# Nonlinear Vibrations of Carbon Nanotubes with Thermal-Electro-Mechanical Coupling
## 3 Results
### Subharmonics Resonance of SWCNTs  
When \(\Omega\) is far away from \(\omega_{j}\), a large vibrational amplitude requires a large exciting amplitude. In this paper, we will study the subharmonic resonant. Therefore, we rewrite Equation (9) as:  
\[\ddot{\eta}_{j}+\omega_{j}^{2}\eta_{j}+2\varepsilon\overline{\epsilon}_{j}\dot{ \eta}_{j}+\varepsilon\overline{k}_{j3}\eta_{1}^{3}=f_{j}\cos\omega\overline{t}, \ j=1,2. \tag{23}\]  
The seeking form of the solution is:  
\[\eta_{j}=\eta_{j1}(T_{0},T_{1})+\varepsilon\eta_{j2}(T_{0},T_{1}) \tag{24}\]  
Substituting Equation (24) into Equation (23) and taking the coefficients of order \(\varepsilon\) equal, we obtain [32]:  
\[\varepsilon^{0}:\ D_{0}^{2}\eta_{j1}+\omega_{j}^{2}\eta_{j1}=f_{j}\cos\bigl{(} \Omega T_{0}+\tau_{j}\bigr{)} \tag{25}\]  
\[\varepsilon:\ D_{0}^{2}\eta_{j2}+\omega_{j}^{2}\eta_{j2}=-2D_{0}D_{1}\eta_{j1 }-2\overline{\epsilon}_{j}D_{0}\eta_{j1}-\overline{k}_{j3}\eta_{j1}^{3} \tag{26}\]  
The form of the solution of Equation (25) is:  
\[\eta_{j1}=A(T_{1})\exp\bigl{(}i\omega_{j}T_{0}\bigr{)}+\Lambda(T_{1})\exp(i \Omega T_{0})+cc \tag{27}\]  
Figure 8: Excitation amplitude response curves of the first mode with \((\varepsilon_{1},\sigma)=(5,15)\). Figure 9: Excitation amplitude response curves of the second mode with \((\varepsilon_{2},\sigma)=(5,20)\). where \(cc\) stands for the complex conjugate of the preceding terms and \(\Lambda=\frac{1}{2}f_{j}\Big{(}\omega_{j}^{2}-\Omega^{2}\Big{)}^{-1}\). When Equation (27) is brought into Equation (26), we can obtain:  
\[ \begin{array}{l}D_{1}^{2}\eta_{j2}+\omega_{j}^{2}\eta_{j2}=-\left[2i\omega_{j} \big{(}A^{\prime}+\overline{c}_{j}A\big{)}+6\overline{k}_{j3}A\Lambda^{2}+3 \overline{k}_{j3}A^{2}\overline{A}\right]\exp\left(i\omega_{j}T_{0}\right)\\ -\overline{k}_{j3}\Big{\{}A^{3}\exp\left(3i\omega_{j}T_{0}\right)+\Lambda^{3} \exp\left(3i\Omega T_{0}\right)+3A^{2}\Lambda\exp\left(i\left(2\omega_{j}+ \Omega\right)T_{0}\right)\\ +3\overline{A}^{2}\Lambda\exp\left(i\left(\Omega-2\omega_{j}\right)T_{0} \right)+3A\Lambda^{2}\exp\left(i\left(\omega_{j}+2\Omega\right)T_{0}\right)\\ +3A\Lambda^{2}\exp\left(i\left(\omega_{j}-2\Omega\right)T_{0}\right)\Big{\}}- \Lambda\Big{[}2i\overline{c}_{j}\Omega+3\overline{k}_{j3}\Lambda^{2}+6 \overline{k}_{j3}A\overline{A}\Big{]}\exp(i\Omega T_{0})+cc\end{array}  \tag{28}\]  
Introducing a dimensionless detuning parameter \(\sigma\), the exciting frequency can be written as \(\Omega=3\omega_{j}+\varepsilon\sigma\), and eliminating the duration terms of \(\exp\left[i\big{(}\Omega-2\omega_{j}\big{)}T_{0}\right]\), we obtain:  
\[2i\omega_{j}\big{(}A^{\prime}+\overline{c}_{j}A\big{)}+6\overline{k}_{j3}A \Lambda^{2}+3\overline{k}_{j3}A^{2}\overline{A}+3\overline{k}_{j3}\Lambda \overline{A}^{2}\exp(i\sigma T_{1})=0 \tag{29}\]  
Introducing the polar form of \(A\), hawse obtain:  
\[A=\frac{1}{2}a\exp\left(i\beta\right), \tag{30}\]  
where \(a\) and \(\beta\) are real functions. Substituting Equation (30) into Equation (29) and separating the real part and imaginary parts, we obtain:  
\[\left\{\begin{array}{l}a^{\prime}=-\overline{c}_{j}a-\frac{3\overline{k}_{j 3}\Lambda}{4\omega_{j}}a^{2}\sin(\sigma T_{1  
The equations are a quadratic function solution of \(a^{2}\), and their solution is:  
\[a^{2}=p\pm\left(p^{2}-q^{2}\right)^{1/2}, \tag{37}\]  
Here,  
\[p=\frac{8\omega_{j}\sigma}{\overline{k}_{j3}}-6\Lambda^{2},\ q=\frac{64\omega_{ j}^{2}}{81\overline{k}_{j3}^{2}}\left[9\overline{c}_{j}^{2}+\left(\sigma- \frac{9\overline{k}_{j3}\Lambda^{2}}{\omega_{j}}\right)^{2}\right]. \tag{38}\]  
We notice that \(q\) is always positive; therefore, there is a nontrivial free vibration amplitude if \(p>0\) and \(p^{2}\geq q\). This means \(\Lambda^{2}<\frac{4\omega_{j}\sigma}{2\overline{k}_{j3}}\) and \(\frac{\overline{k}_{j3}\Lambda^{2}}{\omega_{j}}\bigg{(}\sigma-\frac{6\overline {3}_{j3}\Lambda^{2}}{8\omega_{j}}\bigg{)}-2\overline{c}_{j}^{2}\geq 0\).