## 2 Governing Equations
\right\}\Delta T \tag{6}\]  
In the expression, \(\sigma_{xx}\) and \(\sigma_{yy}\) are \(x\) directional normal stress and \(y\) directional normal stress, respectively; \(\sigma_{xy}\) is in-plane shear stress; \(\varepsilon_{xx}\), \(\varepsilon_{yy}\) are \(x\) directional and \(y\) directional normal strains, respectively; \(\varepsilon_{xy}\) is in-plane shear strain; \(D_{x}\), \(D_{y}\) and \(D_{z}\) are electric displacements in three directions of \(x\), \(y\), \(z\); \(E_{x}\), \(E_{y}\) and \(E_{z}\) are the electric field intensity in the three directions of \(x\), \(y\), \(z\), respectively; \(C_{11}\), \(C_{12}\) and \(C_{66}\) are the elastic moduli; \(e_{31}\) is piezoelectric coefficient; \(S_{11}\) and \(S_{33}\) are dielectric coefficients; \(\beta_{1}\), \(a_{3}\) and \(\Delta T\) are the thermo-elastic modulus, pyroelectric constant and temperature change value, respectively. As shown in Figure 1, the static voltage is applied along the thickness of the plate, and without considering the physical force and free charge, the equilibrium equation of the electric displacement can be expressed as  
\[\frac{\partial D_{z}}{\partial z}=0 \tag{7}\]  
Let the electric potential at any point in the direction of plate thickness be \(\phi(x,y,z,t)\), which can be written from the relationship between electric potential and electric field intensity as  
\[E_{z}=-\frac{\partial\phi}{\partial z} \tag{8}\]According to expressions (6)-(8) and boundary conditions \(\phi(h/2)=v_{0}\) and \(\phi(-h/2)=0\), the expressions of electric potential and electric field intensity can be written as  
\[\phi=\frac{e_{31}}{s_{33}}\bigg{(}\frac{z^{2}}{2}-\frac{h^{2}}{8}\bigg{)}\bigg{(} \frac{\partial^{2}w}{\partial x^{2}}+\frac{\partial^{2}w}{\partial y^{2}} \bigg{)}+\frac{v_{0}}{h}z+\frac{v_{0}}{2} \tag{9}\]  
\[E_{z}=-\frac{e_{31}z}{s_{33}}\bigg{(}\frac{\partial^{2}w}{\partial x^{2}}+ \frac{\partial^{2}w}{\partial y^{2}}\bigg{)}-\frac{v_{0}}{h}, \tag{10}\]  
the strain energy density per unit volume of the rectangular plate is  
\[U=\frac{1}{2}\big{(}\sigma_{xx}\varepsilon_{xx}+\sigma_{yy}\varepsilon_{yy}+ \sigma_{xy}\varepsilon_{xy}-D_{x}E_{x}-D_{z}E_{z}\big{)}, \tag{11}\]  
the kinetic energy of the rectangular plate is  
\[T=\frac{1}{2}\int_{\Omega}\rho\Big{(}\dot{u}^{2}+\dot{v}^{2}+\dot{w}^{2}\Big{)} \mathrm{d}\Omega, \tag{12}\]  
and the external potential energy of the rectangular plate is  
\[W_{F}=-\iint qw\mathrm{d}x\mathrm{d}y+\frac{1}{2}\iint\Bigg{\{}N_{x}\bigg{(} \frac{\partial w}{\partial x}\bigg{)}^{2}+N_{y}\bigg{(}\frac{\partial w}{ \partial y}\bigg{)}^{2}+2N_{xy}\bigg{(}\frac{\partial w}{\partial x}\frac{ \partial w}{\partial y}\bigg{)}+Kw^{2}\Bigg{\}}\mathrm{d}x\mathrm{d}y \tag{13}\]  
where \(q\) is the uniform load imposed on the surface of the rectangular plate. The axial force \(N_{x}\) and \(N_{y}\) of the \(x\) and \(y\) directions can be expressed as  
\[\left\{ \begin{array}{c}N_{x}=N_{Ex}+N_{Tx}\\ N_{y}=N_{Ey}+N_{Ty}\end{array} \right.. \tag{14}\]  
\(\big{(}N_{Ex},N_{Ey}\big{)}\) are the axial force generated in the directions \(x\) and \(y\) under the action of external static voltage; \(\big{(}N_{Tx},N_{Ty}\big{)}\) are the axial force generated by the temperature change in the \(x\) and \(y\) directions. \[N_{Ex}=N_{Ey}=-\int_{-h/2}^{h/2}e_{31}v_{0}\mathrm{d}z,N_{Tx}=N_{Ty}=\beta_{1 }h\Delta T \tag{15}\]  
According to Hamilton principle, the variation in the total energy is divided into zero, i.e.,  
\[\delta\int_{0}^{t}\left(\iint\limits_{\Omega}U\mathrm{d}\Omega+W_{F}-T\right) \mathrm{d}t=0. \tag{16}\]  
Substituting Equations (11)-(13) into Equation (16), the governing equation of the rectangular plate is  
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