## 3 Superharmonic Resonance  
We use the method of multiple scales [55; 56; 57] to solve three superharmonic resonance. The adjusting parameters are as follows: \(\mu_{n}=O(\varepsilon),\Gamma_{mmnn}=O(\varepsilon),k_{ann}=O(\varepsilon),f_{n}= O(\varepsilon),3\Omega=\omega_{0}+\varepsilon\sigma,\sigma=O(1).\) We express the solution of Equation (13) in the form:  
\[q_{n}(t;\varepsilon)=q_{n0}(T_{0},T_{1},...)+\varepsilon q_{n1}(T_{0},T_{1},...)+..., \tag{14}\]  
Substituting Equation (14) into Equation (13) and equating the coefficients of \(\varepsilon^{0}\) and \(\varepsilon^{1}\) on both sides, we obtain:  
\[D_{0}^{2}q_{n0}+\omega_{0}^{2}q_{n0}=f_{n}\cos\Omega T_{0}, \tag{15}\]  
\[D_{0}^{2}q_{n1}+\omega_{0}^{2}q_{n1}=-2D_{0}D_{1}q_{n0}-\mu_{n}D_{0}q_{n0}- \Gamma_{mmnn}q_{n0}^{3}-k_{ann}q_{n0}(t-\tau). \tag{16}\]  
The general solution of Equation (15) can be written as:  
\[q_{n0}=A_{n}(T_{1})\exp(i\omega_{0}T_{0})+\Lambda_{n}\exp(i\Omega T_{0})+cc, \tag{17}\]  
where \(i=\sqrt{-1},\Lambda_{n}=\frac{1}{2}f_{n}(\omega_{0}^{2}-\Omega^{2})^{-1},\) and \(cc\) stands for the complex conjugate of the preceding terms. Substituting \(q_{n0}\) into Equation (16), we obtain:  
\[ \begin{split} D_{0}^{2}q_{n1}+\omega_{0}^{2}q_{n1}=& -[i\omega_{0}(2A_{n}^{\prime}+\mu_{n}A_{n})+6\Gamma_{mmnn}A_{n}^{ 2}+3\Gamma_{mmnn}A_{n}^{2}\bar{A}_{n}\\ &+k_{ann}A_{n}\exp(-i\omega_{0}\tau)]\exp(i\omega_{0}T_{0})- \Gamma_{mmnn}A_{n}^{3}\exp(3i\omega_{0}T_{0})\\ &+\Gamma_{mmn}\Lambda_{n}^{3}\exp(3i\Omega T_{0})+3\Gamma_{mmn}A_ {n}^{2}\Lambda_{n}\exp[i(2\omega_{0}+\Omega)T_{0}]\\ &+3\Gamma_{mmnn}A_{n}^{2}\Lambda_{n}\exp[i(\Omega-2\omega_{0})T_ {0}]+3\Gamma_{mmnn}A_{n}\Lambda_{n}^{2}\exp[i(\omega_{0}+2\Omega)T_{0}]\\ &+3\Gamma_{mmn}A_{n}\Lambda_{n}^{2}\exp[i(\omega_{0}-2\Omega)T_ {0}]-\Lambda_{n}[i\mu_{n}\Omega+3\Gamma_{mmn}\Lambda_{n}^{2}+6\Gamma A_{n} \bar{A}_{n}]\\ &\exp(i\Omega T_{0})-k_{ann}\Lambda_{n}\exp[i\omega_{0}(T_{0}- \tau)]+cc.\end{split}  \tag{18}\]  
Secular terms will be eliminated from the particular solution of Equation (18), if we let:  
\[ \begin{split} i\omega_{0}(2A_{n}^{\prime}+\mu_{n}A_{n})+6\Gamma_ {mmnn}A_{n}\Lambda_{n}^{2}+3\Gamma_{mmn}A_{n}^{2}\bar{A}_{n}+\Gamma_{mmn} \Lambda_{n}^{3}\exp(i\sigma T_{1})\\ +k_{ann}A_{n}\exp(-i\omega_{0}\tau)=0.\end{split}  \tag{19}\]  
To solve Equation (19), we write \(A_{n}\) in the polar form:  
\[A_{n}=\frac{1}{2}a_{n}\exp(i\beta_{n}), \tag{20}\]  
where \(a_{n}\) and \(\beta_{n}\) are real functions of \(T_{1}\). Substituting Equation (20) into Equation (19) and separating the result into real and imaginary parts, we have:  
\[a_{n}^{\prime}=-\frac{1}{2}\mu_{ne}a_{n}-\frac{\Gamma_{mmn}\Lambda_{n}^{3}}{ \omega_{0}}\sin\gamma_{n}, \tag{21}\]  
\[a_{n}\gamma_{n}^{\prime}=\sigma_{e}a_{n}-\frac{3\Gamma_{mmn}\Lambda_{n}^{2}}{ \omega_{0}}a_{n}-\frac{3\Gamma_{mmn}\Lambda_{n}^{3}}{8\omega_{0}}a_{n}^{3}- \frac{\Gamma_{mmn}\Lambda_{n}^{3}}{\omega_{0}}\cos\gamma_{n}, \tag{22}\]  
where \(\gamma_{n}=\sigma T_{1}-\beta_{n},\mu_{ne}=\mu_{n}-k_{ann}\sin(\omega_{0}\tau) /\omega_{0}\) and \(\sigma_{e}=\sigma-k_{ann}\cos(\omega_{0}\tau)/(2\omega_{0}).\) Therefore, for the first approximation:  
\[q_{n}=\frac{1}{2}a_{n}\cos(3\Omega t-\gamma_{n})+f_{n}(\omega_{0}^{2}-\Omega^ {2})^{-1}\cos\Omega t+O(\varepsilon). \tag{23}\]  
where \(a_{n}\) and \(\gamma_{n}\) are defined by Equations (21) and (22).