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Interaction solutions of high-order breathers and lumps for a (3+1)-dimensional conformable fractional potential-YTSF-like model

  • Wen-Yuan Li , Nong-Sen Li , Rui-Gang Zhang , Yi-Lin Tian and Ji-Feng Cui EMAIL logo
Published/Copyright: August 8, 2025

Abstract

In this study, the breather solutions and interaction solutions for the (3+1)-dimensional conformable fractional potential Yu-Toda-Sasa-Fukuyama-like model are conducted. The fractional derivative is described using the conformable derivative. By using certain commutative methods, we derived the breather solutions and the interaction solutions of breathers and lumps in terms of Gramian. Solutions of these forms have never been studied. The features of the solutions were examined. Each interval of the breather has one and only one peak as well as trough, and the interaction between the breather wave and the breather wave or the lump wave is elastic. In addition, the effect of a fractional order change on the shape of the breathers is discussed through 3-dimensional images, and this change is able to be reduced to integer order and different fractional order.

1 Introduction

Fractional differential equations (FDEs) hold a significant place in the field of science, and their frequent use in research and industrial applications has piqued the interest and attention of numerous researchers. These equations have been used in a number of fields, such as telecommunications [1], transportation, optics [2], instruments instrumentation [3], geochemistry geophysics, fluid dynamics [4], and information processing [5]. FDEs represent an extension of integer-order differential equations, suitable for describing materials with historical memory and hereditary properties, and have advantages over integer-order differential models in many fields. Researchers have identified the nonlocality of fractional order differentiation as a key factor in its ability to describe a range of complex mechanical and physical behaviors [6]. Nonlinear fractional partial differential equations (FPDEs), which have been applied in number of fields of physics and engineering including thermodynamics [7], mathematical computational biology [8], optics [9], environmental sciences ecology, and geology [10], require the utilization of scientific fields to identify accurate or approximative solutions to the issues presented by these systems [11]. However, for FPDEs, it is typically impossible to obtain an exact solution. Li and Chen presented Galerkin finite element method, spectral method and the finite difference method as techniques to solve FPDEs [12]. Ammi et al. developed an entirely discrete scheme for solving the time fractional diffusion equation, utilizing time finite difference method and space finite element method [13]. Cen et al. employed the upwind finite difference method to FDEs, offering stability and a post-test error analysis of discrete schemes, in addition to adaptive methods [14]. Bayrak and Demir provided approximations to analytical solutions for arbitrary order space-time FDEs using a semi-analytical method and then obtained fractional power series solutions using residual power series method [15]. Furthermore, they provide stability and error estimates for this scheme. Even when dealing with applications that necessitate the use of algebra or other sciences, it remains challenging to identify exact solutions.

On the other hand, the Bogoyavlenskii-Schiff (BS) equations are widely used to study solitons and nonlinear waves in the region, including complex nonlinear problems in hydrodynamics, weakly dispersive media. Furthermore, the BS equation is deeply related to many of the classical equations in fluid dynamics [16].

(1) 4 υ t + Φ υ υ z = 0 , Φ = 2 + 4 υ + 2 υ x 1 .

Upon making z = x , Eq. (1) becomes the potential Kadomtsev–Petviashvili (KP) equation. Substituting υ y = 0 , Eq. (1) becomes the potential Korteweg–de Vries (KdV) equation. Yu, Toda, Sasa, and Fukuyama in their study of BS equation proposed the following equation [17]:

(2) [ 4 υ t + Φ υ υ z ] x + 3 υ y y = 0 , Φ = 2 + 4 υ + 2 υ x 1 .

As a productive extension of KP equations and BS equations, Yu–Toda–Sasa–Fukuyama (YTSF) equations are commonly used to describe mixed reaction appearing in shallow water equations. They are also employed to study solitons and wave equations or weakly dispersive medium equations in nonlinear dynamics.

The solitary wave solutions of the established model equations are also a topic of great interest in the field of research. In soliton theory, the pursuit of analytical solutions to nonlinear evolution equations represents a pivotal area of inquiry. In the case of high-dimensional equations that describe complex physical phenomena, it is also of great importance to find their solutions. Since Korteweg proposed a solution to the KdV equation, which involved solitary wave solutions, numerous scholars have subsequently proposed various analytical methods with the objective of identifying the nonlinear equations’ exact solutions, such as the Darboux transformation method proposed by Darboux [18], the bilinear transformation method proposed by Hirota [19,20], the simplest equation method [21], the semi-inverse method [22], the Jacobi elliptic function expansion method [23], and the homogeneous balance method [24]. However, the inherent complexity of physical problems, the presence of multiple interacting physical phenomena, and strong nonlinearity make it challenging to devise a method for obtaining unification of the analytical solutions of the resulting nonlinear partial differential equations.

In recent years, scholars have expanded and extended the Hirota bilinear form and its applications to obtain the soliton solutions such as breather solutions and lump solutions [25,26]. For instance, Masayoshi proposed a conjugate parameter method based on the Hirota method, which can transform the N-soliton solution into an N/2-breather solution [27]. Zheng analyzed the breathers and multi-soliton solutions of potential-YTSF equation based on the Hirota bilinear method [28]. Wang studied the soliton, breather and lump solutions of the fractional (2+1)-dimensional Boussinesq equation by the Hirota bilinear method and the variational approach [29]. Ma analyzed the solitons and confirmed the N-soliton condition of the Hirota bilinear form for the B-type KP equation, using the Hirota bilinear formulation [30]. Seadawy presented the application of the Hirota bilinear method, in conjunction with symbolic computation, which results in the following solutions: Y-shape, generalized breather, lump one strip, lump two strip, and lump periodic solution [31]. Jin-Bo et al. presented a bilinear form of the nonisospectral Ablowitz-Kaup-Newell-Segur equation and derived the N-soliton condition solutions by using the Hirota bilinear method [32]. Peng studied the stochastic Schrodinger-Hirota equation in refractive birefringent optical fibers with spatiotemporal chromatic dispersion and nonlinearity of the parametric law [33]. Kudryashov conducted a comprehensive analysis of the compatibility of the overdetermined system of equations, ultimately identifying the existence of optical solitons of the fourth order [34]. Kumar et al. utilized the generalized Kudryashov method and Lie symmetry to transform the specified partial differential equation to a system of ordinary differential equations [35].

The objective of this study is to propose new solutions and method for the resolution of (3+1)-dimensional fractional potential-YTSF equation. The study is structured as follows. Section 2 analyzes the properties of the conformable fractional order and reduces the (3+1)-dimensional conformable fractional equation to integer order. Section 3 obtains the N th order breather solutions based on the KP hierarchy reduction. Section 4 investigates the interaction solutions taking the long-wave limit technique. Section 5 analyses the images of solutions in Sections 3 and 4. Section 6 provides the conclusion.

2 Fractional differential models and Hirota bilinear form

In this study, we focus on the following (3+1)-dimensional fractional potential YTSF-like model which reads [36]

(3) 4 D x t 2 β φ + D x x x z 4 β φ + 4 D x β φ D x z 2 β φ + 2 D x x 2 β φ D z β φ + 3 D y y 2 β φ = 0 ,

where D x β , D y β , D z β , and D t β , order β is a fractional order derivative, 0 < β 1 . φ ( x , y , z , t ) is a differentiable function on four independent variables. x , y , z , and t are independent variables under order β . This model exists in other areas of physics including dispersion relations, plasmas, fluid dynamics, and other areas of physics. A variety of effective methods exist for solving the (3+1)-dimensional YTSF equation, including the extended homoclinic test technique [37], the homoclinic assay, the G′/G-extension method [38], the variational methods based on two-scale fractal complex transforms and variational principles [39], the three-wave method [40], and the modulated phase shift method [41]. In 2014, Khalil and his team initially proposed a novel fractional derivative, designated the “Conformable Derivative” [42]. In Eq. (3), D x t 2 β being a conformable fractional derivative of order 2 β when independent variables x , t are order of 2 β . The same reasoning leads to D x z 4 β being a conformable fractional derivative of order 4 β when x , z are order of 4 β . Unlike classical integer derivative calculus, conformable fractional calculus and conformable fractional derivatives can simulate nonlinear phenomena on the nonsmooth boundaries as well as diffusion phenomenon or heat transmission phenomenon in porous media.

Definition

Assume g : ( 0 , ) R is a function. Conformable fractional derivative is defined as follows when t > 0 and 0 < α 1 :

(4) D t α x ( t ) = lim ε 0 x ( t + ε t 1 α ) x ( t ) ε .

A few properties of conformable fractional derivatives are listed as follows [36].

Assume α ( 0, 1 ] and x and y are α -order differentiable for any independent variable t > 0 . Then

  1. D t α ( a 1 x + a 2 y ) = a 1 D t α x + a 2 D t α y , a 1 , a 2 R .

  2. D t α t r = r t r α for any r R .

  3. D t α λ = 0 , for any constant value functions x ( t ) = λ .

  4. D t α x y = x D t α y + y D t α x .

  5. D t α x y = x D t α y y D t α x y 2 .

  6. If x in this case can be differential, then D t α x ( t ) = t 1 α d x d t .

By using certain commutative methods φ ( x , y , z , t ) = ψ ( x β β , y β β , z β β , t β β ) = ψ ( X , Y , Z , τ ) , and it is used directly to obtain

(5) 4 D X τ 2 ψ + D X X X Z 4 ψ + 4 D X ψ D X Z 2 ψ + 2 D X X 2 ψ D Z ψ + 3 D Y Y 2 ψ = 0 .

Using the transformation ξ = X + ϖ Z in Eq. (5), we obtain

(6) 4 ψ ξ τ + ϖ ψ ξ ξ ξ ξ + 6 ψ ξ ψ ξ ξ + 3 ψ Y Y = 0 .

Through the transformation

(7) ψ = 2 ( ln f ) ξ + ψ 0 ,

substitute Eq. (7) in Eq. (6) to obtain its bilinear form

(8) ( 3 D Y 2 + ϖ D ξ 4 4 D ξ D τ ) f f = 0 ,

where f is a real function, D ξ , D Y , and D τ are the differential bilinear Hirota operators defined by

D ξ k 1 D Y k 2 D τ k 3 ( A B ) = ξ ξ k 1 Y Y k 2 τ τ k 3 A ( ξ , Y , τ ) B ( ξ , Y , τ ) ξ = ξ , Y = Y , τ = τ ,

where A is a function of ξ , Y , and τ ; B is a function of ξ , Y , and τ and k 1 , k 2 , and k 3 are non-negative integers.

3 M th order breather solutions for Eq. (5)

We start with the KP hierarchy of deterministic solutions and use the Gramian equation to deduce a bilinear form of the breathing solution as listed below:

(9) ( D x 1 4 4 D x 1 D x 3 + 3 D x 2 2 ) ϕ ϕ = 0 ,

then the following solution can be obtained in terms of Gramian

(10) ϕ = det 1 r , j N ( m r , j ( n ) )

and the elements of the matrix m r , j ( n ) satisfy the following conditions [43]:

(11) x 1 m r , j ( n ) = Ψ r ( n ) Φ j ( n ) ,

(12) x 2 m r , j ( n ) = Ψ r ( n + 1 ) Φ j ( n ) + Ψ r ( n ) Φ j ( n 1 ) ,

(13) x 3 m r , j ( n ) = Ψ r ( n + 2 ) Φ j ( n ) + Ψ r ( n + 1 ) Φ j ( n 1 ) + Ψ r ( n ) Φ j ( n 2 ) ,

(14) m r , j ( n + 1 ) = m r , j ( n ) + Ψ r ( n ) Φ j ( n ) ,

(15) x i Ψ r ( n ) = Ψ r ( n + i ) , ( i = 1 , 2 , 3 ) ,

(16) x i Φ j ( n ) = Φ j ( n i ) , ( i = 1 , 2 , 3 ) ,

where r and j Z + , Z + is the set of positive integers. And Ψ r ( n ) and Φ j ( n ) satisfy

(17) m r , j ( n ) = δ r j + ( p r + q j ) 1 Ψ r ( n ) Φ j ( n ) ,

(18) Ψ r ( n ) = p r n ( p r + q j ) e ξ r ,

(19) Φ j ( n ) = ( q j ) n e η j ,

(20) ξ r = p r x 1 + p r 2 x 2 + p r 3 x 3 + ξ r 0 ,

(21) η j = q j x 1 q j 2 x 2 + q j 3 x 3 + η j 0 ,

where δ r j is the Kronecker delta notation. Then, compare Eqs (8) and (9), and make f = ϕ , it can be obtained immediately that

(22) x 1 = i ξ , x 2 = i ϖ Y , x 3 = i ϖ τ .

Theorem 1

The breather solution which in bilinear form (8) in the Gramian forms is like

(23) f = δ r j + p r + q r p r + q j e λ r + η j 1 r , j N ,

with

λ r = i p r ξ + i p r 2 ϖ Y i p r 3 ϖ τ + λ r 0 , η j = i q j ξ i q j 2 ϖ Y i q j 3 ϖ τ + η j 0 , p 2 k 1 = Ω k + θ k 2 , p 2 k = Ω k θ k 2 , λ 2 k 0 = λ 2 k 1 0 , q 2 k 1 = Ω k + θ k 2 , q 2 k = Ω k θ k 2 , η 2 k 0 = η 2 k 1 0 ,

where k , r , and j Z + ; N is a positive even number; Ω k , p r , q j , λ r 0 , and η j 0 are plural and θ k are real numbers.

Theorem 2

Equation (7) has a Mth order breather solutions by Theorem 1 of the form

(24) ψ = 2 ( ln f ) ξ + ψ 0 ,

(25) f = Ξ Λ 11 Λ 12 Λ 1 , M Λ 21 Λ 22 Λ 2 , M Λ M , 1 Λ M , 2 Λ M , M ,

with

Λ k , k = 1 θ k e ς k + 1 θ k 1 Ω k Ω k 1 Ω k Ω k 1 θ k e ς k + 1 θ k , Λ k , l = 1 Ω l Ω k + θ k + θ l 2 1 Ω l Ω k + θ k θ l 2 1 Ω k Ω l + θ k θ l 2 1 Ω k Ω l + θ k + θ l 2 ( k l ) , ς k = i θ k ξ 2 i θ k Ω k ϖ Y i θ k ( 3 Ω k 2 + θ k 2 4 ) ϖ τ + ς k 0 , Ξ = e k = 1 M ς k + ς k Π k = 1 M θ k 2 ,

where M = N 2 , k , and l Z + , and ς k 0 = λ k 0 + η k 0 . Because the denominator should not be zero, Ω k is not real. Since 2 ( ln f ) ξ = ( 2 f ξ f ) and Ξ is a function with no dependence on x, so Ξ can be about to be dropped.

Let M = 1 in Eq. (25), to obtain the first-order breather solution

(26) f = 1 + e ς 1 + e ς 1 + [ 1 ( θ 1 Ω 1 Ω 1 ) 2 ] e ς 1 + ς 1 ,

where ς 1 = i θ 1 ξ 2 i θ 1 Ω 1 ϖ Y i θ 1 ( 3 Ω 1 2 + θ 1 2 4 ) ϖ τ + ς 1 0 .

Let M = 2 in Eq. (25), to obtain the second-order breather solution

(27) f = 1 θ 1 e ς 1 + 1 θ 1 1 Ω 1 Ω 1 1 θ 1 + θ 2 2 + Ω 2 Ω 1 1 θ 1 θ 2 2 + Ω 2 Ω 1 1 Ω 1 Ω 1 1 θ 1 e ς 1 + 1 θ 1 1 θ 1 θ 2 2 + Ω 1 Ω 2 1 θ 1 + θ 2 2 + Ω 1 Ω 2 1 θ 1 + θ 2 2 + Ω 1 Ω 2 1 θ 2 θ 1 2 + Ω 1 Ω 2 1 θ 2 e ς 2 + 1 θ 2 1 Ω 2 Ω 2 1 θ 2 θ 1 2 + Ω 2 Ω 1 1 θ 1 + θ 2 2 + Ω 2 Ω 1 1 Ω 2 Ω 2 1 θ 2 e ς 2 + 1 θ 2 ,

where

ς 1 = i θ 1 ξ 2 i θ 1 Ω 1 ϖ Y i θ 1 ( 3 Ω 1 2 + θ 1 2 4 ) ϖ τ + ς 1 0 , ς 2 = i θ 2 ξ 2 i θ 2 Ω 2 ϖ Y i θ 2 ( 3 Ω 2 2 + θ 2 2 4 ) ϖ τ + ς 2 0 .

4 Interaction solutions

Theorem 3

Let N = M ˜ + M ˜ and making θ k 0 , which keeps e ς k 0 = 1 by the way. Then, obtain the M ˜ -order lumps and M ˜ -order breathers solutions of the form

(28) f = A M ˜ × M ˜ B M ˜ × M ˜ C M ˜ × M ˜ D M ˜ × M ˜ ,

where M ˜ and M ˜ Z + .

A k , k = μ k 1 Ω k Ω k 1 Ω k Ω k μ k ,

A k , l = 1 Ω l Ω k 1 Ω l Ω k 1 Ω k Ω l 1 Ω k Ω l ,

B k , l = 1 Ω l Ω k + θ l 2 1 Ω l Ω k θ l 2 1 Ω k Ω l θ l 2 1 Ω k Ω l + θ l 2 ,

C k , l = 1 Ω l Ω k + θ k 2 1 Ω l Ω k + θ k 2 1 Ω k Ω l + θ k 2 1 Ω k Ω l + θ k 2 ,

D k , k = 1 θ k + e ς k + 1 θ k 1 Ω k Ω k 1 Ω k Ω k 1 θ k + e ς k + 1 θ k ,

D k , l = 1 Ω l Ω k + θ k + θ l 2 1 Ω l Ω k + θ k θ l 2 1 Ω k Ω l θ k + θ l 2 1 Ω k Ω l + θ k + θ l 2 ,

where

μ k = i ξ 2 i Ω k ϖ Y 3 i Ω k 2 ϖ τ , ς k = i θ k ξ 2 i θ k Ω k ϖ Y i θ k ( 3 Ω k 2 + θ k 2 4 ) ϖ τ + ς k 0 .

Let M ˜ = M ˜ = 1 , then the solution of Eq. (5) in the form of a first-order breather and one lump will be obtained

(29) f = μ 1 1 Ω 1 Ω 1 1 Ω 2 Ω 1 + θ 2 2 1 Ω 2 Ω 1 θ 2 2 1 Ω 1 Ω 1 μ 1 1 Ω 1 Ω 2 θ 2 2 1 Ω 1 Ω 2 + θ 2 2 1 Ω 1 Ω 2 + θ 2 2 1 Ω 1 Ω 2 + θ 2 2 1 θ 2 e ς 2 + 1 θ 2 1 Ω 2 Ω 2 1 Ω 2 Ω 1 + θ 2 2 1 Ω 2 Ω 1 + θ 2 2 1 Ω 2 Ω 2 1 θ 2 e ς 2 + 1 θ 2 ,

where

μ 1 = i ξ 2 i Ω 1 ϖ Y 3 i Ω 1 2 ϖ τ , ς 2 = i θ 2 ξ 2 i θ 2 Ω 2 ϖ Y i θ 2 ( 3 Ω 2 2 + θ 2 2 4 ) ϖ τ + ς 2 0 .

5 Discussion

Figure 1 presents the dissemination of the first-order breather within the x y plane under the influence of different fractional-order derivative. In each interval, the breathers only have a peak and a trough. Figure 2 shows the dissemination of the first-order breather within the z y plane under the influence of different fractional-order derivative. Similar to the integer-order case, the breather propagates along the x -axis with no movement along the y -axis. The situation within the z y plane is similar to the situation within the x y plane. The further from the origin, the smaller the value of β , the longer the interval between reappearances of the breather. Since the image of the wave is continuous and does not change traits so the part where t < 0 can be easily inferred. Figure 3 presents the dissemination of the first-order breather within the x z plane under the influence of different fractional-order derivative. The breather within the x z plane appear periodically like soliton. Similar to Figures 1 and 2, the value of β and the distance to the origin influence the interval between reappearances of the breather. Since the image of the wave is continuous and does not change traits, so the part where t < 0 can be easily inferred. Figure 4 presents the first-order breather within the x y and z y planes. It can be seen as the case of integer order, but the case of fractional order like x < 0 or z < 0 or t < 0 needs to be discarded. Comparing Figure 4 with Figures 1, 2, and 3, it can be observed that the movement of the breathers is very similar, with only significant differences in the intervals.

Figure 1 
               The impact of different values of 
                     
                        
                        
                           β
                        
                        \beta 
                     
                   on the results of the first-order breather with 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 1
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{1}^{0}=0
                     
                  , 
                     
                        
                        
                           θ
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \theta =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           ϖ
                           =
                           −
                           2
                        
                        \varpi =-2
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 1
                              
                           
                           =
                           i
                        
                        {\Omega }_{1}=i
                     
                  , and 
                     
                        
                        
                           z
                           =
                           0
                        
                        z=0
                     
                  . (a) 
                     
                        
                        
                           β
                           =
                           1
                        
                        \beta =1
                     
                  , (b) 
                     
                        
                        
                           β
                           =
                           
                              
                                 7
                              
                              
                                 8
                              
                           
                        
                        \beta =\frac{7}{8}
                     
                  , (c) 
                     
                        
                        
                           β
                           =
                           
                              
                                 3
                              
                              
                                 5
                              
                           
                        
                        \beta =\frac{3}{5}
                     
                  .
Figure 1

The impact of different values of β on the results of the first-order breather with ς 1 0 = 0 , θ = 1 2 , ϖ = 2 , Ω 1 = i , and z = 0 . (a) β = 1 , (b) β = 7 8 , (c) β = 3 5 .

Figure 2 
               The impact of different values of 
                     
                        
                        
                           β
                        
                        \beta 
                     
                   on the results of the first-order breather with 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 1
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{1}^{0}=0
                     
                  , 
                     
                        
                        
                           θ
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \theta =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           ϖ
                           =
                           −
                           2
                        
                        \varpi =-2
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 1
                              
                           
                           =
                           i
                        
                        {\Omega }_{1}=i
                     
                  , and 
                     
                        
                        
                           x
                           =
                           0
                        
                        x=0
                     
                  . (a) 
                     
                        
                        
                           β
                           =
                           1
                        
                        \beta =1
                     
                  , (b) 
                     
                        
                        
                           β
                           =
                           
                              
                                 7
                              
                              
                                 8
                              
                           
                        
                        \beta =\frac{7}{8}
                     
                  , (c) 
                     
                        
                        
                           β
                           =
                           
                              
                                 3
                              
                              
                                 5
                              
                           
                        
                        \beta =\frac{3}{5}
                     
                  .
Figure 2

The impact of different values of β on the results of the first-order breather with ς 1 0 = 0 , θ = 1 2 , ϖ = 2 , Ω 1 = i , and x = 0 . (a) β = 1 , (b) β = 7 8 , (c) β = 3 5 .

Figure 3 
               The impact of different values of 
                     
                        
                        
                           β
                        
                        \beta 
                     
                   on the results of the first-order breather with 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 1
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{1}^{0}=0
                     
                  , 
                     
                        
                        
                           θ
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \theta =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           ϖ
                           =
                           −
                           2
                        
                        \varpi =-2
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 1
                              
                           
                           =
                           i
                        
                        {\Omega }_{1}=i
                     
                  , and 
                     
                        
                        
                           y
                           =
                           0
                        
                        y=0
                     
                  . (a) 
                     
                        
                        
                           β
                           =
                           1
                        
                        \beta =1
                     
                  , (b) 
                     
                        
                        
                           β
                           =
                           
                              
                                 7
                              
                              
                                 8
                              
                           
                        
                        \beta =\frac{7}{8}
                     
                  , (c) 
                     
                        
                        
                           β
                           =
                           
                              
                                 3
                              
                              
                                 5
                              
                           
                        
                        \beta =\frac{3}{5}
                     
                  .
Figure 3

The impact of different values of β on the results of the first-order breather with ς 1 0 = 0 , θ = 1 2 , ϖ = 2 , Ω 1 = i , and y = 0 . (a) β = 1 , (b) β = 7 8 , (c) β = 3 5 .

Figure 4 
               The first-order breather on the results of the first-order breather with 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 1
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{1}^{0}=0
                     
                  , 
                     
                        
                        
                           θ
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \theta =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           β
                           =
                           1
                        
                        \beta =1
                     
                  , 
                     
                        
                        
                           ϖ
                           =
                           −
                           2
                        
                        \varpi =-2
                     
                  , and 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 1
                              
                           
                           =
                           i
                        
                        {\Omega }_{1}=i
                     
                  . (a) 
                     
                        
                        
                           z
                           =
                           0
                        
                        z=0
                     
                  , (b) 
                     
                        
                        
                           x
                           =
                           0
                        
                        x=0
                     
                  .
Figure 4

The first-order breather on the results of the first-order breather with ς 1 0 = 0 , θ = 1 2 , β = 1 , ϖ = 2 , and Ω 1 = i . (a) z = 0 , (b) x = 0 .

Figure 5 presents the two breathers interaction within the x y plane. Although only half of the breather propagating around the x -axis can be observed, starting from τ = 0 , the breather propagating around the y -axis can be gradually observed. For each interval, the breathers all have a peak and a trough during the interaction process, and the effect of β is similar to that in Figures 1 and 2. It also shows that two breathers holds their shapes, which indicates that interaction is elastic, allowing either to expand to the other side of the image when t < 0 through these images, or to compare it directly with the integer order part of τ < 0 . Figure 6 presents the case of integer order, how the two breathers interact with each other for τ from 3 to 8. Upon comparing Figure 6 with Figure 5, it can be noted that if we ignore the case when τ < 0 , the situation is still similar to Figures 14, but harder to recognize when τ = 0 .

Figure 5 
               The impact of different values of 
                     
                        
                        
                           β
                        
                        \beta 
                     
                   on the results of two breathers with 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 1
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{1}^{0}=0
                     
                  , 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 2
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{2}^{0}=0
                     
                  , 
                     
                        
                        
                           
                              
                                 θ
                              
                              
                                 1
                              
                           
                           =
                           2
                        
                        {\theta }_{1}=2
                     
                  , 
                     
                        
                        
                           
                              
                                 θ
                              
                              
                                 2
                              
                           
                           =
                           1
                        
                        {\theta }_{2}=1
                     
                  , 
                     
                        
                        
                           ϖ
                           =
                           −
                           1
                        
                        \varpi =-1
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 1
                              
                           
                           =
                           −
                           
                              
                                 i
                              
                              
                                 2
                              
                           
                        
                        {\Omega }_{1}=-\frac{i}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 2
                              
                           
                           =
                           
                              
                                 1
                              
                              
                                 3
                              
                           
                           +
                           i
                        
                        {\Omega }_{2}=\frac{1}{3}+i
                     
                  , and 
                     
                        
                        
                           z
                           =
                           0
                        
                        z=0
                     
                  . (a) 
                     
                        
                        
                           β
                           =
                           1
                        
                        \beta =1
                     
                  , (b) 
                     
                        
                        
                           β
                           =
                           
                              
                                 7
                              
                              
                                 8
                              
                           
                        
                        \beta =\frac{7}{8}
                     
                  , (c) 
                     
                        
                        
                           β
                           =
                           
                              
                                 3
                              
                              
                                 5
                              
                           
                        
                        \beta =\frac{3}{5}
                     
                  .
Figure 5

The impact of different values of β on the results of two breathers with ς 1 0 = 0 , ς 2 0 = 0 , θ 1 = 2 , θ 2 = 1 , ϖ = 1 , Ω 1 = i 2 , Ω 2 = 1 3 + i , and z = 0 . (a) β = 1 , (b) β = 7 8 , (c) β = 3 5 .

Figure 6 
               The first-order breather on the results of two breathers with 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 1
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{1}^{0}=0
                     
                  , 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 2
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{2}^{0}=0
                     
                  , 
                     
                        
                        
                           
                              
                                 θ
                              
                              
                                 1
                              
                           
                           =
                           2
                        
                        {\theta }_{1}=2
                     
                  , 
                     
                        
                        
                           
                              
                                 θ
                              
                              
                                 2
                              
                           
                           =
                           1
                        
                        {\theta }_{2}=1
                     
                  , 
                     
                        
                        
                           ϖ
                           =
                           −
                           1
                        
                        \varpi =-1
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 1
                              
                           
                           =
                           −
                           
                              
                                 i
                              
                              
                                 2
                              
                           
                        
                        {\Omega }_{1}=-\frac{i}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 2
                              
                           
                           =
                           
                              
                                 1
                              
                              
                                 3
                              
                           
                           +
                           i
                        
                        {\Omega }_{2}=\frac{1}{3}+i
                     
                  , 
                     
                        
                        
                           β
                           =
                           1
                        
                        \beta =1
                     
                  , and 
                     
                        
                        
                           z
                           =
                           0
                        
                        z=0
                     
                  .
Figure 6

The first-order breather on the results of two breathers with ς 1 0 = 0 , ς 2 0 = 0 , θ 1 = 2 , θ 2 = 1 , ϖ = 1 , Ω 1 = i 2 , Ω 2 = 1 3 + i , β = 1 , and z = 0 .

Figure 7 presents the interactions solutions of a first-order breather and one lump within the x y plane, while the breather propagates in y directions and the lumps propagate at different speeds. In lump solutions, β has a similar effect as β in the breather solutions. Such a interaction is obviously elastic, so allowing either to compare it directly with the case of integer orders, or to expand the other side of the image when t < 0 from the τ > 0 part; however, the second method is somewhat unintuitive. Figure 8 presents the interactions in the case of integer order, for τ from 3 to 8, the interaction of the lumps and the breathers can be observed. The shapes of the lumps and the breathers do not change before contact and after separation.

Figure 7 
               The impact of various values of 
                     
                        
                        
                           β
                        
                        \beta 
                     
                   on the results of a first-order breather and one lump with 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 2
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{2}^{0}=0
                     
                  , 
                     
                        
                        
                           θ
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \theta =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           ϖ
                           =
                           −
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \varpi =-\frac{1}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                           +
                           
                              
                                 i
                              
                              
                                 2
                              
                           
                        
                        {\Omega }_{1}=\frac{1}{2}+\frac{i}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 2
                              
                           
                           =
                           
                              
                                 1
                              
                              
                                 10
                              
                           
                           −
                           i
                        
                        {\Omega }_{2}=\frac{1}{10}-i
                     
                  , and 
                     
                        
                        
                           z
                           =
                           0
                        
                        z=0
                     
                  . (a) 
                     
                        
                        
                           β
                           =
                           1
                        
                        \beta =1
                     
                  , (b) 
                     
                        
                        
                           β
                           =
                           
                              
                                 7
                              
                              
                                 8
                              
                           
                        
                        \beta =\frac{7}{8}
                     
                  , (c) 
                     
                        
                        
                           β
                           =
                           
                              
                                 3
                              
                              
                                 5
                              
                           
                        
                        \beta =\frac{3}{5}
                     
                  .
Figure 7

The impact of various values of β on the results of a first-order breather and one lump with ς 2 0 = 0 , θ = 1 2 , ϖ = 1 2 , Ω 1 = 1 2 + i 2 , Ω 2 = 1 10 i , and z = 0 . (a) β = 1 , (b) β = 7 8 , (c) β = 3 5 .

Figure 8 
               The first-order breather on the results of a first-order breather and one lump with 
                     
                        
                        
                           
                              
                                 ς
                              
                              
                                 2
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\varsigma }_{2}^{0}=0
                     
                  , 
                     
                        
                        
                           θ
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \theta =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           ϖ
                           =
                           −
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \varpi =-\frac{1}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                           +
                           
                              
                                 i
                              
                              
                                 2
                              
                           
                        
                        {\Omega }_{1}=\frac{1}{2}+\frac{i}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 Ω
                              
                              
                                 2
                              
                           
                           =
                           
                              
                                 1
                              
                              
                                 10
                              
                           
                           −
                           i
                        
                        {\Omega }_{2}=\frac{1}{10}-i
                     
                  , 
                     
                        
                        
                           β
                           =
                           1
                        
                        \beta =1
                     
                  , and 
                     
                        
                        
                           z
                           =
                           0
                        
                        z=0
                     
                  .
Figure 8

The first-order breather on the results of a first-order breather and one lump with ς 2 0 = 0 , θ = 1 2 , ϖ = 1 2 , Ω 1 = 1 2 + i 2 , Ω 2 = 1 10 i , β = 1 , and z = 0 .

6 Conclusions

In this work, the (3+1)-dimensional fractional potential-YTSF-like model is studied. The transformation of Eq. (5) was utilized to effect the transformation of the (3+1)-dimensional fractional potential-YTSF-like equation into the (3+1)-dimensional potential-YTSF Eq. (6), thus resulting in the subsequent acquisition of its Hirota bilinear form. By using the KP hierarchy reduction, we have obtained the breather solutions in terms of Gramian and the N -order interaction solutions. From Figures 14, it can be observed that a first-order breather exhibits a single peak and trough. Figures 5 and 6 illustrate the interaction between two breathers that are propagated, respectively, along the x -axis and the y -axis. The long-wave limit technique is used to solve solution Eq. (24), which introduces the interaction solution Eq. (28). Figures 7 and 8 illustrate the interaction of the lumps and breathers, which undergo a shape change before contact and after separation. This indicates that the interaction is elastic. It is our hope that these results will contribute to the further exploration of phenomena in physical systems. The propagation behavior is found to be nonlinear and complex for different conformable fractional order parameters and amplitudes.

The wave solution of fractional order can be seen as a modified result of the potential-YTSF equation. The Gramian solution is distinguished from the conventional travelling wave solution in that the x , y , and t obtained in terms of Gramian are independent of each other. The study of the Gramian solution can facilitate the design process in related industries. Complex propagation behavior which is nonlinear is also able to be observed for different fractional dimensions and amplitudes. The effects of these variables on the structure of the solution are different, and they are better at simulating water waves in real situations. The objective of this study is to contribute to the advancement of research in the field of physical systems. However, some of the other soliton type solution methods for this type of fractional order equations where other soliton-type solution methods are applicable do not allow the use of Gramian methods. We will be looking at the additional new solutions to the fractional order-YTSF equation as well as applying the variational method to other equations with different fractional order definitions.

  1. Funding information: This work was supported by the National Natural Science Foundation of China (Approval Nos. 12062018, 12172333), Program for Young Talents of Science and Technology in Universities of Inner Mongolia Autonomous Region (Approval No. NJYT22075).

  2. Author contributions: Wenyuan Li: writing – original draft preparation, formal analysis and investigation, writing – review and editing. Nongsen Li: methodology, software. Ruigang Zhang: writing – review and editing, funding acquisition, methodology. Yilin Tian: methodology, software. Jifeng Cui: supervision, writing – review and editing, funding acquisition. All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: The authors state no conflict of interest.

  4. Data availability statement: Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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Received: 2024-11-08
Revised: 2025-04-08
Accepted: 2025-06-27
Published Online: 2025-08-08

© 2025 the author(s), published by De Gruyter

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

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