The (𝐺^′/𝐺,1/𝐺)-Expansion Method and Its Applications to Find the Exact Solutions of Nonlinear PDEs for Nanobiosciences

Abstract

The two-variable (𝐺^′/𝐺,1/𝐺)-expansion method is employed to construct exact traveling wave solutions with parameters of nanobiosciences partial differential equation. When the parameters are replaced by special values, the solitary wave solutions and the periodic wave solutions of this equation have been obtained from the traveling waves. This method can be thought of as the generalization of well-known original (𝐺^′/𝐺)-expansion method proposed by M. Wang et al. It is shown that the two-variable (𝐺^′/𝐺,1/𝐺)-expansion method provides a more powerful mathematical tool for solving many other nonlinear PDEs in mathematical physics. Comparison between our results and the well-known results is given.

1. Introduction

In recent years, investigations of exact solutions to nonlinear partial differential equations (PDEs) play an important role in the study of nonlinear physical phenomena. Many powerful methods for finding these exact solutions have been presented, such as the inverse scattering method [1], the Hirota bilinear transform method [2], the truncated Painleve expansion method [3–6], the Backlund transform method [7, 8], the exp-function method [9–13], the tanh-function method [14–17], the Jacobi elliptic function expansion method [18–20], the (𝐺^′/𝐺)-expansion method [21–30], the modified (𝐺^′/𝐺)-expansion method [31], the (𝐺^′/𝐺,1/𝐺)-expansion method [32–35], the modified simple equation method [36], the multiple exp-function algorithm method [37], the transformed rational function method [38], the local fractional variation iteration method [39], and the local fractional series expansion method [40]. Further exact solutions to some real-life physical problems were already given in the recent articles [41–44]. The key idea of the one-variable (𝐺^′/𝐺)-expansion method is that the exact solutions of nonlinear PDEs can be expressed by a polynomial in one variable (𝐺^′/𝐺) in which 𝐺=𝐺(𝜉) satisfies the second-order linear ODE 𝐺^′′(𝜉)+𝜆𝐺^′(𝜉)+𝜇𝐺(𝜉)=0, where 𝜆 and 𝜇 are constants and ^′=𝑑/𝑑𝜉. The key idea of the two variable (𝐺^′/𝐺,1/𝐺)-expansion method is that the exact traveling wave solutions of nonlinear PDEs can be expressed by a polynomial in two variables (𝐺^′/𝐺) and (1/𝐺) in which 𝐺=𝐺(𝜉) satisfies the second-order linear ODE 𝐺^′′(𝜉)+𝜆𝐺(𝜉)=𝜇, where 𝜆 and 𝜇 are constants. The degree of this polynomial can be determined by considering the homogeneous balance between the highest order derivatives and the nonlinear terms appearing in the given nonlinear PDEs. The coefficients of this polynomial can be obtained by solving a set of algebraic equations resulting from the process of using this method. Recently, Li et al. [32] have applied the (𝐺^′/𝐺,1/𝐺)-expansion method and determined the exact solutions of Zakharov equations, while Zayed et al. [33–35] have used this method to find the exact solutions of the combined KdV-mKdV equation, the Kadomtsev-Petviashvili equation, and the potential YTSF equation, respectively.

The objective of this paper is to apply the two-variable (𝐺^′/𝐺,1/𝐺)-expansion method to find the exact traveling wave solutions of the following nonlinear partial differential equation of nanobiosciences [45, 46]: ℓ²3𝑢_𝑥𝑥𝑥+𝑍^3/2ℓ𝜒𝐺₀−2𝛿𝐶₀𝑢𝑢_𝑡+2𝑢_𝑥+𝑍𝐶₀ℓ𝑢_𝑡+1ℓ𝑅𝑍⁻¹−𝐺₀𝑍𝑢=0, (1) where 𝑅=0.34×10⁹Ω is the resistance of the ER with length ℓ=8×10⁻⁹m, 𝐶₀=1.8×10⁻¹⁵F is total maximal capacitance of the ER, 𝐺₀=1.1×10⁻¹³Si is conductance of pertaining NPs, and 𝑍=5.56×10¹⁰Ω is the characteristic impedance of our system. Parameters 𝛿 and 𝜒 describe nonlinearity of ER capacitor and conductance of NPs in RE, respectively.

The rest of this paper is organized as follows: In Section 2, we give the description of the two-variable (𝐺^′/𝐺,1/𝐺)-expansion method. In Section 3, we apply this method to solve (1). In Section 4, some conclusions are given.

2. Description of the Two-Variable (𝐺^′/𝐺,1/𝐺)-Expansion Method

Before we describe the main steps of this method, we need the following remarks (see [32–35]).

Remark 1. If we consider the second-order linear ODE 𝐺^′′(𝜉)+𝜆𝐺(𝜉)=𝜇, (2) and set 𝜙=𝐺^′/𝐺, 𝜓=1/𝐺, then we get 𝜙^′=−𝜙²+𝜇𝜓−𝜆,𝜓^′=−𝜙𝜓, (3) where 𝜆 and 𝜇 are constants.

Remark 2. If 𝜆<0, then the general solution of (2) has the form 𝐺(𝜉)=𝐴₁sinh𝜉√−𝜆+𝐴₂cosh𝜉√+𝜇−𝜆𝜆, (4) where 𝐴₁ and 𝐴₂ are arbitrary constants. Consequently, we have 𝜓²=−𝜆𝜆²𝜎₁+𝜇²𝜙²−2𝜇𝜓+𝜆, (5) where 𝜎₁=𝐴²₁−𝐴²₂.

Remark 3. If 𝜆>0, then the general solution of (2) has the form 𝐺(𝜉)=𝐴₁sin𝜉√𝜆+𝐴₂cos𝜉√𝜆+𝜇𝜆, (6) and hence 𝜓²=𝜆𝜆²𝜎₂−𝜇²𝜙²−2𝜇𝜓+𝜆, (7) where 𝜎₂=𝐴²₁+𝐴²₂.

Remark 4. If 𝜆=0, then the general solution of (2) has the form 𝐺(𝜉)=𝜇2𝜉²+𝐴₁𝜉+𝐴₂, (8) and hence 𝜓²=1𝐴²₁−2𝜇𝐴₂𝜙²−2𝜇𝜓. (9) Suppose we have the following nonlinear evolution equation: 𝐹𝑢,𝑢_𝑡,𝑢_𝑥,𝑢_𝑥𝑥,…=0, (10) where 𝐹 is a polynomial in 𝑢(𝑥,𝑡) and its partial derivatives. In the following, we give the main steps of the (𝐺^′/𝐺,1/𝐺)-expansion method [32–35].

Step 1. The traveling wave transformation 𝑢(𝑥,𝑡)=𝑢(𝜉),𝜉=𝑥−𝑐𝑡, (11) where 𝑐 is a nonzero constant, reduces (10) to an ODE in the form 𝑃𝑢,𝑢^′,𝑢^′′,…=0, (12) where 𝑃 is a polynomial of 𝑢(𝜉) and its total derivatives with respect to 𝜉.

Step 2. Assume that the solution of (12) can be expressed by a polynomial in the two variables 𝜙 and 𝜓 as follows: 𝑢(𝜉)=^𝑁_𝑖=0𝑎_𝑖𝜙^𝑖+^𝑁_𝑖=1𝑏_𝑖𝜙^𝑖−1𝜓, (13) where 𝑎_𝑖(𝑖=0,1,2,…,𝑁) and 𝑏_𝑖(𝑖=1,2,…,𝑁) are constants to be determined later.

Step 3. Determine the positive integer 𝑁 in (13) by using the homogeneous balance between the highest-order derivatives and the nonlinear terms in (12).

Step 4. Substitute (13) into (12) along with (3) and (5); the left-hand side of (12) can be converted into a polynomial in 𝜙 and 𝜓, in which the degree of 𝜓 is not longer than one. Equating each coefficient of this polynomial to zero yields a system of algebraic equations which can be solved by using the Maple or Mathematica to get the values of 𝑎_𝑖, 𝑏_𝑖, 𝑐, 𝜇, 𝐴₁, 𝐴₂, and 𝜆, where 𝜆<0.

Step 5. Similar to Step 4, substituting (13) into (12) along with (3) and (7) for 𝜆>0, (or (3) and (9) for 𝜆=0), we obtain the exact solutions of (12) expressed by trigonometric functions (or by rational functions), respectively.

3. An Application

In this section, we will apply the method described in Section 2 to find the exact traveling wave solutions of (1) which describe models of microtubules (MTs) as nonlinear transmission lines. The physical details of derivation of (1) describing its ionic currents are elaborated in [45].

In order to solve (1) by the two-variable (𝐺^′/𝐺,1/𝐺)-expansion method, we see that the traveling wave transformation (11) with 𝜉=(1/ℓ)𝑥−𝑐(𝑡/𝜏) and the characteristic time of charging ER capacitor is 𝜏=𝑅𝐶₀=0.6×10⁻⁶s. Thus (1) takes the form of ODE 𝑢^′′′+𝐴𝑐𝑢𝑢^′+(6−𝐵𝑐)𝑢^′+𝐶𝑢=0, (14) where 𝐴=3(𝑍^3/2/𝜏)(2𝛿𝐶₀−𝜒𝐺₀), 𝐵=3(𝑍𝐶₀/𝜏), 𝐶=3(𝑅𝑍⁻¹−𝐺₀𝑍).

By balancing 𝑢^′′′ with 𝑢𝑢^′ in (14), we get 𝑁=2. Consequently, we get 𝑢(𝜉)=𝑎₀+𝑎₁𝜙(𝜉)+𝑎₂𝜙²(𝜉)+𝑏₁𝜓(𝜉)+𝑏₂𝜙(𝜉)𝜓(𝜉), (15) where 𝑎₀, 𝑎₁, 𝑎₂, 𝑏₁, and 𝑏₂ are constants to be determined later. There are three cases to be discussed as follows.

Case 1 (hyperbolic function solutions (𝜆<0)). If 𝜆<0, substituting (15) into (14) and using (3) and (5), the left-hand side of (14) becomes a polynomial in 𝜙 and 𝜓. Setting the coefficients of this polynomial to be zero yields a system of algebraic equations in 𝑎₀, 𝑎₁, 𝑎₂, 𝑏₁, 𝑏₂, 𝜇, 𝜆, and 𝑐 as follows: 𝜙⁵∶2𝐴𝑐𝜆𝑏²₂𝜎₁𝜆²+𝜇²−2𝐴𝑐𝑎²₂−24𝑎₂𝜙=0,₄∶3𝐴𝑐𝜆𝑏₁𝑏₂𝜎₁𝜆²+𝜇²−3𝐴𝑐𝑎₁𝑎₂−36𝜆𝜇𝑏₂𝜎₁𝜆²+𝜇²−6𝑎₁−3𝐴𝑐𝜆𝜇𝑎₂𝑏₂𝜎₁𝜆²+𝜇²𝜙=0,₄𝜓∶−24𝑏₂−4𝐴𝑐𝑎₂𝑏₂𝜙=0,₃∶2𝐵𝑐𝑎₂−40𝜆𝑎₂−12𝑎₂−𝐴𝑐𝑎²₁+24𝜆𝜇²𝑎₂𝜎₁𝜆²+𝜇²−2𝐴𝑐𝑎₀𝑎₂−6𝜆𝜇𝑏₁𝜎₁𝜆²+𝜇²−2𝐴𝑐𝜆𝑎²₂+3𝐴𝑐𝜆²𝑏²₂𝜎₁𝜆²+𝜇²+𝐴𝑐𝜆𝑏²₁𝜎₁𝜆²+𝜇²−2𝐴𝑐𝜆²𝜇²𝑏²₂𝜇²+𝜆²𝜎₁²−2𝐴𝑐𝜆𝜇𝑎₁𝑏₂𝜎₁𝜆²+𝜇²−2𝐴𝑐𝜆𝜇𝑎₂𝑏₁𝜎₁𝜆²+𝜇²𝜙=0,₃𝜓∶54𝜇𝑎₂−6𝑏₁−3𝐴𝑐𝑎₁𝑏₂−3𝐴𝑐𝑎₂𝑏₁+2𝐴𝑐𝜇𝑎²₂−5𝐴𝑐𝜆𝜇𝑏²₂𝜎₁𝜆²+𝜇²𝜙=0,₂∶𝐶𝑎₂−6𝑎₁−8𝜆𝑎₁+𝐵𝑐𝑎₁+3𝜆𝜇²𝑎₁𝜎₁𝜆²+𝜇²−47𝜆²𝜇𝑏₂𝜎₁𝜆²+𝜇²−𝐴𝑐𝑎₀𝑎₁−6𝜆𝜇𝑏₂𝜎₁𝜆²+𝜇²+12𝜆²𝜇³𝑏₂𝜎₁𝜆²+𝜇²²−3𝐴𝑐𝜆𝑎₁𝑎₂+4𝐴𝑐𝜆²𝑏₁𝑏₂𝜎₁𝜆²+𝜇²+𝐵𝑐𝜆𝜇𝑏₂𝜎₁𝜆²+𝜇²−2𝐴𝑐𝜆²𝜇²𝑏₁𝑏₂𝜎₁𝜆²+𝜇²²−3𝐴𝑐𝜆²𝜇𝑎₂𝑏₂𝜎₁𝜆²+𝜇²−𝐴𝑐𝜆𝜇𝑎₀𝑏₂𝜎₁𝜆²+𝜇²−𝐴𝑐𝜆𝜇𝑎₁𝑏₁𝜎₁𝜆²+𝜇²𝜙=0,₂𝜓∶12𝜇𝑎₁−12𝑏₂−28𝜆𝑏₂+2𝐵𝑐𝑏₂+78𝜆𝜇²𝑏₂𝜎₁𝜆²+𝜇²−2𝐴𝑐𝑎₀𝑏₂−2𝐴𝑐𝑎₁𝑏₁+3𝐴𝑐𝜇𝑎₁𝑎₂−3𝐴𝑐𝜆𝑎₂𝑏₂+6𝐴𝑐𝜆𝜇²𝑎₂𝑏₂𝜎₁𝜆²+𝜇²−7𝐴𝑐𝜆𝜇𝑏₁𝑏₂𝜎₁𝜆²+𝜇²=0,𝜙∶𝐶𝑎₁−12𝜆𝑎₂−16𝜆²𝑎₂−6𝜆²𝜇𝑏₁𝜎₁𝜆²+𝜇²+2𝐵𝑐𝜆𝑎₂+24𝜆²𝜇²𝑎₂𝜎₁𝜆²+𝜇²−𝐴𝑐𝜆𝑎²₁+𝐴𝑐𝜆²𝑏²₁𝜎₁𝜆²+𝜇²+𝐴𝑐𝜆³𝑏²₂𝜎₁𝜆²+𝜇²−2𝐴𝑐𝜆𝑎₀𝑎₂−2𝐴𝑐𝜆³𝜇²𝑏²₂𝜎₁𝜆²+𝜇²²−2𝐴𝑐𝜆²𝜇𝑎₁𝑏₂𝜎₁𝜆²+𝜇²−2𝐴𝑐𝜆²𝜇𝑎₂𝑏₁𝜎₁𝜆²+𝜇²=0,𝜙𝜓∶𝐶𝑏₂−6𝑏₁−5𝜆𝑏₁+12𝜇𝑎₂+𝐵𝑐𝑏₁+40𝜆𝜇𝑎₂+12𝜆𝜇²𝑏₁𝜎₁𝜆²+𝜇²−48𝜆𝜇³𝑎₂𝜎₁𝜆²+𝜇²−2𝐵𝑐𝜇𝑎₂−𝐴𝑐𝑎₀𝑏₁+𝐴𝑐𝜇𝑎²₁+2𝐴𝑐𝜇𝑎₀𝑎₂−2𝐴𝑐𝜆𝑎₁𝑏₂−2𝐴𝑐𝜆𝑎₂𝑏₁−2𝐴𝑐𝜆𝜇𝑏²₁𝜎₁𝜆²+𝜇²+4𝐴𝑐𝜆²𝜇³𝑏²₂𝜎₁𝜆²+𝜇²²−3𝐴𝑐𝜆²𝜇𝑏²₂𝜎₁𝜆²+𝜇²+4𝐴𝑐𝜆𝜇²𝑎₁𝑏₂𝜎₁𝜆²+𝜇²+4𝐴𝑐𝜆𝜇²𝑎₂𝑏₁𝜎₁𝜆²+𝜇²=0,𝜓∶𝐶𝑏₁+6𝜇𝑎₁−6𝜆𝑏₂−5𝜆²𝑏₂+5𝜆𝜇𝑎₁−6𝜆𝜇³𝑎₁𝜎₁𝜆²+𝜇²+12𝜆𝜇²𝑏₂𝜎₁𝜆²+𝜇²−𝐵𝑐𝜇𝑎₁+𝐵𝑐𝜆𝑏₂+28𝜆²𝜇²𝑏₂𝜎₁𝜆²+𝜇²−24𝜆²𝜇⁴𝑏₂𝜎₁𝜆²+𝜇²²+𝐴𝑐𝜇𝑎₀𝑎₁−𝐴𝑐𝜆𝑎₀𝑏₂−𝐴𝑐𝜆𝑎₁𝑏₁−2𝐵𝑐𝜆𝜇²𝑏₂𝜎₁𝜆²+𝜇²+4𝐴𝑐𝜆²𝜇³𝑏₁𝑏₂𝜎₁𝜆²+𝜇²²+2𝐴𝑐𝜆𝜇²𝑎₀𝑏₂𝜎₁𝜆²+𝜇²+2𝐴𝑐𝜆𝜇²𝑎₁𝑏₁𝜎₁𝜆²+𝜇²−3𝐴𝑐𝜆²𝜇𝑏₁𝑏₂𝜎₁𝜆²+𝜇²𝜙=0,₀∶𝐶𝑎₀−6𝜆𝑎₁−2𝜆²𝑎₁−6𝜆²𝜇𝑏₂𝜎₁𝜆²+𝜇²−11𝜆³𝜇𝑏₂𝜎₁𝜆²+𝜇²+𝐵𝑐𝜆𝑎₁+3𝜆²𝜇²𝑎₁𝜎₁𝜆²+𝜇²+12𝜆³𝜇³𝑏₂𝜎₁𝜆²+𝜇²²−𝐴𝑐𝜆𝑎₀𝑎₁+𝐵𝑐𝜆²𝜇𝑏₂𝜎₁𝜆²+𝜇²+𝐴𝑐𝜆³𝑏₁𝑏₂𝜎₁𝜆²+𝜇²−2𝐴𝑐𝜆³𝜇²𝑏₁𝑏₂𝜎₁𝜆²+𝜇²²−𝐴𝑐𝜆²𝜇𝑎₀𝑏₂𝜎₁𝜆²+𝜇²=0. (16)
On solving the above algebraic equations using the Maple or Mathematica, we get the following results.
Result 1. Consider 𝑎₀=𝐵𝐴±𝑏₂(5𝜆+6)√−𝜆6𝜆²𝜎₁+𝜇²,𝑎₁=0,𝑎₂𝑏=±²√−𝜆𝜆²𝜎₁+𝜇²,𝑏₁=±𝜇𝑏₂√−𝜆𝜆²𝜎₁+𝜇²,6𝑐=±𝑏₂𝐴𝜆²𝜎₁+𝜇²−𝜆,𝑏₂=𝑏₂,𝜇=𝜇,𝐶=0. (17)
From (4), (15), and (17), we deduce the traveling wave solution of (1) as follows: 𝑢(𝜉)=𝐵𝐴±𝑏₂(5𝜆+6)√−𝜆6𝜆²𝜎₁+𝜇²±𝑏₂𝜆√−𝜆𝜆²𝜎₁+𝜇²×𝐴₁cosh𝜉√−𝜆+𝐴₂sinh𝜉√−𝜆𝐴₁sinh𝜉√−𝜆+𝐴₂cosh𝜉√−𝜆+𝜇/𝜆²+√−𝜆𝑏₂𝐴₁sinh𝜉√−𝜆+𝐴₂cosh𝜉√×𝐴−𝜆+𝜇/𝜆₁cosh𝜉√−𝜆+𝐴₂sinh𝜉√−𝜆𝐴₁sinh𝜉√−𝜆+𝐴₂cosh𝜉√±𝜇−𝜆+𝜇/𝜆𝜆²𝜎₁+𝜇², (18) where 𝑏₂ is a nonzero constant, 𝜉=(1/ℓ)𝑥±(6/𝑏₂𝐴)(𝜆²𝜎₁+𝜇²)/−𝜆(𝑡/𝜏).

In particular, by setting 𝐴₁=0, 𝐴₂≠0, and 𝜇=0 in (18), we have the solitary wave solution 𝑢(𝜉)=𝐵𝐴𝑏±𝑖²𝐴₂5𝜆+66√+𝑏−𝜆²√−𝜆𝐴₂×tanh𝜉√−𝜆sech𝜉√−𝜆±𝑖tanh𝜉√,√−𝜆𝑖=−1, (19) while if 𝐴₁≠0, 𝐴₂=0, and 𝜇=0, then we have the solitary wave solution 𝑢(𝜉)=𝐵𝐴±𝑏₂𝐴₁5𝜆+66√+𝑏−𝜆²√−𝜆𝐴₁×coth𝜉√−𝜆csch𝜉√−𝜆±coth𝜉√−𝜆. (20)

Result 2. Consider 𝑎₀=𝑎₀,𝑎₁=0,𝑎₂=6𝐴𝑎₀−𝐵𝐴(),𝑏3+4𝜆₁=0,𝑏₂2()=0,𝑐=−3+4𝜆𝐴𝑎₀,−𝐵𝜇=0,𝐶=0. (21)

In this result, we deduce the traveling wave solution of (1) as follows: 𝑢(𝜉)=𝑎₀−6𝜆𝐴𝑎₀−𝐵𝐴()×𝐴3+4𝜆₁cosh𝜉√−𝜆+𝐴₂sinh𝜉√−𝜆𝐴₁sinh𝜉√−𝜆+𝐴₂cosh𝜉√−𝜆², (22) where 𝑎₀ is an arbitrary constant and 𝜉=(1/ℓ)𝑥+(2(3+4𝜆)/(𝐴𝑎₀−𝐵))(𝑡/𝜏).

In particular, by setting 𝐴₁=0 and 𝐴₂≠0 in (22), we have the solitary wave solution (Figure 1) 𝑢(𝜉)=𝑎₀−6𝜆𝐴𝑎₀−𝐵𝐴(3+4𝜆)tanh²𝜉√−𝜆, (23) while if 𝐴₁≠0 and 𝐴₂=0, then we have the solitary wave solution 𝑢(𝜉)=𝑎₀−6𝜆𝐴𝑎₀−𝐵𝐴(3+4𝜆)coth²𝜉√−𝜆. (24)

Figure 1 

The plot of solution (23) when 𝜆=−1, 𝐴=1, 𝐵=2, 𝑎0=1, 𝜏=ℓ=1/2.

Note that our results (23) and (24) are in agreement with the results 𝑢_𝑖 (𝑖=1,2) obtained in page 1251 of [47] with interchanges 𝜆↔𝑞𝑟 and with the results (28) and (29) in page 1544 of [48] with the interchanges 𝜆↔𝜇.

We close this case with the remark that our results (18)–(20) and (22) are new and are not reported elsewhere.

Case 2 (trigonometric function solution (𝜆>0)). If 𝜆>0, substituting (15) into (14) and using (3) and (7), the left-hand side of (14) becomes a polynomial in 𝜙 and 𝜓. Setting the coefficients of this polynomial to be zero yields a system of algebraic equations in 𝑎₀, 𝑎₁, 𝑎₂, 𝑏₁, 𝑏₂, 𝜇, 𝜆, and 𝑐 as follows: 𝜙⁵∶−24𝑎₂+2𝐴𝑐𝜆𝑏²₂𝜇²−𝜎₂𝜆²−2𝐴𝑐𝑎²₂𝜙=0,₄∶−6𝑎₁−36𝜆𝜇𝑏₂𝜇²−𝜎₂𝜆²+3𝐴𝑐𝜆𝑏₁𝑏₂𝜇²−𝜎₂𝜆²−3𝐴𝑐𝑎₁𝑎₂−3𝐴𝑐𝜆𝜇𝑎₂𝑏₂𝜇²−𝜎₂𝜆²𝜙=0,₄𝜓∶−24𝑏₂−4𝐴𝑐𝑎₂𝑏₂𝜙=0,₃∶2𝐵𝑐𝑎₂−40𝜆𝑎₂−12𝑎₂+24𝜆𝜇²𝑎₂𝜇²−𝜎₂𝜆²−6𝜆𝜇𝑏₁𝜇²−𝜎₂𝜆²+3𝐴𝑐𝜆²𝑏²₂𝜇²−𝜎₂𝜆²−𝐴𝑐𝑎²₁−2𝐴𝑐𝑎₀𝑎₂−2𝐴𝑐𝜆𝑎²₂+𝐴𝑐𝜆𝑏²₁𝜇²−𝜎₂𝜆²−2𝐴𝑐𝜆²𝜇²𝑏²₂𝜎₂𝜆²−𝜇²²−2𝐴𝑐𝜆𝜇𝑎₁𝑏₂𝜇²−𝜎₂𝜆²−2𝐴𝑐𝜆𝜇𝑎₂𝑏₁𝜇²−𝜎₂𝜆²𝜙=0,₃𝜓∶−6𝑏₁+54𝜇𝑎₂+2𝐴𝑐𝜇𝑎²₂−3𝐴𝑐𝑎₂𝑏₁−3𝐴𝑐𝑎₁𝑏₂−5𝐴𝑐𝜆𝜇𝑏²₂𝜇²−𝜎₂𝜆²𝜙=0,₂∶−8𝜆𝑎₁−6𝑎₁+𝐵𝑐𝑎₁+3𝜆𝜇²𝑎₁𝜇²−𝜎₂𝜆²−47𝜆²𝜇𝑏₂𝜇²−𝜎₂𝜆²−6𝜆𝜇𝑏₂𝜇²−𝜎₂𝜆²+12𝜆²𝜇³𝑏₂𝜎₂𝜆²−𝜇²²+𝐵𝑐𝜆𝜇𝑏₂𝜇²−𝜎₂𝜆²+𝐶𝑎₂−𝐴𝑐𝑎₀𝑎₁−3𝐴𝑐𝜆𝑎₁𝑎₂+4𝐴𝑐𝜆²𝑏₁𝑏₂𝜇²−𝜎₂𝜆²−2𝐴𝑐𝜆²𝜇²𝑏₁𝑏₂𝜎₂𝜆²−𝜇²²−3𝐴𝑐𝜆²𝜇𝑎₂𝑏₂𝜇²−𝜎₂𝜆²−𝐴𝑐𝜆𝜇𝑎₀𝑏₂𝜇²−𝜎₂𝜆²−𝐴𝑐𝜆𝜇𝑎₁𝑏₁𝜇²−𝜎₂𝜆²𝜙=0,₂𝜓∶12𝜇𝑎₁−12𝑏₂−28𝜆𝑏₂+2𝐵𝑐𝑏₂+78𝜆𝜇²𝑏₂𝜇²−𝜎₂𝜆²+3𝐴𝑐𝜇𝑎₁𝑎₂−2𝐴𝑐𝑎₁𝑏₁−2𝐴𝑐𝑎₀𝑏₂−3𝐴𝑐𝜆𝑎₂𝑏₂+6𝐴𝑐𝜆𝜇²𝑎₂𝑏₂𝜇²−𝜎₂𝜆²−7𝐴𝑐𝜆𝜇𝑏₁𝑏₂𝜇²−𝜎₂𝜆²=0,𝜙∶2𝐵𝑐𝜆𝑎₂−12𝜆𝑎₂−16𝜆²𝑎₂−6𝜆²𝜇𝑏₁𝜇²−𝜎₂𝜆²+24𝜆²𝜇²𝑎₂𝜇²−𝜎₂𝜆²+𝐶𝑎₁−𝐴𝑐𝜆𝑎²₁+𝐴𝑐𝜆²𝑏²₁𝜇²−𝜎₂𝜆²+𝐴𝑐𝜆³𝑏²₂𝜇²−𝜎₂𝜆²−2𝐴𝑐𝜆𝑎₀𝑎₂−2𝐴𝑐𝜆³𝜇²𝑏²₂𝜎₂𝜆²−𝜇²²−2𝐴𝑐𝜆²𝜇𝑎₁𝑏₂𝜇²−𝜎₂𝜆²−2𝐴𝑐𝜆²𝜇𝑎₂𝑏₁𝜇²−𝜎₂𝜆²=0,𝜙𝜓∶12𝜇𝑎₂−5𝜆𝑏₁−6𝑏₁+𝐵𝑐𝑏₁+40𝜆𝜇𝑎₂+12𝜆𝜇²𝑏₁𝜇²−𝜎₂𝜆²−48𝜆𝜇³𝑎₂𝜇²−𝜎₂𝜆²−2𝐵𝑐𝜇𝑎₂+𝐶𝑏₂−𝐴𝑐𝑎₀𝑏₁+𝐴𝑐𝜇𝑎²₁+2𝐴𝑐𝜇𝑎₀𝑎₂−2𝐴𝑐𝜆𝑎₁𝑏₂−2𝐴𝑐𝜆𝑎₂𝑏₁−2𝐴𝑐𝜆𝜇𝑏²₁𝜇²−𝜎₂𝜆²+4𝐴𝑐𝜆²𝜇³𝑏²₂𝜎₂𝜆²−𝜇²²−3𝐴𝑐𝜆²𝜇𝑏²₂𝜇²−𝜎₂𝜆²+4𝐴𝑐𝜆𝜇²𝑎₁𝑏₂𝜇²−𝜎₂𝜆²+4𝐴𝑐𝜆𝜇²𝑎₂𝑏₁𝜇²−𝜎₂𝜆²=0,𝜓∶6𝜇𝑎₁−6𝜆𝑏₂−5𝜆²𝑏₂+5𝜆𝜇𝑎₁−6𝜆𝜇³𝑎₁𝜇²−𝜎₂𝜆²+12𝜆𝜇²𝑏₂𝜇²−𝜎₂𝜆²−𝐵𝑐𝜇𝑎₁+𝐵𝑐𝜆𝑏₂+28𝜆²𝜇²𝑏₂𝜇²−𝜎₂𝜆²−24𝜆²𝜇⁴𝑏₂𝜎₂𝜆²−𝜇²²−2𝐵𝑐𝜆𝜇²𝑏₂𝜇²−𝜎₂𝜆²+𝐶𝑏₁+𝐴𝑐𝜇𝑎₀𝑎₁−𝐴𝑐𝜆𝑎₀𝑏₂−𝐴𝑐𝜆𝑎₁𝑏₁+4𝐴𝑐𝜆²𝜇³𝑏₁𝑏₂𝜎₂𝜆²−𝜇²²+2𝐴𝑐𝜆𝜇²𝑎₀𝑏₂𝜇²−𝜎₂𝜆²+2𝐴𝑐𝜆𝜇²𝑎₁𝑏₁𝜇²−𝜎₂𝜆²−3𝐴𝑐𝜆²𝜇𝑏₁𝑏₂𝜇²−𝜎₂𝜆²𝜙=0,₀∶𝐵𝑐𝜆𝑎₁−6𝜆𝑎₁−2𝜆²𝑎₁−6𝜆²𝜇𝑏₂𝜇²−𝜎₂𝜆²−11𝜆³𝜇𝑏₂𝜇²−𝜎₂𝜆²+3𝜆²𝜇²𝑎₁𝜇²−𝜎₂𝜆²+12𝜆³𝜇³𝑏₂𝜎₂𝜆²−𝜇²²+𝐵𝑐𝜆²𝜇𝑏₂𝜇²−𝜎₂𝜆²+𝐶𝑎₀−𝐴𝑐𝜆𝑎₀𝑎₁+𝐴𝑐𝜆³𝑏₁𝑏₂𝜇²−𝜎₂𝜆²−2𝐴𝑐𝜆³𝜇²𝑏₁𝑏₂𝜎₂𝜆²−𝜇²²−𝐴𝑐𝜆²𝜇𝑎₀𝑏₂𝜇²−𝜎₂𝜆²−𝐴𝑐𝜆²𝜇𝑎₁𝑏₁𝜇²−𝜎₂𝜆²=0. (25)
On solving the above algebraic equations using the Maple or Mathematica, we get the following results.

Result 1. Consider 𝑎₀=𝐵𝐴±𝑏₂(5𝜆+6)√𝜆6𝜆²𝜎₂−𝜇²,𝑎₁=0,𝑎₂𝑏=±²√𝜆𝜆²𝜎₂−𝜇²,𝑏₁=±𝜇𝑏₂√𝜆𝜆²𝜎₂−𝜇²6,𝑐=±𝑏₂𝐴𝜆²𝜎₂−𝜇²𝜆,𝑏₂=𝑏₂,𝜇=𝜇,𝐶=0. (26)

From (6), (15), and (26), we deduce the traveling wave solution of (1) as follows: 𝑢(𝜉)=𝐵𝐴±𝑏₂(5𝜆+6)√𝜆6𝜆²𝜎₂−𝜇²±𝜆𝑏₂√𝜆𝜆²𝜎₂−𝜇²×𝐴₁cos𝜉√𝜆−𝐴₂sin𝜉√𝜆𝐴₁sin𝜉√𝜆+𝐴₂cos𝜉√𝜆+𝜇/𝜆²+√𝜆𝑏₂𝐴₁sin𝜉√𝜆+𝐴₂cos𝜉√𝜆×𝐴+𝜇/𝜆₁cos𝜉√𝜆−𝐴₂sin𝜉√𝜆𝐴₁sin𝜉√𝜆+𝐴₂cos𝜉√𝜆±𝜇+𝜇/𝜆𝜆²𝜎₂−𝜇², (27) where 𝑏₂ is a nonzero constant, 𝜉=(1/ℓ)𝑥±(6/𝑏₂𝐴)(𝜆²𝜎₂−𝜇²)/𝜆(𝑡/𝜏).

In particular, by setting 𝐴₁=0, 𝐴₂≠0, and 𝜇=0 in (27), we have the periodic solution 𝑢(𝜉)=𝐵𝐴±𝑏₂𝐴₂5𝜆+66√𝜆−√𝜆𝑏₂𝐴₂tan𝜉√𝜆sec𝜉√𝜆±tan𝜉√𝜆, (28) while if 𝐴₁≠0, 𝐴₂=0, and 𝜇=0, then we have the periodic solution 𝑢(𝜉)=𝐵𝐴±𝑏₂𝐴₁5𝜆+66√𝜆+√𝜆𝑏₂𝐴₁cot𝜉√𝜆csc𝜉√𝜆±cot𝜉√𝜆. (29)

Result 2. Consider 𝑎₀=𝑎₀,𝑎₁=0,𝑎₂=6𝐴𝑎₀−𝐵𝐴(),𝑏3+4𝜆₁=0,𝑏₂2()=0,𝜇=0,𝑐=−3+4𝜆𝐴𝑎₀−𝐵,𝐶=0. (30)

In this result, we deduce the traveling wave solution of (1) as follows: 𝑢(𝜉)=𝑎₀+6𝜆𝐴𝑎₀−𝐵𝐴()×𝐴3+4𝜆₁cos𝜉√𝜆−𝐴₂sin𝜉√𝜆𝐴₁sin𝜉√𝜆+𝐴₂cos𝜉√𝜆², (31) where 𝑎₀ is an arbitrary constant and 𝜉=(1/ℓ)𝑥+(2(3+4𝜆)/(𝐴𝑎₀−𝐵))(𝑡/𝜏).

In particular, by setting 𝐴₁=0 and 𝐴₂≠0 in (31), we have the periodic solution (Figure 2) 𝑢(𝜉)=𝑎₀+6𝜆𝐴𝑎₀−𝐵𝐴(3+4𝜆)tan²𝜉√𝜆, (32) while if 𝐴₁≠0 and 𝐴₂=0, then we have the periodic solution 𝑢(𝜉)=𝑎₀+6𝜆𝐴𝑎₀−𝐵𝐴(3+4𝜆)cot²𝜉√𝜆. (33)

Figure 2 

The plot of solution (32) when 𝜆=1, 𝐴=8, 𝐵=7, 𝑎0=1, 𝜏=ℓ=1/2.

Note that our results (32) and (33) are in agreement with the results 𝑢₁₃ and 𝑢₁₄ obtained in page 1251 of [47] with interchanges 𝜆↔𝑞𝑟 and with the results (33) and (34) in page 1544 of [48] with the interchanges 𝜆↔𝜇.

We close this case with the remark that our results (27)–(29) and (31) are new and are not reported elsewhere.

Case 3 (rational function solutions (𝜆=0)). If 𝜆=0, substituting (15) into (14) and using (3) and (9), the left-hand side of (14) becomes a polynomial in 𝜙 and 𝜓. Setting the coefficients of this polynomial to be zero yields a system of algebraic equations in 𝑎₀,𝑎₁, 𝑎₂, 𝑏₁, 𝑏₂, 𝜇, and 𝑐 as follows: 𝜙⁵∶−24𝑎₂−2𝐴𝑐𝑏²₂𝐴²₁−2𝜇𝐴₂−2𝐴𝑐𝑎²₂𝜙=0,₄∶36𝑏₂𝜇𝐴²₁−2𝜇𝐴₂−6𝑎₁−3𝐴𝑐𝑏₁𝑏₂𝐴²₁−2𝜇𝐴₂−3𝐴𝑐𝑎₁𝑎₂+3𝐴𝑐𝑎₂𝑏₂𝜇𝐴²₁−2𝜇𝐴₂𝜙=0,₄𝜓∶−24𝑏₂−4𝐴𝑐𝑎₂𝑏₂𝜙=0,₃∶−24𝑎₂𝜇²𝐴²₁−2𝜇𝐴₂+6𝑏₁𝜇𝐴²₁−2𝜇𝐴₂−2𝐴𝑐𝑏²₂𝜇²𝐴²₁−2𝜇𝐴₂²−2𝐴𝑐𝑎₀𝑎₂−𝐴𝑐𝑏²₁𝐴²₁−2𝜇𝐴₂−𝐴𝑐𝑎²₁+2𝐴𝑐𝑎₁𝑏₂𝜇𝐴²₁−2𝜇𝐴₂+2𝐴𝑐𝑏₁𝑎₂𝜇𝐴²₁−2𝜇𝐴₂−12𝑎₂+2𝐵𝑐𝑎₂𝜙=0,₃𝜓∶−6𝑏₁+54𝑎₂𝜇−3𝐴𝑐𝑎₁𝑏₂+2𝐴𝑐𝑎²₂𝜇−3𝐴𝑐𝑏₁𝑎₂+5𝐴𝑐𝑏²₂𝜇𝐴²₁−2𝜇𝐴₂𝜙=0,₂∶−3𝑎₁𝜇²𝐴²₁−2𝜇𝐴₂+12𝑏₂𝜇³𝐴²₁−2𝜇𝐴₂²−𝐴𝑐𝑎₀𝑎₁+𝐴𝑐𝑎₀𝑏₂𝜇𝐴²₁−2𝜇𝐴₂+𝐴𝑐𝑏₁𝑎₁𝜇𝐴²₁−2𝜇𝐴₂−2𝐴𝑐𝑏₁𝑏₂𝜇²𝐴²₁−2𝜇𝐴₂²−6𝑎₁+6𝑏₂𝜇𝐴²₁−2𝜇𝐴₂−𝐵𝑐𝑏₂𝜇𝐴²₁−2𝜇𝐴₂+𝐵𝑐𝑎₁+𝐶𝑎₂𝜙=0,₂𝜓∶−78𝑏₂𝜇²𝐴²₁−2𝜇𝐴₂+12𝑎₁𝜇−2𝐴𝑐𝑎₀𝑏₂−2𝐴𝑐𝑎₁𝑏₁+7𝐴𝑐𝑏₁𝑏₂𝜇𝐴²₁−2𝜇𝐴₂−6𝐴𝑐𝑎₂𝑏₂𝜇²𝐴²₁−2𝜇𝐴₂+3𝐴𝑐𝑎₁𝑎₂𝜇−12𝑏₂+2𝐵𝑐𝑏₂=0,𝜙∶𝐶𝑎₁=0,𝜙𝜓∶48𝑎₂𝜇³𝐴²₁−2𝜇𝐴₂−12𝑏₁𝜇²𝐴²₁−2𝜇𝐴₂−𝐴𝑐𝑎₀𝑏₁+𝐴𝑐𝑎²₁𝜇−4𝐴𝑐𝑎₁𝑏₂𝜇²𝐴²₁−2𝜇𝐴₂+2𝐴𝑐𝑎₀𝑎₂𝜇−4𝐴𝑐𝑏₁𝑎₂𝜇²𝐴²₁−2𝜇𝐴₂+4𝐴𝑐𝑏²₂𝜇³𝐴²₁−2𝜇𝐴₂²+2𝐴𝑐𝑏²₁𝜇𝐴²₁−2𝜇𝐴₂+12𝑎₂𝜇−6𝑏₁−2𝐵𝑐𝑎₂𝜇+𝐵𝑐𝑏₁+𝐶𝑏₂=0,𝜓∶6𝑎₁𝜇³𝐴²₁−2𝜇𝐴₂−24𝑏₂𝜇⁴𝐴²₁−2𝜇𝐴₂²+𝐴𝑐𝑎₀𝑎₁𝜇−2𝐴𝑐𝑎₀𝑏₂𝜇²𝐴²₁−2𝜇𝐴₂−2𝐴𝑐𝑏₁𝑎₁𝜇²𝐴²₁−2𝜇𝐴₂+4𝐴𝑐𝑏₁𝑏₂𝜇³𝐴²₁−2𝜇𝐴₂²+6𝑎₁𝜇−12𝑏₂𝜇²𝐴²₁−2𝜇𝐴₂−𝐵𝑐𝑎₁𝜇+2𝐵𝑐𝑏₂𝜇²𝐴²₁−2𝜇𝐴₂+𝐶𝑏₁𝜙=0,₀:𝐶𝑎₀=0. (34)
On solving the above algebraic equations using the Maple or Mathematica, we get the following results.
Result 1. Consider 𝑎₀=𝐵𝑐−6𝐴𝑐,𝑎₁=0,𝑎₂=−12,𝑏𝐴𝑐₁=0,𝑏₂=0,𝑐=𝑐,𝜇=0,𝐶=0. (35)
From (8), (15), and (35), we deduce the traveling wave solution of (1) as follows: 𝑢(𝜉)=𝐵𝑐−6−𝐴𝑐12𝐴𝐴𝑐𝐴₁𝜉+¹𝐴₂². (36)
Result 2. Consider 𝑎₀=𝐵𝑐−6𝐴𝑐,𝑎₁=0,𝑎₂6=−,𝑏𝐴𝑐₁=36𝐴²₁−𝐴²𝑐²𝑏²₂12𝑐𝐴₂𝐴,𝑐=𝑐,𝑏₂=𝑏₂,𝜇=36𝐴²₁−𝐴²𝑐²𝑏²₂72𝐴₂,𝐶=0. (37)
In this result, we deduce the traveling wave solution of (1) as follows: 𝑢(𝜉)=𝐵𝑐−6−𝐴𝑐𝜇𝜉+𝐴₁6𝜇𝜉+𝐴₁−𝐴𝑐𝑏₂𝐴𝑐𝜇/2𝜉²+𝐴₁𝜉+𝐴₂²+36𝐴²₁−𝐴²𝑐²𝑏²₂12𝑐𝐴₂𝐴𝜇/2𝜉²+𝐴₁𝜉+𝐴₂, (38) where 𝜉 in (36) and (38) is given by 𝜉=(1/ℓ)𝑥−𝑐(𝑡/𝜏). Our result (38) is new and is not reported elsewhere.

Remark 5. In all the above results, we have noted the term 𝐶=3𝑅𝑍⁻¹−𝐺₀𝑍=0, (39) which is equivalent to the condition (21) of [45]. The condition (39) enables that the ohmic loss could be balanced by fresh ions injected from the nanopores in microtubules since these act ionic pumps in suitable voltage regime. Also, the authors [46] have obtained the condition (39) when they have discussed (1) using the extended tanh-function method. The same remark has been happened in [47, 48] using the improved Riccati equation mapping method and the improved (𝐺^′/𝐺)-expansion method, respectively.

Remark 6. Since 𝐶=0, then (14) becomes 𝑢^′′′+𝐴𝑐𝑢𝑢^′+(6−𝐵𝑐)𝑢^′=0. (40) Integrating (40) once with respect to 𝜉 with zero constant of integration, we get 𝑢^′′+12𝐴𝑐𝑢²+(6−𝐵𝑐)𝑢=0. (41)
Multiplying (41) by 𝑢^′ and integrating once with respect to 𝜉 with zero constant of integration, we have 𝑢^′2+13𝐴𝑐𝑢³+(6−𝐵𝑐)𝑢²=0, (42) and hence we have 𝑢^′𝑢=±√33(𝐵𝑐−6)−𝐴𝑐𝑢. (43) Consequently, we have 𝑑𝑢𝑢3(𝐵𝑐−6)−𝐴𝑐𝑢=±𝜉+𝜉₀√3, (44) where 𝜉₀ is a constant of integration.
From (44) we have the following solution (Figure 3): 𝑢(𝜉)=3(𝐵𝑐−6)𝐴𝑐sech²12√𝐵𝑐−6𝜉+𝜉₀, (45) which is in agreement with the solution (17) obtained in [48] when 𝑎₀=0, 𝜇=0 using the improved (𝐺^′/𝐺)-expansion method. After some simple calculation, the solution (45) is equivalent to the solution (34) obtained in [45].

Figure 3 

The plot of solution (45) when 𝑐=1, 𝐴=1, 𝐵=7, 𝜏=ℓ=1/2, 𝜉0=0.

Remark 7. Integrating (40) twice with respect to 𝜉, we get 𝑢^′2=−𝐴𝑐3𝑢³−(6−𝐵𝑐)𝑢²+𝐾₁𝑢+𝐾₂, (46) where 𝐾₁ and 𝐾₂ are nonzero constants of integration.Setting 𝑢(𝜉)()(𝜉)=−6−𝐵𝑐+12𝑃,𝐴𝑐 (47) we obtain the Weierstrass equation [49, 50] 𝑃^′2=4𝑃³−𝑔₂𝑃−𝑔₃, (48) where 𝑔₂=(𝐴𝑐𝐾₁+(6−𝐵𝑐)²)/12, 𝑔₃=(6−𝐵𝑐)³/216+(𝐴𝑐𝐾₁(6−𝐵𝑐)−𝐴²𝑐²𝐾₂)/144, while 𝑃 is the Weierstrass elliptic 𝑃-function.
If 𝑔₂=𝑔₃=0, (48) reduces to the equation 𝑃^′2=4𝑃³, which admits the elementary solution 𝑃(𝜉)=1𝜉+𝜉₀², (49) where 𝜉₀ is a constant of integration. From (47) and (49), we get 𝑢(𝜉)=(𝐵𝑐−6)−𝐴𝑐12𝐴𝑐𝜉+𝜉₀², (50) which is equivalent to the solution (36) for 𝐴₁=1, 𝐴₂=𝜉₀.
From the well-known solutions of (48), see, for example, [50], we rewrite the solution (47) in terms of the Jacobi elliptic functions as follows: 𝑢₁(𝜉)()=−6−𝐵𝑐−𝐴𝑐12𝑒𝐴𝑐²−𝑒₂−𝑒₃cn²(,𝑢𝑅𝜉;𝑚)₂(𝜉)()=−6−𝐵𝑐−𝐴𝑐12𝑒𝐴𝑐³+𝑒₁−𝑒₃ns²(𝑅𝜉;𝑚), (51) where 𝑅=√𝑒₁−𝑒₃, 𝑚²=(𝑒₂−𝑒₃)/(𝑒₁−𝑒₃) is the modulus of the Jacobi elliptic functions and 𝑒_𝑖, 𝑖=1,2,3, 𝑒₁>𝑒₂>𝑒₃ are three roots of the cubic equation 4𝑧³−𝑔₂𝑧−𝑔₃=0. When 𝑚→1, that is, 𝑒₁→𝑒₂, we get ns(𝑅𝜉;𝑚)→coth(𝑅𝜉) and cn(𝑅𝜉;𝑚)→sech(𝑅𝜉). Finally, we note that the solutions (51) are new and are not reported elsewhere.

4. Conclusions

The two-variable (𝐺^′/𝐺,1/𝐺)-expansion method is used in this paper to obtain some new solutions of a selected nonlinear partial differential equation of nanoionic currents along MTs (1). As the two constants 𝐴₁ and 𝐴₂ take special values, we obtain the solitary wave solutions. When 𝜇=0 and 𝑏_𝑖=0 in (2) and (13), the two-variable (𝐺^′/𝐺,1/𝐺)-expansion method reduces to the (𝐺^′/𝐺)-expansion method. So the two-variable (𝐺^′/𝐺,1/𝐺)-expansion method is an extension of the (𝐺^′/𝐺)-expansion method. The proposed method in this paper is more effective and more general than the (𝐺^′/𝐺)-expansion method because it gives exact solutions in more general forms. In summary, the advantage of the two-variable (𝐺^′/𝐺,1/𝐺)-expansion method over the (𝐺^′/𝐺)-expansion method is that the solutions using the first method recover the solutions using the second one. On comparing our results obtained in this paper with the well-known results obtained in [45, 47, 48] we deduce that some of these results are the same while the others are new which are not reported elsewhere. Finally, all solutions obtained in this paper have been checked with the Maple by putting them back into the original equation (14).

Glossary

ER:	elementary ring
MT:	microtubule
NP:	nanopore
ODE:	ordinary differential equation
PDE:	partial differential equation.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.