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

Authors: E. M. E. Zayed; K. A. E. Alurrfi
Journal: Mathematical Problems in Engineering
 (2014)

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## 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.


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## Body

## 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]:
(1)ℓ23uxxx+Z3/2ℓ(χG0-2δC0)uut+2ux+ZC0ℓut+1ℓ(RZ-1-G0Z)u=0,
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 Section2, 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(2)G′′(ξ)+λG(ξ)=μ,
and set ϕ=G′/G, ψ=1/G, then we get
(3)ϕ′=-ϕ2+μψ-λ,ψ′=-ϕψ,
where λ and μ are constants.Remark 2.
Ifλ<0, then the general solution of (2) has the form
(4)G(ξ)=A1sinh⁡(ξ-λ)+A2cosh⁡(ξ-λ)+μλ,
where A1 and A2 are arbitrary constants. Consequently, we have
(5)ψ2=-λλ2σ1+μ2(ϕ2-2μψ+λ),
where σ1=A12-A22.Remark 3.
Ifλ>0, then the general solution of (2) has the form
(6)G(ξ)=A1sin(ξλ)+A2cos⁡(ξλ)+μλ,
and hence
(7)ψ2=λλ2σ2-μ2(ϕ2-2μψ+λ),
where σ2=A12+A22.Remark 4.
Ifλ=0, then the general solution of (2) has the form
(8)G(ξ)=μ2ξ2+A1ξ+A2,
and hence
(9)ψ2=1A12-2μA2(ϕ2-2μψ).
Suppose we have the following nonlinear evolution equation:
(10)F(u,ut,ux,uxx,…)=0,
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(11)u(x,t)=u(ξ),ξ=x-ct,
where c is a nonzero constant, reduces (10) to an ODE in the form
(12)P(u,u′,u′′,…)=0,
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:
(13)u(ξ)=∑i=0Naiϕi+∑i=1Nbiϕi-1ψ,
where ai(i=0,1,2,…,N) and bi(i=1,2,…,N) are constants to be determined later.Step 3.
Determine the positive integerN 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 ai, bi, c, μ, A1, A2, and λ, where λ<0.Step 5.
Similar to Step4, 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 Section2 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
(14)u′′′+Acuu′+(6-Bc)u′+Cu=0,
where A=3(Z3/2/τ)(2δC0-χG0), B=3(ZC0/τ), C=3(RZ-1-G0Z).By balancingu′′′ with uu′ in (14), we get N=2. Consequently, we get
(15)u(ξ)=a0+a1ϕ(ξ)+a2ϕ2(ξ)+b1ψ(ξ)+b2ϕ(ξ)ψ(ξ),
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)).
Ifλ<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:
(16)ϕ5:2Acλb22σ1λ2+μ2-2Aca22-24a2=0,ϕ4:3Acλb1b2σ1λ2+μ2-3Aca1a2-36λμb2σ1λ2+μ2-6a1-3Acλμa2b2σ1λ2+μ2=0,ϕ4ψ:-24b2-4Aca2b2=0,ϕ3:2Bca2-40λa2-12a2-Aca12+24λμ2a2σ1λ2+μ2-2Aca0a2-6λμb1σ1λ2+μ2-2Acλa22+3Acλ2b22σ1λ2+μ2+Acλb12σ1λ2+μ2-2Acλ2μ2b22(μ2+λ2σ1)2-2Acλμa1b2σ1λ2+μ2-2Acλμa2b1σ1λ2+μ2=0,ϕ3ψ:54μa2-6b1-3Aca1b2-3Aca2b1+2Acμa22-5Acλμb22σ1λ2+μ2=0,ϕ2:Ca2-6a1-8λa1+Bca1+3λμ2a1σ1λ2+μ2-47λ2μb2σ1λ2+μ2-Aca0a1-6λμb2σ1λ2+μ2+12λ2μ3b2(σ1λ2+μ2)2-3Acλa1a2+4Acλ2b1b2σ1λ2+μ2+Bcλμb2σ1λ2+μ2-2Acλ2μ2b1b2(σ1λ2+μ2)2-3Acλ2μa2b2σ1λ2+μ2-Acλμa0b2σ1λ2+μ2-Acλμa1b1σ1λ2+μ2=0,ϕ2ψ:12μa1-12b2-28λb2+2Bcb2+78λμ2b2σ1λ2+μ2-2Aca0b2-2Aca1b1+3Acμa1a2-3Acλa2b2+6Acλμ2a2b2σ1λ2+μ2-7Acλμb1b2σ1λ2+μ2=0,ϕ:Ca1-12λa2-16λ2a2-6λ2μb1σ1λ2+μ2+2Bcλa2+24λ2μ2a2σ1λ2+μ2-Acλa12+Acλ2b12σ1λ2+μ2+Acλ3b22σ1λ2+μ2-2Acλa0a2-2Acλ3μ2b22(σ1λ2+μ2)2-2Acλ2μa1b2σ1λ2+μ2-2Acλ2μa2b1σ1λ2+μ2=0,ϕψ:Cb2-6b1-5λb1+12μa2+Bcb1+40λμa2+12λμ2b1σ1λ2+μ2-48λμ3a2σ1λ2+μ2-2Bcμa2-Aca0b1+Acμa12+2Acμa0a2-2Acλa1b2-2Acλa2b1-2Acλμb12σ1λ2+μ2+4Acλ2μ3b22(σ1λ2+μ2)2-3Acλ2μb22σ1λ2+μ2+4Acλμ2a1b2σ1λ2+μ2+4Acλμ2a2b1σ1λ2+μ2=0,ψ:Cb1+6μa1-6λb2-5λ2b2+5λμa1-6λμ3a1σ1λ2+μ2+12λμ2b2σ1λ2+μ2-Bcμa1+Bcλb2+28λ2μ2b2σ1λ2+μ2-24λ2μ4b2(σ1λ2+μ2)2+Acμa0a1-Acλa0b2-Acλa1b1-2Bcλμ2b2σ1λ2+μ2+4Acλ2μ3b1b2(σ1λ2+μ2)2+2Acλμ2a0b2σ1λ2+μ2+2Acλμ2a1b1σ1λ2+μ2-3Acλ2μb1b2σ1λ2+μ2=0,ϕ0:Ca0-6λa1-2λ2a1-6λ2μb2σ1λ2+μ2-11λ3μb2σ1λ2+μ2+Bcλa1+3λ2μ2a1σ1λ2+μ2+12λ3μ3b2(σ1λ2+μ2)2-Acλa0a1+Bcλ2μb2σ1λ2+μ2+Acλ3b1b2σ1λ2+μ2-2Acλ3μ2b1b2(σ1λ2+μ2)2-Acλ2μa0b2σ1λ2+μ2=0.
On solving the above algebraic equations using the Maple or Mathematica, we get the following results.
Result 1.Consider
(17)a0=BA±b2(5λ+6)-λ6λ2σ1+μ2,a1=0,a2=±b2-λλ2σ1+μ2,b1=±μb2-λλ2σ1+μ2,c=±6b2Aλ2σ1+μ2-λ,b2=b2,μ=μ,C=0.
From (4), (15), and (17), we deduce the traveling wave solution of (1) as follows:
(18)u(ξ)=BA±b2(5λ+6)-λ6λ2σ1+μ2±b2λ-λλ2σ1+μ2×[A1cosh⁡(ξ-λ)+A2sinh⁡(ξ-λ)A1sinh⁡(ξ-λ)+A2cosh⁡(ξ-λ)+μ/λ]2+[-λb2A1sinh⁡(ξ-λ)+A2cosh⁡(ξ-λ)+μ/λ]×[μλ2σ1+μ2A1cosh⁡(ξ-λ)+A2sinh⁡(ξ-λ)A1sinh⁡(ξ-λ)+A2cosh⁡(ξ-λ)+μ/λ±μλ2σ1+μ2],
where b2 is a nonzero constant, ξ=(1/ℓ)x±(6/b2A)(λ2σ1+μ2)/-λ(t/τ).In particular, by settingA1=0, A2≠0, and μ=0 in (18), we have the solitary wave solution
(19)u(ξ)=BA±ib2A2(5λ+66-λ)+b2-λA2×tanh(ξ-λ)[sech(ξ-λ)±itanh(ξ-λ)],i=-1,
while if A1≠0, A2=0, and μ=0, then we have the solitary wave solution
(20)u(ξ)=BA±b2A1(5λ+66-λ)+b2-λA1×coth⁡(ξ-λ)[csch(ξ-λ)±coth⁡(ξ-λ)].Result 2.Consider
(21)a0=a0,a1=0,a2=6(Aa0-B)A(3+4λ),b1=0,b2=0,c=-2(3+4λ)Aa0-B,μ=0,C=0.In this result, we deduce the traveling wave solution of (1) as follows:
(22)u(ξ)=a0-6λ(Aa0-B)A(3+4λ)×[A1cosh⁡(ξ-λ)+A2sinh⁡(ξ-λ)A1sinh⁡(ξ-λ)+A2cosh⁡(ξ-λ)]2,
where a0 is an arbitrary constant and ξ=(1/ℓ)x+(2(3+4λ)/(Aa0-B))(t/τ).In particular, by settingA1=0 and A2≠0 in (22), we have the solitary wave solution (Figure 1)
(23)u(ξ)=a0-6λ(Aa0-B)A(3+4λ)tanh2(ξ-λ),
while if A1≠0 and A2=0, then we have the solitary wave solution
(24)u(ξ)=a0-6λ(Aa0-B)A(3+4λ)coth2(ξ-λ).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 ui (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)).
Ifλ>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:
(25)ϕ5:-24a2+2Acλb22μ2-σ2λ2-2Aca22=0,ϕ4:-6a1-36λμb2μ2-σ2λ2+3Acλb1b2μ2-σ2λ2-3Aca1a2-3Acλμa2b2μ2-σ2λ2=0,ϕ4ψ:-24b2-4Aca2b2=0,ϕ3:2Bca2-40λa2-12a2+24λμ2a2μ2-σ2λ2-6λμb1μ2-σ2λ2+3Acλ2b22μ2-σ2λ2-Aca12-2Aca0a2-2Acλa22+Acλb12μ2-σ2λ2-2Acλ2μ2b22(σ2λ2-μ2)2-2Acλμa1b2μ2-σ2λ2-2Acλμa2b1μ2-σ2λ2=0,ϕ3ψ:-6b1+54μa2+2Acμa22-3Aca2b1-3Aca1b2-5Acλμb22μ2-σ2λ2=0,ϕ2:-8λa1-6a1+Bca1+3λμ2a1μ2-σ2λ2-47λ2μb2μ2-σ2λ2-6λμb2μ2-σ2λ2+12λ2μ3b2(σ2λ2-μ2)2+Bcλμb2μ2-σ2λ2+Ca2-Aca0a1-3Acλa1a2+4Acλ2b1b2μ2-σ2λ2-2Acλ2μ2b1b2(σ2λ2-μ2)2-3Acλ2μa2b2μ2-σ2λ2-Acλμa0b2μ2-σ2λ2-Acλμa1b1μ2-σ2λ2=0,ϕ2ψ:12μa1-12b2-28λb2+2Bcb2+78λμ2b2μ2-σ2λ2+3Acμa1a2-2Aca1b1-2Aca0b2-3Acλa2b2+6Acλμ2a2b2μ2-σ2λ2-7Acλμb1b2μ2-σ2λ2=0,ϕ:2Bcλa2-12λa2-16λ2a2-6λ2μb1μ2-σ2λ2+24λ2μ2a2μ2-σ2λ2+Ca1-Acλa12+Acλ2b12μ2-σ2λ2+Acλ3b22μ2-σ2λ2-2Acλa0a2-2Acλ3μ2b22(σ2λ2-μ2)2-2Acλ2μa1b2μ2-σ2λ2-2Acλ2μa2b1μ2-σ2λ2=0,ϕψ:12μa2-5λb1-6b1+Bcb1+40λμa2+12λμ2b1μ2-σ2λ2-48λμ3a2μ2-σ2λ2-2Bcμa2+Cb2-Aca0b1+Acμa12+2Acμa0a2-2Acλa1b2-2Acλa2b1-2Acλμb12μ2-σ2λ2+4Acλ2μ3b22(σ2λ2-μ2)2-3Acλ2μb22μ2-σ2λ2+4Acλμ2a1b2μ2-σ2λ2+4Acλμ2a2b1μ2-σ2λ2=0,ψ:6μa1-6λb2-5λ2b2+5λμa1-6λμ3a1μ2-σ2λ2+12λμ2b2μ2-σ2λ2-Bcμa1+Bcλb2+28λ2μ2b2μ2-σ2λ2-24λ2μ4b2(σ2λ2-μ2)2-2Bcλμ2b2μ2-σ2λ2+Cb1+Acμa0a1-Acλa0b2-Acλa1b1+4Acλ2μ3b1b2(σ2λ2-μ2)2+2Acλμ2a0b2μ2-σ2λ2+2Acλμ2a1b1μ2-σ2λ2-3Acλ2μb1b2μ2-σ2λ2=0,ϕ0:Bcλa1-6λa1-2λ2a1-6λ2μb2μ2-σ2λ2-11λ3μb2μ2-σ2λ2+3λ2μ2a1μ2-σ2λ2+12λ3μ3b2(σ2λ2-μ2)2+Bcλ2μb2μ2-σ2λ2+Ca0-Acλa0a1+Acλ3b1b2μ2-σ2λ2-2Acλ3μ2b1b2(σ2λ2-μ2)2-Acλ2μa0b2μ2-σ2λ2-Acλ2μa1b1μ2-σ2λ2=0.
On solving the above algebraic equations using the Maple or Mathematica, we get the following results.Result 1. Consider
(26)a0=BA±b2(5λ+6)λ6λ2σ2-μ2,a1=0,a2=±b2λλ2σ2-μ2,b1=±μb2λλ2σ2-μ2,c=±6b2Aλ2σ2-μ2λ,b2=b2,μ=μ,C=0.From (6), (15), and (26), we deduce the traveling wave solution of (1) as follows:
(27)u(ξ)=BA±b2(5λ+6)λ6λ2σ2-μ2±λb2λλ2σ2-μ2×[A1cos⁡(ξλ)-A2sin(ξλ)A1sin(ξλ)+A2cos⁡(ξλ)+μ/λ]2+(λb2A1sin(ξλ)+A2cos⁡(ξλ)+μ/λ)×[μλ2σ2-μ2A1cos⁡(ξλ)-A2sin(ξλ)A1sin(ξλ)+A2cos⁡(ξλ)+μ/λ±μλ2σ2-μ2],
where b2 is a nonzero constant, ξ=(1/ℓ)x±(6/b2A)(λ2σ2-μ2)/λ(t/τ).In particular, by settingA1=0, A2≠0, and μ=0 in (27), we have the periodic solution
(28)u(ξ)=BA±b2A2(5λ+66λ)-λb2A2tan(ξλ)[sec(ξλ)±tan(ξλ)],
while if A1≠0, A2=0, and μ=0, then we have the periodic solution
(29)u(ξ)=BA±b2A1(5λ+66λ)+λb2A1cot(ξλ)[csc(ξλ)±cot(ξλ)].Result 2. Consider
(30)a0=a0,a1=0,a2=6(Aa0-B)A(3+4λ),b1=0,b2=0,μ=0,c=-2(3+4λ)Aa0-B,C=0.In this result, we deduce the traveling wave solution of (1) as follows:
(31)u(ξ)=a0+6λ(Aa0-B)A(3+4λ)×[A1cos⁡(ξλ)-A2sin(ξλ)A1sin(ξλ)+A2cos⁡(ξλ)]2,
where a0 is an arbitrary constant and ξ=(1/ℓ)x+(2(3+4λ)/(Aa0-B))(t/τ).In particular, by settingA1=0 and A2≠0 in (31), we have the periodic solution (Figure 2)
(32)u(ξ)=a0+6λ(Aa0-B)A(3+4λ)tan2(ξλ),
while if A1≠0 and A2=0, then we have the periodic solution
(33)u(ξ)=a0+6λ(Aa0-B)A(3+4λ)cot2(ξλ).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)).
Ifλ=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:
(34)ϕ5:-24a2-2Acb22A12-2μA2-2Aca22=0,ϕ4:36b2μA12-2μA2-6a1-3Acb1b2A12-2μA2-3Aca1a2+3Aca2b2μA12-2μA2=0,ϕ4ψ:-24b2-4Aca2b2=0,ϕ3:-24a2μ2A12-2μA2+6b1μA12-2μA2-2Acb22μ2(A12-2μA2)2-2Aca0a2-Acb12A12-2μA2-Aca12+2Aca1b2μA12-2μA2+2Acb1a2μA12-2μA2-12a2+2Bca2=0,ϕ3ψ:-6b1+54a2μ-3Aca1b2+2Aca22μ-3Acb1a2+5Acb22μA12-2μA2=0,ϕ2:-3a1μ2A12-2μA2+12b2μ3(A12-2μA2)2-Aca0a1+Aca0b2μA12-2μA2+Acb1a1μA12-2μA2-2Acb1b2μ2(A12-2μA2)2-6a1+6b2μA12-2μA2-Bcb2μA12-2μA2+Bca1+Ca2=0,ϕ2ψ:-78b2μ2A12-2μA2+12a1μ-2Aca0b2-2Aca1b1+7Acb1b2μA12-2μA2-6Aca2b2μ2A12-2μA2+3Aca1a2μ-12b2+2Bcb2=0,ϕ:Ca1=0,ϕψ:48a2μ3A12-2μA2-12b1μ2A12-2μA2-Aca0b1+Aca12μ-4Aca1b2μ2A12-2μA2+2Aca0a2μ-4Acb1a2μ2A12-2μA2+4Acb22μ3(A12-2μA2)2+2Acb12μA12-2μA2+12a2μ-6b1-2Bca2μ+Bcb1+Cb2=0,ψ:6a1μ3A12-2μA2-24b2μ4(A12-2μA2)2+Aca0a1μ-2Aca0b2μ2A12-2μA2-2Acb1a1μ2A12-2μA2+4Acb1b2μ3(A12-2μA2)2+6a1μ-12b2μ2A12-2μA2-Bca1μ+2Bcb2μ2A12-2μA2+Cb1=0,ϕ0:Ca0=0.
On solving the above algebraic equations using the Maple or Mathematica, we get the following results.
Result 1. Consider
(35)a0=Bc-6Ac,a1=0,a2=-12Ac,b1=0,b2=0,c=c,μ=0,C=0.
From (8), (15), and (35), we deduce the traveling wave solution of (1) as follows:
(36)u(ξ)=Bc-6Ac-12Ac[A1A1ξ+A2]2.
Result 2. Consider
(37)a0=Bc-6Ac,a1=0,a2=-6Ac,b1=36A12-A2c2b2212cA2A,c=c,b2=b2,μ=36A12-A2c2b2272A2,C=0.
In this result, we deduce the traveling wave solution of (1) as follows:
(38)u(ξ)=Bc-6Ac-(μξ+A1)[6(μξ+A1)-Acb2]Ac((μ/2)ξ2+A1ξ+A2)2+36A12-A2c2b2212cA2A((μ/2)ξ2+A1ξ+A2),
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(39)C=3(RZ-1-G0Z)=0,
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.
SinceC=0, then (14) becomes
(40)u′′′+Acuu′+(6-Bc)u′=0.
Integrating (40) once with respect to ξ with zero constant of integration, we get
(41)u′′+12Acu2+(6-Bc)u=0.
Multiplying (41) by u′ and integrating once with respect to ξ with zero constant of integration, we have
(42)(u′)2+13Acu3+(6-Bc)u2=0,
and hence we have
(43)u′=±u33(Bc-6)-Acu.
Consequently, we have
(44)∫duu3(Bc-6)-Acu=±(ξ+ξ0)3,
where ξ0 is a constant of integration.
From (44) we have the following solution (Figure 3):
(45)u(ξ)=3(Bc-6)Acsech2[12Bc-6(ξ+ξ0)],
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
(46)(u′)2=-Ac3u3-(6-Bc)u2+K1u+K2,
where K1 and K2 are nonzero constants of integration.Setting
(47)u(ξ)=-(6-Bc)+12P(ξ)Ac,
we obtain the Weierstrass equation [49, 50]
(48)P′2=4P3-g2P-g3,
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.
Ifg2=g3=0, (48) reduces to the equation P′2=4P3, which admits the elementary solution
(49)P(ξ)=1(ξ+ξ0)2,
where ξ0 is a constant of integration. From (47) and (49), we get
(50)u(ξ)=(Bc-6)Ac-12Ac(ξ+ξ0)2,
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:
(51)u1(ξ)=-(6-Bc)Ac-12Ac[e2-(e2-e3)cn2(Rξ;m)],u2(ξ)=-(6-Bc)Ac-12Ac[e3+(e1-e3)ns2(Rξ;m)],
where R=e1-e3, m2=(e2-e3)/(e1-e3) is the modulus of the Jacobi elliptic functions and ei, 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 bi=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).



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*Source: 521712-2014-03-25.xml*