﻿ 基于首次积分法研究GDNLSE方程的精确解 The First Integral Method for Solving Exact Solution of GDNLSE

Vol.07 No.04(2018), Article ID:24439,7 pages
10.12677/AAM.2018.74036

The First Integral Method for Solving Exact Solution of GDNLSE

Na Yang, Longwei Chen*

College of Statistics and Mathematics, Yunnan University of Finance and Economics, Kunming Yunnan

Received: Mar. 29th, 2018; accepted: Apr. 12th, 2018; published: Apr. 20th, 2018

ABSTRACT

The first integral method is mainly adopted in this paper to study the nonlinear generalized Schrodinger equation with derivative. By introducing the traveling wave transformation, the original nonlinear generalized Schrodinger equation with derivative has been changed into an ordinary differential equation. Then according to the division theorem of polynomial, exact traveling wave solutions of the nonlinear generalized Schrodinger equation with derivative are obtained.

Keywords:The First Integral Method, Ordinary Differential Equation, Traveling Wave Solutions

1. 引言

$i{u}_{x}={u}_{tt}+2{|u|}^{2}u+i\alpha {\left({|u|}^{2}u\right)}_{t}+i\beta {\left({|u|}^{2}\right)}_{t}u+\gamma {|u|}^{4}u\text{ }.$ (1)

2. 求解过程

2.1. 化简方程

$\left\{\begin{array}{l}\left(\frac{{c}^{2}}{4{k}^{2}}-w\right)f={k}^{2}{f}^{″}+\left(2-\frac{\alpha c}{2k}\right){f}^{3}+\gamma {f}^{5},\\ \left(3\alpha +2\beta \right)k{f}^{2}{f}^{\prime }=0.\end{array}$ (2)

${f}^{″}=-\frac{r}{{k}^{2}}{f}^{5}+\frac{\alpha c-4k}{2{k}^{3}}{f}^{3}+\left(\frac{{c}^{2}}{4{k}^{4}}-\frac{w}{{k}^{2}}\right)f.$ (3)

${f}^{″}=A{f}^{5}+B{f}^{3}+Cf.$ (4)

2.2. 求解方程

$\left\{\begin{array}{l}{X}^{\prime }=Y,\hfill \\ {Y}^{\prime }=A{X}^{5}+B{X}^{3}+CX.\hfill \end{array}$ (5)

$P\left(X,Y\right)=\sum _{i=0}^{m}{a}_{i}\left(X\right){Y}^{i}=0.$ (6)

$\frac{\text{d}P}{\text{d}\xi }=\frac{\partial P}{\partial X}\cdot \frac{\partial X}{\partial \xi }+\frac{\partial P}{\partial Y}\cdot \frac{\partial Y}{\partial \xi }=\left(g\left(X\right)+h\left(X\right)Y\right)\left(\sum _{i=0}^{m}{a}_{i}\left(X\right){Y}^{i}=0\right).$ (7)

2.2.1. 情况一

$m=1$ 时，将(5)和(6)式带入到(7)，并且由 ${Y}^{i}\left(i=0,1,2\right)$ 的系数相等，对应 ${Y}^{2},{Y}^{1},{Y}^{0}$ 的系数，分别得到

${{a}^{\prime }}_{1}\left(X\right)={a}_{1}\left(X\right)h\left(X\right),$ (8)

${{a}^{\prime }}_{0}\left(X\right)={a}_{1}\left(X\right)g\left(X\right)+{a}_{0}\left(X\right)h\left(X\right),$ (9)

${a}_{1}\left(X\right)\left(A{X}^{5}+B{X}^{3}+CX\right)={a}_{0}\left(X\right)g\left(X\right).$ (10)

$g\left(X\right)={A}_{2}{X}^{2}+{A}_{1}X+{A}_{0},$ (11)

${a}_{0}\left(X\right)=\frac{1}{3}{A}_{2}{X}^{3}+\frac{1}{2}{A}_{1}{X}^{2}+{A}_{0}X+{B}_{0},$ (12)

${A}_{2}=±\sqrt{3A},\text{\hspace{0.17em}}{A}_{1}=0,\text{\hspace{0.17em}}{A}_{0}=±\sqrt{C},\text{\hspace{0.17em}}{B}_{0}=0.$ (13)

$±\frac{1}{3}\sqrt{3A}{X}^{3}±\sqrt{C}X+Y=0.$ (14)

${X}^{\prime }=±\frac{1}{3}\sqrt{3A}{X}^{3}±\sqrt{C}X.$ (15)

$X=±\frac{\sqrt{\sqrt{C}{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}\left(3-3\sqrt{3A}{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}\right)}}{-1+\sqrt{3A}{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}},$ (16)

$X=±\frac{\sqrt{3\sqrt{C}{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}-3\sqrt{3AC}}}{\sqrt{3A}-{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}},$ (17)

$u\left(x,t\right)=±\left(\frac{\sqrt{\sqrt{C}{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}\left(3-3\sqrt{3A}{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}\right)}}{-1+\sqrt{3A}{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}}\right){\text{e}}^{i\eta },$ (18)

$u\left(x,t\right)=±\left(\frac{\sqrt{3\sqrt{C}{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}-3\sqrt{3AC}}}{\sqrt{3A}-{\text{e}}^{2\sqrt{C}\left(\xi +{\xi }_{01}\right)}}\right){\text{e}}^{i\eta }.$ (19)

2.2.2. 情况二

$m=2$ 时，即

$P\left(X,Y\right)={a}_{0}\left(X\right)+{a}_{1}\left(X\right)Y+{a}_{2}\left(X\right){Y}^{2}.$ (20)

${{a}^{\prime }}_{2}\left(X\right)={a}_{2}\left(X\right)h\left(X\right),$ (21)

${{a}^{\prime }}_{1}\left(X\right)={a}_{2}\left(X\right)g\left(X\right)+{a}_{1}\left(X\right)h\left(X\right),$ (22)

${{a}^{\prime }}_{0}\left(X\right)+2{a}_{2}\left(X\right){Y}^{\prime }={a}_{1}\left(X\right)g\left(X\right)+{a}_{0}\left(X\right)h\left(X\right),$ (23)

${a}_{1}\left(X\right){Y}^{\prime }={a}_{0}\left(X\right)g\left(X\right)={a}_{1}\left(X\right)\left(A{X}^{5}+B{X}^{3}+CX\right).$ (24)

1) 当 ${a}_{1}\left(X\right)=0,g\left(X\right)=0$ 时，由(22)和(24)知，满足条件。又由(23)知，

${a}_{0}\left(X\right)=-\frac{1}{3}A{X}^{6}-\frac{1}{2}B{X}^{4}-C{X}^{2}+{A}_{0}$${A}_{0}$ 为任意常数。

${a}_{0}\left(X\right),{a}_{1}\left(X\right),{a}_{2}\left(X\right)$ 代入(6)式得 ${Y}^{2}=\frac{1}{3}A{X}^{6}+\frac{1}{2}B{X}^{4}+C{X}^{2}-{A}_{0}$ 。即

${X}^{\prime }=±\sqrt{\frac{1}{3}A{X}^{6}+\frac{1}{2}B{X}^{4}+C{X}^{2}-{A}_{0}}.$ (25)

${\Phi }^{\prime }=2\sqrt{\Phi \left(\frac{1}{3}A{\Phi }^{3}+\frac{1}{2}B{\Phi }^{2}+C\Phi -{A}_{0}\right)}.$ (26)

$f\left(\Phi \right)=\frac{1}{3}A{\Phi }^{3}+\frac{1}{2}B{\Phi }^{2}+C\Phi -{A}_{0}$ ，于是根据三次方程的判别式D及盛金公式，有以下几种情况：

a) 当 ${B}^{2}=4AC,{A}_{0}=-\frac{BC}{6A}$ 时，方程有一个三重根 $\frac{B}{2A}$ ，则 $f\left(\Phi \right)=\frac{1}{3}A{\left(\Phi -\frac{B}{2A}\right)}^{3}$

${\Phi }^{\prime }=2\sqrt{\frac{A}{3}\Phi {\left(\Phi -\frac{B}{2A}\right)}^{3}}$ 。那么此时， $\Phi =-\frac{{B}^{3}\left(\xi +{\xi }_{02}\right)}{24{A}^{2}-2A{B}^{2}\left(\xi +{\xi }_{02}\right)}$ ，其中 ${\xi }_{02}$ 为任意常数。而 $X=±\sqrt{\Phi }$

$u\left(x,t\right)=±{\left(-\frac{{B}^{3}\left(\xi +{\xi }_{02}\right)}{24{A}^{2}-2A{B}^{2}\left(\xi +{\xi }_{02}\right)}\right)}^{\frac{1}{2}}{\text{e}}^{i\eta }.$ (27)

b) 当 $\Delta <0$ 时，方程有三个不相等的实根 ${\Phi }_{1},{\Phi }_{2},{\Phi }_{3}$ ，于是 $f\left(\Phi \right)=\frac{1}{3}A\left(\Phi -{\Phi }_{1}\right)\left(\Phi -{\Phi }_{2}\right)\left(\Phi -{\Phi }_{3}\right)$ ，而 ${\Phi }^{\prime }=2\sqrt{\frac{A}{3}\Phi \left(\Phi -{\Phi }_{1}\right)\left(\Phi -{\Phi }_{2}\right)\left(\Phi -{\Phi }_{3}\right)}$ ，那么此时， $\Phi =\frac{{\Phi }_{2}{\Phi }_{3}s{n}^{2}\left({\omega }_{1}\left(\xi +{\xi }_{03}\right),{k}_{1}\right)}{\left({\Phi }_{2}-{\Phi }_{3}\right)+{\Phi }_{3}s{n}^{2}\left({\omega }_{1}\left(\xi +{\xi }_{03}\right),{k}_{1}\right)}$ ，其中 ${\xi }_{03}$ 为任意常数， ${\omega }_{1}^{2}=\frac{A}{3}{\Phi }_{1}\left({\Phi }_{3}-{\Phi }_{2}\right)$${k}_{1}^{2}=\frac{\left({\Phi }_{1}-{\Phi }_{2}\right){\Phi }_{3}}{\left({\Phi }_{3}-{\Phi }_{2}\right){\Phi }_{1}}$

$u\left(x,t\right)=±{\left(\frac{{\Phi }_{2}{\Phi }_{3}s{n}^{2}\left({\omega }_{1}\left(\xi +{\xi }_{03}\right),{k}_{1}\right)}{\left({\Phi }_{2}-{\Phi }_{3}\right)+{\Phi }_{3}s{n}^{2}\left({\omega }_{1}\left(\xi +{\xi }_{03}\right),{k}_{1}\right)}\right)}^{\frac{1}{2}}{\text{e}}^{i\eta }.$ (28)

c) 当 $\Delta =0$ 时，方程有三个实根，其中有一个两重根，分别为 ${\Phi }_{4},{\Phi }_{5}$ (为两重根)。于是

$f\left(\Phi \right)=\frac{1}{3}A\left(\Phi -{\Phi }_{4}\right){\left(\Phi -{\Phi }_{5}\right)}^{2}$${\Phi }^{\prime }=2\left(\Phi -{\Phi }_{5}\right)\sqrt{\frac{A}{3}\Phi \left(\Phi -{\Phi }_{4}\right)}$ ，此时， $\Phi =-\frac{{\Phi }_{5}{\left({\text{e}}^{\rho \left(\xi +{\xi }_{04}\right)}-A{\Phi }_{4}\right)}^{2}}{4{\Phi }_{5}{\text{e}}^{\rho \left(\xi +{\xi }_{04}\right)}-{\left({\text{e}}^{\rho \left(\xi +{\xi }_{04}\right)}+A{\Phi }_{4}\right)}^{2}}$ ，其中 ${\xi }_{04}$ 为任意常数， $\rho =-\frac{2}{3}\sqrt{3A{\Phi }_{5}\left({\Phi }_{5}-{\Phi }_{4}\right)}$

$u\left(x,t\right)=±{\left(-\frac{{\Phi }_{5}{\left({\text{e}}^{\rho \left(\xi +{\xi }_{04}\right)}-A{\Phi }_{4}\right)}^{2}}{4{\Phi }_{5}{\text{e}}^{\rho \left(\xi +{\xi }_{04}\right)}-{\left({\text{e}}^{\rho \left(\xi +{\xi }_{04}\right)}+A{\Phi }_{4}\right)}^{2}}\right)}^{\frac{1}{2}}{\text{e}}^{i\eta }.$ (29)

d) 当 $\Delta >0$ 时，方程有一个实根 ${\Phi }_{6}$ 和一对共轭虚根 $\Phi =b±ai$ 。于是

$f\left(\Phi \right)=\frac{1}{3}A\left(\Phi -{\Phi }_{6}\right)\left[{\left(\Phi -b\right)}^{2}+{a}^{2}\right]$${\Phi }^{\prime }=2\sqrt{\frac{A}{3}\Phi \left(\Phi -{\Phi }_{6}\right)\left[{\left(\Phi -b\right)}^{2}+{a}^{2}\right]}$ ，此时， $\Phi =\frac{{\Phi }_{6}N\left(1-cn\left({\omega }_{2}\left(\xi +{\xi }_{05}\right),k\right)\right)}{\left(M+N\right)-\left(M+N\right)cn\left({\omega }_{2}\left(\xi +{\xi }_{05}\right),k\right)}$ ，其中 ${\xi }_{05}$ 为任意常数， ${M}^{2}={\left({\Phi }_{6}-b\right)}^{2}+{a}^{2}$${N}^{2}={a}^{2}+{b}^{2}$

(30)

2) 当时，只有当时不会产生矛盾。不

a)，且

(31)

b)，且

(32)

c)，且

，即，解常微分方程可得，其中

(33)

3. 结论

The First Integral Method for Solving Exact Solution of GDNLSE[J]. 应用数学进展, 2018, 07(04): 303-309. https://doi.org/10.12677/AAM.2018.74036

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NOTES

*通讯作者。