In this section, we have investigated three NPDEs using the first integral method for the first time.
3.1. The (2+1)-Dimensional Hyperbolic Nonlinear Schrodinger Equation
Let us consider the (2+1)-dimensional hyperbolic nonlinear Schrodinger (HNLS) equation [25] which reads
(7)iut+12uxx-12uyy+|u|2u=0,
where x is dimensionless variable, y is the propagation coordinate, i=-1, and t is the time. The above equation can be derived from optics [26] and large-scale Rossby waves [27]. Various types of NLS or HNLS equations describing time and space evolutions of slowly varying envelopes have wide applications in various branches of physics [28, 29].
By considering the transformations u(x,y,t)=φ(ξ)exp(iη), and the wave variable ξ=x+ay-ct+ς, η=mx+ny+wt+ε, where, ς, and ε are arbitrary constants, (7) changes into a system of ordinary differential equations as follows:
(8)(c2-1)φ′′(ξ)=2φ3(ξ)+(w2-2n-(a+cw)2)φ(ξ),
where prime denotes the derivative with respect to the same variable ξ.
Using (4) and (5a) and (5b), we can get(9a)X′(ξ)=Y(ξ),(9b)Y′(ξ)=(2c2-1)X3(ξ)+(w2-2n-(a+cw)2c2-1)X(ξ).According to the first integral method, we suppose that X(ξ) and Y(ξ) are nontrivial solutions of (9a) and (9b) and P(X,Y)=∑i=0mai(X)Yi is an irreducible polynomial in the complex domain C[X,Y] such that
(10)P[X(ξ),Y(ξ)]=∑i=0mai(X(ξ))Yi(ξ)=0,
where ai(X), (i=0,1,2,…,m) are polynomials of X and am(X)≠0.
Equation (10) is called the first integral to (9a) and (9b). Due to the division theorem, there exists a polynomial h(X)+g(X)Y in the complex domain C[X,Y] such that
(11)dPdξ=∂P∂XdXdξ+∂P∂YdYdξ=[h(X)+g(X)Y]∑i=0mai(X)Yi.
Here, we have considered two different cases, assuming that m=1 and m=2 in (10).
Case 1.
Suppose that m=1, by equating the coefficients of Yi (i=2,1,0) on both sides of (11), we have(12a)a1′(X)=g(X)a1(X),(12b)a0′(X)=h(X)a1(X)+g(X)a0(X),(12c)a1(X)((2c2-1)X3+(w2-2n-(a+cw)2c2-1)X) =h(X)a0(X).Since ai(X)(i=0,1) are polynomials, then from (12a) we have deduced that a1(X) is constant and g(X)=0. For simplicity, we have a1(X)=1.
Balancing the degrees of h(X) and a0(X), we have concluded that deg(h(X))=1 only. Suppose that h(X)=AX+B and A≠0, we find a0(X)(13)a0(X)=A2X2+BX+D,
where D is an arbitrary integration constant.
Substituting a0(X), a1(X), and h(X) for (12c) and setting all the coefficients of powers X to be zero, we have obtained a system of nonlinear algebraic equations and by solving it, we have obtained
(14)D=∓121-1+c2[2n+(a+(-1+c)w)(a+w+cw)],B=0, A=±2-11-c-11+c.
Using the conditions (14) in (10), we obtain
(15)Y(ξ)=∓22-11-c-11+cX2(ξ)±121-1+c2[2n+(a+(-1+c)w) ×(a+w+cw)],
respectively.
Combining (15) with (9a), we have obtained the exact solutions to (9a) ad (9b). The exact traveling wave solutions to the (2+1)-dimensional hyperbolic nonlinear Schrodinger equation (7) can be written as
(16)u1,2(x,y,t)=-12-2n-(a+(-1+c)w)(a+w+cw)×tan[{(2(-1+c)-1+c)-1}-2n-(a+(-1+c)w)(a+w+cw) ×(±-1+c(x+ay-ct+ς) +2(-1+c)-1+cξ0) ×(2(-1+c)-1+c)-1]×exp[i(mx+ny+wt+ε)],
respectively, where ξ0 is an arbitrary integration constant.
Case 2.
Assume that m=2, by equating the coefficients of Yi( i=3,2,1,0 ) on both sides of (11), we have(17a)a2′(X)=g(X)a2(X),(17b)a1′(X)=h(X)a2(X)+g(X)a1(X),(17c)a0′(X)+2a2(X)[(2c2-1)X3+(w2-2n-(a+cw)2c2-1)X] =h(X)a1(X)+g(X)a0(X),(17d)a1(X)[(2c2-1)X3+(w2-2n-(a+cw)2c2-1)X] =h(X)a0(X).Since, ai(X) (i=0,1,2) are polynomials, then from (17a) it can be deduced that a2(X) is a constant and g(X)=0. For simplicity, let us suppose that a2(X)=1. Balancing the degrees of h(X) and a0(X) it can be concluded that deg(h(X))=1 only.
In this case, let us assume that h(X)=AX+B and A≠0, then we find a1(X) and a0(X) as follows:(18a)a1(X)=(A2)X2+BX+D,(18b)a0(X)=(A28-1c2-1)X4+12(AB)X3+(AD+B22-w2-2n-(a+cw)2c2-1)X2+BDX+F,where A,B,D, and F are arbitrary integration constants.
Substituting a0(X), a1(X), a2(X), and h(X) for (17d) and setting all the coefficients of powers X to be zero, then we have obtained a system of nonlinear algebraic equations and by solving it, we get(19a)F=0, D=0, B=0,w=-ac-a2+2n-2c2n-1+c2,A=∓22-11-c-11+c,(19b)F=0, D=0, B=0,w=-ac+a2+2n-2c2n-1+c2,A=∓22-11-c-11+c.Using the conditions (19a) and (19b) in (10), we have obtained
(20)Y(ξ)=±1-1+c2X2(ξ).
Combining (20) with (9a) we have obtained the exact solutions to (9a) and (9b) and hence the exact traveling wave solutions to the (2+1)-dimensional hyperbolic nonlinear Schrodinger equation (7) can be written as
(21)u3,4(x,y,t) =-1+c1+c∓(x+ay-ct+ς)--1+c1+c ×exp[i(mx+ny+-ac-a2+2n-2c2n-1+c2t+ε)],(22)u5,6(x,y,t) =-1+c1+c∓(x+ay-ct+ς)--1+c1+c ×exp[i(mx+ny+-ac+a2+2n-2c2n-1+c2t+ε)].
Comparing these results with the results obtained in [1], it can be seen that the solutions here are new.
3.2. The Generalized Nonlinear Schrodinger (GNLS) Equation with a Source
Let us Consider the generalized nonlinear Schrodinger (GNLS) equation with a source [30, 31], in the form
(23)iut+auxx+bu|u|2+icuxxx+id(u|u|2)x =kei[χ(ξ)-wt],
where ξ=α(x-vt) is a real function and a, b, c, d, k, α, v, and w are all real.
The GNLS equation (23) plays an important role in many nonlinear sciences. It arises as an asymptotic limit for a slowly varying dispersive wave envelope in a nonlinear medium. For example, its significant application in optical soliton communication plasma physics has been proved.
Furthermore, the GNLS equation enjoys many remarkable properties (e.g., bright and dark soliton solutions, Lax pair, Liouvile integrability, inverse scattering transformation, conservation laws, Backlund transformation, etc.).
We have considered a plane wave transformation in the form
(24)u(x,t)=ψ(ξ)ei[χ(ξ)-wt],
where ψ(ξ) is a real function. For convenience, let χ=βξ+x0 where β and x0 are real constants and ξ=α(x-vt)+ς. Then by replacing (23) and its appropriate derivatives in (22) and separating the real and imaginary parts of the result, we have obtained the following two ordinary differential equations:
(25)cα3ψ′′′+(-αv+2aβα2-3cα3β2)ψ′ +3dαψ2ψ′=0,(26)(aα2-3cψ3β)ψ′′+(αβv+w-aα2β2+cα3β3)ψ +(b-dαβ)ψ3-k=0.
Integrating (25) once, with respect to ξ, we have
(27)cα2ψ′′(ξ)+(-v+2aαβ-3cα2β2)ψ(ξ)-M=0,
where M is an arbitrary integration constant. Since the same function ψ(ξ) satisfies (26) and (27), we have obtained the following constraint condition:
(28)aα2-3cψ3βcα2=αβv+w-aα2β2+cα3β3-v+2aαβ-3cα2β2=b-dαβd=kM.
Using (4) and (5a) and (5b), we can get(29a)X′(ξ)=Y(ξ),(29b)Y′(ξ)=(-dcα2)X3(ξ)+(vcα2-2aβcα+3β2)X(ξ)+Mcα2.According to the first integral method, we suppose that X(ξ) and Y(ξ) are nontrivial solutions of (29a) and (29b) and P(X,Y)=∑i=0mai(X)Yi is an irreducible polynomial in the complex domain C[X,Y] such that
(30)P[X(ξ),Y(ξ)]=∑i=0mai(X(ξ))Y(ξ)i=0,
where ai(X), (i=0,1,2,…,m) are polynomials of X and am(X)≠0.
Equation (30) is called the first integral to (29a) and (29b). Due to the division theorem, there exists a polynomial h(X)+g(X)Y in the complex domain C[X,Y] such that
(31)dPdξ=∂P∂XdXdξ+∂P∂YdPdξ,=[h(X)+g(X)Y]∑i=0mai(X)Yi.
In this example, we have taken two different cases, assuming that m=1 and m=2 in (30).
Case 3.
Suppose that m=1, by equating the coefficients of Yi (i=2,1,0) on both sides of (31), we have(32a)a1′(X)=g(X)a1(X),(32b)a0′(X)=h(X)a1(X)+g(X)a0(X),(32c)a1(X)[[-dcα2]X3+[vcα2-2aβcα+3β2]X+Mcα2] =h(X)a0(X).Since ai(X) (i=0,1) are polynomials, then from (32a) it can be deduced that a1(X) is constant and g(X)=0. For simplicity, it was taken a1(X)=1.
Balancing the degrees of h(X) and a0(X), it can be concluded that deg(h(X))=1 only. Suppose that h(X)=AX+B, and A≠0, then we find
(33)a0(X)=A2X2+BX+D,
where D is an arbitrary integration constant.
Substituting a0(X), a1(X), and h(X) in (32c) and setting all the coefficients of powers X to be zero, we have obtained a system of nonlinear algebraic equations and by solving it, we obtain(34a)v=-i2cdDα+2aαβ-3cα2β2,M=0, A=-i2dcα, B=0,(34b)v=i2cdDα+2aαβ-3cα2β2,M=0, A=i2dcα, B=0.Using the conditions (34a) and (34b) in (30), we obtain
(35)Y(ξ)=(±i2dcα)X2(ξ)-D,
respectively.
Combining (35) with (29a), the exact solutions to (29a) and (29b) were obtained and then the exact traveling wave solutions to the generalized nonlinear Schrodinger (GNLS) equation with a source (23) can be written as
(36)u1(x,t)=i(-2)1/4c1/4Dα×tanh[(23/4c1/4α)-1(1+i)d1/4D(αx-αvt+ς-2cαξ0) ×(23/4c1/4α)-1]×(d1/4)-1×exp[i(β{αx-αvt+ς}-wt)],v=-i2cdDα+2aαβ-3cα2β2,(37)u2(x,t)=i(-2)1/4c1/4Dα×tan[(23/4c1/4α)-1(1+i)d1/4D(αx-αvt+ς-2cαξ0) ×(23/4c1/4α)-1]×(d1/4)-1×exp[i(β{αx-αvt+ς}-wt)],v=i2cdDα+2aαβ-3cα2β2,
where ξ0 is an arbitrary integration constant.
Case 4.
Suppose that m=2, by equating the coefficients of Yi (i=3,2,1,0) on both sides of (31), we have(38a)a2′(X)=g(X)a2(X),(38b)a1′(X)=h(X)a2(X)+g(X)a1(X),(38c)a0′(X)+2a2(X)[[-dcα2]X3+[vcα2-2aβcα+3β2]X+Mcα2] =h(X)a1(X)+g(X)a0(X),(38d)a1(X)[[-dcα2]X3+[vcα2-2aβcα+3β2]X+Mcα2] =h(X)a0(X).Since ai(X) (i=0,1,2) are polynomials, then from (38a) it can be deduced that a2(X) is a constant and g(X)=0. For simplicity, we have taken a2(X)=1. Balancing the degrees of h(X) and a0(X) we have concluded that deg(h(X))=1 only.
In this case, it was assumed that h(X)=AX+B and A≠0; then we find a1(X) and a0(X) as follows:(39a)a1(X)=(A2)X2+BX+D,(39b)a0(X)=(A28+d2cα2)X4+12(AB)X3+(AD+B22-cα2+2aβcα-3β2)X2+(BD-2Mcα2)X+F,where A,B,D, and F are arbitrary integration constants.
Substituting a0(X), a1(X), a2(X), and h(X) for (38d) and setting all the coefficients of powers X to be zero, a system of nonlinear algebraic equations was obtained and by solving it, we got(40a)M=0, v=12[-i2cdDα+4aαβ-6cα2β2],F=D24, A=-2i2dcα, B=0,(40b)M=0, v=12[i2cdDα+4aαβ-6cα2β2],F=D24, A=2i2dcα, B=0.Using the conditions (40a) and (40b) in (30), we obtain
(41)Y(ξ)=±i2dX2(ξ)-cDα2cα,
respectively. Combining (41) with (29a) we have obtained the exact solutions to (29a) and (29b) and thus the exact traveling wave solutions to the generalized nonlinear Schrodinger (GNLS) equation with a source (23) can be written as
(42)u3(x,t)=(-1)3/4c1/4Dα×tanh[(12+i2)d1/4D ×(αx-αvt+ς-2cαξ0) ×(21/4c1/4α)-1(12+i2)]×(21/4d1/4)-1×exp[i(β{αx-αvt+ς}-wt)],v=-12i2cdDα+2aαβ-3cα2β2,(43)u4(x,t)=-(-1)3/4c1/4Dα×tan[(23/4c1/4α)-1(-1)1/4d1/4 ×D(-αx+αvt+ς+2cαξ0) ×(23/4c1/4α)-1]×(21/4d1/4)-1×exp[i(β{αx-αvt+ς}-wt)]v=12i2cdDα+2aαβ-3cα2β2,
respectively, where ξ0 is an arbitrary integration constant.
Equations (36)-(37) and (42)-(43) are new types of exact traveling wave solutions to the generalized nonlinear Schrodinger (GNLS) equation with a source (23). It could not be obtained by the methods presented in [32].
3.3. The Higher-Order Nonlinear Schrodinger Equation in Nonlinear Optical Fibers
The higher-order nonlinear Schrodinger equation describing propagation of ultrashort pulses in nonlinear optical fibers [33–39] reads
(44)ψz=iα1ψtt+iα2ψ|ψ|2+α3ψttt+α4(ψ|ψ|2)t+α5ψ(|ψ|2)t,
where ψ is slowly varying envelope of the electric field, the subscripts z and t are the spatial and temporal partial derivative in retard time coordinates, and α1, α2, α3, α4, α5 are the real parameters related to the group velocity dispersion (GVD), self-phase modulation (SPM), third-order dispersion (TOD), and self-steepening and self-frequency shift arising from simulated Raman scattering, respectively. Some properties of the equation, as well as many versions of it have been studied [33–39]. Up to now, the bright, dark and the combined bright and dark solitary waves and periodic waves were found of (43) and its special case.
To seek traveling wave solutions of (44), we make the gauge transformation
(45)ψ(z,t)=φ(ξ)exp[i(kz-ωt)],ξ=βt-λz+ε,
where β, k, ω, λ, ε are constants. Substituting (45) into (44) yields a complex ODE of φ(ξ), the real and imaginary parts of which, respectively,
(46)(β2α1-3β2α3ω)φ′′+(α3ω3-α1ω2-k)φ +(α2-α4ω)φ3=0,β3α3φ′′′+(2βα1ω-3βα3ω2+λ)φ′ +(3βα4+2βα5)φ2φ′=0.
It is easy to see that (46) becomes an equation
(47)φ′′+2βα1ω-3βα3ω2+λβ3α3φ+3α4+2α53β2α3φ3=0,
under the constraint conditions
(48)ω=3α1α4+2α1α5-3α2α36α3(α4+α5),k=1α3[1β(3α3ω-α1)λ-2ω(α1-2α3ω)2] =(3α2α3-3α1α4-2α1α5)(3α2α3+α!α5)227α32(α4+α5)3 +(α1α4-3α2α3)λ2βα3(α4+α5).
Using (4) and (5a) and (5b), we can get(49a)X′=Y,(49b)Y′=-(3α4+2α53β2α3)X3-(2βα1ω-3βα3ω2+λβ3α3)X.According to the first integral method, we suppose that X(ξ) and Y(ξ) are nontrivial solutions of (49a) and (49b) and P(X,Y)=∑i=0mai(X)Yi is an irreducible polynomial in the complex domain C[X,Y] such that
(50)P[X(ξ),Y(ξ)]=∑i=0mai(X(ξ))Y(ξ)i=0,
where ai(X), (i=0,1,2,…,m) are polynomials of X and am(X)≠0.
Equation (50) is called the first integral to (49a) and (49b) due to the division theorem, there exists a polynomial h(X)+g(X)Y in the complex domain C[X,Y] such that
(51)dPdξ=∂P∂XdXdξ+∂P∂YdPdξ=[h(X)+g(X)Y]∑i=0mai(X)Yi.
In this example, we take two different cases, assuming that m=1 and m=2 in (50).
Case 5.
Suppose that m=1, by equating the coefficients of Yi (i=2,1,0) on both sides of (51), we have(52a)a1′(X)=g(X)a1(X),(52b)a0′(X)=h(X)a1(X)+g(X)a0(X),(52c)a1(X)[-(3α4+2α53β2α3)X3-(2βα1ω-3βα3ω2+λβ3α3)X] =h(X)a0(X).Since ai(X) (i=0,1) are polynomials, then from (52a) it was deduced that a1(X) is constant and g(X)=0. For simplicity, take a1(X)=1.
Balancing the degrees of h(X) and a0(X), it was concluded that deg(h(X))=1 only. Suppose that h(X)=AX+B and A≠0, then we find
(53)a0(X)=A2X2+BX+D,
where D is an arbitrary integration constant.
Substituting a0(X), a1(X), and h(X) in (52c) and setting all the coefficients of powers X to be zero, then we have obtained a system of nonlinear algebraic equations and by solving it, we obtain
(54)D=±3/2(λ+βω(2α1-3ωα3))β2α3-3α4-2α5,B=0,A=∓2/3-3α4-2α5βα3.
Using the conditions (54) in (50), we obtain
(55)Y(ξ)=(±2/3-3α4-2α5βα3)X2(ξ)∓3/2(λ+βω(2α1-3ωα3))β2α3-3α4-2α5,
respectively.
Combining (55) with (49a), the exact solutions to (49a) and (49b) were obtained and then the exact traveling wave solutions to the higher-order nonlinear Schrodinger equation in nonlinear optical fibers can be written as
(56)ψ1,2(z,t) =±1βp3q ×tan[×(2β3/2α3p)-1q((βt-λz+ε)r±23β2ξ0α3p2) ×(2β3/2α3p)-1]×exp[i(kz-ωt)],
respectively, where
(57)p=3α4+2α5, q=λ+βw(2α1-3wα3),r=-6α4-4α5, p2=3α4+2α5,
and ξ0 is an arbitrary integration constant.
Case 6.
Suppose that m=2, by equating the coefficients of Yi (i=3,2,1,0) on both sides of (51), we have(58a)a2′(X)=g(X)a2(X),(58b)a1′(X)=h(X)a2(X)+g(X)a1(X),(58c)a0′(X)+2a2(X)[[-dcα2]X3+[vα2-2aβcα+3β2]X+Mcα2] =h(X)a1(X)+g(X)a0(X),(58d)a1(X)[-(2βα1ω-3βα3ω2+λβ3α3)X-(3α4+2α53β2α3)X3] =h(X)a0(X).Since ai(X) (i=0,1,2) are polynomials, then from (58a) it was deduced that a2(X) is a constant and g(X)=0. For simplicity, we have taken a2(X)=1. Balancing the degrees of h(X) and a0(X) we conclude that deg(h(X))=1 only.
In this case, it was assumed that h(X)=AX+B and A≠0; then we find a1(X) and a0(X) as follows:(59a)a1(X)=(A2)X2+BX+D,(59b)a0(X)=(A28+3α4+2α56β2α3)X4+12(AB)X3+(AD+B22+2βα1ω-3βα3ω2+λβ3α3)X2+BDX+F,where A,B,D, and F are arbitrary integration constants.
Substituting a0(X), a1(X), a2(X), and h(X) into (58d) and setting all the coefficients of powers X to be zero, a system of nonlinear algebraic equations was obtained and by solving it, we get(60a)λ=-2βωα1+3βω2α3, F=0, B=0, D=0,A=∓22/3-3α4-2α5βα3,(60b)λ=-2βωα1+3βω2α3±Dβ2α3-3α4-2α56,F=D24, B=0, A=∓22/3-3α4-2α5βα3.Using the conditions (60a) in (50), we obtain
(61)Y(ξ)=±-3α4-2α56βα3X2(ξ),
respectively. Combining (61) with (49a), the exact solutions to (49a) and (49b) were obtained and thus the exact traveling wave solutions to the higher-order nonlinear Schrodinger equation in nonlinear optical fibers (44) can be written as
(62)ψ3,4(z,t) =6βα3 ×({6(βt-(-2βωα1+3βω2α3)z+ε)}-6βc1α3 ∓6(βt-(-2βωα1+3βω2α3)z+ε)ip)-1 ×exp[i(kz-ωt)],
respectively.
Similarly, in the case of (60b), from (50), we get
(63)Y(ξ)=-D2±-3α4-2α56βα3X2(ξ),
respectively. Combining (63) with (49a), the exact solutions to (49a) and (49b) were obtained and thus the exact traveling wave solutions to the higher-order nonlinear Schrodinger equation in nonlinear optical fibers (44) can be written as
(64)ψ5(z,t)=(3/2)1/4Dβα31/4p×tan[(23/431/4βα31/4(-1)1/4p1/2)-1((-1)3/4p3/2)-1D((βt-λz+ε)-6βξ0α3)p ×(23/431/4βα31/4(-1)1/4p1/2)-1]×((-1)3/4p3/2)-1×exp[i(kz-ωt)],λ=-2βωα1+3βω2α3+Dβ2α3-3α4-2α56,(65)ψ6(z,t)=(3/2)1/4Dβα31/4×ptanh[(23/431/4βα31/4(-1)1/4p1/2)-1D((βt-λz+ε)-6βξ0α3)p ×(23/431/4βα31/4(-1)1/4p1/2)-1]×((-1)3/4p3/2)-1×exp[i(kz-ωt)],λ=-2βωα1+3βω2α3-Dβ2α3-3α4-2α56,
respectively, where p is as defined in (57) and ξ0 is an arbitrary integration constant.
Comparing these results with Liu’s results [39], it can be seen that the solutions here are new.