Home Physical Sciences Bifurcation, chaotic behavior, and traveling wave solutions for the fractional (4+1)-dimensional Davey–Stewartson–Kadomtsev–Petviashvili model
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Bifurcation, chaotic behavior, and traveling wave solutions for the fractional (4+1)-dimensional Davey–Stewartson–Kadomtsev–Petviashvili model

  • Zhao Li EMAIL logo and Yueyong Jiang
Published/Copyright: May 21, 2025

Abstract

This article investigates the traveling wave solution of the fractional (4+1)-dimensional Davey–Stewartson–Kadomtsev–Petviashvili model by using the complete discriminant system method. These solutions not only include rational function solutions, trigonometric function solutions, but also Jacobian function solutions. In order to illustrate the propagation of these solutions in the field of nonlinear optics and water wave models, some three-dimensional, two-dimensional, and contour maps are drawn. Meanwhile, the phase portrait of two-dimensional dynamical systems and its perturbation systems are studied using the planar dynamical system analysis method. By drawing phase diagrams, it is easy to observe the stability, periodicity, and chaotic behavior of two-dimensional dynamical systems through geometric visualization, which can also provide strong basis for researchers to design corresponding control systems.

1 Introduction

The fractional (4+1)-dimensional Davey–Stewartson–Kadomtsev–Petviashvili (DSKP) model is described as follows [1]:

(1.1) 4 D M , x t 2 α , ϒ H D M , x x x y 4 α , ϒ H + D M , x y y y 4 α , ϒ H + 12 D M , x α , ϒ H D M , y α , ϒ H + 12 H D M , x y 2 α , ϒ H 6 D M , z w 2 α , ϒ H = 0 ,

where H = H ( x , y , z , w , t ) is a real-valued function. This equation is a new equation synthesized by the ancient Greek mathematician Fokas by combining the integrable Kadomtsev–Petviashvili (KP) equation and Davey–Stewartson (DS) equation. This equation is an important mathematical model in the fields of nonlinear optics and applied science, such as shallow water wave models, nonlinear phenomena in plasma physics, and ocean wave phenomena. Currently, research on Eq. (1.1) mainly focuses on its traveling wave solution. Alsharidi and Junjuan [1] obtained some soliton solutions of Eq. (1.1) using two different methods. The research in previous studies [26] mainly focuses on the study of traveling wave solutions for (4+1)-dimensional DSKP equation. For example, Ahmad et al. [2] considered the (4+1) DSKP equation using ( g g + g + A ) -expansion method. El-Shorbagy et al. [3] obtained the solitary wave solutions of (4+1)-dimensional DSKP equation using the modified Sardar sub-equation method, the improved F -expansion method. Ahmad et al. [4] studied the exact solutions of the DSKP equation using the modified tanh method along associated with the Ricatti equation. Rehman et al. [5] obtained the solitary wave solution of the (4+1)-dimensional DSKP equation using Sardar sub-equation method, the Kudryashov’s method, and the ( 1 ϑ ( ζ ) , ϑ ( ζ ) ϑ ( ζ ) ) method, respectively. Rabie et al. [6] obtained the soliton solutions of the (4+1)-dimensional DSKP equation using the modified extended mapping method. In recent years, with the rapid development of fractional-order derivatives [7,8], various types of fractional-order derivatives have been proposed, and many experts and scholars are studying the traveling wave solutions [912] and dynamic properties [1316] of fractional partial differential equations. Particularly, the study of fractional (4+1)-dimensional DSKP equation is still a hot topic in current research, and there are still many very important problems that need to be solved urgently. This article will focus on the study of traveling wave solution and dynamic behavior of fractional (4+1)-dimensional DSKP equation.

The remaining sections are described as follows: in Section 2, the phase portraits and chaotic behaviors of two-dimensional dynamical system and its perturbation system are discussed. In Section 3, the solitary wave solutions of the fractional (4+1)-dimensional DSKP equation are constructed. In Section 4, three-dimensional, two-dimensional, and contour maps of some solutions of Eq. (1.1) are drawn. In Section 5, the solution obtained in this article is discussed in relation to existing solutions. In Section 6, a brief conclusion is given.

2 Bifurcation and chaotic behavior

Definition 2.1

[17] Let H : [ 0 , + ) ( , + ) . For 0 < α 1 , a truncated M-fractional derivative is denoted as

D M , x α , ϒ H ( x ) = lim τ 0 H ( x E ϒ ( τ x 1 α ) ) H ( x ) τ , 0 < α 1 , ϒ ( 0 , ) ,

where E ϒ ( ) is a truncated Mittag–Leffler function. The constant α stands for a fractional-order derivative.

Remark 2.2

In definition 2.1, E ϒ ( ) is a truncated Mittag–Leffler function defined as [18]

E ϒ ( z ) = Σ j i x j Γ ( γ j + 1 ) , x [ 0 , + ) .

First, the wave transformation is given as follows:

(2.1) H ( x , y , z , w , t ) = H ( ξ ) , ξ = Γ ( 1 + γ ) α ( a x α + b y α + c z α + τ w α λ t α ) ,

where a , b , c , τ , and λ represent the real constants.

Substituting Eq. (2.1) into Eq. (1.1) and integrating it twice, such that the first integration constant is zero and the second integration constant is non-zero, yields an ordinary differential equation

(2.2) a b ( b 2 a 2 ) H + 6 a b H 2 H ( 4 a λ + 6 c τ ) = C 1 ,

where C 1 is an integral constant.

When a b 3 a 3 b 0 , the two-dimensional dynamic system of Eq. (2.2) can be rewritten as

(2.3) d H d ξ = z , d z d ξ = μ 2 H 2 + μ 1 H + μ 0 ,

with Hamiltonian function

(2.4) F ( H , z ) = 1 2 z 2 1 3 μ 2 H 3 1 2 μ 1 H 2 μ 0 H = f 0 ,

where μ 2 = 6 b 2 a 2 , μ 1 = 4 a λ + 6 c τ a b ( b 2 a 2 ) , μ 0 = C 1 a b ( b 2 a 2 ) . f 0 is an integral constant.

Here, we suppose that G ( H ) = μ 2 H 2 + μ 1 H + μ 0 and G ( H ) = 2 μ 2 H + μ 1 . Furthermore, suppose that M ( H j , 0 ) = 0 1 2 μ 2 H j + μ 1 0 is the coefficient matrix of (2.3) at the equilibrium point, where H j ( j = 1 , 2, 3) is the root of equation G ( H j ) = 0 . Thus, we have

(2.5) det ( M ( H j , 0 ) ) = G ( H j ) = ( 2 μ 2 H j + μ 1 ) .

If μ 1 2 4 μ 0 μ 2 > 0 and μ 2 0 , we can obtain that the equation G ( H ) = 0 has two unequal real roots denoted as H 1 = μ 1 μ 1 2 4 μ 0 μ 2 2 μ 2 and H 2 = μ 1 + μ 1 2 4 μ 0 μ 2 2 μ 2 . If μ 1 2 4 μ 0 μ 2 = 0 and μ 2 0 , we can obtain that the equation G ( H ) = 0 has a real root denoted as H 3 = μ 1 2 μ 2 . According to the theory of planar dynamical systems [1921], we can draw the phase diagram of system (2.3) for different parameters. When μ 1 2 4 μ 0 μ 2 > 0 , we obtain that G ( H 1 ) > 0 , then we have ( H 1 , 0 ) is the saddle point. When μ 1 2 4 μ 0 μ 2 > 0 , we have G ( H 2 ) < 0 ; thus, ( H 2 , 0 ) is the center point. When μ 1 2 4 μ 0 μ 2 = 0 , ( H 3 , 0 ) is the degenerate saddle point. Using Maple software, we have drawn the two-dimensional phase portraits of Eqs (2.3), as shown in Figure 1.

Figure 1 
               2D phase portraits of Eq. (2.3). (a) 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 0
                              
                           
                           =
                           3
                        
                        {\mu }_{0}=3
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 1
                              
                           
                           =
                           6
                        
                        {\mu }_{1}=6
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 2
                              
                           
                           =
                           3
                        
                        {\mu }_{2}=3
                     
                  , (b) 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 0
                              
                           
                           =
                           2
                        
                        {\mu }_{0}=2
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 3
                              
                              
                                 2
                              
                           
                        
                        {\mu }_{1}=\frac{3}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 2
                              
                           
                           =
                           −
                           1
                        
                        {\mu }_{2}=-1
                     
                  , (c) 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 0
                              
                           
                           =
                           
                              
                                 25
                              
                              
                                 8
                              
                           
                        
                        {\mu }_{0}=\frac{25}{8}
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 1
                              
                           
                           =
                           5
                        
                        {\mu }_{1}=5
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 2
                              
                           
                           =
                           2
                        
                        {\mu }_{2}=2
                     
                  , (d) 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 0
                              
                           
                           −
                           
                              
                                 25
                              
                              
                                 16
                              
                           
                        
                        {\mu }_{0}-\frac{25}{16}
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 5
                              
                              
                                 2
                              
                           
                        
                        {\mu }_{1}=\frac{5}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 2
                              
                           
                           =
                           −
                           1
                        
                        {\mu }_{2}=-1
                     
                  .
Figure 1

2D phase portraits of Eq. (2.3). (a) μ 0 = 3 , μ 1 = 6 , μ 2 = 3 , (b) μ 0 = 2 , μ 1 = 3 2 , μ 2 = 1 , (c) μ 0 = 25 8 , μ 1 = 5 , μ 2 = 2 , (d) μ 0 25 16 , μ 1 = 5 2 , μ 2 = 1 .

Next, we directly add a periodic disturbance to the second equation of system (2.3)

(2.6) d H d ξ = z , d z d ξ = μ 2 H 2 + μ 1 H + μ 0 + A sin ( k ξ ) ,

where A and k are the constants (Figure 2).

Figure 2 
               2D and 3D phase portraits of Eq. (2.3). (a) 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 0
                              
                           
                           =
                           2
                        
                        {\mu }_{0}=2
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 1
                              
                           
                           =
                           −
                           
                              
                                 3
                              
                              
                                 2
                              
                           
                        
                        {\mu }_{1}=-\frac{3}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 2
                              
                           
                           =
                           −
                           2
                        
                        {\mu }_{2}=-2
                     
                  , 
                     
                        
                        
                           A
                           =
                           0.38
                        
                        A=0.38
                     
                  , 
                     
                        
                        
                           k
                           =
                           1.35
                        
                        k=1.35
                     
                  , (b) 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 0
                              
                           
                           =
                           2
                        
                        {\mu }_{0}=2
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 1
                              
                           
                           =
                           −
                           
                              
                                 3
                              
                              
                                 2
                              
                           
                        
                        {\mu }_{1}=-\frac{3}{2}
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 2
                              
                           
                           =
                           −
                           2
                        
                        {\mu }_{2}=-2
                     
                  , 
                     
                        
                        
                           A
                           =
                           0.38
                        
                        A=0.38
                     
                  , 
                     
                        
                        
                           k
                           =
                           1.35
                        
                        k=1.35
                     
                  , (c) 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 0
                              
                           
                           =
                           −
                           4
                        
                        {\mu }_{0}=-4
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 1
                              
                           
                           =
                           −
                           5
                        
                        {\mu }_{1}=-5
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 2
                              
                           
                           =
                           2
                        
                        {\mu }_{2}=2
                     
                  , 
                     
                        
                        
                           A
                           =
                           0.38
                        
                        A=0.38
                     
                  , 
                     
                        
                        
                           k
                           =
                           0.6
                        
                        k=0.6
                     
                  , (d) 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 0
                              
                           
                           =
                           −
                           4
                        
                        {\mu }_{0}=-4
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 1
                              
                           
                           =
                           −
                           5
                        
                        {\mu }_{1}=-5
                     
                  , 
                     
                        
                        
                           
                              
                                 μ
                              
                              
                                 2
                              
                           
                           =
                           2
                        
                        {\mu }_{2}=2
                     
                  , 
                     
                        
                        
                           A
                           =
                           0.38
                        
                        A=0.38
                     
                  , 
                     
                        
                        
                           k
                           =
                           0.6
                        
                        k=0.6
                     
                  .
Figure 2

2D and 3D phase portraits of Eq. (2.3). (a) μ 0 = 2 , μ 1 = 3 2 , μ 2 = 2 , A = 0.38 , k = 1.35 , (b) μ 0 = 2 , μ 1 = 3 2 , μ 2 = 2 , A = 0.38 , k = 1.35 , (c) μ 0 = 4 , μ 1 = 5 , μ 2 = 2 , A = 0.38 , k = 0.6 , (d) μ 0 = 4 , μ 1 = 5 , μ 2 = 2 , A = 0.38 , k = 0.6 .

3 Traveling wave solutions of Eq. (1.1)

According to the polynomial complete discriminant system method proposed by Professor Liu [22], we can obtain the single traveling wave solution of Eq. (1.1) in this section. In recent years, many experts [2325] have utilized Professor Liu’s complete discriminant system method to construct single traveling wave solutions for nonlinear partial differential equations.

Multiplying both sides of Eq. (2.2) by H integral simultaneously yields

(3.1) ( H ) 2 = 4 a 2 b 2 H 3 + 2 ( 2 a λ + 3 c τ ) a b ( b 2 a 2 ) H 2 + 2 C 1 a b ( b 2 a 2 ) H + 2 C 2 a b ( b 2 a 2 ) .

Here, we suppose that

(3.2) Ψ = 4 a 2 b 2 1 3 H , χ 2 = 2 ( 2 a λ + 3 c τ ) a b ( b 2 a 2 ) 4 a 2 b 2 2 3 , χ 1 = 2 C 1 a b ( b 2 a 2 ) 4 a 2 b 2 1 3 , χ 0 = 2 C 2 a b ( b 2 a 2 ) .

Substituting Eq. (3.2) into Eq. (3.1), we have

(3.3) ( Ψ ) 2 = Ψ 3 + χ 2 Ψ 2 + χ 1 Ψ + χ 0 ,

here, let us assume that the third-order polynomial is P ( Ψ ) = Ψ 3 + χ 2 Ψ 2 + χ 1 Ψ + χ 0 , and the complete discriminative system of the polynomial is

(3.4) = 27 2 χ 2 3 27 + χ 0 χ 1 χ 2 3 2 4 χ 1 χ 2 2 3 3 , D 1 = χ 1 χ 2 2 3 .

Therefore, the integral expression of Eq. (3.3) is

(3.5) d Ψ Ψ 3 + χ 2 Ψ 2 + χ 1 Ψ + χ 0 = ± 4 a 2 b 2 1 3 ( ξ ξ 0 ) ,

where ξ 0 is an integral constant.

Case 1. = 0 , D 1 < 0 , namely, 27 ( 2 χ 2 3 27 + χ 0 χ 1 χ 2 3 ) 2 4 ( χ 1 χ 2 2 3 ) 3 = 0 , χ 1 χ 2 2 3 < 0 . At this point, the equation P ( Ψ ) = 0 has a double real and a single real root, i.e., P ( Ψ ) = ( Ψ β 1 ) 2 ( Ψ β 2 ) , where β 1 β 2 .

When Ψ > β 2 , the integral Eq. (3.5) can be rewritten as

(3.6) ± 4 a 2 b 2 1 3 ( ξ ξ 0 ) = d Ψ ( Ψ β 1 ) Ψ β 2 = 1 β 1 β 2 ln Ψ β 2 β 1 β 2 Ψ β 2 + β 1 β 2 , ( β 1 > β 2 ) , 2 β 2 β 1 arctan Ψ β 2 β 2 β 1 , ( β 1 < β 2 ) .

From Eqs (2.1) and (3.6), we obtain

(3.7) H 1 ( x , y , z , w , t ) = 4 a 2 b 2 1 3 { ( β 1 β 2 ) tanh 2 β 1 β 2 2 4 a 2 b 2 1 3 × Γ ( 1 + γ ) α ( a x α + b y α + c z α + τ w α λ t α ) ξ 0 + β 2 , β 1 > β 2 ,

(3.8) H 2 ( x , y , z , w , t ) = 4 a 2 b 2 1 3 ( β 1 β 2 ) coth 2 β 1 β 2 2 4 a 2 b 2 1 3 × Γ ( 1 + γ ) α ( a x α + b y α + c z α + τ w α λ t α ) ξ 0 + β 2 , β 1 > β 2 ,

(3.9) H 3 ( x , y , z , w , t ) = 4 a 2 b 2 1 3 { ( β 1 + β 2 ) tan 2 β 1 + β 2 2 4 a 2 b 2 1 3 × Γ ( 1 + γ ) α ( a x α + b y α + c z α + τ w α λ t α ) ξ 0 + β 2 , β 1 < β 2 .

Case 2. = 0 , D 1 = 0 , namely, 27 ( 2 χ 2 3 27 + χ 0 χ 1 χ 2 3 ) 2 4 ( χ 1 χ 2 2 3 ) 3 = 0 , χ 1 χ 2 2 3 = 0 . At this point, the equation P ( Ψ ) = 0 has a triple real root, i.e., P ( Ψ ) = ( Ψ β 3 ) 3 ; and then, we obtain traveling wave solution of Eq. (1.1)

(3.10) H 4 ( x , y , z , w , t ) = 4 4 a 2 b 2 2 3 Γ ( 1 + γ ) α ( a x α + b y α + c z α + τ w α λ t α ) ξ 0 ] 2 + β 3 .

Case 3. > 0 , D 1 < 0 , namely, 27 ( 2 χ 2 3 27 + χ 0 χ 1 χ 2 3 ) 2 4 ( χ 1 χ 2 2 3 ) 3 > 0 , χ 1 χ 2 2 3 < 0 . At this point, the equation P ( Ψ ) = 0 has three different real roots, i.e., P ( Ψ ) = ( Ψ β 4 ) ( Ψ β 5 ) ( Ψ β 6 ) , where β 4 < β 5 < β 6 .

When β 4 < Ψ < β 6 , we make the variable replacement Ψ = β 4 + ( β 5 β 4 ) sin 2 ϑ . From Eq. (3.6), we have

(3.11) ± 4 a 2 b 2 1 3 ( ξ ξ 0 ) = d Ψ P ( Ψ ) = 2 ( β 5 β 4 ) sin ϑ cos ϑ d ϑ β 6 β 4 ( β 5 β 4 ) sin ϑ cos ϑ 1 m 2 sin 2 ϑ = 2 β 6 β 4 d ϑ 1 m 2 sin 2 ϑ ,

where m 2 = β 5 β 4 β 6 β 4 .

(3.12) H 5 ( x , y , z , w , t ) = 4 a 2 b 2 1 3 β 4 + ( β 5 β 4 ) sn 2 β 6 β 4 2 × 4 a 2 b 2 1 3 Γ ( 1 + γ ) α ( a x α + b y α + c z α + τ w α λ t α ) ξ 0 , m .

When Ψ > β 6 , we make the variable replacement Ψ = β 5 sin 2 ϑ + β 6 cos 2 ϑ . From Eq. (3.6), we also obtain the solution of Eq. (1.1)

(3.13) H 6 ( x , y , z , w , t ) = 4 a 2 b 2 1 3 × β 6 β 5 sn 2 β 6 β 4 2 4 a 2 b 2 1 3 ( Γ ( 1 + γ ) α ( a x α + b y α + c z α + τ w α λ t α ) ξ 0 ) , m cn 2 [ β 6 β 4 2 4 a 2 b 2 1 3 ( Γ ( 1 + γ ) α ( a x α + b y α + c z α + τ w α λ t α ) ξ 0 ) , m ] .

Case 4. < 0 , namely, 27 ( 2 χ 2 3 27 + χ 0 χ 1 χ 2 3 ) 2 4 ( χ 1 χ 2 2 3 ) 3 < 0 . At this point, the equation P ( Ψ ) = 0 has one real root, i.e., P ( Ψ ) = ( Ψ β 7 ) ( Ψ 2 + p Ψ + q ) , where p 2 4 q < 0 .

When Ψ > β 7 , we make the variable replacement Ψ = β 7 2 + p β 7 + q tan 2 ϑ 2 . From Eq. (3.6) and the definition of Jacobian elliptic cosine function, we have

(3.14) H 7 ( x , y , z , w , t ) = 4 a 2 b 2 1 3 β 7 + 2 β 7 2 + p β 7 + q 1 + cn 2 ( β 7 2 + p β 7 + q ) 1 4 4 a 2 b 2 1 3 Γ ( 1 + γ ) α ( a x α + b y α + c z α + τ w α λ t α ) ξ 0 , m β 7 2 + p β 7 + q } .

4 Numerical simulation

In this section, we plotted the three-dimensional, two-dimensional, and contour plots of the two solutions H 1 and H 5 of Eq. (1.1) by controlling different parameters, as shown in Figures 3 and 4. From Figure 3, it can be seen that the solution of Eq. (1.1) is a hyperbolic function solution H 1 and a bright soliton solution. From Figure 4, it can be concluded that the solution H 5 of Eq. (1.1) is a periodic wave solution. Figures 3(c) and 4(c) are contour maps indicating the basic contour of the understanding.

Figure 3 
               Graphics of 
                     
                        
                        
                           
                              
                                 H
                              
                              
                                 1
                              
                           
                        
                        {H}_{1}
                     
                   at 
                     
                        
                        
                           a
                           =
                           2
                        
                        a=2
                     
                  , 
                     
                        
                        
                           b
                           =
                           
                              
                                 5
                              
                           
                        
                        b=\sqrt{5}
                     
                  , 
                     
                        
                        
                           c
                           =
                           4
                        
                        c=4
                     
                  , 
                     
                        
                        
                           
                              
                                 χ
                              
                              
                                 2
                              
                           
                           =
                           3
                        
                        {\chi }_{2}=3
                     
                  , 
                     
                        
                        
                           
                              
                                 χ
                              
                              
                                 1
                              
                           
                           =
                           
                              
                                 χ
                              
                              
                                 0
                              
                           
                           =
                           0
                        
                        {\chi }_{1}={\chi }_{0}=0
                     
                  , 
                     
                        
                        
                           λ
                           =
                           3
                        
                        \lambda =3
                     
                  , 
                     
                        
                        
                           τ
                           =
                           
                              
                                 5
                              
                           
                           −
                           1
                        
                        \tau =\sqrt{5}-1
                     
                  , 
                     
                        
                        
                           γ
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \gamma =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           α
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \alpha =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           y
                           =
                           z
                           =
                           w
                           =
                           1
                        
                        y=z=w=1
                     
                  : (a) 3D surface, (b) 2D surface at 
                     
                        
                        
                           t
                           =
                           10
                        
                        t=10
                     
                  , and (c) contour plot.
Figure 3

Graphics of H 1 at a = 2 , b = 5 , c = 4 , χ 2 = 3 , χ 1 = χ 0 = 0 , λ = 3 , τ = 5 1 , γ = 1 2 , α = 1 2 , y = z = w = 1 : (a) 3D surface, (b) 2D surface at t = 10 , and (c) contour plot.

Figure 4 
               Graphics of 
                     
                        
                        
                           
                              
                                 H
                              
                              
                                 5
                              
                           
                        
                        {H}_{5}
                     
                   at 
                     
                        
                        
                           a
                           =
                           
                              
                                 5
                              
                           
                        
                        a=\sqrt{5}
                     
                  , 
                     
                        
                        
                           b
                           =
                           2
                        
                        b=2
                     
                  , 
                     
                        
                        
                           c
                           =
                           −
                           
                              
                                 7
                              
                              
                                 8
                              
                           
                        
                        c=-\frac{7}{8}
                     
                  , 
                     
                        
                        
                           
                              
                                 χ
                              
                              
                                 2
                              
                           
                           =
                           3
                        
                        {\chi }_{2}=3
                     
                  , 
                     
                        
                        
                           
                              
                                 χ
                              
                              
                                 1
                              
                           
                           =
                           0
                        
                        {\chi }_{1}=0
                     
                  , 
                     
                        
                        
                           
                              
                                 χ
                              
                              
                                 0
                              
                           
                           =
                           −
                           2
                        
                        {\chi }_{0}=-2
                     
                  , 
                     
                        
                        
                           λ
                           =
                           1
                        
                        \lambda =1
                     
                  , 
                     
                        
                        
                           τ
                           =
                           
                              
                                 5
                              
                           
                        
                        \tau =\sqrt{5}
                     
                  , 
                     
                        
                        
                           γ
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \gamma =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           α
                           =
                           
                              
                                 1
                              
                              
                                 2
                              
                           
                        
                        \alpha =\frac{1}{2}
                     
                  , 
                     
                        
                        
                           y
                           =
                           z
                           =
                           w
                           =
                           1
                        
                        y=z=w=1
                     
                  : (a) 3D surface, (b) 2D surface at 
                     
                        
                        
                           t
                           =
                           10
                        
                        t=10
                     
                  , and (c) contour plot.
Figure 4

Graphics of H 5 at a = 5 , b = 2 , c = 7 8 , χ 2 = 3 , χ 1 = 0 , χ 0 = 2 , λ = 1 , τ = 5 , γ = 1 2 , α = 1 2 , y = z = w = 1 : (a) 3D surface, (b) 2D surface at t = 10 , and (c) contour plot.

5 Discussion of results

Alsharidi et al. [1] obtained the exact solution of the DSKP equation using both the unified technique and modified extended tanh-expansion function technique, respectively. Compared with Alsharidi et al. [1], this article not only constructs the traveling wave solution of the DSKP equation using the fully discriminative system method, but also obtains the dynamic behavior of the DSKP equation using the planar dynamical system method. The solutions obtained in this article not only include common rational function solutions and trigonometric function solutions, but also more general Jacobian function solutions. In order to obtain the dynamic behavior of the equation more easily, this article also draws three-dimensional and two-dimensional phase diagrams of the two-dimensional dynamical system and its perturbation system. By analyzing the phase diagrams, it is easy to obtain the dynamic behavior of the two-dimensional dynamical system.

6 Conclusion

In this article, we obtained the traveling wave solution of the fractional (4+1)-dimensional DSKP equation. We not only discussed the planar phase portraits of the two-dimensional dynamical system of Eq. (1.1). Moreover, we have added periodic perturbations to the two-dimensional dynamical system and plotted the two-dimensional and three-dimensional phase prrtraits of the perturbed two-dimensional dynamical system. Based on Professor Liu’s complete discriminant system method, we obtained the Jacobian function solution, trigonometric function solution, and hyperbolic function solution of Eq. (1.1). We also simulated the obtained solutions by drawing three-dimensional, two-dimensional, and contour maps. In future work, we will continue to study soliton solutions of fractional-order partial differential equations, such as multi-soliton solutions or experimental validations.

  1. Funding information: This research was supported by the Open Fund of Sichuan Provincial Key Laboratory of Ecological Security and Protection at Mianyang Normal University (Grant No. ESP2404).

  2. Author contributions: Zhao Li: writing – orginal draft; Yueyong Jiang: supervision. 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 datasets were generated or analysed during the current study.

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Received: 2024-11-18
Revised: 2025-03-09
Accepted: 2025-04-26
Published Online: 2025-05-21

© 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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