应用数学进展  >> Vol. 8 No. 2 (February 2019)

利用exp(-G(ξ))方法和拟设函数法求Sharma-Tasso-Olver方程精确解
The Exact Solutions of the Sharma-Tasso-Olver Equation Using exp(-G(ξ)) Method and Ansatz Method

DOI: 10.12677/AAM.2019.82025, PDF, HTML, XML, 下载: 415  浏览: 1,765 

作者: 张逢燕, 王俊梅:山东师范大学,数学与统计学院,山东 济南;李冬雪:德州市第十中学,山东 德州

关键词: Sharma-Tasso-Olver方程exp(-G(ξ))方法孤子解拟设双曲函数法Sharma-Tasso-Olver Equation exp(-G(ξ))-Method Soliton Solution Ansatz Method

摘要: 本文运用行波变换和齐次平衡原理对Sharma-Tasso-Olver方程进行化简,得到约化的常微分方程。利用exp(-G(ξ))方法和拟设双曲函数法求得方程的三角函数解、双曲函数解、有理函数解和孤子解。
Abstract: Using the traveling wave transformation and homogeneous balance simplified the Shar-ma-Tasso-Olver equation to obtain the reduced ordinary differential equations, using the exp(-G(ξ))-method to get the trigonometric function solution, hyperbolic function solution and the rational function solution. In addition, an exact soliton solution is provided by using the Ansatz method.

文章引用: 张逢燕, 李冬雪, 王俊梅. 利用exp(-G(ξ))方法和拟设函数法求Sharma-Tasso-Olver方程精确解[J]. 应用数学进展, 2019, 8(2): 219-226. https://doi.org/10.12677/AAM.2019.82025

1. 引言

非线性发展方程是国内外研究的热点问题,Sharma-Tasso-Olver (STO)方程在数学和物理领域有着重要的作用,很多专家对其有深入研究。例如,文献 [1] [2] 运用Bäcklund变换求精确解。在 [3] 中介绍扩展双曲函数方法的STO方程的显式行波解。楼森岳等人 [4] 通过使用标准的截断Painlevé分析,Hirota双线性方法和Bäcklund变换方法彻底检查了孤子裂变和聚变。在 [5] 中,Yan通过使用两种类型的Cole-Hopf变换,研究了两类(2 + 1)维广义Sharma-Tasso-Olver积分-微分方程的可积性。他证明了这两个GSTO方程都拥有Painlevé属性和双哈密顿结构。另外,利用耦合Riccati方程方法 [6] ,修正最简单方程方法 [7] ,Exp函数方法 [8] [9] 和李对称分析求出STO方程精确解等 [10] [11] [12] 。

我们研究以下STO方程

u t + 3 ε u 2 u x + 3 ε u x 2 + 3 ε u u x x + ε u x x x = 0 (1)

其中 u = u ( x , t ) ε 是任意常数。

本文首先对Sharma-Tasso-Olver方程进行了综述。第二部分运用行波变换,利用 exp ( G ( ξ ) ) 方法 [13] [14] [15] 和齐次平衡原理 [16] [17] 将原方程约化成常微分方程,求出方程的三角函数解,双曲函数解和有理函数解。第三部分利用拟设双曲函数的行波变换得到方程的孤子解 [18] [19] 。

2. e x p ( G ( ξ ) ) 方法

对于方程(1),我们进行行波变换:

u ( x , t ) = u ( ξ ) , ξ = k x w t ,

得到约化的常微分方程:

w u + 3 ε k u 2 u + 3 ε k 2 ( u ) 2 + 3 ε k 2 u u + ε k 3 u = 0 ,

积分一次为:

w u + ε k u 3 + 3 ε k 2 u u + ε k 3 u = 0. (2)

我们假设方程(2)有以下形式的精确解 [14] [15] [16] :

u ( ξ ) = n = 0 m a n ( exp ( G ( ξ ) ) ) n , (3)

其中 a n 为任意常数,m为正整数, G ( ξ ) 满足以下辅助常微分方程:

G ( ξ ) = exp ( G ( ξ ) ) + μ exp ( G ( ξ ) ) + λ , (4)

从辅助方程(4)中我们可以得到不同的精确解。

把(3)和(4)带入(2),利用齐次平衡原理可以得到 m = 1 ,则方程(1)的解可以表示为:

u ( ξ ) = a 0 + a 1 exp ( G ( ξ ) ) . (5)

把(4)和(5)带入(2),收集 exp ( G ( ξ ) ) n n = 0 , 1 , 2 , 3 的系数,令其系数方程等于0,可以得到关于 a 0 , a 1 , μ , λ , w , k , ε 的超定方程组:

exp ( G ( ξ ) ) 3 : 2 k 3 ε a 1 3 ε k 2 a 1 2 + k ε a 1 3 = 0 , (6)

exp ( G ( ξ ) ) 2 : k 3 ε a 1 k 2 ε λ a 1 2 + k 2 ε a 1 2 a 0 = 0 , (7)

exp ( G ( ξ ) ) : k 3 ε λ 2 a 1 + 2 k 3 ε μ a 1 3 k 2 ε μ a 1 2 3 k 2 ε λ a 1 a 0 + 3 k ε a 1 a 0 2 w a 1 = 0 , (8)

exp ( G ( ξ ) ) 0 : k 3 ε λ μ a 1 + 3 k 2 ε μ a 0 + k ε a 0 3 w a 0 = 0. (9)

求解(6)~(9),得到两组解分别为:

情况1: k = k , w = k 3 ε λ 2 4 k 3 ε μ 4 , a 0 = λ k 2 , a 1 = k ,带入方程(5),得到五种不同的解为:

① 当 λ 2 4 μ > 0 ,且 μ 0 时,

u 1.1 = 2 k μ λ 2 4 μ tanh ( λ 2 4 μ / 2 ( ξ + C ) ) λ + λ k 2 (如图1)

② 当 λ 2 4 μ < 0 ,且 μ 0 时,

u 1.2 = 2 k μ 4 μ λ 2 tan ( 4 μ λ 2 / 2 ( ξ + C ) ) λ + λ k 2 (如图2)

③ 当 λ 2 4 μ > 0 ,且 μ = 0 , λ 0 时,

u 1.3 = λ k cosh ( λ ( ξ + C ) ) + sinh ( λ ( ξ + C ) ) 1 + λ k 2 (如图3)

④ 当 λ 2 4 μ = 0 ,且 μ 0 , λ 0 时,

u 1.4 = λ 2 k ( ξ + C ) 2 λ ( ξ + C ) + 4 + λ k 2 (如图4)

⑤ 当 λ 2 4 μ = 0 ,且 μ = 0 , λ = 0 时,

u 1.5 = k ξ + C + λ k 2 (如图5)

其中 ξ = k x k 3 ε λ 2 4 k 3 ε μ 4 t k , C 为任意常数。

Figure 1. Hyperbolic function solution u 1.1

图1. 双曲函数解 u 1.1

Figure 2. Trigonometric function solution u 1.2

图2. 三角函数解 u 1.2

Figure 3. Hyperbolic function solution u 1.3

图3. 双曲函数解 u 1.3

Figure 4. Rational function solution u 1.4

图4. 有理函数解 u 1.4

Figure 5. Rational function solution u 1.5

图5. 有理函数解 u 1.5

情况2: k = k , w = k 3 ε λ 2 4 k 3 ε μ , a 0 = λ k , a 1 = 2 k ,带入方程(7),得到五种不同的解为:

① 当 λ 2 4 μ > 0 ,且 μ 0 时,

u 2.1 = 4 k μ λ 2 4 μ tanh ( λ 2 4 μ / 2 ( ξ + C ) ) λ + λ k

② 当 λ 2 4 μ < 0 ,且 μ 0 时,

u 2.2 = 4 k μ 4 μ λ 2 tan ( 4 μ λ 2 / 2 ( ξ + C ) ) λ + λ k

③ 当 λ 2 4 μ > 0 ,且 μ = 0 , λ 0 时,

u 2.3 = 2 λ k cosh ( λ ( ξ + C ) ) + sinh ( λ ( ξ + C ) ) 1 + λ k

④ 当 λ 2 4 μ = 0 ,且 μ 0 , λ 0 时,

u 2.4 = λ 2 k ( ξ + C ) λ ( ξ + C ) + 2 + λ k

⑤ 当 λ 2 4 μ = 0 ,且 μ = 0 , λ = 0 时,

u 2.5 = 2 k ξ + C + λ k

其中 ξ = k x ( k 3 ε λ 2 4 k 3 ε μ ) t k , C 为任意常数。

3. 拟设双曲函数法

假设方程有以下形式的解 [18] [19] :

u ( x , t ) = A tan h τ τ = B ( x v t ) (10)

其中A和B是自由参数,p是固定参数,v是孤子速度。

由(10)可以得到:

u t = n v A B ( tanh n 1 τ tanh n + 1 τ ) , (11)

u x = n v A B ( tanh n 1 τ tanh n + 1 τ ) , , (12)

u x x = n ( n 1 ) A B 2 tanh n 2 τ 2 n 2 A B 2 tanh n τ + n ( n + 1 ) A B 2 tanh n + 2 τ , (13)

u x x x = n ( n 1 ) ( n 2 ) A B 3 tanh n 3 τ [ n ( n 1 ) ( n 2 ) + 2 n 3 ] A B 3 tanh n 1 τ + [ n ( n + 1 ) ( n + 2 ) + 2 n 3 ] A B 3 tanh n + 1 τ n ( n + 1 ) ( n + 2 ) A B 3 tanh n + 3 τ , (14)

把(11)~(14)带入方程(1),得到:

n v A B ( tanh n 1 τ tanh n + 1 τ ) + 3 ε n v A 2 B ( tanh 2 n 1 τ tanh 2 n + 1 τ ) + 3 ε [ n v A B ( tanh n 1 τ tanh n + 1 τ ) ] 2 + 3 ε A 2 B 2 [ n ( n 1 ) tanh 2 n 2 τ 2 n 2 tanh 2 n τ n ( n + 1 ) tanh 2 n + 2 τ ] + ε [ n ( n 1 ) ( n 2 ) A B 3 tanh n 3 τ ( n ( n 1 ) ( n 2 ) + 2 n 3 ) A B 3 tanh n 1 τ + ( n ( n + 1 ) ( n + 2 ) + 2 n 3 ) A B 3 tanh n + 1 τ n ( n + 1 ) ( n + 2 ) A B 3 tanh n + 3 τ . (15)

从(15)的指数可以得到 2 n + 2 n + 3 相等,得到 n = 1 。代入(29)收集 tanh n τ 的系数,令其系数方程等于0,可以得到:

B = A 或者 B = A 2 v = ε A 2

上式给出了自由参数A和B的关系,扰动孤子的速度v。我们就得到了方程(1)的1-孤子解:

u ( x , t ) = A tanh ( B ( x v t ) )

4. 结论

本文运用行波变换、齐次平衡原理,利用 exp ( G ( ξ ) ) 方法求出Sharma-Tasso-Olver方程的新显式行波解。这些解包括双曲函数解,三角函数解和有理数解。利用拟设双曲函数得到了方程的1-孤子解。

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