﻿ (3+1)维Mel’nikov方程的Lump解研究 Lump Solutions to the (3+1)-Dimensional Mel’nikov Equation

Vol. 08  No. 06 ( 2019 ), Article ID: 30637 , 6 pages
10.12677/AAM.2019.86121

Lump Solutions to the (3+1)-Dimensional Mel’nikov Equation

Xiaoyu Li, Xuelin Yong*, Yehui Huang

School of Mathematical Sciences and Physics, North China Electric Power University, Beijing

Received: May 6th, 2019; accepted: May 28th, 2019; published: Jun. 4th, 2019

ABSTRACT

In this article, lump solutions of the (3+1)-dimensional Mel’nikov are obtained via the Hirota bilinear method and symbolic computation with Maple. A class of lump solutions rationally localized in all directions in the space is obtained. And we propose the conditions for the analyticity and rational localization of the lump solutions. By selecting special value of parameters involved, the dynamic characteristics of the solutions are illustrated.

Keywords:Lump Solution, (3+1)-Dimensional Mel’nikov, Hirota Bilinear Method

(3+1)维Mel’nikov方程的Lump解研究

1. 引言

$\left\{\begin{array}{l}{\left({u}_{t}+6u{u}_{x}+{u}_{xxx}+8{|\varphi |}_{x}^{2}\right)}_{x}-{u}_{yy}+{u}_{zz}=0,\\ i{\varphi }_{y}=2{\varphi }_{xx}+2u\varphi ,\\ i{\varphi }_{z}={\varphi }_{xx}+u\varphi ,\end{array}$ (1)

2. Mel’nikov的Lump解

$\left\{\begin{array}{l}u=2{\left(\mathrm{ln}f\right)}_{xx},\\ \varphi =\frac{g}{f},\end{array}$ (2)

$\left\{\begin{array}{l}\left({D}_{x}{D}_{t}+{D}_{x}^{4}-{D}_{y}^{2}+{D}_{z}^{2}\right)f\cdot f+8\left(g\stackrel{¯}{g}-{f}^{2}\right)=0,\\ \left(i{D}_{y}-2{D}_{x}^{2}\right)g\cdot f=0,\\ \left(i{D}_{z}-{D}_{x}^{2}\right)g\cdot f=0.\end{array}$ (3)

${D}_{x}^{m}{D}_{t}^{n}\left(a,b\right)={{\left(\frac{\partial }{\partial x}-\frac{\partial }{\partial {x}^{\prime }}\right)}^{m}{\left(\frac{\partial }{\partial t}-\frac{\partial }{\partial {t}^{\prime }}\right)}^{n}a\left(x,t\right)b\left({x}^{\prime },{t}^{\prime }\right)|}_{{x}^{\prime }=x,{t}^{\prime }=t}$ (4)

$\left\{\begin{array}{l}f=1+{s}^{2}+{h}^{2},\\ {g}_{R}={b}_{0}+{b}_{1}s+{b}_{2}h+{b}_{3}{s}^{2}+{b}_{4}{h}^{2},\\ {g}_{I}={c}_{0}+{c}_{1}s+{c}_{2}h+{c}_{3}{s}^{2}+{c}_{4}{h}^{2},\end{array}$ (5)

$\left\{\begin{array}{l}s={a}_{1}x+{a}_{2}y+{a}_{3}z+{a}_{4}t+{a}_{5},\\ h={a}_{6}x+{a}_{7}y+{a}_{8}z+{a}_{9}t+{a}_{10},\end{array}$ (6)

$\begin{array}{l}\left\{{a}_{2}=2{a}_{3},{a}_{4}=\frac{\left({a}_{3}^{2}-{a}_{1}^{4}\right)\left[3{\left({a}_{1}^{4}+{a}_{3}^{2}\right)}^{2}+16{a}_{1}^{4}\right]}{{a}_{1}{\left({a}_{1}^{4}+{a}_{3}^{2}\right)}^{2}},{a}_{6}=0,{a}_{7}=2{a}_{1}^{2},{a}_{8}=\frac{1}{2}{a}_{7},{a}_{9}=\frac{2{a}_{1}{a}_{3}\left[3{\left({a}_{1}^{4}+{a}_{3}^{2}\right)}^{2}-16{a}_{1}^{4}\right]}{{\left({a}_{1}^{4}+{a}_{3}^{2}\right)}^{2}},\\ {b}_{0}=\frac{{b}_{3}\left({a}_{3}^{2}-3{a}_{1}^{4}\right)}{{a}_{1}^{4}+{a}_{3}^{2}},{b}_{1}=k{c}_{1},{b}_{2}=k{c}_{2},{b}_{4}={b}_{3},{c}_{0}=-k{b}_{0},{c}_{1}=-\frac{4{a}_{3}{b}_{3}{a}_{1}^{2}}{{a}_{1}^{4}+{a}_{3}^{2}},{c}_{2}=\frac{4{a}_{1}^{4}{b}_{3}}{{a}_{1}^{4}+{a}_{3}^{2}},{c}_{3}=-k{b}_{3},{c}_{4}={c}_{3}\right\}\end{array}$ (7)

${b}_{3}^{2}\left(1+{k}^{2}\right)=1$ (8)

$\left\{\begin{array}{l}u=\frac{4{a}_{1}^{2}\left(1-{s}^{2}+{h}^{2}\right)}{{\left(1+{s}^{2}+{h}^{2}\right)}^{2}},\\ \varphi =\frac{{g}_{R}+i{g}_{I}}{1+{s}^{2}+{h}^{2}},\end{array}$ (9)

$\begin{array}{l}{g}_{R}=\frac{{b}_{3}\left({a}_{3}^{2}-3{a}_{1}^{4}\right)}{{a}_{1}^{4}\text{+}{a}_{3}^{2}}-k\frac{4{a}_{3}{b}_{3}{a}_{1}^{2}}{{a}_{1}^{4}\text{+}{a}_{3}^{2}}s+k\frac{4{a}_{1}^{4}{b}_{3}}{{a}_{1}^{4}\text{+}{a}_{3}^{2}}h+{b}_{3}\left({s}^{2}+{h}^{2}\right),\\ {g}_{I}=-k\frac{{b}_{3}\left({a}_{3}^{2}-3{a}_{1}^{4}\right)}{{a}_{1}^{4}\text{+}{a}_{3}^{2}}-\frac{4{a}_{3}{b}_{3}{a}_{1}^{2}}{{a}_{1}^{4}\text{+}{a}_{3}^{2}}s+\frac{4{a}_{1}^{4}{b}_{3}}{{a}_{1}^{4}\text{+}{a}_{3}^{2}}h-k{b}_{3}\left({s}^{2}+{h}^{2}\right),\end{array}$ (10)

$\begin{array}{l}s={a}_{1}x+2{a}_{3}y+{a}_{3}z+\frac{\left({a}_{3}^{2}-{a}_{1}^{4}\right)\left[3{\left({a}_{1}^{4}+{a}_{3}^{2}\right)}^{2}\text{+}16{a}_{1}^{4}\right]}{{a}_{1}{\left({a}_{1}^{4}+{a}_{3}^{2}\right)}^{2}}t+{a}_{5},\\ h=2{a}_{1}^{2}y+{a}_{1}^{2}z+\frac{2{a}_{1}{a}_{3}\left[3{\left({a}_{1}^{4}+{a}_{3}^{2}\right)}^{2}-16{a}_{1}^{4}\right]}{{\left({a}_{1}^{4}+{a}_{3}^{2}\right)}^{2}}t+{a}_{10}.\end{array}$ (11)

Figure 1. Plots of lump solution for u with ${a}_{1}=1,\text{\hspace{0.17em}}{a}_{3}=2,\text{\hspace{0.17em}}{b}_{3}=-0.8,\text{\hspace{0.17em}}{a}_{5}={a}_{10}=0,\text{\hspace{0.17em}}k=0.75$ when $x=0,\text{\hspace{0.17em}}z=0$

Figure 2. Plots of lump solution for ${|\varphi |}^{2}$ with ${a}_{1}=1,\text{\hspace{0.17em}}{a}_{3}=2,\text{\hspace{0.17em}}{b}_{3}=-0.8,\text{\hspace{0.17em}}{a}_{5}={a}_{10}=0,\text{\hspace{0.17em}}k=0.75$ when $x=0,\text{\hspace{0.17em}}z=0$

Figure 3. Plots of lump solution for u with ${a}_{1}=2,\text{\hspace{0.17em}}{a}_{3}=-2,\text{\hspace{0.17em}}{b}_{3}=0.8,\text{\hspace{0.17em}}{a}_{5}={a}_{10}=0,\text{\hspace{0.17em}}k=0.75$ when $y=0,\text{\hspace{0.17em}}t=0$

Figure 4. Plots of lump solution for ${|\varphi |}^{2}$ with ${a}_{1}=2,\text{\hspace{0.17em}}{a}_{3}=-2,\text{\hspace{0.17em}}{b}_{3}=0.8,\text{\hspace{0.17em}}{a}_{5}={a}_{10}=0,\text{\hspace{0.17em}}k=0.75$ when $y=0,\text{\hspace{0.17em}}t=0$

3. 结论

Lump Solutions to the (3+1)-Dimensional Mel’nikov Equation[J]. 应用数学进展, 2019, 08(06): 1058-1063. https://doi.org/10.12677/AAM.2019.86121

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33. NOTES

*通讯作者。