# Nonlinear Dynamics of an Electromagnetically Actuated
## 6 Analysis of the Stability for the Primary Resonance  
### Study of Stability of Steady-State Motion  
In this section, we reconsider the primary resonance \(\omega=\Omega_{0}+\delta\epsilon\Omega_{0}=\sqrt{1-4a_{0}}\), using the two-variable expansion method [29]. In what follows, we will use the notation \(\xi\), which represents stretched time \(\omega t\), and \(\eta\), which represents slow time \(\epsilon\)t:  
\[\xi=\omega t,\ \eta=\epsilon t \tag{48}\]  
In order to substitute these transformations into Equation (19), we needed the following expressions:  
\[ \begin{array}{l}\frac{d\epsilon}{dt}=\frac{\partial z}{\partial\xi}\frac{d \xi}{dt}+\frac{\partial z}{\partial\eta}\frac{d\eta}{dt}=\omega\frac{\partial z }{\partial\xi}+\epsilon\frac{\partial z}{\partial\eta}\\ \frac{d\epsilon^{2}z}{dt^{2}}=\omega^{2}\frac{\partial^{2}z}{\partial\xi^{2} }+2\omega\epsilon\frac{\partial^{2}z}{\partial\xi\partial\eta}+\epsilon^{2} \frac{\partial^{2}z}{\partial\eta^{2}}\end{array}  \tag{49}\]  
Inserting Equation (49) into Equation (19) and then expanding \(z\) in the power series \(z(\xi,\eta)=z_{0}(\xi,\eta)+\epsilon z_{1}(\xi,\eta)\), we obtained the following partial differential equation:  
\[ \begin{array}{l}(\Omega_{0}+\delta\epsilon)^{2}\Big{(}\frac{\partial^{2}z_ {0}}{\partial\xi^{2}}+\epsilon\frac{\partial^{2}z_{1}}{\partial\xi^{2}}\Big{)} +2(\Omega_{0}+\delta\epsilon)\epsilon\Big{(}\frac{\partial^{2}z_{0}}{ \partial\xi\partial\eta}+\epsilon\frac{\partial^{2}z_{1}}{\partial\xi\partial \eta}\Big{)}+\epsilon^{2}\Big{(}\frac{\partial^{2}z_{0}}{\partial\xi^{2}}+ \epsilon\frac{\partial^{2}z_{1}}{\partial\xi\eta}\Big{)}+\epsilon c\Big{[}( \Omega_{0}+\delta\epsilon)\Big{(}\frac{\partial z_{0}}{\partial\xi}+\epsilon \frac{\partial z_{1}}{\partial\eta}\Big{)}\Big{]}+(z_{0}+\epsilon z_{1})\Omega _{0}^{2}+\epsilon(k-8a_{0})(z_{0}^{3}+3z_{0}^{2}\epsilon)-12a_{0}\epsilon(z_{ 0}^{5}+5z_{0}^{4}z_{1})-16a_{0}\epsilon(z_{0}^{7}+7z_{0}^{6}z_{1}\epsilon)- \epsilon f\cos\xi=0\end{array}  \tag{50}\]  
Neglecting the terms of \(0(\epsilon^{2})\), after collecting terms, we can obtain the following:  
\[\Omega_{0}^{2}\bigg{(}\frac{\partial^{2}z_{0}}{\partial\xi^{2}}+z_{0}\bigg{)}=0 \tag{51}\]  
\[\Omega_{0}^{2}\bigg{[}\frac{\partial^{2}z_{1}}{\partial\xi^{2}}\ +z_{1}+\frac{2\delta}{ \Omega_{0}}\frac{\partial^{2}z_{0}}{\partial\xi^{2}}+\frac{2}{\Omega_{0}} \frac{\partial^{2}z_{0}}{\partial\xi\partial\eta}+\frac{c}{\Omega_{0}}\frac{ \partial z_{0}}{\partial\For nonresonant terms in Equation (52), the coefficients of \(\sin\xi\), and \(\cos\xi\), are needed vanish, such that we have the following slow flow:  
\[2\frac{dA}{d\eta}+cA+2\delta B-\frac{3(k-8a_{0})}{4\Omega_{0}}B(A^{2}+B^{2})+ \frac{15a_{0}}{2\Omega_{0}}B(A^{2}+B^{2})^{2}+\frac{35a_{0}}{4\Omega_{0}}B(A^{ 2}+B^{2})^{3}=0 \tag{54}\]  
\[2\frac{dB}{d\eta}+cB-2\delta A+\frac{3(k-8a_{0})}{4\Omega_{0}}A(A^{2}+B^{2})- \frac{15a_{0}}{2\Omega_{0}}B(A^{2}+B^{2})^{2}-\frac{35a_{0}}{4\Omega_{0}}A(A^{ 2}+B^{2})^{3}=\frac{f}{\Omega_{0}^{2}} \tag{55}\]  
Equilibrium points of the slow flow (54) and (55) correspond to the periodic motion of Equation (19). To determine them, the following is set:  
\[\frac{dA}{d\eta}=\frac{dB}{d\eta}=0 \tag{56}\]  
Using the notation \(B\) = Ay from Equations (54)-(56), it holds that  
\[A=\frac{fy}{\Omega_{0}c(1+y^{2})},\ B=\frac{fy^{2}}{\Omega_{0}c(1+y^{2})} \tag{57}\]  
where \(y\) is obtained from the following algebraic expression:  
\[\left(\frac{y^{2}}{1+y^{2}}\right)^{3}+\frac{6}{7}\left(\frac{\Omega_{0}c}{f} \right)^{2}\left(\frac{y^{2}}{1+y^{2}}\right)^{2}-\frac{3(k-8a_{0})}{35\Omega_ {0}}\left(\frac{\Omega_{0}c}{f}\right)^{4}\frac{y^{2}}{1+y^{2}}+\frac{4c\Omega _{0}}{35a_{0}y}\left(\frac{\Omega_{0}c}{f}\right)^{6}+\frac{8\Omega_{0}}{35a_ {0}}\left(\frac{\Omega_{0}c}{f}\right)^{6}\delta=0 \tag{58}\]  
Figures 10-12 plot the variation of y with respect to \(\delta\) for \(\delta\in[-0.4,\,0)\cup(0,\,0.04]\) and \(A\) and \(B\) with respect to \(\delta\) for \(c=0.08,f=0.0008,\,a_{0}=0.001,\) and \(k=2\).