# Scattering Reduction and Resonant Trapping of Flexural Waves: Two Rings to Rule Them
## 2 Analytical Methods
### Trapped Vibrations in a Double-Ring Cluster of Point Resonators
#### 2.2.2 Evaluation of the Integrals in the Secular Equation  
The integrals in (13) and (14) and in the secular Equation (15) can be evaluated in the closed analytical form.  
We have two rings with radii \(R_{1}\) and \(R_{2}\), and assume \(R_{1}>R_{2}\). Let \(\theta_{1}\) and \(\theta_{2}\) denote polar angles on the two rings. The integral we will evaluate analytically is:  
\[\mathcal{I}_{2,1}(q,\beta)=\int_{0}^{2\pi}\mathrm{e}^{\mathrm{i}q(\theta_{2}- \theta_{1})}G(|R_{1}\mathrm{e}^{\mathrm{i}\theta_{1}}-R_{2}\mathrm{e}^{ \mathrm{i}\theta_{2}}|)\mathrm{d}\theta_{1}. \tag{20}\]  
Here, the Green's function \(G\) for the biharmonic problem is  
\[G(r,\beta)=\frac{-\mathrm{i}}{8\beta^{2}}[H_{0}^{(1)}(\beta r)-K_{0}(\beta r)]. \tag{21}\]  
The tool we use in the evaluation is [PERSON]'s addition theorem (see Section 10.23 of [38] and [39]). This can be written for (length, angle) pairs \((u,\theta_{u})\), \((v,\theta_{v})\), \(w=u\cos(\theta_{u})-v\cos(\theta_{v})+\mathrm{i}[u\sin(\theta_{u})-v\sin( \theta_{v})]\):  
\[H_{v}^{(1)}(|w|)\mathrm{e}^{\mathrm{i}v[\arg(w)-\theta_{u}]}=\sum_{k=-\infty }^{\infty}H_{v+k}^{(1)}(u)J_{k}(v)\exp[\mathrm{i}k(\theta_{u}-\theta_{v})] \tag{22}\]  
and  
\[K_{v}(|w|)\mathrm{e}^{\mathrm{i}v[\arg(w)-\theta_{u}]}=\sum_{k=-\infty}^{ \infty}K_{v+k}(u)J_{k}(v)\exp[\mathrm{i}k(\theta_{u}-\theta_{v})]. \tag{23}\]  
Here, \(v\) is an integer, and \(u>v\) is assumed.  
In applying this to \(G\), we need only the case \(v=0\):  
\[H_{0}^{(1)}(|w|)=\sum_{k=-\infty}^{\infty}H_{k}^{(1)}(u)J_{k}(v)\exp[\mathrm{ i}k(\theta_{u}-\theta_{v})] \tag{24}\]  
and  
\[K_{0}(|w|)=\sum_{k=-\infty}^{\infty}K_{k}(u)J_{k}(v)\exp[\mathrm{i}k(\theta_{ u}-\theta_{v})]. \tag{25}\]  
Figure 3: Plate incorporating a double-ring of masses or spring-mass resonators.  
Using (24) and (25) in (20), we obtain  
\[\begin{split}\mathcal{I}_{2,1}(q,\beta)=\frac{-\mathrm{i}}{8\beta^{2 }}\int_{0}^{2\pi}\mathrm{e}^{\mathrm{i}q(\theta_{2}-\theta_{1})}\mathrm{d} \theta_{1}\\ \sum_{k=-\infty}^{\infty}\mathrm{e}^{\mathrm{i}k(\theta_{1}- \theta_{2}))}\bigg{[}H_{k}^{(1)}(\beta R_{1})J_{k}(\beta R_{2})+\frac{2 \mathrm{i}}{\pi}K_{k}(\beta R_{1})I_{k}(\beta R_{2})\bigg{]}.\end{split} \tag{26}\]  
Carrying out the integral over \(\theta_{1}\), the result is:  
\[\mathcal{I}_{2,1}(k,\beta)=\frac{\pi}{4\beta^{2}}\bigg{[}-\mathrm{i}J_{k}( \beta R_{1})J_{k}(\beta R_{2})+Y_{k}(\beta R_{1})J_{k}(\beta R_{2})+\frac{2 \mathrm{i}}{\pi}K_{k}(\beta R_{1})I_{k}(\beta R_{2})\bigg{]}. \tag{27}\]  
This obeys the symmetry relation \(\mathcal{I}_{2,1}(k,\beta)=\mathcal{I}_{2,1}(-k,\beta)\). We have derived (27) for \(\beta\) real, but it can be extended to \(\beta\) complex by analytic continuation.  
We can also consider  
\[\mathcal{I}_{1,2}(q,\beta)=\int_{0}^{2\pi}\mathrm{e}^{\mathrm{i}q(\theta_{2}- \theta_{1})}G(|R_{1}\mathrm{e}^{\mathrm{i}\theta_{1}}-R_{2}\mathrm{e}^{\mathrm{ i}\theta_{2}}|)\mathrm{d}\theta_{2}. \tag{28}\]  
Then it follows from the symmetry rule just given that \(\mathcal{I}_{1,2}(q,\beta)=\mathcal{I}_{2,1}(q,\beta)\).  
The remaining integrals to calculate are  
\[\mathcal{I}_{1,1}(k,\beta)=\frac{\pi}{4\beta^{2}}\bigg{[}-\mathrm{i}J_{k}( \beta R_{1})^{2}+Y_{k}(\beta R_{1})J_{k}(\beta R_{1})+\frac{2}{\pi}K_{k}( \beta R_{1})I_{k}(\beta R_{1})\bigg{]} \tag{29}\]  
and  
\[\mathcal{I}_{2,2}(k,\beta)=\frac{\pi}{4\beta^{2}}\bigg{[}-\mathrm{i}J_{k}( \beta R_{2})^{2}+Y_{k}(\beta R_{2})J_{k}(\beta R_{2})+\frac{2}{\pi}K_{k}( \beta R_{2})I_{k}(\beta R_{2})\bigg{]}. \tag{30}\]  
These agree with the results quoted in [32].