## 2 Statement of the Energy Harvester Problem
This set of pairwise coupled equations is in contrast to the Euler-Bernoulli formulation in which the transverse dynamics is directly coupled to the electrical dynamics (see [PERSON] [30], [PERSON] and [PERSON] [31; 32], [PERSON] [1; 2; 3], [PERSON] and [PERSON] [4], and [PERSON] et al. [5]). In our first statement, we justify the reformulation of the boundary-value problem in Equations (1)-(5) in the form that does not contain any distributional terms. **Proposition 1**. : _The boundary-value problem given by Equations (1)-(5) is equivalent to the following boundary-value problem:_  
\[\rho(x)W_{tt}(x,t)+\frac{d}{dx}\bigg{(}K(x)\Big{(}\Phi(x,t)-W_{x}(x,t)\Big{)} \bigg{)}=-\rho(x)y_{tt}(t), \tag{6}\]  
\[S\rho\Phi_{tt}(x,t)-\frac{d}{dx}\Big{(}EI(x)\Phi_{x}(x,t)\Big{)}+K(x)\Big{(} \Phi(x,t)-W_{x}(x,t)\Big{)}=0, \tag{7}\]  
\[v_{t}(t)+\frac{\theta}{C}\Phi_{t}(L,t)+\frac{1}{CR}v(t)=0. \tag{8}\]  
_The boundary conditions are:_  
\[W(0,t)=\Phi(0,t)=0; \tag{9}\]  
\[\Phi(L,t)-W_{x}(L,t)=0,\qquad EI(L)\Phi_{x}(L,t)=\theta v(t). \tag{10}\]  
**Remark 1**. : _The only difference between Equations (1)-(3) and Equations (7)-(8) is that the distributional term, \(\theta v(t)\Big{(}\partial(x)-\delta(x-L)\Big{)}\), in Equation (2) is replaced by an additional term \(\theta v(t)\) in the right-hand side of the boundary condition in Equation (10)._  
Proof of Proposition 1.: To show that the two problems are equivalent, we check that their weak (variational) formulations coincide. Consider the following class of functions:  
\[\mathcal{P}=\Big{\{}\varphi\in C^{\infty}[0,L],\ \ \varphi(0)=0.\Big{\}}\]  
Multiplying Equation (1) by \(\varphi\in\mathcal{P}\) and integrating, we have  
\[ \begin{split}&\int_{0}^{L}\rho(\xi)W_{tt}(\xi,t)\varphi(\xi)d \xi+\int_{0}^{L}K(\xi)\Big{(}\Phi(\xi,t)-W_{x}(\xi,t)\Big{)}^{\prime}\varphi (\xi)d\xi=\\ &-\int_{0}^{L}\rho(\xi)y_{tt}\varphi(\xi)d\xi.\end{split}  \tag{11}\]Taking into account that \(\Phi(0,t)=0\), and the first condition of Equation (10), we obtain that  
\[\int_{0}^{L}\Big{(}W_{tt}(\xi,t)+y_{tt}(t)\Big{)}\rho(\xi)\varphi(\xi)d\xi=\int_{ 0}^{L}K(\xi)\Big{(}\Phi(\xi,t)-W^{\prime}(\xi,t)\Big{)}\varphi^{\prime}(\xi)d\xi. \tag{12}\]  
Now, we multiply Equation (2) by \(\psi\in\mathcal{P}\) and integrate to have  
\[ \begin{split}&\int_{0}^{L}S\rho(\xi)\Phi_{tt}(\xi,t)\psi(\xi)d \xi-\int_{0}^{L}\Big{(}EI(\xi)\Phi^{\prime}(\xi,t)\Big{)}^{\prime}\psi(\xi)d \xi+\\ &\int_{0}^{L}K(\xi)\Big{(}\Phi(\xi,t)-W^{\prime}(\xi,t)\Big{)} \psi(\xi)d\xi+\theta v(t)\Big{(}\delta(0)-\delta(L)\Big{)}=0.\end{split}  \tag{13}\]  
Taking into account that \(\psi(0)=0\) and \(\Phi(L,t)=0\), we reduce Equation (13) to  
\[ \begin{split}&\int_{0}^{L}S\rho(\xi)\Phi_{tt}(\xi,t)\psi(\xi)d \xi+\int_{0}^{L}EI(\xi)\Phi^{\prime}(\xi,t)\psi^{\prime}(\xi)d\xi+\\ &\int_{0}^{L}K(\xi)\Big{(}\Phi(\xi,t)-W^{\prime}(\xi,t)\Big{)} \psi(\xi)d\xi-\theta v(t)\delta(L)=0.\end{split}  \tag{14}\]  
Thus, Equations (3), (12) and (14) constitute the weak formulation of the problem given by Equations (1)-(3). To obtain the weak formulation of the problem given by Equations (7)-(8), we multiply Equation (7) by \(\varphi(\xi)\) and integrate, which immediately yields Equation (13). Finally, we multiply Equation (10) by \(\psi(\xi)\) and integrate to have  
\[ \begin{split}&\int_{0}^{L}S\rho(\xi)\Phi_{tt}(\xi,t)\psi(\xi)d \xi+\int_{0}^{L}EI(\xi)\Phi^{\prime}(\xi,t)\psi^{\prime}(\xi)d\xi+\\ &\int_{0}^{L}K(\xi)\Big{(}\Phi(\xi,t)-W^{\prime}(\xi,t)\Big{)} \psi(\xi)d\xi-EI(L)\Phi^{\prime}(L,t)\psi(L)+EI(0)\Phi^{\prime}(0,t)\psi(0)=0. \end{split}  \tag{15}\]  
If we take into account that \(\psi(0)=0\) and the second boundary condition from Equation (10), we immediately reduce Equation (15) to the form of Equation (14). The fact that weak formulation in Equations (14), (15), (4), and (5) is the same for both problems in Equation (1)-(3), (4), (5), and (7)-(8) with Equations (9) and (10) implies that these two problems are equivalent.