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Novel soliton solutions for the fractional three-wave resonant interaction equations

  • Sahar M. Alqaraleh EMAIL logo and Adeeb G. Talafha
Published/Copyright: September 22, 2022
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Demonstratio Mathematica
From the journal Demonstratio Mathematica Volume 55 Issue 1

Abstract

In this article, we obtained new infinite sets of exact soliton solutions for the nonlinear evolution system of three-wave resonant interaction equations. The solved system contains the non-zero second-order dispersion coefficients, the non-zero phase velocity mismatch, and the conformable fractional time derivative of order between zero and one. The solution method is a constructed ansatz that consists of linear combinations of the tan and cotan hyperbolic functions with complex coefficients. We stated clear systematic steps toward writing an exact soliton solution for the studied system. To show the efficiency of this method, we introduced some numerical examples on each obtained set of solutions. The computations showed that similar solutions can be obtained if one replaces the tan and cotan hyperbolic functions with the tan and cotan trigonometric functions. The new obtained fractional solutions could be useful in studying the broad applications of triad resonances in plasma physics and in nonlinear optics.

Keywords: conformable fractional derivative; ansatz method
MSC 2010: 35C08; 76B15; 35Q51

1 Introduction

The system of three-wave-resonant-interaction (TWRI) equations is a mathematical model that consists of three nonlinear evolution partial differential equations (PDEs). It has many applications in applied mathematics, fluid dynamics, optics waves, plasma, and soliton physics. It describes the profile of three propagating waves moving in the same direction with different phases k 1 ≠ k 2 ≠ k 3 , frequencies ω 1 ≠ ω 2 ≠ ω 3 , and velocities v 1 ≠ v 2 ≠ v 3 . Previous numerical and analytical approaches [1] toward studying the behavior of its solutions showed that: under some restrictions the solutions may develop singularities, in this case, we say that the solutions of the system are explosive waves. Or they may produce reflection waves, in this case, we say that the solutions of the system are backscattering waves. Or they may produce waves that maintain similar shapes before and after interacting with each other; in this case, we say that the solutions of the system are soliton waves.

In this article, we will consider a system that utilizes soliton wave solutions. The soliton behavior of the propagated waves is a physical phenomenon that was experimentally and theoretically noted [2]. This behavior is described as follows [1]: we have three disjoint propagating waves moving in the same direction, so that the slower wave is ahead and followed by the faster one. The interaction is turned on when the faster wave catches the slower one, it lasts for some short time interval in a space region called the interaction zone, it becomes over when the faster wave passed the slower one, and it ends when the three-waves are separated again. After that, all waves reconstruct their original shapes, but with small shifts in their phases; they keep moving in the same direction until they disappear after enough time interval.

To obtain a closer look at this behavior we need to obtain the exact solutions for this mathematical model. Many numerical and analytical methods have been used to obtain soliton solutions for the system of TWRI equations, such as the inverse scattering transform method [1], the ansatz method [3], the Darboux dressing transformation method [4], the extended tan and tan hyperbolic methods [5,6,7], the spectral method [8], the Hirota method [9], and the δ -dressing method [10].

To obtain an efficient nonlinear interaction between the propagated waves, we will assume that:

  1. The phase velocity mismatch Δ k = k 3 − k 2 − k 1 of the three-waves is nearly zero. This condition is generally the practical case [11].

  2. The considered waves are short enough so that the effects of the second-order dispersion coefficients g 1 , g 2 , and g 3 are not negligible.

  3. The interaction occurs in a dispersive medium that admits some changes in the waves’ shapes, where the index of refraction depends on the waves’ frequencies.

Examples on interacting waves with a mismatch in their group velocities are as follows: The interaction between short laser waves propagating so that the sum frequency generation and the second harmonic processes are expected [12]. Also, the interaction between the radio waves where the change in their shapes, the interchange in their energies, and the formulation of an intense pulse occur in the interaction zone [13].

Since few decades ago, the applications of fractional derivatives have an extensive area of research in pure and applied mathematics. The various definitions of the fractional and the conformable fractional derivatives have impressed the scientific researchers while they were studying the features of the nonlinear wave interaction phenomenon [14]. There are a lot of articles in the literature which applied these definitions. Most of these articles studied some derived systems of PDEs but with fractional derivatives instead of the classical derivatives. For example, in [15] an obtained result showed that the fractional derivative has a significant effect on the dynamics of the interaction of fractional two-sided Kdv wave equation. In [16], some conservation laws are obtained for a system of coupled fractional Kdv equations. In [17], numerical solutions and interesting properties are obtained for the time fractional Fredholm PDEs of Dirichlet function type. In [18], some soliton solutions for the fractional Schrödinger model are investigated. In [19], some numerical solutions for the fractional Kundu-Eckhaus models are obtained. In [20], some analytical solutions for the fractional Lax, Kaup-Kupershmidt, and Sawada-Kotera-Ito equations are obtained and studied. In [21], more applications of the fractional derivative could be seen and reviewed.

The motivation of this article is to modify the mathematical model which represents the TWRI equations to include the conformable fractional time derivative with order α ∈ ( 0 , 1 ] instead of the partial derivative with order α = 1 , then to see whether the modified system will still have exact soliton solutions or not.

The aim of this article is to obtain exact soliton solutions for the modified system. Achieving this aim will answer many research questions that may arise, such as: Are there some changes in the behavior of the solution? What are the applications of the modified system? Can we note some new physical interesting phenomena for the modified system? What are the differences and the similarities between the new and the previous solutions? To answer all these open questions and probably many more, we need to find the exact solutions for the modified system.

The order of this article is as follows: Section 1 is the introduction. Section 2 is a recall of the definition of the conformable fractional derivative and some of its basic properties. Section 3 is the statement of the problem. Section 4 is the methodology of obtaining the exact soliton solutions for the modified system. Section 5 is the introductory of systematic steps toward getting analytic soliton solutions. Section 6 is a statement of four numerical examples that show the efficiency of the used method. Section 7 is a brief statement of other similar groups of solutions and the future work. Finally, Section 8 is the conclusion.

2 Conformable fractional derivative

In this section, we state the definition and some properties of the conformable fractional derivative which will be used throughout the rest of this article.

Definition

[14,22] Let u ( t ) : R + ⟶ R be a given function of time t , the conformable fractional derivative D t α ( u ( t ) ) of order α ∈ ( 0 , 1 ] is defined by:

(1) D t α ( u ( t ) ) = Lim ε ⟶ 0 u ( t + ε t 1 − α ) − u ( t ) ε .

The following theorem states some basic properties for the conformable fractional derivative:

Theorem

[14,22] If D t α ( u ( t ) ) and D t α ( v ( t ) ) exist ∀ t > 0 , then

  1. D t α ( . ) is a linear operator: D t α ( c 1 u ( t ) + c 2 v ( t ) ) = c 1 D t α ( u ( t ) ) + c 2 D t α ( v ( t ) ) , ∀ c 1 , c 2 ∈ R .

  2. D t α ( . ) satisfies the multiplication rule for the derivative:

    D t α ( u ( t ) v ( t ) ) = u ( t ) D t α ( v ( t ) ) + v ( t ) D t α ( u ( t ) ) .

  3. D t α ( . ) satisfies the quotient rule for the derivative: D t α u ( t ) v ( t ) = v ( t ) D t α ( u ( t ) ) − u ( t ) D t α ( v ( t ) ) ( v ( t ) ) 2 .

  4. D t α ( . ) operates on t n the same way as the first derivative does: D t α ( t n ) = n t n − α .

  5. D t α ( . ) operates on the composition of two functions by the relation:

    D t α ( ( u ∘ v ) ( t ) ) = t 1 − α u ′ ( v ( t ) ) v ′ ( t ) .

Corollary

  1. If we consider n = α , then by (4) we have: D t α ( t α ) = α .

  2. If we consider v ( t ) = t , then by (5) we have: D t α ( u ( t ) ) = t 1 − α u ′ ( t ) .

3 Problem statement

The nonlinear evolution system of TWRI equations with positive second-order dispersion coefficients and with non-zero phase velocity mismatch has been stated in [7,8] based on the Zakharov-Manakov derivation procedure. This system is a mathematical model that describes the propagation of three solitary waves that travel in a dispersive medium. In this article, we generalized the order of the time derivative that appeared in this system from the first order α = 1 to the conformable fractional order α ∈ ( 0 , 1 ] , then solved the resulting system. Throughout this article, we will consider j = 1 , 2 , 3 and r = 0 , 1 , 2 . Let A j ( x , t ) be the complex profile of the propagated j th wave in which absolute value is the slowly varying wave’s packet; the propagation direction is toward the positive x -axis, and Lim t → ± ∞ A j ( x , t ) = 0 . This condition means that the shapes of the three profiles disappear before and after a sufficient time. The modified mathematical model for the nonlinear evolution system of TWRI equations is then given by [3,11,23]:

(2) I ∂ x A 1 + 1 v 1 D t α A 1 − g 1 D t 2 α A 1 = − σ A 3 A 2 ∗ e I Δ k x , I ∂ x A 2 + 1 v 2 D t α A 2 − g 2 D t 2 α A 2 = − σ ω 2 ω 1 A 3 A 1 ∗ e I Δ k x , I ∂ x A 3 + 1 v 3 D t α A 3 − g 3 D t 2 α A 3 = − σ ω 3 ω 1 A 1 A 2 e − I Δ k x ,

where v j is the group velocities of the j th wave, g j is the positive second-order dispersion coefficients, σ is the nonlinear coupling constant given by σ ≈ 2 π χ n l ω 1 2 k 1 c 2 , where χ n l is the nonlinear dielectric susceptibility, ω j is the frequency of the j th wave, k j is the j th wave number given by k j = n j ω j c , n j is the j th wave refractive index, c is the speed of light, Δ k is the non-zero phase mismatch given by Δ k = k 3 − k 1 − k 2 → 0 , D t 2 α is the second-order conformable fractional time derivative, A j ∗ is the complex conjugate of the j th wave, and I 2 = − 1 . If we put

(3) A 1 ( x , t ) = ω 1 σ ω 2 ω 3 U 1 ( x , t ) , A 2 ( x , t ) = ω 1 σ ω 3 U 2 ( x , t ) , A 3 ( x , t ) = − ω 1 σ ω 2 U 3 ∗ ( x , t ) ,

and consider the complex conjugate of the third equation, then the system in (2) becomes

(4) I ∂ x U 1 + 1 v 1 D t α U 1 − g 1 D t 2 α U 1 = U 2 ∗ U 3 ∗ e I Δ k x , I ∂ x U 2 + 1 v 2 D t α U 2 − g 2 D t 2 α U 2 = U 1 ∗ U 3 ∗ e I Δ k x , I ∂ x U 3 + 1 v 3 D t α U 3 + g 3 D t 2 α U 3 = − U 1 ∗ U 2 ∗ e I Δ k x .

Many special cases of the system in (4) were studied previously:

  1. In [3,4, 5], the ansatz method and the generalized tanh method were used to obtain soliton solutions for the case α = 1 , g j ≠ 0 , Δ k ≠ 0 .

  2. In [6,7, 8], many methods were used to obtain soliton solutions for the case α = 1 , g j = 0 = Δ k .

  3. In [23], the unified method was used to obtain soliton solutions for the case α ∈ ( 0 , 1 ] , g j = 0 = Δ k .

  4. In this article, we will use the ansatz method to obtain soliton solutions for the case α ∈ ( 0 , 1 ] , g j ≠ 0 , Δ k ≠ 0 .

The originality of this article is considered in case (4) which is considered the most general case, since in our new soliton solutions: if we put α = 1 , then we will obtain the solutions given in case (1), if we put α = 1 , g j = 0 = Δ k , then we will obtain the solutions given in case (2), and if we put g j = 0 = Δ k , then we will obtain the solutions given in case (3).

4 Methodology: the ansatz method

Many numerical and analytical methods were used to solve some special cases of the system in (4); for example, in [24,25, 26,27,28, 29,30], one can review these methods and see the obtained solutions. In this article, our methodology is to use a solving method called the ansatz method. The main idea of this method is to transform a given system of PDEs which is usually nonlinear, into a system of ordinary differential equations (ODEs). Based on some experience and previous studies, we suggest a form for the solution of the produced system. This form consists of complex constants multiplied by some trigonometric or hyperbolic functions, then trying to find these constants. This method has many advantages such as it is efficient, direct, easily applicable, and usually produces an infinite set of exact solutions. However, many disadvantages may arise, such as the solutions may have different formula than the suggested one, or the produced system of nonlinear algebraic equations may have no solutions.

In this article, we suggest an ansatz that consists of a linear combination of the tan and cotan hyperbolic square functions. We successfully obtain an infinite set of exact soliton solutions for the system given in (4), by using this ansatz. The ansatz method starts by assuming that

(5) ξ ( x , t ) = m 1 x + m 2 t α α , η ( x , t ) = x ,

where m 1 and m 2 are real constants to be found, the derivatives in (4) can be rewritten as:

(6) ∂ x U j = ∂ U j ∂ η + m 1 ∂ U j ∂ ξ , D t α U j = m 2 ∂ U j ∂ ξ , D t 2 α U j = m 2 2 ∂ 2 U j ∂ ξ 2 .

If we substitute (6) in (4), then the system in (4) will be transformed into the following system:

(7) I ∂ U 1 ∂ η + γ 1 ∂ U 1 ∂ ξ − g 1 m 2 2 ∂ 2 U 1 ∂ ξ 2 = U 3 ∗ U 2 ∗ e I Δ k η , I ∂ U 2 ∂ η + γ 2 ∂ U 2 ∂ ξ − g 2 m 2 2 ∂ 2 U 2 ∂ ξ 2 = U 3 ∗ U 1 ∗ e I Δ k η , I ∂ U 3 ∂ η + γ 3 ∂ U 3 ∂ ξ + g 3 m 2 2 ∂ 2 U 3 ∂ ξ 2 = − U 1 ∗ U 2 ∗ e I Δ k η , γ j = m 1 + m 2 v j .

The system in (7) can be transformed into a system of ODEs by defining the following functions:

(8) θ ( ξ , η ) = Ω ξ + L η , θ 1 ( ξ , η ) = Ω 1 ξ + L 1 η , θ 2 ( ξ , η ) = Ω 2 ξ + L 2 η , θ 3 ( ξ , η ) = − ( Ω 1 + Ω 2 ) ξ + ( Δ k − L 1 − L 2 ) η ,

and

(9) U j ( ξ , η ) = Q j ( θ ( ξ , η ) ) e I θ j ( ξ , η ) = Q j ( θ ) e I θ j ,

where { Ω , L , Ω 1 , Ω 2 , L 1 , L 2 } are real constants to be found, and θ 3 ( ξ , η ) is defined so that the term e I Δ k η = e I ( θ 1 + θ 2 + θ 3 ) ; this will make the exponential term in (7) equal to one because when we substitute the above functions in (7); a term in the form e I ( θ 1 + θ 2 + θ 3 − Δ k η ) will appear.

If we use the functions defined in (8) and (9), then the partial derivatives in (7) can be written as:

(10) ∂ U 1 ∂ η = ( L Q 1 ′ ( θ ) + I L 1 Q 1 ( θ ) ) e I θ 1 , ∂ U 2 ∂ η = ( L Q 2 ′ ( θ ) + I L 2 Q 2 ( θ ) ) e I θ 2 , ∂ U 3 ∂ η = ( L Q 3 ′ ( θ ) + I ( Δ k − L 1 − L 2 ) Q 3 ( θ ) ) e I θ 3 , ∂ U 1 ∂ ξ = ( Ω Q 1 ′ ( θ ) + I Ω 1 Q 1 ( θ ) ) e I θ 1 , ∂ U 2 ∂ ξ = ( Ω Q 2 ′ ( θ ) + I Ω 2 Q 2 ( θ ) ) e I θ 2 , ∂ U 3 ∂ ξ = ( Ω Q 3 ′ ( θ ) − I ( Ω 1 + Ω 2 ) Q 3 ( θ ) ) e I θ 3 , ∂ 2 U 1 ∂ ξ 2 = ( Ω 2 Q 1 ″ ( θ ) + 2 I Ω Ω 1 Q 1 ′ ( θ ) − Ω 1 2 Q 1 ( θ ) ) e I θ 1 , ∂ 2 U 2 ∂ ξ 2 = ( Ω 2 Q 2 ″ ( θ ) + 2 I Ω Ω 2 Q 2 ′ ( θ ) − Ω 2 2 Q 2 ( θ ) ) e I θ 2 , ∂ 2 U 3 ∂ ξ 2 = ( Ω 2 Q 3 ″ ( θ ) − 2 I Ω ( Ω 1 + Ω 2 ) Q 3 ′ ( θ ) − ( Ω 1 + Ω 2 ) 2 Q 3 ( θ ) ) e I θ 3 .

If we substitute (10) in (7), then the system in (7) will be transformed into the following system of ODEs:

(11) − τ 1 Q 1 ″ ( θ ) + I χ 1 Q 1 ′ ( θ ) + Φ 1 Q 1 ( θ ) − Q 2 ∗ ( θ ) Q 3 ∗ ( θ ) = 0 , − τ 2 Q 2 ″ ( θ ) + I χ 2 Q 2 ′ ( θ ) + Φ 2 Q 2 ( θ ) − Q 3 ∗ ( θ ) Q 1 ∗ ( θ ) = 0 , τ 3 Q 3 ″ ( θ ) + I χ 3 Q 3 ′ ( θ ) + Φ 3 Q 3 ( θ ) + Q 1 ∗ ( θ ) Q 2 ∗ ( θ ) = 0 ,

where

(12) τ r = Ω 2 g r m 2 2 , { χ 1 , χ 2 , χ 3 } = { L + Ω γ 1 − 2 Ω g 1 m 2 2 Ω 1 , L + Ω γ 2 − 2 Ω g 2 m 2 2 Ω 2 , L + Ω γ 3 − 2 Ω g 3 m 2 2 ( Ω 1 + Ω 2 ) } , { Φ 1 , Φ 2 , Φ 3 } = { − L 1 − Ω 1 ( γ 1 − g 1 m 2 2 Ω 1 ) , − L 2 − Ω 2 ( γ 2 − g 2 m 2 2 Ω 2 ) , − Δ k + L 1 + L 2 + ( Ω 1 + Ω 2 ) ( γ 3 − g 3 m 2 2 ( Ω 1 + Ω 2 ) ) } .

If later we are able to find some values for { τ j , χ j , Φ j } , then we can substitute these values in (12) to obtain { Ω , Ω 1 , Ω 2 , L , L 1 , L 2 , m 1 , m 2 } . To solve the system of ODEs in (11), we use the following ansatz:

(13) Q 1 ( θ ) = A 0 + A 1 tanh 2 ( θ ) + A 2 coth 2 ( θ ) , Q 2 ( θ ) = B 0 + B 1 tanh 2 ( θ ) + B 2 coth 2 ( θ ) , Q 3 ( θ ) = C 0 + C 1 tanh 2 ( θ ) + C 2 coth 2 ( θ ) ,

where { A r , B r , C r } are in general complex numbers to be found, if we substitute (13) in (11), and use the identities tanh n ( θ ) = tanh n − 2 ( θ ) ( 1 − sech 2 ( θ ) ) and coth n ( θ ) = coth n − 2 ( θ ) ( 1 + csch 2 ( θ ) ) , then (11) will become in the form of some algebraic terms multiplied by ( sech 2 ( θ ) , csch 2 ( θ ) , sech 4 ( θ ) , csch 4 ( θ ) , coth ( θ ) csch 2 ( θ ) ), and ( tanh ( θ ) sech 2 ( θ ) ), if we make these coefficients zero, then we will obtain the following algebraic system:

The constant coefficients are as follows:

(14) ( A 0 + A 1 + A 2 ) Φ 1 − ( B 0 ∗ + B 1 ∗ + B 2 ∗ ) ( C 0 ∗ + C 1 ∗ + C 2 ∗ ) = 0 , ( B 0 + B 1 + B 2 ) Φ 2 − ( A 0 ∗ + A 1 ∗ + A 2 ∗ ) ( C 0 ∗ + C 1 ∗ + C 2 ∗ ) = 0 , ( C 0 + C 1 + C 2 ) Φ 3 + ( A 0 ∗ + A 1 ∗ + A 2 ∗ ) ( B 0 ∗ + B 1 ∗ + B 2 ∗ ) = 0 .

The coefficients of sech 2 ( θ ) are as follows:

(15) A 1 ( 4 τ 1 − Φ 1 ) + B 0 ∗ C 1 ∗ + B 1 ∗ ( C 0 ∗ + 2 C 1 ∗ ) = 0 , B 1 ( 4 τ 2 − Φ 2 ) + A 0 ∗ C 1 ∗ + A 1 ∗ ( C 0 ∗ + 2 C 1 ∗ ) = 0 , C 1 ( 4 τ 3 + Φ 3 ) + A 0 ∗ B 1 ∗ + A 1 ∗ ( B 0 ∗ + 2 B 1 ∗ ) = 0 .

The coefficients of csch 2 ( θ ) are as follows:

(16) A 2 ( 4 τ 1 − Φ 1 ) + B 0 ∗ C 2 ∗ + B 2 ∗ ( C 0 ∗ + 2 C 2 ∗ ) = 0 , B 2 ( 4 τ 2 − Φ 2 ) + A 0 ∗ C 2 ∗ + A 2 ∗ ( C 0 ∗ + 2 C 2 ∗ ) = 0 , C 2 ( 4 τ 3 + Φ 3 ) + A 0 ∗ B 2 ∗ + A 2 ∗ ( B 0 ∗ + 2 B 2 ∗ ) = 0 .

The coefficients of sech 4 ( θ ) are as follows:

(17) 6 A 1 τ 1 + B 1 ∗ C 1 ∗ = 0 , 6 B 1 τ 2 + A 1 ∗ C 1 ∗ = 0 , 6 C 1 τ 3 + A 1 ∗ B 1 ∗ = 0 .

The coefficients of csch 4 ( θ ) are as follows:

(18) 6 A 2 τ 1 + B 2 ∗ C 2 ∗ = 0 , 6 B 2 τ 2 + A 2 ∗ C 2 ∗ = 0 , 6 C 2 τ 3 + A 2 ∗ B 2 ∗ = 0 .

The coefficients of coth ( θ ) csch 2 ( θ ) and tanh ( θ ) sech 2 ( θ ) are as follows:

(19) 2 I A 2 χ 1 = 0 , 2 I B 2 χ 2 = 0 , − 2 I C 2 χ 3 = 0 , 2 I A 1 χ 1 = 0 , 2 I B 1 χ 2 = 0 , 2 I C 1 χ 3 = 0 .

5 Soliton solutions

The algebraic system in (14)–(19) is a nonlinear system of 21 equations with 17 unknowns; namely { A r , B r , C r } and { m 1 , m 2 , Ω , L , Ω 1 , Ω 2 , L 1 , L 2 } which are embedded in { Φ j , χ j , τ j } . By Mathematica software, we found some sets of solutions for the system in (14)–(19), these solutions can be obtained by applying the following steps:

Step 1: Choose some values for the inputs { v j , g j , Δ k } of the system in (4), and choose some arbitrary real values for { m 1 , m 2 } .

Step 2: Put { γ j , χ j , τ j } = { m 1 + m 2 v j , 0 , Ω 2 m 2 2 g j : ∀ j } , where Ω is a real constant to be found.

Step 3: Consider Φ j as given from one of the following groups of solutions stated in (23)–(58).

Step 4: Substitute the values for γ j , χ j , and Φ j in (12), to obtain:

(20) ( L + Ω γ 1 − 2 Ω g 1 m 2 2 Ω 1 ) = χ 1 = 0 , ( L + Ω γ 2 − 2 Ω g 2 m 2 2 Ω 2 ) = χ 2 = 0 , ( L + Ω γ 3 − 2 Ω g 3 m 2 2 ( Ω 1 + Ω 2 ) ) = χ 3 = 0 , ( − L 1 − Ω 1 ( γ 1 − g 1 m 2 2 Ω 1 ) ) = Φ 1 , ( − L 2 − Ω 2 ( γ 2 − g 2 m 2 2 Ω 2 ) ) = Φ 2 , ( − Δ k + L 1 + L 2 + ( Ω 1 + Ω 2 ) ( γ 3 − g 3 m 2 2 ( Ω 1 + Ω 2 ) ) ) = Φ 3 ,

then solve (20) for { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } .

Step 5: Substitute the functions defined in (5), and the values of { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } obtained from Step 4, in (8) to obtain the following functions of x and t :

(21) θ ( x , t ) = Ω ξ + L η = Ω m 1 x + m 2 t α α + L x = ( L + m 1 Ω ) x + m 2 Ω α t α , θ 1 ( x , t ) = Ω 1 ξ + L 1 η = Ω 1 m 1 x + m 2 t α α + L 1 x = ( L 1 + m 1 Ω 1 ) x + m 2 Ω 1 α t α , θ 2 ( x , t ) = Ω 2 ξ + L 2 η = Ω 2 m 1 x + m 2 t α α + L 2 x = ( L 2 + m 1 Ω 2 ) x + m 2 Ω 2 α t α , θ 3 ( x , t ) = − ( Ω 1 + Ω 2 ) ξ + ( Δ k − L 1 − L 2 ) η = − ( Ω 1 + Ω 2 ) m 1 x + m 2 t α α + ( Δ k − L 1 − L 2 ) x = ( Δ k − L 1 − L 2 − m 1 ( Ω 1 + Ω 2 ) ) x − m 2 ( Ω 1 + Ω 2 ) α t α .

Step 6: Choose δ j so that δ 1 + δ 2 + δ 3 = 2 n π , n ∈ N , and put

(22) A r = a r τ 2 τ 3 e I δ 1 , B r = b r τ 1 τ 3 e I δ 2 , C r = c r τ 1 τ 2 e I δ 3 ,

where { a r , b r , c r } are the values given in the same set of solutions which was considered to obtain Φ j in Step 3.

Step 7: Substitute the function θ ( x , t ) from (21), and the values of { A r , B r , C r } obtained in Step 6, in (13) to obtain Q j ( x , t ) .

Step 8: Substitute the functions θ j ( x , t ) from (21), and the functions Q j ( x , t ) obtained in Step 7, in (9) to obtain U j ( x , t ) , which are the solution of the system in (4).

5.1 Groups of solutions

Group 1: { a 0 , b 0 } = { 0 , 0 } , and

(23) { c 0 = − 12 , a 1 = 6 , b 1 = − 6 , c 1 = 6 , a 2 = − 6 , b 2 = 6 , c 2 = 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = 8 τ 3 } ,

(24) { c 0 = − 12 , a 1 = − 6 , b 1 = 6 , c 1 = 6 , a 2 = 6 , b 2 = − 6 , c 2 = 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = 8 τ 3 } ,

(25) { c 0 = 12 , a 1 = 6 , b 1 = 6 , c 1 = − 6 , a 2 = − 6 , b 2 = − 6 , c 2 = − 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = 8 τ 3 } ,

(26) { c 0 = 12 , a 1 = − 6 , b 1 = − 6 , c 1 = − 6 , a 2 = 6 , b 2 = 6 , c 2 = − 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = 8 τ 3 } .

Group 2: { a 0 , c 0 } = { 0 , 0 } , and

(27) { b 0 = − 12 , a 1 = 6 , b 1 = 6 , c 1 = − 6 , a 2 = − 6 , b 2 = 6 , c 2 = 6 , Φ 1 = 4 τ 1 , Φ 2 = − 8 τ 2 , Φ 3 = − 4 τ 3 } ,

(28) { b 0 = − 12 , a 1 = − 6 , b 1 = 6 , c 1 = 6 , a 2 = 6 , b 2 = 6 , c 2 = − 6 , Φ 1 = 4 τ 1 , Φ 2 = − 8 τ 2 , Φ 3 = − 4 τ 3 } ,

(29) { b 0 = 12 , a 1 = 6 , b 1 = − 6 , c 1 = 6 , a 2 = − 6 , b 2 = − 6 , c 2 = − 6 , Φ 1 = 4 τ 1 , Φ 2 = − 8 τ 2 , Φ 3 = − 4 τ 3 } ,

(30) { b 0 = 12 , a 1 = − 6 , b 1 = − 6 , c 1 = − 6 , a 2 = 6 , b 2 = − 6 , c 2 = 6 , Φ 1 = 4 τ 1 , Φ 2 = − 8 τ 2 , Φ 3 = − 4 τ 3 } .

Group 3: { b 0 , c 0 } = { 0 , 0 } , and

(31) { a 0 = − 12 , a 1 = 6 , b 1 = − 6 , c 1 = 6 , a 2 = 6 , b 2 = 6 , c 2 = − 6 , Φ 1 = − 8 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(32) { a 0 = − 12 , a 1 = 6 , b 1 = 6 , c 1 = − 6 , a 2 = 6 , b 2 = − 6 , c 2 = 6 , Φ 1 = − 8 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(33) { a 0 = 12 , a 1 = − 6 , b 1 = − 6 , c 1 = − 6 , a 2 = − 6 , b 2 = 6 , c 2 = 6 , Φ 1 = − 8 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(34) { a 0 = 12 , a 1 = − 6 , b 1 = 6 , c 1 = 6 , a 2 = − 6 , b 2 = − 6 , c 2 = − 6 , Φ 1 = − 8 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } .

Group 4: { a 1 , b 1 , c 1 } = { 0 , 0 , 0 } , and

(35) { a 0 = − 6 , b 0 = − 6 , c 0 = 6 , a 2 = 6 , b 2 = 6 , c 2 = − 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(36) { a 0 = − 6 , b 0 = 6 , c 0 = − 6 , a 2 = 6 , b 2 = − 6 , c 2 = 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(37) { a 0 = 6 , b 0 = − 6 , c 0 = − 6 , a 2 = − 6 , b 2 = 6 , c 2 = 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(38) { a 0 = 6 , b 0 = 6 , c 0 = 6 , a 2 = − 6 , b 2 = − 6 , c 2 = − 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(39) { a 0 = − 2 , b 0 = − 2 , c 0 = 2 , a 2 = 6 , b 2 = 6 , c 2 = − 6 , Φ 1 = − 4 τ 1 , Φ 2 = − 4 τ 2 , Φ 3 = 4 τ 3 } ,

(40) { a 0 = − 2 , b 0 = 2 , c 0 = − 2 , a 2 = 6 , b 2 = − 6 , c 2 = 6 , Φ 1 = − 4 τ 1 , Φ 2 = − 4 τ 2 , Φ 3 = 4 τ 3 } ,

(41) { a 0 = 2 , b 0 = − 2 , c 0 = − 2 , a 2 = − 6 , b 2 = 6 , c 2 = 6 , Φ 1 = − 4 τ 1 , Φ 2 = − 4 τ 2 , Φ 3 = 4 τ 3 } ,

(42) { a 0 = 2 , b 0 = 2 , c 0 = 2 , a 2 = − 6 , b 2 = − 6 , c 2 = − 6 , Φ 1 = − 4 τ 1 , Φ 2 = − 4 τ 2 , Φ 3 = 4 τ 3 } .

Group 5: { a 2 , b 2 , c 2 } = { 0 , 0 , 0 } , and

(43) { a 0 = − 6 , b 0 = − 6 , c 0 = 6 , a 1 = 6 , b 1 = 6 , c 1 = − 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(44) { a 0 = − 6 , b 0 = 6 , c 0 = − 6 , a 1 = 6 , b 1 = − 6 , c 1 = 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(45) { a 0 = 6 , b 0 = − 6 , c 0 = − 6 , a 1 = − 6 , b 1 = 6 , c 1 = 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(46) { a 0 = 6 , b 0 = 6 , c 0 = 6 , a 1 = − 6 , b 1 = − 6 , c 1 = − 6 , Φ 1 = 4 τ 1 , Φ 2 = 4 τ 2 , Φ 3 = − 4 τ 3 } ,

(47) { a 0 = − 2 , b 0 = − 2 , c 0 = 2 , a 1 = 6 , b 1 = 6 , c 1 = − 6 , Φ 1 = − 4 τ 1 , Φ 2 = − 4 τ 2 , Φ 3 = 4 τ 3 } ,

(48) { a 0 = − 2 , b 0 = 2 , c 0 = − 2 , a 1 = 6 , b 1 = − 6 , c 1 = 6 , Φ 1 = − 4 τ 1 , Φ 2 = − 4 τ 2 , Φ 3 = 4 τ 3 } ,

(49) { a 0 = 2 , b 0 = − 2 , c 0 = − 2 , a 1 = − 6 , b 1 = 6 , c 1 = 6 , Φ 1 = − 4 τ 1 , Φ 2 = − 4 τ 2 , Φ 3 = 4 τ 3 } ,

(50) { a 0 = 2 , b 0 = 2 , c 0 = 2 , a 1 = − 6 , b 1 = − 6 , c 1 = − 6 , Φ 1 = − 4 τ 1 , Φ 2 = − 4 τ 2 , Φ 3 = 4 τ 3 } .

Group 6:

(51) { a 0 = − 12 , b 0 = − 12 , c 0 = 12 , a 1 = 6 , b 1 = 6 , c 1 = − 6 , a 2 = 6 , b 2 = 6 , c 2 = − 6 , Φ 1 = 16 τ 1 , Φ 2 = 16 τ 2 , Φ 3 = − 16 τ 3 } ,

(52) { a 0 = − 12 , b 0 = 12 , c 0 = − 12 , a 1 = 6 , b 1 = − 6 , c 1 = 6 , a 2 = 6 , b 2 = − 6 , c 2 = 6 , Φ 1 = 16 τ 1 , Φ 2 = 16 τ 2 , Φ 3 = − 16 τ 3 } ,

(53) { a 0 = 12 , b 0 = − 12 , c 0 = − 12 , a 1 = − 6 , b 1 = 6 , c 1 = 6 , a 2 = − 6 , b 2 = 6 , c 2 = 6 , Φ 1 = 16 τ 1 , Φ 2 = 16 τ 2 , Φ 3 = − 16 τ 3 } ,

(54) { a 0 = 12 , b 0 = 12 , c 0 = 12 , a 1 = − 6 , b 1 = − 6 , c 1 = − 6 , a 2 = − 6 , b 2 = − 6 , c 2 = − 6 , Φ 1 = 16 τ 1 , Φ 2 = 16 τ 2 , Φ 3 = − 16 τ 3 } ,

(55) { a 0 = − 4 , b 0 = − 4 , c 0 = − 4 , a 1 = − 6 , b 1 = − 6 , c 1 = − 6 , a 2 = − 6 , b 2 = − 6 , c 2 = − 6 , Φ 1 = − 16 τ 1 , Φ 2 = − 16 τ 2 , Φ 3 = 16 τ 3 } ,

(56) { a 0 = − 4 , b 0 = 4 , c 0 = 4 , a 1 = − 6 , b 1 = 6 , c 1 = 6 , a 2 = − 6 , b 2 = 6 , c 2 = 6 , Φ 1 = − 16 τ 1 , Φ 2 = − 16 τ 2 , Φ 3 = 16 τ 3 } ,

(57) { a 0 = 4 , b 0 = − 4 , c 0 = 4 , a 1 = 6 , b 1 = − 6 , c 1 = 6 , a 2 = 6 , b 2 = − 6 , c 2 = 6 , Φ 1 = − 16 τ 1 , Φ 2 = − 16 τ 2 , Φ 3 = 16 τ 3 } ,

(58) { a 0 = 4 , b 0 = 4 , c 0 = − 4 , a 1 = 6 , b 1 = 6 , c 1 = − 6 , a 2 = 6 , b 2 = 6 , c 2 = − 6 , Φ 1 = − 16 τ 1 , Φ 2 = − 16 τ 2 , Φ 3 = 16 τ 3 } .

6 Examples

In this section, we will use some random numbers for the inputs of the system in (4) and the arbitrary constants, then substitute them in the other variables according to their formula and follow the previous steps successively.

Example 1

If we follow the steps of obtaining the exact soliton solution for the system in (4), we will obtain the following solution:

  1. Choose { m 1 , m 2 , v 1 , v 2 , v 3 , g 1 , g 2 , g 3 , Δ k } randomly as:

    (59) { m 1 , m 2 , v 1 , v 2 , v 3 , g 1 , g 2 , g 3 , Δ k } = { 1.47353 , − 0.816706 , 0.152579 , 2.76826 , 3.72437 , 1.75837 , 2.73009 , 3.24216 , 0.001 } .

  2. Put { γ j , χ j , τ j } as:

    (60) { γ 1 , γ 2 , γ 3 , χ 1 , χ 2 , χ 3 , τ 1 , τ 2 , τ 3 } = { − 3.87914 , 1.1785 , 1.25424 , 0 , 0 , 0 , 1.17285 Ω 2 , 1.82099 Ω 2 , 2.16255 Ω 2 } .

  3. Consider Φ j from the solutions obtained in (34):

    (61) { Φ 1 , Φ 2 , Φ 3 } = { − 8 τ 1 , 4 τ 2 , − 4 τ 3 } = { − 9.38279 Ω 2 , 7.28398 Ω 2 , − 8.6502 Ω 2 } .

  4. Substitute the values of { γ j , χ j , Φ j } in (20) and solve the resulting system to obtain:

    (62) { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } = { 0.684335 , 0.19839 , − 0.183188 , 1.27072 , − 0.30196 , 1.15618 } .

    Note that if we substitute the obtained value of Ω in (60), then we will obtain:

    (63) { τ 1 , τ 2 , τ 3 } = { 0.0461617 , 0.0716718 , 0.085115 } .

  5. Substitute the obtained values of { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } in (21) to obtain θ ( x , t ) and θ j ( x , t ) .

  6. Choose δ j as { δ 1 , δ 2 , δ 3 } = π 7 , π 6 , 71 π 42 and substitute (34) and (63) in (22) to obtain:

    (64) { A 0 , A 1 , A 2 , B 0 , B 1 , B 2 , C 0 , C 1 , C 2 } = { 0.844439 + 0.40666 I , − 0.422219 − 0.20333 I , − 0.422219 − 0.20333 I , 0 . , 0.325706 + 0.188046 I , − 0.325706 − 0.188046 I , 0 . , 0.194411 − 0.285149 I , − 0.194411 + 0.285149 I } .

  7. Substitute (64) and the function θ ( x , t ) in (13) to obtain Q j ( x , t ) .

  8. Substitute the functions θ j ( x , t ) and Q j ( x , t ) in (9) to obtain the following solution for the system in (4):

    (65) U 1 ( x , t ) = e I − 0.571893 x + 0.149611 t α α ( ( 0.844439 + 0.40666 I ) − ( 0.422219 + 0.20333 I ) × coth 2 0.976668 x − 0.162026 t α α − ( 0.422219 + 0.20333 I ) tanh 2 0.976668 x − 0.162026 t α α , U 2 ( x , t ) = e I 3.02861 x − 1.0378 t α α 0 . − ( 0.325706 + 0.188046 I ) coth 2 0.976668 x − 0.162026 t α α + ( 0.325706 + 0.188046 I ) tanh 2 0.976668 x − 0.162026 t α α , U 3 ( x , t ) = e I − 2.45572 x + 0.888193 t α α 0 . − ( 0.194411 − 0.285149 I ) coth 2 0.976668 x − 0.162026 t α α + ( 0.194411 − 0.285149 I ) tanh 2 0.976668 x − 0.162026 t α α .

Example 2

If we follow the steps of obtaining the exact soliton solution for the system in (4), we will obtain the following solution:

  1. Choose { m 1 , m 2 , v 1 , v 2 , v 3 , g 1 , g 2 , g 3 , Δ k } randomly as:

    (66) { m 1 , m 2 , v 1 , v 2 , v 3 , g 1 , g 2 , g 3 , Δ k } = { 1.15207 , 1.8392 , 0.709484 , 1.28149 , 2.45191 , 0.457125 , 1.41295 , 2.1826 , 0.03 } .

  2. Put { γ j , χ j , τ j } as:

    (67) { γ 1 , γ 2 , γ 3 , χ 1 , χ 2 , χ 3 , τ 1 , τ 2 , τ 3 } = { 3.74438 , 2.58728 , 1.90218 , 0 , 0 , 0 , 1.54629 Ω 2 , 4.77951 Ω 2 , 7.38298 Ω 2 } .

  3. Consider Φ j from the solutions obtained in (42):

    (68) { Φ 1 , Φ 2 , Φ 3 } = { − 4 τ 1 , − 4 τ 2 , 4 τ 3 } = { − 6.18517 Ω 2 , − 19.1181 Ω 2 , 29.5319 Ω 2 } .

  4. Substitute the values of { γ j , χ j , Φ j } in (20) and solve the resulting system to obtain:

    (69) { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } = { 0.222893 , − 0.059364 , − 0.00333145 , − 0.122126 , 0.0342884 , 0.454632 } .

    Note that if we substitute the obtained value of Ω in (67), then we will obtain:

    (70) { τ 1 , τ 2 , τ 3 } = { 0.00544926 , 0.0168434 , 0.0260183 } .

  5. Substitute the obtained values of { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } in (21) to obtain θ ( x , t ) and θ j ( x , t ) .

  6. Choose δ j as { δ 1 , δ 2 , δ 3 } = π , π 8 , 7 π 8 , and substitute (42) and (70) in (22) to obtain:

    (71) { A 0 , A 1 , A 2 , B 0 , B 1 , B 2 , C 0 , C 1 , C 2 } = { − 0.0418682 , 0 . , 0.125605 , 0.0220016 + 0.00911334 I , 0 . , − 0.0660047 − 0.02734 I , − 0.0177023 + 0.00733253 I , 0 . , 0.0531069 − 0.0219976 I } .

  7. Substitute (71) and the function θ ( x , t ) in (13) to obtain Q j ( x , t ) .

  8. Substitute the functions θ j ( x , t ) and Q j ( x , t ) in (9) to obtain the following solution for the system in (4):

    (72) U 1 ( x , t ) = e I 0.0304503 x − 0.00612719 t α α − 0.0418682 + 0.125605 coth 2 0.154501 x − 0.109182 t α α , U 2 ( x , t ) = e I 0.313934 x − 0.224613 t α α ( 0.0220016 + 0.00911334 I ) − ( 0.0660047 + 0.02734 I ) coth 2 0.154501 x − 0.109182 t α α , U 3 ( x , t ) = e I − 0.314384 x + 0.230741 t α α ( − 0.0177023 + 0.00733253 I ) + ( 0.0531069 − 0.0219976 I ) coth 2 0.154501 x − 0.109182 t α α .

Example 3

If we follow the steps of obtaining the exact soliton solution for the system in (4), we will obtain the following solution:

  1. Choose { m 1 , m 2 , v 1 , v 2 , v 3 , g 1 , g 2 , g 3 , Δ k } randomly as:

    (73) { m 1 , m 2 , v 1 , v 2 , v 3 , g 1 , g 2 , g 3 , Δ k } = { 1.71221 , 2.72379 , 0.216092 , 2.57857 , 3.57755 , 1.75174 , 2.36382 , 3.43641 , 0.001 } .

  2. Put { γ j , χ j , τ j } as:

    (74) { γ 1 , γ 2 , γ 3 , χ 1 , χ 2 , χ 3 , τ 1 , τ 2 , τ 3 } = { 14.317 , 2.76853 , 2.47357 , 0 , 0 , 0 , 12.9962 Ω 2 , 17.5372 Ω 2 , 25.4948 Ω 2 } .

  3. Consider Φ j from the solutions obtained in (50):

    (75) { Φ 1 , Φ 2 , Φ 3 } = { − 4 τ 1 , − 4 τ 2 , 4 τ 3 } = { − 51.9848 Ω 2 , − 70.1489 Ω 2 , 101.979 Ω 2 } .

  4. Substitute the values of { γ j , χ j , Φ j } in (20) and solve the resulting system to obtain:

    (76) { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } = { − 1.78691 , 0.145636 , 0.0787654 , − 0.270886 , 0.055531 , 3.52466 } .

    Note that if we substitute the obtained value of Ω in (74), then we will obtain:

    (77) { τ 1 , τ 2 , τ 3 } = { 0.275647 , 0.37196 , 0.540738 } .

  5. Substitute the obtained values of { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } in (21) to obtain θ ( x , t ) and θ j ( x , t ) .

  6. Choose δ j as { δ 1 , δ 2 , δ 3 } = { π 3 , π 4 , 17 π 12 } , and substitute (50) and (77) in (22) to obtain:

    (78) { A 0 , A 1 , A 2 , B 0 , B 1 , B 2 , C 0 , C 1 , C 2 } = { 0.448479 + 0.776788 I , − 1.34544 − 2.33036 I , 0 . , 0.54599 , + 0.54599 I − 1.63797 − 1.63797 I , 0 . , − 0.165749 − 0.618584 I , 0.497247 + 1.85575 I , 0 . } .

  7. Substitute (78) and the function θ ( x , t ) in (13) to obtain Q j ( x , t ) .

  8. Substitute the functions θ j ( x , t ) and Q j ( x , t ) in (9) to obtain the following solution for the system in (4):

    (79) U 1 ( x , t ) = e I 0.190394 x + 0.21454 t α α ( 0.448479 + 0.776788 I ) − ( 1.34544 + 2.33036 I ) tanh 2 1.53755 x − 0.396681 t α α , U 2 ( x , t ) = e I 3.06085 x − 0.737835 t α α ( 0.54599 + 0.54599 I ) − ( 1.63797 + 1.63797 I ) tanh 2 1.53755 x − 0.396681 t α α , U 3 ( x , t ) = e I − 3.25024 x + 0.523295 t α α ( − 0.165749 − 0.618584 I ) + ( 0.497247 + 1.85575 I ) tanh 2 1.53755 x − 0.396681 t α α .

Example 4

If we follow the steps of obtaining the exact soliton solution for the system in (4), we will obtain the following solution:

  1. Choose { m 1 , m 2 , v 1 , v 2 , v 3 , g 1 , g 2 , g 3 , Δ k } randomly as:

    (80) { m 1 , m 2 , v 1 , v 2 , v 3 , g 1 , g 2 , g 3 , Δ k } = { − 0.599109 , 1.27332 , 0.413179 , 2.16132 , 3.34004 , 2.33643 , 3.19925 , 3.9334 , 0.001 } .

  2. Put { γ j , χ j , τ j } as:

    (81) { γ 1 , γ 2 , γ 3 , χ 1 , χ 2 , χ 3 , τ 1 , τ 2 , τ 3 } = { 2.48266 , − 0.00996802 , − 0.21788 , 0 , 0 , 0 , 3.78815 Ω 2 , 5.18708 Ω 2 , 6.37739 Ω 2 } .

  3. Consider Φ j from the solutions obtained in (58):

    (82) { Φ 1 , Φ 2 , Φ 3 } = { − 16 τ 1 , − 16 τ 2 , 16 τ 3 } = { − 60.6104 Ω 2 , − 82.9933 Ω 2 , 102.038 Ω 2 } .

  4. Substitute the values of { γ j , χ j , Φ j } in (20) and solve the resulting system to obtain:

    (83) { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } = { 0.0387038 , − 0.016884 , 0.0251207 , − 0.221927 , − 0.0426974 , 0.276918 } .

    Note that if we substitute the obtained value of Ω in (81), then we will obtain:

    (84) { τ 1 , τ 2 , τ 3 } = { 0.00107989 , 0.00147868 , 0.001818 } .

  5. Substitute the obtained values of { L , Ω , Ω 1 , Ω 2 , L 1 , L 2 } in (21) to obtain θ ( x , t ) and θ j ( x , t ) .

  6. Choose δ j as { δ 1 , δ 2 , δ 3 } = π , π 3 , 2 π 3 , and substitute (58) and (84) in (22) to obtain:

    (85) { A 0 , A 1 , A 2 , B 0 , B 1 , B 2 , C 0 , C 1 , C 2 } = { − 0.00655835 , − 0.00983753 , − 0.00983753 , 0.00280231 + 0.00485375 I , 0.00420347 + 0.00728063 I , 0.00420347 + 0.00728063 I , 0.0025273 − 0.00437741 I , 0.00379095 − 0.00656612 I , 0.00379095 − 0.00656612 I } .

  7. Substitute (85) and the function θ ( x , t ) in (13) to obtain Q j ( x , t ) .

  8. Substitute the functions θ j ( x , t ) and Q j ( x , t ) in (9) to obtain the following solution for the system in (4):

    (86) U 1 ( x , t ) = e I − 0.0577474 x + 0.0319867 t α α − 0.00655835 − 0.00983753 coth 2 0.0488192 x − 0.0214988 t α α − 0.00983753 tanh 2 0.0488192 x − 0.0214988 t α α , U 2 ( x , t ) = e I 0.409876 x − 0.282583 t α α ( ( 0.00280231 + 0.00485375 I ) + ( 0.00420347 + 0.00728063 I ) × coth 2 ( 0.0488192 x − 0.0214988 t α α ) + ( 0.00420347 + 0.00728063 I ) tanh 2 0.0488192 x − 0.0214988 t α α , U 3 ( x , t ) = e I − 0.351128 x + 0.250597 t α α ( ( 0.0025273 − 0.00437741 I ) + ( 0.00379095 − 0.00656612 I ) × coth 2 0.0488192 x − 0.0214988 t α α + ( 0.00379095 − 0.00656612 I ) tanh 2 0.0488192 x − 0.0214988 t α α .

7 Similar solutions and future work

If we use any of the following assumptions for { A r , B r , C r } , instead of those introduced in Step (6), we will obtain similar groups of solutions as those obtained in (23)–(58):

(87) A r = a r τ 2 τ 3 e I δ 1 , B r = I b r τ 1 τ 3 e I δ 2 , C r = I c r τ 1 τ 2 e I δ 3 ,

or

(88) A r = I a r τ 2 τ 3 e I δ 1 , B r = b r τ 1 τ 3 e I δ 2 , C r = I c r τ 1 τ 2 e I δ 3 ,

or

(89) A r = I a r τ 2 τ 3 e I δ 1 , B r = I b r τ 1 τ 3 e I δ 2 , C r = c r τ 1 τ 2 e I δ 3 ,

or

(90) A r = a r τ 2 τ 3 , B r = b r τ 1 τ 3 , C r = c r τ 1 τ 2 ,

or

(91) A r = a r τ 2 τ 3 , B r = I b r τ 1 τ 3 , C r = I c r τ 1 τ 2 ,

or

(92) A r = I a r τ 2 τ 3 , B r = b r τ 1 τ 3 , C r = I c r τ 1 τ 2 ,

or

(93) A r = I a r τ 2 τ 3 , B r = I b r τ 1 τ 3 , C r = c r τ 1 τ 2 .

Also, if we use the following ansatz for Q j ( θ ) , instead of the ansatz introduced in (13), and repeat the computations, then we will obtain groups of solutions similar to all groups of solutions obtained in (23)–(58):

(94) Q 1 ( θ ) = A 0 + A 1 tan 2 ( θ ) + A 2 cot 2 ( θ ) , Q 2 ( θ ) = B 0 + B 1 tan 2 ( θ ) + B 2 cot 2 ( θ ) , Q 3 ( θ ) = C 0 + C 1 tan 2 ( θ ) + C 2 cot 2 ( θ ) .

The future work will be to study some physical behavior of the obtained solutions, such as the sensitivity of the three-waves if their phases are amplified or deamplified. In addition to the mentioned applications here and in the references we will look for more interested applications from the point of view that the power of the time is now a fraction. More than that, we are very sure that the used methodology could also be applied to a general form of the main mathematical model given in (2), which is the modified TWRI equations in (3+1) dimensions instead of the ( 1 + 1 ) dimensions, this more general model will be [31,32]:

(95) I c 11 ∂ x A 1 + c 12 ∂ y A 1 + c 13 ∂ z A 1 + 1 v 1 D t α A 1 − g 1 D t 2 α A 1 = − σ A 3 A 2 ∗ e I Δ k x , I c 21 ∂ x A 2 + c 22 ∂ y A 2 + c 23 ∂ z A 2 + 1 v 2 D t α A 2 − g 2 D t 2 α A 2 = − σ ω 2 ω 1 A 3 A 1 ∗ e I Δ k x , I c 31 ∂ x A 3 + c 32 ∂ y A 3 + c 33 ∂ z A 3 + 1 v 3 D t α A 3 − g 3 D t 2 α A 3 = − σ ω 3 ω 1 A 1 A 2 e − I Δ k x .

8 Conclusion

We successfully constructed an ansatz consists of a linear combination of tan and cotan hyperbolic functions with complex coefficients. We solved the conformable fractional time derivative TWRI equations. We obtained infinite sets of exact soliton solutions for this system. We stated systematic steps toward writing a soliton solution for the main problem, and we followed these steps to write some numerical examples. We got the result that: if one replaces the tan and cotan hyperbolic functions with the tan and cotan trigonometric functions, then other infinite sets of exact soliton solutions will be obtained similarly. Eventually, we clearly stated the future work to develop the obtained results and to obtain more interested physical applications.

Acknowledgments

We gratefully acknowledge Al-Hussein Bin Talal University for supporting this article. The authors appreciate the unknown referees’ valuable and profound comments.

  1. Funding information: Authors state no funding involved.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission

  3. Conflict of interest: Authors state no conflict of interest.

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2022-05-02
Revised: 2022-07-07
Accepted: 2022-07-27
Published Online: 2022-09-22

© 2022 Sahar M. Alqaraleh and Adeeb G. Talafha, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/dema-2022-0153
Keywords for this article
conformable fractional derivative; ansatz method
Creative Commons
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Articles in the same Issue

  1. Regular Articles
  2. On some summation formulas
  3. A study of a meromorphic perturbation of the sine family
  4. Asymptotic behavior of even-order noncanonical neutral differential equations
  5. Unconditionally positive NSFD and classical finite difference schemes for biofilm formation on medical implant using Allen-Cahn equation
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