Advancing analytical solutions: Novel wave insights and methodologies for beta fractional Kuralay-II equations

Serife Muge Ege

doi:10.1515/dema-2024-0088

Article Open Access

Advancing analytical solutions: Novel wave insights and methodologies for beta fractional Kuralay-II equations

Serife Muge Ege

Published/Copyright: May 10, 2025

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From the journal Demonstratio Mathematica Volume 58 Issue 1

Abstract

This article investigates new analytical wave solutions within the beta ( β ) fractional framework (F κ IIAE and F κ IIBE) of the Kuralay II equations, which are significant in the field of nonlinear optics. To achieve this, we employ the improved Kudryashov method, the R method, and the Sardar sub-equation method. The study successfully derives a variety of soliton solutions, including dark, bright, singular, and others, some of which are illustrated through 2D and 3D graphics. For the first time, this research graphically demonstrates the impact of the beta fractional derivative on these solutions. The findings provide valuable insights that may aid in the advancement of future models. The methodologies applied here are not only effective and straightforward to implement but also robust for addressing other fractional, nonlinear partial differential equations.

Keywords: fractional Kuralay-II equations (FκIIAE and FκIIBE); improved Kudryashov method; R scheme; Sardar’s subequation scheme

MSC 2010: 35A01; 65L10; 65L12; 65L20; 65L70

1 Introduction

In many studies, methods that obtain soliton-like solutions of the non-linear partial differential model appear. Particularly in studies conducted in recent years, it can be seen that fractional order models are considered. Researchers have developed solution approaches for these models using different fractional derivative definitions. The most commonly used fractional order derivatives include Jumarie’s Riemann Liouville fractional derivative, M fractional derivative, and beta fractional derivative [1–13]. The subequation method [14–16], the generalized Jacobi elliptic function method [17], the modified exponential function method [18,19], the generalized exp − ϕ ( ξ ) − expansion and improved F-expansion method [20], the extended fractional D ξ α G ′ ⁄ G -expansion method [21], the extended rational sinh-cosh method and the modified extended tanh-function method [22,23], Sine-Gordon expansion method [23–25], the exp a function method [26], the general Riccati equation method [27], generalized Kudryashov method [28,29], and Kudryashov R method [30] are some of the methods used to obtain semi-analytical solutions of nonlinear models containing fractional derivatives.

In this work, it is aimed to find dark, bright, singular, and other types of soliton solutions of the fractional derivative version of the Kuralay II model, which is used in various fields such as nonlinear optics, ferromagnetic materials, and optical fibers [26]. Although previous studies have used methods like the new auxiliary equation method [31], the modified expansion method, the new extended auxiliary equation method [32], and the extended simple equation method [33] to explore various soliton structures of the Kuralay II model, there remain unexplored avenues. Solutions with kink wave solitons, peakon solitons,periodic wave solitons, mixed bright and dark solitons, periodic traveling waves, singular solitons, and solitary waves structure have been obtained. Recently, Zafar et al. have obtained soliton-like solutions of the M fractional derivative Kuralay II model with the exp a function method; extended sinh-Gordon equation expansion method and generalized Kudryashov’s method [26].

The objective of this work is to investigate two new types of fractional Kuralay II models, namely, F κ IIAE and F κ IIBE, incorporating beta derivatives. These models have not been examined in prior research and offer the potential to reveal new soliton solutions. In the first part of the article, beta derivative definition and properties are given. In the second part, the algorithms of improved Kudryashov scheme (IKS) [34], Kudryashov’s R scheme (KRS) [35,36], and Sardar’s subequation scheme (SSS) [37] are given, and then the structures of the models and solutions are presented. The final section presents the results of the obtained solutions, highlighting their implications for advancing the study of nonlinear fractional models.

2 Preliminaries

2.1 Beta derivative definition and properties

Let U be a function, U : [ a , ∞ ] → R . The β derivative is defined as follows [2]:

(1) D x β 0 A [ U ( x ) ] = lim ζ → 0 U x + ζ x + 1 Γ ( β ) 1 − β − U [ x ] ζ ,

for all x ≥ a , 0 < β ≤ 1 . The existence of the limit in equation (1) signifies that the function f is considered β -differentiable. Notably, the definition of the β derivative remains independent of the interval. If the function is differentiable, at a point zero, equation (1) is not equal to zero.

Assuming that U and V ≠ 0 are two functions β -differentiable with 0 < β ≤ 1 , there exist some properties:

i) D x β 0 A [ r U ( x ) + s V ( x ) ] = a D x β 0 A [ U ( x ) ] + b D x β 0 A [ V ( x ) ] ,

for all r and s real number.

ii) D x β 0 A [ a ] = 0 ,

a is given constant.

iii) D x β 0 A [ U ( x ) V ( x ) ] = V ( x ) D x β 0 A [ U ( x ) ] + U ( x ) D x β 0 A [ V ( x ) ] ,

iv) D x β 0 A U ( x ) V ( x ) = V ( x ) D x β 0 A [ U ( x ) ] − U ( x ) D x β 0 A [ V ( x ) ] V 2 ( x ) .

Considering equation (1) Z = x + 1 Γ ( β ) 1 − β h , h → 0 , when Z → 0 , therefore, we obtain

D x β 0 A [ U ( x ) ] = x + 1 Γ ( β ) 1 − β d U ( x ) d x ,

with

n = K β x + 1 Γ ( β ) β ,

where K is constant. In that,

D x β 0 A [ U ( n ) ] = K d U ( n ) d n

is satisfied.

3 Mathematical methodologies

In this section, we give basic steps of IKS, KRS, and SSS.

3.1 Methodology of IKS

The main items of the proposed method are represented as follows [34]:

Step 1: First, we consider the general form of nonlinear fractional partial differential equations as follows:

(2) φ ( u , u x , u y , … , u t , u x x , u x y , … , D y β u , D z β u … , D t β u , … , D y 2 β u , D z 2 β U … , D t 2 β u , … ) = 0 , 0 < β ≤ 1 ,

where u ( x , y , z , w , t ) is an unknown function and φ is polynomial in u and its various partial derivatives in that the highest-order derivatives and nonlinear terms are involved, and in addition, D t β u , D y β u , D z β u , D w β u …are beta derivatives of u .

Step 2: Second, by using the following traveling wave transformation

u ( x , y , z , w , t ) = U ( ψ ) , ψ = ω x − α β y + 1 Γ ( β ) β + γ β z + 1 Γ ( β ) β + σ β w + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ,

we can reduce the fractional order differential equation equation (2) into ordinary differential equation. Here, ω , α , γ , σ , and λ are arbitrary constants. Then, by using the chain rule, we have,

(3) μ ( U , U ′ , U ″ , … ) = 0 .

Specific items of the method can be applied after that reduction.

Step 3: In this step, according to the proposed method, we assume equation (3) has a solution in the form:

(4) U ( ψ ) = ∑ j = − 1 B − 1 a j + 1 χ j + 1 ( ψ ) ,

where

(5) χ ( ψ ) = ± ( 1 ± exp ( 2 ψ ) ) − 1 ⁄ 2 ,

and χ ( ψ ) satisfies the following ordinary differential equation:

(6) χ ψ 2 = ( χ 2 ( χ 4 − 2 χ 2 + 1 ) ) 1 ⁄ 2 .

After taking equations (5) and (6) and considering the solution equation (4), we can write the first, second, and third order derivatives of equation (4) as follows:

(7) U ψ = ∑ j = 0 B − 1 a j + 1 ( j + 1 ) ( χ j + 3 ( ψ ) − χ j + 1 ( ψ ) ) , U ψ ψ = ∑ j = 0 B − 1 a j + 1 [ ( j + 1 ) ( j + 3 ) χ s + 5 ( ψ ) − 2 ( j + 1 ) ( j + 2 ) χ j + 3 ( ψ ) + ( j + 1 ) 2 χ j + 1 ( ψ ) ] , U ψ ψ ψ = ∑ j = 0 B − 1 a j + 1 [ ( j + 1 ) ( j + 3 ) ( j + 5 ) χ j + 7 ( ψ ) − 3 ( j + 1 ) ( j + 3 ) 2 χ j + 5 ( ψ ) + ( 3 ( j + 1 ) 2 ( j + 3 ) + 4 ( j + 1 ) ) χ j + 3 ( ψ ) − ( j + 1 ) 3 χ j + 1 ( ψ ) ] .

If necessary, other higher order derivatives can also be written.

Step 4: By substituting equation (4), along with equations (7), into equation (3) and subsequently employing the homogeneous balance method, we can distinctly define the pole order B . Following this procedure, we are able to solve the algebraic system based on the degrees of χ , thereby obtaining the coefficients of the polynomial equation (4). Ultimately, by substituting these coefficients, along with the parameters and the traveling wave transformation, into the derived polynomial, we arrive at the solutions for equation (2).

3.2 Methodology of KRS

The main items of the method are represented as follows [35]:

Step 1: First, we consider the general form of nonlinear fractional partial differential equations as follows:

(8) φ ( u , u x , u y , … , u t , u x x , u x y , … , D y β u , D z β u … , D t β u , … , D y 2 β u , D z 2 β U … , D t 2 β u , … ) = 0 ,

where 0 < β ≤ 1 , u ( x , y , z , w , t ) is an unknown function and φ is polynomial in u and its various partial derivatives in that the highest-order derivatives and nonlinear terms are involved, and in addition, D t β u , D y β u , D z β u , D w β u … are beta derivatives of u .

Step 2: Second, we presume that the following traveling wave transformation:

u ( x , y , z , w , t ) = U ( ψ ) , ψ = ω x − α β y + 1 Γ ( β ) β + γ β z + 1 Γ ( β ) β + σ β w + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ,

reduce the fractional order differential equation equation (8) into ordinary differential equation. Here, ω , α , γ , σ , and λ are arbitrary constants. Then we have,

(9) μ ( U , U ′ , U ″ , … ) = 0 .

Step 3: According to the proposed method, we assume equation (9) has a solution in the form:

(10) U ( ψ ) = ∑ j = − 1 B − 1 a j + 1 χ j + 1 ( ψ ) ,

where

(11) χ ( ψ ) = 4 k 4 k 2 e ψ + m e − ψ ,

where m = 4 k h and χ ( ψ ) adopts the given ordinary differential equation:

(12) χ ψ 2 = χ 2 ( 1 − m χ 2 ) .

Then, we can find the relations:

(13) U ψ = ∑ j = 0 B − 1 a j + 1 χ s ( ψ ) χ ψ ( ψ ) , U ψ ψ = ∑ j = 0 B − 1 a j + 1 [ ( j + 1 ) 2 χ j + 1 ( ψ ) − ( j + 1 ) 2 m χ j + 3 ( ψ ) + ( j + 1 ) m χ j + 4 ( ψ ) ] , U ψ ψ ψ = ∑ j = 0 B − 1 a j + 1 [ ( j + 1 ) 3 χ s ( ψ ) − m ( j + 1 ) 2 ( j + 3 ) χ j + 2 ( ψ ) − m ( j + 1 ) ( j + 3 ) χ j + 2 ( ψ ) ] χ ψ ( ψ ) .

Step 4: By equating the degrees of the highest order linear term ( U ( p ) ( χ ) ) r and the highest degree nonlinear term U l ( χ ) U ( s ) ( χ ) , we can readily define the pole order B . Subsequently, the coefficients of the polynomial given in equation (10) and the associated parameters can be determined. Finally, by substituting these coefficients, parameters, and the traveling wave transformation into the polynomial obtained, we derive the solutions of equation (8).

3.3 Methodology of SSS

The major items of the method proposed above are indicated as follows [37]:

Step 1: Initially, we consider the general form of nonlinear fractional partial differential equations given by

(14) φ ( u , u x , u y , … , u t , u x x , u x y , … , D y β u , D z β u … , D t β u , … , D y 2 β u , D z 2 β U … , D t 2 β u , … ) = 0 , 0 < β ≤ 1 ,

Step 2: Second, we presume that the following traveling wave transformation:

u ( x , y , z , w , t ) = U ( ψ ) , ψ = ω x − α β y + 1 Γ ( β ) β + γ β z + 1 Γ ( β ) β + σ β w + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ,

reduce the fractional order differential equation (14) into ordinary differential equation. Here, ω , α , γ , σ , and λ are arbitrary constants. Then we have,

(15) μ ( U , U ′ , U ″ , … ) = 0 .

In this step, according to the proposed method, we assume equation (15) has a solution in the following form:

(16) U ( ψ ) = ∑ j = − 1 B − 1 a j + 1 χ j + 1 ( ψ ) ,

where χ ( ψ ) is the solution of the following differential equation:

(17) χ ψ 2 = n + m χ 2 ( ψ ) + χ 4 ( ψ ) ,

and the solutions of equation (17) has four sets of solutions:

Set 1: If m > 0 and n = 0 , then

(18) χ 1 ( ψ ) = ± − k h m sech k h ( m ψ ) , χ 2 ( ψ ) = ± k h m csch k h ( m ψ ) ,

where sech k h ( ψ ) = 2 k e ψ + h e − ψ and csch k h ( ψ ) = 2 k e ψ − h e − ψ .

Set 2: If m < 0 and n = 0 , then

(19) χ 3 ( ψ ) = ± − k h m sec k h ( − m ψ ) , χ 4 ( ψ ) = ± − k h m csc k h ( − m ψ ) ,

where sec k h ( ψ ) = 2 k e i ψ + h e − i ψ and csc k h ( ψ ) = 2 k e i ψ − h e − i ψ .

Set 3: If m < 0 and n = m 2 4 h , then

(20) χ 5 ( ψ ) = ± − m 2 tanh k h − m 2 ψ , χ 6 ( ψ ) = ± − m 2 coth k h − m 2 ψ ,

χ 7 ( ψ ) = ± − m 2 ( tanh k h ( − 2 m ψ ) ± i k h sech k h ( − 2 m ψ ) ) , χ 8 ( ψ ) = ± − m 2 ( coth k h ( − 2 m ψ ) ± k h csch k h ( − 2 m ψ ) ) , χ 9 ( ψ ) = ± − m 8 tanh k h − m 8 ψ + coth k h − m 8 ψ ,

where tanh k h ( ψ ) = k e ψ − h e − ψ k e ψ + h e ψ and coth k h ( ψ ) = k e ψ + h e − ψ k e ψ − h e ψ .

Set 4: If m > 0 and n = m 2 4 , then

(21) χ 10 ( ψ ) = ± m 2 tan k h m 2 ψ , χ 11 ( ψ ) = ± m 2 cot k h m 2 ψ , χ 12 ( ψ ) = ± m 2 ( tan k h ( 2 m ψ ) ± k h sec k h ( 2 m ψ ) ) , χ 13 ( ψ ) = ± m 2 ( cot k h ( 2 m ψ ) ± k h csc k h ( 2 m ψ ) ) , χ 14 ( ψ ) = ± m 8 tan k h m 8 ψ + cot k h m 8 ψ ,

where tanh k h ( ψ ) = k e ψ − h e − ψ k e ψ + h e ψ and coth k h ( ψ ) = k e ψ + h e − ψ k e ψ − h e ψ .

Step 3: Inserting equation (16) into equation (15) and using the following derivations:

(22) U ψ = ∑ j = 0 B − 1 ( j + 1 ) a j + 1 χ j ( ψ ) χ ψ ( ψ ) , U ψ ψ = ∑ j = 0 B − 1 a j + 1 [ j ( j + 1 ) n χ j − 1 ( ψ ) + ( j + 1 ) ( j m + 1 ) χ j + 1 ( ψ ) + ( j + 1 ) ( j + 2 ) χ j + 3 ( ψ ) ] , U ψ ψ ψ = ∑ j = 0 B − 1 a j + 1 [ ( j 2 − 1 ) j n χ j − 2 ( ψ ) + ( j + 1 ) 2 ( j m + 1 ) χ j ( ψ ) + ( j + 1 ) ( j + 2 ) ( j + 3 ) χ j + 2 ( ψ ) ] χ ψ ( ψ ) ,

we can define the pole order B clearly. After that, we obtain the algebraic system.

Step 4: By solving the algebraic system based on the degrees of χ , we can determine the coefficients of the polynomial given in equation (16) as well as the associated parameters. Subsequently, by substituting these coefficients, parameters, and the traveling wave transformation into the derived polynomial, we obtain the solutions of equation (14).

4 Solutions of fractional κ II equation (F κ IIE) model

β fractional Kuralay II equation is:

(23) i D t β 0 A u + D x β 0 A ( D t β 0 A u ) − v u = 0 , D x β 0 A v − 2 E D t β 0 A ( ∣ u ∣ 2 ) = 0 ,

where u = u ( x , t ) is a complex valued function, while v = v ( x , t ) is a real valued function. The conjugate of the complex valued function u is u * , E = ± 1 , where x and t are the spatio temporal real variables. There are two types of the fractional Kuralay equation.

(24) i D t β 0 A u − D x β 0 A ( D t β 0 A u ) − v u = 0 , i D t β 0 A r + D x β 0 A ( D t β 0 A r ) − v r = 0 , D x β 0 A v + 2 ( d 2 ) D x β 0 A ( r u ) = 0

is called beta fractional Kuralay IIA equation (F κ IIAE) and

(25) i D t β 0 A u + D x β 0 A ( D t β 0 A u ) − v u = 0 , i D t β 0 A r − D x β 0 A ( D t β 0 A r ) + v r = 0 , D x β 0 A v − 2 D x β 0 A ( r u ) = 0

is called beta fractional Kuralay IIB equation (F κ IIBE).

4.1 Mathematical treatment of F κ IIE

4.1.1 F κ IIAE

Consider the F κ IIAE for d = 1 and r = E u * given as follows:

(26) i D t β 0 A u − D x β 0 A ( D t β 0 A u ) − v u = 0 , D x β 0 A v − 2 E D t β 0 A ( ∣ u ∣ 2 ) = 0 .

Consider the traveling wave transformation:

(27) u ( x , t ) = U ( ψ ) e i Φ ( x , t ) , v ( x , t ) = υ ( ψ ) ,

and

(28) ψ = ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β , Φ ( x , t ) = τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c ,

where ω , λ , τ , and μ are parameters. By inserting equation (27) into equations (26) and, we obtain the real and imaginary parts of the ordinary differential equation, respectively:

(29) 2 λ U 3 E − U ω ( μ ( τ − 1 ) − c ) + λ ω 2 U ″ = 0 ,

and

(30) ( λ − τ λ − μ ω ) U ′ = 0 .

From equation (30), we find the speed of the soliton as follows:

(31) λ = μ ω 1 − τ .

4.1.2 F κ IIBE

Consider the F κ IIBE for d = 1 and r = E u * given as follows:

(32) i D t β 0 A u + D x β 0 A ( D t β 0 A u ) − v u = 0 , D x β 0 A v + 2 E D t β 0 A ( ∣ u ∣ 2 ) = 0 .

Consider the traveling wave transformation:

(33) u ( x , t ) = U ( ψ ) e i Φ ( x , t ) , v ( x , t ) = υ ( ψ )

and

(34) ψ = ω β x + 1 β β + λ β t + 1 β β , Φ ( x , t ) = τ β x + 1 β β + μ β t + 1 β β + c ,

where ω , λ , τ , and μ are parameters. By inserting equation (33) into equations (32), we obtain the real and imaginary parts, respectively:

(35) 2 U 3 ω E − λ U ω ( μ ( τ + 1 ) − c ) + λ 2 ω U ″ = 0 ,

and

(36) ( λ + τ λ + μ ω ) U ′ = 0 .

From equation (36), we obtain the speed of the soliton as follows:

(37) λ = − μ ω 1 + τ .

4.2 Solutions of F κ IIAE

4.2.1 By using the IKS approach

According to the IKS, we assume equation (29) has a polynomial solution as follows:

(38) U ( ψ ) = ∑ j = − 1 B − 1 a j + 1 χ j + 1 ( ψ ) ,

where

(39) χ ( ψ ) = ± ( 1 + exp ( 2 ψ ) ) − 1 ⁄ 2 ,

and U ( ψ ) adopts the given ordinary differential equation:

(40) χ ψ 2 = ( χ 2 ( χ 4 − 2 χ 2 + 1 ) ) 1 ⁄ 2 .

Then according to the previous adoption and using the homogenous balance principle, we obtain pole order N = 2 , and then we can clearly write as follows:

(41) U ( ψ ) = a 0 + a 1 χ ( ψ ) + a 2 χ 2 ( ψ ) .

After determining the quadratic polynomial, we can easily taking the following derivative:

(42) U ″ ( ψ ) = a 1 χ ( ψ ) + 4 a 2 χ 2 ( ψ ) − 4 a 1 χ 3 ( ψ ) − 12 a 2 χ 4 ( ψ ) + 3 a 1 χ 5 ( ψ ) + 8 a 2 χ 6 ( ψ ) .

By substituting equations (41) and (42) into equation (29), we obtain two solutions of F κ IIAE (Figure 1).

$Figure 1 3D surface plot of bright and dark solitary wave solution of ∣ u 1 , 2 ( x , t ) ∣ | {u}_{1,2}\left(x,t)| in equations (43) and (44) for ω = ‒ 0.1 , λ = 0.1 , E = ‒ 0.01 \omega =‒0.1,\lambda =0.1,E=‒0.01 , and β = 0.25 \beta =0.25 ; 2D plot drawn for ‒ 10 < t < 10 ‒10\lt t\lt 10 , x = 0 , x = 1 x=0,x=1 , and x = 2 x=2 ; contour plot for 0 < x < 10 0\lt x\lt 10 and 0 < t < 10 0\lt t\lt 10 .$

Figure 1

3D surface plot of bright and dark solitary wave solution of ∣ u 1 , 2 ( x , t ) ∣ in equations (43) and (44) for ω = ‒ 0.1 , λ = 0.1 , E = ‒ 0.01 , and β = 0.25 ; 2D plot drawn for ‒ 10 < t < 10 , x = 0 , x = 1 , and x = 2 ; contour plot for 0 < x < 10 and 0 < t < 10 .

Case 1:

a 0 = i ω E , a 1 = 0 , a 2 = − 2 i ω E , μ = c − 2 λ ω τ − 1 .

(43) u 1 ( x , t ) = i ω E tanh ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

(44) u 2 ( x , t ) = i ω E coth ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

Case 2:

a 0 = − i ω E , a 1 = 0 , a 2 = 2 i ω E , μ = c − 2 λ ω τ + 1 .

(45) u 3 ( x , t ) = − i ω E tanh ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β × e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

(46) u 4 ( x , t ) = − i ω E coth ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β × e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

4.2.2 By using the KRS approach

Supposing the solution of equation (29) in the form:

(47) U ( ψ ) = ∑ j = − 1 B − 1 a j + 1 χ j + 1 ( ψ ) ,

where

(48) χ ( ψ ) = 4 k 4 k 2 e ψ + m e − ψ ,

where m = 4 k h and U ( ψ ) adopts the given ordinary differential equation:

(49) χ ψ 2 = χ 2 ( 1 − m χ 2 ) .

Then according to the previous adoption and using the homogenous balance principle, we obtain pole order N = 1 , and then, we can clearly write as follows:

(50) U ( ψ ) = a 0 + a 1 χ ( ψ ) .

Then by determining the first degree polynomial, we can easily taking the following derivative:

(51) U ″ ( ψ ) = a 1 χ ( ψ ) ( 1 − 2 m χ 2 ( ψ ) ) .

By substituting equations (74) and (75) into equation (29), we obtain two cases solutions of F κ IIAE (Figure 2).

$Figure 2 3D surface plot of the dark soliton solution of ∣ u ( x , t ) ∣ | u\left(x,t)| in equation (52) for ω = ‒ 0.1 , λ = 0.1 , E = 0.01 , a = 1 , b = 1 \omega =‒0.1,\lambda =0.1,E=0.01,a=1,b=1 , and β = 0.25 \beta =0.25 ; 2D plot drawn for ‒ 10 < t < 10 ‒10\lt t\lt 10 , x = 0 , x = 1 x=0,x=1 , and x = 2 x=2 ; contour plot for 0 < x < 10 0\lt x\lt 10 and 0 < t < 10 0\lt t\lt 10 .$

Figure 2

3D surface plot of the dark soliton solution of ∣ u ( x , t ) ∣ in equation (52) for ω = ‒ 0.1 , λ = 0.1 , E = 0.01 , a = 1 , b = 1 , and β = 0.25 ; 2D plot drawn for ‒ 10 < t < 10 , x = 0 , x = 1 , and x = 2 ; contour plot for 0 < x < 10 and 0 < t < 10 .

Case 1:

a 0 = 0 , a 1 = − ω m E , μ = c + λ ω τ − 1 .

(52) u 5 ( x , t ) = − ω m E 4 a 4 a 2 e ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β + m e − ω β x + 1 Γ ( β ) β − λ β t + 1 Γ ( β ) β × e τ β x + 1 β β + μ β t + 1 β β + c .

Case 2:

a 0 = 0 , a 1 = ω m E , μ = c + λ ω τ − 1 .

(53) u 6 ( x , t ) = ω m E 4 a 4 a 2 e ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β + m e − ω β x + 1 Γ ( β ) β − λ β t + 1 Γ ( β ) β × e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

4.2.3 By using the SSS approach

(54) U ( ψ ) = ∑ j = − 1 B − 1 a j + 1 χ j + 1 ( ψ ) ,

where χ ( ψ ) is the solution of the following differential equation:

(55) χ ψ 2 = n + m χ 2 ( ψ ) + χ 4 ( ψ ) .

Then according to the previous adoption and by using the homogenous balance principle, we obtain pole order N = 1 , and then we can clearly write as follows (Figure 3):

(56) U ( ψ ) = a 0 + a 1 χ ( ψ ) .

Then by determining the first degree polynomial, we can easily take the following derivative:

(57) U ″ ( ψ ) = a 1 m χ ( ψ ) + 2 a 1 χ 3 ( ψ ) .

By substituting equations (56) and (57) into equation (29), we obtain four cases solutions of change as per above red highlighted.

$Figure 3 3D surface plot of the dark soliton solution of ∣ u ( x , t ) ∣ | u\left(x,t)| in equation (52) for ω = ‒ 0.1 , λ = 0.1 , E = 0.01 , a = 1 , b = 1 \omega =‒0.1,\lambda =0.1,E=0.01,a=1,b=1 , and β = 1.0 \beta =1.0 ; 2D plot drawn for ‒ 10 < t < 10 ‒10\lt t\lt 10 , x = 0 , x = 1 x=0,x=1 , and x = 2 x=2 ; contour plot for 0 < x < 10 0\lt x\lt 10 and 0 < t < 10 0\lt t\lt 10 .$

Figure 3

3D surface plot of the dark soliton solution of ∣ u ( x , t ) ∣ in equation (52) for ω = ‒ 0.1 , λ = 0.1 , E = 0.01 , a = 1 , b = 1 , and β = 1.0 ; 2D plot drawn for ‒ 10 < t < 10 , x = 0 , x = 1 , and x = 2 ; contour plot for 0 < x < 10 and 0 < t < 10 .

Cases:

a 0 = 0 , a 1 = ± i ω E , μ = c + m λ ω τ − 1 .

Set 1: If λ ω > 0 , μ ( τ − 1 ) > c and n = 0 or λ < 0 , μ ( 1 + τ ) < c and n = 0 , then

(58) u 1 , 1 ( x , y ) = ± i ω − k h m E sech k h m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 2 ( x , y ) = ± i ω k h m E cosh k h m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

Set 2: If λ > 0 , μ ( 1 + τ ) < c and n = 0 or λ < 0 , μ ( 1 + τ ) > c , and n = 0 , then (Figure 4)

(59) u 1 , 3 ( x , y ) = ± i ω − k h m E sec k h − m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 4 ( x , y ) = ± i ω − k h m E csc k h − m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

$Figure 4 3D surface plot of the singular 1 ‒ 1‒ soliton solution of ∣ u 1 , 1 ( x , t ) ∣ | {u}_{1,1}\left(x,t)| in equation (58) for ω = 0.1 , λ = 1 , E = 0.01 \omega =0.1,\lambda =1,E=0.01 , β = 0.25 \beta =0.25 and ω = 0.1 , λ = 1 , E = 0.01 \omega =0.1,\lambda =1,E=0.01 , β = 1.0 \beta =1.0 ; 2D plot drawn for ‒ 10 < t < 10 ‒10\lt t\lt 10 , x = 0 , x = 1 x=0,x=1 and x = 2 x=2 ; contour plot for ‒ 5 < x < 5 ‒5\lt x\lt 5 and ‒ 1 < t < 1 ‒1\lt t\lt 1 .$

Figure 4

3D surface plot of the singular 1 ‒ soliton solution of ∣ u 1 , 1 ( x , t ) ∣ in equation (58) for ω = 0.1 , λ = 1 , E = 0.01 , β = 0.25 and ω = 0.1 , λ = 1 , E = 0.01 , β = 1.0 ; 2D plot drawn for ‒ 10 < t < 10 , x = 0 , x = 1 and x = 2 ; contour plot for ‒ 5 < x < 5 and ‒ 1 < t < 1 .

Set 3: If λ > 0 , μ ( 1 + τ ) < c and n = μ ( 1 + τ ) − c 2 λ 2 or λ < 0 , μ ( 1 + τ ) > c and n = μ ( 1 + τ ) − c 2 λ 2 , then (Figure 5)

(60) u 1 , 5 ( x , y ) = ± i ω − m 2 E tanh k h − m 2 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 6 ( x , y ) = ± i ω − m 2 E coth k h − m 2 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 7 ( x , y ) = ± i ω − m 2 E tanh k h − 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ± k h sech k h − 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

u 1 , 8 ( x , y ) = ± i ω − m 2 E coth k h − 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ± k h csch k h − 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 9 ( x , y ) = ± i ω m 8 E tanh k h − m 8 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β + cot k h − m 8 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

Set 4: If λ > 0 , μ ( 1 + τ ) > c and n = μ ( 1 + τ ) − c 2 λ 2 or λ < 0 , μ ( 1 + τ ) < c and n = μ ( 1 + τ ) − c 2 λ 2 , then

(61) u 1 , 10 ( x , y ) = ± i ω m 2 E tan k h m 2 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 11 ( x , y ) = ± i ω m 2 E cot k h − m 2 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 12 ( x , y ) = ± i ω − m 2 E tan k h 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ± k h sec k h 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 13 ( x , y ) = ± i ω m 2 E cot k h 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ± k h csc k h 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

$Figure 5 3D surface plot of the periodic solitary wave solution of ∣ u 1 , 1 ( x , t ) ∣ | {u}_{1,1}\left(x,t)| in equation (59) for ω = 0.1 , λ = 1 , E = 0.01 \omega =0.1,\lambda =1,E=0.01 , β = 0.25 \beta =0.25 and ω = 0.1 , λ = 1 , E = 0.01 \omega =0.1,\lambda =1,E=0.01 , β = 1.0 \beta =1.0 ; 2D plot drawn for ‒ 10 < t < 10 ‒10\lt t\lt 10 , x = 0 , x = 1 x=0,x=1 , and x = 2 x=2 ; contour plot for ‒ 5 < x < 5 ‒5\lt x\lt 5 and ‒ 1 < t < 1 ‒1\lt t\lt 1 .$

Figure 5

3D surface plot of the periodic solitary wave solution of ∣ u 1 , 1 ( x , t ) ∣ in equation (59) for ω = 0.1 , λ = 1 , E = 0.01 , β = 0.25 and ω = 0.1 , λ = 1 , E = 0.01 , β = 1.0 ; 2D plot drawn for ‒ 10 < t < 10 , x = 0 , x = 1 , and x = 2 ; contour plot for ‒ 5 < x < 5 and ‒ 1 < t < 1 .

4.3 Solutions of F κ IIBE

4.3.1 By using the IKS approach

According to the IKS, we assume equation (35) has a polynomial solution as follows:

(62) U ( ψ ) = ∑ j = − 1 B − 1 a j + 1 χ j + 1 ( ψ ) ,

where

(63) χ ( ψ ) = ± ( 1 + exp ( 2 ψ ) ) − 1 ⁄ 2 ,

and U ( ψ ) adopts the given ordinary differential equation:

(64) χ ψ 2 = ( χ 2 ( χ 4 − 2 χ 2 + 1 ) ) 1 ⁄ 2 .

On the basis of the preceding approach and employing the homogeneous balance principle, we determine a pole order of N = 2 , enabling us to express the following explicitly:

(65) U ( ψ ) = a 0 + a 1 χ ( ψ ) + a 2 χ 2 ( ψ ) .

Once the quadratic polynomial has been determined, we can readily proceed with the following derivative:

(66) U ″ ( ψ ) = a 1 χ ( ψ ) + 4 a 2 χ 2 ( ψ ) − 4 a 1 χ 3 ( ψ ) − 12 a 2 χ 4 ( ψ ) + 3 a 1 χ 5 ( ψ ) + 8 a 2 χ 6 ( ψ ) .

By substituting equations (65) and (66) into equation (35), we obtain two solutions of F κ IIBE.

Case 1:

a 0 = i λ E , a 1 = 0 , a 2 = − 2 i λ E , μ = c − 2 λ ω τ + 1 .

(67) u 1 b ( x , t ) = i λ E tanh ω β x + 1 β β + λ β t + 1 β β e τ β x + 1 β β + μ β t + 1 β β + c .

(68) u 2 b ( x , t ) = − i λ E coth ω β x + 1 β β + λ β t + 1 β β e τ β x + 1 β β + μ β t + 1 β β + c .

Case 2:

a 0 = − i λ E , a 1 = 0 , a 2 = 2 i λ E , μ = c − 2 λ ω τ + 1 .

(69) u 3 b ( x , t ) = − i λ E tanh ω β x + 1 β β + λ β t + 1 β β e τ β x + 1 β β + μ β t + 1 β β + c .

(70) u 4 b ( x , t ) = i λ E coth ω β x + 1 β β + λ β t + 1 β β e τ β x + 1 β β + μ β t + 1 β β + c .

4.3.2 By using the KRS approach

Supposing the solution of equation (35) in the form:

(71) U ( ψ ) = ∑ j = − 1 B − 1 a j + 1 χ j + 1 ( ψ ) ,

where

(72) χ ( ψ ) = 4 k 4 k 2 e ψ + m e − ψ ,

where m = 4 k h and U ( ψ ) adopts the given ordinary differential equation:

(73) χ ψ 2 = χ 2 ( 1 − m χ 2 ) .

Then according to the previous adoption and by using the homogenous balance principle, we obtain pole order N = 1 and then we can clearly write as follows:

(74) U ( ψ ) = a 0 + a 1 χ ( ψ ) .

Then by determining the first degree polynomial, we can easily taking the following derivative:

(75) U ″ ( ψ ) = a 1 χ ( ψ ) ( 1 − 2 m χ 2 ( ψ ) ) .

By substituting equations (74) and (75) into equation (35), we obtain two cases solutions of F κ IIBE.

Case 1:

a 0 = 0 , a 1 = − λ m E , μ = c + λ ω 1 + τ .

(76) u 5 b ( x , t ) = − λ m E 4 a 4 a 2 e ω β x + 1 β β + λ β t + 1 β β + m e − ω β x + 1 β β − λ β t + 1 β β e τ β x + 1 β β + μ β t + 1 β β + c .

Case 2:

a 0 = 0 , a 1 = λ m E , μ = c + λ ω 1 + τ .

(77) u 6 b ( x , t ) = λ m E 4 a 4 a 2 e ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β + m e − ω β x + 1 Γ ( β ) β − λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

4.3.3 By using the SSS approach

(78) U ( ψ ) = ∑ j = − 1 B − 1 a j + 1 χ j + 1 ( ψ ) ,

where χ ( ψ ) is the solution of the following differential equation:

(79) χ ψ 2 = n + m χ 2 ( ψ ) + χ 4 ( ψ ) .

On the basis of the preceding adoption and the application of the homogenous balance principle, we deduce a pole order of N = 1 , enabling us to articulate the following succinctly:

(80) U ( ψ ) = a 0 + a 1 χ ( ψ ) .

Upon establishing the first-degree polynomial, we can readily proceed with the following derivative:

(81) U ″ ( ψ ) = a 1 m χ ( ψ ) + 2 a 1 χ 3 ( ψ ) .

By substituting equations (80) and (81) into equation (35), we derive solutions for two distinct cases of F κ IIBE.

Cases:

a 0 = 0 , a 1 = ± i λ E ω , μ = c + m λ τ + 1 .

Set 1: If λ > 0 , μ ( 1 + τ ) > c and n = 0 or λ < 0 , μ ( 1 + τ ) < c and n = 0 , then

(82) u 1 , 1 b ( x , y ) = ± i λ − k h m E ω sech k h m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 2 b ( x , y ) = ± i λ k h m E ω cosh k h m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

Set 2: If λ > 0 , μ ( 1 + τ ) < c and n = 0 or λ < 0 , μ ( 1 + τ ) > c and n = 0 , then

(83) u 1 , 3 b ( x , y ) = ± i λ − k h m E ω sec k h − m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

u 1 , 4 b ( x , y ) = ± i λ − k h m E ω csc k h − m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

Set 3: If λ > 0 , μ ( 1 + τ ) < c and n = μ ( 1 + τ ) − c 2 λ 2 or λ < 0 , μ ( 1 + τ ) > c and n = μ ( 1 + τ ) − c 2 λ 2 , then (Figures 6 and 7)

(84) u 1 , 5 b ( x , y ) = ± i λ − m 2 E ω tanh k h − m 2 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 6 b ( x , y ) = ± i λ − m 2 E ω coth k h − m 2 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 7 b ( x , y ) = ± i λ − m 2 E ω tanh k h − 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ± k h sech k h − 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 8 b ( x , y ) = ± i λ − m 2 E ω coth k h − 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ± k h csch k h − 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

u 1 , 9 b ( x , y ) = ± i λ m 8 E ω tanh k h − m 8 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β + cot k h − m 8 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

Set 4: If λ > 0 , μ ( 1 + τ ) > c and n = μ ( 1 + τ ) − c 2 λ 2 or λ < 0 , μ ( 1 + τ ) < c and n = μ ( 1 + τ ) − c 2 λ 2 , then

(85) u 1 , 10 b ( x , y ) = ± i λ m 2 E ω tan k h m 2 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 11 b ( x , y ) = ± i λ m 2 E ω cot k h − m 2 ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 12 b ( x , y ) = ± i λ − m 2 E ω tan k h 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ± k h sec k h 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c . u 1 , 13 b ( x , y ) = ± i λ m 2 E ω cot k h 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β ± k h csc k h 2 m ω β x + 1 Γ ( β ) β + λ β t + 1 Γ ( β ) β e τ β x + 1 Γ ( β ) β + μ β t + 1 Γ ( β ) β + c .

$Figure 6 3D surface plot of the periodic solitary wave solution of ∣ u 1 , 3 b ( x , t ) ∣ | {u}_{1,3b}\left(x,t)| in equation (83) for ω = ‒ 0.1 , λ = ‒ 0.1 , E = 0.1 \omega =‒0.1,\lambda =‒0.1,E=0.1 , β = 0.25 \beta =0.25 ; 2D plot drawn for ‒ 10 < t < 10 ‒10\lt t\lt 10 , x = 0 , x = 1 x=0,x=1 and x = 2 x=2 ; contour plot for 0 < x < 10 0\lt x\lt 10 and 0 < t < 10 0\lt t\lt 10 .$

Figure 6

3D surface plot of the periodic solitary wave solution of ∣ u 1 , 3 b ( x , t ) ∣ in equation (83) for ω = ‒ 0.1 , λ = ‒ 0.1 , E = 0.1 , β = 0.25 ; 2D plot drawn for ‒ 10 < t < 10 , x = 0 , x = 1 and x = 2 ; contour plot for 0 < x < 10 and 0 < t < 10 .

$Figure 7 3D surface plot of the dark soliton solution of ∣ u 1 , 4 b ( x , t ) ∣ | {u}_{1,4b}\left(x,t)| in equation (83) for ω = ‒ 0.1 , λ = ‒ 0.1 , E = 0.1 \omega =‒0.1,\lambda =‒0.1,E=0.1 , and β = 1 \beta =1 ; 2D plot drawn for ‒ 10 < t < 10 ‒10\lt t\lt 10 , x = 0 , x = 1 x=0,x=1 , and x = 2 x=2 ; contour plot for 0 < x < 10 0\lt x\lt 10 and 0 < t < 10 0\lt t\lt 10 .$

Figure 7

3D surface plot of the dark soliton solution of ∣ u 1 , 4 b ( x , t ) ∣ in equation (83) for ω = ‒ 0.1 , λ = ‒ 0.1 , E = 0.1 , and β = 1 ; 2D plot drawn for ‒ 10 < t < 10 , x = 0 , x = 1 , and x = 2 ; contour plot for 0 < x < 10 and 0 < t < 10 .

5 Conclusion

This study represents a significant advancement in the realm of analytical wave solutions for the Kuralay-II equations. By harnessing the power of the β -fractional derivative in conjunction with sophisticated methodologies such as the IKS, KRS, and SSS, this research has unveiled a new frontier of analytical exploration. The robustness and validity of these solutions have been rigorously verified through meticulous parameter analysis and visualization techniques utilizing Mathematica.

Moreover, the introduction of fractional derivatives, particularly the truncated beta-fractional derivative, marks a groundbreaking innovation in this model, offering unparalleled accuracy and relevance when compared to traditional derivatives. The close alignment of the analytical wave solutions derived via the β -fractional derivative with experimental results underscores their practical significance and potential for real-world applications.

Looking ahead, the avenues for future research are vast and promising. From exploring applications in optical couplers, metamaterials, and fractional order integrable Kuralay equations to delving deeper into the implications for nonlinear optics, ferromagnetic materials, and optical fibers, this work sets the stage for further advancements in diverse scientific domains. Furthermore, the demonstrated efficacy and versatility of the improved Kudryashov, KRS, and SSSs in tackling nonlinear PDEs highlight their potential as invaluable tools for driving progress in related fields. This article not only contributes significantly to the theoretical understanding of the fractional KuralayII equations but also paves the way for practical innovations with far-reaching implications.

Overall, the strengths of this work lie in its innovative use of fractional calculus, advanced analytical methods, rigorous validation, alignment with experimental data, and broad applicability, making it a significant contribution to both theoretical research and practical advancements. This research enhances the mathematical understanding of soliton solutions in fractional models, offers new computational techniques, and broadens the potential for applying these methods in various scientific and engineering domains.

Acknowledgement

This research is produced from the Project of Scientific Research in Ege University (BAP), Project Number: 24004 – FM-GAP-2022-24004.

Funding information: The author declares that no funding was received for the manuscript.
Author contributions: The author confirms the sole responsibility for the conception and design of this manuscript. The author read and approved the final manuscript.
Conflict of interest: The author declare no conflict of interest.
Data availability statement: The authors confirm that the data supporting the findings of this study are available within the manuscript.

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Received: 2024-04-26

Revised: 2024-09-16

Accepted: 2024-11-18

Published Online: 2025-05-10

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

Articles in the same Issue

https://doi.org/10.1515/dema-2024-0088

Keywords for this article

fractional Kuralay-II equations (FκIIAE and FκIIBE); improved Kudryashov method; R scheme; Sardar’s subequation scheme

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