ZAMM - Journal of Applied Mathematics and Mechanics / Zeitschrift für Angewandte Mathematik und Mechanik
Volume 105, Issue 2 e202401292
ORIGINAL PAPER
Open Access

Two non-circular compressible liquid inclusions in an infinite elastic matrix

Xu Wang, 

Corresponding Author

Xu Wang

School of Mechanical and Power Engineering, East China University of Science and Technology, Shanghai, China

Correspondence

Xu Wang, School of Mechanical and Power Engineering, East China University of Science and Technology, 130 Meilong Road, Shanghai 200237, China.

Email: [email protected]

Peter Schiavone, Department of Mechanical Engineering, University of Alberta, 10–203 Donadeo Innovation Centre for Engineering, Edmonton, Alberta Canada T6G 1H9.

Email: [email protected]

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Peter Schiavone, 

Corresponding Author

Peter Schiavone

Department of Mechanical Engineering, 10-203 Donadeo Innovation Centre for Engineering, University of Alberta, Edmonton, Alberta, Canada

Correspondence

Xu Wang, School of Mechanical and Power Engineering, East China University of Science and Technology, 130 Meilong Road, Shanghai 200237, China.

Email: [email protected]

Peter Schiavone, Department of Mechanical Engineering, University of Alberta, 10–203 Donadeo Innovation Centre for Engineering, Edmonton, Alberta Canada T6G 1H9.

Email: [email protected]

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First published: 21 February 2025
https://doi.org/10.1002/zamm.202401292

Abstract

We derive an analytical solution to the plane strain problem associated with two interacting identical non-circular compressible liquid inclusions embedded in an infinite isotropic elastic matrix subjected to uniform remote in-plane normal stresses. The pair of analytic functions characterizing the elastic field in the matrix is derived using a conformal mapping function for quadrature domains by Crowdy (2015) and the technique of analytic continuation. The internal uniform hydrostatic stress field within each one of the two equal symmetric liquid inclusions and the hoop stress along each liquid-solid interface on the matrix side are studied in detail.

1 INTRODUCTION

In the past decade, the micro-mechanics analysis of a composite composed of a solid matrix and liquid inclusions has attracted focused and sustained attention with the objective being to gain a better understanding of this kind of novel composite (see, for example [1−17]). The majority of previous studies have been confined to an isolated liquid inclusion of general shape. In our recent investigation on two interacting liquid inclusions [16], each liquid inclusion is of circular shape and is incompressible. In many practical problems, however, an elastic body of pre-existing and closely-spaced non-circular pores can be filled with a liquid having its own compressibility.

In this paper, we carry out a rigorous analysis of the plane strain problem of two interacting identical non-circular compressible liquid inclusions embedded in an infinite isotropic elastic matrix subjected to uniform remote in-plane stresses. An analytical solution is derived with the aid of the conformal mapping function for the doubly connected quadrature domain occupied by the matrix [18, 19] and analytic continuation [20]. A general solution to the internal uniform hydrostatic stress field within each liquid inclusion is presented. The problem is completely solved once the three regular integrals appearing in the general solution are evaluated by the trapezoidal rule. Numerical results for the internal uniform hydrostatic stress field within each liquid inclusion and hoop stress along each liquid-solid interface on the matrix side are presented to demonstrate the solution. When the conformal modulus in the conformal mapping function approaches zero and unity, our solution recovers the corresponding result for an isolated elliptical (including circular) liquid inclusion.

2 COMPLEX VARIABLE FORMULATION

A Cartesian coordinate system { x i } ( i = 1 , 2 , 3 ) $\{ {{{x}_i}} \} (i = 1,2,3)$ is established. For plane strain deformation of an isotropic elastic material, the three in-plane stresses ( σ 11 , σ 22 , σ 12 ) $({{\sigma }_{11}}, {{\sigma }_{22}}, {{\sigma }_{12}})$ , two in-plane displacements ( u 1 , u 2 ) $({{u}_1}, {{u}_2})$ and two stress functions ( φ 1 , φ 2 ) $({{\varphi }_1}, {{\varphi }_2})$ are given in terms of two analytic functions ϕ ( z ) $\phi (z)$ and ψ ( z ) $\psi (z)$ of the complex variable z = x 1 + i x 2 $z = {{x}_1} + {\mathrm{i}}{{x}_2}$ as [21]
σ 11 + σ 22 = 2 ϕ ′ ( z ) + ϕ ′ ( z ) ¯ , σ 22 − σ 11 + 2 i σ 12 = 2 z ¯ ϕ ′ ′ ( z ) + ψ ′ ( z ) ; $$\begin{equation} \begin{aligned} & {{\sigma }_{11}} + {{\sigma }_{22}} = 2\left[ {\phi ^{\prime}(z) + \overline {\phi ^{\prime}(z)} } \right],\\ & {{\sigma }_{22}} - {{\sigma }_{11}} + 2{\mathrm{i}}{{\sigma }_{12}} = 2\left[ {\bar{z}\phi ^{\prime\prime}(z) + \psi ^{\prime}(z)} \right]; \end{aligned} \end{equation}$$ (1)
and
2 μ ( u 1 + i u 2 ) = κ ϕ ( z ) − z ϕ ′ ( z ) ¯ − ψ ( z ) ¯ , φ 1 + i φ 2 = i ϕ ( z ) + z ϕ ′ ( z ) ¯ + ψ ( z ) ¯ , $$\begin{equation} \begin{aligned} & 2\mu ({{u}_1} + {\mathrm{i}}{{u}_2}) = \kappa \phi (z) - z\overline {\phi ^{\prime}(z)} - \overline {\psi (z)} ,\\ & {{\varphi }_1} + {\mathrm{i}}{{\varphi }_2} = {\mathrm{i}}\left[ {\phi (z) + z\overline {\phi ^{\prime}(z)} + \overline {\psi (z)} } \right], \end{aligned} \end{equation}$$ (2)
where κ = 3 − 4 ν $\kappa = 3 - 4\nu $ , μ and ν 0 ≤ ν ≤ 1 / 2 $\nu \text{ }\left( 0\le \nu \le 1/2 \right)$ are the shear modulus and Poisson's ratio, respectively. In addition, the three in-plane stress components are related to the two stress functions through [22]
σ 11 = − φ 1 , 2 , σ 12 = φ 1 , 1 , σ 21 = − φ 2 , 2 , σ 22 = φ 2 , 1 . $$\begin{equation} \begin{aligned} & {{\sigma }_{11}} = - {{\varphi }_{1,2}},\quad {{\sigma }_{12}} = {{\varphi }_{1,1}},\\ & {{\sigma }_{21}} = - {{\varphi }_{2,2}},\quad {{\sigma }_{22}} = {{\varphi }_{2,1}}. \end{aligned} \end{equation}$$ (3)

3 ANALYTICAL SOLUTION

As shown in Figure 1, we consider two identical non-circular compressible liquid inclusions embedded in an infinite isotropic elastic matrix. The two liquid inclusions are perfectly bonded to the matrix through the left and right liquid-solid interfaces L 1 ${{L}_1}$ and L 2 ${{L}_2}$ . When the matrix is subjected only to uniform remote shear stress σ 12 ∞ $\sigma _{12}^\infty $ , the two non-circular interfaces will become traction-free. This case has been studied by Crowdy [18]. Thus, we will focus on the case in which the matrix is subjected to uniform remote in-plane normal stresses σ 11 ∞ $\sigma _{11}^\infty $ and σ 22 ∞ $\sigma _{22}^\infty $ .

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FIGURE 1
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Two identical non-circular compressible liquid inclusions embedded in an infinite isotropic elastic matrix subjected to uniform remote in-plane normal stresses σ 11 ∞ $\sigma _{11}^\infty $ and σ 22 ∞ $\sigma _{22}^\infty $ .
We introduce the following conformal mapping function for the matrix [18, 19]
z = ω ( ξ ) = R P ( − ξ − ξ ρ ρ ) P ( − ξ ρ ) P ( ξ ρ ) P ( ξ ξ ρ ρ ) P ( ξ ρ e i θ ) P ( ξ ρ e − i θ ) , ρ ≤ ξ ≤ 1 , $$\begin{equation} z = \omega (\xi ) = R\frac{{P({{ - \xi } \mathord{\left/ {\vphantom {{ - \xi } {\sqrt \rho }}} \right. \kern-\nulldelimiterspace} \!{\sqrt \rho }})P( - \xi \sqrt \rho )P(\xi \sqrt \rho )}}{{P({\xi \mathord{\left/ {\vphantom {\xi {\sqrt \rho }}} \right. \kern-\nulldelimiterspace} \!{\sqrt \rho }})P(\xi \sqrt \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }})P(\xi \sqrt \rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})}},\quad \rho \le \left| \xi \right| \le 1, \end{equation}$$ (4)
where R , θ $R, \theta $ and the conformal modulus ρ governing the interaction of the two liquid inclusions are real constants (R is further assumed to be positive), and
P ( ξ ) = ( 1 − ξ ) P ̂ ( ξ ) , P ̂ ( ξ ) = ∏ k = 1 + ∞ ( 1 − ρ 2 k ξ ) ( 1 − ρ 2 k ξ − 1 ) . $$\begin{equation}P(\xi ) = (1 - \xi )\hat{P}(\xi ),\quad \hat{P}(\xi ) = \prod\limits_{k = 1}^{ + \infty } {(1 - {{\rho }^{2k}}\xi )(1 - {{\rho }^{2k}}{{\xi }^{ - 1}})} .\end{equation}$$ (5)

As shown in Figure 2, using the mapping function in Equation (4), the doubly connected quadrature domain occupied by the matrix is mapped onto the annulus ρ ≤ | ξ | ≤ 1 $\rho \le | \xi | \le 1$ ; the left liquid-solid interface L 1 ${{L}_1}$ is mapped onto | ξ | = 1 $| \xi | = 1$ and the right liquid-solid interface L 2 ${{L}_2}$ is mapped onto | ξ | = ρ $| \xi | = \rho $ ; the point z = ∞ $z = \infty $ is mapped onto ξ = ρ $\xi = \sqrt \rho $ and the point z = 0 $z = 0$ is mapped onto ξ = − ρ $\xi = - \sqrt \rho $ .

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FIGURE 2
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The image ξ-plane.
It follows from Equation (2)2 that the continuity conditions of tractions across the two perfect liquid-solid interfaces L 1 ${{L}_1}$ and L 2 ${{L}_2}$ can be expressed as
ϕ ( z ) + z ϕ ′ ( z ) ¯ + ψ ( z ) ¯ = σ 0 z + γ , z ∈ L 1 ; ϕ ( z ) + z ϕ ′ ( z ) ¯ + ψ ( z ) ¯ = σ 0 z − γ , z ∈ L 2 , $$\begin{equation} \begin{aligned} & \phi (z) + z\overline {\phi ^{\prime}(z)} + \overline {\psi (z)} = {{\sigma }_0}z + \gamma , z \in {{L}_1};\\ & \phi (z) + z\overline {\phi ^{\prime}(z)} + \overline {\psi (z)} = {{\sigma }_0}z - \gamma , z \in {{L}_2}, \end{aligned} \end{equation}$$ (6)
where σ 0 ${{\sigma }_0}$ is the unknown internal uniform hydrostatic tension within each liquid inclusion and γ is an unknown complex constant. The internal stress field within each liquid inclusion is σ 11 = σ 22 = σ 0 , σ 12 = 0 . ${{\sigma }_{11}} = {{\sigma }_{22}} = {{\sigma }_0}, {{\sigma }_{12}} = 0.$
We can write the pair of analytic functions ϕ ( z ) $\phi (z)$ and ψ ( z ) $\psi (z)$ defined in the matrix in the following form:
ϕ ( z ) = ϕ 1 ( z ) + σ 11 ∞ + σ 22 ∞ 4 z , ψ ( z ) = ψ 1 ( z ) + σ 22 ∞ − σ 11 ∞ 2 z . $$\begin{equation} \begin{aligned} & \phi (z) = {{\phi }_1}(z) + \frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{4}z,\\ & \psi (z) = {{\psi }_1}(z) + \frac{{\sigma _{22}^\infty - \sigma _{11}^\infty }}{2}z. \end{aligned} \end{equation}$$ (7)
where ϕ 1 ( z ) ≅ O ( 1 ) , ψ 1 ( z ) ≅ O ( 1 ) ${{\phi }_1}(z) \cong O(1), {{\psi }_1}(z) \cong O(1)$ as | z | → ∞ $| z | \to \infty $ .
Substitution of Equations (7) into (6) yields
ϕ 1 ( z ) + z ϕ ′ 1 ( z ) ¯ + ψ 1 ( z ) ¯ = σ 0 − σ 11 ∞ + σ 22 ∞ 2 z − σ 22 ∞ − σ 11 ∞ 2 z ¯ + γ , z ∈ L 1 ; ϕ 1 ( z ) + z ϕ ′ 1 ( z ) ¯ + ψ 1 ( z ) ¯ = σ 0 − σ 11 ∞ + σ 22 ∞ 2 z − σ 22 ∞ − σ 11 ∞ 2 z ¯ − γ , z ∈ L 2 . $$\begin{equation} \begin{aligned} & {{\phi }_1}(z) + z\overline {{{{\phi ^{\prime}}}_1}(z)} + \overline {{{\psi }_1}(z)} = \left( {{{\sigma }_0} - \frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{2}} \right)z - \frac{{\sigma _{22}^\infty - \sigma _{11}^\infty }}{2}\bar{z} + \gamma , z \in {{L}_1};\\ & {{\phi }_1}(z) + z\overline {{{{\phi ^{\prime}}}_1}(z)} + \overline {{{\psi }_1}(z)} = \left( {{{\sigma }_0} - \frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{2}} \right)z - \frac{{\sigma _{22}^\infty - \sigma _{11}^\infty }}{2}\bar{z} - \gamma , z \in {{L}_2}. \end{aligned} \end{equation}$$ (8)
Considering the mapping function in Equation (4), the interface conditions in Equation (8) can be expressed in the ξ-plane as
ϕ ¯ 1 1 ξ + ω ¯ 1 ξ ϕ ′ 1 ( ξ ) ω ′ ( ξ ) + ψ 1 ( ξ ) = σ 0 − σ 11 ∞ + σ 22 ∞ 2 ω ¯ 1 ξ − σ 22 ∞ − σ 11 ∞ 2 ω ( ξ ) + γ ¯ , ξ = 1 ; ϕ ¯ 1 ρ 2 ξ + ω ¯ ρ 2 ξ ϕ ′ 1 ( ξ ) ω ′ ( ξ ) + ψ 1 ( ξ ) = σ 0 − σ 11 ∞ + σ 22 ∞ 2 ω ¯ ρ 2 ξ − σ 22 ∞ − σ 11 ∞ 2 ω ( ξ ) − γ ¯ , ξ = ρ , $$\begin{equation} \begin{aligned} {{{\bar{\phi }}}_1}\left( {\frac{1}{\xi }} \right) + \bar{\omega }\left( {\frac{1}{\xi }} \right)\frac{{{{{\phi ^{\prime}}}_1}(\xi )}}{{\omega ^{\prime}(\xi )}} + {{\psi }_1}(\xi ) = \left( {{{\sigma }_0} - \frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{2}} \right)\bar{\omega }\left( {\frac{1}{\xi }} \right) - \frac{{\sigma _{22}^\infty - \sigma _{11}^\infty }}{2}\omega (\xi ) + \bar{\gamma }, \left| \xi \right| = 1;\\ {{{\bar{\phi }}}_1}\left( {\frac{{{{\rho }^2}}}{\xi }} \right) + \bar{\omega }\left( {\frac{{{{\rho }^2}}}{\xi }} \right)\frac{{{{{\phi ^{\prime}}}_1}(\xi )}}{{\omega ^{\prime}(\xi )}} + {{\psi }_1}(\xi ) = \left( {{{\sigma }_0} - \frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{2}} \right)\bar{\omega }\left( {\frac{{{{\rho }^2}}}{\xi }} \right) - \frac{{\sigma _{22}^\infty - \sigma _{11}^\infty }}{2}\omega (\xi ) - \bar{\gamma }, \left| \xi \right| = \rho , \end{aligned} \end{equation}$$ (9)
where ϕ 1 ( ξ ) = ϕ 1 ( ω ( ξ ) ) ${{\phi }_1}(\xi ) = {{\phi }_1}(\omega (\xi ))$ and ψ 1 ( ξ ) = ψ 1 ( ω ( ξ ) ) ${{\psi }_1}(\xi ) = {{\psi }_1}(\omega (\xi ))$ . In writing Equation (9), the technique of analytic continuation [20] has been applied.
Subtracting the two conditions in Equation (9) and making use of the following property for quadrature domains [18, 19]
ω ¯ ρ 2 ξ = ω ¯ 1 ξ , $$\begin{equation}\bar{\omega }\left( {\frac{{{{\rho }^2}}}{\xi }} \right) = \bar{\omega }\left( {\frac{1}{\xi }} \right),\end{equation}$$ (10)
we arrive at
ϕ ¯ 1 ρ 2 ξ − ϕ ¯ 1 1 ξ = − 2 γ ¯ , $$\begin{equation}{{\bar{\phi }}_1}\left( {\frac{{{{\rho }^2}}}{\xi }} \right) - {{\bar{\phi }}_1}\left( {\frac{1}{\xi }} \right) = - 2\bar{\gamma },\end{equation}$$ (11)
or equivalently
ϕ 1 ( ρ 2 ξ ) = ϕ 1 ( ξ ) − 2 γ . $$\begin{equation}{{\phi }_1}({{\rho }^2}\xi ) = {{\phi }_1}(\xi ) - 2\gamma .\end{equation}$$ (12)
It follows from Equation (9)1 that
ϕ 1 ( ξ ) = σ 0 − σ 11 ∞ + σ 22 ∞ 2 ω ( ξ ) − σ 22 ∞ − σ 11 ∞ 2 ω ¯ 1 ξ − ω ( ξ ) ϕ ¯ ′ 1 1 ξ ω ¯ ′ 1 ξ − ψ ¯ 1 1 ξ + γ , $$\begin{equation} {{\phi }_1}(\xi ) = \left( {{{\sigma }_0} - \frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{2}} \right)\omega (\xi ) - \frac{{\sigma _{22}^\infty - \sigma _{11}^\infty }}{2}\bar{\omega }\left( {\frac{1}{\xi }} \right) - \omega (\xi )\frac{{{{{\bar{\phi }^{\prime}}}_1}\left( {\frac{1}{\xi }} \right)}}{{\bar{\omega }^{\prime}\left( {\frac{1}{\xi }} \right)}} - {{\bar{\psi }}_1}\left( {\frac{1}{\xi }} \right) + \gamma ,\end{equation}$$ (13)
which serves as an analytic continuation of ϕ 1 ( ξ ) ${{\phi }_1}(\xi )$ across | ξ | = 1 $| \xi | = 1$ .
Considering Equation (13), ϕ 1 ( ξ ) ${{\phi }_1}(\xi )$ takes the form:
ϕ 1 ( ξ ) = σ 11 ∞ + σ 22 ∞ 2 − σ 0 A K ( ξ ρ e i θ ) + B K ( ξ ρ e − i θ ) + σ 11 ∞ − σ 22 ∞ 2 C K ( ξ ρ ) + D K ( ξ ρ e i θ ) + E K ( ξ ρ e − i θ ) + F , ρ ≤ ξ ≤ 1 ρ , $$\begin{equation} \begin{aligned} {{\phi }_1}(\xi ) &= \left( {\frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{2} - {{\sigma }_0}} \right)\left[ {AK(\xi \sqrt \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) + BK(\xi \sqrt \rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})} \right]\\ &\quad + \frac{{\sigma _{11}^\infty - \sigma _{22}^\infty }}{2}\left[ {CK(\xi \sqrt \rho ) + DK(\xi \sqrt \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) + EK(\xi \sqrt \rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})} \right] + F, \rho \le \left| \xi \right| \le \frac{1}{\rho }, \end{aligned} \end{equation}$$ (14)
where A , B , C , D , E $A, B, C, D, E$ and F are unknown complex constants to be determined, and
K ( ξ ) = ξ P ′ ( ξ ) P ( ξ ) = − ξ 1 − ξ + ∑ n = 1 + ∞ − ρ 2 n ξ 1 − ρ 2 n ξ + ρ 2 n ρ 2 n ξ ξ 1 − ρ 2 n ρ 2 n ξ ξ . $$\begin{equation}K(\xi ) = \frac{{\xi P^{\prime}(\xi )}}{{P(\xi )}} = - \frac{\xi }{{1 - \xi }} + \sum\limits_{n = 1}^{ + \infty } {\left( { - \frac{{{{\rho }^{2n}}\xi }}{{1 - {{\rho }^{2n}}\xi }} + \frac{{{{{{\rho }^{2n}}} \mathord{\left/ {\vphantom {{{{\rho }^{2n}}} \xi }} \right. \kern-\nulldelimiterspace} \xi }}}{{1 - {{{{\rho }^{2n}}} \mathord{\left/ {\vphantom {{{{\rho }^{2n}}} \xi }} \right. \kern-\nulldelimiterspace} \xi }}}} \right)} .\end{equation}$$ (15)
By substituting Equations (14) into (13) and equating the residues of the three simple poles at ξ = e ± i θ / e ± i θ ρ ρ , 1 / 1 ρ ρ $\xi = {{{{{\mathrm{e}}}^{ \pm {\mathrm{i}}\theta }}} \mathord{/ {\vphantom {{{{{\mathrm{e}}}^{ \pm {\mathrm{i}}\theta }}} {\sqrt \rho }}} \kern-\nulldelimiterspace} {\sqrt \rho }}, {1 \mathord{/ {\vphantom {1 {\sqrt \rho }}} \kern-\nulldelimiterspace} {\sqrt \rho }}$ , we obtain
A = B = − 1 L ( ρ ) + L ( ρ e − 2 i θ ) X + e i θ ρ a ¯ , C = ρ b , D = E = ρ b L ( ρ e − i θ ) X A , $$\begin{equation} \begin{aligned} & A = B = - \frac{1}{{\frac{{L(\rho ) + L(\rho {{{\mathrm{e}}}^{ - 2{\mathrm{i}}\theta }})}}{X} + \frac{{{{{\mathrm{e}}}^{{\mathrm{i}}\theta }}}}{{\sqrt \rho \bar{a}}}}},\\ & C = \sqrt \rho b,\quad D = E = \frac{{\sqrt \rho bL(\rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})}}{X}A, \end{aligned} \end{equation}$$ (16)
where
L ( ξ ) = ξ d K ( ξ ) d ξ = − ∑ n = 0 + ∞ ρ 2 n ξ ( 1 − ρ 2 n ξ ) 2 − ∑ n = 1 + ∞ ρ 2 n ρ 2 n ξ ξ ( 1 − ρ 2 n ρ 2 n ξ ξ ) 2 , a = − R ρ e i θ P ( − e − i θ − e − i θ ρ ρ ) P ( e − i θ ) P ( − e − i θ ) P ̂ ( 1 ) P ( e − i θ e − i θ ρ ρ ) P ( e − 2 i θ ) , b = R P ( − 1 ) P ( − ρ ) P ( ρ ) ρ P ̂ ( 1 ) P ( ρ e i θ ) P ( ρ e − i θ ) , X = X ¯ = R P ( − e i θ ) P ( ρ e i θ ) P ( − ρ e i θ ) P ( e i θ ) P ( ρ e 2 i θ ) P ( ρ ) K ( − e i θ ) + K ( ρ e i θ ) + K ( − ρ e i θ ) − K ( e i θ ) − K ( ρ e 2 i θ ) − K ( ρ ) . $$\begin{equation} \begin{aligned} & L(\xi ) = \xi \frac{{{\mathrm{d}}K(\xi )}}{{{\mathrm{d}}\xi }} = - \sum\limits_{n = 0}^{ + \infty } {\frac{{{{\rho }^{2n}}\xi }}{{{{{(1 - {{\rho }^{2n}}\xi )}}^2}}}} - \sum\limits_{n = 1}^{ + \infty } {\frac{{{{{{\rho }^{2n}}} \mathord{\left/ {\vphantom {{{{\rho }^{2n}}} \xi }} \right. \kern-\nulldelimiterspace} \xi }}}{{{{{(1 - {{{{\rho }^{2n}}} \mathord{\left/ {\vphantom {{{{\rho }^{2n}}} \xi }} \right. \kern-\nulldelimiterspace} \xi })}}^2}}}} ,\\ & a = - \frac{R}{{\sqrt \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}}}\frac{{P({{ - {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }}} \mathord{\left/ {\vphantom {{ - {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }}} \rho }} \right. \kern-\nulldelimiterspace} \rho })P({{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})P( - {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})}}{{\hat{P}(1)P({{{{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }}} \mathord{\left/ {\vphantom {{{{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }}} \rho }} \right. \kern-\nulldelimiterspace} \rho })P({{{\mathrm{e}}}^{ - 2{\mathrm{i}}\theta }})}},\quad b = \frac{{RP( - 1)P( - \rho )P(\rho )}}{{\sqrt \rho \hat{P}(1)P(\rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }})P(\rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})}},\\ & X = \bar{X} = \frac{{RP( - {{{\mathrm{e}}}^{{\mathrm{i}}\theta }})P(\rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }})P( - \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }})}}{{P({{{\mathrm{e}}}^{{\mathrm{i}}\theta }})P(\rho {{{\mathrm{e}}}^{2{\mathrm{i}}\theta }})P(\rho )}}\left[ {K( - {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) + K(\rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) + K( - \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) - K({{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) - K(\rho {{{\mathrm{e}}}^{2{\mathrm{i}}\theta }}) - K(\rho )} \right]. \end{aligned} \end{equation}$$ (17)
It is seen from Equation (16) that the five constants A , B , C , D $A, B, C, D$ and E are in fact real valued. In view of Equation (16), we can rewrite Equation (14) in the following form
ϕ 1 ( ξ ) = A σ 11 ∞ + σ 22 ∞ 2 + C L ( ρ e − i θ ) ( σ 11 ∞ − σ 22 ∞ ) 2 X − σ 0 K ( ξ ρ e i θ ) + K ( ξ ρ e − i θ ) + C ( σ 11 ∞ − σ 22 ∞ ) 2 K ( ξ ρ ) + F , ρ ≤ ξ ≤ 1 ρ . $$\begin{equation} \begin{aligned} {{\phi }_1}(\xi ) &= A\left[ {\frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{2} + \frac{{CL(\rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})(\sigma _{11}^\infty - \sigma _{22}^\infty )}}{{2X}} - {{\sigma }_0}} \right]\left[ {K(\xi \sqrt \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) + K(\xi \sqrt \rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})} \right]\\ &\quad + \frac{{C(\sigma _{11}^\infty - \sigma _{22}^\infty )}}{2}K(\xi \sqrt \rho ) + F, \rho \le \left| \xi \right| \le \frac{1}{\rho }. \end{aligned} \end{equation}$$ (18)
Substituting Equation (18) into Equation (12) and using the following property [18, 19]
K ( ρ 2 ξ ) = K ( ξ ) − 1 , $$\begin{equation}K({{\rho }^2}\xi ) = K(\xi ) - 1,\end{equation}$$ (19)
we arrive at the constant γ as
γ = A 2 σ 11 ∞ + σ 22 ∞ + C L ( ρ e − i θ ) ( σ 11 ∞ − σ 22 ∞ ) X − 2 σ 0 + C ( σ 11 ∞ − σ 22 ∞ ) 4 , $$\begin{equation}\gamma = \frac{A}{2}\left[ {\sigma _{11}^\infty + \sigma _{22}^\infty + \frac{{CL(\rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})(\sigma _{11}^\infty - \sigma _{22}^\infty )}}{X} - 2{{\sigma }_0}} \right] + \frac{{C(\sigma _{11}^\infty - \sigma _{22}^\infty )}}{4},\end{equation}$$ (20)
which indicates that the constant γ is also real valued.
Using Equation (18) to enforce the following condition that
ϕ 1 ( z ) z = 0 = ϕ 1 ( ξ ) ξ = − ρ = 0 , $$\begin{equation}{{\left. {{{\phi }_1}(z)} \right|}_{z = 0}} = {{\left. {{{\phi }_1}(\xi )} \right|}_{\xi = - \sqrt \rho }} = 0,\end{equation}$$ (21)
we arrive at the constant F as
F = − A σ 11 ∞ + σ 22 ∞ 2 + C L ( ρ e − i θ ) ( σ 11 ∞ − σ 22 ∞ ) 2 X − σ 0 K ( − ρ e i θ ) + K ( − ρ e − i θ ) + C ( σ 22 ∞ − σ 11 ∞ ) 2 K ( − ρ ) = C ( σ 22 ∞ − σ 11 ∞ ) 2 K ( − ρ ) , $$\begin{equation} \begin{aligned} F &= - A\left[ {\frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{2} + \frac{{CL(\rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})(\sigma _{11}^\infty - \sigma _{22}^\infty )}}{{2X}} - {{\sigma }_0}} \right]\left[ {K( - \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) + K( - \rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})} \right] + \frac{{C(\sigma _{22}^\infty - \sigma _{11}^\infty )}}{2}K( - \rho )\\ & = \frac{{C(\sigma _{22}^\infty - \sigma _{11}^\infty )}}{2}K( - \rho ), \end{aligned} \end{equation}$$ (22)
which indicates that the constant F is also real valued.
The change in area Δ A 1 M $\Delta {{A}_{1M}}$ of the interface L 1 ${{L}_1}$ on the side of the matrix is
Δ A 1 M = − 1 2 μ Im ∫ ξ = 1 κ ϕ ( ξ ) − ω ( ξ ) ϕ ′ ( ξ ) ¯ ω ′ ( ξ ) ¯ − ψ ( ξ ) ¯ ¯ ω ′ ( ξ ) d ξ = − 1 2 μ Im ∫ ξ = 1 ( κ + 1 ) ϕ 1 ( ξ ) ¯ − γ + ( κ + 1 ) ( σ 11 ∞ + σ 22 ∞ ) 4 − σ 0 ω ( ξ ) ¯ ω ′ ( ξ ) d ξ , $$\begin{equation} \begin{aligned} \Delta {{A}_{1M}} &= - \frac{1}{{2\mu }}{\mathrm{Im}} \left\{ {\int_{{\left| \xi \right| = 1}}{{\left[ {\overline {\kappa \phi (\xi ) - \frac{{\omega (\xi )\overline {\phi ^{\prime}(\xi )} }}{{\overline {\omega ^{\prime}(\xi )} }} - \overline {\psi (\xi )} } } \right]\omega ^{\prime}(\xi ){\mathrm{d}}\xi }}} \right\}\\ & = - \frac{1}{{2\mu }}{\mathrm{Im}} \left\{ {\int_{{\left| \xi \right| = 1}}{{\left[ {(\kappa + 1)\overline {{{\phi }_1}(\xi )} - \gamma + \left[ {\frac{{(\kappa + 1)(\sigma _{11}^\infty + \sigma _{22}^\infty )}}{4} - {{\sigma }_0}} \right]\overline {\omega (\xi )} } \right]\omega ^{\prime}(\xi ){\mathrm{d}}\xi }}} \right\}, \end{aligned} \end{equation}$$ (23)
where the integral is in the counter-clockwise direction, ϕ ( ξ ) = ϕ ( ω ( ξ ) ) , ψ ( ξ ) = ψ ( ω ( ξ ) ) $\phi (\xi ) = \phi (\omega (\xi )), \psi (\xi ) = \psi (\omega (\xi ))$ , and according to Crowdy [18]
ω ′ ( ξ ) = ω ( ξ ) ξ K ( − ξ − ξ ρ ρ ) + K ( ξ ρ ) + K ( − ξ ρ ) − K ( ξ ξ ρ ρ ) − K ( ξ ρ e i θ ) − K ( ξ ρ e − i θ ) . $$\begin{equation}\omega ^{\prime}(\xi ) = \frac{{\omega (\xi )}}{\xi }\left[ {K({{ - \xi } \mathord{\left/ {\vphantom {{ - \xi } {\sqrt \rho }}} \right. \kern-\nulldelimiterspace} {\sqrt \rho }}) + K(\xi \sqrt \rho ) + K( - \xi \sqrt \rho ) - K({\xi \mathord{\left/ {\vphantom {\xi {\sqrt \rho }}} \right. \kern-\nulldelimiterspace} {\sqrt \rho }}) - K(\xi \sqrt \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) - K(\xi \sqrt \rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})} \right].\end{equation}$$ (24)
In view of the fact that the initial area A 1 ${{A}_1}$ of the domain enclosed by L 1 ${{L}_1}$ is
A 1 = − 1 2 i ∫ ξ = 1 ω ( ξ ) ¯ ω ′ ( ξ ) d ξ , $$\begin{equation}{{A}_1} = - \frac{1}{{2{\mathrm{i}}}}\int_{{\left| \xi \right| = 1}}{{\overline {\omega (\xi )} \omega ^{\prime}(\xi ){\mathrm{d}}\xi }},\end{equation}$$ (25)
with the integral in the counter-clockwise direction, Equation (23) can be equivalently written as
Δ A 1 M = − κ + 1 2 μ Im ∫ ξ = 1 ϕ 1 ( ξ ) ω ′ ( ξ ) d ξ + 1 μ ( κ + 1 ) ( σ 11 ∞ + σ 22 ∞ ) 4 − σ 0 A 1 . $$\begin{equation}\Delta {{A}_{1M}} = - \frac{{\kappa + 1}}{{2\mu }}{\mathrm{Im}} \left\{ {\int_{{\left| \xi \right| = 1}}{{{{\phi }_1}(\xi )\omega ^{\prime}(\xi ){\mathrm{d}}\xi }}} \right\} + \frac{1}{\mu }\left[ {\frac{{(\kappa + 1)(\sigma _{11}^\infty + \sigma _{22}^\infty )}}{4} - {{\sigma }_0}} \right]{{A}_1}.\end{equation}$$ (26)
On the other hand, the change in area Δ A 1 I $\Delta {{A}_{1I}}$ of the left compressible liquid inclusion with its boundary L 1 ${{L}_1}$ due to internal uniform hydrostatic stresses is given by
Δ A 1 I = σ 0 A 1 K , $$\begin{equation}\Delta {{A}_{1I}} = \frac{{{{\sigma }_0}{{A}_1}}}{K},\end{equation}$$ (27)
where K is the plane-strain bulk modulus of each liquid inclusion.
In order to ensure the continuity of displacements across the perfect liquid-solid interface L 1 ${{L}_1}$ , it is necessary that
Δ A 1 I = Δ A 1 M . $$\begin{equation}\Delta {{A}_{1I}} = \Delta {{A}_{1M}}.\end{equation}$$ (28)
Substituting Equations (26) and (27) into Equation (28), we arrive at the following relationship
σ 0 ( 1 + β ) + κ + 1 2 A 1 Im ∫ ξ = 1 ϕ 1 ( ξ ) ¯ ω ′ ( ξ ) d ξ = ( κ + 1 ) ( σ 11 ∞ + σ 22 ∞ ) 4 , $$\begin{equation}{{\sigma }_0}(1 + \beta ) + \frac{{\kappa + 1}}{{2{{A}_1}}}{\mathrm{Im}} \left\{ {\int_{{\left| \xi \right| = 1}}{{\overline {{{\phi }_1}(\xi )} \omega ^{\prime}(\xi ){\mathrm{d}}\xi }}} \right\} = \frac{{(\kappa + 1)(\sigma _{11}^\infty + \sigma _{22}^\infty )}}{4},\end{equation}$$ (29)
where
β = μ K . $$\begin{equation}\beta = \frac{\mu }{K}.\end{equation}$$ (30)
Substituting the expression for ϕ 1 ( ξ ) ${{\phi }_1}(\xi )$ in Equation (18) into Equation (29), a general solution to the internal uniform hydrostatic tension within each liquid inclusion can be finally derived as
σ 0 = ( κ + 1 ) σ 11 ∞ + σ 22 ∞ − C I 2 ( σ 11 ∞ − σ 22 ∞ ) A 1 − A I 1 A 1 σ 11 ∞ + σ 22 ∞ + C L ( ρ e − i θ ) ( σ 11 ∞ − σ 22 ∞ ) X 4 ( 1 + β ) − 2 A I 1 ( κ + 1 ) A 1 , $$\begin{equation}{{\sigma }_0} = \frac{{(\kappa + 1)\left\{ {\sigma _{11}^\infty + \sigma _{22}^\infty - \frac{{C{{I}_2}(\sigma _{11}^\infty - \sigma _{22}^\infty )}}{{{{A}_1}}} - \frac{{A{{I}_1}}}{{{{A}_1}}}\left[ {\sigma _{11}^\infty + \sigma _{22}^\infty + \frac{{CL(\rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})(\sigma _{11}^\infty - \sigma _{22}^\infty )}}{X}} \right]} \right\}}}{{4(1 + \beta ) - \frac{{2A{{I}_1}(\kappa + 1)}}{{{{A}_1}}}}},\end{equation}$$ (31)
where the two real coefficients I 1 ${{I}_1}$ and I 2 ${{I}_2}$ are given by
I 1 = Im ∫ ξ = 1 K ( ξ ρ e i θ ) ¯ + K ( ξ ρ e − i θ ) ¯ ω ′ ( ξ ) d ξ , I 2 = Im ∫ ξ = 1 K ( ξ ρ ) ¯ ω ′ ( ξ ) d ξ . $$\begin{equation} \begin{aligned} & {{I}_1} = {\mathrm{Im}} \left\{ {\int_{{\left| \xi \right| = 1}}{{\left[ {\overline {K(\xi \sqrt \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }})} + \overline {K(\xi \sqrt \rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})} } \right]\omega ^{\prime}(\xi ){\mathrm{d}}\xi }}} \right\},\\ & {{I}_2} = {\mathrm{Im}} \left\{ {\int_{{\left| \xi \right| = 1}}{{\overline {K(\xi \sqrt \rho )} \omega ^{\prime}(\xi ){\mathrm{d}}\xi }}} \right\}. \end{aligned} \end{equation}$$ (32)

Thus, in order to obtain σ 0 ${{\sigma }_0}$ , it is necessary to perform the three regular integrals in Equations (25) and (32). These integrals can be evaluated by the trapezoidal rule [18]. We illustrate in Figures 3 and 4 the internal uniform hydrostatic tension within each liquid inclusion for different values of β and ρ. It is seen from Figure 3 that when the matrix is subjected to a remote deviatoric load ( σ 11 ∞ = − σ 22 ∞ $\sigma _{11}^\infty = - \sigma _{22}^\infty $ ), (i) σ 0 / σ 0 σ 11 ∞ σ 11 ∞ ${{{{\sigma }_0}} \mathord{/ {\vphantom {{{{\sigma }_0}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }}$ is a decreasing function of both β and ρ; (ii) as ρ → 0 $\rho \to 0$ for two elliptical liquid inclusions each with an aspect ratio of 1/3 set far apart from each other, σ 0 / σ 0 σ 11 ∞ σ 11 ∞ = m ( κ + 1 ) / m ( κ + 1 ) [ 1 + m 2 κ + β ( 1 − m 2 ) ] [ 1 + m 2 κ + β ( 1 − m 2 ) ] ${{{{\sigma }_0}} \mathord{/ {\vphantom {{{{\sigma }_0}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }} = {{m(\kappa + 1)} \mathord{/ {\vphantom {{m(\kappa + 1)} {[ {1 + {{m}^2}\kappa + \beta (1 - {{m}^2})} ]}}} \kern-\nulldelimiterspace} {[ {1 + {{m}^2}\kappa + \beta (1 - {{m}^2})} ]}}$ with m = 0.5 $m = 0.5$ , which is in agreement with the result for an isolated elliptical compressible liquid inclusion under a remote deviatoric load; (iii) as ρ → 1 $\rho \to 1$ (the circumscribed boundary tends to a circle), σ 0 / σ 0 σ 11 ∞ σ 11 ∞ = 0 ${{{{\sigma }_0}} \mathord{/ {\vphantom {{{{\sigma }_0}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }} = 0$ as expected; (iv) the effect of the conformal modulus ρ on σ 0 / σ 0 σ 11 ∞ σ 11 ∞ ${{{{\sigma }_0}} \mathord{/ {\vphantom {{{{\sigma }_0}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }}$ is apparent. It is seen from Figure 4 that when the matrix is subjected to a remote hydrostatic load ( σ 11 ∞ = σ 22 ∞ $\sigma _{11}^\infty = \sigma _{22}^\infty $ ), (i) as ρ → 0 $\rho \to 0$ for two elliptical liquid inclusions each with an aspect ratio of 1/3 set far apart from each other, σ 0 / σ 0 σ 11 ∞ σ 11 ∞ = 0.5 ( κ + 1 ) ( 1 + m 2 ) / ( κ + 1 ) ( 1 + m 2 ) [ 1 + m 2 κ + β ( 1 − m 2 ) ] [ 1 + m 2 κ + β ( 1 − m 2 ) ] ${{{{\sigma }_0}} \mathord{/ {\vphantom {{{{\sigma }_0}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }} = 0.5{{(\kappa + 1)(1 + {{m}^2})} \mathord{/ {\vphantom {{(\kappa + 1)(1 + {{m}^2})} {[ {1 + {{m}^2}\kappa + \beta (1 - {{m}^2})} ]}}} \kern-\nulldelimiterspace} {[ {1 + {{m}^2}\kappa + \beta (1 - {{m}^2})} ]}}$ with m = 0.5 $m = 0.5$ , which is in agreement with the result for an isolated elliptical compressible liquid inclusion under a remote hydrostatic load; (iii) as ρ → 1 $\rho \to 1$ (the circumscribed boundary tends to a circle), σ 0 / σ 0 σ 11 ∞ σ 11 ∞ = ( κ + 1 ) / ( κ + 1 ) [ β + 0.5 ( κ + 3 ) ] [ β + 0.5 ( κ + 3 ) ] ${{{{\sigma }_0}} \mathord{/ {\vphantom {{{{\sigma }_0}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }} = {{(\kappa + 1)} \mathord{/ {\vphantom {{(\kappa + 1)} {[ {\beta + 0.5(\kappa + 3)} ]}}} \kern-\nulldelimiterspace} {[ {\beta + 0.5(\kappa + 3)} ]}}$ , which is in agreement with the expectation for a circular compressible liquid inclusion containing two circular elastic inclusions under a remote hydrostatic load; (iv) σ 0 = σ 11 ∞ ${{\sigma }_0} = \sigma _{11}^\infty $ when β = ( κ − 1 ) / ( κ − 1 ) 2 2 = 0.5 $\beta = {{(\kappa - 1)} \mathord{/ {\vphantom {{(\kappa - 1)} 2}} \kern-\nulldelimiterspace} 2} = 0.5$ (i.e., the plane strain bulk modulus of each liquid inclusion is equal to that of the elastic matrix) for any value of ρ; (v) the effect of the conformal modulus ρ on σ 0 / σ 0 σ 11 ∞ σ 11 ∞ ${{{{\sigma }_0}} \mathord{/ {\vphantom {{{{\sigma }_0}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }}$ under a remote hydrostatic load is not as apparent as that under a remote deviatoric load.

Details are in the caption following the image
FIGURE 3
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Variations of σ 0 / σ 0 σ 11 ∞ σ 11 ∞ ${{{{\sigma }_0}} \mathord{/ {\vphantom {{{{\sigma }_0}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }}$ as a function of β and ρ with θ = 2 π / 2 π 3 3 , κ = 2 , σ 11 ∞ = − σ 22 ∞ $\theta = {{2\pi } \mathord{/ {\vphantom {{2\pi } 3}} \kern-\nulldelimiterspace} 3}, \kappa = 2, \sigma _{11}^\infty = - \sigma _{22}^\infty $ .
Details are in the caption following the image
FIGURE 4
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Variations of σ 0 / σ 0 σ 11 ∞ σ 11 ∞ ${{{{\sigma }_0}} \mathord{/ {\vphantom {{{{\sigma }_0}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }}$ as a function of β and ρ with θ = 2 π / 2 π 3 3 , κ = 2 , σ 11 ∞ = σ 22 ∞ $\theta = {{2\pi } \mathord{/ {\vphantom {{2\pi } 3}} \kern-\nulldelimiterspace} 3}, \kappa = 2, \sigma _{11}^\infty = \sigma _{22}^\infty $ .
Now ϕ 1 ( ξ ) ${{\phi }_1}(\xi )$ in Equation (18) has been completely determined. The other analytic function ψ 1 ( ξ ) ${{\psi }_1}(\xi )$ can then be determined by using the analytic continuation in Equation (13) as
ψ 1 ( ξ ) = σ 0 − σ 11 ∞ + σ 22 ∞ 2 ω ¯ 1 ξ + σ 11 ∞ − σ 22 ∞ 2 ω ( ξ ) − ϕ ¯ 1 1 ξ − ω ¯ 1 ξ ϕ ′ 1 ( ξ ) ω ′ ( ξ ) + γ , ρ ≤ ξ ≤ 1 . $$\begin{equation}{{\psi }_1}(\xi ) = \left( {{{\sigma }_0} - \frac{{\sigma _{11}^\infty + \sigma _{22}^\infty }}{2}} \right)\bar{\omega }\left( {\frac{1}{\xi }} \right) + \frac{{\sigma _{11}^\infty - \sigma _{22}^\infty }}{2}\omega (\xi ) - {{\bar{\phi }}_1}\left( {\frac{1}{\xi }} \right) - \bar{\omega }\left( {\frac{1}{\xi }} \right)\frac{{{{{\phi ^{\prime}}}_1}(\xi )}}{{\omega ^{\prime}(\xi )}} + \gamma ,\quad \rho \le \left| \xi \right| \le 1.\end{equation}$$ (33)
Thus, the stress field in the matrix is obtained. For example, the hoop stress σ t t ${{\sigma }_{tt}}$ along the liquid–solid interface L 1 ${{L}_1}$ on the matrix side is given by
σ t t = σ 11 ∞ + σ 22 ∞ − σ 0 + Re A 2 ( σ 11 ∞ + σ 22 ∞ ) + 2 C L ( ρ e − i θ ) ( σ 11 ∞ − σ 22 ∞ ) X − 4 σ 0 L ( ξ ρ e i θ ) + L ( ξ ρ e − i θ ) + 2 C ( σ 11 ∞ − σ 22 ∞ ) L ( ξ ρ ) ξ ω ′ ( ξ ) , ξ = e i δ , $$\begin{align} {{\sigma }_{tt}} &= \sigma _{11}^\infty + \sigma _{22}^\infty - {{\sigma }_0}\nonumber\\ &\quad + {\mathrm{Re}} \left\{ {\frac{{\left( A\left[ {2(\sigma _{11}^\infty + \sigma _{22}^\infty ) + \frac{{2CL(\rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})(\sigma _{11}^\infty - \sigma _{22}^\infty )}}{X} - 4{{\sigma }_0}} \right]\left[ {L(\xi \sqrt \rho {{{\mathrm{e}}}^{{\mathrm{i}}\theta }}) + L(\xi \sqrt \rho {{{\mathrm{e}}}^{ - {\mathrm{i}}\theta }})} \right]\ + 2C(\sigma _{11}^\infty - \sigma _{22}^\infty )L(\xi \sqrt \rho ) \right)}}{{\xi \omega ^{\prime}(\xi )}}} \right\},\quad \xi = {{{\mathrm{e}}}^{{\mathrm{i}}\delta }}, \end{align}$$ (34)
which is illustrated in Figures 5 and 6. It is seen from Figures 5 and 6 that: (i) the effect of β on the hoop stress distribution is apparent, especially under a remote hydrostatic load; (ii) σ 0 ≡ σ 11 ∞ ${{\sigma }_0} \equiv \sigma _{11}^\infty $ when β = ( κ − 1 ) / ( κ − 1 ) 2 2 = 0.5 $\beta = {{(\kappa - 1)} \mathord{/ {\vphantom {{(\kappa - 1)} 2}} \kern-\nulldelimiterspace} 2} = 0.5$ (i.e., the plane-strain bulk modulus of each liquid inclusion is equal to that of the matrix) under a remote hydrostatic load, which is as expected.
Details are in the caption following the image
FIGURE 5
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Distribution of σ t t / σ t t σ 11 ∞ σ 11 ∞ ${{{{\sigma }_{tt}}} \mathord{/ {\vphantom {{{{\sigma }_{tt}}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }}$ along the left liquid-solid interface L 1 ${{L}_1}$ for different values of β with ρ = 0.2 , θ = 2 π / 2 π 3 3 , κ = 2 , σ 11 ∞ = − σ 22 ∞ $\rho = 0.2, \theta = {{2\pi } \mathord{/ {\vphantom {{2\pi } 3}} \kern-\nulldelimiterspace} 3}, \kappa = 2, \sigma _{11}^\infty = - \sigma _{22}^\infty $ .
Details are in the caption following the image
FIGURE 6
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Distribution of σ t t / σ t t σ 11 ∞ σ 11 ∞ ${{{{\sigma }_{tt}}} \mathord{/ {\vphantom {{{{\sigma }_{tt}}} {\sigma _{11}^\infty }}} \kern-\nulldelimiterspace} {\sigma _{11}^\infty }}$ along the left liquid-solid interface L 1 ${{L}_1}$ for different values of β with ρ = 0.2 , θ = 2 π / 2 π 3 3 , κ = 2 , σ 11 ∞ = σ 22 ∞ $\rho = 0.2, \theta = {{2\pi } \mathord{/ {\vphantom {{2\pi } 3}} \kern-\nulldelimiterspace} 3}, \kappa = 2, \sigma _{11}^\infty = \sigma _{22}^\infty $ .

4 CONCLUSIONS

We have analytically solved the problem of two identical non-circular compressible liquid inclusions in an infinite elastic matrix under uniform remote normal stresses. The pair of analytic functions ϕ 1 ( ξ ) ${{\phi }_1}(\xi )$ and ψ 1 ( ξ ) ${{\psi }_1}(\xi )$ is obtained in Equations (18) and (33) after the introduction of a conformal mapping function in Equation (4) for quadrature domains [18, 19]. The general solution of the internal uniform hydrostatic stress field within each non-circular liquid inclusion is obtained in Equation (31) which contains three integrals. The problem is completely solved once these integrals are accomplished. As ρ → 0 $\rho \to 0$ and ρ → 1 $\rho \to 1$ , the present solution recovers the result for an isolated elliptical compressible liquid inclusion (see Figures 3 and 4).

In principle the method based on the standard Laurent series expansion in ref. [16] can be applied to solve the present problem since the geometry in the image ξ-plane corresponds to that of a circular annulus. However, the resulting solution using the method in ref. [16] is not as concise and explicit as that in terms of prime functions obtained in the current paper: we are required to solve a set of coupled linear algebraic equations to identify the coefficients appearing in the Laurent series expansion.

The methodology in refs. [18, 19] applied in this paper can be extended to find the solution to the plane problem of an infinite elastic matrix containing any finite number of liquid inclusions. The multiply connected domain occupied by the matrix is a quadrature domain. In our understanding this extension is not straightforward and requires further significant considerations.

Although the conformal mapping function for a doubly connected domain occupied by the infinite z-plane excluding two arbitrarily shaped liquid inclusions (e.g., two circular, elliptical or square liquid inclusions) can be established, in general the resulting mapping function lacks the property for quadrature domains in Equation (10). In this case, the solution cannot be expressed in explicit closed-form as in this paper, and only series-form solutions based on Laurent series expansions [16] can be derived.

ACKNOWLEDGEMENTS

The authors are grateful to two anonymous reviewers for their constructive comments and helpful suggestions. This work is supported by a Discovery Grant from the Natural Sciences and Engineering Research Council of Canada (Grant No. RGPIN-2023-03227 Schiavo).

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