# The Classification of Traveling Wave Solutions to Space-Time Fractional Resonance Nonlinear Schrödinger Type Equation with a Special Kind of Fractional Derivative

Authors: Lianguang Jia; Li Tang
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this paper, the classification of single traveling wave solutions to a special kind of space-time fractional Schrödinger type equation with high-order nonlinearities is obtained by the complete discrimination system for polynomial method, where the fractional derivative is defined by Atangana’s conformable definition. The descriptions of the method and the fractional derivative are given, and the concrete procedure of the method is also presented in the paper. Especially, some new solutions such as hyperelliptic functions solution could be found and all the existing results about traveling wave solutions to the equation could also be seen. Moreover, concrete examples indicate that all the solutions obtained in the paper can be realized.


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## Body

## 1. Introduction

In order to portray anomalous phenomena in science and engineering, nonlinear equation with fractional order derivative is proposed and has been widely applied in the fields such as anomalous diffusion [1, 2], and thus, finding exact solutions to such a special kind of equation is an important and significant task. And fractional derivative also has many definitions, for example, Caputo derivative [3] and Riemann-Liouville derivative [4]. However, some of the definitions are very complicated that it is too difficult to find expected exact solutions to the corresponding equation and some definitions such as modified Riemann-Liouville derivative have already been proved wrong [5, 6]. Recently, Atangana et al. proposed a special kind of definition called Atangana’s fractional derivative [7], which is easy to apply due to the reason that it obeys almost all the properties satisfied by the conventional Newtonian concept of derivative such as chain rules. The definition of the derivative is given as follows:
(1)D0Atαϕt=limE⟶0ϕt+Et+1/Γα1−α−ϕtE.If the above limitation holds, then we would have the following properties [7].
(2)10ADtαϕt±λt=D0Atαϕt±D0Atαλt,2D0Atαϕtλt=λtD0Atαϕt+ϕtD0A0Aλt,3D0Atαϕtλt=λtD0Atαϕt−ϕtD0Atαλtλ2t,4D0Atαc=0,where ϕt and λt are the two β-differentiable functions with β∈0,1. Via this definition, many interesting and important results have been proposed [8–10].In this paper, we consider the following space-time fractional resonance nonlinear Schrödinger type equation with Atangana’s derivative [11].
(3)i0ADtαuz,t0ADtα+ρ0AD02αuz,t+αu+βu2+γu3uz,t+r0ADz2αu2uuz,t=0,where z is the longitudinal coordinate along the optical fibers, uz,t represents the normalized electric-field envelope, 0ADtα and 0ADz2α are the Atangana’s fractional derivative with respect to t and z, r is the coefficient of resonant nonlinearity, and α, β, and γ are the corresponding coefficients of high-order nonlinearities. This equation is widely used in fluid dynamics, especially the Madelung’s fluids in several nonlinear systems [12–14]. And in other fields such as nonlinear optics [15, 16] and plasma and nuclear physics [17–20], the resonance nonlinear Schrödinger equation also plays a vital role. In [11], Yepezmartinez and Gomezaguilar obtained soliton solutions to some forms of deformation of Equation (3) by the subequation method. His results are interesting.Nowadays, many famous and important methods are proposed to achieve the goal of finding the exact solutions to the nonlinear fractional PDEs due to the significance of them, for example, subequation method [21, 22], (G′/G) expansion method [23, 24], and Sine-Gordon equation expansion method [25, 26]. Other results about exact solutions to nonlinear partial differential equations can be seen in [27–31]. However, though the mentioned methods are very powerful, they cannot find all the exact traveling wave solutions to the expected equation, and some kinds of solutions such as elliptic function double periodic solutions could hardly be found. In order to overcome these difficulties, Liu proposed a powerful method called complete discrimination system for polynomial method [32–35]. This method could not only be used to solve integer-order differential equations [36–38] but can also get exact solutions to nonlinear fractional-order equations [39–44]. Moreover, besides obtaining the exact solutions, Kai et al. also find that this method could be used to get qualitative properties to nonlinear equations [45, 46]. So, in this paper, we firstly transform Equation (3) into an ordinary differential equation by a special kind of fractional complex traveling wave transformation and then use the complete discrimination system for polynomial method of 5th order to get the classification of the traveling wave solutions to it. Some new solutions such as hyperelliptic function solutions are given, and concrete examples with concrete parameters are also presented to ensure the existence of each solution obtained in the paper. This indicates that the complete discrimination system for polynomial method is efficient and powerful in handing the nonlinear problems arising in mathematics and physics.The construction of the present paper is as follows. We first give a brief introduction to the complete discrimination system for polynomial method in Section2, and then transform the original equation into the ordinary differential equation and solve it in Section 3. Moreover, concrete examples are also presented in this section. In the final, a conclusion is drawn in Section 4.


## 2. Complete Discrimination System for Polynomial Method

The main idea of the complete discrimination system for polynomial method can be concluded as follows. When we encounter with a nonlinear differential equation(4)Fu,ut,uz,utt,⋯=0,then we need to take a traveling wave transformation
(5)ξ=kx+ωt,to transform it into
(6)Fu,u′,u′′,⋯=0.If Equation (6) could be written into the following integral form,
(7)ξ−ξ0=∫duGu,where Gu could be any type of function such as rational function. Then, we can just obtain the classification of single traveling wave solutions to the original equation by discussing the relationship between the roots and the parameters of Gu. We can see it clearly that the steps of using complete discrimination system for polynomial method can be given as follows:Step 1.
Take the traveling wave transformation to transform the original equation into the ordinary equation.Step 2.
Rewrite the equation into the integral form.Step 3.
Obtain the exact solutions to the original equation by discussing the relationship between the roots and the coefficients.In the next section, we will use this method to obtain the classification of traveling wave solutions to Equation (3).


## 3. The Classification of the Traveling Wave Solutions to Equation (3)

To solve Equation (3), we first take the following fractional complex transformation:
(8)uz,t=Fξexpiθ,where
(9)ξ=kαz+1Γαα+υt+1Γαα,θ=−καz+1Γαα+ωt+1Γαα,andθ, κ, ω, v, and k are all real constants. Then by substituting Equation (9) into Equation (3), we can get the following:
(10)v=2κρ,for the imaginary part and
(11)−κ2ρ+ωF+αF2+βF3+γF4+k2r+ρF′′=0,for the real part. Multiplying both sides with F′ and integrating it once yields
(12)−κ2ρ+ω2F2+α3F3+β4F4+γ5F5+k2r+ρ2F′2=c,then we have the following:
(13)F′2=a5F5+a4F4+a3F3+a2F2+a0,where a5=−2γ/5k2r+ρ, a4=−β/2k2r+ρ, a3=−2α/3k3r+ρ, a2=−κ2ρ+ω/k2r+ρ, and a0=2c/k2r+ρ. Via the transformation
(14)φ=a51/5F+a45a5,ξ1=a51/5ξ,we can obtain the following equation:
(15)φ′2=φ5+b3φ2+b2φ2+b1φ+b0=Gφ,where b3=−2/5a42a4−5/8+a3a5−3/5, b2=4/25a43a5−12/5+a2a5−2/5, b1=−3/125a44a5−16/5+a2a5−2/5, and b0=4/3125a44a5−5+1/25a2a42a5−2+a0. Thus, we can get the following:
(16)±ξ1−ξ0=∫dφGφ,and then the corresponding complete discrimination system is given as follows:
(17)D2=−b3,D3=40b1b3−12b33−45b22,D4=−4b22b33+12b1b34+117b1b22b3−88b12b32−40b0b2b32−27b13b24−300b0b1b2+160b13,D5=−1600b0b2b13−3750b2b3b03+2000b3b02b12−4b12b22b33+16b0b23b33−900b1b02b33+825b02b22b32+144b3b22b13+2250b1b02b22+16b13b34+108b02b35−128b14b22−27b12b24+108b0b25+256b15+3125b04−72b0b1b2b34+560b0b2b12b32−630b3b1b0b23,E2=160b12b33+900b12b22−48b1b35+60b1b22b32+1500b0b1b2b3+16b22b34−1100b0b2b33+625b02b32−3375b0b23,F2=3b22−8b1b3.Now we can apply the complete discrimination system for polynomial method to get the classification of single traveling wave solutions to the original equation, and twelve cases need to be discussed here. What has to be illustrated here is that the values of parameters are not from literature; they are just estimated to show the corresponding solutions directly.Case 1.
D3>0, D4=0, D5=0, and E2≠0. Gφ is given as follows:
(18)G=φ=φ−β12φ−β22φ−β3,where β1, β2, and β3 are the corresponding three roots satisfying 2β1+2β2+β3=0. Then, the solutions in different conditions to the original equation shall be shown as follows:
(19)±β2−β12ξ1−ξ0=β3−β1arctanφ−β3β3−β1−β3−β2arctanφ−β3β3−β2β3>β1,β3>β2,(20)±β2−β12ξ1−ξ0=β3−β1arctanφ−β3β3−β1−1β3−β2lnφ−β3−β2−β3φ−β2+β2−β3β1<β3<β2,(21)±β2−β12ξ1−ξ0=−β3−β2arctanφ−β3β3−β1+1β1−β3lnφ−β3−β1−β3φ−β3+β1−β3β2<β3<β1,and
(22)±β2−β12ξ1−ξ0=12β1−β3lnφ−β3−β1−β3φ−β3+β1−β3−12β2−β3lnφ−β3−β2−β3φ−β3+β2−β3β3<β1,β3<β2.
For example, whenk=1, κ=1, r=−2, ρ=1, γ=5, α=−9, β=0, ω=−5, and c=0, namely, b3=−3, b2=2, and b1=b0=0, then we have β1=0, β2=1, and β3=−2 and the solution is as follows:
(23)±ξ−ξ0=122lnφ+2−2φ+2+2−123lnφ+2−3φ+2+3.Case 2.
D2≠0, D3=0, D4=0, D5=0, and F2≠0. Then,
(24)Gφ=φ−β13φ−β22,where β1 and β2 satisfy the relation 3β1+2β2=0. We have the following:
(25)±β2−β12ξ1−ξ0=−1φ−β1−β1−β2arctanφ−β1β1−β2β1>β2,(26)±β2−β12ξ1−ξ0=−1φ−β1−12β2−β1lnφ−β1−β2−β1φ−β1+β1−β3β1<β2.
For instance, whenk=1, κ=10, r=−2, ρ=1, γ=5, α=−15, β=−20, ω=−10, and c=108, namely, b3=−15, b2=10, b1=60, and b0=−72, then we have β1=2, β2=−3, and the following solution:
(27)±52ξ−ξ0=−1φ−2−5arctanφ−25.Case 3.
WhenD2≠0, D3=0, D4=0, D5=0, and F2=0, we have the following:
(28)Fφ=φ−β14φ−β2,where β1 and β2 are the two different real constants satisfying 4β1+β2=0. Then, the solution of Equation (3) can be shown as follows:
(29)±β2−β12ξ1−ξ0=−φ−β22φ−β1−14β1−β2lnφ−β2−β1−β2φ−β2+β1−β2β1>β2,(30)±β2−β12ξ1−ξ0=−φ−β22φ−β1−12β2−β1arctanφ−β2β2−β1β1<β2.
Take the concrete parametersk=1, κ=0, r=−2, ρ=1, γ=5, α=0, β=20, ω=0, and c=0 as an illustrative example, which yields b3=−10, b=20, b1=−15, and b0=4. Thus, when β1=1 and β2=−4, then we can get the following:
(31)±52=ξ−ξ0=−φ+42φ−1−125lnφ+4−5φ+4+5.Case 4.
D2=D3=D4=D5=0. Then,
(32)Fφ=φ5.
This leads to the following:(33)±ξ1−ξ0=−23φ−3/2,which yields
(34)φ=32ξ1−ξ02/3.
For example, whenk=0, r=−2, ρ=1, and κ=γ=α=β=ω=c=0, then the corresponding solution is given by Equation (34).Case 5.
D3<0, D4=0, D5=0, and E2=0. Gφ is given as follows:
(35)Gφ=φ−β1φ2+β2φ+β32,where β1, β2, and β3 are the three different roots satisfying β22−4β3<0 and 2β2=β1; thus, we have the following:
(36)±ξ1+ξ0=−2ρ4β3−β22cosδarctan2ρsinδφ−β1φ−β1−ρ2+sinδ2lnφ−β1−ρ2−2ρcosδφ−β1φ−β1−ρ2+2ρcosδφ−β1,where ρ=β12+β1β2+β31/4 and δ=−1/2arctan4β3−β22/2β1+β2.
Take the concrete parametersk=1, κ=1, r=−2, ρ=1, γ=5, α=87/16, β=0, ω=21/8, and c=−1/8 as an illustrative example; then, b3=−29/16, b2=−17/32, b1=0, and b0=−1/8; thus, β1=1/2 and β2=β3=1/4; then, we can get the following:
(37)±ξ−ξ0=−8ρ17cosδarctan2ρsinδφ−1/2φ−1/2−ρ2+sinδ2lnφ−1/2−ρ2−2ρcosδφ−1/2φ−1/2−ρ2+2ρcosδφ−1/2,where ρ=5/81/4 and δ=−1/2arctan17/5.Case 6.
WhenD4>0 and D5=0, we have the following:
(38)Gφ=φ−β12φ−β2φ−β3φ−β4,where βii=1,2,3,4 are the four distinct real roots satisfying 2β1+β2+β3+β4=0. Thus, we can get the following:
(39)±ξ1−ξ0=−2β1−β3β3−β4Fη,l−β2−β3β2−β1∏η,β2−β3β2−β1,l,where
(40)Fη,l=∫0ηdη1−l2sin2η,(41)∏η,n,l=∫0ηdη1+nsin2η1−l2sin2η.
For instance, whenk=1, κ=−4, r=0, ρ=−2, γ=5, α=−21, β=0, ω=4, and c=0, we have b3=−7, b2=−6, and b1=b0=0; thus, β1=0, β2=1, β3=2, and β4=−3; then, we can attain the solution to the original equation as follows:
(42)±ξ−ξ0=∫dφφ2φ−1φ−2φ+3.
The explicit form of the solution could not be represented by any common function including the Jacobian elliptic function, so we only are given its integral expression here.Case 7.
D3<0 and D4=D5=E2=0. Then, we have the following:
(43)Gφ=φ−β13φ−β22+β32,where β1, β2, and β3 are the corresponding three roots satisfying 3β1+2β2=0.Thus, the solution of Equation (3) can be obtained as follows:
(44)±ξ1−ξ0=tanθ+cotθ2β3tanθ−β2−β1β3/sin32θFη,s−β3tanθ+β3cotθβ3cotθ+β1+β2×tanθ+β1+β2β3cotθ+β2−β1sinη1−k2sin2η+Fη,l−Eη,sβ1≠β2+β3,(45)±ξ1−ξ0=sin32θ4β331sarcsinssinη−Fη,sβ1=β2+β3,where Fη,s is defined by Equation (40), tan2θ=β3/β1−β2, s=sinθ0<θ<π/2, and
(46)Eη,s=∫0η1−s2sin2ζdζ.For instance, whenk=1, κ=0, r=0, ρ=−2, γ=0, α=3, β=0, ω=0, and c=0, then b3=1 and b2=b1=b0=0, namely, β1=β2=0 and β3=1; thus, we have the following:
(47)±ξ−ξ0=∫dφφ3/2φ2+1.Case 8.
WhenD4<0 and D5=0, we have the following:
(48)Gφ=φ−β12φ−β2φ−β32+β42,where βii=1,2,3,4 are the real numbers satisfying 2β1+2β3+β2=0. So, we have the following solutions:
(49)±ξ1−ξ0=tanθ+cotθ2β3tanθ−β2−β1β3/sin32θFη,l−β3tanθ+β3cotθβ3cotθ+β1+β2×tanθ+β1+β2β3cotθ+β2−β1sinη1−s2sin2η+Fη,l−Eη,sβ1≠β3−β4tanθ,β1≠β3+β4cotθ,(50)±ξ1−ξ0=sin32θ4β4312arcsinlsinη−Fη,sβ1=β3−β4tanθ,(51)±ξ1−ξ0=sin32θ4β43Fη,s−11−s2ln1−s2sin2η+1−s2sinηcosηβ1=β3+β4cotθ,where Fη,l is also defined by (40), tan2θ=β4/β2−β1, s=sinθ0<θ<π/2, and Eη,s is given by Equation (46).For instance, whenk=1, κ=0, r=0, ρ=−2, γ=5, α=0, β=0, ω=8, and c=0, then b3=0, b2=−4, and b1=b0=0, namely, β1=0, β2=2, β3=−1, and β4=1; thus, we can get the following solution:
(52)±ξ−ξ0=∫dφφφ3−4.Case 9.
D3>0 and D4=D5=E2=0. Gφ is given as follows:
(53)Gφ=φ−β13φ−β2φ−β3,where β1, β2, and β3 satisfy the relation of 3β1+β2+β3=0. Then, the solution shall be given as follows:
(54)±ξ1−ξ0=1β1−β3Earcsinβ1−β3φ−β3,β2−β3β1−β3−φ−β2φ−β1φ−β2,where E is defined by Equation (46).For instance, whenk=1, κ=0, r=0, ρ=−2, γ=5, α=−3, β=0, ω=0, and c=0, then b3=−1 and b2=b1=b0=0, namely, β1=0, β2=−1, and β3=1; thus, we have the following:
(55)±ξ−ξ0=∫dφφ3/2φ2−1.Case 10.
D2,D3,D4,D5>0. Fφ can be written as follows:
(56)Fφ=φ−β1φ−β2φ−β3φ−β4φ−β5,where βii=1,2,⋯,5 are the different real constants satisfying ∑i=15βi=0. Then, the solution in hyperelliptic function form is given as follows:
(57)±ξ1−ξ0=dφφ−β1φ−β2φ−β3φ−β4φ−β5.
For instance, whenk=1, κ=1, r=−2, ρ=1, γ=5, α=−3, β=−76/5, ω=−2, and c=−36/5, then β1=6/25β2=31/25β3=−44/25β4=−49/25, and β5=56/25; thus, we have the following:
(58)±ξ−ξ0=∫3125dφφ−6φ−31φ+44φ+49φ−56.Case 11.
WhenD2,D3,D4,D5>0, we have the following:
(59)Fφ=φ−β1φ−β2φ−β3φ−β42+β52,where βii=1,2,⋯,5 are the constants satisfying β1+β2+β3+2β4=0 and β1≠β2≠β3. Then, the hyperelliptic function type solution is given as follows:
(60)±ξ1−ξ0=dφφ−β1φ−β2φ−β3φ−β42+β52.
For instance, whenk=1, κ=1, r=−2, ρ=1, γ=5, α=−3, β=−76/5, ω=−2, and c=−36/5, then β1=8/15, β2=23/15, β3=−17/15, β4=−7/15, and β5=1, so the solution to the original equation is given as follows:
(61)±ξ=ξ0=∫dφφ5−7φ4/3+φ3−φ2+4/3.Case 12.
D5>0∧D4≤0∨D3≤0∨D2≤0, where ∧ means “and” and ∨ means “or”. We have the following:
(62)Gφ=φ−β1φ−β22+β32φ−β42+β52,where βii=1,2,⋯,5 are the roots satisfying β1+2β2+2β4=0. Then, we can get the following hyperelliptic functions solution:
(63)±ξ1−ξ0=∫dφφ−β1φ−β22+β32φ−β42+β52.For instance, whenk=1, κ=1, r=−2, ρ=1, γ=5, α=3, β=−4, ω=−3, and c=2, then β1=−6/5, β2=−1/5, β3=1, β4=−4/5, and β5=1, so we have the following solution:
(64)±ξ−ξ0=∫dφφ5+3φ3/5+36φ2/25+62φ/125+6396/3125.From the results above, we can see that all the single traveling wave solutions, namely, the classification of single traveling wave solutions, are obtained. Besides the implicit solutions, Equation (34) is the rational function solution, and Equations (57), (61), and (64) are the hyperelliptic function solutions. Some of the results are new (i.e., Equation (61)), which shows the novelty of the present paper. However, we only show the single traveling wave solutions to Equation (3); other kinds of solutions such as multisoliton solution need to be further studied. Moreover, qualitative analysis to this equation should also be a future direction.


## 4. Graphical Representations

In this section, we give graphical representations to the solutions above. Since almost all solutions are implicit except Equation (34), and we really do not know how to present implicit function, thus we only show Equation (34) here.From the figures above, we can see that the solutions given in the paper could be truly realized. From Figure1, we can see that the amplitude increases with the time and the direction. Thus, the modulus becomes larger and larger with the process of evolution.Figure 1
(a) The 3D graph of rational function solutions (Equation (34)) when α=0.5. (b) The corresponding 2D plot.
(a)(b)


## 5. Conclusion

In this paper, the space-time fractional resonance nonlinear Schrödinger type equation with Atangana’s derivative is considered by the complete discrimination system for polynomial method, and the present work would enrich the results of the fractional derivative, due to the reason that to the best of our knowledge, this is the first time that the hyperelliptic function solutions are obtained to the nonlinear fractional Schrödinger type equation. And this special kind of solution could hardly be obtained by other methods, which means new solutions are found. Moreover, the results presented in this paper show that the space-time fractional resonance nonlinear Schrödinger equation also has triangle function type periodic solutions and rational function type solutions. For each solution obtained in the paper, we have constructed concrete examples with concrete parameters so that it can be seen clearly that all the solutions given in the paper could be realized and the correctness of the results is also proved. Furthermore, considering the significance of the equation, our results would have practical applications in many scientific fields such as fluid dynamics.



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