RECONSTRUCTION HARMONIC BALANCE METHOD AND ITS APPLICATION IN SOLVING COMPLEX NONLINEAR DYNAMICAL SYSTEMS
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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.
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关键词:
- 非多项式型非线性系统 /
- 重构谐波平衡法 /
- 微分方程重铸 /
- 非线性单摆 /
- 拟周期响应
Abstract: The harmonic balance method is the most commonly used method for solving periodic solutions of nonlinear dynamic systems, but the high-order approximation of nonlinear terms requires sophisticated formula derivations, which limits its ultra-high accuracy computation. The authors' team proposed the reconstruction harmonic balance (RHB) method through the equivalent reconstruction of the frequency domain nonlinear quantity in the time domain, which successfully conquered the problem of ultra-high-order calculation of the classical harmonic balance method. However, both two methods require the dynamical system to be polynomial nonlinear, and cannot be directly used to solve the quasi-periodic solution of the nonlinear system. In view of the above problems, this paper proposes a novel method that combines the RHB method and the recast technique for complex nonlinear systems. First, the general nonlinear problem is non-destructively recast into a polynomial nonlinear system, and then the RHB method is used for high-precision solutions. Aiming at computing the quasi-periodic response, which is hard to obtain relatively accurate results by considering only one base frequency, the RHB method based on the idea of "supplemental frequency" is derived. By optimizing and selecting base frequencies, the fast and accurate capture of quasi-periodic response is achieved in this paper. The typical systems such as the nonlinear pendulum, and the nonlinear coupling asymmetric pendulum are selected for simulation and algorithm verification. Simulation results show that the accuracy of the proposed RHB-recast method achieves an accuracy up to 10−12 when solving the non-polynomial type nonlinear systems, reaching the computer round-off-error accuracy, far exceeding the state-of-the-art methods. The supplemental frequency RHB method has been shown to produce efficient solutions for quasi-periodic problems, expanding the scope of the RHB-like methods for solving complex physical responses. -
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 
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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
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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
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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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