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## 留言板 引用本文: 代洪华, 王其偲, 严子朴, 岳晓奎. 重构谐波平衡法及其求解复杂非线性问题应用. 力学学报, 2024, 56(1): 1-13 Dai Honghua, Wang Qisi, Yan Zipu, Yue Xiaokui. Reconstruction harmonic balance method and its application in solving complex nonlinear dynamical systems. Chinese Journal of Theoretical and Applied Mechanics, 2024, 56(1): 1-13 doi: 10.6052/0459-1879-23-369
 Citation: Dai Honghua, Wang Qisi, Yan Zipu, Yue Xiaokui. Reconstruction harmonic balance method and its application in solving complex nonlinear dynamical systems. Chinese Journal of Theoretical and Applied Mechanics, 2024, 56(1): 1-13 ## 重构谐波平衡法及其求解复杂非线性问题应用

##### doi: 10.6052/0459-1879-23-369

###### 通讯作者: 代洪华, 教授, 主要研究方向为在轨服务动力学仿真与高性能计算、航天器动力学与控制. E-mail: hhdai@nwpu.edu.cn
• 中图分类号: V412.4+1

## RECONSTRUCTION HARMONIC BALANCE METHOD AND ITS APPLICATION IN SOLVING COMPLEX NONLINEAR DYNAMICAL SYSTEMS

• 摘要: 谐波平衡法是求解非线性动力学系统周期解的最常用方法, 但对非线性项进行高阶近似需要庞杂的公式推导, 限制了该方法的超高精度解算. 通过对频域非线性量的时域等价重构, 提出了重构谐波平衡法(RHB法), 解决了经典谐波平衡法超高阶次计算难题. 然而, 上述两种方法均要求动力学系统为多项式型非线性, 且无法直接用来求解非线性系统的拟周期解. 针对上述问题, 文章提出一种将RHB法和复杂非线性系统等价重铸法相结合的计算方法, 首先将一般非线性问题无损重铸为多项式型非线性系统, 然后用RHB法进行高精度求解; 针对拟周期响应求解问题, 提出基于“补频”思想的RHB方法, 通过基频的优化筛选, 实现拟周期响应的快速精准捕捉. 选取非线性单摆、相对论谐振子和非线性耦合非对称摆等典型系统进行仿真计算, 仿真结果表明, 所提出的RHB-重铸法在解非多项式型非线性系统的稳态响应时精度保持为${10^{ - 12}}$量级, 达计算机精度, 远超现有方法水平. 补频RHB法则实现了对拟周期问题的高效解算, 拓宽了方法对真实物理响应的求解范围.

• 图  1  HB法和HDHB法计算杜芬方程对比(N = 2)

Figure  1.  Comparison of the Duffing equation computed by the HB and HDHB methods (N = 2)

图  2  Mickens变换解与真实物理解对比

Figure  2.  Comparison of Mickens transformation result and real physical solution

图  3  55阶RHB-重铸法求解相对论谐振子的计算结果

Figure  3.  The computing result of the RHB-recast method ($N = 55$) for solving relativistic harmonic oscillator

图  4  非线性方程组求解算法对计算精度的影响(相对论谐振子)

Figure  4.  Influence of nonlinear equation algorithm on calculation accuracy (relativistic harmonic oscillator)

图  5  改变代数方程对计算精度影响

Figure  5.  Influence of changed algebraic equation on calculation accuracy

图  6  不同初速度下RHB-重铸法求解相对论谐振子

Figure  6.  Solving relativistic harmonic oscillator by the RHB-recast method under different initial velocities

图  7  增加时域配点对抑制混淆误差的效果(使用HB-AFT法)

Figure  7.  The effect of increasing time domain collocation on suppressing aliasing error (the HB-AFT method)

图  8  25阶RHB-重铸法求解非线性单摆

Figure  8.  The computing result of the RHB-recast method ($N = 25$) for solving nonlinear pendulum

图  9  非线性方程组求解算法对计算精度的影响(非线性单摆)

Figure  9.  Influence of nonlinear equation algorithm on calculation accuracy (nonlinear pendulum)

图  10  不同代数方程组合方案对计算精度影响

Figure  10.  Influence of different combination schemes of nonlinear algebraic equation on calculation accuracy

图  11  3种方法求解非线性单摆对应误差曲线对比

Figure  11.  Comparison of corresponding error curves of three methods for solving nonlinear pendulum

图  12  RHB-重铸法求解非线性单摆问题: 相平面图

Figure  12.  The RHB-recast method for solving nonlinear pendulum: phase plot

图  13  多时间尺度法与修正Chebyshev-Picard迭代法(参考解)求解轨迹对比

Figure  13.  Comparison of the method of multiple scales and the MCPI method (reference solution) for solving trajectory diagram

图  14  5阶RMHB法求解非线性耦合非对称摆

Figure  14.  Solving nonlinear coupling asymmetric pendulum by the RMHB5

表  1  RHB法与HDHB法在分岔处的解分布

Table  1.   Statistical distribution of solution by the RHB and the HDHB method at bifurcation point

 Method Upper branch/% Lower branch/% Unstable branch/% Non-physical/% RHB 57.13 19.71 23.16 0 HDHB 29.14 14.21 15.96 48.69

表  2  各阶RHB-重铸法求解相对论谐振子

Table  2.   Solving relativistic harmonic oscillator by the RHB-recast method with different orders

 Order of method Amplitude error Computing time/s 25 $4.36 \times {10^{ - 7}}$ 0.73 35 $1.95 \times {10^{ - 9}}$ 0.77 55 $3.87 \times {10^{ - 12}}$ 1.07

表  3  3种方法计算非线性单摆结果对比

Table  3.   Comparison of corresponding results of three methods for solving nonlinear pendulum

 Methods $M$ Average error Computing time/s RHB-recast 76 1.9 × 10−12 0.70 RHB-Taylor 401 8.9 × 10−10 1.82 HB-AFT 210 1.5 × 10−9 0.69

A1  常见初等超越函数的重铸

A1.   Recast form of the most common elementary transcendental functions

 Original function Differential relationship Companion variable Recast equation $u\left( t \right) = \exp {x\left( t \right)}$ $\dot u\left( t \right) = \dot x\left( t \right) \cdot u\left( t \right)$ $\dot u\left( t \right) = \dot x\left( t \right) \cdot u\left( t \right)$ $u\left( t \right) = {\log _a} {x\left( t \right)}$ $\dot u\left( t \right) = {{\dot x\left( t \right)} \mathord{\left/ {\vphantom {{\dot x\left( t \right)} {\left( {x\left( t \right) \cdot \ln a} \right)}}} \right. } {\left( {x\left( t \right) \cdot \ln a} \right)}}$ $v\left( t \right) = {1 \mathord{\left/ {\vphantom {1 {x\left( t \right)}}} \right. } {x\left( t \right)}}$ \left. \begin{aligned}& {\dot u\left( t \right) = {{\left( {v\left( t \right) \cdot \dot x\left( t \right)} \right)} / {\ln a}}} \\ & {x\left( t \right)v\left( t \right) - 1 = 0} \end{aligned}\right\} \left. \begin{aligned}& {u\left( t \right) = \sin {x\left( t \right)} } \\ & {v\left( t \right) = \cos {x\left( t \right)} } \end{aligned} \right\} \left.\begin{aligned}& {\dot u\left( t \right) = \dot x\left( t \right) \cdot \cos {x\left( t \right)}} \\ & {\dot v\left( t \right) = - \dot x\left( t \right) \cdot \sin {x\left( t \right)} } \end{aligned} \right\} \left. \begin{aligned}& {\dot u\left( t \right) = \dot x\left( t \right) \cdot v\left( t \right)} \\ & {\dot v\left( t \right) = - \dot x\left( t \right) \cdot u\left( t \right)} \end{aligned}\right\} $u\left( t \right) = \tan {x\left( t \right)}$ $\dot u\left( t \right) = \dot x\left( t \right) \cdot \left( {1 + {u^2}\left( t \right)} \right)$ $\dot u\left( t \right) = \dot x\left( t \right) + \dot x\left( t \right) \cdot {u^2}\left( t \right)$ $u\left( t \right) = {\text{arc}}\sin {x\left( t \right)}$ $\dot u\left( t \right) = {{\dot x\left( t \right)} \mathord{\left/ {\vphantom {{\dot x\left( t \right)} {\sqrt {1 - {x^2}\left( t \right)} }}} \right. } {\sqrt {1 - {x^2}\left( t \right)} }}$ \left. \begin{aligned}& {v\left( t \right) = \sqrt {1 - {x^2}\left( t \right)} } \\ & {w\left( t \right) = {1 / {v\left( t \right)}}} \end{aligned} \right\} \left. \begin{aligned}& {\dot u\left( t \right) = \dot x\left( t \right)w\left( t \right)} \\ & {w\left( t \right)v\left( t \right) - 1 = 0} \\ & {{v^2}\left( t \right) + {x^2}\left( t \right) - 1 = 0} \end{aligned} \right\} $u\left( t \right) = {\text{arccos}} {x\left( t \right)}$ $\dot u\left( t \right) = {{ - \dot x\left( t \right)} \mathord{\left/ {\vphantom {{ - \dot x\left( t \right)} {\sqrt {1 - {x^2}\left( t \right)} }}} \right. } {\sqrt {1 - {x^2}\left( t \right)} }}$ \left.\begin{aligned}& {v\left( t \right) = \sqrt {1 - {x^2}\left( t \right)} } \\ & {w\left( t \right) = - {1 / {v\left( t \right)}}} \end{aligned} \right\} \left. \begin{aligned}& {\dot u\left( t \right) = \dot x\left( t \right)w\left( t \right)} \\ & {w\left( t \right)v\left( t \right) + 1 = 0} \\ & {{v^2}\left( t \right) + {x^2}\left( t \right) - 1 = 0} \end{aligned} \right\} $u\left( t \right) = {\text{arctan}}{x\left( t \right)}$ $\dot u\left( t \right) = {{\dot x\left( t \right)} \mathord{\left/ {\vphantom {{\dot x\left( t \right)} {\left( {1 + {x^2}\left( t \right)} \right)}}} \right. } {\left( {1 + {x^2}\left( t \right)} \right)}}$ \left. \begin{aligned}& {v\left( t \right) = 1 + {x^2}\left( t \right)} \\ & {w\left( t \right) = {1 / {v\left( t \right)}}} \end{aligned}\right\} \left.\begin{aligned}& {\dot u\left( t \right) = \dot x\left( t \right)w\left( t \right)} \\ & {w\left( t \right)v\left( t \right) - 1 = 0} \\ & {v\left( t \right) - {x^2}\left( t \right) - 1 = 0} \end{aligned} \right\}
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##### 出版历程
• 收稿日期:  2023-08-01
• 录用日期:  2023-10-17
• 网络出版日期:  2023-10-18

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