# Analytical Solutions to the Caudrey-Dodd-Gibbon-Sawada-Kotera Equation via Symbol Calculation Approach

Authors: Yongyi Gu
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we derive analytical solutions of the Caudrey-Dodd-Gibbon-Sawada-Kotera (CDGSK) equation via symbol calculation approach. Applying the exp−φz-expansion method, we achieve the trigonometric, exponential, hyperbolic, and rational function solutions for the CDGSK equation. By choosing the appropriate parameters, we give some computer simulation to the analytical solutions of the CDGSK equation.


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## Body

## 1. Introduction

Nonlinear fractional and integer order differential equations are widely utilized in fluid dynamics, solid state physics, plasma physics, biology, nonlinear optics, chemistry, etc. It has aroused the strong interest of researchers to these differential equations [1–18]. The study to exact solutions of various NLDEs is extremely important in modern mathematics with ramifications to some areas of physics, mathematics, and other sciences. There are many systematic methods to seek exact solutions of nonlinear differential equations, for example, the Hirota bilinear method [19], Tanh method [20], Lie symmetry [21], modified Kudryashov method [22], Exp-function method [23], sine-Gordon expansion method [24], complex method [25–29], and exp−φz-expansion method [30–32].Sawada and Kotera [33] proposed one of basic models in soliton theory as follows:
(1)uxxxxx+15uuxxx+15uxuxx+45u2ux+ut=0,which is also introduced by Caudrey et al. independently [34]; so in literature, Equation (1) is called the Caudrey-Dodd-Gibbon-Sawada-Kotera equation. Many years have passed by, lots of research results for the CDGSK equation have been developed. As to this equation, the finite dimensional reduction was investigated by Enolski et al. [35], and N-soliton solutions were discovered by Parker [36] via the dressing method. Darboux transformation [37] and Bäcklund transformation in bilinear forms [38] were applied to study the CDGSK equation. Riemann theta function solutions of the CDGSK equation were also established [39]. In this paper, we employ the exp−φz-expansion method [30–32] to obtain the exact solutions of the CDGSK equation.


## 2. The exp−φz-Expansion Method

In this section, we give the main steps of the mentioned method.Step 1.
Inserting traveling wave transform(2)ux,t=uz,z=λx+ωtinto a nonlinear PDE
(3)Pu,ux,uy,ut,uxx,uyy,utt,⋯=0yields
(4)Fu,u′,u″,⋯=0,in which F is the polynomial of u along with its derivatives.Step 2.
Assume that Equation (4) has analytical solutions as follows:
(5)uz=∑τ=0mBτexp−φzτ,where Bτ0≤τ≤m are constants which will be ascertained subsequently, such that Bτ≠0 and φ=φz satisfy
(6)φ′z=γ+exp−φz+ϑexpφz.
The solutions of Equation (6) are given in the following:
Whenγ2−4ϑ>0,ϑ≠0,
(7)φz=ln−γ2−4ϑtanhγ2−4ϑ/2z+c−γ2ϑ,(8)φz=ln−γ2−4ϑcothγ2−4ϑ/2z+c−γ2ϑ.
Whenγ2−4ϑ<0,ϑ≠0,
(9)φz=ln4ϑ−γ2tan4ϑ−γ2/2z+c−γ2ϑ,φz=ln4ϑ−γ2cot4ϑ−γ2/2z+c−γ2ϑ.
Whenγ2−4ϑ>0,γ≠0,ϑ=0,
(10)φz=−lnγexpγz+c−1.
Whenγ2−4ϑ=0,γ≠0,ϑ≠0,
(11)φz=ln−2γz+c+2γ2z+c.
Whenγ2−4ϑ=0,γ=0,ϑ=0,
(12)φz=lnz+c.
In Equations (7)–(12), Bτ≠0,γ,ϑ,c are constants. Taking the homogeneous balance between the nonlinear terms and highest order derivatives of Equation (4) yields the positive integer m.Step 3.
Insert Equation (5) into Equation (4), and collect the function exp−φz to yield the polynomial to exp−φz. Let all coefficients with the same power of exp−φz be zero to obtain a system of algebraic equations. In solving these equations, we achieve the values of Bτ≠0,γ,ϑ and substitute them into Equation (5) as well as Equations (7)–(12) to accomplish the determination for analytical solutions of the original PDE.


## 3. Exact Solutions of the CDGSK Equation

Inserting(13)ux,t=uz,z=λx+ωtinto Equation (1) and then integrating it, we obtain
(14)λ5u″″+15λ3uu″+15λu3+ωu+δ=0,where δ is the integration constant.Taking the homogeneous balance betweenu3 and u″″ of Equation (14) yields
(15)uz=B0+B1exp−φz+B2exp−φz2,where B2≠0 and B1 and B0 are constants.Substitutingu″″,uu″,u3,u into Equation (14) and letting the coefficients about exp−φz be zero, we obtain
(16)e0−φz:λ5B1γ3ϑ+14λ5B2γ2ϑ2+8λ5B1γϑ2+16λ5B2ϑ3+15λ3B0B1ϑγ+30λ3B0B2ϑ2+15λB03+ωB0+δ=0,e1−φz:B1ω+B1γ4λ5+30B2γ3λ5ϑ+22B1γ2λ5ϑ+120B2γλ5ϑ2+30λ3B0B1ϑ+15B0B1γ2λ3+90B0B2γλ3ϑ+15B12γλ3ϑ+30B1B2λ3ϑ2+16B1λ5ϑ2+45B02B1λ=0,e2−φz:16B2γ4λ5+15λ5B1γ3+232B2γ2λ5ϑ+60B1γλ5ϑ+136B2λ5ϑ2+15B12γ2λ3+60B0B2γ2λ3+105B1B2γλ3ϑ+30B22λ3ϑ2+45B0B1γλ3+120B0B2λ3ϑ+30B12λ3ϑ+45B02B2λ+45B0B12λ+B2ω=0,e3−φz:130B2γ3λ5+50B1γ2λ5+440B2γλ5ϑ+75B1B2γ2λ3+40B1λ5ϑ+15B13λ+150B0B2γλ3+45B12γλ3+150B1B2λ3ϑ+30B0B1λ3+90B0B1B2λ+90B22γλ3ϑ=0,e4−φz:330B2γ2λ5+60B1γλ5+60B22γ2λ3+240B2λ5ϑ+195B1B2γλ3+30B12λ3+120B22λ3ϑ+90B0B2λ3+45B0B22λ+45B12B2λ=0,e5−φz:336B2γλ5+24B1λ5+150B22γλ3+120B1B2λ3+45B1B22λ=0,e6−φz:120B2λ5+90B22λ3+15B23λ=0.We solve the above algebraic equations and then obtain two different cases:Case 1.
(17)B2=−2λ2,B1=−2γλ2,B0=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ,δ=λ7γ2−4ϑ39−5λλ5γ2−4ϑ2−4ω2λ5γ2−4ϑ2+ω45λ,where ϑ and γ are arbitrary constants.
Substituting Equation (17) into Equation (15) yields
(18)uz=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ−2γλ2exp−φz−2λ2exp−φz2.
Applying Equations (7)–(12) into Equation (18), respectively, yields analytical solutions of the CDGSK equation as follows:Case 1.1.
Whenγ2−4ϑ>0,ϑ≠0,
(19)u11z=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ+4λ2γϑγ2−4ϑtanhγ2−4ϑ/2z+c+γ−8λ2ϑ2γ2−4ϑtanhγ2−4ϑ/2z+c+γ2,u12z=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ+4λ2γϑγ2−4ϑcothγ2−4ϑ/2z+c+γ−8λ2ϑ2γ2−4ϑcothγ2−4ϑ/2z+c+γ2.Case 1.2.
Whenγ2−4ϑ<0,ϑ≠0,
(20)u13z=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ−4λ2γϑ4ϑ−γ2tan4ϑ−γ2/2z+c−γ−8λ2ϑ24ϑ−γ2tan4ϑ−γ2/2z+c−γ2,u14z=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ−4λ2γϑ4ϑ−γ2cot4ϑ−γ2/2z+c−γ−8λ2ϑ24ϑ−γ2cot4ϑ−γ2/2z+c−γ2.Case 1.3.
Whenγ2−4ϑ>0,γ≠0,ϑ=0,
(21)u15z=−5λ3γ2+5λλ5γ4−4ω30λ−2λ2γ2expγz+c−1−2λ2γ2expγz+c−12.Case 1.4.
Whenγ2−4ϑ=0,γ≠0,ϑ≠0,
(22)u16z=−30λ3ϑ+−5λω15λ+λ2γ3z+cγz+c+2−λ2γ4z+c22γz+c+22.Case 1.5.
Whenγ2−4ϑ=0,γ=0,ϑ=0,
(23)u17z=−5λω15λ−2λ2z+c2.Case 2.
(24)B2=−2λ2,B1=−2γλ2,B0=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ,δ=λ7γ2−4ϑ39+5λλ5γ2−4ϑ2−4ω2λ5γ2−4ϑ2+ω45λ,where ϑ and γ are arbitrary constants.
Substituting Equation (24) into (15) yields
(25)uz=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ−2λ2γexp−φz−2λ2exp−φz2.
Applying Equations (7)–(12) into Equation (25), respectively, yields analytical solutions of the CDGSK equation as follows:Case 2.1.
Whenγ2−4ϑ>0,ϑ≠0,
(26)u21z=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ,+4λ2γϑγ2−4ϑtanhγ2−4ϑ/2z+c+γ−8λ2ϑ2γ2−4ϑtanhγ2−4ϑ/2z+c+γ2,u22z=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ,+4λ2γϑγ2−4ϑcothγ2−4ϑ/2z+c+γ−8λ2ϑ2γ2−4ϑcothγ2−4ϑ/2z+c+γ2.Case 2.2.
Whenγ2−4ϑ<0,ϑ≠0,
(27)u23z−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ,−4λ2γϑ4ϑ−γ2tan4ϑ−γ2/2z+c−γ−8λ2ϑ24ϑ−γ2tan4ϑ−γ2/2z+c−γ2,u24z=−5λ3γ2+8ϑ+5λ6γ2−4ϑ2−20λω30λ,−4λ2γϑ4ϑ−γ2cot4ϑ−γ2/2z+c−γ−8λ2ϑ24ϑ−γ2cot4ϑ−γ2/2z+c−γ2.Case 2.3.
Whenγ2−4ϑ>0,γ≠0,ϑ=0,
(28)u25z=−5λ3γ2+5λλ5γ4−4ω30λ−2λ2γ2expγz+c−1−2λ2γ2expγz+c−12.Case 2.4.
Whenγ2−4ϑ=0,γ≠0,ϑ≠0,
(29)u26z=−30λ3ϑ+−5λω15λ+λ2γ3z+cγz+c+2−λ2γ4z+c22γz+c+22.Case 2.5.
Whenγ2−4ϑ=0,γ=0,ϑ=0,
(30)u27z=−−5λω15λ−2λ2z+c2.Figures1–5 show the properties of the solutions.Figure 1
The 3D and 2D graphics ofu11z by choosing the values γ=5, ϑ=6, λ=1, ω=1/4, c=1, and t=0 for the 2D graphic.Figure 2
The 3D and 2D graphics ofu12z by choosing the values γ=5, ϑ=6, λ=1, ω=1/4, c=1, and t=0 for the 2D graphic.Figure 3
The 3D and 2D graphics ofu13z by choosing the values γ=4, ϑ=5, λ=1, ω=1, c=1, and t=0 for the 2D graphic.Figure 4
The 3D and 2D graphics ofu14z by choosing the values γ=4, ϑ=5, λ=1, ω=1, c=1, and t=0 for the 2D graphic.Figure 5
The 3D and 2D graphics ofu15z by choosing the values γ=2, ϑ=0, λ=1, ω=1, c=1, and t=0 for the 2D graphic.


## 4. Conclusions

In this paper, abundant analytical solutions of the CDGSK equation are obtained via symbol calculation approach. The properties of the solutions are shown by some graphs. It shows that the exp−φz-expansion method is an effective method to seek analytical solutions for nonlinear differential equations.



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*Source: 5042724-2020-05-05.xml*