# New Exact Traveling Wave Solutions of the Time Fractional Complex Ginzburg-Landau Equation via the Conformable Fractional Derivative

Authors: Zhao Li; Tianyong Han
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this study, the exact traveling wave solutions of the time fractional complex Ginzburg-Landau equation with the Kerr law and dual-power law nonlinearity are studied. The nonlinear fractional partial differential equations are converted to a nonlinear ordinary differential equation via a traveling wave transformation in the sense of conformable fractional derivatives. A range of solutions, which include hyperbolic function solutions, trigonometric function solutions, and rational function solutions, is derived by utilizing the new extendedG′/G-expansion method. By selecting appropriate parameters of the solutions, numerical simulations are presented to explain further the propagation of optical pulses in optic fibers.


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## Body

## 1. Introduction

The Ginzburg-Landau (GL) equation [1–7] is one of the most important partial differential equations in the field of mathematics and physics, which was introduced into the study of superconductivity phenomenology theory in the 20th century by Ginzburg and Landau. The GL equation usually describes the optical soliton [8–15] propagation through optical fibers over longer distances. Therefore, it is very important to analyze the dynamic behavior of the GL equation. In particular, searching for the approximate analytical solutions or exact traveling solutions of the GL equation has been more and more widely followed with interest because they can help to explain the complexity of nonlinear optics. So, in recent years, many powerful methods were used for finding the exact traveling solutions of the GL equation, which include the first integral method [16], dynamical system method [17], G′/G-expansion method [18], Hirota’s method [19], and harmonic balance method [20].In recent years, the fractional GL equation [21–24] has been more and more widely followed with interest because it can be used to accurately model some nonlinear optical phenomena. In particular, constructing the exact traveling solutions of the fractional GL equation is very important work because it can be better explained the dynamics of soliton propagation through optical fibers over longer distances in nonlinear optics. So far, many important methods have been proposed to find the exact solutions of the fractional GL equation. In [25], Arshed extracted the hyperbolic, trigonometric, and rational function solutions of the fractional GL equation by the exp−ϕξ-expansion method. In [26], Raza obtained the periodic and hyperbolic soliton solutions of the conformable time fractional GL equation by using the solitary wave ansatz method and exp−φχ-expansion method. In [27], Abdou et al. employed Jacobi’s elliptic function expansion approach to retrieve doubly periodic function solutions for the time-space fractional GL equation. In [28], Sirisubtawee et al. investigated the cubic-quintic GL equation by the modified Kudryashov method and G′/G,1/G-expansion method. In [29], Ghanbari and Gómez-Aguilar obtained the periodic and hyperbolic soliton solutions of the conformable GL equation by using the generalized exponential rational function method. Although exact traveling wave solutions of the fractional GL equation have been successfully obtained by utilizing the above methods, it is far from enough on searching for the exact solution of a more general fractional complex GL equation. The main purpose of this paper is to construct exact traveling wave solutions of the time fractional complex GL equation with the Kerr law and dual-power law nonlinearity by using the new extended G′/G-expansion method, and a range of solutions which include hyperbolic function solutions, trigonometric function solutions, rational function solutions, and negative power solutions is derived.In this paper, we will consider the following time fractional complex GL equation [26, 29]:
(1)iDtδu+auxx+bFu2u=αu2u2xx−βu2x2u2u∗+γu,0<δ≤1,where x represents distance across the fiber in dimensionless form. a and b denote the coefficient of the group velocity dispersion and the coefficient of nonlinearity. The terms with constants α and β arise from the perturbation effects. In particular, γ comes from the detuning effect. a, b, α, β, and γ are real-valued constants. u∗ is the complex conjugate numbers of u. Equation (1) is used to model various physical phenomena, such as nonlinear waves, Bose-Einstein condensation, second-order phase transitions, superconductivity, superfluidity, liquid crystals, and strings.In Equation (1), Fu2u:C→C is a real-valued algebraic function. Let Fu2u be k times continuously differentiable. Then,
(2)Fu2u∈⋃m,n=1∞Ck−n,n×−m,m;R2,where C is a complex plane of two-dimensional linear space in R2.The rest of this paper is organized as follows. In Section2, we will give the definition of the conformable fractional derivative and its properties. In Section 3, we will introduce the extended G′/G-expansion method. In Section 4, we will discuss the exact traveling solutions of the fractional complex GL equation with the Kerr law and dual-power law nonlinearity by using the extended G′/G-expansion method and plot the profiles of several representative exact traveling solutions. In Section 5, we will present the concluding remarks.


## 2. Preliminaries

Definition 1 [28].
Letf:0,∞→R. Then, the conformable fractional derivative of f of order α is defined as
(3)Dtδft=limε→0ft+εt1−δ−ftε,forallt>0,0<δ≤1.If the limit in Equation (3) exists, then we say that f is δ-conformable differentiable at a point t.Theorem 2 [28].
Letα∈0,1 and ft and gt be δ-conformable differentiable at a point t>0. Then,
(4)Dtδtμ=μtμ−δ,forallμ∈R,Dtδaft+bgt=aDtδft+bDtδgt,foralla,b∈R,Dtδftgt=ftDtδgt+gtDtδft,Dtδftgt=gtDtδft−ftDtδgtg2t,Dtδft=t1−δdftdt,providedthatftisdifferentiable.Theorem 3 [28].
Letf:0,∞→R be a function such that f is differentiable and δ-conformable differentiable. Further, let g be a differentiable function defined in the range of f. Then,
(5)Dtαf∘gt=t1−αgtα−1g′tDtαftt=gt,where the prime notation ′ denotes the classical derivative.


## 3. Overview of the ExtendedG′/G-Expansion Method

Let us consider a general fractional partial differential equation in the form(6)Pu,ut,ux,⋯,Dtδu,Dxδu,DtδDxδ,⋯=0,0<δ≤1,where u is an unknown function and P is a polynomial of u and its partial fractional order derivatives, in which the highest order derivatives and nonlinear terms are involved.Step 1.
By using the traveling wave transformationut,x=Uξ, where ξ=x−ltδ/δ, l is a nonzero arbitrary constant, 0<δ≤1, we can rewrite Equation (6) as the following nonlinear ordinary differential equation (ODE) of the integer order with respect to ξ:
(7)QU,U′,U′′,⋯=0.If it is possible, we should integrate Equation (7) term by term one or more times; the constants of integration are considered to be zero.Step 2.
Assume that the traveling wave solution of Equation (7) can be expressed as follows:
(8)Uξ=∑i=0NaiG′Gi+∑i=1NbiG′G−i,where either aN or bN may be zero, but both aN and bN cannot be zero at a time, and aii=0,1,2,⋯,N and bii=1,2,⋯,N are arbitrary constants, where G=Gξ satisfies the following nonlinear ODE:
(9)GG′′=AG′2+BGG′+CG2,where A, B, and C are real parameters.Step 3.
The value of the positive integerN in Equation (8) can be computed by using the homogeneous balance between the highest order nonlinear term and the highest order derivative in Uξ appearing in Equation (8).Step 4.
Substituting Equations (8) and (9) into Equation (7) with the value of N obtained in Step 3, we obtain polynomials in G′/GNN=0,1,2,⋯ and G′/G−NN=0,1,2,⋯. Then, we collect all coefficients of the resulted polynomial to zero, and we obtain a set of algebraic equations for aii=0,1,2,⋯ and bii=1,2,⋯,N.Step 5.
By solving the algebraic equations obtained in Step4, we can obtain the values of the constants aii=0,1,2,⋯ and bii=1,2,⋯,N. Replacing the values into Equation (8), we construct a more general type and new exact traveling wave solutions of Equation (3).The solutions of Equation (9) can be categorized into the following three cases.Case 1 (hyperbolic function solutions).
WhenB2−4A−1C>0 and A≠1, then
(10)G′ξGξ=B2+4C−4AC21−AC1sinhB2+4C−4AC/2ξ+C2coshB2+4C−4AC/2ξC1coshB2+4C−4AC/2ξ+C2sinhB2+4C−4AC/2ξ+B21−A.Case 2 (trigonometric function solutions).
When4A−1C−B2>0 and A≠1, then
(11)G′ξGξ=4AC−4C−B221−A−C1sin4AC−4C−B2/2ξ+C2cos4AC−4C−B2/2ξC1cos4AC−4C−B2/2ξ+C2sin4AC−4C−B2/2ξ+B21−A.Case 3 (rational function solutions).
WhenB2−4A−1C=0 and A≠1, then
(12)G′ξGξ=11−AC1C1ξ+C2+B2.Remark 4.
In fact, by takingA=0, B=−λ, and C=−μ in Equation (9), the second nonlinear ODE reduces into the following second-order linear ODE:
(13)G′′+λG′+μG=0.


## 4. Exact Solutions of the Time Fractional Complex Ginzburg-Landau Equation

In order to obtain the traveling wave solution of Equation (1), we assume that the exact solution has the following form:
(14)ut,x=Uξeiη,where ξ=x−vtδ/δ and η=−kx+wtδ/δ+θ.Substituting (14) into Equation (1), then collecting the real and imaginary parts, we get the following equations:
(15)Re:−wU+aU′′−k2U+bFU2U=2α−2βU′2U+2αU′′+γU,(16)Im:v=−2ak,where U′′=d2U/dξ2.By takingα=2β, Equation (15) becomes
(17)−wU+aU′′−k2U+bFU2U=2αU′′+γU.


### 4.1. Kerr Law Nonlinearity

The Kerr law nonlinearity is the case whenFU=U; then, Equation (17) reduces to
(18)2α−aU′′−bU3+γ+w+ak2U=0.By using the homogeneous balance principle, we selectN=1. Then, we suppose that the solution of Equation (18) satisfies
(19)Uξ=a1G′G+a0+b1G′G−1,where a1, a0, and b1 are constants to be determined.Substituting (19) together with (9) into (18) yields a polynomial in G′/GN (N=0,1,2,⋯) and G′/G−N (N=0,1,2,⋯). Collecting the coefficient of the resulted polynomial and setting them to zero, we obtain a nonlinear algebraic equation in a1, a0, and b1. Solving the algebraic equation using the computer algebraic software of Maple, we get the following results:Case 1.
a1=±22α−aA−12/b, a0=±2α−aB2/2b, b1=0, k=k, and w=2α−aB2+4C−4AC/2−γ−ak2.Case 2.
a1=0, a0=±2α−aB2/2b, b1=±22α−aC2/b, k=k, and w=2α−aB2+4C−4AC/2−γ−ak2, where A, B, C, a, b, α, and γ are free parameters.Substituting these solutions into (19), using (14), we obtain the traveling wave solution as follows:
(20)ut,x=±2α−aB22b±22α−aA−12bG′ξGξei−kx+wtδ/δ+θ,ut,x=±2α−aB22b±22α−aC2bG′ξGξ−1ei−kx+wtδ/δ+θ,where G′ξ/Gξ is defined in Section 3, and ξ=x+2aktδ/δ.Family 1.
The hyperbolic function solutions can be obtained asB2−4A−1C>0 and A≠1. Then,
(21)u1t,x=±2α−aB22b±22α−aA−12bB21−A+B2+4C−4AC21−AH1ei−kx+wtδ/δ+θ,(22)u2t,x=±2α−aB22b±22α−aC2bB21−A+B2+4C−4AC21−AH1−1ei−kx+wtδ/δ+θ,where H1=C1sinhB2+4C−4AC/2x+2aktδ/δ+C2coshB2+4C−4AC/2x+2aktδ/δ/C1coshB2+4C−4AC/2x+2aktδ/δ+C2sinhB2+4C−4AC/2x+2aktδ/δ, and C1 and C2 are arbitrary constants.In particular, if we assume thatC2=0, A>1, B>0, and C1≠0 in (21), then we have
(23)u11t,x=∓2α−aB2+4C−4AC2btanhB2+4C−4AC2x+2aktδ/δei−kx+wtδ/δ+θ,and if C1=0, A>1, B>0, and C2≠0 in (21), then we obtain
(24)u12t,x=∓2α−aB2+4C−4AC2bcothB2+4C−4AC2x+2aktδ/δei−kx+wtδ/δ+θ.The fixed values areα=2, a=1, b=1, A=2, B=4, C=3, and k=1, and the fractional parameter δ∈1/5,1/2,1 is used to plot the positive real part of the exact solution u11t,x as in Figures 1 and 2.Family 2.
The trigonometric function solutions can be obtained as4A−1C−B2>0 and A≠1. Then,
(25)u3t,x=±2α−aB22b±22α−aA−12bB21−A+4AC−4C−B221−AH2ei−kx+wtδ/δ+θ,(26)u4t,x=±2α−aB22b±22α−aC2bB21−A+4AC−4C−B221−AH2−1ei−kx+wtδ/δ+θ,where H2=−C1sin4AC−B2−4C/2x+2aktδ/δ+C2cos4AC−B2−4C/2x+2aktδ/δ/C1cos4AC−B2−4C/2x+2aktδ/δ+C2sin4AC−B2−4C/2x+2aktδ/δ, and C1 and C2 are arbitrary constants.Figure 1
The positive real part ofu11t,x in Equation (23) for differential values of the fractional parameter δ.
(a)
δ=1/5(b)
δ=1/2(c)
δ=4/5Figure 2
The positive real part ofu11t,x at x=1.0 in Equation (23) for differential values of the fractional parameter δ.Specifically, if we suppose thatC2=0, A>1, B>0, and C1≠0 in (25), then we obtain
(27)u31t,x=±2α−a4AC−4C−B22btan4AC−4C−B22x+2aktδδei−kx+wtδ/δ+θ,and if C1=0, A>1, B>0, and C2≠0 in (25), then we have
(28)u32t,x=∓2α−a4AC−4C−B22bcot4AC−4C−B22x+2aktδδei−kx+wtδ/δ+θ.The fixed values areα=2, a=1, b=1, A=4, B=3, C=3, and k=1, and the fractional parameter δ∈1/5,1/2,4/5 is used to plot the positive real part of the exact solution u31t,x as in Figures 3 and 4.Family 3.
The rational function solutions can be obtained asB2−4A−1C=0 and A≠1. Then,
(29)u5t,x=±2α−aB22b±11−A22α−aA−12bC1C1x+2aktδ/δ+C2+B2ei−kx+wtδ/δ+θ,u6t,x=±2α−aB22b±11−A22α−aC2bC1C1x+2aktδ/δ+C2+B2−1ei−kx+wtδ/δ+θ,where C1 and C2 are arbitrary constants.Figure 3
The positive real part ofu31t,x in Equation (27) for differential values of the fractional parameter δ.
(a)
δ=1/5(b)
δ=1/2(c)
δ=4/5Figure 4
The positive real part ofu31t,x at x=1.0 in Equation (27) for differential values of the fractional parameter δ


### 4.2. Dual-Power Law

In this case,FU=Un+lU2n; then, Equation (17) takes the following form:
(30)2α−aU′′+γ+w+ak2U−bU2n+1+lU4n+1=0,where l is a constant.Using transformation(31)U=V1/2n,Equation (30) will be reduced into the following ODE:
(32)2α−a1−2nV′2+2nVV′′+4n2γ+w+ak2V2−4bn2V2V+lV2=0.Now, applying the balancing principle betweenV4 and VV′′ in the above equation, we get N=1. Therefore, we have
(33)Vξ=a1G′G+a0+b1G′G−1,where a1, a0, and b1 are constants to be determined.Substituting (33) together with (9) into (32) yields a polynomial in G′/GN (N=0,1,2,⋯) and G′/G−N (N=0,1,2,⋯). Collecting the coefficient of the resulted polynomial and setting them to zero, we obtain a nonlinear algebraic equation in a1, a0, and b1. Solving the algebraic equation using the computer algebraic software of Maple, we get the following results:Case 1.
a1=±1/2n2n+12α−aA−12/bl, a0=−2n+1/4ln+1±B/4nA−12n+12α−aA−12/bl, b1=0, k=k, and w=2α−aB2+4C−4AC/8n2−3b2n+1/8ln+12−γ−ak2.Case 2.
a1=0, a0=−2n+1/4ln+1±B/4nC2n+12α−aC2/bl, b1=±1/2n2n+12α−aC2/bl, k=k, and w=2α−aB2+4C−4AC/8n2−3b2n+1/8ln+12−γ−ak2, where A, B, C, a, b, α, and γ are free parameters.Substituting these solutions into (33), using (14) and (31), we obtain the traveling wave solution as follows:
(34)ut,x=−2n+14ln+1±B4nA−12n+12α−aA−12bl±12n2n+12α−aA−12blG′ξGξ1/2nei−kx+wtδ/δ+θ,ut,x=−2n+14ln+1±B4nC2n+12α−aC2bl±12n2n+12α−aC2blG′ξGξ−11/2nei−kx+wtδ/δ+θ,where G′ξ/Gξ is defined in Section 3, and ξ=x+2aktδ/δ.Family 1.
The hyperbolic function solutions can be obtained asB2−4A−1C>0 and A≠1. Then,
(35)u1t,x=−2n+14ln+1±B4nA−12n+12α−aA−12bl±12n2n+12α−aA−12blB21−A+B2+4C−4AC21−AH11/2nei−kx+wtδ/δ+θ,(36)u2t,x=−2n+14ln+1±B4nC2n+12α−aC2bl±12n2n+12α−aC2blB21−A+B2+4C−4AC21−AH1−11/2nei−kx+wtδ/δ+θ,where H1=C1sinhB2+4C−4AC/2x+2aktδ/δ+C2coshB2+4C−4AC/2x+2aktδ/δ/C1coshB2+4C−4AC/2x+2aktδ/δ+C2sinhB2+4C−4AC/2x+2aktδ/δ, and C1 and C2 are arbitrary constants.In particular, if we assume thatC2=0, A>1, and C1≠0 in (35), then we have
(37)u11t,x=−2n+14ln+1∓14n2n+12α−aB2+4C−4ACbltanhB2+4C−4AC2x+2aktδδ1/2nei−kx+wtδ/δ+θ,and if C1=0, A>1, and C2≠0 in (35), then we obtain
(38)u12t,x=−2n+14ln+1∓14n2n+12α−aB2+4C−4ACblcothB2+4C−4AC2x+2aktδδ1/2nei−kx+wtδ/δ+θ.The fixed values areα=2, a=1, b=1, A=2, B=4, C=3, k=1, l=3, and n=2, and the fractional parameter δ∈1/5,1/2,1 is used to plot the real part of the exact solution u11t,x as in Figures 5 and 6.Family 2.
The trigonometric function solutions can be obtained as4A−1C−B2>0 and A≠1. Then,
(39)u3t,x=−2n+14ln+1±B4nA−12n+12α−aA−12bl±12n2n+12α−aA−12blB21−A+4AC−4C−B221−AH21/2nei−kx+wtδ/δ+θ,(40)u4t,x=−2n+14ln+1±B4nC2n+12α−aC2bl±12n2n+12α−aC2blB21−A+4AC−4C−B221−AH2−11/2nei−kx+wtδ/δ+θ,where H2=−C1sin4AC−B2−4C/2x+2aktδ/δ+C2cos4AC−B2−4C/2x+2aktδ/δ/C1cos4AC−B2−4C/2x+2aktδ/δ+C2sin4AC−B2−4C/2x+2aktδ/δ, and C1 and C2 are arbitrary constants.Figure 5
The positive real part ofu11t,x in Equation (37) for differential values of the fractional parameter δ.
(a)
δ=1/5(b)
δ=1/2(c)
δ=4/5Figure 6
The positive real part ofu11t,x at x=1.0 in Equation (37) for differential values of the fractional parameter δSpecifically, if we suppose thatC2=0, A>1, and C1≠0 in (39), then we obtain
(41)u31t,x=−2n+14ln+1±14n2n+12α−a4AC−4C−B2bl·tan4AC−4C−B22x+2aktδδ1/2nei−kx+wtδ/δ+θ,and if C1=0, A>1, and C2≠0 in (39), then we obtain
(42)u32t,x=−2n+14ln+1∓14n2n+12α−a4AC−4C−B2bl·cot4AC−4C−B22x+2aktδδ1/2nei−kx+wtδ/δ+θ.The fixed values areα=2, a=1, b=1, A=4, B=3, C=3, k=1, l=3, and n=2, and the fractional parameter δ∈1/5,1/2,1 is used to plot the real part of the exact solution u31t,x as in Figures 7 and 8.Family 3.
The rational function solutions can be obtained asB2−4A−1C=0 and A≠1. Then,
(43)u5t,x=−2n+14ln+1±B4nA−12n+12α−aA−12bl±12n1−A2n+12α−aA−12blC1C1ξ+C2+B21/2nei−kx+wtδ/δ+θ,u6t,x=−2n+14ln+1±B4nC2n+12α−aC2bl±12n1−A2n+12α−aC2blC1C1ξ+C2+B2−11/2nei−kx+wtδ/δ+θ.Figure 7
The positive real part ofu31t,x in Equation (41) for differential values of the fractional parameter δ.
(a)
δ=1/5(b)
δ=1/2(c)
δ=4/5Figure 8
The positive real part ofu31t,x at x=1.0 in Equation (41) for differential values of the fractional parameter δ


## 4.1. Kerr Law Nonlinearity

The Kerr law nonlinearity is the case whenFU=U; then, Equation (17) reduces to
(18)2α−aU′′−bU3+γ+w+ak2U=0.By using the homogeneous balance principle, we selectN=1. Then, we suppose that the solution of Equation (18) satisfies
(19)Uξ=a1G′G+a0+b1G′G−1,where a1, a0, and b1 are constants to be determined.Substituting (19) together with (9) into (18) yields a polynomial in G′/GN (N=0,1,2,⋯) and G′/G−N (N=0,1,2,⋯). Collecting the coefficient of the resulted polynomial and setting them to zero, we obtain a nonlinear algebraic equation in a1, a0, and b1. Solving the algebraic equation using the computer algebraic software of Maple, we get the following results:Case 1.
a1=±22α−aA−12/b, a0=±2α−aB2/2b, b1=0, k=k, and w=2α−aB2+4C−4AC/2−γ−ak2.Case 2.
a1=0, a0=±2α−aB2/2b, b1=±22α−aC2/b, k=k, and w=2α−aB2+4C−4AC/2−γ−ak2, where A, B, C, a, b, α, and γ are free parameters.Substituting these solutions into (19), using (14), we obtain the traveling wave solution as follows:
(20)ut,x=±2α−aB22b±22α−aA−12bG′ξGξei−kx+wtδ/δ+θ,ut,x=±2α−aB22b±22α−aC2bG′ξGξ−1ei−kx+wtδ/δ+θ,where G′ξ/Gξ is defined in Section 3, and ξ=x+2aktδ/δ.Family 1.
The hyperbolic function solutions can be obtained asB2−4A−1C>0 and A≠1. Then,
(21)u1t,x=±2α−aB22b±22α−aA−12bB21−A+B2+4C−4AC21−AH1ei−kx+wtδ/δ+θ,(22)u2t,x=±2α−aB22b±22α−aC2bB21−A+B2+4C−4AC21−AH1−1ei−kx+wtδ/δ+θ,where H1=C1sinhB2+4C−4AC/2x+2aktδ/δ+C2coshB2+4C−4AC/2x+2aktδ/δ/C1coshB2+4C−4AC/2x+2aktδ/δ+C2sinhB2+4C−4AC/2x+2aktδ/δ, and C1 and C2 are arbitrary constants.In particular, if we assume thatC2=0, A>1, B>0, and C1≠0 in (21), then we have
(23)u11t,x=∓2α−aB2+4C−4AC2btanhB2+4C−4AC2x+2aktδ/δei−kx+wtδ/δ+θ,and if C1=0, A>1, B>0, and C2≠0 in (21), then we obtain
(24)u12t,x=∓2α−aB2+4C−4AC2bcothB2+4C−4AC2x+2aktδ/δei−kx+wtδ/δ+θ.The fixed values areα=2, a=1, b=1, A=2, B=4, C=3, and k=1, and the fractional parameter δ∈1/5,1/2,1 is used to plot the positive real part of the exact solution u11t,x as in Figures 1 and 2.Family 2.
The trigonometric function solutions can be obtained as4A−1C−B2>0 and A≠1. Then,
(25)u3t,x=±2α−aB22b±22α−aA−12bB21−A+4AC−4C−B221−AH2ei−kx+wtδ/δ+θ,(26)u4t,x=±2α−aB22b±22α−aC2bB21−A+4AC−4C−B221−AH2−1ei−kx+wtδ/δ+θ,where H2=−C1sin4AC−B2−4C/2x+2aktδ/δ+C2cos4AC−B2−4C/2x+2aktδ/δ/C1cos4AC−B2−4C/2x+2aktδ/δ+C2sin4AC−B2−4C/2x+2aktδ/δ, and C1 and C2 are arbitrary constants.Figure 1
The positive real part ofu11t,x in Equation (23) for differential values of the fractional parameter δ.
(a)
δ=1/5(b)
δ=1/2(c)
δ=4/5Figure 2
The positive real part ofu11t,x at x=1.0 in Equation (23) for differential values of the fractional parameter δ.Specifically, if we suppose thatC2=0, A>1, B>0, and C1≠0 in (25), then we obtain
(27)u31t,x=±2α−a4AC−4C−B22btan4AC−4C−B22x+2aktδδei−kx+wtδ/δ+θ,and if C1=0, A>1, B>0, and C2≠0 in (25), then we have
(28)u32t,x=∓2α−a4AC−4C−B22bcot4AC−4C−B22x+2aktδδei−kx+wtδ/δ+θ.The fixed values areα=2, a=1, b=1, A=4, B=3, C=3, and k=1, and the fractional parameter δ∈1/5,1/2,4/5 is used to plot the positive real part of the exact solution u31t,x as in Figures 3 and 4.Family 3.
The rational function solutions can be obtained asB2−4A−1C=0 and A≠1. Then,
(29)u5t,x=±2α−aB22b±11−A22α−aA−12bC1C1x+2aktδ/δ+C2+B2ei−kx+wtδ/δ+θ,u6t,x=±2α−aB22b±11−A22α−aC2bC1C1x+2aktδ/δ+C2+B2−1ei−kx+wtδ/δ+θ,where C1 and C2 are arbitrary constants.Figure 3
The positive real part ofu31t,x in Equation (27) for differential values of the fractional parameter δ.
(a)
δ=1/5(b)
δ=1/2(c)
δ=4/5Figure 4
The positive real part ofu31t,x at x=1.0 in Equation (27) for differential values of the fractional parameter δ


## 4.2. Dual-Power Law

In this case,FU=Un+lU2n; then, Equation (17) takes the following form:
(30)2α−aU′′+γ+w+ak2U−bU2n+1+lU4n+1=0,where l is a constant.Using transformation(31)U=V1/2n,Equation (30) will be reduced into the following ODE:
(32)2α−a1−2nV′2+2nVV′′+4n2γ+w+ak2V2−4bn2V2V+lV2=0.Now, applying the balancing principle betweenV4 and VV′′ in the above equation, we get N=1. Therefore, we have
(33)Vξ=a1G′G+a0+b1G′G−1,where a1, a0, and b1 are constants to be determined.Substituting (33) together with (9) into (32) yields a polynomial in G′/GN (N=0,1,2,⋯) and G′/G−N (N=0,1,2,⋯). Collecting the coefficient of the resulted polynomial and setting them to zero, we obtain a nonlinear algebraic equation in a1, a0, and b1. Solving the algebraic equation using the computer algebraic software of Maple, we get the following results:Case 1.
a1=±1/2n2n+12α−aA−12/bl, a0=−2n+1/4ln+1±B/4nA−12n+12α−aA−12/bl, b1=0, k=k, and w=2α−aB2+4C−4AC/8n2−3b2n+1/8ln+12−γ−ak2.Case 2.
a1=0, a0=−2n+1/4ln+1±B/4nC2n+12α−aC2/bl, b1=±1/2n2n+12α−aC2/bl, k=k, and w=2α−aB2+4C−4AC/8n2−3b2n+1/8ln+12−γ−ak2, where A, B, C, a, b, α, and γ are free parameters.Substituting these solutions into (33), using (14) and (31), we obtain the traveling wave solution as follows:
(34)ut,x=−2n+14ln+1±B4nA−12n+12α−aA−12bl±12n2n+12α−aA−12blG′ξGξ1/2nei−kx+wtδ/δ+θ,ut,x=−2n+14ln+1±B4nC2n+12α−aC2bl±12n2n+12α−aC2blG′ξGξ−11/2nei−kx+wtδ/δ+θ,where G′ξ/Gξ is defined in Section 3, and ξ=x+2aktδ/δ.Family 1.
The hyperbolic function solutions can be obtained asB2−4A−1C>0 and A≠1. Then,
(35)u1t,x=−2n+14ln+1±B4nA−12n+12α−aA−12bl±12n2n+12α−aA−12blB21−A+B2+4C−4AC21−AH11/2nei−kx+wtδ/δ+θ,(36)u2t,x=−2n+14ln+1±B4nC2n+12α−aC2bl±12n2n+12α−aC2blB21−A+B2+4C−4AC21−AH1−11/2nei−kx+wtδ/δ+θ,where H1=C1sinhB2+4C−4AC/2x+2aktδ/δ+C2coshB2+4C−4AC/2x+2aktδ/δ/C1coshB2+4C−4AC/2x+2aktδ/δ+C2sinhB2+4C−4AC/2x+2aktδ/δ, and C1 and C2 are arbitrary constants.In particular, if we assume thatC2=0, A>1, and C1≠0 in (35), then we have
(37)u11t,x=−2n+14ln+1∓14n2n+12α−aB2+4C−4ACbltanhB2+4C−4AC2x+2aktδδ1/2nei−kx+wtδ/δ+θ,and if C1=0, A>1, and C2≠0 in (35), then we obtain
(38)u12t,x=−2n+14ln+1∓14n2n+12α−aB2+4C−4ACblcothB2+4C−4AC2x+2aktδδ1/2nei−kx+wtδ/δ+θ.The fixed values areα=2, a=1, b=1, A=2, B=4, C=3, k=1, l=3, and n=2, and the fractional parameter δ∈1/5,1/2,1 is used to plot the real part of the exact solution u11t,x as in Figures 5 and 6.Family 2.
The trigonometric function solutions can be obtained as4A−1C−B2>0 and A≠1. Then,
(39)u3t,x=−2n+14ln+1±B4nA−12n+12α−aA−12bl±12n2n+12α−aA−12blB21−A+4AC−4C−B221−AH21/2nei−kx+wtδ/δ+θ,(40)u4t,x=−2n+14ln+1±B4nC2n+12α−aC2bl±12n2n+12α−aC2blB21−A+4AC−4C−B221−AH2−11/2nei−kx+wtδ/δ+θ,where H2=−C1sin4AC−B2−4C/2x+2aktδ/δ+C2cos4AC−B2−4C/2x+2aktδ/δ/C1cos4AC−B2−4C/2x+2aktδ/δ+C2sin4AC−B2−4C/2x+2aktδ/δ, and C1 and C2 are arbitrary constants.Figure 5
The positive real part ofu11t,x in Equation (37) for differential values of the fractional parameter δ.
(a)
δ=1/5(b)
δ=1/2(c)
δ=4/5Figure 6
The positive real part ofu11t,x at x=1.0 in Equation (37) for differential values of the fractional parameter δSpecifically, if we suppose thatC2=0, A>1, and C1≠0 in (39), then we obtain
(41)u31t,x=−2n+14ln+1±14n2n+12α−a4AC−4C−B2bl·tan4AC−4C−B22x+2aktδδ1/2nei−kx+wtδ/δ+θ,and if C1=0, A>1, and C2≠0 in (39), then we obtain
(42)u32t,x=−2n+14ln+1∓14n2n+12α−a4AC−4C−B2bl·cot4AC−4C−B22x+2aktδδ1/2nei−kx+wtδ/δ+θ.The fixed values areα=2, a=1, b=1, A=4, B=3, C=3, k=1, l=3, and n=2, and the fractional parameter δ∈1/5,1/2,1 is used to plot the real part of the exact solution u31t,x as in Figures 7 and 8.Family 3.
The rational function solutions can be obtained asB2−4A−1C=0 and A≠1. Then,
(43)u5t,x=−2n+14ln+1±B4nA−12n+12α−aA−12bl±12n1−A2n+12α−aA−12blC1C1ξ+C2+B21/2nei−kx+wtδ/δ+θ,u6t,x=−2n+14ln+1±B4nC2n+12α−aC2bl±12n1−A2n+12α−aC2blC1C1ξ+C2+B2−11/2nei−kx+wtδ/δ+θ.Figure 7
The positive real part ofu31t,x in Equation (41) for differential values of the fractional parameter δ.
(a)
δ=1/5(b)
δ=1/2(c)
δ=4/5Figure 8
The positive real part ofu31t,x at x=1.0 in Equation (41) for differential values of the fractional parameter δ


## 5. Conclusion

The time fractional complex GL equation with the Kerr law and dual-power law nonlinearity, which depicts the dynamics of soliton propagation through optical fibers over longer distances, is studied by using the extendedG′/G-expansion method. A series of new exact solutions are constructed, including hyperbolic function solutions, trigonometric function solutions, and rational function solutions. Comparing our results with the solutions in the previous literature, we get many new exact solutions. In order to further understand the dynamic behaviors of these results, with the help of software Maple 18, we have provided some graphical representation of the exact solutions of the time fractional complex GL equation with the Kerr law and dual-power law nonlinearity by setting the appropriate parameters. We believe that the obtained solutions may be of cardinal significance in the field of nonlinear optics. Moreover, in the paper, we have confirmed that the extended G′/G-expansion method is powerful, reliable, and systematic, and they could be applied for solving more extensive and complicated fractional partial differential equations.



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