1eθ1+a2eθ2+a1−a22/a1+a22eθ1+θ2/1+eθ1+eθ2+a1−a22/a1+a22eθ1+θ2. It is easier in this format to study the asymptotic of the two soliton solutions. As in the BT discussion, first, we choose a coordinate system with the origin at the center of the around i=1,2 soliton; taking t⟶ti+δt with δx=x−x0−viti, where vi=ai2 and let ti⟶±∞, we have θi=vi/2δx−viδt(75)θj≠i=vj2δx−vjδt+vj2vi−vjti.
Near thei soliton, we have (by definition) δx,δt≈0, so that θi is small but θj≠i⟶±∞ as follows: (t⟶−∞, if i=1 , θ1 is fixed and θj=2⟶−∞, and if i=2, θj=1⟶−∞), (t⟶+∞, if i=1, θj=1⟶−∞, and if i=2, θj=2⟶+∞). When θj≠i⟶−∞ the f clearly reduces to the single soliton case, f=1+eθi. When θj≠i⟶+∞, we have to do a little more work. To lead order we have(76)f=eθj1+a1−a22a1+a22eθi+…..where the ellipsis indicates terms of order e−θ2. But then w=2∂xlogf(77)w=2aj+2∂xlog1+a1−a22a1+a22eθi+…..
Sincew⟶w+const gives equivalent solutions we might ignore the constant piece. The remaining solution is equivalent to an f of the for(78)f=1+a1−a22a1+a22eθi.
This is the original singlei-soliton solution but with a phase shift of(79)δx0−2vilogv1−v2v1−v2.
Precisely what we found for the 2 KdV soliton solution using the much more complicated BT method. Finally, it was just presented the general solution which might be proved by induction. Define anNxN matrix SN with elements. Considering the 2-soliton solution for equation (68), f=1+εeθ1+εeθ2+ε2a1−a22/a1+a22eθ1+θ2(80)=1+εeθ11+εeθ2−ε2eθ1+θ2+ε2a1−a2a1+a22eθ1+θ2=1+εeθ11+εeθ2−ε24a1a2a1+a22eθ1+θ2=1+εeθ1ε2a1a1+a2eθ2ε2a2a1+a2eθ11+εeθ2.f=detS,whereSNij=δij+ε2aiai+ajeθj.and i,j=1,2. Then, the N-soliton solution U=2∂x2logf, where f=detSN.Figure 5
The result of the two-soliton solution fora1=1,a2=1.5,c=0.5.Figure 6
The result of the two-soliton solution fora1=−1,a2=−0.5,c=−1.5.Figure 7
The result of the two-soliton solution fora1=−3,a2=−2.5,c=−3.5.Figure 8
The result of the two-soliton solution fora1=−5,a2=−3.5,c=−4.5.


## 2.3. Illustrative Examples

Example 1.
Consider the fractional-order nonlinear KdV system is given as follows:(81)∂ρμ∂τρ=−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,∂ρv∂τρ=−a∂3v∂ζ3−3aμ∂v∂ζ′.0<ρ<1.with the initial condition,(82)μζ,0=η2sech2α2+μζ2,vζ,0=α2η2sech2α2+μζ2.
Forρ=1, the exact results of the KdV scheme (82) are given by(83)μζ,τ=η2sech2α2+ηζ2−aη3τ2,vζ,τ=α2η2sech2α2+ηζ2−aη3τ2.
Using the ELzaki transform to (83), we obtain,(84)1sρEμζ,τ−∑k=0m−1μkζ,0s2−ρ+k=E−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,1sρEvζ,τ−∑k=0m−1vkζ,0s2−ρ+k=E−a∂3v∂ζ3−3aμ∂v∂ζ,(85)1sρEμζ,τ=μkζ,0s2−ρ+E−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,1sρEvζ,τ=vkζ,0s2−ρ+E−a∂3v∂ζ3−3aμ∂v∂ζ,Eμζ,τ=s2μζ,0+Esρ−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,Evζ,τ=s2vζ,0+Esρ−a∂3v∂ζ3−3aμ∂v∂ζ.
Applying the inverse ELzaki transform of (85), we have(86)μζ,τ=E−1s2μζ,0+E−1sρE−a∂3μ∂ζ3−6aμ∂μ∂ζ+6v∂v∂ζ,vζ,τ=E−1s2vζ,0+E−1sρE−a∂3v∂ζ3−3aμ∂v∂ζ.
Now, by using the suggested analytical method, we get(87)μ0ζ,τ=η2sech2α2+μζ2,v0ζ,τ=α2η2sech2α2+μζ2,μ1ζ,τ=E−1sρE−a∂3μ0∂ζ3−6aμ0∂μ0∂ζ+6v0∂v0∂ζ,v1ζ,τ=E−1sρE−a∂3v0∂ζ3−3aμ0∂v0∂ζ,μ1ζ,τ=η5atanhα2+ηζ2sech2α2+ηζ2τρΓρ+1,v1ζ,τ=η5a3/22tanhα2+ηζ2sech2α2+ηζ2τρΓρ+1,μ2ζ,τ=E−1sρE−a∂3μ1∂ζ3−6aμ1∂μ1∂ζ+6v1∂v1∂ζ,v2ζ,τ=E−1sρE−a∂3v1∂ζ3−3aμ1∂v1∂ζ,μ2ζ,τ=η8a222cosh2α2+ηζ2−3sech4α2+ηζ2τ2ρΓ2ρ+1,v2ζ,τ=η5a5/2242cosh2α2+ηζ2−3sech4α2+ηζ2τ2ρΓ2ρ+1,μ3ζ,τ=E−1sρE−a∂3μ2∂ζ3−6aμ2∂μ2∂ζ+6v2∂v2∂ζ,v3ζ,τ=E−1sρE−a∂3v2∂ζ3−3aμ2∂v2∂ζ,μ3ζ,τ=sinhα/2+ηζ/2τ3ρa3η42Γ3ρ+1Γρ+12cosh7α/2+ηζ/22Γρ+12cosh4α2+ηζ2−18Γρ+12cosh2α2+ηζ2+6Γ2ρ+1cosh2α2+ηζ2−18Γρ+12−9Γ2ρ+1,v3ζ,τ=2sinhα/2+ηζ/2τ3ρa7/2η114Γ3ρ+1Γρ+12cosh7α/2+ηζ/22Γρ+12cosh4α2+ηζ2−18Γρ+12cosh2α2+ηζ2+6Γ2ρ+1cosh2α2+ηζ2−18Γρ+12−9Γ2ρ+1.
Generally, thenth solution result was obtained as follows:(88)μnζ,τ=E−1sρE−a∂3μn−1∂ζ3−6aμn−1∂μn−1∂ζ+6vn−1∂vn−1∂ζ,vnζ,τ=E−1sρE−a∂3vn−1∂ζ3−3aμn−1∂vn−1∂ζ,n≥0.
The series form result is as follows:(89)μζ,τ=μ0ζ,τ+μ1ζ,τ+μ2ζ,τ+μ3ζ,τ+…..μnζ,τ,vζ,τ=v0ζ,τ+v1ζ,τ+v2ζ,τ+v3ζ,τ+…..vnζ,τ,μζ,τ=η2sech2α2+μζ2+E−1sρE−a∂3μ0∂ζ3−6aμ0∂μ0∂ζ+6v0∂v0∂ζ+E−1sρE−a∂3μ1∂ζ3−6aμ1∂μ1∂ζ+6v1∂v1∂ζ+sinhα/2+ηζ/2τ3ρa3η42Γ3ρ+1Γρ+12cosh7α/2+ηζ/22Γρ+12cosh4α2+ηζ2−18Γρ+12cosh2α2+ηζ2+6Γ2ρ+1cosh2α/2+ηζ/2−18Γρ+12−9Γ2ρ+1+….vζ,τ=α2η2sech2α2+μζ2+E−1sρE−a∂3v0∂ζ3−3aμ0∂v0∂ζ+E−1sρE−a∂3v1∂ζ3−3aμ1∂v1∂ζ+2sinhα/2+ηζ/2τ3ρa7/2η114Γ3ρ+1Γρ+12cosh7α/2+ηζ/22Γρ+12cosh4α2+ηζ2−18Γρ+12cosh2α2+ηζ2+6Γ2ρ+1cosh2α2+ηζ2−18Γρ+12−9Γ2ρ+1+….
Forρ=1, the exact results of the KdV scheme (81) are given by(90)μζ,τ=η2sech2α2+ηζ2−aη3τ2,vζ,τ=α2η2sech2α2+ηζ2−aη3τ2.Example 2.
Consider the fractional-order nonlinear dispersive long wave scheme.(91)∂ρμ∂τρ=−∂v∂ζ−12∂μ2∂ζ,∂ρμ∂τρ=−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ,0<ρ<1,and the initial condition;(92)μζ,0=atanhη2+αζ2+1,vζ,0=−1+12sech2η2+αζ2.
While using ELzaki transform of (91), we obtained(93)1sρEμζ,τ−∑k=0m−1μkζ,0s2−ρ+k=E−∂v∂ζ−12∂μ2∂ζ,1sρEvζ,τ−∑k=0m−1vkζ,0s2−ρ+k=E−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ,1sρEμζ,τ=∑k=0m−1μkζ,0s2−ρ+k+E−∂v∂ζ−12∂μ2∂ζ,1sρEvζ,τ=∑k=0m−1vkζ,0s2−ρ+k+E−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ,1sρEμζ,τ=μ0ζ,0s2−ρ+E−∂v∂ζ−12∂μ2∂ζ,1sρEvζ,τ=v0ζ,0s2−ρ+E−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ,Eμζ,τ=s2μζ,0+sρE−∂v∂ζ−12∂μ2∂ζ,Evζ,τ=s2vζ,0+sρE−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ.
Applying the inverse ELzaki transform of equation (93), we get(94)μζ,τ=E−1s2μζ,0+E−1sρE−∂v∂ζ−12∂μ2∂ζ,vζ,τ=E−1s2vζ,0+E−1sρE−∂3μ∂ζ3−∂μ∂ζ−∂μv∂ζ.
Now, by using the current analytical method, we get(95)μ0ζ,0=atanhη2+αζ2+1,v0ζ,0=−1+12a2sech2η2+αζ2μ1ζ,τ=E−1sρE−∂v0∂ζ−12∂μ02∂ζ,v1ζ,τ=E−1sρE−∂3μ0∂ζ3−∂μ0∂ζ−∂μ0v0∂ζμ1ζ,τ=−a22sech2η2+αζ2τρΓρ+1v1ζ,τ=a32sinhη2+αζ2sech3η2+αζ2τρΓρ+1μ2ζ,τ=E−1sρE−∂v1∂ζ−12∂μ12∂ζ,v2ζ,τ=E−1sρE−∂3μ1∂ζ3−∂μ1∂ζ−∂μ1v1∂ζμ2ζ,τ=−a54sech2η2+αζ2τ2ρΓ2ρ+1+3a54sinh2η2+αζ2+a74sinhη2+αζ2sech4η2+αζ2τ2ρΓ2ρ+1+sech5η2+αζ2Γ2ρ+1τ3ρΓ3ρ+1Γρ+12v2ζ,τ=a642cosh2η2+αζ2−3sech4η2+αζ2τ2ρΓ2ρ+1μnζ,τ=E−1sρE−∂vn−1∂ζ−12∂μn−12∂ζ,vnζ,τ=E−1sρE−∂3μn−1∂ζ3−∂μn−1∂ζ−∂μn−1vn−1∂ζn≥0
Thus, the series form of the result is as follows:(96)μζ,τ=μ0ζ,τ+μ1ζ,τ+μ2ζ,τ+⋯+μnζ,τ,vζ,τ=v0ζ,τ+v1ζ,τ+v2ζ,τ+⋯+vnζ,τ,μζ,τ=atanhη2+αζ2+1−a22sech2η2+αζ2τρσρ+1−a54sech2η2+αζ2τ2ρΓ2ρ+1,+3a54sinh2η2+αζ2sech4η2+αζ2τ2ρΓ2ρ+1+a74sinhη2+αζ2sech5η2+αζ2Γ2ρ+1τ3ρσ3ρ+1Γρ+12+⋯,vζ,τ=−1+12a2sech2η2+αζ2+a32sinhη2+αζ2sech3η2+αζ2τρΓρ+1+a642cosh2η2+αζ2−3sech4η2+αζ2τ2ρΓ2ρ+1+⋯.
Forρ=1, the exact results of the KdV scheme (91)is as follows:(97)μζ,τ=atanhη2+αζ2−a2τ2+1,vζ,τ=−1+12sech2η2+αζ2−a2τ2.Example 3.
Considering the nonlinear fractional-order new coupled modified KdV system,we get(98)∂ρμ∂τρ=12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,∂ρv∂τρ=−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,∂ρw∂τρ=−∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ,and the initial conditions are(99)μζ,0=1+12tanhζ,vζ,0=12−14tanhζ,wζ,0=2−tanhζ.
Using the ELzaki transform to (98), we get(100)1sρEμζ,τ−∑k=0m−1μkζ,0s2−ρ+k=E12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,1sρEvζ,τ−∑k=0m−1vkζ,0s2−ρ+k=E−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,1sρEwζ,τ−∑k=0m−1wkζ,0s2−ρ+k=E∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ,1sρEμζ,τ=μ0ζ,0s2−ρ+E12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,1sρEvζ,τ=v0ζ,0s2−ρ+E−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,1sρEwζ,τ=w0ζ,0s2−ρ+E∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ,(101)Eμζ,τ=s2μζ,0+sρE12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,Evζ,τ=s2vζ,0+sρE−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,Ewζ,τ=s2wζ,0+sρE∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ.
Applying the inverse ELzaki transform of (101)(102)μζ,τ=E−1s2μζ,0+E−1sρE12∂3μ∂τ3−3μ2∂μ∂ζ+32w∂2v∂ζ2+3∂v∂ζ∂w∂ζ+32v∂2w∂ζ2+3vw∂μ∂ζ+3μw∂v∂ζ+3μv∂w∂ζ,vζ,τ=E−1s2vζ,0+E−1sρE−∂3v∂τ3−3∂v∂ζ∂μ∂ζ−3v∂2μ∂ζ2−3v2∂w∂ζ+6μv∂μ∂ζ+3μ2∂v∂ζ,wζ,τ=E−1s2wζ,0+E−1sρE∂3w∂τ3−3∂w∂ζ∂μ∂ζ−3w∂2μ∂ζ2−3w2∂v∂ζ+6μw∂μ∂ζ+3μ2∂w∂ζ.
Now, by using the suggested analytical method, we get(103)μζ,0=1+12tanhζ,vζ,0=12−14tanhζ,wζ,0=2−tanhζ,μ1ζ,τ=E−1sρE12∂3μ0∂τ3−3μ02∂μ0∂ζ+32w0∂2v0∂ζ2+3∂v0∂ζ∂w0∂ζ+32v0∂2w0∂ζ2+3v0w0∂μ0∂ζ+3μ0w0∂v0∂ζ+3μ0v0∂w0∂ζ,v1ζ,τ=E−1sρE−∂3v0∂τ3−3∂v0∂ζ∂μ0∂ζ−3v0∂2μ0∂ζ2−3v02∂w0∂ζ+6μ0v0∂μ0∂ζ+3μ02∂v0∂ζ,w1ζ,τ=E−1sρE∂3w0∂τ3−3∂w0∂ζ∂μ0∂ζ−3w0∂2μ0∂ζ2−3w02∂v0∂ζ+6μ0w0∂μ0∂ζ+3μ02∂w0∂ζ,μ1ζ,τ=112sech2ζηρΓρ+1,v1ζ,τ=−118sech2ζηρΓρ+1,w1ζ,τ=−112sech2ζηρΓρ+1,μ2ζ,τ=E−1sρE12∂3μ1∂τ3−3μ12∂μ1∂ζ+32w1∂2v1∂ζ2+3∂v1∂ζ∂w1∂ζ+32v1∂2w1∂ζ2+3v1w1∂μ1∂ζ+3μ1w1∂v1∂ζ+3μ1v1∂w1∂ζ,v2ζ,τ=E−1sρE−∂3v1∂τ3−3∂v1∂ζ∂μ1∂ζ−3v1∂2μ1∂ζ2−3v12∂w1∂ζ+6μ1v1∂μ1∂ζ+3μ12∂v1∂ζ,w2ζ,τ=E−1sρE∂3w1∂τ3−3∂w1∂ζ∂μ1∂ζ−3w1∂2μ1∂ζ2−3w12∂v1∂ζ+6μ1w1∂μ1∂ζ+3μ12∂w1∂ζ,μ2ζ,τ=−1218tanhζsech2ζη2ρΓ2ρ+1,v2ζ,τ=1218tanhζsech2ζη2ρΓ2ρ+1,w2ζ,τ=1214tanhζsech2ζη2ρΓ2ρ+1,μ3ζ,τ=133148cosh2ζ−2sech4ζη3ρΓ3ρ+1,v3ζ,τ=−133196cosh2ζ−2sech4ζη3ρΓ3ρ+1,w3ζ,τ=−133124cosh2ζ−2sech4ζη3ρΓ3ρ+1,μnζ,τ=E−1sρE12∂3μn−1∂τ3−3μn−12∂μn−1∂ζ+32wn−1∂2vn−1∂ζ2+3∂vn−1∂ζ∂wn−1∂ζ+32vn−1∂2wn−1∂ζ2+3vn−1wn−1∂μn−1∂ζ+3μn−1wn−1∂vn−1∂ζ+3μn−1vn−1∂wn−1∂ζ,vnζ,τ=E−1sρE−∂3vn−1∂τ3−3∂vn−1∂ζ∂μn−1∂ζ−3vn−1∂2μn−1∂ζ2−3vn−12∂wn−1∂ζ+6μn−1vn−1∂μn−1∂ζ+3μn−12∂vn−1∂ζ,wnζ,τ=E−1sρE∂3wn−1∂τ3−3∂wn−1∂ζ∂μn−1∂ζ−3wn−1∂2μn−1∂ζ2−3wn−12∂vn−1∂ζ+6μn−1wn−1∂μn−1∂ζ+3μn−12∂wn−1∂ζ.
The series form result is given as follows:(104)μζ,τ=μ0ζ,τ+μ1ζ,τ+μ2ζ,τ+μ3ζ,τ+⋯+μnζ,τ,vζ,τ=v0ζ,τ+v1ζ,τ+v2ζ,τ+v3ζ,τ+⋯+vnζ,τ,wζ,τ=w0ζ,τ+w1ζ,τ+w2ζ,τ+w2ζ,τ+⋯+wnζ,τ,μζ,τ=1+12tanhζ+112sech2ζηρΓρ+1−1218tanhζsech2ζη2ρΓ2ρ+1+133148cosh2ζ−2sech4ζη3ρΓ3ρ+1+⋯,vζ,τ=12−14tanhζ−118sech2ζηρΓρ+1+1218tanhζsech2ζη2ρΓ2ρ+1−133196cosh2ζ−2sech4ζη3ρΓ3ρ+1+⋯wζ,τ=2−tanhζ−112sech2ζηρΓρ+1+1214tanhζsech2ζη2ρΓ2ρ+1−133124cosh2ζ−2sech4ζη3ρΓ3ρ+1+⋯.
The exact result of (98) is as follows:(105)μζ,τ=1+12tanhζ−112η,vζ,τ=12−14tanhζ−112η,wζ,τ=2−tanhζ−112η.


## 3. Conclusion

Nonlinear partial differential equations can be solved using different methods based on their complexities to determine the exact solutions. In this work, the balance method is applied to solve the modified KdV of the NLPDEs of third order kind analytically. The modified Korteweg–de Vries equation is solved using the balanced method considered with the bilinear and Hirota methods. The obtained soliton solutions of the first and second kinds with certain parameters illustrated their physical models. The obtained results were illustrated graphically to show the geometrical interpretation of models and a few illustrative examples were presented to check the applicability of the method. The obtained solution will serve as a very useful event in the study of nonlinear partial differential equations. This work reveals that the balance method is sufficient, effective, and convenient to simplify the complexities in solving NLPDEs and can be applied appropriately for solving other modified KdV equations of solitary wave type.



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*Source: 7663326-2023-02-21.xml*