## 2 Dynamic Model and Parameters Optimization
### Parameters Optimization  
The primary structure in Figure 1 is subjected to harmonic excitation \(F\cos(\omega t)\); hence, the H\({}_{\infty}\) optimization method is used to tuning the optimal parameters. For the given mass ratio \(\mu\), the amplification ratio \(L\), and inerter-mass ratio \(\beta\), the aim is to determine four optimal parameters \(\
u\), \(\alpha_{1}\), \(\alpha_{2}\), and \(\zeta\) relating to \(\mu\), \(L\), and \(\beta\), which minimize the maximum of \(A\) in Equation (6). The mathematical formula is as follows  
\[\min\Big{(}\max_{\
u,\alpha_{1},\alpha_{2},\zeta}\ A(\lambda)\Big{)}\]  
According to Equation (6), for given values of \(\mu\), \(\beta\), \(\
u\), \(\alpha_{1}\), \(\alpha_{2}\), \(L\), the amplitude amplification factor \(A\) of the primary system can be plotted with respect to \(\lambda\). In Figure 2, the normalized amplitude-frequency curves of model with several different damping ratios of 0.2, 0.5, and 0.8 are given. The curves all pass through three fixed points \(P\), \(Q\), and \(R\), which are independent of the damping ratio. The position of three fixed points in the amplitude frequency curve can be determined by the damping ratio \(\zeta=0\) and \(\zeta\rightarrow\infty\) in Equation (6). Aided by the idea of fixed-point theory, the response values are equal when the damping ratio approaches zero or infinity in Equation (6)  
\[\left|\frac{D_{1}}{D_{3}}\right|=\left|\frac{D_{2}}{D_{4}}\right| \tag{9}\]  
We further obtain a cubic equation of \(\lambda^{2}\)  
\[\lambda^{6}+p_{1}\lambda^{4}+p_{2}\lambda^{2}+p_{3}=0 \tag{10}\]  
where  
\[ \begin{split} p_{1}=&-1-\mu\
u^{2}L^{2}-\frac{( \alpha_{1}+2\alpha_{2}+2)\
u^{2}}{1+\beta}\\ p_{2}=&\frac{\big{[}(1+\mu\
u^{2}L^{2})(1+\alpha_ {1}+2\alpha_{2})+1]\
u^{2}}{1+\beta}+\frac{(1+\alpha_{2})h_{1}\
u^{4}}{(1+ \beta)^{2}}\\ p_{3}=&-\frac{\mu L^{2}[\alpha_{1}(1+2\alpha_{2}) +2\alpha_{2}(1+\alpha_{2})]\
u^{6}+2h_{1}(1+\alpha_{2})\
u^{4}}{2(1+\beta)^{2 }}\end{split}  \tag{11}\]  
Figure 2: The normalized amplitude–frequency curves of system with \(\mu=0.1\), \(\beta=1.5\), \(\
u=2\), \(\alpha_{1}=1\), \(\alpha_{2}=0.7\), \(L=3.5\). When \(\xi=0\), we can obtain  
\[A=\left|\frac{X_{1}}{\delta_{st}}\right|=\left|\frac{(1+\alpha_{2})v^{2}-(1+\beta )\lambda^{2}}{(1+\beta)\lambda^{4}-l_{1}\lambda^{2}+h_{2}}\right| \tag{12}\]  
when \(\xi\rightarrow\infty\), we can obtain  
\[A=\left|\frac{X_{1}}{\delta_{st}}\right|=\left|\frac{h_{1}v^{2}-(1+\beta) \lambda^{2}}{(1+\beta)\lambda^{4}-(l_{1}+\alpha_{1}v^{2})\lambda^{2}+l_{2}}\right| \tag{13}\]  
where  
\[l_{1}=(1+\beta)\Big{(}1+\mu v^{2}L^{2}\Big{)}+(1+\alpha_{2})v^{2},\quad l_{2}= h_{1}v^{2}+\mu L^{2}(\alpha_{1}+\alpha_{2})v^{4}\]  
When \(\xi=0\) and \(\xi\rightarrow\infty\), the amplitude amplification factor of the primary system corresponding to \(P\), \(Q\) and \(R\) are equal on the amplitude-frequency response curves, but the phase difference is \(180^{\circ}\), which is a positive and negative sign difference. Hence by Equations (12) and (13), we have  
\[A=\left|\frac{X_{1}}{\delta_{st}}\right|=\left|\frac{(\alpha_{1}+2\alpha_{2}+2 )v^{2}-2(1+\beta)\lambda^{2}}{\alpha_{1}v^{2}(1+\mu L^{2}v^{2}-\lambda^{2})}\right|\]  
Let \(\lambda_{P}^{2}\), \(\lambda_{Q}^{2}\) and \(\lambda_{R}^{2}\) be the three roots of Equation (10), and the ordinates of \(P\), \(Q\) and \(R\) are as follows.