Khaled El-Rashidy , Aly R. Seadawy , Saad Althobaiti and M. M. Makhlouf

Investigation of interactional phenomena and multi wave solutions of the quantum hydrodynamic Zakharov–Kuznetsov model

De Gruyter | Published online: March 9, 2021

Abstract

The symbolic computation with the ansatz function and the logarithmic transformation method are used to obtain a formula for certain exact solutions of the ( 3 + 1 ) Zakharov–Kuznetsov (Z–K) equation. We use homoclinic breather, three waves method, and double exponential. There is a conflict of results with considerably known results, which indicates the solutions found in this study are new. By selecting appropriate parameter values, 3d representations are plotted to establish W-shaped, multi-peak, and kinky breathers solutions.

1 Introduction

Many physical phenomena originate in various fields of engineering and science, such as fluid dynamics, fiber optics, quantum mechanics, plasma, and physics. These physical phenomena were constructed in the form of nonlinear equations [1,2, 3,4,5, 6,7]. To understand the behavior of a phenomenon, we need to solve nonlinear equations that describe the phenomenon, which is often challenging. Many methods have been developed over the years to solve such nonlinear equations, many of which are based on assumptions. Investigating the exact solutions for these nonlinear equations is a major area of concern for many mathematicians and physicists because of their important role in understanding the behavior of nonlinear physical phenomena. The Zakharov–Kuznetsov (Z–K) equation supports stable solitary waves, see ref. [8,9]. This makes the Z–K equation a very attractive model equation for investigating of vortices in geophysical owe. The Z–K equation determines weak non-linear behavior of ion sound waves that contain electrons with hot and cold temperatures in a regular magnetic field, see ref. [9]. With regard to the exact solutions of the Z–K and ( 3 + 1 ) -dimensional Z–K equations, many solving methods have been used to solve it, see [2,9, 10,11,12, 13,14], and have many applications in numerical and analytical approaches in physical sciences [15,16, 17,18,19, 20,21]. An efficient analytical technique for fractional partial differential equations occurs in ion-acoustic waves in plasma and warm plasma [37,38].

Some new methods and important developments in the search for analysis have been done to investigate wave solutions for partial non-linear differential equations. The results of this manuscript may be completely complementary to in existence manuscripts of literature such as: direct algebraic method that is extended and modified; techniques’ Seadawy and extended mapping method to find solutions for some nonlinear partial differential equations [5]; showing the bi-directional propagation of small-amplitude long-capillary gravity waves on the surface of shallow water [22]; bright and dark solitons, solitary wave solutions of higher-order non-linear differential equations and the elliptic function [23]; the higher-order non-linear differential equations with cubic quintic nonlinearity [24]; solitary wave solutions to the nonlinear-modified KdV dynamical equation [25]; modified equal–width equations and dispersive traveling wave solutions of the equal–width [26]; the exact travelling wave solutions of the SRLW equation and modified Liouville equation [27]; fourth-order nonlinear Ablowitz–Kaup–Newell–Segur water wave dynamical equation [28]; nonlinear wave solutions of the Kudryashov–Sinelshchikov dynamical equation [29]; the third-order NLS equation [30]; approximate solutions of nonlinear Parabolic equation by using modified variational iteration algorithm [31]; new solitary wave solutions of nonlinear Nizhnik–Novikov–Vesselov equation [32]; and Hirota bilinear method was used to study multiple soliton interactions [33].

The purpose of this work is to find traveling wave solutions of ( 3 + 1 ) equation-dimensional Z–K using logarithmic transformation and symbolic computation by using the function method, see [10,34, 35,36]:

(1) u t + μ u u x + 1 2 u x x x + 1 2 ( δ + 1 ) u x y y + u x z z = 0 ,
the function u is a function of the variables x , y , z and the temporal variable t which is the electrostatic wave potential in plasmas, where μ and δ are real constants coefficient.

Consider the solution in its complex form as follows:

(2) u ( x , y , z , t ) = ϕ ( ξ ) , ξ = k x + ν y + λ z + ω t ,
where k , ν , λ , and ω are real constants wave numbers and frequency.

Substituting from equation (2) into equation (1) and separating the imaginary and real parts lead to the ordinary differential equation as:

(3) k ( δ + 1 ) ν 2 + k 2 + 2 λ 2 ϕ ( ξ ) + k μ ϕ ( ξ ) 2 + 2 ω ϕ ( ξ ) = 0

For exact solutions to equation (1), we use the logarithmic transformation and function method for equation (3). Introduction to the set of basic equations is procured in Section 3. A conclusion is carried out in Section 4.

2 Interactional phenomena and multi-waves solutions

Solutions for multiple waves and fractional phenomena logarithmic transformation can be shown as

(4) ϕ ( ξ ) = 2 log ( f ( ξ ) ) ξ
equation ( 3) has the form:
(5) k ( δ + 1 ) ν 2 + k 2 f ( ξ ) 2 f ( 3 ) ( ξ ) + 2 k ( δ + 1 ) ν 2 + k 2 + 2 λ 2 f ( ξ ) 3 + 2 k μ f ( ξ ) f ( ξ ) 2 + 2 ω f ( ξ ) 2 f ( ξ ) 3 k ( δ + 1 ) ν 2 + k 2 + 2 λ 2 f ( ξ ) f ( ξ ) f ( ξ ) = 0

2.1 Type (I): Three waves hypothesis

In this subsection, for the next three waves hypothesis, multiple wave solutions can be expressed for nonlinear equation (5) as follows:

(6) f ( ξ ) = b 0 cosh ( a 1 ξ + a 2 ) + b 1 cos ( a 3 ξ + a 4 ) + b 2 cosh ( a 5 ξ + a 6 ) ,
where the real constants a j , ( j = 1 , , 6 ) are determined later. Substituting ( 6) into ( 5) with the help of collecting all the parameters for all the exponents of cosh ( a 1 ξ + a 2 ) , cos ( a 3 ξ + a 4 ) , cosh ( a 5 ξ + a 6 ) , sinh ( a 1 ξ + a 2 ) , sin ( a 3 ξ + a 4 ) and sinh ( a 5 ξ + a 6 ) functions be zero, which establishes a set of algebraic equations with respect to above constants. With help of the Wolfram Mathematica program software, solving the determination algebraic equations, we conclude the following cases:

Case 1

(7) a 1 = ω 2 λ k 2 6 λ 2 4 , a 5 = a 1 , a 3 = 0 , b 1 = b 0 2 + b 2 2 , ν = 4 λ 2 k 2 δ + 1 ,
substituting from ( 6), ( 7), and ( 4). By taking γ 1 = ω 2 λ k 2 6 λ 2 4 and τ = 2 ω λ k 2 6 λ 2 4 , we obtain,
(8) u 11 ( x , y , z , t ) = τ b 0 sinh a 2 + γ ( k x + t ω + ν y + λ z ) + b 2 sinh a 6 + γ ( k x + t ω + ν y + λ z ) b 0 cosh a 2 + γ ( k x + t ω + ν y + λ z ) + b 2 cosh a 6 + γ ( k x + t ω + ν y + λ z ) + b 0 2 + b 2 2 cos a 4

Figure 1

Figure 1 
                  This figure for wave solutions (8) using parameters: 
                        
                           
                           
                              k
                              =
                              3
                           
                           k=3
                        
                     , 
                        
                           
                           
                              λ
                              =
                              1
                           
                           \lambda =1
                        
                     , 
                        
                           
                           
                              δ
                              =
                              5
                           
                           \delta =5
                        
                     , 
                        
                           
                           
                              ω
                              =
                              4
                           
                           \omega =4
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    2
                                 
                              
                              =
                              1
                           
                           {a}_{2}=1
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    4
                                 
                              
                              =
                              30
                           
                           {a}_{4}=30
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    6
                                 
                              
                              =
                              2
                           
                           {a}_{6}=2
                        
                     , 
                        
                           
                           
                              
                                 
                                    b
                                 
                                 
                                    0
                                 
                              
                              =
                              1
                           
                           {b}_{0}=1
                        
                     , 
                        
                           
                           
                              
                                 
                                    b
                                 
                                 
                                    2
                                 
                              
                              =
                              1
                           
                           {b}_{2}=1
                        
                     , 
                        
                           
                           
                              z
                              =
                              1
                           
                           z=1
                        
                     , 
                        
                           
                           
                              y
                              =
                              
                                 
                                    
                                       
                                          3
                                       
                                       
                                          5
                                       
                                    
                                 
                              
                           
                           y=\sqrt{\frac{3}{5}}
                        
                     . (a) and (b) Represented wave solutions in three dimensions, (c) and (d) showed wave solutions in two dimensions and (e) and (f) a contour solutions.

Figure 1

This figure for wave solutions (8) using parameters: k = 3 , λ = 1 , δ = 5 , ω = 4 , a 2 = 1 , a 4 = 30 , a 6 = 2 , b 0 = 1 , b 2 = 1 , z = 1 , y = 3 5 . (a) and (b) Represented wave solutions in three dimensions, (c) and (d) showed wave solutions in two dimensions and (e) and (f) a contour solutions.

Case 2

(9) a 1 = 1 2 , a 3 = i 2 , a 5 = 0 , ω = λ 2 k 2 6 λ 2 , δ = k 2 + 4 λ 2 ν 2 ν 2 , b 0 = i b 1 ,
substituting from ( 6), ( 9), and ( 4), we have:
(10) u 12 ( x , y , z , t ) = 2 b 1 i sinh a 2 + k x + t ω + ν y + λ z 2 + sinh k x + t ω + ν y + λ z 2 + i a 4 b 2 cosh a 6 + b 1 i cosh a 2 + k x + t ω + ν y + λ z 2 + cosh k x + t ω + ν y + λ z 2 + i a 4
Figure 2

Figure 2 
                  This figure for wave solutions (10) using parameters: 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    2
                                 
                              
                              =
                              1
                           
                           {a}_{2}=1
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    4
                                 
                              
                              =
                              1
                           
                           {a}_{4}=1
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    6
                                 
                              
                              =
                              2
                           
                           {a}_{6}=2
                        
                     , 
                        
                           
                           
                              
                                 
                                    b
                                 
                                 
                                    2
                                 
                              
                              =
                              2
                              
                                 
                                    2
                                 
                              
                           
                           {b}_{2}=2\sqrt{2}
                        
                     , 
                        
                           
                           
                              
                                 
                                    b
                                 
                                 
                                    1
                                 
                              
                              =
                              2
                              
                                 
                                    2
                                 
                              
                           
                           {b}_{1}=2\sqrt{2}
                        
                     , 
                        
                           
                           
                              y
                              =
                              2
                           
                           y=2
                        
                     , 
                        
                           
                           
                              k
                              =
                              4
                           
                           k=4
                        
                     , 
                        
                           
                           
                              ν
                              =
                              2
                           
                           \nu =2
                        
                     . Wave solutions appear in (a) and (b) as three dimensions but (c) and (d) represented wave solutions in two dimensions and (e) and (f) a contour solutions.

Figure 2

This figure for wave solutions (10) using parameters: a 2 = 1 , a 4 = 1 , a 6 = 2 , b 2 = 2 2 , b 1 = 2 2 , y = 2 , k = 4 , ν = 2 . Wave solutions appear in (a) and (b) as three dimensions but (c) and (d) represented wave solutions in two dimensions and (e) and (f) a contour solutions.

Case 3

(11) a 1 = 0 , a 3 = k δ λ 2 ν 2 + 2 λ 4 + λ 2 ν 2 , a 5 i k δ λ 2 ν 2 + 2 λ 4 + λ 2 ν 2 , b 1 = i b 2 , ω = 4 k 2 δ ν 2 + 2 λ 2 + ν 2
substituting from ( 6), ( 11), and ( 4). By taking γ 2 = k ( δ + 1 ) λ 2 ν 2 + 2 λ 4 , we obtain
(12) u 13 ( x , y , z , t ) 2 i b 2 γ 3 sin a 4 + γ 2 ( k x + t ω + ν y + λ z ) i sin γ 2 ( k x + t ω + ν y + λ z ) i a 6 b 0 cosh a 2 + b 2 i cos a 4 + γ 2 ( k x + t ω + ν y + λ z ) + cos γ 2 ( k x + t ω + ν y + λ z ) i a 6
Figure 3

Figure 3 
                  This figure for wave solutions (12) using parameters: 
                        
                           
                           
                              k
                              =
                              −
                              4
                           
                           k=-4
                        
                     , 
                        
                           
                           
                              λ
                              =
                              2
                           
                           \lambda =2
                        
                     , 
                        
                           
                           
                              δ
                              =
                              1
                           
                           \delta =1
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    2
                                 
                              
                              =
                              2
                           
                           {a}_{2}=2
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    4
                                 
                              
                              =
                              1
                           
                           {a}_{4}=1
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    6
                                 
                              
                              =
                              2
                           
                           {a}_{6}=2
                        
                     , 
                        
                           
                           
                              
                                 
                                    b
                                 
                                 
                                    2
                                 
                              
                              =
                              2
                           
                           {b}_{2}=2
                        
                     , 
                        
                           
                           
                              z
                              =
                              1
                           
                           z=1
                        
                     , 
                        
                           
                           
                              y
                              =
                              2
                           
                           y=2
                        
                     , 
                        
                           
                           
                              ν
                              =
                              2
                           
                           \nu =2
                        
                     , 
                        
                           
                           
                              
                                 
                                    b
                                 
                                 
                                    0
                                 
                              
                              =
                              1
                           
                           {b}_{0}=1
                        
                     . (a) and (b) Ploted as wave solutions in three dimensions, (c) and (d) showed wave solutions in two dimensions and (e) and (f) a contour solutions.

Figure 3

This figure for wave solutions (12) using parameters: k = 4 , λ = 2 , δ = 1 , a 2 = 2 , a 4 = 1 , a 6 = 2 , b 2 = 2 , z = 1 , y = 2 , ν = 2 , b 0 = 1 . (a) and (b) Ploted as wave solutions in three dimensions, (c) and (d) showed wave solutions in two dimensions and (e) and (f) a contour solutions.

2.2 Type (II): Interactional phenomena and exponential form

The hypothesis of a double exponential function takes the form:

(13) f ( ξ ) = b 1 e a 1 ξ + a 2 + b 2 e a 3 ξ + a 4 ,
where the real constants a j ( j = 1 , , 6 ) are determined later. By substituting ( 13) into ( 5) with the help of symbolic accounts having all the coefficients of all powers of exponential functions zero, we have a system of algebraic equations. By resolving this system, we get
(14) a 1 = 2 3 , a 3 = 2 3 , μ = 6 3 ν 2 + 12 ν 2 11 3 + 22 ,
substitution from ( 13), ( 14) and ( 4), we obtain
(15) u 2 ( x , y , z , t ) = 2 2 e a 2 b 1 + e a 4 b 2 e 2 ( k x + t ω + ν y + λ z ) 3 3 e a 2 b 1 + e a 4 b 2 e 2 ( k x + t ω + ν y + λ z ) 3
Figure 4

Figure 4 
                  This figure for wave solutions (15) using parameters: 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    1
                                 
                              
                              =
                              
                                 
                                    2
                                 
                                 
                                    
                                       
                                          3
                                       
                                    
                                 
                              
                           
                           {a}_{1}=\frac{2}{\sqrt{3}}
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    3
                                 
                              
                              =
                              
                                 
                                    2
                                 
                                 
                                    
                                       
                                          3
                                       
                                    
                                 
                              
                           
                           {a}_{3}=\frac{2}{\sqrt{3}}
                        
                     .  (a) and (b) Represented wave solutions in three dimensions, (c) and (d) showed wave solutions in two dimensions and (e) and (f) a contour solutions.

Figure 4

This figure for wave solutions (15) using parameters: a 1 = 2 3 , a 3 = 2 3 . (a) and (b) Represented wave solutions in three dimensions, (c) and (d) showed wave solutions in two dimensions and (e) and (f) a contour solutions.

2.3 Type (III): Homoclinic breather approach

The Homoclinic breather approach function takes the form:

(16) f ( ξ ) = e ξ 1 ( p ) + b 0 cos ξ 3 p 1 + b 1 e ξ 2 p ,
where the real constants a j ( j = 1 , , 6 ) are determined later. Substituting ( 16) into ( 4) and by resolving the algebraic equations, we obtain:
(17) a 5 i 6 μ λ 2 p 1 , a 1 3 μ λ 2 p , a 3 3 μ λ 2 p , ω 9 k μ 2 λ 2 ,
substitution from ( 16) and ( 17) using ( 4). By taking γ 3 = μ λ 2 , the wave solution takes the form:
(18) u 3 ( x , y , z , t ) = 2 γ 3 6 b 0 e a 2 p + 3 γ 3 ( k x + t ω + ν y + λ z ) sinh 6 γ 3 ( k x + t ω + ν y + λ z ) i a 6 p 1 3 b 1 e a 2 + a 4 p 3 b 0 e a 2 p + 3 γ 3 ( k x + t ω + ν y + λ z ) cosh 6 γ 3 ( k x + t ω + ν y + λ z ) i a 6 p 1 + b 1 e a 2 + a 4 p + 1
Figure 5

Figure 5 
                  This figure for wave solutions (18) using parameters: 
                        
                           
                           
                              k
                              =
                              1
                           
                           k=1
                        
                     , 
                        
                           
                           
                              z
                              =
                              1
                           
                           z=1
                        
                     , 
                        
                           
                           
                              y
                              =
                              1
                           
                           y=1
                        
                     , 
                        
                           
                           
                              
                                 
                                    p
                                 
                                 
                                    1
                                 
                              
                              =
                              1
                           
                           {p}_{1}=1
                        
                     , 
                        
                           
                           
                              p
                              =
                              1
                           
                           p=1
                        
                     , 
                        
                           
                           
                              λ
                              =
                              1
                           
                           \lambda =1
                        
                     , 
                        
                           
                           
                              μ
                              =
                              
                                 
                                    1
                                 
                                 
                                    3
                                 
                              
                           
                           \mu =\frac{1}{3}
                        
                     , 
                        
                           
                           
                              ν
                              =
                              1
                           
                           \nu =1
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    2
                                 
                              
                              =
                              1
                           
                           {a}_{2}=1
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    4
                                 
                              
                              =
                              1
                           
                           {a}_{4}=1
                        
                     , 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    6
                                 
                              
                              =
                              −
                              i
                           
                           {a}_{6}=-i
                        
                     , 
                        
                           
                           
                              
                                 
                                    b
                                 
                                 
                                    0
                                 
                              
                              =
                              1
                           
                           {b}_{0}=1
                        
                     , 
                        
                           
                           
                              
                                 
                                    b
                                 
                                 
                                    1
                                 
                              
                              =
                              1
                           
                           {b}_{1}=1
                        
                     . (a) and (b) Represented wave solutions in three dimensions, (c) and (d) showed wave solutions in two dimensions and (e) and (f) a contour solutions.

Figure 5

This figure for wave solutions (18) using parameters: k = 1 , z = 1 , y = 1 , p 1 = 1 , p = 1 , λ = 1 , μ = 1 3 , ν = 1 , a 2 = 1 , a 4 = 1 , a 6 = i , b 0 = 1 , b 1 = 1 . (a) and (b) Represented wave solutions in three dimensions, (c) and (d) showed wave solutions in two dimensions and (e) and (f) a contour solutions.

3 Conclusion

In this work, the multi-wave solution method has been used successfully to obtain solution the ( 3 + 1 ) -dimensions (Z–K) equation by helping of logarithmic transformation and symbolic computation. This methods is important from an applied and theoretical point of view, and an additional family of the solutions has been found, which is the best thing about this method. Novel solutions include the generalized solutions to ( 2 + 1 ) -dimensional (Z–K) equation and represent their graphs by determining the appropriate values of the parameters involved.

Acknowledgments

Taif University Researchers Supporting Project number (TURSP-2020/165), Taif University, Taif, Saudi Arabia.

    Conflict of interest: Authors state no conflict of interest.

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Received: 2020-09-30
Revised: 2021-01-28
Accepted: 2021-02-09
Published Online: 2021-03-09

© 2021 Khaled El-Rashidy et al., published by De Gruyter

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