Second-Order Asymptotic Analysis and Computations of Axially and Spherically Symmetric Piezoelectric Problems for Composite Structures

Abstract

The second-order coupled piezoelectric models based on the multi-scale asymptotic expansion method are developed for the axially and spherically symmetric composites. The governing piezoelectric equations are compactly formulated in cylindrical and spherical coordinates, and the composite domains are assumed to be periodically occupied by the representative cells. The multi-scale asymptotic expansions for the displacement and the electric potential are formally defined and the effective elastic, piezoelectric, and dielectric coefficients are expressed in terms of the microscopic functions defined on the cell domain. Particularly, the cell solutions and the homogenized solutions for the plane axisymmetric and spherically symmetric problem are derived analytically. The corresponding finite element procedure is proposed, in which the Newmark algorithm is applied to construct the computational scheme in the temporal domain. Numerical experiments are carried out to simulate both the static and dynamic asymptotic behavior of the space axisymmetric and the one-dimensional plane axisymmetric structures. It is validated from the numerical examples that the asymptotic models proposed in the current work are effective to capture the macroscopic performance of the piezoelectric structures and the second-order expansions of the solutions is essential for obtaining the correct distributions of the stress and electric displacement.

Appendices

Appendix A: Cell Solutions for the Plane Axisymmetric and Spherically Symmetric Problems

For the plane axisymmetric problem (6), the first-order periodic cell functions $N_1$, F, M, $R_1$, H and T satisfy the following boundary value problems

$$\begin{aligned}&\left\{ \begin{array}{l} - \dfrac{d}{{d{y_1}}}\left( {G_1}\dfrac{{d{N_1}}}{{d{y_1}}}\right) - \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{dH}}{{d{y_1}}}\right) = \dfrac{{d{G_1}}}{{d{y_1}}}\quad \text {in}\ [0,1],\\ - \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}\dfrac{{dH}}{{d{y_1}}}\right) + \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{N_1}}}{{d{y_1}}}\right) = - \dfrac{{d{e_1}}}{{d{y_1}}}\quad \text {in}\ [0,1],\\ {N_1}\left( 0\right) = {N_1}\left( 1\right) = H\left( 0\right) = H\left( 1\right) = 0, \end{array} \right. \end{aligned}$$

(47)

$$\begin{aligned}&\left\{ \begin{array}{l} - \dfrac{d}{{d{y_1}}}\left( {G_1}\dfrac{{dF}}{{d{y_1}}}\right) - \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{R_1}}}{{d{y_1}}}\right) = \dfrac{{d{e_1}}}{{d{y_1}}}\quad \text {in}\ [0,1],\\ - \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}\dfrac{{d{R_1}}}{{d{y_1}}}\right) + \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{dF}}{{d{y_1}}}\right) = \dfrac{{d{\epsilon _1}}}{{d{y_1}}}\quad \text {in}\ [0,1],\\ F\left( 0\right) = F\left( 1\right) = {R_1}\left( 0\right) = {R_1}\left( 1\right) = 0, \end{array} \right. \end{aligned}$$

(48)

$$\begin{aligned}&\left\{ \begin{array}{l} - \dfrac{d}{{d{y_1}}}\left( {G_1}\dfrac{{dM}}{{d{y_1}}}\right) - \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d T}}{{d{y_1}}}\right) = \dfrac{{d\lambda }}{{d{y_1}}}\quad \text {in}\ [0,1],\\ - \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}\dfrac{{d T}}{{d{y_1}}}\right) + \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{dM}}{{d{y_1}}}\right) = - \dfrac{{d{e_\theta }}}{{d{y_1}}}\quad \text {in}\ [0,1],\\ M\left( 0\right) = M\left( 1\right) = T\left( 0\right) = T\left( 1\right) = 0, \end{array} \right. \end{aligned}$$

(49)

respectively. Based on these problems and the homogenized coefficients in Eq. (40), we have

$$\begin{aligned} \begin{array}{l} G_1^0 = \dfrac{{{V_G}(1)}}{{{V_0}}},\epsilon _1^0 = \dfrac{{{V_\epsilon }(1)}}{{{V_0}}},e_1^0 = \dfrac{{{V_e}(1)}}{{{V_0}}},\\ {\lambda ^0} = e_1^0{V_1}(1) + G_1^0{V_2}(1) = \displaystyle {\int _0^1} {\dfrac{{\lambda (G_1^0{\epsilon _1} + e_1^0{e_1}) + {e_\theta }(G_1^0{e_1} - {G_1}e_1^0)}}{{{G_1}{\epsilon _1} + {{({e_1})}^2}}}dy}, \\ e_\theta ^0 = e_1^0{V_2}(1) - \epsilon _1^0{V_1}(1) = \displaystyle {\int _0^1} {\dfrac{{{e_\theta }({G_1}\epsilon _1^0 + e_1^0{e_1}) + \lambda (e_1^0{\epsilon _1} - {e_1}\epsilon _1^0)}}{{{G_1}{\epsilon _1} + {{({e_1})}^2}}}dy} ,\\ G_\theta ^0 = \displaystyle {\int _0^1} {\left[ {{G_\theta } + \dfrac{{({G_1}{e_\theta } - \lambda {e_1})({e_\theta } - e_\theta ^0) - (\lambda {\epsilon _1} + {e_\theta }{e_1})(\lambda - {\lambda ^0})}}{{{G_1}{\epsilon _1} + {{({e_1})}^2}}}} \right] dy}, \end{array} \end{aligned}$$

(50)

where

$$\begin{aligned} {V_e}({y_1})= & {} \displaystyle {\int _0^{{y_1}}} {\dfrac{{{e_1}}}{{{G_1}{\epsilon _1} + {{({e_1})}^2}}}dy} ,\quad {V_G}({y_1}) = \displaystyle {\int _0^{{y_1}}} {\dfrac{{{G_1}}}{{{G_1}{\epsilon _1} + {{({e_1})}^2}}}dy} ,\\ {V_\epsilon }({y_1})= & {} \displaystyle {\int _0^{{y_1}}} {\dfrac{{{\epsilon _1}}}{{{G_1}{\epsilon _1} + {{({e_1})}^2}}}dy} ,\\ {V_1}({y_1})= & {} \displaystyle {\int _0^{{y_1}}} {\dfrac{{\lambda {e_1} - {G_1}{e_\theta }}}{{{G_1}{\epsilon _1} + {{({e_1})}^2}}}dy} , \quad {V_2}({y_1}) = \displaystyle {\int _0^{{y_1}}} {\dfrac{{\lambda {\epsilon _1} + {e_\theta }{e_1}}}{{{G_1}{\epsilon _1} + {{({e_1})}^2}}}dy} , \\ {V_0}= & {} {V_G}(1){V_\epsilon }(1) + V_e^2(1). \end{aligned}$$

As the polarized piezoelectric constituent materials are not isotropic, it is of few interests to compare the magnitude of the coefficients in different directions as was done in [51,52,53].

Moreover, the explicit solutions of all the first-order cell functions are

$$\begin{aligned} \begin{array}{l} {N_1}({y_1}) = G_1^0{V_\epsilon }({y_1}) + e_1^0{V_e}({y_1}) - {y_1},\quad H({y_1}) = G_1^0{V_e}({y_1}) - e_1^0{V_G}({y_1}),\\ F({y_1}) = e_1^0{V_\epsilon }({y_1}) - \epsilon _1^0{V_e}({y_1}), \quad {R_1}({y_1}) =\epsilon _1^0{V_G}({y_1}) + e_1^0{V_e}({y_1}) - {y_1},\\ M({y_1}) = e_\theta ^0{V_e}({y_1}) + {\lambda ^0}{V_\epsilon }({y_1}) - {V_2}({y_1}),\quad T({y_1}) = {\lambda ^0}{V_e}({y_1}) - e_\theta ^0{V_G}({y_1}) - {V_1}({y_1}), \end{array} \end{aligned}$$

(51)

respectively, and we also have

$$\begin{aligned} \begin{array}{l} G_1^0 = {G_1} + {G_1}\dfrac{{d{N_1}}}{{d{y_1}}} + {e_1}\dfrac{{dH}}{{d{y_1}}}, \quad e_1^0 = {e_1} + {e_1}\dfrac{{d{R_1}}}{{d{y_1}}} + {G_1}\dfrac{{dF}}{{d{y_1}}} = {e_1} + {e_1}\dfrac{{d{N_1}}}{{d{y_1}}} - {\epsilon _1}\dfrac{{dH}}{{d{y_1}}}, \\ {\lambda ^0} = \lambda + {G_1}\dfrac{{dM}}{{d{y_1}}} + {e_1}\dfrac{{dS}}{{d{y_1}}}, \quad e_\theta ^0 = {e_\theta } - {\epsilon _1}\dfrac{{dS}}{{d{y_1}}} + {e_1}\dfrac{{dM}}{{d{y_1}}}, \quad \epsilon _1^0 = {\epsilon _1} + {\epsilon _1}\dfrac{{d{R_1}}}{{d{y_1}}} - {e_1}\dfrac{{dF}}{{d{y_1}}}. \end{array} \end{aligned}$$

By considering these equalities above, the boundary value problems for the second-order cell functions $N_{11}$, $H_{11}$, $\bar{N}_1$ , $\bar{H}_1$ , $\bar{M}_1$ , $\bar{T}$ , $F_{11}$ , $H_{11}$ ,$F_1$ , $\bar{R}_1$ , $\bar{P}_1$ and $\bar{W}$ can be simplified as

$$\begin{aligned}&{\left\{ \begin{array}{ll} \begin{aligned} - \dfrac{d}{{d{y_1}}}\left( {G_1}\dfrac{{d{N_{11}}}}{{d{y_1}}}\right) - \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{H_{11}}}}{{d{y_1}}}\right) =\, &{} \dfrac{d}{{d{y_1}}}\left( {G_1}{N_1} + {e_1}H\right) + G_1^0\left( 1 - \dfrac{\rho }{{{\rho ^0}}}\right) \quad \text {in}\ [0,1],\\ - \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}\dfrac{{d{H_{11}}}}{{d{y_1}}}\right) + \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{N_{11}}}}{{d{y_1}}}\right) =\, &{} \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}H - {e_1}{N_1}\right) - e_1^0\left( 1 - \dfrac{{{\rho _\phi }}}{{\rho _\phi ^0}}\right) \quad \text {in}\ [0,1],\\ \end{aligned}\\ {N_{11}}(0) = {N_{11}}(1) = {H_{11}}(0) = {H_{11}}(1) = 0, \end{array}\right. } \end{aligned}$$

(52)

$$\begin{aligned}&{\left\{ \begin{array}{ll} \begin{aligned} - \dfrac{d}{{d{y_1}}}\left( {G_1}\dfrac{{d{{\bar{N}}_1}}}{{d{y_1}}}\right) - \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{{\bar{H}}_1}}}{{d{y_1}}}\right) =\, &{} \dfrac{d}{{d{y_1}}}\left( {G_1}M + \lambda {N_1} + {e_1}T\right) + G_1^0\left( 1 - \dfrac{\rho }{{{\rho ^0}}}\right) \\ &{} - \lambda \dfrac{{d{N_1}}}{{d{y_1}}} - {e_\theta }\dfrac{{dH}}{{d{y_1}}} - \lambda + {\lambda ^0}\quad \text {in}\ [0,1],\\ - \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}\dfrac{{d\bar{H}}}{{d{y_1}}}\right) + \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{{\bar{N}}_1}}}{{d{y_1}}}\right) =\, &{}\dfrac{d}{{d{y_1}}}\left( {\epsilon _1}T - {e_1}M - {e_\theta }{N_1}\right) \\ &{} - \left( e_1^0 + e_\theta ^0\right) \left( 1 - \dfrac{{{\rho _\phi }}}{{\rho _\phi ^0}}\right) \quad \text {in}\ [0,1],\\ \end{aligned}\\ {{\bar{N}}_1}(0) = {{\bar{N}}_1}(1) = \bar{H}(0) = \bar{H}(1) = 0, \end{array}\right. } \end{aligned}$$

(53)

$$\begin{aligned}&{\left\{ \begin{array}{ll} \begin{aligned} - \dfrac{d}{{d{y_1}}}\left( {G_1}\dfrac{{d{{\bar{M}}_1}}}{{d{y_1}}}\right) - \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d\bar{T}}}{{d{y_1}}}\right) =\ &{} \dfrac{d}{{d{y_1}}}\left( \left( \lambda - {G_1}\right) M - {e_1}T\right) \\ &{} - \lambda \dfrac{{dM}}{{d{y_1}}} - {e_\theta }\dfrac{{dT}}{{d{y_1}}} - {G_\theta } + \dfrac{{\rho G_\theta ^0}}{{{\rho ^0}}}\quad {\text {in}}\ [0,1],\\ - \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}\dfrac{{d\bar{T}}}{{d{y_1}}}\right) + \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{{\bar{M}}_1}}}{{d{y_1}}}\right) =\ &{} \dfrac{d}{{d{y_1}}}\left( \left( {e_1} - {e_\theta }\right) M - {\epsilon _1}T\right) \quad \text {in}\ [0,1], \end{aligned}\\ {{\bar{M}}_1}(0) = {{\bar{M}}_1}(1) = \bar{T}(0) = \bar{T}(1) = 0, \end{array}\right. } \end{aligned}$$

(54)

$$\begin{aligned}&{\left\{ \begin{array}{ll} - \dfrac{d}{{d{y_1}}}\left( {G_1}\dfrac{{d{F_{11}}}}{{d{y_1}}}\right) - \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{R_{11}}}}{{d{y_1}}}\right) = \dfrac{d}{{d{y_1}}}\left( {G_1}F + {e_1}{R_1}\right) + e_1^0\left( 1 - \dfrac{\rho }{{{\rho ^0}}}\right) \quad {\text {in}}\ [0,1],\\ - \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}\dfrac{{d{R_{11}}}}{{d{y_1}}}\right) + \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{F_{11}}}}{{d{y_1}}}\right) = \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}{R_1} - {e_1}F\right) + \epsilon _1^0\left( 1 - \dfrac{{{\rho _\phi }}}{{\rho _\phi ^0}}\right) \quad {\text {in}}\ [0,1],\\ {F_{11}}\left( 0\right) = {F_{11}}\left( 1\right) = {R_{11}}\left( 0\right) = {R_{11}}\left( 1\right) = 0, \end{array}\right. } \end{aligned}$$

(55)

$$\begin{aligned}&{\left\{ \begin{array}{ll} - \dfrac{d}{{d{y_1}}}\left( {G_1}\dfrac{{d{F_1}}}{{d{y_1}}}\right) - \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{{\bar{R}}_1}}}{{d{y_1}}}\right) = \dfrac{d}{{d{y_1}}}\left( \lambda F\right) - \lambda \dfrac{{dF}}{{d{y_1}}}\\ \qquad \qquad - {e_\theta }\dfrac{{d{R_1}}}{{d{y_1}}} - {e_\theta } + e_\theta ^0\dfrac{\rho }{{{\rho ^0}}} + e_1^0\left( 1 - \dfrac{\rho }{{{\rho ^0}}}\right) \quad {\text {in}}\ [0,1],\\ - \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}\dfrac{{d{{\bar{R}}_1}}}{{d{y_1}}}\right) + \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{F_1}}}{{d{y_1}}}\right) = - \dfrac{d}{{d{y_1}}}\left( {e_\theta }F\right) + \epsilon _1^0\left( 1 - \dfrac{{{\rho _\phi }}}{{\rho _\phi ^0}}\right) \quad {\text {in}}\ [0,1],\\ {F_1}\left( 0\right) = {F_1}\left( 1\right) = {{\bar{R}}_1}\left( 0\right) = {{\bar{R}}_1}\left( 1\right) = 0, \end{array}\right. } \end{aligned}$$

(56)

$$\begin{aligned}&{\left\{ \begin{array}{ll} - \dfrac{d}{{d{y_1}}}\left( {G_1}\dfrac{{d{{\bar{P}}_1}}}{{d{y_1}}}\right) - \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d\bar{W}}}{{d{y_1}}}\right) = {f_1} - \dfrac{{\rho f_1^0}}{{{\rho ^0}}}\quad \text {in}\ [0,1],\\ - \dfrac{d}{{d{y_1}}}\left( {\epsilon _1}\dfrac{{d\bar{W}}}{{d{y_1}}}\right) + \dfrac{d}{{d{y_1}}}\left( {e_1}\dfrac{{d{{\bar{P}}_1}}}{{d{y_1}}}\right) = 0\quad \text {in}\ [0,1],\\ {{\bar{P}}_1}\left( 0\right) = {{\bar{P}}_1}\left( 1\right) = \bar{W}\left( 0\right) = \bar{W}\left( 1\right) = 0, \end{array}\right. } \end{aligned}$$

(57)

respectively. Combining all the first-order cell solutions in Eq. (51) and the homogenized coefficients in Eq. (40), the explicit expressions of all the second-order cell functions can also be obtained successively.

For the second-order correctors of the spherically symmetric problem, the only pair of cell functions that is different from the one in the axisymmetric case is $(\bar{M}_1, \bar{T})$, which satisfies the cell problem in Q as

$$\begin{aligned} \left\{ \begin{array}{lll} - \dfrac{d }{{d {y_1}}}\left( {G_1}\dfrac{{d {{\bar{M}}_1}}}{{d {y_1}}}\right) - \dfrac{d }{{d {y_1}}}\left( {e_1}\dfrac{{d \bar{T}}}{{d {y_1}}}\right) = \dfrac{d }{{d {y_1}}}\left[ (2\lambda - {G_1})M - {e_1}T)\right] + {\lambda ^0}\left( 1 - \dfrac{\rho }{{{\rho ^0}}}\right) \\ \quad \quad - 2\lambda \dfrac{{d M}}{{d {y_1}}} - 2{e_\theta }\dfrac{{d T}}{{d {y_1}}} - \left( {G_\theta } + {\lambda _\theta }\right) + {\left( {G_\theta } + {\lambda _\theta }\right) ^0}\dfrac{\rho }{{{\rho ^0}}}\quad {\text {in}}\ [0,1],\\ - \dfrac{d }{{d {y_1}}}\left( {\epsilon _1}\dfrac{{d \bar{T}}}{{d {y_1}}}\right) + \dfrac{d }{{d {y_1}}}\left( {e_1}\dfrac{{d {{\bar{M}}_1}}}{{d {y_1}}}\right) = \dfrac{d }{{d {y_1}}}(({e_1} - 2{e_\theta })M - {\epsilon _1}T) - e_\theta ^0\left( 1 - \dfrac{{{\rho _\phi }}}{{\rho _\phi ^0}}\right) \quad {\text {in}}\ [0,1],\\ {{\bar{M}}_1}(0) = {{\bar{M}}_1}(1) = \bar{T}(0) = \bar{T}(1) = 0. \end{array} \right. \end{aligned}$$

Appendix B: Homogenized Solutions for the Plane Asixymmetric and Spherically Symmetric Cases

The static plane axisymmetric piezoelectric problem without the source term is written as

$$\begin{aligned} \left\{ \begin{array}{lll} G_1^0\dfrac{{{d^2}u_1^0}}{{dx_1^2}} + G_1^0\dfrac{1}{{{x_1}}}\dfrac{{du_1^0}}{{d{x_1}}} - G_\theta ^0\dfrac{{u_1^0}}{{x_1^2}} + e_1^0\dfrac{{{d^2}{\phi ^0}}}{{dx_1^2}} + (e_1^0 - e_\theta ^0)\dfrac{1}{{{x_1}}}\dfrac{{d{\phi ^0}}}{{d{x_1}}} = 0\quad \text {in}\ [a,b],\\ - \epsilon _1^0\dfrac{{{d^2}{\phi ^0}}}{{dx_1^2}} - \epsilon _1^0\dfrac{1}{{{x_1}}}\dfrac{{d{\phi ^0}}}{{d{x_1}}} + e_1^0\dfrac{{{d^2}u_1^0}}{{dx_1^2}} + (e_1^0 + e_\theta ^0)\dfrac{1}{{{x_1}}}\dfrac{{du_1^0}}{{d{x_1}}} = 0\quad \text {in}\ [a,b],\\ \sigma _1^0(a) = {p_1},\sigma _1^0(b) = {p_2},{\phi ^0}(a) = D_1^0(b) = 0. \end{array} \right. \end{aligned}$$

(58)

Let $u_1^0=\alpha x_1^k$ and $\phi = \beta x_1^k$, where $\alpha$ , $\beta$ and k are real numbers. Substituting them into the equations in (58), we have

$$\begin{aligned} \begin{aligned}&(G_1^0{k^2} - G_\theta ^0)\alpha + (e_1^0{k^2} - e_\theta ^0k)\beta = 0, \\&(e_1^0{k^2} + e_\theta ^0k)\alpha - \epsilon _1^0{k^2}\beta = 0. \end{aligned} \end{aligned}$$

Since the nontrivial solutions are sought, $\alpha$ and $\beta$ cannot be both 0. Thus, the determinant of the coefficient matrix above has to vanish, i.e.,

$$\begin{aligned} \left| {\begin{array}{*{20}{c}} {G_1^0{k^2} - G_\theta ^0}&{}{e_1^0{k^2} - e_\theta ^0k}\\ {e_1^0{k^2} + e_\theta ^0k}&{}{ - \epsilon _1^0{k^2}} \end{array}} \right| = 0. \end{aligned}$$

Then

$$\begin{aligned} k = 0,\ k = \pm \gamma ,\ \gamma = \sqrt{{V_0}(G_\theta ^0\epsilon _1^0 + {{(e_\theta ^0)}^2})}, \end{aligned}$$

and we obtain the general solution for $u_1^0$ and $\phi ^0$ as

$$\begin{aligned}&u_1^0 = {C_1}\epsilon _1^0{\gamma ^2}{x^\gamma } + {C_2}\epsilon _1^0{\gamma ^2}{x^{ - \gamma }} + {C_3},\nonumber \\&{\phi ^0} = {C_1}(e_1^0{\gamma ^2} + e_\theta ^0\gamma ){x^\gamma } +\, {C_2}(e_1^0{\gamma ^2} - e_\theta ^0\gamma ){x^{ - \gamma }} - {C_3}\dfrac{{G_\theta ^0}}{{e_\theta ^0}}\ln x + {C_4}. \end{aligned}$$

(59)

Together with the boundary conditions, the unknown $C_1$ , $C_2$ , $C_3$ and $C_4$ can be determined.

The homogenized static piezoelectric problem with spherical symmetry reads as

$$\begin{aligned} \left\{ \begin{array}{lll} G_1^0\dfrac{{{d^2}u_1^0}}{{dx_1^2}} + G_1^0\dfrac{2}{{{x_1}}}\dfrac{{du_1^0}}{{d{x_1}}} + ({\lambda ^0} - {({G_\theta } + {\lambda _\theta })^0})\dfrac{{2u_1^0}}{{x_1^2}} + e_1^0\dfrac{{{d^2}{\phi ^0}}}{{dx_1^2}} + (e_1^0 - e_\theta ^0)\dfrac{2}{{{x_1}}}\dfrac{{d{\phi ^0}}}{{d{x_1}}} = 0\quad \text {in}\ [a,b],\\ - \epsilon _1^0\dfrac{{{d^2}{\phi ^0}}}{{dx_1^2}} - \epsilon _1^0\dfrac{2}{{{x_1}}}\dfrac{{d{\phi ^0}}}{{d{x_1}}} + e_1^0\dfrac{{{d^2}u_1^0}}{{dx_1^2}} + (e_1^0 + e_\theta ^0)\dfrac{2}{{{x_1}}}\dfrac{{du_1^0}}{{d{x_1}}} + e_\theta ^0\dfrac{{2u_1^0}}{{x_1^2}} = 0\quad \text {in}\ [a,b]. \end{array} \right. \end{aligned}$$

(60)

By the same argument as the axisymmetric case, we obtain the general solutions written by

$$\begin{aligned} u_1^0= & {} {C_1}_1^0V_kx_1^{{k_1}} + {C_2}\epsilon _1^0V_k x_1^{{k_2}} + {C_3}e_\theta ^0x_1^{ - 1},\\ {\phi ^0}= & {} {C_1}(e_1^0{k_1} + 2e_\theta ^0)({k_1} + 1)x_1^{{k_1}} + {C_2}(e_1^0{k_2} + 2e_\theta ^0)({k_2} + 1)x_1^{{k_2}} \\&+\, {C_3}({\lambda ^0} -\, {({G_\theta } + {\lambda _\theta })^0})x_1^{ - 1} + {C_4}, \end{aligned}$$

where $V_k = 2{V_0}(\epsilon _1^0({({G_\theta } + {\lambda _\theta })^0} - {\lambda ^0}) - e_\theta ^0(2e_\theta ^0 - e_1^0))$, and $k_1$ and $k_2$ are the roots of the following quadratic equation

$$\begin{aligned} {k^2} + k - V_k = 0. \end{aligned}$$

The unknown constants $C_1$, $C_2$, $C_3$ and $C_4$ are obtained by applying the boundary conditions.