# Thickness-Stretch Vibration of an Infinite Piezoelectric Plate with Flexoelectricity
## 2 Mathematical Modeling
### One-Dimensional Model for Thickness-Stretch Vibration  
The three-dimensional model is reduced to a one-dimensional problem by neglecting the displacements \(u_{1}\) and \(u_{2}\) for the configuration shown in Figure 1. We consider the following displacement fields and electric field for the one-dimensional thickness-stretch vibration analysis of an infinite piezoelectric plate. \[ \begin{array}{ll}u_{1}=0,&u_{2}=0,\quad\ u_{3}=u_{3}(x_{3})\exp(i\omega t)\\ &\varphi=\varphi(x_{3})\exp(i\omega t)\end{array}  \tag{8}\]  
Substituting Equation (13) into the first equation of Equation (12), we obtain the following equation. \[\left(c_{33}+\frac{e_{33}^{2}}{\varepsilon_{33}}\right)u_{3,33}-\frac{\mu_{3333} ^{2}}{\varepsilon_{33}}u_{3,3333}=\rho\ddot{u}_{3} \tag{14}\]  
For the fourth-order differential equation, the general solution of Equation (14) is assumed as the following:  
\[u_{3}=Ae^{\lambda x_{3}} \tag{15}\]  
Substituting Equation (15) into Equation (14) results in the following characteristic equation. \[a_{1}\lambda^{4}-a_{2}\lambda^{2}-a_{3}=0 \tag{16}\]  
where  
\[a_{1}=\frac{\mu_{3333}^{2}}{\varepsilon_{33}},\ \ \ a_{2}=c_{33}+\frac{e_{33}^{2}} {\varepsilon_{33}},\ \ \ a_{3}=\omega^{2}\rho\]  
By defining \(k=\lambda^{2}\), Equation (16) becomes the following:  
\[a_{1}k^{2}-a_{2}k-a_{3}=0 \tag{17}\]  
The solution of Equation (17) is expressed by:  
\[k=\frac{a_{2}\pm\sqrt{a_{2}^{2}+4a_{1}a_{3}}}{2a_{1}} \tag{18}\]  
Thus, four solutions of the characteristic equation, Equation (16), are expressed as below. \[\lambda_{1,2}=\pm\sqrt{\frac{a_{2}+\sqrt{a_{2}^{2}+4a_{1}a_{3}}}{2a_{1}}},\ \lambda_{3,4}=\pm i\sqrt{\frac{\sqrt{a_{2}^{2}+4a_{1}a_{3}-a_{2}}}{2a_{1}}} \tag{19}\]  
The solutions in Equation (19) are a pair of positive values and a pair of conjugate imaginary values. Thus, the general solution of Equation (14) can now be assumed as the following combination of hyperbolic and sinusoidal functions. \[u_{3}=A_{1}\text{sinh}(\alpha x_{3})+A_{2}\cosh(\alpha x_{3})+A_{3}\sin(\beta x _{3})+A_{4}\cos(\beta x_{3}) \tag{20}\]  
where  
\[\alpha=\sqrt{\frac{a_{2}+\sqrt{a_{2}^{2}+4a_{1}a_{3}}}{2a_{1}}},\ \beta=\sqrt{\frac{\sqrt{a_{2}^{2}+4a_{1}a_{3}-a_{2}}}{2a_{1}}}\]  
It should be noted that the frequency is contained in \(a_{3}\), which will be solved by the following frequency functions with the help of the boundary conditions. The next step is to obtain the frequency function from the boundary conditions. We consider two electric boundary conditions: the open circuit condition, and the short circuit condition in this work. (a) Open circuit condition (OP)  
The following traction-free boundary condition, boundary condition for higher order stress and electric displacement condition have to be satisfied for the open circuit condition:  
\[T_{33}-\tau_{333,3}=0,\ \ \ \tau_{333}=0\ \ \ \ \ (x=\pm h) \tag{21a}\]  
\[u_{3}=Ae^{\lambda x_{3}} \tag{15}\]0Substituting the expression of \(D_{3}\), Equation (10d), into the third equation of Equation (21), it gives the following equation. \[u_{3}=Ae^{\lambda x_{3}} \tag{15}\]1  
The above equation implies that \(B_{1}=0\) for the open circuit boundary condition. By substituting the constitutive equations, Equation (10) into Equation (21), we obtain the following four boundary equations at two surfaces. \[u_{3}=Ae^{\lambda x_{3}} \tag{15}\]2 \[u_{3}=Ae^{\lambda x_{3}} \tag{15}\]3 (23b) \[u_{3}=Ae^{\lambda x_{3}} \tag{15}\]4 (23c) \[u_{3}=Ae^{\lambda x_{3}} \tag{15}\]5  
Substituting the general expression of \(u_{3}\) into Equation (23), we rearrange them in the following form. \[ \begin{array}{l}A_{1}\big{[}\alpha\cosh(\alpha h)-k_{1}\alpha^{3}\cosh( \alpha h)\big{]}+A_{2}\big{[}\alpha\sinh(\alpha h)-k_{1}\alpha^{3}\sinh( \alpha h)\big{]}\\ +A_{3}\big{[}\beta\cos(\beta h)+k_{1}\beta^{3}\cos(\beta h)\big{]}-A_{4} \big{[}\beta\sin(\beta h)+k_{1}\beta^{3}\sin(\beta h)\big{]}=0\end{array}  \tag{24a}\] \[