Startseite Analytical and numerical study for the generalized q-deformed sinh-Gordon equation
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Analytical and numerical study for the generalized q-deformed sinh-Gordon equation

  • Khalid K. Ali EMAIL logo
Veröffentlicht/Copyright: 14. Februar 2023
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Abstract

In this article, we study the generalized q -deformed sinh-Gordon equation analytically using the new general form of Kudryashov’s approach and numerically using the finite difference method. We develop a general form of the Kudryashov method that contains more than one constant that is used to give more explanations for the solutions that are obtained. The numerical results are also presented using the finite difference approach. We also provide numerous figures to demonstrate the various solitons propagation patterns. The proposed equation has opened up new options for describing physical systems that have lost their symmetry. The equation under study has not been studied extensively, so we completed the lesson that started a short time ago on it.

MSC 2010: 34B18; 34A30; 34k10

1 Introduction

Despite the fact that dynamical models are the cornerstone of many scientific disciplines, they are understudied in the literature. They are important research areas in mathematics and applied sciences. Nonlinear ordinary or partial differential equations are used in these models. Non-linearity has a big impact on explaining dynamics in microscopical systems that are governed by quantum physics principles [1,2]. Many partial differential equations with important applications in different fields have been studied by many researchers [314].

Quantum calculus, often known as calculus without bounds, is the same as regular infinitesimal calculus but without the concept of bounds. It distinguishes between “ q -calculus” and “ h -calculus,” with h denoting Planck’s constant and q denoting quantum. The following formula connects the two parameters [15]:

(1.1) q = e i h = e 2 i π ,

where = h 2 π . I think that the study of these equations with the use of q -calculus has not been done to a great extent. However, some articles have appeared recently that deal with this topic [16,17].

q -calculus: This field of study has a variety of applications. In mathematical physics, the theory of relativity, and special functions, the q -calculus has been developed and applied in numerous ways [1822].

The generalized q -deformed sinh-Gordon equation (Eleuch equation) is the subject of this essay [16,17]. It explains

(1.2) 2 Φ x 2 2 Φ τ 2 = [ sinh q ( Φ κ ) ] s ϖ ,

where κ and s are constants and 0 < q < 1 is a function. For q = 1 , the standard sinh-Gordon equation is obtained.

We investigate the optical soliton solutions of (1.2) using a novel approach. Furthermore, we use the finite difference approach to solve this equation numerically [2429]. For this equation, we look at two scenarios.

The Eq. (1.2), as well as several solution techniques, were known in the 19th century, but the equation grew greatly in importance when it was realized that it led to solutions (“kink” and “antikink”) with the collision properties of solitons. The sinh-Gordon equation also appears in a number of other physical applications including the propagation of fluxing in Josephson junctions (a junction between two superconductors), the motion of rigid pendulum attached to a stretched wire, and dislocations in crystals.

Studying this type of equation opens the way for researchers to explain some phenomena in quantum mechanics, engineering, and other sciences.

This article has been organized as follows: The second section introduces several q -calculus concepts and characteristics. The analysis of the proposed equation is presented in Section 3. The methodology of the new method is presented in Section 4. The solutions are presented in Section 5. The numerical solutions to the equation are shown in Section 6. We provide several different figures of solutions in Section 7. We introduce a conclusion to what we have done in the work in Section 8.

2 q -Calculus

The following are some fundamental definitions of the q -calculus [23]:

  1. Let a real parameter q ( 0 , 1 ) , we define a q -real number [ l ] q by

    [ l ] q = 1 q l 1 q , l R .

  2. Differential of function is defined in the q -calculus as follows:

    d q ( y ( τ ) ) = y ( q τ ) y ( τ ) .

  3. The q -gamma function is defined by

    Γ q ( τ ) = K ( q τ ) ( 1 q ) τ 1 K ( q ) , τ R { 0 , 1 , 2 , } ,

    where K ( q τ ) = 1 ( q τ ; q ) .

  4. Or, equivalently,

    Γ q ( τ ) = 1 q ( τ 1 ) 1 q τ 1

    and satisfies Γ q ( τ + 1 ) = [ τ ] q Γ q ( τ ) .

  5. The q -derivative of a function y is given by

    ( D q y ) ( τ ) = d q ( y ( τ ) ) d q ( τ ) = y ( τ ) y ( τ q ) τ q τ , ( D q y ) ( 0 ) = lim τ ( D q y ) ( τ ) .

  6. The sinh q is defined by:

    sinh q ( τ ) = e τ q e τ 2 .

  7. The cosh q is defined by:

    cosh q ( τ ) = e τ + q e τ 2 .

  8. Below are several simple and useful q -deformed function relations.

    ( cosh q ( τ ) ) 2 ( sinh q ( τ ) ) 2 = q , sinh q ( τ ) = q sinh 1 q ( τ ) , cosh q ( τ ) = q cosh 1 q ( τ ) , d d τ cosh q ( τ ) = sinh q ( τ ) , d d τ sinh q ( τ ) = cosh q ( τ ) ,

3 The mathematical examination of the model

The following transformation is used to find the traveling wave solution of (1.2).

(3.1) Φ ( x , τ ) = u ( ς ) ,

where

(3.2) ς = ϑ x τ ,

where is the traveling wave speed. From (3.1) and (3.2), then (1.2) becomes,

(3.3) ( ϑ 2 2 ) u ( ς ) + ϖ [ sinh q ( u ( ς ) κ ) ] s = 0 .

Now we will look at two cases for (3.3).

  • Case 1: Suppose s = κ = 1 , ϖ = 0 .

Eq. (3.3) can then be written as follows:

(3.4) ( ϑ 2 2 ) u ( ς ) sinh q ( u ( ς ) ) = 0 .

We can multiply both sides of (3.4) by u ( ς ) and obtain the following equation after integration.

(3.5) 1 2 ( ϑ 2 2 ) ( u ( ς ) ) 2 cosh q ( u ( ς ) ) C 1 = 0 ,

where C 1 is the integration constant.

Let

(3.6) u ( ς ) = ln ( v ( ς ) ) .

As a result, (3.5) becomes,

(3.7) ( 2 ϑ 2 ) v 2 ( ς ) + 2 C 1 v 2 ( ς ) + q v ( ς ) + v 3 ( ς ) = 0 .

Thus, we can solve (3.7) with our approach, and we can extract the solution of (1.2) in the first case from (3.6) and (3.1).

  • Case 2: Assume that κ = 1 , s = 2 , and ϖ = q 2 .

Then, (3.3) can be written as follows:

(3.8) ( ϑ 2 2 ) u ( ς ) ( sinh q ( u ( ς ) ) ) 2 q 2 = 0 .

We obtain the following equation after simplifying (3.8).

(3.9) ( ϑ 2 2 ) u ( ς ) 1 2 cosh q 2 ( 2 u ( ς ) ) = 0 .

Let

(3.10) u ( ς ) = 1 2 ln ( v ( ς ) ) .

As a result, (3.10) becomes,

(3.11) 2 ( ϑ 2 2 ) v 2 ( ς ) 2 ( ϑ 2 2 ) v ( ς ) v ( ς ) + q 2 v ( ς ) + v 3 ( ς ) = 0 .

As a result, we can solve (3.11) using a proposed method, and we can solve (1.2) in the second case using (3.10) and (3.1).

4 The strategy of the new technique

Form the governing equation as follows:

(4.1) M ( Φ , Φ x x , Φ τ τ , ) = 0 ,

where M is the partial derivatives of the polynomial Φ = Φ ( x , τ ) . To convert (3.1) to an ordinary differential equation, use a traveling wave transformation:

(4.2) G ( u , u , ) = 0 .

The following are the key steps in the new generalization of the Kudryashov method:

  1. Suppose the exact solutions of (4.2) can be expressed as follows:

    (4.3) v ( ς ) = i = 0 N ( E i ( Q ( ς ) ) i ,

    where E i ( i = 0 , 1 , 2 , , N ) , E N 0 are constants that may be found using ( E i 0 . The homogeneous balancing principle will decide N .

  2. Q ( ς ) has the following definition:

    (4.4) Q ( ς ) = Λ ϕ ( ς ) Λ + γ ( ϕ ( ς ) 1 ) , γ 0 ,

    where E 0 , E 1 , E 2 , , E N , Λ and γ are determined arbitrary constants. The ordinary differential equation is fulfilled by the function ϕ ( ς ) :

    (4.5) d ϕ d ς = χ ( ϕ ( ς ) 1 ) ϕ ( ς ) .

    The solution to (4.5) is as follows:

    (4.6) ϕ ( ς ) = b b + d e χ ς .

  3. We obtain a polynomial of Q ( ς ) by putting (4.3) into (4.2). We obtain a set of algebraic equations by combining all terms with like powers of Q ( ς ) and setting each coefficient to zero.

  4. We acquire the solution of (4.2) by solving this system with the Mathematica program.

5 The mathematical solution to the model

The generalization of the Kudryashov method is used in this part to obtain the analytic solution to the two situations specified for the problem (1.2).

  • Case 1’s analytical solution at s = κ = 1 , ϖ = 0 :

Applying the balance principle in (3.7) between v 2 and v 3 we obtain 2 N + 2 = 3 N N = 2 . The solution of (3.7) can be written as follows:

(5.1) v ( ς ) = i = 0 2 E i ( Q ( ς ) ) i .

We obtain the following system by substituting (5.1) for (3.7) and setting the coefficient of like power of Q ( ς ) to zero:

2 E 0 2 C 1 + E 0 q + E 0 3 = 0 , 4 E 1 E 0 C 1 + E 1 q + 3 E 1 E 0 2 = 0 , ϑ 2 χ 2 E 1 2 + χ 2 E 1 2 2 + 4 E 2 E 0 C 1 + 2 E 1 2 C 1 + E 2 q + 3 E 2 E 0 2 + 3 E 1 2 E 0 = 0 , 2 ϑ 2 χ 2 E 1 2 4 ϑ 2 χ 2 E 2 E 1 2 χ 2 E 1 2 2 + 4 χ 2 E 2 E 1 2 + 4 E 2 E 1 C 1 + E 1 3 + 6 E 0 E 2 E 1 = 0 , ϑ 2 χ 2 E 1 2 4 ϑ 2 χ 2 E 2 2 + 8 ϑ 2 χ 2 E 1 E 2 + χ 2 E 1 2 2 + 4 χ 2 E 2 2 2 8 χ 2 E 1 E 2 2 + 2 E 2 2 C 1 + 3 E 0 E 2 2 + 3 E 1 2 E 2 = 0 , 8 ϑ 2 χ 2 E 2 2 4 ϑ 2 χ 2 E 1 E 2 8 χ 2 E 2 2 2 + 4 χ 2 E 1 E 2 2 + 3 E 1 E 2 2 = 0 , 4 ϑ 2 χ 2 E 2 2 + 4 χ 2 E 2 2 2 + E 2 3 = 0 .

We obtain the following set of solutions by using the Mathematica program to solve the previous set of equations (Appendix 1).

  • Class 1:

(5.2) E 0 = q , E 1 = 4 q , E 2 = 4 q , ϑ = χ 2 2 + q χ , C 1 = q .

We can find the solutions of (1.2) at s = κ = 1 , ϖ = 0 by substituting (5.2) into (5.1) using (3.6) and (3.1).

(5.3) Φ 1 , 2 ( x , τ ) = ln E 2 b Λ ( b + d exp ( χ ( ϑ x τ ) ) ) γ b b + d exp ( χ ( ϑ x τ ) ) 1 + Λ 2 + E 1 ( b Λ ) ( b + d exp ( χ ( ϑ x τ ) ) ) γ b b + d exp ( χ ( ϑ x τ ) ) 1 + Λ + E 0 .

  • Class 2:

(5.4) E 0 = q , E 1 = 4 q , E 2 = 4 q , ϑ = χ 2 2 q χ , C 1 = q .

We can derive the solutions of (1.2) at s = κ = 1 , ϖ = 0 by substituting (5.4) into (5.1) using (3.6) and (3.1).

(5.5) Φ 3 , 4 ( x , τ ) = ln E 2 b Λ ( b + d exp ( χ ( ϑ x τ ) ) ) γ b b + d exp ( χ ( ϑ x τ ) ) 1 + Λ 2 + E 1 ( b Λ ) ( b + d exp ( χ ( ϑ x τ ) ) ) γ b b + d exp ( χ ( ϑ x τ ) ) 1 + Λ + E 0 .

  • Case 2’s analytical solution can be found at s = 2 , κ = 1 , and ϖ = q 2 :

We obtain 2 N + 2 = 3 N N = 2 by using the balance principle in (3.11) between v v and v 3 . The solution of (3.7) can be derived from (4.3) as follows:

(5.6) v ( ς ) = i = 0 2 E i ( Q ( ς ) ) i .

We obtain the following system by substituting (5.6) into (3.11) and setting the coefficient of like power of Q ( ς ) to zero:

E 0 q 2 + E 0 3 = 0 , 2 ϑ 2 χ 2 E 0 E 1 + 2 χ 2 E 0 E 1 2 + E 1 q 2 + 3 E 0 2 E 1 = 0 , 6 ϑ 2 χ 2 E 0 E 1 8 ϑ 2 χ 2 E 0 E 2 6 χ 2 E 0 E 1 2 + 8 χ 2 E 0 E 2 2 + E 2 q 2 + 3 E 0 E 1 2 + 3 E 0 2 E 2 = 0 , 2 ϑ 2 χ 2 E 1 2 4 ϑ 2 χ 2 E 0 E 1 2 ϑ 2 χ 2 E 2 E 1 + 20 ϑ 2 χ 2 E 0 E 2 2 χ 2 E 1 2 2 + 4 χ 2 E 0 E 1 2 + 2 χ 2 E 2 E 1 2 20 χ 2 E 0 E 2 2 + E 1 3 + 6 E 0 E 2 E 1 = 0 , 2 ϑ 2 χ 2 E 1 2 12 ϑ 2 χ 2 E 0 E 2 + 10 ϑ 2 χ 2 E 1 E 2 + 2 χ 2 E 1 2 2 + 12 χ 2 E 0 E 2 2 10 χ 2 E 1 E 2 2 + 3 E 0 E 2 2 + 3 E 1 2 E 2 = 0 , 4 ϑ 2 χ 2 E 2 2 8 ϑ 2 χ 2 E 1 E 2 4 χ 2 E 2 2 2 + 8 χ 2 E 1 E 2 2 + 3 E 1 E 2 2 = 0 , 4 ϑ 2 χ 2 E 2 2 + 4 χ 2 E 2 2 2 + E 2 3 = 0 .

We obtain the following set of solutions by using the Mathematica program to solve the previous set of equations.

  • Class 1:

(5.7) E 0 = i q , E 1 = 4 i q , E 2 = 4 i q , ϑ = χ 2 2 i q χ .

We can find the solutions of (1.2) at s = 2 , κ = 1 , and ϖ = q 2 by substituting (5.7) into (5.6) using (3.10) and (3.1).

(5.8) Φ 1 , 2 ( x , τ ) = 1 2 ln E 2 b Λ ( b + d exp ( χ ( ϑ x τ ) ) ) γ b b + d exp ( χ ( ϑ x τ ) ) 1 + Λ 2 + E 1 ( b Λ ) ( b + d exp ( χ ( ϑ x τ ) ) ) γ b b + d exp ( χ ( ϑ x τ ) ) 1 + Λ + E 0 .

  • Class 2:

(5.9) E 0 = i q , E 1 = 4 i q , E 2 = 4 i q , ϑ = χ 2 2 + i q χ .

By inserting (5.9) into (5.6) and utilizing (3.10) with (3.1), we can determine the solutions of (1.2) at s = 2 , κ = 1 , and ϖ = q 2 .

(5.10) Φ 3 , 4 ( x , τ ) = 1 2 ln E 2 b Λ ( b + d exp ( χ ( a x τ ) ) ) γ b b + d exp ( χ ( a x τ ) ) 1 + Λ 2 + E 1 ( b Λ ) ( b + d exp ( χ ( a x τ ) ) ) γ b b + d exp ( χ ( a x τ ) ) 1 + Λ + E 0 .

6 The numerical solution to the model

As cited in [24,25], we employ approximations for space x and time τ derivatives in this section.

(6.1) u τ τ U i , j + 1 2 U i , j + U i , j 1 Δ τ 2 , u x x U i + 1 , j 2 U i , j + U i 1 , j Δ x 2 .

We assume that Φ is the precise solution at the grid point ( x i , τ j ) and that U i , j is the numerical solution at the same location. We obtain the system of difference equations by substituting (6.1) into (1.2), which may be solved to obtain the numerical answers.

6.1 The numerical outcomes

Now we will look at some numerical results for the q -deformed sinh-Gordon equation in general. We also offer the numerical solution for two particular situations of the generalized q -deformed sinh-Gordon equation, as we examined the analytic solution for these two cases:

  • Case 1’s numerical results: s = κ = 1 , ϖ = 0 .

We compare the numerical findings with the analytical solution (5.3) for (1.2) at Δ x = 0.2 , Δ τ = 0.1 , χ = 0.4 , b = 0.6 , = 0.3 , γ = 0.3 , Λ = 0.2 , d = 0.1 , q = 0.2 in Table 1 and Figure 1.

Table 1

A comparison of the numerical and analytical solutions

x Numerical solution Analytical solution Absolute error
10.0 3.0825 3.0825 0.00000
8.0 2.9267 2.9267 8.66856 × 1 0 9
6.0 3.0799 3.0799 9.39291 × 1 0 9
4.0 3.1733 3.1733 1.03583 × 1 0 8
4.0 3.1733 3.1733 1.03583 × 1 0 8
6.0 3.0799 3.0799 9.39291 × 1 0 9
8.0 2.9267 2.9267 8.66856 × 1 0 9
10.0 3.0825 3.0825 0.00000
Figure 1 
                  At 
                        
                           
                           
                              Δ
                              x
                              =
                              0.2
                           
                           \Delta x=0.2
                        
                     , 
                        
                           
                           
                              Δ
                              τ
                              =
                              0.1
                           
                           \Delta \tau =0.1
                        
                     , 
                        
                           
                           
                              χ
                              =
                              0.4
                           
                           \chi =0.4
                        
                     , 
                        
                           
                           
                              b
                              =
                              0.6
                           
                           b=0.6
                        
                     , 
                        
                           
                           
                              ℧
                              =
                              0.3
                           
                           \mho =0.3
                        
                     , 
                        
                           
                           
                              γ
                              =
                              0.3
                           
                           \gamma =0.3
                        
                     , 
                        
                           
                           
                              Λ
                              =
                              0.2
                           
                           \Lambda =0.2
                        
                     , 
                        
                           
                           
                              d
                              =
                              0.1
                           
                           d=0.1
                        
                     , 
                        
                           
                           
                              q
                              =
                              0.2
                           
                           q=0.2
                        
                     , compare the numerical findings of (1.2) with the analytical solution of (5.3).
Figure 1

At Δ x = 0.2 , Δ τ = 0.1 , χ = 0.4 , b = 0.6 , = 0.3 , γ = 0.3 , Λ = 0.2 , d = 0.1 , q = 0.2 , compare the numerical findings of (1.2) with the analytical solution of (5.3).

  • Case 2’s numerical results are as follows: s = 2 , κ = 1 , ϖ = q 2 .

Table 2 and Figure 2 provide a comparison of the numerical results with the analytical solution (5.10) for (1.2) at Δ x = 0.2 , Δ τ = 0.1 , χ = 0.1 , b = 0.8 , = 0.8 , γ = 0.1 , Λ = 0.2 , d = 0.2 , d = 0.2 , q = 0.1 .

Table 2

A comparison of numerical and analytical solutions is shown

x Numerical solution Analytical solution Absolute error
10.0 1.17815 1.17815 0.00000
8.0 1.13277 1.13277 2.02469 × 1 0 7
6.0 1.37286 1.37286 2.47561 × 1 0 7
4.0 1.56705 1.56705 1.91849 × 1 0 7
4.0 1.48885 1.48885 1.47740 × 1 0 8
6.0 1.44165 1.44165 9.18621 × 1 0 9
8.0 1.41067 1.41067 5.86270 × 1 0 9
10.0 1.39399 1.39399 0.00000
Figure 2 
                  At 
                        
                           
                           
                              Δ
                              x
                              =
                              0.2
                           
                           \Delta x=0.2
                        
                     , 
                        
                           
                           
                              Δ
                              τ
                              =
                              0.1
                           
                           \Delta \tau =0.1
                        
                     , 
                        
                           
                           
                              χ
                              =
                              0.1
                           
                           \chi =0.1
                        
                     , 
                        
                           
                           
                              b
                              =
                              0.8
                           
                           b=0.8
                        
                     , 
                        
                           
                           
                              ℧
                              =
                              0.8
                           
                           \mho =0.8
                        
                     , 
                        
                           
                           
                              γ
                              =
                              0.1
                           
                           \gamma =0.1
                        
                     , 
                        
                           
                           
                              Λ
                              =
                              0.2
                           
                           \Lambda =0.2
                        
                     , 
                        
                           
                           
                              d
                              =
                              0.2
                           
                           d=0.2
                        
                     , 
                        
                           
                           
                              q
                              =
                              0.1
                           
                           q=0.1
                        
                     , the numerical findings of (1.2) are compared to the analytical solution of (5.10).
Figure 2

At Δ x = 0.2 , Δ τ = 0.1 , χ = 0.1 , b = 0.8 , = 0.8 , γ = 0.1 , Λ = 0.2 , d = 0.2 , q = 0.1 , the numerical findings of (1.2) are compared to the analytical solution of (5.10).

7 Illustrations with graphics

Here, we show some two-dimensional and three-dimensional figures to help clarify the solutions we presented. Figures 15 depict some of the analytical and numerical solutions. In Figure 3, we use our method to introduce the graph of (5.5) at χ = 0.5 , b = 0.2 , = 0.3 , γ = 0.3 , Λ = 0.2 , d = 0.1 , and q = 0.5 and we notice that the wave rises upward over time. In addition, Figure 4 shows the graph of (5.8) χ = 0.3 , b = 0.8 , = 0.5 , γ = 0.1 , Λ = 0.6 , d = 0.2 , and q = 0.09 , and we notice that the wave goes up and down over time. In addition, Figure 5 shows the graph of (5.10) at χ = 0.3 , b = 0.8 , = 0.5 , γ = 0.1 , Λ = 0.6 , d = 0.2 , q = 0.09 we notice that the wave goes up and down over time. In addition, we compare the numerical results of (1.2) with the analytical solution (5.3) at Δ x = 0.2 , Δ τ = 0.1 , χ = 0.4 , b = 0.6 , = 0.3 , γ = 0.3 , Λ = 0.2 , d = 0.1 , q = 0.2 in Figure 1. Finally, Figure 2 shows a comparison of the numerical findings of (1.2) with the analytical solution of (5.10) at Δ x = 0.2 , Δ τ = 0.1 , χ = 0.1 , b = 0.8 , = 0.8 , γ = 0.1 , Λ = 0.2 , d = 0.2 , q = 0.1 . From Figures 1 and 2, we notice that the analytical and numerical solutions are very close, which means the accuracy of the methods used.

Figure 3 
               Structure of (5.5) at 
                     
                        
                        
                           χ
                           =
                           0.5
                        
                        \chi =0.5
                     
                  , 
                     
                        
                        
                           b
                           =
                           0.2
                        
                        b=0.2
                     
                  , 
                     
                        
                        
                           ℧
                           =
                           0.3
                        
                        \mho =0.3
                     
                  , 
                     
                        
                        
                           γ
                           =
                           0.3
                        
                        \gamma =0.3
                     
                  , 
                     
                        
                        
                           Λ
                           =
                           0.2
                        
                        \Lambda =0.2
                     
                  , 
                     
                        
                        
                           d
                           =
                           0.1
                        
                        d=0.1
                     
                  , 
                     
                        
                        
                           q
                           =
                           0.5
                        
                        q=0.5
                     
                  .
Figure 3

Structure of (5.5) at χ = 0.5 , b = 0.2 , = 0.3 , γ = 0.3 , Λ = 0.2 , d = 0.1 , q = 0.5 .

Figure 4 
               Structure of (5.8) at 
                     
                        
                        
                           χ
                           =
                           0.3
                           ,
                           b
                           =
                           0.8
                           ,
                           ℧
                           =
                           0.5
                        
                        \chi =0.3,b=0.8,\mho =0.5
                     
                  , 
                     
                        
                        
                           γ
                           =
                           0.1
                        
                        \gamma =0.1
                     
                  , 
                     
                        
                        
                           Λ
                           =
                           0.6
                        
                        \Lambda =0.6
                     
                  , 
                     
                        
                        
                           d
                           =
                           0.2
                        
                        d=0.2
                     
                  , 
                     
                        
                        
                           q
                           =
                           0.09
                           .
                        
                        q=0.09.
Figure 4

Structure of (5.8) at χ = 0.3 , b = 0.8 , = 0.5 , γ = 0.1 , Λ = 0.6 , d = 0.2 , q = 0.09 .

Figure 5 
               
                  
                     
                        
                        
                           χ
                           =
                           0.3
                        
                        \chi =0.3
                     
                  , 
                     
                        
                        
                           b
                           =
                           0.8
                        
                        b=0.8
                     
                  , 
                     
                        
                        
                           ℧
                           =
                           0.5
                        
                        \mho =0.5
                     
                  , 
                     
                        
                        
                           γ
                           =
                           0.1
                        
                        \gamma =0.1
                     
                  , 
                     
                        
                        
                           Λ
                           =
                           0.6
                        
                        \Lambda =0.6
                     
                  , 
                     
                        
                        
                           d
                           =
                           0.2
                        
                        d=0.2
                     
                  , 
                     
                        
                        
                           q
                           =
                           0.09
                        
                        q=0.09
                     
                   graph of (5.10) using the new Kudryashov’s approach.
Figure 5

χ = 0.3 , b = 0.8 , = 0.5 , γ = 0.1 , Λ = 0.6 , d = 0.2 , q = 0.09 graph of (5.10) using the new Kudryashov’s approach.

8 Conclusion

Finally, we showed how to explore the generalized q -deformed sinh-Gordon equation using a novel general form of the Kudryashov method. We developed the Kudryashov approach and put it in a broad form that comprises more than one adjustable constant based on the premise that we want to improve solutions and describe physical phenomena optimally. In this way, we examined the model and gave figures demonstrating the validity of the results we planned to achieve using the proposed strategy. We also presented a comprehensive numerical analysis of this model using the finite differences method. Furthermore, the analytical and numerical solutions were compared. Looking at the results that we obtained numerically or analytically, we find that they are good results and are a clear contribution to solving this type of equation. The forms that we presented for some of the solutions are completely compatible with the numerical results in terms of accuracy, the stability of the wave propagation, and the preservation of its properties. We present a future strategy for our study at the end of this section, as we can solve more than one model this way and discover alternative techniques to solve different models. The proposed equation has opened up new options for describing physical systems that have lost their symmetry.

Acknowledgment

The author thanks the editor-in-chief of the journal and all those in charge of it.

  1. Funding information: There is no funding.

  2. Author contributions: The author has accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: There is no conflict of interest between the author or anyone else regarding this manuscript.

  4. Ethics approval and consent to participate: The author confirms that all the results they obtained are new and there is no conflict of interest with anyone.

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

Appendix

Solve [ { 2 A 0 2 c 1 + A 0 q + A 0 3 = 0 , 4 A 1 A 0 c 1 + A 1 q + 3 A 1 A 0 2 = 0 , a 2 α 2 A 1 2 + α 2 A 1 2 c 2 + 4 A 2 A 0 c 1 + 2 A 1 2 c 1 + A 2 q + 3 A 2 A 0 2 + 3 A 1 2 A 0 = 0 , 2 a 2 α 2 A 1 2 4 a 2 α 2 A 2 A 1 2 α 2 A 1 2 c 2 + 4 α 2 A 2 A 1 c 2 + 4 A 2 A 1 c 1 + A 1 3 + 6 A 0 A 2 A 1 = 0 , a 2 α 2 A 1 2 4 a 2 α 2 A 2 2 + 8 a 2 α 2 A 1 A 2 + α 2 A 1 2 c 2 + 4 α 2 A 2 2 c 2 8 α 2 A 1 A 2 c 2 + 2 A 2 2 c 1 + 3 A 0 A 2 2 + 3 A 1 2 A 2 = 0 , 8 a 2 α 2 A 2 2 4 a 2 α 2 A 1 A 2 8 α 2 A 2 2 c 2 + 4 α 2 A 1 A 2 c 2 + 3 A 1 A 2 2 = 0 , 4 a 2 α 2 A 2 2 + 4 α 2 A 2 2 c 2 + A 2 3 = 0 } , { A 0 , A 1 , A 2 , a , c 1 } ] .

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Received: 2022-06-13
Revised: 2022-07-31
Accepted: 2022-09-02
Published Online: 2023-02-14

© 2023 the author(s), published by De Gruyter

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

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