Vol.07 No.07(2018), Article ID:26153,8 pages
10.12677/AAM.2018.77113

The First Integral Method for Solving Exact Solutions of Two Nonlinear Schrodinger Equations

Qingmei Zhang, Mei Xiong, Longwei Chen*

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

Received: Jul. 5th, 2018; accepted: Jul. 23rd, 2018; published: Jul. 30th, 2018

ABSTRACT

The first integral method proposed by Feng is very reliable integral method for solving nonlinear partial differential equations, which is based on the ring theory of commutative algebra. In this paper, exact travelling wave solutions of the generalized nonlinear Schrodinger equation and the high order dispersion nonlinear Schrodinger equation are studied by using the first integral method. By introducing the travelling wave transformations, two nonlinear Schrodinger equations have been transformed into ordinary differential equations. Then according to the division theorem of polynomial, exact travelling wave solutions of two nonlinear Schrodinger equations are obtained.

Keywords:The First Integral Method, The Generalized Nonlinear Schrodinger Equation, The High Order Dispersion Nonlinear Schrodinger Equation, Travelling Wave Solutions

1. 引言

$i{u}_{t}-{r}_{2}{u}_{xx}+{c}_{3}{|u|}^{2}u=i{\left[\left({s}_{0}+{s}_{2}{|u|}^{2}\right)u\right]}_{x}-{c}_{5}{|u|}^{4}u,$ (1)

$i{q}_{t}-a{q}_{xx}-b{q}_{xxxx}+c\left({|q|}^{2}+d{|q|}^{4}\right)q=0.$ (2)

2. 首次积分法的基本介绍

$P\left(u,{u}_{t},{u}_{x},{u}_{xx},{u}_{tt},{u}_{xt},{u}_{xxx},\cdots \right)=0.$ (3)

$u\left(x,t\right)=u\left(\xi \right),\text{\hspace{0.17em}}\xi =x-ct.$ (4)

$\frac{\partial }{\partial t}\left(.\right)=-c\frac{\partial }{\partial \xi }\left(.\right),\text{\hspace{0.17em}}\frac{\partial }{\partial x}\left(.\right)=\frac{\partial }{\partial \xi }\left(.\right),\text{\hspace{0.17em}}\frac{{\partial }^{2}}{\partial {t}^{2}}\left(.\right)={c}^{2}\frac{{\partial }^{2}}{\partial {\xi }^{2}}\left(.\right),\cdots$ (5)

$Q\left(u,{u}_{\xi },{u}_{\xi \xi },\cdots \right)=0.$ (6)

$u\left(x,t\right)=u\left(\xi \right).$ (7)

$X\left(\xi \right)=u\left(\xi \right),\text{\hspace{0.17em}}Y\left(\xi \right)={u}_{\xi }\left(\xi \right).$ (8)

$\left\{\begin{array}{l}{X}_{\xi }\left(\xi \right)=Y\left(\xi \right),\hfill \\ {Y}_{\xi }\left(\xi \right)=F\left(X\left(\xi \right),Y\left(\xi \right)\right).\hfill \end{array}$ (9)

3. 应用

$u\left(x,t\right)=\phi \left(\xi \right){\text{e}}^{i\left(x-\omega t\right)}.$ (10)

$\left(2{r}_{2}-\lambda -{s}_{0}\right){\phi }^{\prime }-3{s}_{2}{\phi }^{2}{\phi }^{\prime }=0,$ (11)

${r}_{2}{\phi }^{″}+\left(\omega +{s}_{0}-{r}_{2}\right)\phi -\left({c}_{3}+{s}_{2}\right){\phi }^{3}+{c}_{5}{\phi }^{5}=0.$ (12)

${\phi }^{″}+\frac{\omega +{r}_{2}-\lambda }{{r}_{2}}\phi +\frac{{c}_{3}}{{r}_{2}}{\phi }^{3}+\frac{{c}_{5}}{{r}_{2}}{\phi }^{5}=0.$ (13)

$\left\{\begin{array}{l}{X}^{\prime }\left(\xi \right)=Y\left(\xi \right),\hfill \\ {Y}^{\prime }\left(\xi \right)={k}_{1}X\left(\xi \right)+{k}_{2}X{\left(\xi \right)}^{3}+{k}_{3}X{\left(\xi \right)}^{5}.\hfill \end{array}$ (14)

$q\left(x,t\right)=\varphi \left(\xi \right){\text{e}}^{i\left(kx-\omega t\right)}.$ (15)

$-4bk{\beta }^{3}{\varphi }^{″}+\left(-\lambda +2a\beta k+4b\beta {k}^{3}\right){\varphi }^{\prime }=0.$ (16)

$-b{\beta }^{4}{\varphi }^{\left(4\right)}+\left(a{\beta }^{2}+b{\beta }^{2}{k}^{2}\right){\varphi }^{″}+\left(\omega -a{k}^{2}-b{k}^{4}\right)\varphi +c{\varphi }^{3}+dc{\varphi }^{5}=0.$ (17)

${\varphi }^{″}-\frac{4bk{p}_{3}{\beta }^{3}}{{p}_{1}b{\beta }^{4}-4{p}_{2}bk{\beta }^{3}}\varphi -\frac{4bck{\beta }^{3}}{{p}_{1}b{\beta }^{4}-4{p}_{2}bk{\beta }^{3}}{\varphi }^{3}-\frac{4bcdk{\beta }^{3}}{{p}_{1}b{\beta }^{4}-4{p}_{2}bk{\beta }^{3}}{\varphi }^{5}=0.$ (18)

$\left\{\begin{array}{l}{X}^{\prime }\left(\xi \right)=Y\left(\xi \right),\hfill \\ {Y}^{\prime }\left(\xi \right)=\frac{4bk{p}_{3}{\beta }^{3}}{{p}_{1}b{\beta }^{4}-4{p}_{2}bk{\beta }^{3}}X\left(\xi \right)+\frac{4bck{\beta }^{3}}{{p}_{1}b{\beta }^{4}-4{p}_{2}bk{\beta }^{3}}X{\left(\xi \right)}^{3}+\frac{4bcdk{\beta }^{3}}{{p}_{1}b{\beta }^{4}-4{p}_{2}bk{\beta }^{3}}X{\left(\xi \right)}^{5}.\hfill \end{array}$ (19)

${k}_{1}=\frac{4bk{p}_{3}{\beta }^{3}}{{p}_{1}b{\beta }^{4}-4{p}_{2}bk{\beta }^{3}},{k}_{2}=\frac{4bck{\beta }^{3}}{{p}_{1}b{\beta }^{4}-4{p}_{2}bk{\beta }^{3}},{k}_{3}=\frac{4bcdk{\beta }^{3}}{{p}_{1}b{\beta }^{4}-4{p}_{2}bk{\beta }^{3}}$ 方程(19)和(14)有相同的首次积分，要找到它们的精确解，我们仅需讨论其中一个方程即可，下面我们将给出对方程(14)的求解过程。

$P\left(X,Y\right)=\underset{i=0}{\overset{m}{\sum }}{a}_{i}\left(X\right){Y}^{i}.$ (20)

$P\left(X\left(\xi \right),Y\left(\xi \right)\right)=\underset{i=0}{\overset{m}{\sum }}{a}_{i}\left(X\left(\xi \right)\right)Y{\left(\xi \right)}^{i}=0.$ (21)

${\frac{\text{d}P}{\text{d}\xi }|}_{\left(20\right)}={\left(\frac{\text{d}P}{\text{d}X}\frac{\text{d}X}{\text{d}\xi }+\frac{\text{d}P}{\text{d}Y}\frac{\text{d}Y}{\text{d}\xi }\right)|}_{\left(20\right)}=\left(h\left(X\right)+g\left(X\right)Y\right)\left(\underset{i=0}{\overset{m}{\sum }}{a}_{i}\left(X\right){Y}^{i}\right).$ (22)

3.1. 情形一

$\underset{i=0}{\overset{1}{\sum }}{{a}^{\prime }}_{i}\left(X\right){Y}^{i+1}+\underset{i=0}{\overset{1}{\sum }}i{a}_{i}\left(X\right){Y}^{i-1}\left({Y}^{\prime }\left(\xi \right)\right)=\left(h\left(X\right)+g\left(X\right)Y\right)\left(\underset{i=0}{\overset{1}{\sum }}{a}_{i}\left(X\right){Y}^{i}\right).$ (23)

${{a}^{\prime }}_{1}\left(X\right)=g\left(X\right){a}_{1}\left(X\right),$ (24)

${{a}^{\prime }}_{0}\left(X\right)=h\left(X\right){a}_{1}\left(X\right)+g\left(X\right){a}_{0}\left(X\right),$ (25)

${a}_{1}\left(X\right)\left({k}_{1}X+{k}_{2}{X}^{3}+{k}_{5}{X}^{5}\right)=h\left(X\right){a}_{0}\left(X\right).$ (26)

${a}_{0}\left(X\right)=\frac{1}{3}A{X}^{3}+\frac{1}{2}B{X}^{2}+CX+D.$ (27)

$\left\{\begin{array}{l}\begin{array}{l}\frac{1}{3}{A}^{2}={k}_{3}\hfill \\ \frac{5}{6}AB=0\hfill \\ \frac{4}{3}AC+\frac{1}{2}{B}^{2}={k}_{2}\hfill \end{array}\hfill \\ AD+\frac{3}{2}BC=0\hfill \\ BD+{C}^{2}={k}_{1}\hfill \\ CD=0\hfill \end{array}$ (28)

$D=0,B=0,A=±\sqrt{3{k}_{3}},C=±\sqrt{{k}_{1}},$ (29)

${k}_{1}{k}_{3}=\frac{3}{16}{k}_{2}^{2}.$ (30)

$Y=-\frac{1}{3}A{X}^{3}-CX.$ (31)

${\phi }^{\prime }\left(\xi \right)=-\frac{1}{3}A\phi {\left(\xi \right)}^{3}-C\phi \left(\xi \right).$ (32)

${Z}^{\prime }=aZ+b{Z}^{p},$ (33)

$Z\left(\xi \right)={\left[\frac{-a/b}{{\xi }_{{}_{0}}{\text{e}}^{a\left(1-p\right)}+1}\right]}^{\frac{1}{p-1}},$ (34)

$=\left\{\begin{array}{l}{\left\{\frac{-a}{2b}\left[1+\mathrm{tanh}\left(\frac{a\left(p-1\right)}{2}\xi -\frac{\mathrm{ln}{\xi }_{0}}{2}\right)\right]\right\}}^{\frac{1}{p-1}}\text{if}\text{\hspace{0.17em}}{\xi }_{0}>0,\hfill \\ {\left\{\frac{-a}{2b}\left[1+\mathrm{coth}\left(\frac{a\left(p-1\right)}{2}\xi -\frac{\mathrm{ln}\left(-{\xi }_{0}\right)}{2}\right)\right]\right\}}^{\frac{1}{p-1}}\text{if}\text{\hspace{0.17em}}{\xi }_{0}<0,\hfill \\ {\left\{\frac{-a}{b}\right\}}^{\frac{1}{p-1}}\text{}\text{if}\text{\hspace{0.17em}}{\xi }_{0}=0.\hfill \end{array}$ (35)

$u\left(\xi \right)=\left\{\begin{array}{l}{\left\{±\frac{1}{2}\sqrt{\frac{3{k}_{1}}{{k}_{3}}}\left[1+\mathrm{tanh}\left(\sqrt{{k}_{1}}\xi -\frac{\mathrm{ln}{\xi }_{0}}{2}\right)\right]\right\}}^{\frac{1}{2}}\text{if}\text{\hspace{0.17em}}{\xi }_{0}>0,\hfill \\ {\left\{±\frac{1}{2}\sqrt{\frac{3{k}_{1}}{{k}_{3}}}\left[1+\mathrm{coth}\left(\sqrt{{k}_{1}}\xi -\frac{\mathrm{ln}\left(-{\xi }_{0}\right)}{2}\right)\right]\right\}}^{\frac{1}{2}}\text{if}\text{\hspace{0.17em}}{\xi }_{0}<0,\hfill \\ {\left\{±\sqrt{\frac{3{k}_{1}}{{k}_{3}}}\right\}}^{\frac{1}{2}}\text{if}\text{\hspace{0.17em}}{\xi }_{0}=0.\hfill \end{array}$ (36)

$u\left(x,t\right)=\left\{\begin{array}{l}{\left\{±\frac{1}{2}\sqrt{\frac{3{k}_{1}}{{k}_{3}}}\left[1+\mathrm{tanh}\left(\sqrt{{k}_{1}}\left(x-\lambda t\right)-\frac{\mathrm{ln}{\xi }_{0}}{2}\right)\right]\right\}}^{\frac{1}{2}}{\text{e}}^{\left(x-\omega t\right)}\text{if}\text{\hspace{0.17em}}{\xi }_{0}>0,\hfill \\ {\left\{±\frac{1}{2}\sqrt{\frac{3{k}_{1}}{{k}_{3}}}\left[1+\mathrm{coth}\left(\sqrt{{k}_{1}}\left(x-\lambda t\right)-\frac{\mathrm{ln}\left(-{\xi }_{0}\right)}{2}\right)\right]\right\}}^{\frac{1}{2}}{\text{e}}^{\left(x-\omega t\right)}\text{if}\text{\hspace{0.17em}}{\xi }_{0}<0,\hfill \\ {\left\{±\sqrt{\frac{3{k}_{1}}{{k}_{3}}}\right\}}^{\frac{1}{2}}{\text{e}}^{\left(x-\omega t\right)}\text{if}\text{\hspace{0.17em}}{\xi }_{0}=0.\hfill \end{array}$ (37)

3.2. 情形二

${{a}^{\prime }}_{2}\left(X\right)=g\left(X\right){a}_{2}\left(X\right),$ (38)

${{a}^{\prime }}_{1}\left(X\right)=h\left(X\right){a}_{2}\left(X\right)+g\left(X\right){a}_{1}\left(X\right),$ (39)

${{a}^{\prime }}_{0}\left(X\right)+2{a}_{2}\left(X\right)\left({k}_{1}X+{k}_{2}{X}^{3}+{k}_{5}{X}^{5}\right)=h\left(X\right){a}_{1}\left(X\right)+g\left(X\right){a}_{0}\left(X\right),$ (40)

${a}_{1}\left(X\right)\left({k}_{1}X+{k}_{2}{X}^{3}+{k}_{5}{X}^{5}\right)=h\left(X\right){a}_{0}\left(X\right).$ (41)

${a}_{1}\left(X\right)=\frac{1}{3}A{X}^{3}+\frac{1}{2}B{X}^{2}+CX+D.$ (42)

$\begin{array}{c}{a}_{0}\left(X\right)=\left(\frac{1}{18}{A}^{2}-\frac{1}{3}{k}_{3}\right){X}^{6}+\left(\frac{1}{6}AB\right){X}^{5}+\left(\frac{1}{3}AC+\frac{1}{8}{B}^{2}+\frac{1}{2}{k}_{2}\right){X}^{4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(\frac{1}{3}AD+\frac{1}{2}BC\right){X}^{3}+\left(\frac{1}{2}BD+\frac{1}{2}{C}^{2}-{k}_{1}\right){X}^{2}+DCX+E.\end{array}$ (43)

$\left\{\begin{array}{l}\begin{array}{l}\begin{array}{l}\frac{1}{18}{A}^{3}=\frac{2}{3}A{k}_{3}\hfill \\ \frac{2}{9}{A}^{2}B=\frac{5}{6}B{k}_{3}\hfill \\ \frac{7}{18}{A}^{2}C+\frac{7}{24}A{B}^{2}=\frac{5}{6}A{k}_{2}+\frac{4}{3}C{k}_{3}\hfill \\ \frac{1}{3}{A}^{2}D+ABC+\frac{1}{8}{B}^{2}=B{k}_{2}+D{k}_{3}\hfill \end{array}\hfill \\ \frac{7}{6}ABD+\frac{5}{6}A{C}^{2}+\frac{1}{8}{B}^{2}C=\frac{4}{3}A{k}_{1}+\frac{3}{2}C{k}_{2}\hfill \\ \frac{4}{3}ACD+\frac{1}{2}D{B}^{2}+B{C}^{2}=\frac{3}{2}B{k}_{1}+2D{k}_{2}\hfill \end{array}\hfill \\ \frac{3}{2}BCD+\frac{1}{2}{C}^{2}+AE=2C{k}_{1}\hfill \\ D{C}^{2}+BE=D{k}_{1}\hfill \\ EC=0,\hfill \end{array}$ (44)

$D=0,B=0,A=±2\sqrt{3{k}_{3}},C=±2\sqrt{{k}_{1}},$ (45)

${k}_{1}{k}_{3}=\frac{3}{16}{k}_{2}^{2}.$ (46)

$Y=-\frac{1}{6}A{X}^{3}-\frac{1}{2}CX.$ (47)

${\phi }^{\prime }\left(\xi \right)=-\frac{1}{6}A\phi {\left(\xi \right)}^{3}-\frac{1}{2}C\phi \left(\xi \right).$ (48)

4. 总结

The First Integral Method for Solving Exact Solutions of Two Nonlinear Schrodinger Equations[J]. 应用数学进展, 2018, 07(07): 962-969. https://doi.org/10.12677/AAM.2018.77113

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