## 3 Time Delayed Velocity Feedback Control
### 1/3 Subharmonic Resonance  
To analyze subharmonic resonance for Equation (3), we introduce the detuning parameter \(\sigma\) according to  
\[\omega_{0}=\frac{1}{3}\omega+\varepsilon\sigma. \tag{15}\]  
Substituting Equations (4) and (15) into Equation (3) and equating coefficients of like powers of e yields the following equations:  
\[D_{0}^{2}q_{0}+\frac{1}{9}\omega^{2}q_{0}=f\sin\omega T_{0}, \tag{16}\]  
\[D_{0}^{2}q_{1}+\frac{1}{9}\omega^{2}q_{1}= -2D_{0}D_{1}q_{0}-\frac{2}{3}\omega\sigma q_{0}-\mu D_{0}q_{0}+ \alpha q_{0}^{3}+pD_{0}q_{0}(t-\tau). \tag{17}\]  
By solving Equation (16) for \(q_{0}\), we have  
\[q_{0}=A(T_{1})e^{i\omega T_{0}/3}+\frac{9}{16\omega^{2}}Fie^{i\omega T_{0}}+cc, \tag{18}\]  
where 'cc' denotes the conjugate term. Substituting Equation (18) into Equation (17) yields  
\[ \begin{split} D_{0}^{2}q_{1}+\frac{1}{9}\omega^{2}q_{1}=& \left[-\frac{2}{3}D_{1}i\omega A-\frac{2}{3}\omega\sigma A-\frac{i}{3}\mu \omega A+3\alpha A^{2}\overline{A}+\frac{243}{128\omega^{2}}\alpha F^{2}A+ \frac{27}{16\omega^{2}}i\alpha F\overline{A}\right.\\ &\left.-\frac{i}{3}\omega p\left(\cos\frac{1}{3}\omega\tau-i\sin \frac{1}{3}\omega\tau\right)A\right]e^{i\omega T_{0}/3}+cc.\end{split}  \tag{19}\]  
The secular term of Equation (19) vanishes if and only if  
\[ \begin{split}&-\frac{2}{3}D_{1}i\omega A-\frac{2}{3}\omega \sigma A-\frac{i}{3}\mu\omega A+3\alpha A^{2}\overline{A}+\frac{243}{128 \omega^{2}}\alpha F^{2}A\\ &+\frac{27}{16\omega^{2}}i\alpha F\overline{A}-\frac{i}{3}\omega p \left(\cos\frac{1}{3}\omega\tau-i\sin\frac{1}{3}\omega\tau\right)A=0.\end{split}  \tag{20}\]  
Letting \(A=a\exp(i\varphi)\) into Equation (20) and separating the real part and the imaginary part, we obtain the autonomous differential equations governing the amplitude and the phase  
\[ \begin{split} a^{\prime}&=-\frac{1}{2}(\mu+p\cos \frac{1}{3}\omega\tau)a-\frac{81}{32}\alpha Fa^{2}\cos\phi,\\ a\phi^{\prime}&=(\sigma+\frac{p}{2\omega}\sin\frac {1}{3}\omega\tau)a+\frac{729}{256\omega}\alpha Fa-\frac{9\alpha}{2\omega}a^{3} -\frac{81}{32}\alpha Fa^{2}\sin\phi,\end{split}  \tag{21}\]  
where \(\phi(T_{1})=\sigma T_{1}-3\varphi(T_{1})\). From Equation (21), we get a set of algebraic equations that governs the amplitude \(a\) and the phase \(\varphi\) of the steady-state \(1/3\) subharmonic resonance  
Figure 9: The primary resonance response with time delay \(\tau=\pi\), (**a**) The time history curves, (**b**) the phase diagram and (**c**) frequency power spectrum. \[-\frac{1}{2}(\mu+p\cos\frac{1}{3}\omega\tau)a =\frac{81}{32}\alpha Fa^{2}\cos\phi, \tag{22}\] \[(\sigma+\frac{p}{2\omega}\sin\frac{1}{3}\omega\tau)a+\frac{729}{256 \omega}\alpha Fa-\frac{9\alpha}{2\omega}a^{3} =\frac{81}{32}\alpha Fa^{2}\sin\phi,\]  
whereby we have the frequency response relation between \(a\) and \(\sigma\) and that between \(\varphi\) and \(\sigma\)  
\[\frac{1}{4}\mu_{e}^{2}a^{2}+\left(\sigma_{e}-\frac{729}{256\omega} \alpha F^{2}-\frac{9\alpha}{2\omega}a^{2}\right)^{2}a^{2}-(\frac{81}{32} \alpha F)^{2}a^{4} =0, \tag{23}\] \[\tan\phi+\frac{2\sigma_{e}+729\alpha F/(256\omega)-9\alpha a^{2} /\omega}{\mu_{e}} =0,\]  
where  
\[\mu_{e}=\mu+p\cos\frac{1}{3}\omega\tau,\sigma_{e}=\sigma+\frac{p}{2\omega} \sin\frac{1}{3}\omega\tau. \tag{24}\]  
The first equation in Equations (23) shows that either \(a=0\) or  
\[\frac{1}{4}\mu_{e}^{2}+\left(\sigma_{e}-\frac{729}{256\omega}\alpha F^{2}- \frac{9\alpha}{2\omega}a^{2}\right)^{2}=(\frac{81}{32}\alpha F)^{2}a^{2}, \tag{25}\]  
which is quadratic in \(a^{2}\).