Startseite Solitary wave solutions of the ionic currents along microtubule dynamical equations via analytical mathematical method
Artikel Open Access

Solitary wave solutions of the ionic currents along microtubule dynamical equations via analytical mathematical method

  • Noufe H. Aljahdaly , Amjad F. Alyoubi und Aly R. Seadawy EMAIL logo
Veröffentlicht/Copyright: 8. September 2021

Abstract

In this article, a new generalized exponential rational function method (GERFM) is employed to extract new solitary wave solutions for the ionic currents along microtubules dynamical equations, which is very interested in nanobiosciences. In this article, the stability of the solutions is also studied. As a result, a variety of solitary waves are obtained with free parameters such as periodic wave solution and dark and bright solitary wave solutions. The solutions are plotted and used to describe physical phenomena of the problem. The work shows the power of GERFM. We found that the proposed method is reliable and effective and gives analytical and exact solutions.

1 Introduction

A wide range of physical phenomena and engineering, biological, and medical applications are described by nonlinear evaluation equations (NLEEs). In order to well understand the dynamics of these applications, it is important to find analytical solutions [1,2, 3,4,5, 6,7,8, 9,10]. Recently, analytical computational methods have attracted attention of researchers and mathematicians [11,12, 13,14,15, 16,17,18, 19,20]. They have developed very powerful and useful methods such as the ( G / G ) -expansion method [21,22], and ( G / G 2 ) -expansion method [23], the sine-cosine method [24], the tanh method [25], the direct algebraic method [26,27,28], the modified Jacobi elliptic function method [29], the extended modified auxiliary equation mapping method [30], the modified Kudryashov method [31], and the exp ( ϕ ( η ) ) -expansion method [32]. In 2018, Ghanbari and Inc provided a novel technique that is the generalized exponential rational function method (GERFM). The method introduces a variety of new travelling wave solutions for some NLEEs. In this study, GERFM is used to examine the ionic currents along microtubule dynamical equations describing microtubule dynamic.

Microtubules are the largest component of the cellular structure cylindrical polymers and they are formed from tubulin protein. A microtubule contains 13 parallel protofilaments. It plays an important role in cellular movement and transport [33,34]. The dynamic system of ion currents in microtubes involves ion transportation throughout the environment within cells. The function of microtubule based on their ability to switch between states of growth and shrinkage [35,36]. GTP caps at their ends stabilizes growing microtubules. Nanowire cable in electrostatics of nanosystems is considered as one of the applications of microtubules, and it plays a vital role in morphogenesis [37,38].

In this article, the transmission line models of nano-ionic currents along microtubules are studied that is given by ref. [39]

(1) 2 f x + L 2 3 3 f x 3 + Z C 0 L f t + β Z 3 / 2 G 0 L 2 b Z 3 / 2 C 0 L f f t = 0 ,

where L = 8 × 1 0 9 m is the length of an elementary ring (ER), C 0 = 1.8 × 1 0 15 F is the total maximal capacitance of the ER, Z = 5.56 × 1 0 10 Ω is the characteristic impedance of the system, and G 0 = 1.1 × 1 0 13 Si is the conductance of pertaining nano-pores (NPs). The parameters β and b describe the nonlinearity of conductance of NPs in ER and ER capacitor, respectively. Our objective is studying the effect of positive and negative nonlinear term and dispersion term travelling wave solution.

The article is arranged as follows: In Section 2, the steps of GERFM are described briefly and for more details, the readers are referred to in ref. [40,41]. In Section 3, the exact solutions of the transmission line models of nano-ionic currents along microtubules via GERFM are derived. In Section 4, the new solutions are discussed and compared with the results in previous work via different methods. In Section 5, the stability of the system is investigated. Section 6 is a short summary.

2 Description of the GERFM

This section briefly explains the fundamental steps of the GERFM:

  1. Consider the nonlinear partial differential equation expressed by

    (2) N ( f , f t , f x , f x x , ) = 0 .

  2. Transforming the function f ( x , t ) to f ( ξ ) by applying the travelling wave transformation ξ = x L k τ t , where k is the velocity. The system of equation (2) is transferred to the following ODE:

    (3) N f , k τ f ξ , 1 L f ξ , 1 L 2 f ξ ξ , = 0 .

  3. Suppose that the solution of equation (3) has the following form:

    (4) f ( ξ ) = A 0 + i = 1 N A i ϕ ( ξ ) i + i = 1 N B i ϕ ( ξ ) i ,

    where the coefficients A 0 , A i , B i ( 1 i N ) are constants and ϕ is expressed as follows:

    (5) ϕ ( ξ ) = p 1 exp ( ξ q 1 ) + p 2 exp ( ξ q 2 ) p 3 exp ( ξ q 3 ) + p 4 exp ( ξ q 4 ) ,

    where p 1 , p 2 , p 3 , p 4 and q 1 , q 2 , q 3 , q 4 are real (or complex) constants. In addition, the positive integer N can be determined by the balancing principle between the higher derivative and the nonlinear terms in equation (3).

  4. Substituting function (4) into equation (3) and collecting all terms. Then, the left-hand side of equation (3) is converted to an algebraic equation P ( G 1 , G 2 , G 3 , G 4 ) = 0 in terms of G n = e q j ξ for n = 1 , , 4 .

  5. Equating every coefficient of P to zero to construct a system of nonlinear equations.

  6. Solving the algebraic equations in step 3 with the help of symbolic computation, such as Maple or Mathematica, to determine the values of p n , q n ( 1 n 4 ), and 1 L , k τ , A 0 , A 1 , A 2 , B 1 , B 2 .

3 Application of GERFM to the ionic currents along microtubule dynamical equation

The GERFM is applied to find new traveling wave solutions of equation (1). First step, we apply the following travelling wave transformation:

(6) f ( x , t ) = f ( ξ ) , ξ = x L k τ t ,

where τ = R 1 C 0 = 0.6 × 1 0 6 s and k is the dimensionless velocity of the wave.

Equation (1) transforms into the following nonlinear ordinary differential equation (ODE):

(7) ( 3 k ) ( 2 b C 0 Z 3 / 2 β G 0 Z 3 / 2 ) 2 τ f ( ξ ) 2 3 C 0 k Z τ 2 f ( ξ ) + 2 f ( ξ ) ξ 2 + D = 0 ,

where D is the constant. Balancing f 2 ( ξ ) with f ( ξ ) in equation (7) gives N = 2 . Therefore, the solution of aforementioned equations takes the form

(8) f ( ξ ) = A 0 + A 1 ϕ ( ξ ) + A 2 ϕ ( ξ ) 2 + B 1 ϕ ( ξ ) + B 2 ϕ ( ξ ) 2 ,

where ϕ ( ξ ) is given by equation (5). Substituting equation (8) into equation (7) and following the method described in Section 2 leads to the following results.

Family 1: For p = [ i , i , 1 , 1 ] and q = [ i , i , i , i ] , in equation (5) which gives

(9) ϕ ( ξ ) = sin ( ξ ) cos ( ξ ) ,

we obtain the following solutions:

Case 1

A 0 = 3 C 0 k Z 14 τ 3 k Z 3 / 2 ( 2 b C 0 β G 0 ) , A 1 = 0 , A 2 = 4 τ k Z 3 / 2 ( 2 b C 0 β G 0 ) , B 1 = 0 , B 2 = 0 , D = 1 6 20 τ k Z 3 / 2 ( 2 b C 0 β G 0 ) + 9 C 0 2 k Z τ ( 2 b C 0 β G 0 ) 36 C 0 Z ( 2 b C 0 β G 0 ) .

Using the aforementioned values and substituting equation (8) and equation (9), the exact solution of equation (1) is obtained as follows:

ϕ 1 ( x , t ) = 3 C 0 k Z 2 τ 6 tan 2 x L k t τ + 7 3 k Z 2 ( 2 b C 0 β G 0 ) .

Case 2

A 0 = 24 b C 0 τ Z 2 b C 0 β G 0 + 12 β G 0 τ Z 2 b C 0 β G 0 + 3 C 0 k Z 3 / 2 2 τ Z 3 ( 2 b C 0 k Z 2 β G 0 k Z 2 ) , A 1 = 0 , A 2 = 0 , B 1 = 0 , B 2 = 4 τ k Z 3 / 2 ( 2 b C 0 β G 0 ) , D = 1 6 τ 9 C 0 2 k 2 Z 5 / 2 2 b C 0 k Z 2 β G 0 k Z 2 120 τ 2 k Z 3 / 2 ( 2 b C 0 β G 0 ) + 20 τ 2 Z 2 b C 0 k Z 2 β G 0 k Z 2 + 240 b C 0 τ 2 Z ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) 120 β G 0 τ 2 Z ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) × 36 C 0 k τ Z 3 / 2 2 b C 0 k Z 2 β G 0 k Z 2 72 b C 0 2 k τ Z 3 / 2 ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) + 36 β C 0 G 0 k τ Z 3 / 2 ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) + 36 C 0 τ Z ( 2 b C 0 β G 0 ) .

Using the aforementioned values and substituting equation (8) and equation (9), the exact solution of equation (1) is obtained:

ϕ 2 ( x , t ) = 3 C 0 k Z 2 τ 6 cot 2 x L k t τ + 7 3 k Z 2 ( 2 b C 0 β G 0 ) .

Case 3

A 0 = 1 3 28 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 + 14 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 + 3 C 0 k Z 2 b C 0 k Z β G 0 k Z , A 1 = 0 , B 1 = 0 , A 2 = 4 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 , B 2 = 4 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 , D = 1 6 9 C 0 2 k 2 Z ( 2 b C 0 k Z β G 0 k Z ) 2 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 440 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 + 220 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 36 C 0 k Z 2 b C 0 k Z β G 0 k Z .

Using the aforementioned values and substituting equation (8) and equation (9), the exact solution of equation (1) is obtained:

ϕ 3 ( x , t ) = 3 C 0 k Z + 2 τ 6 tan 2 x L k t τ 6 cot 2 x L k t τ 7 3 k Z 2 ( 2 b C 0 β G 0 ) .

Family 2: For p = [ 1 , 1 , 1 , 1 ] and q = [ 1 , 1 , 1 , 1 ] , in equation (5), we have

(10) ϕ ( ξ ) = tanh ( ξ )

and the solutions take the following cases:

Case 1

A 0 = 3 C 0 k Z + 2 τ 3 k Z 3 / 2 ( 2 b C 0 β G 0 ) , A 1 = 0 , A 2 = 4 τ k Z 3 / 2 ( 2 b C 0 β G 0 ) , B 1 = 0 , B 2 = 0 , D = 1 6 20 τ k Z 3 / 2 ( 2 b C 0 β G 0 ) + 9 C 0 2 k Z τ ( 2 b C 0 β G 0 ) 36 C 0 Z ( 2 b C 0 β G 0 ) .

Using the aforementioned values and substituting equation (8) and equation (10), the exact solution of equation (1) is obtained:

ϕ 4 ( x , t ) = 3 C 0 k Z + 2 τ 1 6 tanh 2 x L k t τ 3 k Z 2 ( 2 b C 0 β G 0 ) .

Case 2

A 0 = 24 b C 0 τ Z 2 b C 0 β G 0 12 β G 0 τ Z 2 b C 0 β G 0 + 3 C 0 k Z 3 / 2 10 τ Z 3 ( 2 b C 0 k Z 2 β G 0 k Z 2 ) , A 1 = 0 , A 2 = 0 , B 1 = 0 , B 2 = 4 τ k Z 3 / 2 ( 2 b C 0 β G 0 ) , D = 1 6 τ 9 C 0 2 k 2 Z 5 / 2 2 b C 0 k Z 2 β G 0 k Z 2 + 24 τ 2 k Z 3 / 2 ( 2 b C 0 β G 0 ) + 20 τ 2 Z 2 b C 0 k Z 2 β G 0 k Z 2 48 b C 0 τ 2 Z ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) + 24 β G 0 τ 2 Z ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) 36 C 0 k τ Z 3 / 2 2 b C 0 k Z 2 β G 0 k Z 2 + 72 b C 0 2 k τ Z 3 / 2 ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) 36 β C 0 G 0 k τ Z 3 / 2 ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) 36 C 0 τ Z ( 2 b C 0 β G 0 ) .

Using the aforementioned values and substituting equation (8) and equation (10), the exact solution of equation (1) is obtained:

ϕ 5 ( x , t ) = 3 C 0 k Z + 2 τ 1 6 coth 2 x L k t τ 3 k Z 2 ( 2 b C 0 β G 0 ) .

Case 3

A 0 = 1 3 4 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 + 3 C 0 k Z 2 b C 0 k Z β G 0 k Z , A 1 = 0 , A 2 = 4 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 , B 1 = 0 , B 2 = 4 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 , D = 1 6 9 C 0 2 k 2 Z ( 2 b C 0 k Z β G 0 k Z ) 2 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 440 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 + 220 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 36 C 0 k Z 2 b C 0 k Z β G 0 k Z .

Using the aforementioned values and substituting equation (8) and equation (10), the exact solution of equation (1) is obtained:

ϕ 6 ( x , t ) = 3 C 0 k Z + 2 τ 6 tanh 2 x L k t τ 6 coth 2 x L k t τ + 1 3 k Z 2 ( 2 b C 0 β G 0 ) .

Family 3: For p = [ 1 , 1 , 1 , 1 ] and q = [ 1 , 1 , 1 , 1 ] , in equation (5) and

(11) ϕ ( ξ ) = cosh ( ξ ) sinh ( ξ ) .

The obtained solutions are:

Case 1

A 0 = 3 C 0 k Z + 2 τ 3 k Z 3 / 2 ( 2 b C 0 β G 0 ) , A 1 = 0 , A 2 = 4 τ k Z 3 / 2 ( 2 b C 0 β G 0 ) , B 1 = 0 , B 2 = 0 , D = 1 6 20 τ k Z 3 / 2 ( 2 b C 0 β G 0 ) + 9 C 0 2 k Z τ ( 2 b C 0 β G 0 ) 36 C 0 Z ( 2 b C 0 β G 0 ) .

Using the aforementioned values and substituting equation (8) and equation (11), the exact solution of equation (1) is obtained:

ϕ 7 ( x , t ) = 3 C 0 k Z + 2 τ 1 6 coth 2 x L k t τ 3 k Z 2 ( 2 b C 0 β G 0 ) .

Case 2

A 0 = 24 b C 0 τ Z 2 b C 0 β G 0 12 β G 0 τ Z 2 b C 0 β G 0 + 3 C 0 k Z 3 / 2 10 τ Z 3 ( 2 b C 0 k Z 2 β G 0 k Z 2 ) , A 1 = 0 , A 2 = 0 , B 1 = 0 , B 2 = 4 τ k Z 3 / 2 ( 2 b C 0 β G 0 ) , D = 1 6 τ 9 C 0 2 k 2 Z 5 / 2 2 b C 0 k Z 2 β G 0 k Z 2 + 24 τ 2 k Z 3 / 2 ( 2 b C 0 β G 0 ) + 20 τ 2 Z 2 b C 0 k Z 2 β G 0 k Z 2 48 b C 0 τ 2 Z ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) + 24 β G 0 τ 2 Z ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) 36 C 0 k τ Z 3 / 2 2 b C 0 k Z 2 β G 0 k Z 2 + 72 b C 0 2 k τ Z 3 / 2 ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) 36 β C 0 G 0 k τ Z 3 / 2 ( 2 b C 0 β G 0 ) ( 2 b C 0 k Z 2 β G 0 k Z 2 ) 36 C 0 τ Z ( 2 b C 0 β G 0 ) .

Using the aforementioned values and substituting equation (8) and equation (11), the exact solution of equation (1) is obtained:

ϕ 8 ( x , t ) = 3 C 0 k Z + 2 τ 1 6 tanh 2 x L k t τ 3 k Z 2 ( 2 b C 0 β G 0 ) .

Case 3

A 0 = 1 3 4 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 + 3 C 0 k Z 2 b C 0 k Z β G 0 k Z , A 1 = 0 , A 2 = 4 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 , B 1 = 0 B 2 = 4 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 , D = 1 6 9 C 0 2 k 2 Z ( 2 b C 0 k Z β G 0 k Z ) 2 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 2 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 440 b C 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 + 220 β G 0 k τ Z ( 2 b C 0 k Z β G 0 k Z ) 2 36 C 0 k Z 2 b C 0 k Z β G 0 k Z .

Using the aforementioned values and substituting equation (8) and equation (11), the exact solution of equation (1) is obtained:

ϕ 9 ( x , t ) = 3 C 0 k Z + 2 τ 6 tanh 2 x L k t τ 6 coth 2 x L k t τ + 1 3 k Z 2 ( 2 b C 0 β G 0 ) .

4 Dissection and comparison

In order to study the obtained solutions, the figures of the solutions are plotted and the parameters take the following value: τ = 0.6 × 1 0 6 , β = 16.36 × 1 0 3 , b = 0.5 , L = 8 × 1 0 9 , G 0 = 1.1 × 1 0 13 , C 0 = 1.8 × 1 0 15 , Z = 5.56 × 1 0 10 , and k , which is the arbitrary constant [39]. The figures in the current study are new and unique compared to the solutions in previous works that are found by methods of boundary conditions [39,42], the exp ( ϕ ( ξ ) ) -expansion method [43,44], modified extended tanh-function method, hyperbolic function method [45], the modified Khater method [46], the Bernoulli sub-equation function method [47], and the exponential rational function method [48]. Figures 1, 2, and 3 are periodic wave solution; Figures 5, 7, and 9 are bright solitary wave solution; and Figures 4, 6, and 8 are dark solitary wave solution. The solutions show localized propagation of ions as shown in Figures 5, 7, and 9 or oscillated propagation of ions along the microtubules as shown in Figures 1, 2, and 3. The obtained solutions may depict the applications of microtubules of similar phenomena in natural science.

Figure 1 
               (a) 2D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 1
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{1}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  . (b) 3D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 1
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{1}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  .
Figure 1

(a) 2D plot of the solution of ϕ 1 ( x , t ) with k = π 4 . (b) 3D plot of the solution of ϕ 1 ( x , t ) with k = π 4 .

Figure 2 
               (a) 2D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 2
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{2}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           π
                        
                        k=\pi 
                     
                  . (b) 3D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 2
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{2}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           π
                        
                        k=\pi 
                     
                  .
Figure 2

(a) 2D plot of the solution of ϕ 2 ( x , t ) with k = π . (b) 3D plot of the solution of ϕ 2 ( x , t ) with k = π .

Figure 3 
               (a) 2D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 3
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{3}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  . (b) 3D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 3
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{3}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  .
Figure 3

(a) 2D plot of the solution of ϕ 3 ( x , t ) with k = π 4 . (b) 3D plot of the solution of ϕ 3 ( x , t ) with k = π 4 .

Figure 4 
               (a) 2D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 4
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{4}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  . (b) 3D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 4
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{4}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  .
Figure 4

(a) 2D plot of the solution of ϕ 4 ( x , t ) with k = π 4 . (b) 3D plot of the solution of ϕ 4 ( x , t ) with k = π 4 .

Figure 5 
               (a) 2D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 5
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{5}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  . (b) 3D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 5
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{5}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  .
Figure 5

(a) 2D plot of the solution of ϕ 5 ( x , t ) with k = π 4 . (b) 3D plot of the solution of ϕ 5 ( x , t ) with k = π 4 .

Figure 6 
               (a) 2D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 6
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{6}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=\frac{\pi }{4}
                     
                  . (b) 3D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 6
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{6}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=\frac{\pi }{4}
                     
                  .
Figure 6

(a) 2D plot of the solution of ϕ 6 ( x , t ) with k = π 4 . (b) 3D plot of the solution of ϕ 6 ( x , t ) with k = π 4 .

Figure 7 
               (a) 2D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 7
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{7}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  . (b) 3D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 7
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{7}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  .
Figure 7

(a) 2D plot of the solution of ϕ 7 ( x , t ) with k = π 4 . (b) 3D plot of the solution of ϕ 7 ( x , t ) with k = π 4 .

Figure 8 
               (a) 2D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 8
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{8}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  . (b) 3D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 8
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{8}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  .
Figure 8

(a) 2D plot of the solution of ϕ 8 ( x , t ) with k = π 4 . (b) 3D plot of the solution of ϕ 8 ( x , t ) with k = π 4 .

Figure 9 
               (a) 2D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 9
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{9}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  . (b) 3D plot of the solution of 
                     
                        
                        
                           
                              
                                 ϕ
                              
                              
                                 9
                              
                           
                           
                              (
                              
                                 x
                                 ,
                                 t
                              
                              )
                           
                        
                        {\phi }_{9}\left(x,t)
                     
                   with 
                     
                        
                        
                           k
                           =
                           −
                           
                              
                                 π
                              
                              
                                 4
                              
                           
                        
                        k=-\frac{\pi }{4}
                     
                  .
Figure 9

(a) 2D plot of the solution of ϕ 9 ( x , t ) with k = π 4 . (b) 3D plot of the solution of ϕ 9 ( x , t ) with k = π 4 .

5 Stability analysis of the ionic currents along microtubule dynamical equations

In this section, we investigate the connection for the dispersion in equation (1). In order to determine whether the outcomes grow infinitely or disappear in given period, we investigate the real part of σ ( k ) when the solution takes the form e i σ t + i k x . If the real part of σ ( k ) is negative for k , the solution may disappear subsequently and the form of the solution in stable case. If it is positive, the solution will lead to infinity and the form of the solution is unstable. Consider the solution of the form

(12) f ( x , t ) = ε w ( x , t ) + p 0 .

Substituting equation (12) into equation (1), we get the following results:

(13) 6 b C 0 p 0 Z 3 / 2 ε w ( 0 , 1 ) ( x , t ) 6 b C 0 Z 3 / 2 ε 2 w ( x , t ) w ( 0 , 1 ) ( x , t ) + 3 β G 0 p 0 Z 3 / 2 ε w ( 0 , 1 ) ( x , t ) + 3 β G 0 Z 3 / 2 ε 2 w ( x , t ) w ( 0 , 1 ) ( x , t ) + L 3 ε w ( 3 , 0 ) ( x , t ) + 6 L ε w ( 1 , 0 ) ( x , t ) + 3 ZC 0 ε w ( 0 , 1 ) ( x , t ) = 0 ,

where p 0 is the constant. Using the linear term ε equation (13) yields

(14) 6 b C 0 p 0 Z 3 / 2 ε w ( 0 , 1 ) ( x , t ) + 3 β G 0 p 0 Z 3 / 2 ε w ( 0 , 1 ) ( x , t ) + L 3 ε w ( 3 , 0 ) ( x , t ) + 6 L ε w ( 1 , 0 ) ( x , t ) + 3 ZC 0 ε w ( 0 , 1 ) ( x , t ) = 0 .

Suppose that equation (14) has a solution of the form

(15) w ( x , t ) = e i ( k 1 x + σ t ) ,

where k is the normalized wave number.

Substituting equation (15) into equation (14), we get the following results:

(16) 6 b C 0 p 0 σ Z 3 / 2 + 3 β G 0 p 0 σ Z 3 / 2 k 1 3 L 3 + 6 k 1 L + 3 σ ZC 0 = 0 .

Solving for σ from equation (16) yields

(17) σ ( k 1 ) = 6 k 1 L k 1 3 L 3 3 ( 2 b C 0 p 0 Z 3 / 2 β G 0 p 0 Z 3 / 2 ZC 0 ) .

The system is stable and σ ( k 1 ) < 0 when p 0 > 0 and k < 0 or p 0 < 0 and k > 0 .

6 Summary

In this article, the GERFM was successfully applied to find new solutions to the transmission line models of nano-ionic currents along microtubules. We find that the proposed method is reliable and effective and gives analytical and exact solutions. In future work, GERFM will be applied to create new solutions for other types of nonlinear equations.



  1. Conflict of interest: Authors state no conflict of interest.

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Received: 2021-04-06
Revised: 2021-07-13
Accepted: 2021-07-15
Published Online: 2021-09-08

© 2021 Noufe H. Aljahdaly et al., published by De Gruyter

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

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