# Optimizing Energy Conversion in a Piezo Disk Using a Controlled Supply of Electrical Load
## 2 Non-Axisymmetric Planar Oscillations of a Piezoceramic Thin Disk
\begin{array}{l}\text{grad}\psi=\vec{e}_{r}\frac{\partial\psi}{\partial r}+ \frac{\tilde{e}_{r}}{r}\frac{\partial\psi}{\partial\varphi}+\vec{k}\frac{ \partial\psi}{\partial z},\\ \text{div}\vec{u}=\frac{1}{\partial r}\frac{\partial\psi}{\partial r}+\frac{1} {r}\frac{\partial\psi}{\partial r}+\frac{\partial\psi}{\partial z},\\ \text{rot}\vec{u}=\vec{e}_{r}\Big{(}\frac{1}{r}\frac{\partial u_{r}}{\partial \varphi}-\frac{\partial u_{\varphi}}{\partial z}\Big{)}+\vec{e}_{\varphi} \Big{(}\frac{\partial u_{r}}{\partial z}-\frac{\partial u_{\varphi}}{ \partial r}\Big{)}+\vec{k}\Big{(}\frac{1}{r}\frac{\partial(rn_{\varphi})}{ \partial r}-\frac{1}{r}\frac{\partial u_{\varphi}}{\partial\varphi}\Big{)}. \end{array}  \tag{9}\]  
Let us write vector Equation (3) in the form of a system of two scalar differential equations with respect to the divergence \(\text{div}\vec{u}\) of displacements and nonzero component \(\omega_{z}\) of the vortex field of displacements:  
\[\frac{2}{1-\
u}\frac{\partial}{\partial r}(\text{div}\vec{u})-\frac{1}{r} \frac{\partial}{\partial\varphi}\omega_{z}-\frac{\partial}{\partial r}\left(2 d_{31}\frac{1+\
u}{1-\
u}E_{z}\right)=2\rho s_{11}^{E}(1+\
u)\frac{\partial^{2}u_{ r}}{\partial t^{2}}, \tag{10}\]  
\[\frac{2}{(1-\
u)r}\frac{\partial(\text{div}\vec{u})}{\partial\varphi}-\frac{ \partial\omega_{z}}{\partial r}-\frac{1}{r}\frac{\partial\left(2d_{31}\frac{ 1+\
u}{1-\
u}E_{z}\right)}{\partial\varphi}=2\rho s_{11}^{E}(1+\
u)\frac{ \partial^{2}u_{\varphi}}{\partial t^{2}}. \tag{11}\]  
Then, from (7) and (9), we obtain  
\[ \begin{array}{l}\text{div}\vec{u}=\Big{[}\Big{(}\frac{du_{0}}{dr}+\frac{u_ {0}}{r}\Big{)}+\sum_{k=1}^{\infty}\Big{(}\frac{du_{k}}{dr}+\frac{u_{k}}{r}+k \frac{v_{k}}{r}\Big{)}\cos k\varphi\Big{]}\exp^{i\omega t}\\ =\left[\Phi_{0}(r)+\sum_{k=1}^{\infty}\Phi_{k}(r)\cos k\varphi\right]\exp^{i \omega t},\\ \omega_{z}=\left[\sum_{k=1}^{\infty}\Big{(}\frac{du_{k}}{dr}+\frac{u_{k}}{r}-k \frac{v_{k}}{r}\Big{)}\sin k\varphi\right]\exp^{i\omega t}=\left[\sum_{k=1}^{ \infty}\Psi_{k}(r)\right]\exp^{i\omega t}.\end{array} \]  
We perform time differentiation in the right-hand side of (10)-(11), and after reducing the non-zero factors \(\exp^{i\omega t}\), \(\sin(k\varphi)\), \(\cos k\varphi\), we obtain the following system of ordinary differential equations:  
\[ \begin{cases}\frac{2}{1-\
u}\frac{d\Phi_{k}}{dr}-\frac{k}{r}\psi_{k}=-2\rho s _{11}^{E}(1+\
u)\omega^{2}u_{k},\\ -\frac{2}{1-\
u}\frac{k}{r}\Phi_{k}-\frac{d\Psi_{k}}{dr}+d_{31}\frac{1+\
u}{1- \
u}a_{k}=-2\rho s_{11}^{E}(1+\
u)\omega^{2}\
u_{k}.\end{cases}  \tag{12}\]  
To determine the unknown potentials \(\Phi_{k}\) and \(\Psi_{k}\), we obtain a system of two Bessel ordinary differential equations  
\[ \begin{cases}r^{2}\frac{d^{2}\[PERSON]_{k}}{dr^{2}}+r\frac{d\[PERSON]_{k}}{dr}+\Big{(}(k _{1}r)^{2}-k^{2}\Big{)}\[PERSON]_{k}=-k^{2}d_{31}(1+\
u)a_{k}=A_{k},\\ \\ r^{2}\frac{d^{2}\Psi_{k}}{dr^{2}}+r\frac{d\Psi_{k}}{dr}+\Big{(}(k_{2}r)^{2}-k^{ 2}\Big{)}\Psi_{k}=0.\end{cases}  \tag{13}\]  
In the case of odd function \(E(\varphi)\), the first equation is homogeneous, and the second--heterogeneous. In the case of the general form of the electric load function, one can consider the superposition of solutions, or the solution of two inhomogeneous Bessel differential equations. In these equations, respectively, radial and circular wave numbers are denoted with \[k_{1}=\frac{\omega}{c},\quad c=\frac{1}{\sqrt{\rho s_{11}^{E}(1-\
u^{2})}},\quad k _{2}=\sqrt{\frac{2}{1-\
u}}k_{1}.\]  
The general integrals of Equation (13) have the following form:  
\[ \begin{cases}\Phi_{k}=C_{k}J_{k}(k,r)+A_{k}\frac{\pi}{2}\bigg{(}\gamma_{k}(k,r) \int_{0}^{r}J_{k}(k_{1}\rho)\frac{d\rho}{\rho}+J_{k}(k,r)\int_{0}^{r}\gamma_{k }(k_{1}\rho)\frac{d\rho}{\rho}\bigg{)}\\ \Psi_{k}=D_{k}J_{k}(k_{2}r)\end{cases} . \tag{14}\]  
The absence of a Bessel function of the second kind in the solutions of (12) is due to the presence of a feature of this function at \(r=0\). It is necessary to determine the components of the coupled electromechanical field through the obtained potentials.