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On solitary wave solutions of a peptide group system with higher order saturable nonlinearity

  • Mustafa Inc , Ahmed Elhassanein , Mohamed Aly Mohamed Abdou and Yu-Ming Chu EMAIL logo
Published/Copyright: December 10, 2020

Abstract

In this article, a new continuous model of a peptide group system with higher order saturable nonlinearity is derived. The model is constructed up on discrete Schrodinger equation with higher order saturable nonlinearity. New exact solutions of the proposed model are investigated. Via the generalized Riccati equation mapping method travelling wave and soliton solutions are derived. For certain values of parameters kink, antikink, breathers, and dark and bright solitons are presented.

1 Introduction

Recently, nonlinear systems have a great attention [1,2,3,4,5,6,7]. Solitons are robust localized modes generated by this type of nonlinear evolution equations and have a great deal of interest. They have been observed in many such systems. Two soliton solutions for a fifth-order nonlinear evolution equation have been derived [8]. Exact soliton solutions for a discrete nonlinear electrical transmission line in (2 + 1) dimension have been derived [9]. Variational approximation is used to predict the form of the discrete solitons of a generalized version of the Ablowitz–Ladik model with a power-law nonlinearity [10]. Soliton solutions of the Camassa–Holm equation are studied [11]. Optical solitons in birefringent fibers have been discussed [12]. Riccati equation mapping method is one of the most important methods of constructing travelling wave solutions [13,14,15,16,17]. Using Davydov’s idea about energy transfer along protein molecules [18], discrete nonlinear Schrodinger equation (DNLSE) with saturable nonlinearity has been used to model the state of amide-I excitations in proteins [19,20]. In this work, the generalized Riccati equation mapping algorithm is used to investigate the solitary wave solutions of the system introduced by ref. [21].

The article is organized as follows. Generalized Riccati algorithm is given in Section 2. The mathematical model is derived in Section 3,, and Section 4 is devoted to present solitary solutions and numerical example. Conclusion is given in Section 5.

2 Generalized Riccati algorithm

Considering the partial differential equation:

(2.1) L ( ξ , ξ x , ξ t , ξ x x , ξ x t , ζ t t ) = 0 ,

where ξ = ξ ( x , t ) , the algorithm can be summarized in the following steps [17].

  1. Using traveling wave transformations ω = γ x + ϰ t , ξ ( x , t ) = Ϝ ( ω ) , equation (2.1) can be transformed into an ordinary differential equation:

    (2.2) G ( Ϝ , Ϝ , Ϝ , ) = 0 ,

    where primes denote the derivatives with respect to δ .

  2. The solution Ϝ ( δ ) of equation (2.2) is considered in the following form:

    (2.3) Ϝ ( ω ) = j = 0 m ϵ j ϰ j ( ω ) ,

    where m and ϵ j , j = 1 , 2 , , m , are parameters to be evaluated and the function ϰ ( δ ) be such that

    (2.4) d ϰ ( ω ) d ω = a ϰ 2 ( ω ) + b ϰ ( ω ) + c .

  3. Substituting equation (2.3) along with equation (2.4) into equation (2.2), we get a system of algebraic equations corresponding to the powers of ϰ ( δ ) to be solved in order to obtain constants. Exact solutions are depending on the discriminant b 2 4 a c .

3 Mathematical system

The one-dimensional DNLSE lattice model with higher order saturable nonlinearity [21] is investigated:

(3.1) i ξ n t = ( ξ n + 1 + ξ n 1 2 ξ n ) + α 1 ξ n 1 + | ξ n | 2 α 2 | ξ n | 2 ξ n ( 1 + | ξ n | 2 ) 2 ,

where ξ n is the dimensionless state in the n th lattice ( n = 0 , 1 , 2 , , N 1 ) , α 1 and α 2 are real nonlinearity parameters, and t corresponds to temporal coordinates.

Considering the stationary solution of equation (2.1) in the form ξ n = γ n e i β t , we get the system:

(3.2) β γ n + ( γ n + 1 + γ n 1 2 γ n ) α 1 γ n 1 + γ n 2 + α 2 γ n 3 ( 1 + γ n 2 ) 2 = 0 .

Following [17] for a continuous variable x the system (2.2) can be approximated as:

(3.3) d 2 γ d x 2 ( α 1 β ) γ + ( α 1 + α 2 ) γ 3 ( α 1 + 2 α 2 ) γ 5 = 0 .

4 Exact solutions

Here we investigate solitary wave solutions of the system (3.3) using the generalized Riccati algorithm. Considering the traveling wave transformations

(4.1) δ = l ( x α t ) , γ ( x , t ) = Γ ( δ ) , Γ ( δ ) = ± H ( δ ) .

Equation (3.3) can be written as:

(4.2) l 4 H d H d δ 2 + l 2 d 2 H d δ 2 ( α 1 β ) H + ( α 1 + α 2 ) H 2 ( α 1 + 2 α 2 ) H 3 = 0 .

From (2.3) and (2.4), we obtain the following system of algebraic equations:

ϰ 0 : ( α 1 + α 2 ) ϵ 0 3 ( α 1 + 2 α 2 ) ϵ 0 4 + ( β α 1 ) ϵ 0 2 l 4 c 2 ϵ 1 2 + l 2 b c ϵ 0 ϵ 1 = 0 , ϰ 1 : 2 ( β α 1 ) ϵ 0 ϵ 1 + 3 ( α 1 + α 2 ) ϵ 0 2 ϵ 1 4 ( α 1 + 2 α 2 ) ϵ 0 3 ϵ 1 + l 2 ( b 2 + 2 a c ) ϵ 0 ϵ 1 = 0 , ϰ 2 : ( β α 1 ) ϵ 1 2 + 3 ( α 1 + α 2 ) ϵ 0 ϵ 1 2 6 ( α 1 + 2 α 2 ) ϵ 0 2 ϵ 1 2 + l 4 ( b 2 + 2 a c ) ϵ 1 2 + 3 l 2 a c ϵ 0 ϵ 1 = 0 , ϰ 3 : ( α 1 + α 2 ) ϵ 1 3 4 ( α 1 + 2 α 2 ) ϵ 0 ϵ 1 3 + l ( a 2 ϵ 0 + a b ϵ 1 ) ϵ 1 = 0 , ϰ 4 : 3 l 4 a 2 ϵ 1 2 ( α 1 + 2 α 2 ) ϵ 1 4 = 0 .

By solving the preceding system under the condition β = α 1 3 16 ( α 1 + α 2 ) 2 ( α 1 + α 2 ) , we get the needed constants:

(4.3) l = 3 4 ( α 1 + α 2 ) 2 ( α 1 + 2 α 2 ) 1 b 2 4 a c , ϵ 0 = 3 8 ( α 1 + α 2 ) ( α 1 + 2 α 2 ) + 3 8 b 2 b 2 4 a c , ϵ 1 = 3 8 ( α 1 + α 2 ) ( α 1 + 2 α 2 ) a 2 b 2 4 a c , α 1 2 α 2 .

New types of solutions of equation (3.3) are given in the following [12].

Family 1. For υ = b 2 4 a c > 0 , δ = l ( x α t ) and a b 0 (or a c 0 ).

(4.4) γ 1 ( x , t ) = ± ϵ 0 ϵ 1 2 a b + υ tanh υ 2 δ 1 / 2 ,

(4.5) γ 2 ( x , t ) = ± ϵ 0 ϵ 1 2 a b + υ coth υ 2 δ 1 / 2 ,

(4.6) γ 3 ( x , t ) = ± ϵ 0 ϵ 1 4 a 2 b + υ tanh υ 4 δ ± coth υ 4 δ 1 / 2 ,

(4.7) γ 4 ( x , t ) = ± ϵ 0 + ϵ 1 2 a b + ( a 1 2 a 0 2 ) υ + a 0 υ cosh ( υ δ ) a 0 sinh ( υ δ ) + a 1 1 / 2 ,

where a 0 and a 1 are real constants and a 1 2 a 0 2 > 0 .

(4.8) γ 5 ( x , t ) = ± ϵ 0 + 2 c ϵ 1 cosh υ 2 δ υ sinh υ 2 δ b cosh υ 2 δ 1 / 2 ,

(4.9) γ 6 ( x , t ) = ± ϵ 0 2 c ϵ 1 sinh υ 2 δ b sinh υ 2 δ υ cosh υ 2 δ 1 / 2 ,

(4.10) γ 7 ( x , t ) = ± ϵ 0 + 2 c ϵ 1 cosh υ 2 δ υ sinh υ δ b cosh υ δ ± i υ 1 / 2 ,

(4.11) γ 8 ( x , t ) = ± ϵ 0 + 2 c ϵ 1 sinh υ 2 δ b sinh υ δ + υ cosh υ δ ± υ 1 / 2 ,

(4.12) γ 9 ( x , t ) = ± ϵ 0 + 4 c ϵ 1 sinh υ 4 δ cosh υ 4 δ 2 b sinh υ 4 δ cosh υ 4 δ + 2 υ cosh 2 υ 4 δ υ 1 / 2 .

Family 2. For b 2 4 a c < 0 and a b 0 (or a c 0 ).

(4.13) γ 10 ( x , t ) = ± ϵ 0 + ϵ 1 2 a b + υ tan υ 2 δ 1 / 2 ,

(4.14) γ 11 ( x , t ) = ± ϵ 0 ϵ 1 2 a b + υ cot υ 2 δ 1 / 2 ,

(4.15) γ 12 ( x , t ) = ± ϵ 0 + ϵ 1 4 a 2 b + υ tan υ 4 δ ± cot υ 4 δ 1 / 2 ,

(4.16) γ 13 ( x , t ) = ± ϵ 0 + ϵ 1 2 a b + ( b 1 2 b 0 2 ) ( υ ) a 0 υ cos υ δ b 1 sin υ δ + a 1 1 / 2 ,

where b 0 and b 1 are real constants and b 1 2 b 0 2 < 0 .

(4.17) γ 14 ( x , t ) = ± ϵ 0 2 c ϵ 1 cos υ 2 δ υ sin υ 2 δ + b cos υ 2 δ 1 / 2 ,

(4.18) γ 15 ( x , t ) = ± ϵ 0 2 c ϵ 1 sin υ 2 δ b sin υ 2 δ υ cos υ 2 δ 1 / 2 ,

(4.19) γ 16 ( x , t ) = ± ϵ 0 2 c ϵ 1 cos υ 2 δ υ sin υ δ + b cos υ δ ± υ 1 / 2 ,

(4.20) γ 17 ( x , t ) = ± ϵ 0 + 2 c ϵ 1 sin υ 2 δ b sin υ δ + υ cos υ δ ± υ 1 / 2 ,

(4.21) γ 18 ( x , t ) = ± ϵ 0 + 4 c ϵ 1 sinh υ 4 δ cosh υ 4 δ 2 b sin υ 4 δ cos υ 4 δ + 2 υ cos 2 υ 4 δ υ 1 / 2 .

Family 3. For c = 0 , and a b 0 .

(4.22) γ 19 ( x , t ) = ± ϵ 0 d b ϵ 1 a ( d + cosh ( b δ ) sinh ( b δ ) ) 1 / 2 ,

(4.23) γ 20 ( x , t ) = ± ϵ 0 + b ϵ 1 ( cosh ( b δ ) + sinh ( b δ ) ) a ( d + cosh ( b δ ) sinh ( b δ ) ) ) 1 / 2 ,

where d is a constant.

4.1 Numerical example

For α 1 = 10 , α 2 = 21 , a = 1.5 , b = 4 , and c = 2 : (1) kink and antikink solutions of equation (4.4) are plotted in Figure 1(a) and (b), respectively; (2) with a 1 = 2 and a 0 = 1 , breather of equation (4.7) is given in Figure 2(a) and antibreather is given in Figure 2(b); and (3) dark solutions for equation (4.11) are plotted in Figure 3(a) and (d) and bright solutions are plotted in Figure 3(b) and (d).

Figure 1 
                  For equation (4.4) with 
                        
                           
                           
                              
                                 
                                    α
                                 
                                 
                                    1
                                 
                              
                              =
                              10
                           
                           {\alpha }_{1}=10
                        
                     , 
                        
                           
                           
                              
                                 
                                    α
                                 
                                 
                                    2
                                 
                              
                              =
                              21
                           
                           {\alpha }_{2}=21
                        
                     , 
                        
                           
                           
                              a
                              =
                              1.5
                           
                           a=1.5
                        
                     , 
                        
                           
                           
                              b
                              =
                              4
                           
                           b=4
                        
                     , and 
                        
                           
                           
                              c
                              =
                              2
                           
                           c=2
                        
                     , (a) kink solution and (b) antikink solution.
Figure 1

For equation (4.4) with α 1 = 10 , α 2 = 21 , a = 1.5 , b = 4 , and c = 2 , (a) kink solution and (b) antikink solution.

Figure 2 
                  For equation (4.7) with 
                        
                           
                           
                              
                                 
                                    α
                                 
                                 
                                    1
                                 
                              
                              =
                              10
                           
                           {\alpha }_{1}=10
                        
                     , 
                        
                           
                           
                              
                                 
                                    α
                                 
                                 
                                    2
                                 
                              
                              =
                              21
                           
                           {\alpha }_{2}=21
                        
                     , 
                        
                           
                           
                              a
                              =
                              1.5
                           
                           a=1.5
                        
                     , 
                        
                           
                           
                              b
                              =
                              4
                              ,
                              c
                              =
                              2
                              ,
                              
                                 
                                    a
                                 
                                 
                                    0
                                 
                              
                              =
                              1
                           
                           b=4,c=2,{a}_{0}=1
                        
                     , and 
                        
                           
                           
                              
                                 
                                    a
                                 
                                 
                                    1
                                 
                              
                              =
                              2
                           
                           {a}_{1}=2
                        
                     , (a) breather solution and (b) antibreather solution.
Figure 2

For equation (4.7) with α 1 = 10 , α 2 = 21 , a = 1.5 , b = 4 , c = 2 , a 0 = 1 , and a 1 = 2 , (a) breather solution and (b) antibreather solution.

Figure 3 
                  For equation (4.11) with 
                        
                           
                           
                              
                                 
                                    α
                                 
                                 
                                    1
                                 
                              
                              =
                              10
                           
                           {\alpha }_{1}=10
                        
                     , 
                        
                           
                           
                              
                                 
                                    α
                                 
                                 
                                    2
                                 
                              
                              =
                              21
                           
                           {\alpha }_{2}=21
                        
                     , 
                        
                           
                           
                              a
                              =
                              1.5
                           
                           a=1.5
                        
                     , 
                        
                           
                           
                              b
                              =
                              4
                           
                           b=4
                        
                     , and 
                        
                           
                           
                              c
                              =
                              2
                           
                           c=2
                        
                     , (a) dark solution, (b) dark solution, (c) dark solution, and (d) dark solution.
Figure 3

For equation (4.11) with α 1 = 10 , α 2 = 21 , a = 1.5 , b = 4 , and c = 2 , (a) dark solution, (b) dark solution, (c) dark solution, and (d) dark solution.

5 Conclusion and discussion

Using Davydov’s idea about energy transfer along protein molecules, DNLSE with higher order saturable nonlinearity has been used to model the state of amide-I excitations in proteins. Here equation (3.3) is a new continuous derivation of DNLSE and Riccati equation mapping method is successfully employed for constructing the exact solutions for this system. The exact solutions of equation (3.3) have not been considered by any methods till now. Simulation is used to show kink, antikink, breather, antibreather, and dark solutions for equation (3.3).

  1. Funding source: This work was supported by the Natural Science Foundation of China (Grant No. 61673169, 11301127, 11701176, 11626101, and 11601485).

  2. Author contribution: The authors contributed equally to the writing of this article. They read and approved the final manuscript.

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Received: 2020-09-01
Revised: 2020-10-04
Accepted: 2020-10-25
Published Online: 2020-12-10

© 2020 Mustafa Inc et al., published by De Gruyter

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

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