# The Improvedexp⁡-Φξ-Expansion Method and New Exact Solutions of Nonlinear Evolution Equations in Mathematical Physics

Authors: Guiying Chen; Xiangpeng Xin; Hanze Liu
Journal: Advances in Mathematical Physics
 (2019)

---

## Abstract

Theexp(-Φ(ξ))-expansion method is improved by presenting a new auxiliary ordinary differential equation for Φ(ξ). By using this method, new exact traveling wave solutions of two important nonlinear evolution equations, i.e., the ill-posed Boussinesq equation and the unstable nonlinear Schrödinger equation, are constructed. The obtained solutions contain Jacobi elliptic function solutions which can be degenerated to the hyperbolic function solutions and the trigonometric function solutions. The present method is very concise and effective and can be applied to other types of nonlinear evolution equations.


---

## Body

## 1. Introduction

Nonlinear evolution equations have widely applied in various areas of science and engineering, such as nonlinear optics, fluid dynamics, biophysics, and plasma physics. Since the exact solutions of nonlinear evolution equations play a determining role in comprehension the nonlinear phenomena, many effective methods have been proposed, such as the direct algebraic method [1], the extended direct algebraic method [2–5], the function transformation method [1, 6–8], the ansatz function method [9], the extended simple equation method [10, 11], the variational method [12–14], and the extended auxiliary equation method [15].Recently, Khan and Akbar introduced theexp(-Φ(ξ))-expansion method to obtain new traveling wave solutions for the Modified Benjamin-Bona-Mahony Equation in [16]. Afterwards, many nonlinear partial differential equations were solved by means of this method, such as the Vakhnenko-Parkes equation [17], the Zakharov Kuznetsov-Benjamin-Bona-Mahony (ZK-BBM) equation and the ill-posed Boussinesq equation [18], the space time-fractional (3+1)-dimensional nonlinear Jimbo-Miwa equation and nonlinear Hirota-Satsuma coupled KdV system [19], the time-fractional modified BBM equation and the time-fractional Cohn-Hilliard equation [20], the Pochhammer- Chree equation [21], and the Gerdjikov-Ivanov equation [22].The traveling wave solutions of the nonlinear evolution equations solved by theexp(-Φ(ξ))-expansion method have the form(1)uξ=∑i=0nliexp⁡-Φξi,where li(i=1,…,n) are undetermined constants, ln≠0, and Φ(ξ) satisfies the auxiliary equation:(2)Φ′ξ=exp⁡-Φξ+λ+μ·exp⁡Φξ.In this paper, we improve the exp(-Φ(ξ))-expansion method by presenting a new auxiliary equation:(3)Φ′ξ2=k2r·exp⁡-2Φξ+a+b·exp2Φξ.From (2), we get(4)Φ′ξ2=exp⁡-2Φξ+2λexp⁡-Φξ+λ2+2μ+2λμexp⁡Φξ+μ2exp⁡2Φξ.Obviously, if λ=0, k=1, r=1, a=2μ, and b=μ2, then (4) becomes (3). That is, in special case of λ=0, the improved exp(-Φ(ξ))-expansion method is a generalization of the exp(-Φ(ξ))-expansion method.In [16], the hyperbolic function solutions, the trigonometric solutions, and the rational function solutions of (2) were given, while, in this paper, the Jacobi elliptic function solutions of (3) are shown in Section 2. We know that when m→1 (m→0) (where m is the modulus of the Jacobi elliptic function), the Jacobi elliptic functions degenerate to hyperbolic functions (trigonometric functions), so the improved exp(-Φ(ξ))-expansion method extend the tanh function method [23], the solitary wave ansatz method [24], the sine-cosine method [25], etc.In addition, besides the Jacobi elliptic function solutions, (3) also has many combined Jacobi elliptic function solutions which we omit for simplicity in Section 2. In fact, substituting these solutions into (2), we get the power series of the reciprocal of Jacobi elliptic functions or combined Jacobi elliptic functions. Hence, the improved exp(-Φ(ξ))-expansion method is more general than the Jacobi elliptic function method [26], the F-function method [27], and the indirect F-function method [28].We implement the improvedexp(-Φ(ξ))-expansion method to construct new traveling wave solutions for two nonlinear partial differential equations. The first one is the ill-posed Boussinesq equation:(5)utt=uxx+u2xx+uxxxx.This equation was first introduced by the French Scientist J. Boussinesq, which can describe the propagation of long wave in shallow water with a small amplitude in one-dimension nonlinear lattices and in nonlinear strings [29–31]. Equation (5) differs from the well-posed Boussinesq equation only in the sign of the last dispersive term. Recently, the numerical and analytical solutions of (5) were extensively reported in the literature. For instance, in [31], the approximate solutions were obtained by applying filtering and regularization techniques. In [32], the exact homoclinic orbits solutions with periodic boundary condition and even constraint were studied. In [33], Jafari et al. gave the solitary wave solutions by using the sine-cosine and extended tanh function method. In [34], Seadawy A derived the dark and bright solitary wave solutions by means of the direct algebraic method. In [35, 36], the authors used the theory of Lie group to obtain the symmetry reductions and the group invariant solutions. In [24, 37], the authors obtained the new traveling wave solutions and the nonlinear self-adjoint substitution to construct new conservation laws.The second nonlinear partial differential equation is the unstable nonlinear Schrödinger equation:(6)iut+uxx+2λu2u-2γu=0.This equation is a type of nonlinear Schrödinger equation with space and time exchanged. Its behavior occurs in case of lossless symmetric two-stream plasma instability [38] and the two-layer baroclinic instability [39]. Many traveling wave solutions have been derived by different methods, such as the semi-inverse variational principle [40], the sine-cosine method [25], the new Jacobi elliptic function rational expansion method and exponential rational function method [26], the (G′/G)-expansion method [41], the tan⁡(Φ(ξ)/2)-expansion method [42], and the extended simple equation method [43]. To the best of our knowledge, no previous work has been done by the improved exp(-Φ(ξ))-expansion method.The organization of this paper is as follows: firstly, we give the methodology of the improvedexp(-Φ(ξ)) expansion method. Then we apply this method to the ill-posed Boussinesq equation and the unstable nonlinear Schrödinger equation for finding new exact traveling wave solutions. Also we discuss the obtained solutions with the already existing results in the previous literature. At last, a conclusion is given.


## 2. Description of the Improvedexp⁡⁡-Φξ-Expansion Method

In this section, we describe the improvedexp⁡(-Φ(ξ))-expansion method. Consider a nonlinear partial differential equation with independent variables x and t(7)Pu,ut,ux,utt,uxt,uxx,…=0,where u=u(x,t) is an unknown function and P is a polynomial in u=u(x,t) and its partial derivatives.Firstly, applying the traveling wave transformationu(x,t)=u(ξ),ξ=x-vt to (7), we reduce (7) to an ordinary differential equation (ODE):(8)Qu,u′,u′′,…=0,where prime denotes the derivative with respect to ξ. If possible, integrate (8) term by term one or more times. For simplicity, we set the integration constant(s) to zero.Secondly, suppose that the solution of (8) can be expressed as(9)uξ=∑i=0nliexp⁡-Φξi,where Φ(ξ) satisfies the following auxiliary ODE:(10)Φ′ξ2=k2r·exp⁡-2Φξ+a+b·exp⁡2Φξ.li (i=1,2,…,n), v,k,r,a and b are constants to be determined later, ln≠0, and n is a positive integer which can be determined by homogenous balance in (8).We know that the auxiliary equation (10) has many different Jacobi elliptic function solutions. For simplicity, we only list several typical solutions:(1)
Ifr=m2,a=-1-m2,b=1, then Φ1=ln⁡ns(kξ,m).(2)
Ifr=-m2,a=-1+2m2,b=1-m2, then Φ2=ln⁡nc(kξ,m).(3)
Ifr=1,a=-(m2+1),b=m2, then Φ3=ln⁡sn(kξ,m).(4)
Ifr=1-m2,a=2m2-1,b=-m2, then Φ4=ln⁡cn(kξ,m).(5)
Ifr=1-m2,a=2-m2,b=1, then Φ5=ln⁡cs(kξ,m).(6)
Ifr=1,a=2-m2,b=1-m2, then Φ6=ln⁡sc(kξ,m),wherem is the modulus of the Jacobi elliptic function.Thirdly, substituting (9) along with (10) into (8) and equating coefficients of the same power in (exp(-Φ(ξ))) to zero yields a set of algebraic equations for li(i=1,2,…,n),k and v. Solving these equations, we can construct a variety of exact solutions for (7).


## 3. Application of the Improvedexp⁡⁡-Φξ-Expansion Method

In this section, two examples are presented to illustrate how the method works.


### 3.1. The Ill-Posed Boussinesq Equation

Using the traveling wave transformation we can reduce (5) to the following ODE:(11)v2-1u′′-2u′2-2uu′′-u′′′′=0,where prime denotes the derivation with respect to ξ. Integrating (11) twice with respect to ξ, and taking the integration constants as zero, we get(12)v2-1u-u2-u′′=0.By homogeneous balance, we get n=2. So the solution of (12) can be expressed as(13)uξ=l0+l1exp⁡-Φξ+l2exp⁡-Φξ2.Substituting (13) along with (10) into (12) and setting each coefficient of exp(-iΦ(ξ)), (i=0,1,2,3,4) to zero, we obtain a set of algebraic equations as follows:(14)exp⁡-4Φξ:-6l2k2r-l22=0,exp⁡-3Φξ:-2l1l2-2l1k2r=0,exp⁡-2Φξ:v2l2-l2-4l2k2a-2l0l2-l12=0,exp⁡-Φξ:v2l1-l1-l1k2a-2l0l1=0,exp0:v2l0-l02-l0-2l2k2b=0.Solving the above system, we get(15)k=k,v2=1±4k2a2-3rb,l0=2k2-a±a2-3rb,l1=0,l2=-6k2r.Consequently, we have the following different cases for the exact traveling solutions of the ill-posed Boussinesq equation:(1)
Ifr=m2,a=-1-m2,b=1, then(16)u1x,t=2k21+m2±1-m2+m4-6k2m2sn2kx∓1±4k21-m2+m4t,m;(2)
Ifr=-m2,a=-1+2m2,b=1-m2, then(17)u2x,t=2k21-2m2±1-m2+m4+6k2m2cn2kx∓1±4k21-m2+m4t,m;(3)
Ifr=1,a=-(m2+1),b=m2, then(18)u3x,t=2k2m2+1±1-m2+m4-6k2ns2kx∓1±4k21-m2+m4t,m;(4)
Ifr=1-m2,a=2m2-1,b=-m2, then(19)u4x,t=2k21-2m2±1-m2+m4+6k2m2-1nc2kx∓1±4k21-m2+m4t,m;(5)
Ifr=1-m2,a=2-m2,b=1, then(20)u5x,t=2k2m2-2±1-m2+m4-6k21-m2sc2kx∓1±4k21-m2+m4t,m;(6)
Ifr=1,a=2-m2,b=1-m2, then(21)u6x,t=2k2m2-2±1-m2+m4-6k2cs2kx∓1±4k21-m2+m4t,m;Remark 1.
When the modulusm→1, we have sn(ξ,m)→tanh⁡(ξ), cn(ξ,m)→sech⁡(ξ), ns→coth⁡(ξ), cs→csch⁡(ξ). Taking k=λ/2 and m→1 in u1(x,t) and u4(x,t), where λ is an arbitrary constant, we get the following hyperbolic function solutions, respectively:(22)u11x,t=32λ2-32λ2tanh2λ2x±1+λ2t;u12x,t=12λ2-32λ2tanh2λ2x±1-λ2t.(23)u31x,t=32λ2-32λ2coth2λ2x±1+λ2t;u32x,t=12λ2-32λ2coth2λ2x±1-λ2t.
Takingk=B and m→1 in u2(x,t) and u8(x,t), where B is an arbitrary constant, we get(24)u21x,t=6B2sech2Bx∓1+4k2t;u22x,t=-4B2+6B2sech2kx∓1-4k2t.(25)u61x,t=-6B2csch2kx∓1+4k2t;u62x,t=-4B2-6B2csch2kx∓1-4k2t.The above solutions coincide with solutions of Tchier et al. [24]: (i)
our solutionsu11(x,t) and u12(x,t) match with solutions (36) and (34) in [24], respectively;(ii)
obtained solutionu21(x,t) is identical with solution (8) in [24];(iii)
the solutionsu31(x,t) and u32(x,t) turn out solutions (37) and (35) in [24], respectively;(iv)
obtained solutionu61(x,t) comes into solution (15) in [24].When the modulusm→0, we get ns(ξ,m)→csc(ξ), nc(ξ,m)→sec(ξ), sc(ξ,m)→tan(ξ), and cs(ξ,m)→cot(ξ); from u3(x,t), u4(x,t), u5(x,t), and u6(x,t) we obtain the following trigonometric function solutions, respectively:(26)u33x,t=4k2-6k2csc2kx∓1+4k2t.(27)u41x,t=4k2-6k2sec2kx∓1+4k2t.(28)u51x,t=-2k2-6k2tan2kx∓1+4k2t;u52x,t=-6k2-6k2tan2kx∓1-4k2t.(29)u63x,t=-2k2-6k2cot2kx∓1+4k2t;u64x,t=-6k2-6k2cot2kx∓1-4k2t.To the best of our knowledge, the solutionsu33(x,t),u41(x,t),u51(x,t),u52(x,t),u63(x,t), and u64(x,t) have not been reported in the previous literature.


### 3.2. The Unstable Nonlinear Schrödinger Equation

Using the following traveling wave transformations(30)ux,t=Uξeiθ,ξ=x-vtandθ=βx+r1tto (6), we obtain the ODE(31)U′′-β2+r1+2γU+2λU3=0,where v=2β and β is a constant which will be determined later, while γ,r1 and λ are arbitrary constants. We get the balancing number as n=1. Thus the solution can be written as(32)Uξ=l0+l1exp-Φξ.Substituting (32) along with (10) into (31) and collecting the coefficients of each power exp(-iΦ(ξ))(i=0,1,2,3) and setting each of the coefficients to zero, we get a set of algebraic equations of l0,l1,β, and k as follows:(33)exp⁡-Φξ3:2l1k2r+2λl13=0,exp⁡-Φξ2:6λl0l12=0,exp-Φξ:l1ak2-β2l1-r1l1-2γl1+6λl02l1=0,exp0:r1l0-β2l0+2λl13-2γl0=0.Solving the above system with the aid of Maple, we get(34)k=±β2+r1+2γa,β=β,l0=0,l1=±-rβ2+r1+2γλa.Consequently, we have the following different cases for the exact solutions of Schrödinger equation:(1)
Ifr=m2,a=-1-m2,b=1, then(35)u1x,t=±mβ2+r1+2γλ1+m2sn±β2+r1+2γ-1-m2x-2βt,meiβx+r1t;(2)
Ifr=-m2,a=-1+2m2,b=1-m2, then(36)u2x,t=±mβ2+r1+2γλ-1+2m2cn±β2+r1+2γ-1+2m2x-2βt,meiβx+r1t;(3)
Ifr=1,a=-(m2+1),b=m2, then(37)u3x,t=±β2+r1+2γλ1+m2ns±β2+r1+2γ-1-m2x-2βt,meiβx+r1t;(4)
Ifr=1-m2,a=2m2-1,b=-m2, then(38)u4x,t=±m2-1β2+r1+2γλ2m2-1nc±β2+r1+2γ2m2-1x-2βt,meiβx+r1t;(5)
Ifr=1-m2,a=2-m2,b=1, then(39)u5x,t=±m2-1β2+r1+2γλ2-m2sc±β2+r1+2γ2-m2x-2βt,meiβx+r1t;(6)
Ifr=1,a=2-m2,b=1-m2, then(40)u6x,t=±β2+r1+2γλm2-2cs±β2+r1+2γ2-m2x-2βt,meiβx+r1t;Remark 2.
When the modulusm→1, our solutions u1(x,t),u2(x,t),u3(x,t) and u6(x,t) turn into the following hyperbolic function solutions, respectively:(41)u11x,t=±β2+r1+2γ2λtanh⁡β2+r1+2γ-2x-2βteiβx+r1t;(42)u21x,t=±β2+r1+2γλsech⁡β2+r1+2γx-2βteiβx+r1t;(43)u31x,t=±β2+r1+2γ2λcoth⁡β2+r1+2γ-2x-2βteiβx+r1t;(44)u61x,t=±β2+r1+2γ-λcsch⁡β2+r1+2γx-2βteiβx+r1t;.The above solutions coincide with solutions in [25, 40, 41]: (i)
our solutionu11(x,t) matches with solutions u31(x,t) and u35(x,t) in [41];(ii)
obtained solutionu21(x,t) identical with solutions (45) and (46) in [40];(iii)
our solutionu31(x,t) matches with solutions u32(x,t) and u36(x,t) in [41];(iv)
the solutionu61(x,t) turns out solution (52) in [25].When the modulusm→0, our solutions u3(x,t),u4(x,t),u5(x,t) and u6(x,t) come into the following trigonometric function solutions, respectively:(45)u32x,t=±β2+r1+2γλcsc⁡-β2+r1+2γx-2βteiβx+r1t;(46)u41x,t=±β2+r1+2γλsec⁡-β2+r1+2γx-2βteiβx+r1t;(47)u51x,t=±β2+r1+2γ-2λtan⁡β2+r1+2γ2x-2βteiβx+r1t;(48)u62x,t=±β2+r1+2γ-2λcot⁡β2+r1+2γ2x-2βteiβx+r1t;.The above solutions coincide with solutions in [25, 41]: (i)
the solutionsu32(x,t) and u41(x,t) match with solutions (50) and (51), respectively, in [25];(ii)
the solutionsu51(x,t) comes into solutions u33(x,t) and u37(x,t) in [41];(iii)
the solutionu62(x,t) is identical with solutions u34(x,t) and u38(x,t) in [41].Remark 3.
If we takek=1 in the auxiliary equation (10), then the solution of (33) becomes(49)k=1,β=±a-r1-2γ,l0=0,l1=±-rλ.Hence we get the following solutions of (6).(1)
Ifr=m2,a=-1-m2,b=1, then(50)u7x,t=±m-1λsnx∓2-1-m2-r1-2γt,mei±-1-m2-r1-2γx+r1t;(2)
Ifr=-m2,a=-1+2m2,b=1-m2, then(51)u8x,t=±m1λcnx∓2-1+2m2-r1-2γt,mei±-1+2m2-r1-2γx+r1t;(3)
Ifr=-m2,a=2-m2,b=m2-1, then(52)u9x,t=±m1λdnx∓22-m2-r1-2γt,mei±2-m2-r1-2γx+r1t.Our solutionsu7(x,t),u8(x,t) and u9(x,t) match with (10), (15), and (22) in [26], respectively.


## 3.1. The Ill-Posed Boussinesq Equation

Using the traveling wave transformation we can reduce (5) to the following ODE:(11)v2-1u′′-2u′2-2uu′′-u′′′′=0,where prime denotes the derivation with respect to ξ. Integrating (11) twice with respect to ξ, and taking the integration constants as zero, we get(12)v2-1u-u2-u′′=0.By homogeneous balance, we get n=2. So the solution of (12) can be expressed as(13)uξ=l0+l1exp⁡-Φξ+l2exp⁡-Φξ2.Substituting (13) along with (10) into (12) and setting each coefficient of exp(-iΦ(ξ)), (i=0,1,2,3,4) to zero, we obtain a set of algebraic equations as follows:(14)exp⁡-4Φξ:-6l2k2r-l22=0,exp⁡-3Φξ:-2l1l2-2l1k2r=0,exp⁡-2Φξ:v2l2-l2-4l2k2a-2l0l2-l12=0,exp⁡-Φξ:v2l1-l1-l1k2a-2l0l1=0,exp0:v2l0-l02-l0-2l2k2b=0.Solving the above system, we get(15)k=k,v2=1±4k2a2-3rb,l0=2k2-a±a2-3rb,l1=0,l2=-6k2r.Consequently, we have the following different cases for the exact traveling solutions of the ill-posed Boussinesq equation:(1)
Ifr=m2,a=-1-m2,b=1, then(16)u1x,t=2k21+m2±1-m2+m4-6k2m2sn2kx∓1±4k21-m2+m4t,m;(2)
Ifr=-m2,a=-1+2m2,b=1-m2, then(17)u2x,t=2k21-2m2±1-m2+m4+6k2m2cn2kx∓1±4k21-m2+m4t,m;(3)
Ifr=1,a=-(m2+1),b=m2, then(18)u3x,t=2k2m2+1±1-m2+m4-6k2ns2kx∓1±4k21-m2+m4t,m;(4)
Ifr=1-m2,a=2m2-1,b=-m2, then(19)u4x,t=2k21-2m2±1-m2+m4+6k2m2-1nc2kx∓1±4k21-m2+m4t,m;(5)
Ifr=1-m2,a=2-m2,b=1, then(20)u5x,t=2k2m2-2±1-m2+m4-6k21-m2sc2kx∓1±4k21-m2+m4t,m;(6)
Ifr=1,a=2-m2,b=1-m2, then(21)u6x,t=2k2m2-2±1-m2+m4-6k2cs2kx∓1±4k21-m2+m4t,m;Remark 1.
When the modulusm→1, we have sn(ξ,m)→tanh⁡(ξ), cn(ξ,m)→sech⁡(ξ), ns→coth⁡(ξ), cs→csch⁡(ξ). Taking k=λ/2 and m→1 in u1(x,t) and u4(x,t), where λ is an arbitrary constant, we get the following hyperbolic function solutions, respectively:(22)u11x,t=32λ2-32λ2tanh2λ2x±1+λ2t;u12x,t=12λ2-32λ2tanh2λ2x±1-λ2t.(23)u31x,t=32λ2-32λ2coth2λ2x±1+λ2t;u32x,t=12λ2-32λ2coth2λ2x±1-λ2t.
Takingk=B and m→1 in u2(x,t) and u8(x,t), where B is an arbitrary constant, we get(24)u21x,t=6B2sech2Bx∓1+4k2t;u22x,t=-4B2+6B2sech2kx∓1-4k2t.(25)u61x,t=-6B2csch2kx∓1+4k2t;u62x,t=-4B2-6B2csch2kx∓1-4k2t.The above solutions coincide with solutions of Tchier et al. [24]: (i)
our solutionsu11(x,t) and u12(x,t) match with solutions (36) and (34) in [24], respectively;(ii)
obtained solutionu21(x,t) is identical with solution (8) in [24];(iii)
the solutionsu31(x,t) and u32(x,t) turn out solutions (37) and (35) in [24], respectively;(iv)
obtained solutionu61(x,t) comes into solution (15) in [24].When the modulusm→0, we get ns(ξ,m)→csc(ξ), nc(ξ,m)→sec(ξ), sc(ξ,m)→tan(ξ), and cs(ξ,m)→cot(ξ); from u3(x,t), u4(x,t), u5(x,t), and u6(x,t) we obtain the following trigonometric function solutions, respectively:(26)u33x,t=4k2-6k2csc2kx∓1+4k2t.(27)u41x,t=4k2-6k2sec2kx∓1+4k2t.(28)u51x,t=-2k2-6k2tan2kx∓1+4k2t;u52x,t=-6k2-6k2tan2kx∓1-4k2t.(29)u63x,t=-2k2-6k2cot2kx∓1+4k2t;u64x,t=-6k2-6k2cot2kx∓1-4k2t.To the best of our knowledge, the solutionsu33(x,t),u41(x,t),u51(x,t),u52(x,t),u63(x,t), and u64(x,t) have not been reported in the previous literature.


## 3.2. The Unstable Nonlinear Schrödinger Equation

Using the following traveling wave transformations(30)ux,t=Uξeiθ,ξ=x-vtandθ=βx+r1tto (6), we obtain the ODE(31)U′′-β2+r1+2γU+2λU3=0,where v=2β and β is a constant which will be determined later, while γ,r1 and λ are arbitrary constants. We get the balancing number as n=1. Thus the solution can be written as(32)Uξ=l0+l1exp-Φξ.Substituting (32) along with (10) into (31) and collecting the coefficients of each power exp(-iΦ(ξ))(i=0,1,2,3) and setting each of the coefficients to zero, we get a set of algebraic equations of l0,l1,β, and k as follows:(33)exp⁡-Φξ3:2l1k2r+2λl13=0,exp⁡-Φξ2:6λl0l12=0,exp-Φξ:l1ak2-β2l1-r1l1-2γl1+6λl02l1=0,exp0:r1l0-β2l0+2λl13-2γl0=0.Solving the above system with the aid of Maple, we get(34)k=±β2+r1+2γa,β=β,l0=0,l1=±-rβ2+r1+2γλa.Consequently, we have the following different cases for the exact solutions of Schrödinger equation:(1)
Ifr=m2,a=-1-m2,b=1, then(35)u1x,t=±mβ2+r1+2γλ1+m2sn±β2+r1+2γ-1-m2x-2βt,meiβx+r1t;(2)
Ifr=-m2,a=-1+2m2,b=1-m2, then(36)u2x,t=±mβ2+r1+2γλ-1+2m2cn±β2+r1+2γ-1+2m2x-2βt,meiβx+r1t;(3)
Ifr=1,a=-(m2+1),b=m2, then(37)u3x,t=±β2+r1+2γλ1+m2ns±β2+r1+2γ-1-m2x-2βt,meiβx+r1t;(4)
Ifr=1-m2,a=2m2-1,b=-m2, then(38)u4x,t=±m2-1β2+r1+2γλ2m2-1nc±β2+r1+2γ2m2-1x-2βt,meiβx+r1t;(5)
Ifr=1-m2,a=2-m2,b=1, then(39)u5x,t=±m2-1β2+r1+2γλ2-m2sc±β2+r1+2γ2-m2x-2βt,meiβx+r1t;(6)
Ifr=1,a=2-m2,b=1-m2, then(40)u6x,t=±β2+r1+2γλm2-2cs±β2+r1+2γ2-m2x-2βt,meiβx+r1t;Remark 2.
When the modulusm→1, our solutions u1(x,t),u2(x,t),u3(x,t) and u6(x,t) turn into the following hyperbolic function solutions, respectively:(41)u11x,t=±β2+r1+2γ2λtanh⁡β2+r1+2γ-2x-2βteiβx+r1t;(42)u21x,t=±β2+r1+2γλsech⁡β2+r1+2γx-2βteiβx+r1t;(43)u31x,t=±β2+r1+2γ2λcoth⁡β2+r1+2γ-2x-2βteiβx+r1t;(44)u61x,t=±β2+r1+2γ-λcsch⁡β2+r1+2γx-2βteiβx+r1t;.The above solutions coincide with solutions in [25, 40, 41]: (i)
our solutionu11(x,t) matches with solutions u31(x,t) and u35(x,t) in [41];(ii)
obtained solutionu21(x,t) identical with solutions (45) and (46) in [40];(iii)
our solutionu31(x,t) matches with solutions u32(x,t) and u36(x,t) in [41];(iv)
the solutionu61(x,t) turns out solution (52) in [25].When the modulusm→0, our solutions u3(x,t),u4(x,t),u5(x,t) and u6(x,t) come into the following trigonometric function solutions, respectively:(45)u32x,t=±β2+r1+2γλcsc⁡-β2+r1+2γx-2βteiβx+r1t;(46)u41x,t=±β2+r1+2γλsec⁡-β2+r1+2γx-2βteiβx+r1t;(47)u51x,t=±β2+r1+2γ-2λtan⁡β2+r1+2γ2x-2βteiβx+r1t;(48)u62x,t=±β2+r1+2γ-2λcot⁡β2+r1+2γ2x-2βteiβx+r1t;.The above solutions coincide with solutions in [25, 41]: (i)
the solutionsu32(x,t) and u41(x,t) match with solutions (50) and (51), respectively, in [25];(ii)
the solutionsu51(x,t) comes into solutions u33(x,t) and u37(x,t) in [41];(iii)
the solutionu62(x,t) is identical with solutions u34(x,t) and u38(x,t) in [41].Remark 3.
If we takek=1 in the auxiliary equation (10), then the solution of (33) becomes(49)k=1,β=±a-r1-2γ,l0=0,l1=±-rλ.Hence we get the following solutions of (6).(1)
Ifr=m2,a=-1-m2,b=1, then(50)u7x,t=±m-1λsnx∓2-1-m2-r1-2γt,mei±-1-m2-r1-2γx+r1t;(2)
Ifr=-m2,a=-1+2m2,b=1-m2, then(51)u8x,t=±m1λcnx∓2-1+2m2-r1-2γt,mei±-1+2m2-r1-2γx+r1t;(3)
Ifr=-m2,a=2-m2,b=m2-1, then(52)u9x,t=±m1λdnx∓22-m2-r1-2γt,mei±2-m2-r1-2γx+r1t.Our solutionsu7(x,t),u8(x,t) and u9(x,t) match with (10), (15), and (22) in [26], respectively.


## 4. Conclusion

Theexp(-Φ(ξ))-expansion method has been successfully generalized to establish new exact solutions for the ill-posed Boussinesq equation and the unstable nonlinear Schrödinger equation. In special cases, our new Jacobi elliptic function solutions can be degenerated to hyperbolic function solutions and trigonometric function solutions which have been reported in the previous literature. That shows the improved exp(-Φ(ξ))-expansion method is effective and powerful. This method can be used to solve many other nonlinear evolution equations in mathematical physics, engineering, and other nonlinear sciences.



---

*Source: 4354310-2019-04-04.xml*