The First Integral Method to the Nonlinear Schrodinger Equations in Higher Dimensions

Abstract

The first integral method introduced by Feng is adopted for solving some important nonlinear partial differential equations, including the (2 + 1)-dimensional hyperbolic nonlinear Schrodinger (HNLS) equation, the generalized nonlinear Schrodinger (GNLS) equation with a source, and the higher-order nonlinear Schrodinger equation in nonlinear optical fibers. This method provides polynomial first integrals for autonomous planar systems. Through the established first integrals, exact traveling wave solutions are formally derived in a concise manner.

1. Introduction

It is well known that nonlinear complex physical phenomena are related to nonlinear partial differential equations (NLPDEs) which are involved in many fields from physics to biology, chemistry, mechanics, etc. As mathematical models of the phenomena, the investigation of exact solutions of NLPDEs will help us to understand these phenomena better. Many effective methods for obtaining exact solutions of NLPDEs have been established and developed, such as the Lie point symmetries method [1], the exp-function method [2, 3], the sine-cosine method [4, 5], the extended tanh-coth method [6, 7], the projective Riccati equation method [8, 9], and so on.

The first integral method was first proposed by Feng in [10] in solving Burgers-KdV equation which is based on the ring theory of commutative algebra. Recently, this useful method has been widely used by many such as in [11–21] and by the references therein. In Section 2, we have described this method for finding exact travelling wave solutions of nonlinear evolution equations. In Section 3, we have illustrated this method in detail with the (2+1)-dimensional hyperbolic nonlinear Schrodinger (HNLS) equation, the generalized nonlinear Schrodinger (GNLS) equation with a source, and the higher-order nonlinear Schrodinger equation in nonlinear optical fibers. In Section 4, we have given some conclusions.

2. The First Integral Method

Consider a general nonlinear PDE in the form 𝑃𝑢,𝑢_𝑡,𝑢_𝑥,𝑢_𝑥𝑥,𝑢_𝑡𝑡,𝑢_𝑥𝑡,𝑢_𝑥𝑥𝑥,…=0, (1) where 𝑃 is a polynomial in its arguments.

Raslan in [22] has summarized the first integral method in the following steps.

Step 1. Using a wave variable 𝜉=𝑥−𝑐𝑡+𝜀, where 𝜀 is an arbitrary constant, (1) can be written in the following nonlinear ordinary differential equation (ODE): 𝑄𝑈,𝑈^′,𝑈^′′,𝑈^′′′,…=0, (2) where the prime denotes the derivation with respect to 𝜉.

Step 2. Assume that the solution of ODE (2) can be written as 𝑢(𝑥,𝑡)=𝑢(𝜉). (3)

Step 3. We introduced new independent variables 𝑋(𝜉)=𝑢(𝜉),𝑌=𝑢_𝜉(𝜉), (4) which leads a system of nonlinear ODEs𝑋_𝜉(𝜉)(𝜉),𝑌=𝑌_𝜉(𝜉)=𝐹(𝑋(𝜉),𝑌(𝜉)). (5a)(5b)

Step 4. According to the qualitative theory of ODEs [23], if we can find the integrals to (5a) and (5b) under the same conditions, then the general solution to (5a) and (5b) can be found directly. However, in general, it is really difficult to realize this even for one first integral, because for a given plane autonomous system, there is no systematic theory that can tell us how to find its first integrals, nor is there a logical way for telling us what these first integrals are.
We will apply the division theorem to obtain one first integral to (5a) and (5b) which reduces (2) to a first-order integrable ODE.
An exact solution to (1) is then obtained by solving this equation.
Let us now recall the division theorem for two variables in the complex domain 𝐶(𝑤,𝑧).

Theorem 1 (division theorem). Suppose that 𝑃(𝑤,𝑧) and 𝑄(𝑤,𝑧) are polynomials in 𝐶(𝑤,𝑧), and 𝑃(𝑤,𝑧) is irreducible in 𝐶(𝑤,𝑧). If 𝑄(𝑤,𝑧) vanishes at all zero points of 𝑃(𝑤,𝑧), then there exists a polynomial 𝐺(𝑤,𝑧) in 𝐶(𝑤,𝑧) such that 𝑄(𝑤,𝑧)=𝑃(𝑤,𝑧)𝐺(𝑤,𝑧). (6)

The division theorem follows immediately from the Hilbert-Nullstellensatz Theorem [24].

Theorem 2 (Hilbert-Nullstellensatz theorem). Let 𝑘 be a field and 𝐿 an algebraic closure of 𝑘.(1)Every ideal 𝛾 of 𝑘[𝑋₁,…,𝑋_𝑛] not containing 1 admits at least one zero in 𝐿^𝑛. (2)Let 𝑥=(𝑥₁,…,𝑥_𝑛), 𝑦=(𝑦₁,…,𝑦_𝑛) be two elements of 𝐿^𝑛; for the set of polynomials of 𝑘[𝑋₁,…,𝑋_𝑛] zero at 𝑥 to be identical with the set of polynomials of 𝑘[𝑋₁,…,𝑋_𝑛] zero at 𝑦 it is necessary and sufficient that there exists a 𝑘-automorphism 𝑠 of 𝐿 such that 𝑦_𝑖=𝑠(𝑥_𝑖) for 1≤𝑖≤𝑛.(3)For an ideal 𝛼 of 𝑘[𝑋₁,…,𝑋_𝑛] to be maximal, it is necessary and sufficient that there exists an 𝑥 in 𝐿^𝑛 such that 𝛼 is the set of polynomials of 𝑘[𝑋₁,…,𝑋_𝑛] to be zero at 𝑥.(4)For a polynomial 𝑄 of 𝑘[𝑋₁,…,𝑋_𝑛] to be zero on the set of zeros in 𝐿^𝑛 of an ideal 𝛾 of 𝑘[𝑋₁,…,𝑋_𝑛], it is necessary and sufficient that there exists an integer 𝑚≻0 such that 𝑄^𝑚∈𝛾.

3. Applications

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 𝑖𝑢_𝑡+12𝑢_𝑥𝑥−12𝑢_𝑦𝑦+|𝑢|²𝑢=0, (7) 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: 𝑐²−1𝜑^′′(𝜉)=2𝜑³(𝜉)+𝑤²−2𝑛−(𝑎+𝑐𝑤)²𝜑(𝜉), (8) where prime denotes the derivative with respect to the same variable 𝜉.

Using (4) and (5a) and (5b), we can get𝑋^′(𝜉)(𝜉),𝑌=𝑌_′(𝜉)=2𝑐²𝑋−1³(𝜉)+𝑤²−2𝑛−(𝑎+𝑐𝑤)²𝑐²𝑋(𝜉).−1 (9a)(9b)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)=∑^𝑚_𝑖=0𝑎_𝑖(𝑋)𝑌^𝑖 is an irreducible polynomial in the complex domain 𝐶[𝑋,𝑌] such that 𝑃𝑋(𝜉),𝑌(𝜉)=^𝑚_𝑖=0𝑎_𝑖(𝑋(𝜉))𝑌^𝑖(𝜉)=0, (10) where 𝑎_𝑖(𝑋),(𝑖=0,1,2,…,𝑚) are polynomials of 𝑋 and 𝑎_𝑚(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 𝑑𝑃=𝑑𝜉𝜕𝑃𝜕𝑋𝑑𝑋+𝑑𝜉𝜕𝑃𝜕𝑌𝑑𝑌=ℎ(𝑋)(𝑋)𝑌𝑑𝜉+𝑔^𝑚_𝑖=0𝑎_𝑖(𝑋)𝑌^𝑖. (11) 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 𝑌^𝑖(𝑖=2,1,0) on both sides of (11), we have𝑎^′₁(𝑋)=𝑔(𝑋)𝑎₁(𝑋),𝑎^′₀(𝑋)=ℎ(𝑋)𝑎₁(𝑋)+𝑔(𝑋)𝑎₀(𝑋),𝑎₁(𝑋)2𝑐²𝑋−1³+𝑤²−2𝑛−(𝑎+𝑐𝑤)²𝑐²𝑋(𝑋)𝑎−1=ℎ₀(𝑋). (12a)(12b)(12c)Since 𝑎_𝑖(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)𝑎₀(𝑋)=𝐴2𝑋²+𝐵𝑋+𝐷, (13) 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 1𝐷=∓21−1+𝑐²[(()𝑤,√2𝑛+𝑎+−1+𝑐)(𝑎+𝑤+𝑐𝑤)]𝐵=0,𝐴=±2−1−11−𝑐.1+𝑐 (14) Using the conditions (14) in (10), we obtain 𝑌(𝜉)√=∓22−1−11−𝑐𝑋1+𝑐²(𝜉)±121−1+𝑐²[(()𝑤)×(,2𝑛+𝑎+−1+𝑐𝑎+𝑤+𝑐𝑤)] (15) 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 𝑢_1,21𝑥,𝑦,𝑡=−√2−2𝑛−(𝑎+(−1+𝑐)𝑤)(𝑎+𝑤+𝑐𝑤)×tan(()𝑤)×±√−2𝑛−𝑎+−1+𝑐)(𝑎+𝑤+𝑐𝑤()√−1+𝑐𝑥+𝑎𝑦−𝑐𝑡+𝜍+2−1+𝑐−1+𝑐𝜉₀×√2(−1+𝑐)√−1+𝑐⁻¹×exp𝑖𝑚𝑥+𝑛𝑦+𝑤𝑡+𝜀, (16) respectively, where 𝜉₀ is an arbitrary integration constant.

Case 2. Assume that 𝑚=2, by equating the coefficients of 𝑌^𝑖(𝑖=3,2,1,0) on both sides of (11), we have𝑎^′₂(𝑋)=𝑔(𝑋)𝑎₂(𝑋),𝑎^′₁(𝑋)=ℎ(𝑋)𝑎₂(𝑋)+𝑔(𝑋)𝑎₁(𝑋),𝑎^′₀(𝑋)+2𝑎₂(𝑋2)𝑐²𝑋−1³+𝑤²−2𝑛−(𝑎+𝑐𝑤)²𝑐²𝑋(𝑋)𝑎−1=ℎ₁(𝑋)+𝑔(𝑋)𝑎₀(𝑋),𝑎₁(𝑋2)𝑐²𝑋−1³+𝑤²−2𝑛−(𝑎+𝑐𝑤)²𝑐²𝑋(𝑋)𝑎−1=ℎ₀(𝑋). (17a)(17b)(17c)(17d)Since, 𝑎_𝑖(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:𝑎₁(𝑋)=𝐴2𝑋²𝑎+𝐵𝑋+𝐷,₀(𝑋)=𝐴²8−1𝑐²𝑋−1⁴+12(𝐴𝐵)𝑋³+𝐴𝐷+𝐵²2−𝑤²−2𝑛−(𝑎+𝑐𝑤)²𝑐²𝑋−1²+𝐵𝐷𝑋+𝐹, (18a)(18b)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√𝐹=0,𝐷=0,𝐵=0,𝑤=−𝑎𝑐−𝑎²+2𝑛−2𝑐²𝑛−1+𝑐²,𝐴=∓2√2−1−11−𝑐,√1+𝑐𝐹=0,𝐷=0,𝐵=0,𝑤=−𝑎𝑐+𝑎²+2𝑛−2𝑐²𝑛−1+𝑐²,𝐴=∓2√2−1−11−𝑐.1+𝑐 (19a)(19b)Using the conditions (19a) and (19b) in (10), we have obtained 𝑌(𝜉)1=±√−1+𝑐²𝑋²(𝜉). (20) 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 𝑢_3,4=√𝑥,𝑦,𝑡−1+𝑐√1+𝑐∓𝑥+𝑎𝑦−𝑐𝑡+𝜍−√−1+𝑐√𝑖√1+𝑐×exp𝑚𝑥+𝑛𝑦+−𝑎𝑐−𝑎²+2𝑛−2𝑐²𝑛−1+𝑐²,𝑢𝑡+𝜀_5,6=√𝑥,𝑦,𝑡−1+𝑐√1+𝑐∓𝑥+𝑎𝑦−𝑐𝑡+𝜍−√−1+𝑐√𝑖√1+𝑐×exp𝑚𝑥+𝑛𝑦+−𝑎𝑐+𝑎²+2𝑛−2𝑐²𝑛−1+𝑐²𝑡+𝜀. (21)(22) 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 𝑖𝑢_𝑡+𝑎𝑢_𝑥𝑥+𝑏𝑢|𝑢|²+𝑖𝑐𝑢_𝑥𝑥𝑥+𝑖𝑑𝑢|𝑢|²_𝑥=𝑘𝑒^{𝑖[𝜒(𝜉)−𝑤𝑡]}, (23) 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 𝑢(𝑥,𝑡)=𝜓(𝜉)𝑒^{𝑖[𝜒(𝜉)−𝑤𝑡]}, (24) 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: 𝑐𝛼³𝜓^′′′+−𝛼𝑣+2𝑎𝛽𝛼²−3𝑐𝛼³𝛽²𝜓^′+3𝑑𝛼𝜓²𝜓^′=0,𝑎𝛼²−3𝑐𝜓³𝛽𝜓^′′+𝛼𝛽𝑣+𝑤−𝑎𝛼²𝛽²+𝑐𝛼³𝛽³𝜓+𝑏−𝑑𝛼𝛽𝜓³−𝑘=0. (25)(26) Integrating (25) once, with respect to 𝜉, we have 𝑐𝛼²𝜓^′′(𝜉)+−𝑣+2𝑎𝛼𝛽−3𝑐𝛼²𝛽²𝜓(𝜉)−𝑀=0, (27) where 𝑀 is an arbitrary integration constant. Since the same function 𝜓(𝜉) satisfies (26) and (27), we have obtained the following constraint condition: 𝑎𝛼²−3𝑐𝜓³𝛽𝑐𝛼²=𝛼𝛽𝑣+𝑤−𝑎𝛼²𝛽²+𝑐𝛼³𝛽³−𝑣+2𝑎𝛼𝛽−3𝑐𝛼²𝛽²=𝑏−𝑑𝛼𝛽𝑑=𝑘𝑀. (28) Using (4) and (5a) and (5b), we can get𝑋^′(𝜉)(𝜉),𝑌=𝑌_′(𝜉)=−𝑑𝑐𝛼²𝑋³(𝜉)+𝑣𝑐𝛼²−2𝑎𝛽𝑐𝛼+3𝛽²𝑋(𝜉)+𝑀𝑐𝛼². (29a)(29b)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)=∑^𝑚_𝑖=0𝑎_𝑖(𝑋)𝑌^𝑖 is an irreducible polynomial in the complex domain 𝐶[𝑋,𝑌] such that 𝑃𝑋(𝜉),𝑌(𝜉)=^𝑚_𝑖=0𝑎_𝑖(𝑋(𝜉))𝑌(𝜉)^𝑖=0, (30) where 𝑎_𝑖(𝑋),(𝑖=0,1,2,…,𝑚) are polynomials of 𝑋 and 𝑎_𝑚(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 𝑑𝑃=𝑑𝜉𝜕𝑃𝜕𝑋𝑑𝑋+𝑑𝜉𝜕𝑃𝜕𝑌𝑑𝑃,=ℎ(𝑋)(𝑋)𝑌𝑑𝜉+𝑔^𝑚_𝑖=0𝑎_𝑖(𝑋)𝑌^𝑖. (31) 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 𝑌^𝑖(𝑖=2,1,0) on both sides of (31), we have𝑎^′₁(𝑋)=𝑔(𝑋)𝑎₁(𝑋),𝑎^′₀(𝑋)=ℎ(𝑋)𝑎₁(𝑋)+𝑔(𝑋)𝑎₀(𝑋),𝑎₁(𝑋−𝑑)𝑐𝛼²𝑋³+𝑣𝑐𝛼²−2𝑎𝛽𝑐𝛼+3𝛽²𝑀𝑋+𝑐𝛼²=ℎ(𝑋)𝑎₀(𝑋). (32a)(32b)(32c)Since 𝑎_𝑖(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 𝑎₀(𝑋)=𝐴2𝑋²+𝐵𝑋+𝐷, (33) 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 obtain𝑣=−𝑖√2√𝑐√𝑑𝐷𝛼+2𝑎𝛼𝛽−3𝑐𝛼²𝛽²,𝑖√𝑀=0,𝐴=−2√𝑑√√𝑐𝛼,𝐵=0,𝑣=𝑖2√𝑐√𝑑𝐷𝛼+2𝑎𝛼𝛽−3𝑐𝛼²𝛽²,𝑖√𝑀=0,𝐴=2√𝑑√𝑐𝛼,𝐵=0. (34a)(34b)Using the conditions (34a) and (34b) in (30), we obtain 𝑌(𝜉)=±𝑖√2√𝑑√𝑋𝑐𝛼²(𝜉)−𝐷, (35) 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 𝑢₁(𝑥,𝑡)=𝑖(−2)^1/4𝑐^1/4√𝐷√𝛼×tanh(1+𝑖)𝑑^1/4√𝐷𝛼𝑥−𝛼𝑣𝑡+𝜍−2√𝑐𝛼𝜉₀×2^3/4𝑐^1/4√𝛼⁻¹×𝑑^1/4−1𝑖𝛽{},√×exp𝛼𝑥−𝛼𝑣𝑡+𝜍−𝑤𝑡𝑣=−𝑖2√𝑐√𝑑𝐷𝛼+2𝑎𝛼𝛽−3𝑐𝛼²𝛽²,𝑢₂(𝑥,𝑡)=𝑖(−2)^1/4𝑐^1/4√𝐷√𝛼×tan(1+𝑖)𝑑^1/4√𝐷𝛼𝑥−𝛼𝑣𝑡+𝜍−2√𝑐𝛼𝜉₀×2^3/4𝑐^1/4√𝛼⁻¹×𝑑^1/4−1𝑖𝛽{},√×exp𝛼𝑥−𝛼𝑣𝑡+𝜍−𝑤𝑡𝑣=𝑖2√𝑐√𝑑𝐷𝛼+2𝑎𝛼𝛽−3𝑐𝛼²𝛽², (36)(37) where 𝜉₀ is an arbitrary integration constant.

Case 4. Suppose that 𝑚=2, by equating the coefficients of 𝑌^𝑖(𝑖=3,2,1,0) on both sides of (31), we have𝑎^′₂(𝑋)=𝑔(𝑋)𝑎₂(𝑋),𝑎^′₁(𝑋)=ℎ(𝑋)𝑎₂(𝑋)+𝑔(𝑋)𝑎₁(𝑋),𝑎^′₀(𝑋)+2𝑎₂(𝑋−𝑑)𝑐𝛼²𝑋³+𝑣𝑐𝛼²−2𝑎𝛽𝑐𝛼+3𝛽²𝑀𝑋+𝑐𝛼²=ℎ(𝑋)𝑎₁(𝑋)+𝑔(𝑋)𝑎₀(𝑋),𝑎₁(𝑋−𝑑)𝑐𝛼²𝑋³+𝑣𝑐𝛼²−2𝑎𝛽𝑐𝛼+3𝛽²𝑀𝑋+𝑐𝛼²=ℎ(𝑋)𝑎₀(𝑋). (38a)(38b)(38c)(38d)Since 𝑎_𝑖(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:𝑎₁(𝑋)=𝐴2𝑋²𝑎+𝐵𝑋+𝐷,₀(𝑋)=𝐴²8+𝑑2𝑐𝛼²𝑋⁴+12(𝐴𝐵)𝑋³+𝐴𝐷+𝐵²2−𝑐𝛼²+2𝑎𝛽𝑐𝛼−3𝛽²𝑋²+𝐵𝐷−2𝑀𝑐𝛼²𝑋+𝐹, (39a)(39b)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 got1𝑀=0,𝑣=2−𝑖√2√𝑐√𝑑𝐷𝛼+4𝑎𝛼𝛽−6𝑐𝛼²𝛽²,𝐷𝐹=²4√,𝐴=−2𝑖2√𝑑√1𝑐𝛼,𝐵=0,𝑀=0,𝑣=2𝑖√2√𝑐√𝑑𝐷𝛼+4𝑎𝛼𝛽−6𝑐𝛼²𝛽²,𝐷𝐹=²4√,𝐴=2𝑖2√𝑑√𝑐𝛼,𝐵=0. (40a)(40b)Using the conditions (40a) and (40b) in (30), we obtain 𝑌(𝜉)=±𝑖√2√𝑑𝑋²(𝜉)−√𝑐𝐷𝛼2√,𝑐𝛼 (41) 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 𝑢₃(𝑥,𝑡)=(−1)^3/4𝑐^1/4√𝐷√𝛼1×tanh2+𝑖2𝑑^1/4√𝐷×𝛼𝑥−𝛼𝑣𝑡+𝜍−2√𝑐𝛼𝜉₀×2^1/4𝑐^1/4√𝛼⁻¹×2^1/4𝑑^1/4−1𝑖𝛽{},1×exp𝛼𝑥−𝛼𝑣𝑡+𝜍−𝑤𝑡𝑣=−2𝑖√2√𝑐√𝑑𝐷𝛼+2𝑎𝛼𝛽−3𝑐𝛼²𝛽²,𝑢₄(𝑥,𝑡)=−(−1)^3/4𝑐^1/4√𝐷√𝛼×tan(−1)^1/4𝑑^1/4×√𝐷−𝛼𝑥+𝛼𝑣𝑡+𝜍+2√𝑐𝛼𝜉₀×2^3/4𝑐^1/4√𝛼⁻¹×2^1/4𝑑^1/4−1𝑖𝛽{}1×exp𝛼𝑥−𝛼𝑣𝑡+𝜍−𝑤𝑡𝑣=2𝑖√2√𝑐√𝑑𝐷𝛼+2𝑎𝛼𝛽−3𝑐𝛼²𝛽², (42)(43) 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 𝜓_𝑧=𝑖𝛼₁𝜓_𝑡𝑡+𝑖𝛼₂𝜓𝜓²+𝛼₃𝜓_𝑡𝑡𝑡+𝛼₄𝜓𝜓²_𝑡+𝛼₅𝜓𝜓²_𝑡, (44)

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 𝜓(𝑧,𝑡)=𝜑(𝜉)exp[𝑖(𝑘𝑧−𝜔𝑡)],𝜉=𝛽𝑡−𝜆𝑧+𝜀, (45) 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, 𝛽²𝛼₁−3𝛽²𝛼₃𝜔𝜑^′′+𝛼₃𝜔³−𝛼₁𝜔²𝜑+𝛼−𝑘₂−𝛼₄𝜔𝜑³𝛽=0,₃𝛼₃𝜑^′′′+2𝛽𝛼₁𝜔−3𝛽𝛼₃𝜔²+𝜆𝜑^′+3𝛽𝛼₄+2𝛽𝛼₅𝜑²𝜑^′=0. (46) It is easy to see that (46) becomes an equation 𝜑^′′+2𝛽𝛼₁𝜔−3𝛽𝛼₃𝜔²+𝜆𝛽³𝛼₃𝜑+3𝛼₄+2𝛼₅3𝛽²𝛼₃𝜑³=0, (47) under the constraint conditions 𝜔=3𝛼₁𝛼₄+2𝛼₁𝛼₅−3𝛼₂𝛼₃6𝛼₃𝛼₄+𝛼₅,1𝑘=𝛼₃1𝛽3𝛼₃𝜔−𝛼₁𝜆−2𝜔𝛼₁−2𝛼₃𝜔²=3𝛼₂𝛼₃−3𝛼₁𝛼₄−2𝛼₁𝛼₅3𝛼₂𝛼₃+𝛼_!𝛼₅²27𝛼²₃𝛼₄+𝛼₅³+𝛼₁𝛼₄−3𝛼₂𝛼₃𝜆2𝛽𝛼₃𝛼₄+𝛼₅. (48) Using (4) and (5a) and (5b), we can get𝑋^′𝑌=𝑌,_′=−3𝛼₄+2𝛼₅3𝛽²𝛼₃𝑋³−2𝛽𝛼₁𝜔−3𝛽𝛼₃𝜔²+𝜆𝛽³𝛼₃𝑋. (49a)(49b)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)=∑^𝑚_𝑖=0𝑎_𝑖(𝑋)𝑌^𝑖 is an irreducible polynomial in the complex domain 𝐶[𝑋,𝑌] such that 𝑃𝑋(𝜉),𝑌(𝜉)=^𝑚_𝑖=0𝑎_𝑖(𝑋(𝜉))𝑌(𝜉)^𝑖=0, (50) where 𝑎_𝑖(𝑋),(𝑖=0,1,2,…,𝑚) are polynomials of 𝑋 and 𝑎_𝑚(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 𝑑𝑃=𝑑𝜉𝜕𝑃𝜕𝑋𝑑𝑋+𝑑𝜉𝜕𝑃𝜕𝑌𝑑𝑃=ℎ(𝑋)(𝑋)𝑌𝑑𝜉+𝑔^𝑚_𝑖=0𝑎_𝑖(𝑋)𝑌^𝑖. (51) 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 𝑌^𝑖(𝑖=2,1,0) on both sides of (51), we have𝑎^′₁(𝑋)=𝑔(𝑋)𝑎₁(𝑋),𝑎^′₀(𝑋)=ℎ(𝑋)𝑎₁(𝑋)+𝑔(𝑋)𝑎₀(𝑋),𝑎₁(𝑋)−3𝛼₄+2𝛼₅3𝛽²𝛼₃𝑋³−2𝛽𝛼₁𝜔−3𝛽𝛼₃𝜔²+𝜆𝛽³𝛼₃𝑋=ℎ(𝑋)𝑎₀(𝑋). (52a)(52b)(52c)Since 𝑎_𝑖(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 𝑎₀(𝑋)=𝐴2𝑋²+𝐵𝑋+𝐷, (53) 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 𝐷=±3/2𝜆+𝛽𝜔2𝛼₁−3𝜔𝛼₃𝛽²√𝛼₃−3𝛼₄−2𝛼₅,𝐵=0,𝐴=∓2/3−3𝛼₄−2𝛼₅𝛽√𝛼₃. (54) Using the conditions (54) in (50), we obtain 𝑌(𝜉)=±2/3−3𝛼₄−2𝛼₅𝛽√𝛼₃𝑋²(𝜉)∓3/2𝜆+𝛽𝜔2𝛼₁−3𝜔𝛼₃𝛽²√𝛼₃−3𝛼₄−2𝛼₅, (55) 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 𝜓_1,2()1𝑧,𝑡=±√𝛽𝑝𝑞√3𝑞×tan𝛽𝑡−𝜆𝑧+𝜀𝑟±23𝛽²𝜉₀√𝛼₃𝑝²×2𝛽^3/2√𝛼₃𝑝⁻¹×exp[𝑖(𝑘𝑧−𝜔𝑡)], (56) respectively, where 𝑝=3𝛼₄+2𝛼₅,𝑞=𝜆+𝛽𝑤2𝛼₁−3𝑤𝛼₃,𝑟=−6𝛼₄−4𝛼₅,𝑝²=3𝛼₄+2𝛼₅, (57) and 𝜉₀ is an arbitrary integration constant.

Case 6. Suppose that 𝑚=2, by equating the coefficients of 𝑌^𝑖(𝑖=3,2,1,0) on both sides of (51), we have𝑎^′₂(𝑋)=𝑔(𝑋)𝑎₂(𝑋),𝑎^′₁(𝑋)=ℎ(𝑋)𝑎₂(𝑋)+𝑔(𝑋)𝑎₁(𝑋),𝑎^′₀(𝑋)+2𝑎₂(𝑋−𝑑)𝑐𝛼²𝑋³+𝑣𝛼²−2𝑎𝛽𝑐𝛼+3𝛽²𝑀𝑋+𝑐𝛼²=ℎ(𝑋)𝑎₁(𝑋)+𝑔(𝑋)𝑎₀(𝑋),𝑎₁(𝑋)−2𝛽𝛼₁𝜔−3𝛽𝛼₃𝜔²+𝜆𝛽³𝛼₃𝑋−3𝛼₄+2𝛼₅3𝛽²𝛼₃𝑋³=ℎ(𝑋)𝑎₀(𝑋). (58a)(58b)(58c)(58d)Since 𝑎_𝑖(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:𝑎₁(𝑋)=𝐴2𝑋²𝑎+𝐵𝑋+𝐷,₀(𝑋)=𝐴²8+3𝛼₄+2𝛼₅6𝛽²𝛼₃𝑋⁴+12(𝐴𝐵)𝑋³+𝐴𝐷+𝐵²2+2𝛽𝛼₁𝜔−3𝛽𝛼₃𝜔²+𝜆𝛽³𝛼₃𝑋²+𝐵𝐷𝑋+𝐹, (59a)(59b)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 get𝜆=−2𝛽𝜔𝛼₁+3𝛽𝜔²𝛼₃2,𝐹=0,𝐵=0,𝐷=0,𝐴=∓2/3−3𝛼₄−2𝛼₅𝛽√𝛼₃,𝜆=−2𝛽𝜔𝛼₁+3𝛽𝜔²𝛼₃±𝐷𝛽²√𝛼₃−3𝛼₄−2𝛼₅√6,𝐷𝐹=²42,𝐵=0,𝐴=∓2/3−3𝛼₄−2𝛼₅𝛽√𝛼₃. (60a)(60b)Using the conditions (60a) in (50), we obtain 𝑌(𝜉)=±−3𝛼₄−2𝛼₅√6𝛽√𝛼₃𝑋²(𝜉), (61) 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 𝜓_3,4()√𝑧,𝑡=6𝛽𝛼₃×−6𝛽𝑐₁√𝛼₃∓√6𝛽𝑡−−2𝛽𝜔𝛼₁+3𝛽𝜔²𝛼₃𝑧+𝜀𝑖𝑝⁻¹×exp[𝑖(𝑘𝑧−𝜔𝑡)], (62) respectively.

Similarly, in the case of (60b), from (50), we get 𝑌(𝜉)𝐷=−2±−3𝛼₄−2𝛼₅√6𝛽√𝛼₃𝑋²(𝜉), (63) 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 𝜓₅(𝑧,𝑡)=(3/2)^1/4√𝐷𝛽𝛼₃^1/4𝑝×tan√𝐷𝛽𝑡−𝜆𝑧+𝜀−6𝛽𝜉₀√𝛼₃𝑝×2^3/43^1/4𝛽𝛼₃^1/4(−1)^1/4𝑝^1/2−1×(−1)^3/4𝑝^3/2−1×exp[𝑖(𝑘𝑧−𝜔𝑡)],𝜆=−2𝛽𝜔𝛼₁+3𝛽𝜔²𝛼₃+𝐷𝛽²√𝛼₃−3𝛼₄−2𝛼₅√6,𝜓₆(𝑧,𝑡)=(3/2)^1/4√𝐷𝛽𝛼₃^1/4×𝑝tanh√𝐷𝛽𝑡−𝜆𝑧+𝜀−6𝛽𝜉₀√𝛼₃𝑝×2^3/43^1/4𝛽𝛼₃^1/4(−1)^1/4𝑝^1/2−1×(−1)^3/4𝑝^3/2−1×exp[𝑖(𝑘𝑧−𝜔𝑡)],𝜆=−2𝛽𝜔𝛼₁+3𝛽𝜔²𝛼₃−𝐷𝛽²√𝛼₃−3𝛼₄−2𝛼₅√6, (64)(65) 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.

4. Conclusion

Searching for first integrals of nonlinear ODEs is one of the most important problems since they permit us to solve a nonlinear differential equation by quadratures. Applying the first integral method, which is based on the ring theory of commutative algebra, some new exact traveling wave solutions to the (2+1)-dimensional hyperbolic nonlinear Schrodinger (HNLS) equation, generalized nonlinear Schrodinger (GNLS) equation with a source and higher-order nonlinear Schrodinger equation in nonlinear optical fibers were established.

These solutions may be important for the explanation of some practical physical problems.

The first integral method described herein is not only efficient but also has the merit of being widely applicable.

Therefore, this method can be applied to other nonlinear evolution equations and this will be done elsewhere.

Acknowledgment

The author would like to thank the referees for their useful comments which led to some improvements of the current paper.