## 3 Applying Multiple Scales Lindstedt-Poincare
### Applying MSLP on the Damper with Cubic Nonlinearity
 \tag{27}\]  
Setting \(D_{2}A\) equal to zero leads to a complex expression for \(\omega_{n}\), as described above in this case, \(\omega_{2}\) is set to zero, leading to  
\[2 iD_{2}A_{c}=-\frac{id^{*}A_{c}}{\omega}-\frac{3\alpha^{*^{2}}A_{c}^{3} \overline{A_{c}}^{2}}{8\omega^{4}}+\frac{\varphi_{i,j}q^{*}}{2}exp(i\sigma\tau _{2}). \tag{28}\]  
Using the Nayfehs method of recombination [27]:  
\[\frac{d}{dt}A_{c}=\frac{\partial}{\partial\tau_{0}}\frac{d\tau_{0}}{dt}A_{c}+ \frac{\partial}{\partial\tau_{1}}\frac{d\tau_{1}}{dt}A_{c}+\frac{\partial}{ \partial\tau_{2}}\frac{d\tau_{2}}{dt}A_{c}=0+\varepsilon D_{1}A_{c}+\varepsilon ^{2}D_{2}A_{c} \tag{29}\]  
the time derivative of the complex amplitude \(A_{c}=\frac{\partial}{2}exp(i\gamma)\) is  
\[ \begin{array}{ll}2i\Big{(}\frac{q^{*}}{2}exp(i\gamma)+\frac{i\gamma^{\prime}q }{2}exp(i\gamma)\Big{)}\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad=\varepsilon^{2}\Big{(}-\frac{id^ {*}\partial}{2\omega}exp(i\gamma)-\frac{3\alpha^{*^{2}}2\delta^{5}}{256\omega^ {4}}exp(i\gamma)+\frac{\varphi_{i,j}q^{*}}{2}exp(i\sigma\tau_{2})\Big{)}\end{array}  \tag{30}\]  
and introducing \(\beta=\sigma\tau_{2}-\gamma\) can rearrange the equation after a little manipulation  
\[\hat{u}^{\prime}-\gamma^{\prime}\hat{u}=\epsilon^{2}\bigg{(}-\frac{id^{*}\hat {u}}{2\omega}-\frac{3\alpha^{*^{2}}\hat{u}^{5}}{256\omega^{4}}+\frac{\varphi_{i,j}q^{*}}{2}exp(i\beta)\bigg{)} \tag{31}\]where a split into real and imaginary parts conducted for the steady state  
\[Im:\dot{u}^{\prime}=\varepsilon^{2}\bigg{(}-\frac{d^{*}}{2\omega}\dot{u}+\frac{ \varphi_{i,i}\dot{q}^{*}}{2}sin(\beta)\bigg{)}=0 \tag{32}\]  
\[Re:\;\beta^{\prime}=\varepsilon^{2}\bigg{(}\sigma-\frac{3\alpha^{*2}\dot{u}^{4 }}{256\omega^{4}}+\frac{\varphi_{i,i}\dot{q}^{*}}{2\dot{u}}cos(\beta)\bigg{)}=0. \tag{33}\]  
By using mathematical identities:  
\[sin(\beta)=\frac{d^{*}\dot{u}}{\varphi_{i,i}\dot{q}^{*}\omega} \tag{34}\]  
\[\sigma=\frac{3\alpha^{*2}\dot{u}^{4}}{256\omega^{4}}-\frac{\varphi_{i,j}\dot{q }^{*}}{2\dot{u}}cos(\beta)=\frac{3\alpha^{*2}\dot{u}^{4}}{256\omega^{4}}\pm \sqrt{\frac{\left(\varphi_{i,i}\dot{q}^{*}\right)^{2}}{4\dot{u}^{2}}-\frac{d^{ *2}}{4\omega^{2}}} \tag{35}\]  
is obtained and yields the frequency-response curve. \[\Omega=\omega\Big{(}1+\varepsilon^{2}\sigma\Big{)}=\omega\left(1+\varepsilon ^{2}\left(\frac{3\alpha^{*2}\dot{u}^{4}}{256\omega^{4}}\pm\sqrt{\frac{\left( \varphi_{i,j}\dot{q}^{*}\right)^{2}}{4\dot{u}^{2}}-\frac{d^{*2}}{4\omega^{2}} }\right)\right) \tag{36}\]  
with  
\[\omega^{2}=\omega_{0}^{2}+\varepsilon\omega_{1}\rightarrow\omega=\sqrt{ \omega_{0}^{2}+\varepsilon\frac{3\alpha^{*}\dot{u}^{2}}{4}}. \tag{37}\]  
Without \(\varepsilon\) the frequency-response curve is  
\[\Omega=\omega\Big{(}1+\varepsilon^{2}\sigma\Big{)}=\omega\left(1+\frac{3 \alpha^{2}\dot{u}^{4}}{256\omega^{4}\dot{J}^{2}}\pm\sqrt{\frac{\left(\varphi_ {i,i}\dot{q}\right)^{2}}{4\dot{u}^{2}}-\frac{d^{2}}{4\omega^{2}\dot{J}^{2}}} \right);\;\omega=\sqrt{\omega_{0}^{2}+\frac{3\alpha\dot{u}^{2}}{4\dot{J}}}. \tag{38}\]  
The solution of the equation of motion is obtained by using the solutions of the differential equations of \(O(1)\) and \(O(\varepsilon)\). \[u(t)=\dot{u}\cos(\Omega t-\beta)+\frac{\alpha\dot{u}^{3}}{32\omega^{2}\dot{J} }\cos(3(\Omega t-\beta)). \tag{39}\]