# 一类转子系统的非线性随机稳定性及随机Hopf分岔Nonlinear Stochastic Stability and Stochastic Hopf Bifurcation of a Class of Rotor Systems

DOI: 10.12677/AAM.2020.94061, PDF, HTML, XML, 下载: 96  浏览: 154

Abstract: This paper mainly analyzes the nonlinear random stability and random Hopf bifurcation of a four-dimensional high-speed rotor system under a random parameter excitation. The study of rotor system dynamics has made great progress in theory and practice. The system is affected by internal factors and external random wind and replaced by Gaussian color noise. By using the principle of random average, the quasi-hamiltonian system is convergent to a one-dimensional random ITO diffusion process, and then the maximum lyapunov exponential method is used to judge the local stability of the system, and the conditions for the local stability of the system are obtained. Then the Hopf bifurcation is simulated by the solution of FPK equation, namely the stationary probability density.

1. 引言

2. 模型的建立

$\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{l}\stackrel{¨}{x}\\ \stackrel{¨}{y}\end{array}\right)+\chi \left(\begin{array}{cc}{|\stackrel{˙}{x}|}^{a}+{|x|}^{a}& 0\\ 0& {|\stackrel{˙}{y}|}^{b}+{|y|}^{b}\end{array}\right)\left(\begin{array}{l}\stackrel{˙}{x}\\ \stackrel{˙}{y}\end{array}\right)+\left(\begin{array}{cc}{k}_{1}/m& 0\\ 0& {k}_{2}/m\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(00\right)$

$a=b=2$，所以方程可以表示为：

$\left\{\begin{array}{l}\stackrel{¨}{x}+\chi \left({\stackrel{˙}{x}}^{2}+{x}^{2}\right)\stackrel{˙}{x}+\left({k}_{1}/m\right)x={\eta }_{1}\left(t\right)\\ \stackrel{¨}{y}+\chi \left({\stackrel{˙}{y}}^{2}+{y}^{2}\right)\stackrel{˙}{y}+\left({k}_{1}/m\right)y={\eta }_{2}\left(t\right)\end{array}$

$〈{\eta }_{i}\left(t\right)〉=0$$〈{\eta }_{i}\left(t\right){\eta }_{i}\left(s\right)〉=\frac{D}{\tau }\mathrm{exp}\left(-\frac{|t-s|}{\tau }\right)$$i=1,2$

2.1. 下面将色噪声模型进行等效白化 [6]

$\left\{\begin{array}{l}\stackrel{¨}{x}+\chi \left({\stackrel{˙}{x}}^{2}+{x}^{2}\right)\stackrel{˙}{x}+\left(\frac{{k}_{1}}{m}\right)x={\eta }_{1}\left(t\right)\\ \stackrel{¨}{y}+\chi \left({\stackrel{˙}{y}}^{2}+{y}^{2}\right)\stackrel{˙}{y}+\left(\frac{{k}_{2}}{m}\right)y={\eta }_{2}\left(t\right)\\ {\stackrel{˙}{\eta }}_{i}\left(t\right)=-\frac{1}{\tau }{\eta }_{i}\left(t\right)+\frac{1}{\tau }{\xi }_{i}\left(t\right)\end{array}$

$x={q}_{1},\stackrel{˙}{x}={p}_{1},y={q}_{2},\stackrel{˙}{y}={p}_{2},{\omega }_{1}={k}_{1}/m,{\omega }_{2}={k}_{2}/m,\alpha =\chi$，可以得到

$\left\{\begin{array}{l}{\stackrel{˙}{q}}_{1}={p}_{1}\\ {\stackrel{˙}{p}}_{1}=-{\omega }_{1}{q}_{1}-\alpha \left({p}_{1}^{2}+{q}_{1}^{2}\right){p}_{1}+{\eta }_{1}\left(t\right)\\ {\stackrel{˙}{q}}_{2}={p}_{2}\\ {\stackrel{˙}{p}}_{2}=-{\omega }_{2}{q}_{2}-\alpha \left({p}_{2}^{2}+{q}_{2}^{2}\right){p}_{2}+{\eta }_{2}\left(t\right)\\ {\stackrel{˙}{\eta }}_{i}\left(t\right)=-\left(1/\tau \right){\eta }_{i}\left(t\right)+\left(1/\tau \right){\xi }_{i}\left(t\right)\end{array}$

${f}_{1}=-{\omega }_{1}{q}_{1}-\alpha \left({p}_{1}^{2}+{q}_{1}^{2}\right){p}_{1}$${g}_{1}=1$${f}_{2}=-{\omega }_{2}{q}_{2}-\alpha \left({p}_{2}^{2}+{q}_{2}^{2}\right){p}_{2}$${g}_{2}=1$

${\stackrel{˙}{p}}_{i}={f}_{i}+{g}_{i}\eta \left(t\right)$$i=1,2$

${\stackrel{¨}{p}}_{i}+c\left({p}_{i},\tau \right)\frac{{\stackrel{˙}{p}}_{i}}{\tau }-\frac{{f}_{i}}{\tau }=\frac{{g}_{i}\xi \left(t\right)}{\tau }$

${c}_{i}\left({p}_{i},\tau \right)=1+\tau \frac{{G}_{i}}{{g}_{i}}$ , ${G}_{i}={f}_{i}{{g}^{\prime }}_{i}-{{f}^{\prime }}_{i}{g}_{i}$ , $i=1,2$

$\left\{\begin{array}{l}{\stackrel{˙}{q}}_{1}={p}_{1}\\ {\stackrel{˙}{p}}_{1}=\frac{-{\omega }_{1}{\tau }^{-1}}{{c}_{1}\left({p}_{1},\tau \right)}{q}_{1}+\frac{-{\tau }^{-1}\alpha \left({p}_{1}^{2}+{q}_{1}^{2}\right)}{{c}_{1}\left({p}_{1},\tau \right)}{p}_{1}+\frac{{\tau }^{-1}}{{c}_{1}\left({p}_{1},\tau \right)}{\xi }_{1}\left(t\right)\\ {\stackrel{˙}{q}}_{2}={p}_{2}\\ {\stackrel{˙}{p}}_{12}=\frac{-{\omega }_{2}{\tau }^{-1}}{{c}_{2}\left({p}_{2},\tau \right)}{q}_{1}+\frac{-{\tau }^{-1}\alpha \left({p}_{2}^{2}+{q}_{2}^{2}\right)}{{c}_{2}\left({p}_{2},\tau \right)}{p}_{2}+\frac{{\tau }^{-1}}{{c}_{2}\left({p}_{2},\tau \right)}{\xi }_{2}\left(t\right)\end{array}$

$\left\{\begin{array}{l}{\stackrel{˙}{q}}_{1}={p}_{1}\\ {\stackrel{˙}{p}}_{1}=-{\tau }^{-1}\left({k}_{11}+{k}_{11}{p}_{1}\right)\left[{\omega }_{1}{q}_{1}+\alpha \left({p}_{1}^{2}+{q}_{1}^{2}\right){p}_{1}-{\xi }_{1}\left(t\right)\right]\\ {\stackrel{˙}{q}}_{2}={p}_{2}\\ {\stackrel{˙}{p}}_{2}=-{\tau }^{-1}\left({k}_{21}+{k}_{21}{p}_{2}\right)\left[{\omega }_{2}{q}_{2}+\alpha \left({p}_{2}^{2}+{q}_{2}^{2}\right){p}_{2}-{\xi }_{2}\left(t\right)\right]\end{array}$

${E}_{i}=-{\tau }^{-1}\left[\frac{1}{{c}_{i}\left({p}_{i},\tau \right)}-\left({k}_{i1}+{k}_{i1}{p}_{i}\right)\right]\left({\omega }_{i}{q}_{i}+\alpha \left({p}_{i}^{2}+{q}_{i}^{2}\right){p}_{i}-2{\xi }_{i}\left(t\right)\right),i=1,2$

$-{\tau }^{-1}\left[\frac{1}{{c}_{i}\left({p}_{i},\tau \right)}-\left({k}_{i1}+{k}_{i1}{p}_{i}\right)\right]\left({\omega }_{i}{q}_{i}+\alpha \left({p}_{i}^{2}+{q}_{i}^{2}\right){p}_{i}\right)=0,i=1,2$

$\alpha {k}_{i2}{p}^{4}+\alpha {k}_{i1}{p}^{3}+\left({\tau }^{-1}{k}_{i2}+\omega {k}_{i2}+\alpha +\alpha {q}^{2}{k}_{i1}{k}_{i2}\right)p+{\tau }^{-1}{k}_{i1}+\omega {k}_{i1}+\alpha {q}^{2}{k}_{i1}-1=0$

$\left({\tau }^{-1}{k}_{i2}+\omega {k}_{i2}+\alpha \right)p+{\tau }^{-1}{k}_{i1}+\omega {k}_{i1}-1=0$

$\left\{\begin{array}{l}{\tau }^{-1}{k}_{i2}+\omega {k}_{i2}+\alpha =0\\ {\tau }^{-1}{k}_{i1}+\omega {k}_{i1}-1=0\end{array}$

$\left\{\begin{array}{l}{k}_{i1}=\frac{1}{{\tau }^{-1}+\omega }\\ {k}_{i2}=-\frac{\alpha }{{\tau }^{-1}+\omega }\end{array}$

$\left\{\begin{array}{l}{\stackrel{˙}{q}}_{1}={p}_{1}\\ {\stackrel{˙}{p}}_{1}=-{a}_{11}{q}_{1}+{b}_{11}\left({p}_{1}^{2}+{q}_{1}^{2}\right){p}_{1}+{c}_{11}{\xi }_{1}\left(t\right)\\ {\stackrel{˙}{q}}_{2}={p}_{2}\\ {\stackrel{˙}{p}}_{2}=-{a}_{22}{q}_{2}+{b}_{22}\left({p}_{2}^{2}+{q}_{2}^{2}\right){p}_{2}+{c}_{22}{\xi }_{2}\left(t\right)\end{array}$

$H=\frac{1}{2}\left({p}_{1}{}^{2}+{p}_{2}{}^{2}\right)+\frac{1}{2}\left({a}_{11}{q}_{1}^{2}+{a}_{22}{q}_{2}^{2}\right)$

$\text{d}\left(H\right)=\stackrel{¯}{m}\left(H\right)\text{d}t+\stackrel{¯}{\sigma }\left(H\right)\text{d}B\left(t\right)$

${q}_{1}=\frac{R}{\sqrt{{a}_{11}}}\mathrm{cos}\theta ,{q}_{\text{2}}=\frac{R}{\sqrt{{a}_{22}}}\mathrm{sin}\theta$

$\begin{array}{c}\stackrel{¯}{m}\left(H\right)=\frac{\text{2}\pi }{T\left(H\right)}{\int }_{\Omega }\left[\left({b}_{11}+{b}_{22}\right)A\left(H,\theta \right)-\left(\frac{{b}_{11}}{{a}_{11}}{\mathrm{cos}}^{\text{2}}\theta \text{+}\frac{{b}_{\text{22}}}{{a}_{22}}{\mathrm{sin}}^{\text{2}}\theta \right)B\left(H,\theta \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{{R}^{4}}{2}\left(\frac{{c}_{11}^{2}\pi {D}_{1}}{{a}_{11}}{\mathrm{cos}}^{2}\theta +\frac{{c}_{22}^{2}\pi {D}_{2}}{{a}_{22}}{\mathrm{sin}}^{2}\theta \right)\right]\text{d}\theta \end{array}$

${\stackrel{¯}{\sigma }}^{2}\left(H\right)=\frac{4\pi }{T\left(H\right)}{\int }_{0}^{\pi }\left(\frac{{c}_{11}^{2}\pi {D}_{1}}{{a}_{11}}{\mathrm{cos}}^{2}\theta +\frac{{c}_{22}^{2}\pi {D}_{2}}{{a}_{22}}{\mathrm{sin}}^{2}\theta \right)B\left(H,\theta \right)\text{d}\theta$

$T\left(H\right)=\left(\frac{2\pi }{\sqrt{{a}_{11}{a}_{22}}}\right){\int }_{0}^{\pi }{R}^{2}\text{d}\theta$

$\Omega =\left\{\left({Q}_{1},\cdots ,{Q}_{n},{P}_{2},\cdots ,{P}_{n}\right)|H\left({Q}_{1},\cdots ,{Q}_{n},0,{P}_{2},\cdots ,P\right)\le H\right\}$

$A\left(H,\theta \right)=H{R}^{2}-\frac{{R}^{4}}{4}\left(1+\frac{{a}_{11}+{a}_{22}}{2\sqrt{{a}_{11}{a}_{22}}}\mathrm{sin}2\theta \right)-\frac{{b}_{11}+{b}_{22}}{24}{R}^{6}{\left(\frac{\mathrm{cos}\theta }{\sqrt{{a}_{11}}}-\frac{\mathrm{sin}\theta }{\sqrt{{a}_{22}}}\right)}^{4}$

$B\left(H,\theta \right)=\frac{H{R}^{2}}{2}-\frac{{R}^{6}}{6}\left(1+\frac{{a}_{11}+{a}_{22}}{2\sqrt{{a}_{11}{a}_{22}}}\mathrm{sin}2\theta \right)-\frac{{b}_{11}+{b}_{22}}{32}{R}^{8}{\left(\frac{\mathrm{cos}\theta }{\sqrt{{a}_{11}}}-\frac{\mathrm{sin}\theta }{\sqrt{{a}_{22}}}\right)}^{4}$

${R}^{2}$ 是下列方程之解：

$H-\frac{{R}^{2}}{2}\left(1+\frac{{a}_{11}+{a}_{22}}{2\sqrt{{a}_{11}{a}_{22}}}\mathrm{sin}2\theta \right)-\frac{{b}_{11}+{b}_{22}}{8}{R}^{4}{\left(\frac{\mathrm{cos}\theta }{\sqrt{{a}_{11}}}-\frac{\mathrm{sin}\theta }{\sqrt{{a}_{22}}}\right)}^{4}=0$

2.2. 接下来主要讨论随机稳定性 [8]

$\text{d}H={\mu }_{1}H\text{d}t+{\mu }_{2}^{1/2}H\text{d}B\left(t\right)$

${\mu }_{1}=-\frac{1}{2}\left({b}_{11}+{b}_{22}\right)+\frac{1}{2}\left(\frac{{c}_{11}^{2}{D}_{1}}{{a}_{11}}+\frac{{c}_{22}^{2}{D}_{2}}{{a}_{22}}\right)\zeta$

${\mu }_{2}=\frac{1}{3}\left(\frac{{c}_{11}^{2}{D}_{1}}{{a}_{11}}+\frac{{c}_{22}^{2}{D}_{2}}{{a}_{22}}\right)\zeta$

$\zeta ={\int }_{0}^{\pi }{\left(1+\frac{{a}_{11}+{a}_{22}}{2\sqrt{{a}_{11}{a}_{22}}}\mathrm{sin}\theta \right)}^{-2}\text{d}\theta /{\int }_{0}^{\pi }{\left(1+\frac{{a}_{11}+{a}_{22}}{2\sqrt{{a}_{11}{a}_{22}}}\mathrm{sin}\theta \right)}^{-1}\text{d}\theta$

$\stackrel{¯}{m}\left(H\right)={\mu }_{1}H+ο\left(H\right)$${\stackrel{¯}{\sigma }}^{2}\left(H\right)={\mu }_{2}{H}^{2}+ο\left(H2\right)$

$\begin{array}{l}\lambda =\underset{t\to \infty }{\mathrm{lim}}\frac{1}{t}\mathrm{ln}{H}^{\frac{1}{2}}\left(t\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left[{m}^{\prime }\left(0\right)-{\left({\sigma }^{2}\left(0\right)\right)}^{2}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{1}{4}\left({b}_{11}+{b}_{22}\right)+\frac{1}{6}\left(\frac{{c}_{11}^{2}{D}_{1}}{{a}_{11}}+\frac{{c}_{22}^{2}{D}_{2}}{{a}_{22}}\right)\zeta \end{array}$

$\frac{2}{3}\left(\frac{{c}_{11}^{2}{D}_{1}}{{a}_{11}}+\frac{{c}_{22}^{2}{D}_{2}}{{a}_{22}}\right)\zeta <\left({b}_{11}+{b}_{22}\right)$

3. 随机分岔分析

$\frac{\partial p}{\partial t}=-\frac{\partial }{\partial H}\left[a\left(H\right)p\right]+\frac{1}{2}\frac{{\partial }^{2}}{\partial {H}^{2}}\left[b\left(H\right)p\right]$

$a\left(H\right)=\stackrel{¯}{m}\left(H\right),b\left(H\right)={\stackrel{¯}{\sigma }}^{2}\left(H\right)$

${\alpha }_{H}=1$${\beta }_{H}=2$

${c}_{H}=\underset{x\to {x}_{H}^{+}}{\mathrm{lim}}\frac{2a\left(x\right){\left(x-{x}_{H}\right)}^{{\alpha }_{H}-{\beta }_{H}}}{b\left(x\right)}=-\frac{3}{2\zeta }\left({b}_{11}+{b}_{22}\right)+\frac{3}{2}\left(\frac{{c}_{11}^{2}{D}_{1}}{{a}_{11}}+\frac{{c}_{22}^{2}{D}_{2}}{{a}_{22}}\right)$

$f\left(H\right)=C{H}^{{v}_{1}}\mathrm{exp}\left[{u}_{1}H\right]$

${v}_{1}={c}_{H}-{\alpha }_{H}=-2+\left(-\frac{3}{2\zeta }\left({b}_{11}+{b}_{22}\right)+\frac{3}{2}\left(\frac{{c}_{11}^{2}{D}_{1}}{{a}_{11}}+\frac{{c}_{22}^{2}{D}_{2}}{{a}_{22}}\right)\right)$

${u}_{1}=\frac{3}{2}\left(\frac{{c}_{11}^{2}{D}_{1}}{{a}_{11}}+\frac{{c}_{22}^{2}{D}_{2}}{{a}_{22}}\right)$

$\begin{array}{l}f\left({q}_{1},{q}_{2},{p}_{1},{p}_{2}\right)=C{\left(\frac{1}{2}\left({p}_{1}^{2}+{p}_{2}^{2}\right)+\frac{1}{2}\left({a}_{11}{q}_{1}^{2}+{a}_{22}{q}_{2}^{2}\right)\right)}^{{v}_{1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}×\mathrm{exp}\left[{u}_{1}\left(\frac{1}{2}\left({p}_{1}^{2}+{p}_{2}^{2}\right)+\frac{1}{2}\left({a}_{11}{q}_{1}^{2}+{a}_{22}{q}_{2}^{2}\right)\right)\right]\end{array}$

4. 数值模拟

Figure 1. When ${b}_{11}+{b}_{22}=-9.67$ , ${v}_{1}=0.01$ system stationary probability density function and joint probability density function

Figure 2. When ${b}_{11}+{b}_{22}=-9.89$ , ${v}_{1}=0.335$ system stationary probability density function and joint probability density function

Figure 3. When ${b}_{11}+{b}_{22}=-11.67$ , ${v}_{1}=3$ system stationary probability density function and joint probability density function

5. 小结

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