一个求非线性差分方程所有多项式解的算法

YU Jiangtao; LIU Yinping

doi:10.3969/j.issn.1000-5641.201811037

一个求非线性差分方程所有多项式解的算法

doi: 10.3969/j.issn.1000-5641.201811037

余江涛,
柳银萍^,

华东师范大学计算机科学与技术系, 上海　200062

详细信息

中图分类号: O175.7; TP311.1
计量
- 文章访问数: 94
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- PDF下载量: 1
- 被引次数: 0
出版历程
- 收稿日期: 2018-09-19
- 网络出版日期: 2019-12-26
- 刊出日期: 2020-01-01

An algorithm for finding all polynomial solutions of nonlinear difference equations

YU Jiangtao,
LIU Yinping^,

Department of Computer Science and Technology, East China Normal University, Shanghai　200062, China

More Information

Corresponding author: ypliu@cs.ecnu.edu.cn

摘要

摘要: 差分方程是计算机代数中一个重要的研究内容, 但是目前很少有关于一般非线性差分方程求解方法的研究. 受到在非线性微分方程中广泛应用的齐次平衡原则的启发, 用其求解大部分非线性差分方程的多项式解. 同时, 提出了一个新的n阶展开方法, 用于求解齐次平衡原则无法求解的情况. 结合这两个方法提出了能够找到非线性差分方程所有多项式解的算法. 该算法基于Maple实现, 实验表明该算法是有效且高效的.
- 非线性差分方程 /
- 多项式解 /
- 齐次平衡原则 /
- n阶展开方法
Abstract: Difference equations are a major aspect of computer algebra; yet, there are currently few studies on solving general nonlinear difference equations. Inspired by the homogeneous balance principle that works well for solving nonlinear differential equations, we use it to find polynomial solutions for a wide range of nonlinear difference equations, in which a new n-order expansion method is proposed to process the powerless cases of the homogeneous balance principle. They are combined together as an algorithm that can be used to find all polynomial solutions of nonlinear difference equations. The algorithm is implemented in Maple, and the experiments show that it is effective and efficient.
- nonlinear difference equation /
- polynomial solution /
- homogeneous balance principle /
- n-order expansion method

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Fig. 1 Visualization of balance points

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Fig. 2 Time complexity

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Fig. 3 Distribution of balance points

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Fig. 4 Distribution of minimum expansion order

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Tab. 1 A brief analysis of the distributed consensus protocols

Type	Order equation(s)	Order solution(s)
$ BP_1 $	$ 2m = m+5\geqslant 6 $	$ m = 5 $
$ BP_1 $	$ 2m = 6\geqslant m+5 $	$ m\in \varnothing $
$ BP_1 $	$ m+5 = 6\geqslant 2m $	$ m = 1 $
$ BP_2 $	$ m+5>\max\{2m,6\} $	$ m\in \{2,3,4\} $

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