# Exact Solutions of the Nonlinear Space-Time Fractional Partial Differential Symmetric Regularized Long Wave (SRLW) Equation by Employing Two Methods

Authors: Qinghao Zhu; Jianming Qi
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

In this article, with the aid of Maple software, the exact solutions to the space-time fractional symmetric regularized long wave (SRLW) equation are successfully examined byG′/G2-expansion and extended complex methods. Consequently, three types of traveling wave solutions are found such as Weierstrass double periodic elliptic functions, simply periodic functions, and the rational function solutions. The obtained results will play an important role in understanding and studying SRLW equation. It is easy to see that the extended complex and G′/G2-expansion methods are reliable and will be used extensively to seek exact solutions of any other fractional nonlinear partial differential equations (FNPDE).


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## Body

## 1. Introduction

Fractional calculus is a 300-year-old mathematical problem. Despite its long history, its research has focused on the purely theoretical field of mathematics. However, in the recent decades, with the continuous expansion of fractional calculus applications, such as having memory and genetic characteristics, rheology, material and mechanical systems, electrical engineering, electromagnetism, signal processing and system identification, ANN (neural networks), and fractal and chaos, fractional partial differential equations have been developed into effective methods to unfold a series of strange events and processes [1]. Considerable physical phenomena are well modelled as FNPDE such as acoustic waves, acoustic gravity waves, hydromagnetic waves, fluid flow, chemistry, and other areas [2–6]. Due to the fast rapid development of computer computing power and with the help of symbolic computation software like Maple, Mathematica, and Matlab, the study of explicit solutions about FNPDE is deeply studied by both mathematicians and physicists. A great number of analytical and numerical solutions have been well established and applied to solve these FNPDE. Here are many methods as follows: homotopy analysis [7, 8], fractional G′/G2-expansion [9, 10], tanh-function [11–13], extended tanh [14, 15], the fractional sub-equation [16–18], first integral [19, 20], functional variable [21], modified trial equation [22], finite-difference [23], and so on. In 2021, W. X. Ma et al. systematically studied N-soliton solutions to integrable equations with the help of Hirota direct method for both (1+1)-dimensional and (2+1)-dimensional integrable equations [24–26]. These have greatly promoted the study of nonlinear wave phenomena.In 1984, inspired by weak nonlinear ionic acoustic and spatially charged wave models, Seyler and Fenstermacher [27] summed up the space-time fractional symmetric regularized long wave (SRLW) equation. This equation can summarize many physical phenomena, for instance: ion-acoustic waves in plasma and solitary waves with shallow water waves, shallow water waves. Since these equations are important in physics interpretation, a great number of ascendant and powerful methods have been proposed to obtain exact solutions of SRLW. In 2014, O. Gunerl and D. Eser applied functional variable, exp-function, and G′/G2-expansion three methods to obtain the exact solutions of SRLW in the sense of the modified Riemann-Liouville derivative [28]. In 2015, M. Shakeel and S. T. Mohyud-Din [29] used the fractional novel G′/G2-expansion method to look for the exact solutions of SRLW and obtained many exact analytical solutions like hyperbolic function solutions, trigonometric function solutions, and rational solutions. In 2018, O. Guner and A. Bekir [30] presented exact analytical solutions of SRLW equation by using solitay wave ansatz method. In 2019, D. Yaroa et al. [31] used fractional complex transform and the revised Riemann-Liouville derivative to change the SRLW equation into the ordinary differential equations, and applied the improved F-expansion method to obtain the exact solutions which include hyperbolic and trigonometric solutions. In 2021, M. A. Khan, M. A. Akbar, and N. A. Hamid [32] applied the new auxiliary method to solve for the SRLW. In 2021, N. Maarouf et al. [33] investigated the Lie group analysis method of the SRLW equation and obtained the vector fields and similarity reductions of the equation. It showed that we can use a new independent variable to transform the governing FNPDE into a fractional nonlinear ordinary differential equation (FNODE). In 2021, N. Maarouf et al. [33] also obtained the exact solutions by using the power series expansion method. In 2021, S. C. Ünal et al. [34] got the exact solutions of SRLW equation inspired by a direct method based on the Jacobi elliptic functions ideas; furthermore, S. C. Ünal et al. [34] also obtained some general form solutions which include rational, trigonometric, and hyperbolic functions.In our paper, the conformable fractional derivative of a functionf:0,∞⟶R of order α is defined as
(1)Dzαfz=limε⟶0fz+εz1−α−fzε,z>0,α∈0,1.If the above limitations exist, thenf is called α-differentiable. Let α∈0,1 and f,g be α at point z>0 differentiable, then Dα satisfies the following properties:
(2)1Dαaf+bg=aDαf+bDαg,foreverya,b∈R2Dαzn=nzn−α,foralln∈R3Dαλ=0,thentheconstantfunctionfz=λ4Dαfg=fDαg+gDαf,5Dαf/g=gDαf−fDαg/g2,6If,inaddition,fisdifferentiable,thenDαfz=z1−αdfdzz.The nonlinear SRLW equation appears in several physical applications containing ion sound waves in plasma. This equation is an interesting model to describe ion-acoustic and space change waves and the real-valuedux,t with weak nonlinearity. Now, let us study the SRLW equation ([35, 36])
(3)Dt2αu+Dx2αu+uDtαDxαu+DxαuDtαu+Dt2αDx2αu=0,0<α≤1.Using the fractional transformation(4)ux,t=uξ,ξ=kxαΓ1+α+ctαΓ1+α,here k,c are constants, then equation (7) turns into
(5)c2u″+k2u″+ckuu″+cku′2+c2k2uiv=0.Integration (4) twice yields(6)c2k2u″+c2+k2+ru+12cku2+s=0,here r,s are integral constants.In order to obtain the exact solution of SRLW equation, the extended complex method is employed by us to seek the exact solutions of equation (6), and another way we use the G′/G2-expansion method to get some exact solutions of SRLW equation.Considering the following form of a NFPDE(7)Pu,ux,ut,uxt,uxα,utα,uxx2α,utt2α,uxt2α⋯=0,0<α≤1.in equation (7), P is a polynomial with an unknown function μx,t and its fractional derivatives, involving non-nonlinear terms and the highest order derivatives.In order to verify the accuracy of the obtained results, we define accuracy analysis of error function (AAEF) as follows(8)AAEF=c2k2u″+c2+k2+ru+12cku2+s.It should be noted that the smaller the AAEF, the closer the approximate solution is to the exact solution. When the error infinitely approaches zero, we consider the result to be closer to the true value. Therefore, we use AAEF to indicate the accuracy of the results.The structure of this article is as follows: in Section2, we probe the modified G′/G2-expansion method to the SRLW equation; in Section 3, we discuss the extended complex method and main result; in Sections 4 and 5, we give the detail proof of the main results in this paper; in Section 6, we give the specific form of accuracy analysis of error function to show the accuracy of our results; in Section 7, we compared our methods and results with others in detail; in Section 8, through computer simulation images, we further analyze the nature of the solutions that we have obtained; finally, the conclusion and future recommendations in Section 9.


## 2. Employing of the ModifiedG′/G2-Expansion Method to the SRLW Equation

Next, we will use the recently established modifiedG′/G2-expansion method [37, 38] to provide new and more general traveling wave solutions for the above equation (6).Step 1. Inserting the traveling wave transformT:ux,t=uξ,ξ=kxα/Γ1+α+ctα/Γ1+α into equation (7), changing it to the following integer order ordinary differential equation (IOODE):
(9)Ku,u′,u″,u′′′,⋯=0,in the above equation (9), K is a polynomial composed of uξ and its integer order derivatives.Step 2. Regarding that the solution of equation (9) can be expressed by a polynomial in G′/G2 as follows:
(10)uξ=α0+∑n=1mαmG′G2m+βmG′G2−m,in equation (10), G=Gξ satisfies the following differential equation as follows:
(11)G′G2′=σ+μG′G2+ρG′G22,where μ,σ, and ρ are free constants. The positive integer m can be determined by considering the uniform equilibrium between the highest order derivatives and the nonlinear term appearing in the ODE (8).Step 3. Taking equation (10) into equation (9), and using equation (11), the left side (8) of the formula was converted to another polynomial in the G′/G2. Computing all the coefficients of the polynomial to zero yields the algebraic equations for αm, ⋯, λ, and μ.Step 4. The undetermined constantsαm,⋯,α0,β1,⋯,βmcan be obtained by solving the system of algebraic equations obtained in Step 3, since the general form solutions of (10) have five possible solutions as following. Hence, the exact solutions of the given Eq. (7) can be obtained as follows.Here:(12)Case1:Ifσρ>0,μ=0,thenG′G2ξ=σρσC1cosσρξ+C2sinσρξC2cosσρξ−C1sinσρξ;Case2:Ifσρ<0,μ=0,thenG′G2ξ=−σρσC1sinh2σρξ+C2cosh2σρξ+C2C1cosh2σρξ+C1sinh2σρξ−C2;Case3:Ifσ=0,ρ≠0,μ=0,thenG′G2ξ=−C1ρC1ξ+C2;Case4:Ifμ≠0,Δ≥0,thenG′G2ξ=−μ2ρ−ΔC1coshΔ/2ξ+C2sinhΔ/2ξ2ρC2coshΔ/2ξ+C1sinhΔ/2ξ;Case5:Ifμ≠0,Δ<0,thenG′G2ξ=−μ2ρ−−ΔC1cos−Δ/2ξ−C2sin−Δ/2ξ2ρC2cos−Δ/2ξ+C1sin−Δ/2ξ;where C1,C2 are arbitrary constants and Δ=μ2−4ρσ.Step 5. Putting the inverse transformT−1 into the solutions uξξ=kxα/Γ1+α+ctα/Γ1+α, we can get all exact solutions ux,t of the original FNPDE.
(13)u11ξ=−c2+k2+rck±8σρk−12ρ3ckσC1cosσρξ+C2sinσρξC1sinσρξ−C2cosσρξ2−12σ3ckρC1cosσρξ+C2sinσρξC1sinσρξ−C2cosσρξ−2,Remark 1.
From the above five cases, (11) contains trigonometric (Cases1 and 5), rational (Case 3), and hyperbolic (Cases 2 and 4) three forms solutions.Theorem 2.
ByG′/G2-expansion method, we have found the following five cases solutions of Eq. (6):Case 1.
Ifσρ>0, μ=0, thenandc2+k2+r2−2cks=256σ2ρ2c2k4.
(14)u12ξ=−c2+k2+rck±8σρk+12ρ3ckσC1sinh2−σρξ+C2cosh2−σρξ+C2C1cosh2−σρξ+C1sinh2−σρξ−C22+12σ3ckρC1sinh2−σρξ+C2cosh2−σρξ+C2C1cosh2−σρξ+C1sinh2−σρξ−C2−2,Case 2.
Ifσρ<0, μ=0, thenandc2+k2+r2−2cks=256σ2ρ2c2k4.
(15)u13ξ=−c2+k2+rck−12ckC12C1ξ+C22Case 3.
Ifσ=0 and ρ≠0, μ=0, thenandc2+k2+r2=2cks.
(16)u14ξ=2μ2ck−c2+k2+rck−3μ2ckC1cosh±μ/2ξ+C2sinh±μ/2ξC2cosh±μ/2ξ+C1sinh±μ/2ξ2Case 4.
Ifμ≠0 and Δ=μ2−4σρ≥0, thenandc2+k2+r2−2cks=u4c2k2, σ=0 and ρ≠0.
(17)u15ξ=2μ2ck−c2+k2+rck+3μ2ckC1cos±iμ/2ξ−C2sin±iμ/2ξC2cos±iμ/2ξ+C1sin±iμ/2ξ2Case 5.
Ifμ≠0 and Δ=μ2−4σρ<0, thenandc2+k2+r2−2cks=u4c2k2, σ=0 and ρ≠0.Hereξ=kxα/Γ1+α+ctα/Γ1+α. C1 and C2 are arbitrary constants.


## 3. Introduction of the Extended Complex Method and Main Result

The extended complex number method involves the knowledge related to the Weierstrass elliptic function. First, we give the brief introduction of Weierstrass elliptic function:℘ξ≔℘ξ,g2,g3 is a meromorphic function in the complex plane ℂ with double periods ω1,ω2[39–41] and defined as
(18)℘ξ;w1,w2≔1ξ2+∑μ,ν∈Z,μ2+ν2≠01ξ+μw1+νw22−1μw1+νw22,which satisfies the following equation
(19)℘′ξ2=4℘ξ3−g2℘ξ−g3,in equation (19), g2=60s4,g3=140s6, and Δg2,g3≠0, and has the another formula
(20)℘ξ−ξ0=−℘ξ−℘ξ0+14℘′ξ+℘′ξ0℘ξ−℘ξ02.Next, we will employ the recently established extended complex method [42–45] to provide new and more general traveling wave solutions to the equations mentioned above.Step 1. Inserting the traveling wave transformT: ux,t=uξ,ξ=kxα/Γ1+α+ctα/Γ1+α into equation (7), alternating it to the following (IOODE):
(21)Ku,u′,u″,u′′′,⋯=0,here K is a polynomial of uξ and its derivatives.Step 2. Next, we will find out weakp,q condition. To find out the weak p,q condition of equation (21), the Laurent series,
(22)uξ=∑k=−q∞ckξk,q>0,c−q≠0,are replaced into equation (21), then the p distinct Laurent singular parts are obtained as below:
(23)∑k=−q−1ckξk.Herep indicates that there are p distinct meromorphic solutions in the equation, and q means that their poles at ξ=0 have q multiple roots in the equation.Step 3. Take the following pending forms(24)uξ=∑i=1l−1∑j=2qi−1jc−ijj−1!dj−2dzj−214℘′ξ+Bi℘ξ−Ai2−℘ξ+∑i=1l−1c−i12℘′ξ+Bi℘ξ−Ai+∑j=2ql−1jc−ljj−1!dj−2dzj−2℘ξ+c0,(25)uξ=∑i=1l∑j=1qcijξ−ξij+c0,(26)uζ=∑i=1l∑j=1qcijζ−ζij+c0,hereζ=eϑξϑ∈ℂ.In equation (24), c−ij are given as shown in equation (21), and Bi2=4Ai3−g2Ai−g3, Σi=1lc−i1=0, and equations (23), (24), and (25) have l≤p distinct poles of multiplicity q.Step 4. We can get meromorphic solutions and the above addition formulas. Putting the inverse transformT−1 into the solutions uξ, we can obtain all exact solutions ux,t of the original FNPDE.
(a)
The rational function solutions(27)u21ξ=−12ckξ−ξ02−c2+k2+rck,where 2cks=c2+k2+r2, ξ0∈ℂ.(b)
The simply periodic solutions are obtained by(28)u22ξ=−12ckϑ2coth2ϑ2ξ−ξ0−ckϑ2−c2+k2+rck,where 2cks=−c4k4ϑ4+c2+k2+r2, ξ0∈ℂ, ϑ∈ℂ.(c)
The Weierstrass elliptic solutions are as follows:(29)u23ξ=−12ck−℘ξ+14℘′ξ+F℘ξ−E2+12ckE−c2+k2+r2ck,here 2cks=−12c4k4g2+c2+k2+r2, F2=4E3−g2E−g3, g3 and E are arbitrary constants.Theorem 3.
Supposeck≠0, by the extended complex method, we have found the following three cases solutions of Eq. (6):All in above,ξ=kxα/Γ1+α+ctα/Γ1+α.


## 4. Proof of Theorem2

Considering the homogeneous equilibrium term betweenu′′ and u2 in (5), we deduce m=2. So we can infer the solution of (5) as follows:
(30)uξ=α2G′G22+α1G′G2+α0+β2G′G2−2+β1G′G2−1,here α0,α1,α2 are the upcoming constants that will be determined later.Next, we will use (29) and (10) to collect and sort out all terms with the same power ofG′/G2 together.First, from (29), we get(31)u′ξ=α1σ−ρβ1+α1μ+2σα2G′G2+ρα1+2μα2G′G22+2ρα2G′G23−β1σ+2β2μG′G2−2−μβ1+2β2μG′G2−1−2β2σG′G2−3.Substituting (10) into (30), we obtain(32)u″ξ=α1σμ+2σ2α2+ρμβ1+2β2ρ2+α1μ2+2μσα2+2ρα1σ+2μα2σG′G2+α1μρ+2σα2ρ+2ρα1μ+4μ2α2+6ρσα2G′G22+2ρ2α1+4μρα2+6ρα2μG′G23+6ρ2α2G′G24+2β1σ2+4β2μσ+6β2σμG′G2−3+2β1σμ+4β2μ2+μβ1σ+2β2ρσ+6ρβ2σG′G2−2+2ρβ1σ+4β2ρμ+μ2β1+2β2μρG′G2−1+6β2σ2G′G2−4.Now, we put (29), (30), and (31) into (5), and sort out all terms with the same power ofG′ξ/G2ξ together,
(33)6ρ2α2c2k2+12α22ckG′ξG2ξ4+2ρ2α1c2k2+10μρα2c2k2+α1α2ckG′ξG2ξ3+3ρμα1c2k2+8σα2ρc2k2+4μ2α2c2k2+c2α2+k2α2+rα2+12ckα12+α0α2ckG′ξG2ξ2+α1μ2c2k2+4μσα2c2k2+2ρα1σc2k2+c2α1+k2α1+rα1+α0α1ck+α2β1ckG′ξG2ξ+c2k2α1σμ+2σ2α2c2k2+ρμβ1c2k2+2β2ρ2c2k2+12ckα02+α1β1ck+α2β2ck+c2α0+k2α0+rα0+s+6c2k2β2σ2+12ckβ22G′ξG2ξ−4+2β1σ2c2k2+10β2μσc2k2+β1β2ckG′ξG2ξ−3+3β1σμc2k2+4β2μ2c2k2+8β2ρσc2k2+c2β2+k2β2+rβ2+12ckβ12+α0β2ckG′ξG2ξ−2+2ρβ1σc2k2+6β2ρμc2k2+μ2β1c2k2+c2β1+k2β1+rβ1+α0β1ck+α1β2ckG′ξG2ξ−1=0.For theG′ξ/G2ξ functions of the same terms power, extracting its undetermined coefficients and set to zero, and the following equations can be obtained:
(34)c2k2α1σμ+2σ2α2c2k2+ρμβ1c2k2+2β2ρ2c2k2+12ckα02+α1β1ck+α2β2ck+c2α0+k2α0+rα0+s=0,1.1α1μ2c2k2+4μσα2c2k2+2ρα1σc2k2+c2α1+k2α1+rα1+α0α1ck+α2β1ck=0,1.23ρμα1c2k2+8σα2ρc2k2+4μ2α2c2k2+c2α2+k2α2+rα2+12ckα12+α0α2ck=0,1.32ρ2α1c2k2+10μρα2c2k2+α1α2ck=0,1.46ρ2α2c2k2+12α22ck=0,1.56c2k2β2σ2+12ckβ22=0,1.62β1σ2c2k2+10β2μσc2k2+β1β2ck=0,1.73β1σμc2k2+4β2μ2c2k2+8β2ρσc2k2+c2β2+k2β2+rβ2+12ckβ12+α0β2ck=0,1.82ρβ1σc2k2+6β2ρμc2k2+μ2β1c2k2+c2β1+k2β1+rβ1+α0β1ck+α1β2ck=0.1.9.Case 1.
Ifσρ>0, μ=0, under this assumption, solving the above system equation (33), we get
(35)α0=−c2+k2+rck±8σρk,α1=0,α2=−12ρ2ck,β1=0,β2=−12σ2ck,here
(36)c2+k2+r2−2cks=256σ2ρ2c2k4G′G2ξ=σρσC1cosσρξ+C2sinσρξC1sinσρξ−C2cosσρξ.
Now, we give the forms ofuξ:(37)u11ξ=α0+α2G′G22+β2G′G2−2=−c2+k2+rck±8σρk−12ρ3ckσC1cosσρξ+C2sinσρξC1sinσρξ−C2cosσρξ2−12σ3ckρC1cosσρξ+C2sinσρξC1sinσρξ−C2cosσρξ−2,and c2+k2+r2−2cks=256σ2ρ2c2k4.Case 2.
Ifσρ<0, μ=0, under this assumption, solving the above system equation (33), we get
(38)α0=−c2+k2+rck±8σρk,α1=0,α2=−12ρ2ck,β1=0,β2=−12σ2ck,here
(39)c2+k2+r2−2cks=256σ2ρ2c2k4and then the result is similar with Case 1.
(40)G′G2ξ=−σρσC1sinh2σρξ+C2cosh2σρξ+C2C1cosh2σρξ+C1sinh2σρξ−C2.
Now, we give the forms ofuξ:(41)u12ξ=α0+α2G′G22+β2G′G2−2=−c2+k2+rck±8σρk+12ρ3ckσC1sinh2−σρξ+C2cosh2−σρξ+C2C1cosh2−σρξ+C1sinh2−σρξ−C22+12σ3ckρC1sinh2−σρξ+C2cosh2−σρξ+C2C1cosh2−σρξ+C1sinh2−σρξ−C2−2,
andc2+k2+r2−2cks=256σ2ρ2c2k4.Case 3.
Ifσ=0 and ρ≠0, μ=0, under this assumption, solving the above system equation (33), we obtain
(42)α0=−c2+k2+rck,α1=0,α2=−12ρ2ck,β1=0,β2=0,here
(43)c2+k2+r2=2cks,G′G2ξ=−C1ρC1ξ+C2.
Now, we give the forms ofuξ:(44)u13ξ=α0+α2G′G22=−c2+k2+rck−12ckC12C1ξ+C22.Case 4.
Ifμ≠0 and Δ=μ2−4σρ≥0, now we will divide into two subcases.
Subcase 4.1. Ifσ=0 and ρ=0, and
(45)G′G2ξ=−μ2ρ−ΔC1coshΔ/2ξ+C2sinhΔ/2ξ2ρC2coshΔ/2ξ+C1sinhΔ/2ξ.
This contradicts the truth that the denominator cannot be zero. So this case cannot happen.
Ifσ=0 and ρ≠0, under this assumption, solving the above system equation (33), we get
(46)α0=−u2ck−c2+k2+rck,α1=−12ρμck,α2=−12ρ2ck,β1=0,β2=0,here
(47)c2+k2+r2−2cks=u4c2k2,G′G2ξ=−μ2ρ−ΔC1coshΔ/2ξ+C2sinhΔ/2ξ2ρC2coshΔ/2ξ+C1sinhΔ/2ξ.
Now, we give the forms ofuξ:(48)u14ξ=α0+α1G′G2+α2G′G22=−u2ck−c2+k2+rck−12ρμck−μ2ρ−ΔC1coshΔ/2ξ+C2sinhΔ/2ξ2ρC2coshΔ/2ξ+C1sinhΔ/2ξ−12ρ2ck−μ2ρ−ΔC1coshΔ/2ξ+C2sinhΔ/2ξ2ρC2coshΔ/2ξ+C1sinhΔ/2ξ2=2μ2ck−c2+k2+rck−3μ2ckC1cosh±μ/2ξ+C2sinh±μ/2ξC2cosh±μ/2ξ+C1sinh±μ/2ξ2.
Subcase 4.2. Ifσ≠0, during the system equations, by (1.5), we deduce α2=−12ρ2ck; by (1.6), we deduce β2=−12σ2ck; and then from (1.4) and (1.7), we get α1=−12ρμck,β1=−12μσck, and then from (1.8), we deduce
(49)α0=−μ2ck+8ρσck−c2+k2+rck,and then from (1.9), we deduce
(50)α0=−μ2ck+4ρσck−c2+k2+rck,so ρ must be equal to 0. Finally
(51)α0=−u2ck−c2+k2+rckα1=0α2=0β1=−12μσckβ2=−12σ2ck,here
(52)c2+k2+r2−2cks=u4c4k4.G′G2ξ=−μ2ρ−ΔC1coshΔ/2ξ+C2sinhΔ/2ξ2ρC2coshΔ/2ξ+C1sinhΔ/2ξ.
This contradicts the truth that the denominator cannot be zero. So this case cannot happen.Case 5.
Ifμ≠0 and Δ=μ2−4σρ<0, and then the result is the same as Case 4.Only the case thatσ=0 and ρ≠0, under this assumption, solving the above system equation (33), we get
(53)α0=−u2ck−c2+k2+rckα1=−12ρμckα2=−12ρ2ckβ1=0β2=0,here
(54)c2+k2+r2−2cks=u4c2k2,G′G2ξ=−μ2ρ−−ΔC1cos−Δ/2ξ−C2sin−Δ/2ξ2ρC2cos−Δ/2ξ+C1sin−Δ/2ξ.Now, we give the forms ofuξ:(55)u15ξ=α0+α1G′G2+α2G′G22=−u2ck−c2+k2+rck−12ρμck−μ2ρ−−ΔC1cos−Δ/2ξ−C2sin−Δ/2ξ2ρC2cos−Δ/2ξ+C1sin−Δ/2ξ−12ρ2ck−μ2ρ−−ΔC1cos−Δ/2ξ−C2sin−Δ/2ξ2ρC2cos−Δ/2ξ+C1sin−Δ/2ξ2=2μ2ck−c2+k2+rck+3μ2ckC1cos±iμ/2ξ+C2sin±iμ/2ξC2cos±iμ/2ξ+C1sin±iμ/2ξ2.All in the above cases,ξ=kxα/Γ1+α+ctα/Γ1+α and C1 and C2 are arbitrary constants.The proof is complete.


## 5. Proof of Theorem3

Noting that putting (21) into (5), we haveq=2, p=1, c−2=−12ck, c−1=0, c0=−c2+k2+r/ck, c1=0, c2=sck−2c2k2−c2+k2+r2/20c3k3, c3=0, where c4 is arbitrary.Hence, (5) satisfies week1,2 condition and we will then give the form of all the meromorphic solutions for (5).By (14), it is easy to deduce the rational solutions of equation (6) with pole at ξ=0 that
(56)uaξ=c11ξ2+c12ξ+c.Substitutinguaξ into (5), we obtain the following
(57)ua1ξ=−12ckξ2−c2+k2+rck,where c2+k2+r2=2cks. The rational solutions of (5) are as follows:
(58)u21ξ=−12ckξ−ξ02−c2+k2+rck,where c2+k2+r2=2cks, ξ0∈ℂ.To search for simple periodic solutions, settingζ=expϑξ, ϑ∈ℂ, putting u=Rζ into (5), then
(59)c2k2ϑ2ζR′+ζ2R″+c2+k2+rR+12ckR2+s=0,and putting
(60)ubζ=c2ζ−12+c1ζ−ζ1+c10into (55), we get that
(61)ub1ζ=−12ckϑ2ζ−12+−12ckϑ2ζ−1−ckϑ2−c2+k2+rck,here c2+k2+r2=2cks+c4k4ϑ4. Substituting ζ=eϑξ, ϑ∈ℂ into the above form, we get all simply periodic solutions of (5) in pole at ξ=0 as follows:
(62)ub2ξ=−12ckϑ2eϑξ−12−12ckϑ2eϑξ−1−ckϑ2−c2+k2+rck=−12ckϑ2eϑξeϑξ−12−ckϑ2−c2+k2+rck=−12ckϑ2coth2ϑ2ξ−ckϑ2−c2+k2+rck,So all simply periodic solutions of (5) are obtained by(63)u22ξ=−12ckα2coth2ϑ2ξ−ξ0−ckϑ2−c2+k2+rck,where c2+k2+r2=2cks+c4k4ϑ4, ξ0∈ℂ.From the above part, we are about to obtain the form of the elliptic solution to be determined of (5) with pole atξ=0 as follows:
(64)ucξ=c−2℘ξ+c0.Substitutingucξ into the (5), then we obtain that
(65)uc1ξ=−12ck℘ξ−c2+k2+rck,here 2cks=−12c4k4g2+c2+k2+r2.Therefore, all elliptic solutions of (5) are as follows:(66)uc2ξ=−12ck℘ξ−ξ0−c2+k2+rck,here ξ0∈ℂ. We rewrite it to the form
(67)u23ξ=−12ck−℘ξ+14℘′ξ+F℘ξ−E2+12ckE−c2+k2+rck,here 2cks=−12c4k4g2+c2+k2+r2, F2=4E3−g2E−g3, g3 and E are arbitrary.The proof is finished.


## 6. Accuracy Analysis of Error Function (AAEF)

From eq. (8), it is easy to see the smallness of the AAEF indicating that the approximate solution and exact solution are very close. Thus, AAEF can be used to make an estimate of the accuracy of the approximate solution.Now, we take some specific parameters in some solutions of eq. (6) and use the Maple software to reckon the AAEF as follows:For the part solutions in eq. (6), some values of the AAEF are given in Tables 1, 2, and 3, respectively. It is obvious to find that the maximum values of the AAEF are 10−9. It means that the obtained solutions are very near the exact solution.Table 1
AAEF (C1≠0, C2=0) for the solution u11ξ in eq. (13).
αckρσxtREF0.250.0010.001π4π4196249196249−6.39625984∗10−110.50.0010.001π4π4196249196249−3.660149852∗10−110.750.0010.001π4π4196249196249−3.720239068∗10−110.250.0010.001π4π4150000150000−6.79117167∗10−110.50.0010.001π4π4150000150000−3.728142221∗10−110.750.0010.001π/4π/4150000150000−3.668526651∗10−110.250.0010.001π/4π/4250000250000−6.083233454∗10−110.50.0010.001π/4π/4250000250000−3.660763965∗10−110.750.0010.001π/4π/4250000250000−3.704605275∗10−11Table 2
AAEF (C1=0, C2≠0) for the solution u12ξ in eq. (14).
αckρσxtREF0.250.0010.0012−2196249196249−1.541853209∗10−90.50.0010.0012−2196249196249−2.024770373∗10−90.250.0010.0012−2150000150000−1.541838967∗10−90.50.0010.0012−2150000150000−1.898189859∗10−90.250.0010.0012−2250000250000−1.541867807∗10−90.50.0010.0012−22500002500005.931281180∗10−9Table 3
AAEF for the solutionu22ξ in eq. (63).
αckξ0ϑxtREF0.250.0010.001ξ0=1−2071/71962491962491.714265321∗10−180.50.0010.001ξ0=1−2071/71962491962491.714265334∗10−180.750.0010.001ξ0=1−2071/71962491962491.714265297∗10−180.250.0010.001ξ0=1−2071/71500001500001.714265321∗10−180.50.0010.001ξ0=1−2071/71500001500001.714265334∗10−180.750.0010.001ξ0=1−2071/71500001500001.714265297∗10−180.250.0010.001ξ0=1−2071/72500002500001.714265321∗10−180.50.0010.001ξ0=1−2071/72500002500001.714265334∗10−180.750.0010.001ξ0=1−2071/72500002500001.714265297∗10−18Obviously, selecting the parameters is important in the calculation of AAEF. From the three tables, AAEF for the solutionu22ξis the closest to the exact solution, because the value of the AAEF is10−18.If we take another parameter, it is possible to obtain exact solutions close to the other parameters.


## 7. Comparison

In this work, with the aid of Maple software, using the extended complex method, we found double periods and single period function solutions of SRLW equation. We also can find more solutions by theG′/G2-expansion method. In 2021, M. A. Khan et al. [32] applied the new auxiliary method to solve for the SRLW. M. A. Khan et al. [32] found the trigonometric and hyperbolic function solutions for the SRLW. However, we can obtain double periods Weierstrass elliptic function solutions just by complex method. Within our cognitive range, the double periods Weierstrass elliptic function solutions got for SRLW in this article mostly have not been reported in the literature. From the level of computational complexity, the extended complex method is more concise and clear. Although G′/G2-expansion method is relatively complex in computation, more formal solutions are obtained.In 2021, S. C. Ünal et al. [34] got the exact solutions of SRLW equation inspired by a direct method based on the ideas of Jacobi elliptic functions; furthermore, they also get some general form solutions which include rational, single periodic trigonometric and hyperbolic, and double periodics Jacobi elliptic functions. It is well known that the connection between the Jacobian elliptic functions cnξ,m and Weierstrass elliptic function solutions ℘ξ is
(68)℘ξ,g2,g3=e2−e2−e3cn2e1−e3;m,here m2=e2−e3/e1−e3 is the modulus number of Jacobian elliptic functions, and eii=1,2,3,e1≥e2≥e3 are the roots of equation 4ξ2−g2ξ−g3=0. It is well known that if m⟶1 and e2⟶e1, then cnξ;m⟶sechξ, or if m⟶0, then cnξ;m⟶cosξ. If 0<m<1, then Jacobian elliptic function is the double periodic meromorphic function and cannot degenerate into single periodic trigonometric and hyperbolic meromorphic functions. Equation (68) is the important bridge linking among the Weierstrass elliptic function, hyperbolic function, trigonometric function, and Jacobian elliptic functions.Although the means, methods, and ideas of the above methods for studying the exact solutions of SRLW equation vary, the final results obtained still have important relevance and also play important implications for indicating the deep mechanisms of physical phenomena and giving tedious solutions to FNPDE. The extended method enriches the study of the described equations for FNPDE.


## 8. Computer Simulations

In this section, we are trying to explain the results obtained according to two different methods through computer simulation images, and further analyze the nature of the simple periodic solutions (Figures1–3) u11ξ, (Figures 4 and 5) u12ξ, (Figures 6 and 7) u14ξ, (Figures 8 and 9) u15ξ, and (Figures 10 and 11) u22ξ in the SRLW equation.Figure 1
The three-dimensional images ofu11ξ by considering the values c=0.001, k=0.001, ρ=π/4, σ=π/4, s=0, C2=0; from (a) to (c), α take the following three different values: α=0.25, α=0.5, α=0.75. Three graphs demonstrate the multisoliton profiles of u11ξ on the domain.
(a)(b)(c)Figure 2
The two-dimensional images ofu11ξ by considering the values c=0.001, k=0.001, ρ=π/4, σ=π/4, s=0, C2=0; from (a) to (c), α take the following three different values: α=0.25, α=0.5, α=0.75. The above three graphs show corresponding wave propagation.
(a)(b)(c)Figure 3
The corresponding contour images ofu11ξ by considering the values c=0.001, k=0.001, ρ=π/4, σ=π/4, s=0, C2=0; from (a) to (c), α take the following three different values: α=0.25, α=0.5, α=0.75.
(a)(b)(c)Figure 4
The three-dimensional images ofu12ξ by considering the values c=0.001, k=0.001, ρ=2, σ=−2, s=0, C1=0 take the following three different values: α=0.25, α=0.5, α=0.75. The images show annihilation of solitary wave profiles for u12ξ accompanied with different α.
(a)(b)(c)Figure 5
The two-dimensional images ofu12ξ by considering the values c=0.001, k=0.001, ρ=2, σ=−2, s=0, C1=0 take the following three different values: α=0.25, α=0.5, α=0.75. The three images reveal the wave propagation along x-axis accompanied with different α.
(a)(b)(c)Figure 6
The three-dimensional images ofu14ξ by considering the values c=0.001, k=0.001, ρ=2, σ=0, s=0, μ=5, C1=3, C2=4; from (a) to (c), α take the following three different values: α=0.25, α=0.5, α=0.75.
(a)(b)(c)Figure 7
The two-dimensional images ofu14ξ by considering the values c=0.001, k=0.001, ρ=2, σ=0, s=0, μ=5, C1=3, C2=4; from (a) to (c), α take the following three different values: α=0.25, α=0.5, α=0.75. The three graphs show the curved shape solitary profiles with different α of u14ξ.
(a)(b)(c)Figure 8
The three-dimensional images ofu15ξ by considering the values c=0.001, k=0.001, ρ=2, σ=0, s=0, μ=i, C1=3, C2=4; from (a) to (c), α take the following three different values: α=0.25, α=0.5, α=0.75. Three graphs demonstrate the multisoliton profiles of u15ξ on the domain.
(a)(b)(c)Figure 9
The two-dimensional images ofu15ξ by considering the values c=0.001, k=0.001, ρ=2, σ=0, s=0, μ=i, C1=3, C2=4; from (a) to (c), α take the following three different values: α=0.25, α=0.5, α=0.75. Three graphs show corresponding wave propagation profiles of u15ξ on the domain.
(a)(b)(c)Figure 10
The three-dimensional images ofu22ξ by considering the values c=0.001, k=0.001, ξ0=1−207, ϑ=1/7; from (a) to (c), α take the following three different values: α=0.25, α=0.5, α=0.75. Three graphs demonstrate the annihilation of soliton profiles of u22ξ on the domain.
(a)(b)(c)Figure 11
The two-dimensional images ofu22ξ by considering the values c=0.001, k=0.001, ξ0=1−207, ϑ=1/7, y=10000; from (a) to (c), α take the following three different values: α=0.25, α=0.5, α=0.75. Three graphs show corresponding wave propagation profiles of u22ξ on the domain.
(a)(b)(c)The three graphs show the cuspon’s shape with differentα of u14ξ.


## 9. Conclusions and Future Outlook of the Study

The extended complex method and theG′/G2-expansion method are very effective tools for seeking the exact solutions of FNPDE. By traveling wave transformation, many of the FNPDE can be converted into IOODE similar to equation (9). In this paper, we employed the extended complex method and the G′/G2-expansion method to seek the exact solutions about SRLW equation. By traveling wave transformation, we can reduce the dimensions of the FNPDE to IOODE related to mathematical physics and engineering. The results of the full text eloquently prove the above methods are very efficient and powerful in solving the exact solutions of SRLW equation now and in the future. We can apply these ideas and methods of this research to other FNPDE.It is well known that [46] ℘z≔℘z,g2,g3 have one successive degeneracy to simple periodic functions according to
(69)℘ξ,3d2−d3=2d−3dd2coth23d2ξ,if one root ej is double (Δg2,g3=0). Also ℘z≔℘z,g2,g3 have another one successive degeneracy to rational function ξ according to
(70)℘ξ,0,0=1z2,if one root ej is triple g2=g3=0. It shows that there exist simple periodic solutions u22ξ which is not degenerated successively to the double periods elliptic function.We are confident that the extended complex methods andG′/G2-extension methods still have broad and bright applications in the future. Both of these methods are effective tools for obtaining FNPDE exact solutions.In the future, we may consider the simplified improvedtanϕξ/2-expansion method (SITEM) [47] and generalized direct algebraic method [48] to further investigate the SRLW equation. In 2021, Md. Rezwan Ahamed Fahim et al. [49] used the sine-Gordon-expansion approach for finding the exact solutions of the traveling waves about (3+1)-dimensional Kadomtsev-Petviashvili and the modified KdV-Zakharov-Kuznetsov equations. In 2021, W. X. Ma et al. analyzed soliton solutions and verified the Hirota N-soliton condition for the B-type Kadomtsev-Petviashvili equation, the (2+1)-dimensional KdV equation, the Kadomtsev-Petviashvili equation, and the (2+1)-dimensional Hirota-Satsuma-Ito equation [24, 26], within the Hirota bilinear formulation. In 2021, M. Hafiz Uddin et al. [50] employ the double (G′/G, 1/G)-expansion method to investigate the new exact solutions to some fractional nonlinear evolution solutions. This is also another research direction that we will focus on and have great interest in the future.



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