and hence P$_{x}$ = 1 a$_{11}$ ( − E$_{x}$ − d$_{15}$ ( w ′ − ψ ) + f$_{1111}$zψ $^{′′}$) , P$_{z}$ = 1 a$_{33}$ ( − E$_{z}$ + d$_{31}$zψ ′ + f$_{1133}$ψ ′ − f$_{1313}$ ( w ′′ − ψ $^{′}$)) , (21) T$_{k}$ = 1 2 ∫ V b ϱ { ˙ u } $^{T}${ ˙ u } dV$_{b}$ = 1 2 ∫ V b ϱ [ z ˙ ψ 2 + ˙ w $^{2}$] dV$_{b}$ = 1 2 ϱ ∫ l 0 [ I ( ˙ ψ $^{2}$) + A ( ˙ w ) $^{2}$] dx, H = 1 2 a$_{11}$P 2 x + 1 2 a$_{33}$P 2 z + 1 2 c$_{11}$ε 2 x + 1 2 c$_{55}$γ 2 xz + d$_{15}$γ$_{xz}$P$_{x}$ + d$_{31}$ε$_{x}$P$_{z}$ + f$_{1111}$P$_{1}$u$_{xx}$ + P$_{3}$ ( f$_{1133}$u$_{xz}$ + f$_{1313}$u$_{zx}$ + f$_{3113}$w$_{xx}$ ) − 1 2 ε$_{0}$ ( ϕ ′ $_{x}$) 2 − 1 2 ε$_{0}$ ( ϕ ′ $_{z}$) 2 + ϕ ′ $_{x}$P$_{x}$ + ϕ ′ $_{z}$P$_{z}$ = 1 2 a$_{11}$ ( ϕ ′ x − d$_{15}$ ( w ′ − ψ ) + f$_{1111}$zψ $^{′′}$) 2 + 1 2 a$_{33}$ ( ϕ ′ z + d$_{31}$zψ ′ + f$_{1133}$ψ ′ − f$_{1313}$ ( w ′′ − ψ $^{′}$)) 2 + 1 2 c$_{11}$z $^{2}$( ψ $^{′}$) 2 + 1 2 c$_{55}$ ( w ′ − ψ ) 2 + d$_{15}$ ( w ′ − ψ ) 1 a$_{11}$ ( ϕ ′ x − d$_{15}$ ( w ′ − ψ ) + f$_{1111}$zψ $^{′′}$) − d$_{31}$zψ ′ a$_{33}$ ( ϕ ′ z + d$_{31}$zψ ′ + f$_{1133}$ψ ′ − f$_{1313}$ ( w ′′ − ψ $^{′}$)) − f$_{1111}$ zψ ′′ a$_{33}$ ( ϕ ′ x − d$_{15}$ ( w ′ − ψ ) + f$_{1111}$zϕ $^{′′}$) f$_{1133}$ ψ ′ a$_{33}$ ( ϕ ′ z + d$_{31}$zψ ′ + f$_{1133}$ψ ′ − f$_{1313}$ ( w ′′ − ψ $^{′}$)) + f$_{1313}$ w ′′ − ψ ′ a$_{33}$ ( ϕ ′ z + d$_{31}$zψ ′ + f$_{1133}$ψ ′ − f$_{1313}$ ( w ′′ − ψ $^{′}$)) − 1 2 ε$_{0}$ ( ϕ ′ $_{x}$) 2 − 1 2 ε$_{0}$ ( ϕ ′ $_{z}$) 2 + ϕ ′ x a$_{11}$ ( ϕ ′ x − d$_{15}$ ( w ′ − ψ ) + f$_{1111}$zψ $^{′′}$) + ϕ ′ z a$_{33}$ ( ϕ ′ z + d$_{31}$zψ ′ + f$_{1133}$ψ ′ − f$_{1313}$ ( w ′′ − ψ $^{′}$)) After substituting the equations (20), (21) into (3) and (4) and ignoring the polarization gradient, we have The kinetic energy is equal to The work done is described by the equation (12). Applying the Hamilton’s principle similar to previous section, we obtain the equations of 9