(2 + 1)维色散的长波方程的新精确解 New Exact Solutions for (2 + 1)-Dimensional Dispersive Long Wave Equation

doi:10.12677/PM.2018.86079

Pure Mathematics
Vol. 08 No. 06 ( 2018 ), Article ID: 27465 , 7 pages
10.12677/PM.2018.86079

New Exact Solutions for (2 + 1)-Dimensional Dispersive Long Wave Equation

Haiming Fu

●How to Cite this Article

Department of Basic Courses, Guangzhou Hua Xia Vocational College, Guangzhou Guangdong

Received: Oct. 15^th, 2018; accepted: Oct. 27^th, 2018; published: Nov. 8^th, 2018

ABSTRACT

The algebraic method, based on the symbolic computation, has been applied to study new traveling wave solutions for (2 + 1)-dimensional dispersive long wave equations by means of Epsilon package in Maple. More new explicit travelling wave solutions are obtained, which contain solitons, hyperbola function solutions and triangular periodic solutions.

Keywords:(2 + 1)-Dimensional Dispersive Long Wave Equations, F-Expansion Method, Hyperbola Function Solutions, Triangular Periodic Solutions

(2 + 1)维色散的长波方程的新精确解

傅海明

广州华夏职业学院基础部，广东广州

收稿日期：2018年10月15日；录用日期：2018年10月27日；发布日期：2018年11月8日

摘要

利用一种基于符号计算的代数方法，结合Maple环境中的Epsilon软件包，求解(2 + 1)维色散的长波方程，获得了若干其它方法不曾给出的形式更为丰富的新的显式行波解，其中包括孤波解、三角函数解、双曲函数解。

关键词 :(2 + 1)维色散的长波方程，F-展开法，三角函数解，双曲函数解

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

1. 引言

孤立子的研究工作在全世界范围掀起了一股热潮，不仅仅在流体物理，固体物理，基本粒子物理和量子物理等领域中，在凝聚太物理，超导物理等领域中都开展了孤立子的研究。近年来，学者们发现了求解非线性偏微分方程的大量方法，这些方法主要有Banach不动点理论、逆散射法 [1] 、算子半群理论、Backlund法 [2] 、调和分析法、Darboux变换法 [3] 、Galerkin方法、Hirota双线性法 [4] [5] [6] 和Painlevé展开法 [7] 等。结合计算机数学软件(如MATLAB等)，又发展了许多新的求解非线性偏微分波方程的方法，例如双曲函数法 [8] 、代数几何法、齐次平衡法 [9] 、特殊包络变换法 [10] [11] 、双指数函数法 [12] 和F-展开法 [13] [14] [15] 等。

2. (2 + 1)维色散的长波方程的新精确解

为了求得(2 + 1)维色散的长波方程

${\begin{cases} u_{y t} + v_{x x} + \frac{1}{2} {(u^{2})}_{x y} = 0 \\ v_{t} + {(u v + u + u_{x y})}_{x} = 0 \end{cases},$ (1)

的行波解，我们先设该方程的有如下形式的行波变换：

$u (x, y, t) = u (ξ), v (x, y, t) = v (ξ) (ξ = k x + h y + w t + ξ_{0}),$ (2)

其中 $k, h, w$ 为代定实常数， $ξ_{0}$ 为任意实常数。把式(2)代入(2 + 1)维色散的长波方程，并且积分两次，积分常数设为零，可得

${\begin{cases} h w u + k^{2} v + \frac{1}{2} k h u^{2} = 0 \\ w v + k u v + k u + k^{2} h u_{ξ ξ} = 0 \end{cases} .$ (3)

设上面方程中的 $u (ξ)$ 和 $v (ξ)$ 为的有限项级数形式，如下：

$u (ξ) = \sum_{i = 1}^{n} a_{i} F^{i} (ξ) + a_{0}$ , $v (ξ) = \sum_{i = 1}^{m} b_{i} F^{i} (ξ) + b_{0},$ (4)

而 $F (ξ)$ 满足一阶非线性常微分方程：

${(F^{'})}^{2} = h_{6} F^{6} + h_{4} F^{4} + h_{2} F^{2} + h_{0}$ . (5)

平衡非线性常微分方程(3)中的 $u (ξ)$ 和最高阶导数项与最高次非线性项，得 $m = 2, n = 4$ ，即

(6)

把式(6)代入非线性常微分方程(3)，并利用常微分方程(5)可以得到关于 $F^{i} (ξ), (i = 0, 1, 2, \dots, 6)$ 的方程，设其各次方的系数为零，得到如下超越代数方程组：

$k h a_{2}^{2} + 2 k^{2} b_{4} = 0,$ (7)

$k h a_{1} a_{2} + k^{2} b_{3} = 0,$ (8)

$k h (a_{1}^{2} + 2 a_{0} a_{2}) + 2 k^{2} b_{2} + 2 h w a_{2} = 0,$ (9)

$k h a_{0} a_{1} + k^{2} b_{1} + h w a_{1} = 0,$ (10)

$k h a_{0}^{2} + 2 k^{2} b_{0} + 2 h w a_{0} = 0,$ (11)

$8 k^{2} h a_{2} h_{6} + k a_{2} b_{4} = 0,$ (12)

$3 k^{2} h a_{1} h_{6} + k (a_{1} b_{4} + a_{2} b_{3}) = 0,$ (13)

$6 k^{2} h a_{2} h_{4} + k (a_{0} b_{4} + a_{1} b_{3} + a_{2} b_{2}) + w b_{4} = 0,$ (14)

$2 k^{2} h a_{1} h_{4} + k (a_{0} b_{3} + a_{1} b_{2} + a_{2} b_{1}) + w b_{3} = 0,$ (15)

$4 k^{2} h a_{2} h_{2} + k (a_{0} b_{2} + a_{1} b_{1} + a_{2} b_{0}) + w b_{2} + k a_{2} = 0,$ (16)

$k^{2} h a_{1} h_{2} + k (a_{0} b_{1} + a_{1} b_{0}) + w b_{1} + k a_{1} = 0,$ (17)

$2 k^{2} h a_{2} h_{0} + k a_{0} b_{0} + w b_{0} + k a_{0} = 0,$ (18)

利用Maple求解以上超越代数方程组，得到下面这组解：

$a_{1} = 0, b_{1} = 0, b_{3} = 0, a_{2} = 4 k \sqrt{h_{6}}, a_{0} = k B, b_{4} = \frac{8 (b_{0} + 1) h_{6}^{2}}{4 h_{2} h_{6} - h_{4}^{2}}, b_{2} = \frac{4 (b_{0} + 1) h_{4} h_{6}}{4 h_{2} h_{6} - h_{4}^{2}},$

$h = \frac{(b_{0} + 1) h_{6}}{k (h_{4}^{2} - 4 h_{2} h_{6})}, w = \frac{k^{2}}{h_{6}} (h_{4} \sqrt{h_{6}} - B)$ , k为任意常数, (19)

而 $b_{0}$ 由下式给出：

$8 (b_{0} + 1) h_{0} h_{6}^{2} \sqrt{h_{6}} = (4 h_{2} h_{6} - h_{4}^{2}) (B + b_{0} h_{4} \sqrt{h_{6}}),$ (20)

其中， $B = h_{4} \sqrt{h_{6}} \pm \sqrt{\frac{h_{4}^{2} h_{6} (3 b_{0} + 1) - 8 b_{0} h_{2} h_{6}^{2}}{b_{0} + 1}}$ 。

情形1：当 $h_{2} > 0$ 时，非线性常微分方程(5)有如下双曲函数解： $F_{1} = {(\frac{- h_{2} h_{4} s e c h^{2} (\sqrt{h_{2}} ξ)}{h_{4}^{2} - h_{2} h_{6} {(1 \pm \tanh (\sqrt{h_{2}} ξ))}^{2}})}^{\frac{1}{2}}$ ， $F_{2} = {(\frac{h_{2} h_{4} c s c h^{2} (\sqrt{h_{2}} ξ)}{h_{4}^{2} - h_{2} h_{6} {(1 \pm \coth (\sqrt{h_{2}} ξ))}^{2}})}^{\frac{1}{2}}$ ， $F_{3} = 4 {(\frac{h_{2} \exp (\pm 2 \sqrt{h_{2}} ξ)}{{(\exp (\pm 2 \sqrt{h_{2}} ξ) - 4 h_{4})}^{2} - 64 h_{2} h_{6}})}^{\frac{1}{2}}$ 。将这些解代入式(6)，得到(2 + 1)维色散的长波方程的行波解为：

${\begin{array}{l} u_{1} = k B + 4 k \sqrt{h_{6}} \frac{- h_{2} h_{4} s e c h^{2} (\sqrt{h_{2}} ξ)}{h_{4}^{2} - h_{2} h_{6} {(1 \pm \tanh (\sqrt{h_{2}} ξ))}^{2}} \\ v_{1} = b_{0} - \frac{4 (b_{0} + 1)}{4 h_{2} h_{6} - h_{4}^{2}} {\frac{h_{2} h_{4}^{2} h_{6} s e c h^{2} (\sqrt{h_{2}} ξ)}{h_{4}^{2} - h_{2} h_{6} {(1 \pm \tanh (\sqrt{h_{2}} ξ))}^{2}} - 2 {(\frac{h_{2} h_{4} h_{6} s e c h^{2} (\sqrt{h_{2}} ξ)}{h_{4}^{2} - h_{2} h_{6} {(1 \pm \tanh (\sqrt{h_{2}} ξ))}^{2}})}^{2}} \end{array},$

${\begin{array}{l} u_{2} = k B + 4 k \sqrt{h_{6}} \frac{h_{2} h_{4} c s c h^{2} (\sqrt{h_{2}} ξ)}{h_{4}^{2} - h_{2} h_{6} {(1 \pm \coth (\sqrt{h_{2}} ξ))}^{2}} \\ v_{2} = b_{0} + \frac{4 (b_{0} + 1)}{4 h_{2} h_{6} - h_{4}^{2}} {\frac{h_{2} h_{4}^{2} h_{6} c s c h^{2} (\sqrt{h_{2}} ξ)}{h_{4}^{2} - h_{2} h_{6} {(1 \pm \coth (\sqrt{h_{2}} ξ))}^{2}} + 2 {(\frac{h_{2} h_{4} h_{6} c s c h^{2} (\sqrt{h_{2}} ξ)}{h_{4}^{2} - h_{2} h_{6} {(1 \pm \coth (\sqrt{h_{2}} ξ))}^{2}})}^{2}} \end{array},$

${\begin{array}{l} u_{3} = k B + 64 k \sqrt{h_{6}} \frac{h_{2} \exp (\pm 2 \sqrt{h_{2}} ξ)}{{(\exp (\pm 2 \sqrt{h_{2}} ξ) - 4 h_{4})}^{2} - 64 h_{2} h_{6}} \\ v_{3} = b_{0} + \frac{64 (b_{0} + 1)}{4 h_{2} h_{6} - h_{4}^{2}} {\frac{h_{2} h_{4} h_{6} \exp (\pm 2 \sqrt{h_{2}} ξ)}{{(\exp (\pm 2 \sqrt{h_{2}} ξ) - 4 h_{4})}^{2} - 64 h_{2} h_{6}} + {(\frac{4 \sqrt{2} h_{2} h_{6} \exp (\pm 2 \sqrt{h_{2}} ξ)}{{(\exp (\pm 2 \sqrt{h_{2}} ξ) - 4 h_{4})}^{2} - 64 h_{2} h_{6}})}^{2}} \end{array},$