Electric and heat conduction across an elliptic cavity in an anisotropic medium

Abstract

It is well known that the anisotropy of materials will significantly affect heat conduction, and the corresponding results have been applied to the thermal analysis of materials. An elliptic cavity in a nonlinearly coupled anisotropic medium, on the other hand, is much more difficult to analyze. Based on the complex variable method, the problem of a two-dimensional elliptical cavity in an anisotropic material is analyzed in this paper, and the field distributions have been obtained in closed-form. The field intensity factors are discussed in detail. The results show that both the temperature and electric potential gradients at a crack tip are always perpendicular to the crack surface, regardless of the anisotropy and the nonlinearity in the constitutive equations and the arbitrariness of loading direction. These results provide a powerful tool to analyze the effective behavior and reliability of anisotropic materials with cavities.

Keywords

Elliptical cavity anisotropic medium field intensity factors electric current thermal flux

1. Introduction

The flow of electric current carrying energy and information in various components is one of the cornerstones of modern science and technology. However, as a by-product, Joule heat is inevitably produced due to the existence of resistance, which changes the temperature distribution and affects the performance as well as the reliability of the device [1 –3]. Much effort has been devoted to the analysis of the distribution along with the effect of Joule heat. For example, Norio et al. [4,5] analyzed the Joule heat induced by electromagnetic fields in a two-dimensional (2D) plane. Han and Norio [6] analyzed the thermal expansion and thermal stress around an elliptic hole; Ansari and Cho [7] analyzed the Joule heating in piezoresistive microcantilever sensors; Igor et al. [8] studied the effective thermal conductivity of composite materials with inclusions. Cao et al. [9] investigated the thermal properties of in situ grown grapheme-reinforced copper matrix laminated composites.

On the other hand, the anisotropy of materials significantly affects electric and heat conduction, and a large number of literatures have reported electromagnetic fields in anisotropic materials [10 –15]. In addition, the problem of anisotropic thermoelasticity has also attracted attention. Beom [16] studied the transformed function representations of plane solutions for anisotropic elasticity and thermoelasticity; Zhu [17] developed an approximate method to solve anisotropic elliptic problems. These studies highlight the importance of anisotropic analysis, and quite a few theoretical models as well as experimental studies have also been developed to obtain the anisotropic parameters of various materials [18 –23].

For a nonlinear anisotropic medium, however, the field distributions will be significantly different from that of linear or isotropic systems, and the analysis becomes considerably more complicated. To the best of our knowledge, the problem of electric and heat conduction in an anisotropic medium has rarely been analyzed, despite its importance and implications to anisotropic materials. This motivates our current work.

In this work, the 2D problem of electric and heat conduction across an elliptic cavity is studied, with the nonlinear and the anisotropic fully accounted for. We attempt to derive closed-form solutions for the anisotropic medium, for which an elliptical cavity can be used to approximate a wide variety of defects as the aspect ratio is varied, ranging from cracks, to circular holes, to free surfaces. It is shown that both the temperature and electric potential gradients at a crack (the mini axis of an elliptical hole approaches zero) tip are always perpendicular to the crack surface, regardless of the anisotropy and the nonlinearity in the constitutive equations, and the arbitrariness of the loading direction. These results are quite remarkable for nonlinearly coupled anisotropic systems, and they provide a powerful tool to analyze the fracture behavior and reliability of anisotropic materials with cavities.

2. Two-dimensional problems

2.1. Governing equations

Considering the coupled transport of heat and electrons in a 2D medium, the transport equations governing electric current density $J$ and thermal flux $J_{Q}$ are

- J_{Q} = σ \nabla ϕ

(1)

- J_{Q} = κ \nabla T

(2)

where $σ, κ, ϕ$ and T are the electric conductivity, heat conductivity, electric potential and temperature, respectively. Since the total energy is transported by both electrons and thermal conduction, the energy flux $J_{E}$ can be expressed as

J_{E} = J_{Q} + ϕ J

(3)

In the following analysis, we assume that both charge and energy are conserved in the system, such that both current density and energy flux are divergence-free

\nabla \cdot J = 0

(4)

\nabla \cdot J_{E} = \nabla \cdot (J_{Q} + ϕ J) = 0

(5)

what we can infer from Equation (3) is that the thermal flux is not divergence-free due to Joule heating.

2.2. General solutions of two-dimensional problems

2.2.1. Electric fields

For a 2D anisotropic material, we expand Equations (1)–(3) as

- J_{x} = σ_{x} \frac{\partial ϕ}{\partial x} + σ_{xy} \frac{\partial ϕ}{\partial y}

(6.a)

- J_{y} = σ_{y} \frac{\partial ϕ}{\partial y} + σ_{xy} \frac{\partial ϕ}{\partial x}

(6.b)

- J_{Qx} = κ_{x} \frac{\partial T}{\partial x} + κ_{xy} \frac{\partial T}{\partial y}

(7.a)

- J_{Qy} = κ_{y} \frac{\partial T}{\partial y} + κ_{xy} \frac{\partial T}{\partial x}

(7.b)

J_{Ex} = ϕ J_{x} - κ_{x} \frac{\partial T}{\partial x} - κ_{xy} \frac{\partial T}{\partial y}

(8.a)

J_{Ey} = ϕ J_{y} - κ_{y} \frac{\partial T}{\partial y} - κ_{xy} \frac{\partial T}{\partial x}

(8.b)

Substituting Equation (6) into electric charge conservation equation Equation (4), we have

σ_{x} \frac{\partial^{2} ϕ}{\partial x^{2}} + 2 σ_{xy} \frac{\partial^{2} ϕ}{\partial x \partial y} + σ_{y} \frac{\partial^{2} ϕ}{\partial y^{2}} = 0

(9)

According to Gao et al. [24], electric potential $ϕ$ can be expressed by the real part of the analytic complex function $g (z_{t})$

ϕ = 2 Re [g (z_{t})] = g (z_{t}) + \bar{g (z_{t})}

(10)

where

z_{t} = x + μ_{t} y

(11.a)

μ_{t} = \frac{(- σ_{xy} + i λ_{t})}{σ_{y}}

(11.b)

λ_{t} = {(σ_{x} σ_{y} - σ_{xy}^{2})}^{\frac{1}{2}}

(11.c)

Then the electric current density components can be derived from Equation (6) as

J_{x} = - (σ_{x} + σ_{xy} μ_{t}) g' (z_{t}) - (σ_{x} + σ_{xy} {\bar{μ}}_{t}) \bar{g' (z_{t})} = 2 Re [i μ_{t} λ_{t} g' (z_{t})]

(12.a)

J_{y} = (σ_{xy} + σ_{y} {\bar{μ}}_{t}) g' (z_{t}) + (σ_{xy} + σ_{y} μ_{t}) \bar{g' (z_{t})} = - 2 Re [i λ_{t} g' (z_{t})]

(12.b)

and they can be combined to yield

J_{x} + {\bar{μ}}_{t} J_{y} = - \frac{2 λ_{t}^{2}}{σ_{y}} g' (z_{t})

(13)

2.2.2. Temperature fields

The influence of Joule heating makes it easier to analyze the energy flux than to analyze the thermal flux. So we substitute Equations (8), (10) and (13) into the energy conservation equation (5) and then we have

κ_{x} \frac{\partial^{2} T}{\partial x^{2}} + 2 κ_{xy} \frac{\partial^{2} T}{\partial x \partial y} + κ_{y} \frac{\partial^{2} T}{\partial y^{2}} = - 4 \frac{λ_{t}^{2}}{σ_{y}} g' (z_{t}) \bar{g' (z_{t})}

(14)

Equation (14) is a second-order nonhomogeneous partial differential equation, whose solution is composed of general solution $T_{s}$ and particular solution $T_{p}$

T = T_{s} + T_{p}

(15)

Being similar to the solution of electric potential, general solution $T_{s}$ can be expressed by the real part of another analytic complex function $φ (z_{s})$

T_{s} = 2 Re [φ (z_{s})] = φ (z_{s}) + \bar{φ (z_{s})} .

(16)

where

z_{s} = x + μ_{s} y

(17.a)

μ_{s} = \frac{(- κ_{x y} + i λ_{s})}{κ_{y}}

(17.b)

λ_{s} = {(κ_{x} κ_{y} - κ_{x y}^{2})}^{\frac{1}{2}}

(17.c)

Substituting Equations (15)–(17) into Equation (14), the particular solution $T_{p}$ satisfies

\frac{λ_{s}^{2}}{κ_{y}} \frac{\partial^{2} T_{p}}{\partial z_{s} {\bar{z}}_{s}} = - \frac{λ_{t}^{2}}{σ_{y}} g' (z_{t}) \bar{g' (z_{t})}

(18)

Integrating Equation (18), we obtain

T_{p} = - \frac{κ_{y} λ_{t}^{2}}{σ_{y} λ_{s}^{2}} \int \int g^{'} (z_{t}) \bar{g^{'} (z_{t})} d z_{s} d {\bar{z}}_{s}

(19)

We seek to solve Equation (19) for the particular solution of temperature and the corresponding thermal field distributions. To this end, we list the following algebraic relation between $z_{t}$ and $z_{s}$

z_{t} = U_{1} z_{s} + U_{2} {\bar{z}}_{s}

(20)

z_{s} = U_{3} z_{t} + U_{4} {\bar{z}}_{t}

(21)

where $U_{1} = \frac{μ_{t} - {\bar{μ}}_{s}}{μ_{s} - {\bar{μ}}_{s}}$ , $U_{2} = \frac{μ_{s} - μ_{t}}{μ_{s} - {\bar{μ}}_{s}}$ , $U_{3} = \frac{μ_{s} - {\bar{μ}}_{t}}{μ_{t} - {\bar{μ}}_{t}}$ , $U_{4} = \frac{μ_{t} - μ_{s}}{μ_{t} - {\bar{μ}}_{t}}$ .

Then we designate an intermediate function $P_{1}$ as the first-order partial derivatives of $T_{P}$

P_{1} = \frac{\partial T_{P}}{\partial z_{s}} = - \frac{κ_{y} λ_{t}^{2}}{σ_{y} λ_{s}^{2}} \int g' (z_{t}) \bar{g' (z_{t})} d {\bar{z}}_{s}

(22)

Substituting Equations (19)–(21) into Equation (22), the first-order partial differential equations for $P_{1}$ can be expressed as

U_{1} \frac{\partial P_{1}}{\partial z_{t}} + {\bar{U}}_{2} \frac{\partial P_{1}}{\partial {\bar{z}}_{t}} + \frac{κ_{y} λ_{t}^{2}}{σ_{y} λ_{s}^{2}} g' (z_{t}) \bar{g' (z_{t})} = 0

(23)

$P_{1}$ can be determined through the above equation after $g (z_{t})$ is obtained. Then the thermal flux can be written as

J_{Qx} = 2 Re [i μ_{s} λ_{s} (φ' (z_{s}) + P_{1})]

(24.a)

J_{Qy} = - 2 Re [i λ_{s} (φ' (z_{s}) + P_{1})]

(24.b)

or J_{Q x} + {\bar{μ}}_{s} J_{Q y} = - \frac{2 λ_{s}^{2}}{κ_{y}} [φ^{'} (z_{s}) + P_{1}]

(25)

and the energy flux is

J_{Ex} = ϕ J_{x} + 2 Re [i μ_{s} λ_{s} (φ' (z_{s}) + P_{1})]

(26.a)

J_{Ey} = ϕ J_{y} - 2 Re [i λ_{s} (φ' (z_{s}) + P_{1})]

(26.b)

As such, if the two analytic complex functions $g [z_{t}]$ and $φ [z_{s}]$ can be derived, the fields of temperature, electric potential, heat flow, energy flux and electric current can then be determined, and the problem is completely solved.

2.3. General treatment of boundary conditions

For the convenience of application, we attempt to present boundary conditions in terms of $g [z_{t}]$ and $φ [z_{s}]$ . To achieve this, we integrating the normal electric current and express the resultant electric current as

\int_{P}^{Q} J_{r} ds = 2 Re {[i λ_{t} g (z_{t})]}_{P}^{Q}

(27)

where P and Q are arbitrary points at the elliptic cavity surface, $s$ is the arc-length and $r$ denotes the direction perpendicular to the boundary.

Similarly, by integrating the normal energy flux, we have

\begin{matrix} \int J_{Er} ds = \int (ϕ J_{x} + 2 Re [i μ_{s} λ_{s} (φ' (z_{s}) + P_{1})]) dy \\ - (ϕ J_{y} - 2 Re [i λ_{s} (φ' (z_{s}) + P_{1})]) dx \end{matrix}

(28)

After some analysis and manipulation as detailed in the Supplement Information SI-I, Equation (28) can be simplified to

\int J_{Er} ds = \frac{i λ_{t}}{2} [g^{2} (z_{t}) - \bar{g^{2} (z_{t})}] + 2 Re [i λ_{s} φ (z_{s})]

(29)

For an impermeable boundary, the boundary conditions can be finally expressed as

Re {[i λ_{t} g (z_{t})]}_{P}^{Q} = 0

(30)

\frac{i λ_{t}}{2} {[g^{2} (z_{t}) - \bar{g^{2} (z_{t})}]}_{P}^{Q} + 2 Re {[i λ_{s} φ (z_{s})]}_{P}^{Q} = 0

(31)

So far, from Equation (12) we know that $g (z_{t})$ is only related to electric current distribution; thus, it can be determined by the boundary condition shown in Equation (30). Then the remaining unknown function $φ (z_{s})$ can be worked out through the other boundary condition expressed in Equation (31).

3. Solutions for an infinite matrix containing an elliptic cavity

To be specific, we consider an insulated elliptical cavity embedded in an infinite anisotropy material, as shown in Figure 1(a), where a and b denote the lengths of the major and minor semi-axes of the ellipse. At the far field, the material is subjected to in-plane electric current density $J_{x}^{\infty}$ and $J_{y}^{\infty}$ as well as thermal flow $J_{Qx}^{\infty}$ and $J_{Qy}^{\infty}$ . It is worth noting that $J_{x}^{\infty}$ , $J_{y}^{\infty}$ also produce extra thermal flow $J'_{Qx}$ , $J'_{Qy}$ at infinity. Here in the following research we define $J_{Qx}^{\infty}$ , $J_{Qy}^{\infty}$ as thermal loads that are independent of electric current loads. In fact, the total heat flow at infinity is $J_{Qx}^{\infty} + J_{Qx}^{'}$ and $J_{Qy}^{\infty} + J_{Qy}^{'}$ . We also point out that the remote energy fluxes are no longer constants in this nonlinearly coupled anisotropic system (as detailed in SI-II) and cannot be used as loading conditions.

3.1. Solutions for electric fields

According to Equation (13) as well as the remote electric current density, $g (z_{t})$ and $g' (z_{t})$ can be expressed as

g (z_{t}) = - \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) z_{t} + g_{0} (z_{t}) + Cons

(32)

g' (z_{t}) = - \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + {g_{0}}^{'} (z_{t})

(33)

where $g_{0} (z_{t})$ represents the disturbance function caused by the elliptical cavity, which can be considered as zero at infinity without loss of generality

lim_{zt \to \infty} g_{0} (z_{t}) = 0

(34)

Our goal is to determine the distribution of current density and electric field inside the medium. To this end, we denote $R_{t} = \frac{a - i μ_{t} b}{2}, m_{t} = \frac{a + i μ_{t} b}{a - i μ_{t} b}, w = r_{1} \times e^{- i θ_{1}}$ and adopt the following transform for the subsequent analysis [24]

z_{t} = R_{t} (w + \frac{m_{t}}{w})

(35)

and

w = \frac{z_{t} + \sqrt{z_{t}^{2} - 4 m_{t} R_{t}}}{2 R_{t}}

(36)

The analytic function $g [w]$ transformed from Equation (32) using Equation (35) can be expressed as

g (w) = - \frac{σ_{y} R_{t}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w + \sum_{k = 1}^{\infty} A_{- k} w^{- k} + Cons

(37)

Substituting Equation (37) into boundary condition Equation (30), the only non-zero coefficient $A_{- 1}$ can be obtained

A_{- 1} = - \frac{σ_{y} {\bar{R}}_{t}}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty})

(38)

then $g (z_{t})$ and $g' (z_{t})$ can be expressed as

g (z_{t}) = - \frac{R_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w - \frac{{\bar{R}}_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty}) w^{- 1} + Cons

(39)

g' (z_{t}) = - \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + \frac{{bJ}_{x}^{\infty} + {iaJ}_{y}^{\infty}}{2 λ_{t} R_{t} (w^{2} - m_{t})}

(40)

Then the intermediate function $P_{1}$ can be obtained by Equation (23), as detailed in SI-III. The corresponding electric potential and electric current density can be obtained as

ϕ = 2 Re [- \frac{R_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w - \frac{{\bar{R}}_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty}) w^{- 1}] + Cons

(41)

J_{x} = 2 Re [i μ_{t} λ_{t} (- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + \frac{{bJ}_{x}^{\infty} + {iaJ}_{y}^{\infty}}{2 λ_{t} R_{t} (w^{2} - m_{t})})]

(42.a)

J_{y} = - 2 Re [i λ_{t} (- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + \frac{{bJ}_{x}^{\infty} + {iaJ}_{y}^{\infty}}{2 λ_{t} R_{t} (w^{2} - m_{t})})]

(42.b)

3.2. Solutions for temperature fields

To solve the problem of temperature fields, we start by analyzing the remote thermal flux condition. Firstly, the remote thermal flux without electric current should be equal to $J_{Qx}^{\infty}$ and $J_{Qy}^{\infty}$ . Secondly, the direction of Joule thermal flux generated by the remote electric current must also be determined. In order to satisfy the above conditions, the expression of $φ (z_{s})$ is chosen as

φ (z_{s}) = b_{2} z_{s}^{2} + b_{1} z_{s} + \sum_{k = - 1}^{- \infty} b_{- k} {z_{s}}^{- k} + Cons

(43)

Here we require that the direction of the remote thermal flux be uniquely determined, whether it is conductive thermal flux or Joule thermal flux, that is, the direction of Joule thermal flux is consistent with that of conductive thermal flux. Thus, $b_{1}$ and $b_{2}$ can be determined as

b_{1} = - \frac{κ_{y}}{2 λ_{s}^{2}} (J_{Qx}^{\infty} + {\bar{μ}}_{s} J_{Qy}^{\infty})

(44)

b_{2} = \frac{β (J_{Qx}^{\infty} + {\bar{μ}}_{s} J_{Qy}^{\infty})}{2 (J_{Qx}^{\infty} + μ_{s} J_{Qy}^{\infty})}

(45)

where $β = - \frac{κ_{y} σ_{y}}{4 λ_{t}^{2} λ_{s}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) (J_{x}^{\infty} + μ_{t} J_{y}^{\infty})$ .

Now we can start calculating the distribution of the thermal flux and temperature field. To this end, we denote $R_{s} = (a - i μ_{s} b) / (a - i μ_{s} b) 2 2, m_{s} = (a + i μ_{s} b) / (a + i μ_{s} b) (a - i μ_{s} b) (a - i μ_{s} b), ξ = r_{2} \times e^{- i θ_{2}}$ and adopt the following transforms for the subsequent analysis

z_{s} [ξ] = R_{s} (ξ + \frac{m_{s}}{ξ})

(46)

Thus, the analytic functions $φ (ξ)$ transformed from Equation (43) can be expressed as

φ (ξ) = B_{2} ξ^{2} + B_{1} ξ + \sum_{k = - 1}^{- \infty} B_{- k} ξ^{- k} + Cons

(47)

Substituting Equation (47) into the boundary condition Equation (31) and comparing coefficients of the same power of $ξ^{k}$ , we obtain the following non-zero coefficients

B_{- 1} = \bar{b_{1} R_{s}}

(48)

B_{- 2} = \bar{b_{2} {R_{s}}^{2}}

(49)

This set of equations solves the problems completely, leading to the full determination of the field distributions around the elliptic cavity in closed-form. In particular, the temperature and thermal flux are determined as

T = 2 Re [\sum_{k = 1}^{\infty} (B_{k} ξ^{k} + B_{- k} ξ^{- k})] - \frac{κ_{y} λ_{t}^{2}}{σ_{y} λ_{s}^{2}} \int \int g' (z_{t}) \bar{g' (z_{t})} d z_{s} d {\bar{z}}_{s} + Cons

(50)

J_{Qx} = 2 Re [i μ_{s} λ_{s} (\sum_{k = 1}^{2} k b_{k} z_{s}^{k - 1} - \sum_{k = 1}^{2} \frac{k B_{- k}}{R_{s} (ξ^{k + 1} - m_{s} ξ^{k - 1})} + P_{1})]

(51.a)

J_{Qy} = - 2 Re [i λ_{s} (\sum_{k = 1}^{2} k b_{k} z_{s}^{k - 1} - \sum_{k = 1}^{2} \frac{k B_{- k}}{R_{s} (ξ^{k + 1} - m_{s} ξ^{k - 1})} + P_{1})]

(51.b)

J_{Qx} + {\bar{μ}}_{s} J_{Qy} = - \frac{2 λ_{s}^{2}}{κ_{y}} [\sum_{k = 1}^{2} k b_{k} z_{s}^{k - 1} - \sum_{k = 1}^{2} \frac{k B_{- k}}{R_{s} (ξ^{k + 1} - m_{s} ξ^{k - 1})} + P_{1}]

(52)

and the energy fluxes are given by

\begin{matrix} J_{Ex} = 2 Re [- \frac{R_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w - \frac{σ_{y} R_{t}}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty}) w^{- 1}] \\ \times 2 Re [i μ_{t} λ_{t} [- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + \frac{{bJ}_{x}^{\infty} + {iaJ}_{y}^{\infty}}{2 λ_{t} R_{t} (w^{2} - m_{t})}]] \\ + 2 Re [i μ_{s} λ_{s} [\sum_{k = 1}^{2} k b_{k} z_{s}^{k - 1} - \sum_{k = 1}^{2} \frac{k B_{- k}}{R_{s} (ξ^{k + 1} - m_{s} ξ^{k - 1})} + P_{1}]] \end{matrix}

(53.a)

\begin{matrix} J_{Ey} = 2 Re [- \frac{R_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w - \frac{σ_{y} R_{t}}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty}) w^{- 1}] \\ \times 2 Re [- i λ_{t} [- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + \frac{{bJ}_{x}^{\infty} + {iaJ}_{y}^{\infty}}{2 λ_{t} R_{t} (w^{2} - m_{t})}]] \\ - 2 Re [i λ_{s} [\sum_{k = 1}^{2} k b_{k} z_{s}^{k - 1} - \sum_{k = 1}^{2} \frac{k B_{- k}}{R_{s} (ξ^{k + 1} - m_{s} ξ^{k - 1})} + P_{1}]] \end{matrix}

(53.b)

4. Thermo-electric fields for special cases

4.1. Crack

4.1.1. Field distribution

For the special case of a crack with $b / a \to 0$ , $m_{t} = m_{s} \to 1$ and $R_{t} = R_{s} \to a / 2$ , the non-zero coefficients derived in Section 3 can be simplified as

A_{- 1}^{(1)} = - \frac{σ_{y} a}{4 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty})

(54)

B_{- k}^{(1)} = {(\frac{a}{2})}^{k} {\bar{b}}_{k} k = 1, 2

(55)

Thus, the distribution of temperature and electrochemical potential under the condition of a crack can be written as

T = 2 Re (\sum_{k = 1}^{2} (\bar{B_{- k}^{(1)}} ξ^{k} + B_{- k}^{(1)} ξ^{- k})) - \frac{κ_{y} λ_{t}^{2}}{σ_{y} λ_{s}^{2}} \int \int g' (z_{t}) \bar{g' (z_{t})} d z_{s} d {\bar{z}}_{s} + Cons

(56)

ϕ = 2 Re [- \frac{a σ_{y}}{4 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w + A_{- 1}^{(1)} w^{- 1}] + Cons

(57)

Also, the electric current density, thermal flux and energy flux are as follows

J_{x} + \bar{μ_{t}} J_{y} = (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) - \frac{2 λ_{t}^{2}}{σ_{y}} [\frac{{iJ}_{y}^{\infty}}{λ_{t} (w^{2} - 1)}]

(58)

J_{Qx} + {\bar{μ}}_{s} J_{Qy} = - \frac{2 λ_{s}^{2}}{κ_{y}} (\sum_{k = 1}^{2} k b_{k} z_{s}^{k - 1} - \sum_{k = 1}^{n} \frac{{kB}_{- k}^{(1)}}{R_{s} (ξ^{k + 1} - m_{s} ξ^{k - 1})} + P_{1})

(59)

\begin{matrix} J_{Ex} = 2 Re (- \frac{a σ_{y}}{4 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w - \frac{a σ_{y}}{4 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty}) w^{- 1}) \\ \times 2 Re [i μ_{t} λ_{t} (- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + \frac{{iJ}_{y}^{\infty}}{λ_{t} (w^{2} - 1)})] \\ + 2 Re [i μ_{s} λ_{s} (\sum_{k = 1}^{2} k b_{k} {z_{s}}^{k - 1} - \sum_{k = 1}^{2} \frac{{kB}_{- k}^{(1)}}{R_{s} (ξ^{k + 1} - ξ^{k - 1})} + P_{1})] \end{matrix}

(60.a)

\begin{matrix} J_{Ey} = 2 Re (- \frac{a σ_{y}}{4 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w - \frac{a σ_{y}}{4 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty}) w^{- 1}) \\ \times 2 Re [- i λ_{t} (- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + \frac{{iJ}_{y}^{\infty}}{λ_{t} (w^{2} - 1)})] \\ - 2 Re [i λ_{s} (\sum_{k = 1}^{2} k b_{k} {z_{s}}^{k - 1} - \sum_{k = 1}^{2} \frac{{kB}_{- k}^{(1)}}{R_{s} (ξ^{k + 1} - ξ^{k - 1})} + P_{1})] \end{matrix}

(60.b)

4.1.2. Intensity factor

According to Equation (58), the electric current density at the right tip of the crack can be expressed as

J_{x} + {\bar{μ}}_{t} J_{y} = J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty} - \frac{i λ_{t} J_{y}^{\infty}}{σ_{y} (w - 1)} = J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty} - \frac{i λ_{t} \sqrt{a} J_{y}^{\infty}}{σ_{y} \sqrt{2 (z_{t} - a)}}

(61)

It can be seen that both the electric current density components $J_{x}$ and $J_{y}$ at the crack tip have $r^{- 1 / 2}$ singularity, even though only current load $J_{y}^{\infty}$ exists at the far field. This is significantly different from the isotropic case. As a result, we define the combined field intensity factors of electric current density at the right tip of the crack as

K_{Jx} + {\bar{μ}}_{t} K_{Jy} = lim_{zt \to a} \sqrt{2 π (z_{t} - a)} (J_{x} + {\bar{μ}}_{t} J_{y}) = - \frac{i \sqrt{π a} λ_{t}}{σ_{y}} J_{y}^{\infty}

(62)

Similarly, the thermal flux at the right tip of the crack can be expressed as

J_{Qx} + {\bar{μ}}_{s} J_{Qy} = Q_{1} \sqrt{\frac{a}{2 (z_{s} - a)}} + Q_{2} \sqrt{\frac{a}{2 (z_{t} - a)}} + Q_{3}

(63)

where

Q_{1} = \frac{λ_{s}^{2}}{κ_{y}} [a ({\bar{b}}_{2} - b_{2}) + ({\bar{b}}_{1} - b_{1})]

(64.a)

Q_{2} = - \frac{{iaJ}_{y}^{\infty}}{4 λ_{t}} [2 J_{x}^{\infty} + (μ_{t} + {\bar{μ}}_{t}) J_{y}^{\infty}]

(64.b)

Q_{3} = \frac{a σ_{y}}{4 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) [2 J_{x}^{\infty} + (μ_{t} + {\bar{μ}}_{t}) J_{y}^{\infty}] - \frac{2 λ_{s}^{2}}{κ_{y}} [2 b_{2} a + b_{1}]

(64.c)

Thus, the thermal flux intensity factor is

K_{JQx} + {\bar{μ}}_{s} K_{JQy} = lim_{zs \to a} \sqrt{2 π (z_{s} - a)} (J_{Qx} + {\bar{μ}}_{s} J_{Qy}) = \sqrt{π a} (Q_{1} + Q_{2})

(65)

Similar to the electric current field, both the thermal flux components $J_{Qx}$ and $J_{Qy}$ at the crack tip have $r^{- 1 / 2}$ singularity, even though only thermal load $J_{Qy}^{\infty}$ exists at the far field. In order to further study this situation, we continue to analyze the field gradient at the right tip of the crack. The temperature and electric potential gradients along the $x$ and $y$ directions at the right tip of the crack are obtained as

\frac{\partial ϕ}{\partial x} = g' (z_{t}) + \bar{g' (z_{t})} = 2 Re [- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty})]

(66)

\frac{\partial ϕ}{\partial y} = 2 Re [- \frac{μ_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty})] + \frac{2}{a (w - 1)} Re (\frac{ia μ_{t} J_{y}^{\infty}}{2 λ_{t}})

(67)

\begin{matrix} \frac{\partial T}{\partial x} = 2 Re [2 b_{2} a + b_{1}] \\ - \frac{σ_{y}}{κ_{y}} \times 2 Re [- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) \times 2 Re [\frac{a σ_{y}}{4 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty})]] \end{matrix}

(68)

\begin{matrix} \frac{\partial T}{\partial y} = μ_{s} \frac{\partial T_{S}}{\partial z_{s}} + {\bar{μ}}_{s} \frac{\partial T_{S}}{\partial \bar{z_{s}}} + μ_{t} \frac{\partial T_{p}^{0}}{\partial z_{t}} + {\bar{μ}}_{t} \frac{\partial T_{p}^{0}}{\partial \bar{z_{t}}} \\ = 2 Re [μ_{s} (2 b_{2} a + b_{1})] - \frac{2 Re [\frac{a μ_{s}}{2} (a ({\bar{b}}_{2} - b_{2}) + ({\bar{b}}_{1} - b_{1}))]}{a (ξ - 1)} - \\ \frac{a σ_{y}^{2}}{4 κ_{y} λ_{t}^{2}} [2 J_{x}^{\infty} + (μ_{t} + {\bar{μ}}_{t}) J_{y}^{\infty}] [2 Re [- \frac{μ_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty})] + \frac{2 Re [\frac{i μ_{t} {aJ}_{y}^{\infty}}{2 λ_{t}}]}{a (w - 1)}] \end{matrix}

(69)

It can be seen from Equations (66)–(69) that

\frac{\partial ϕ}{\partial x} / \frac{\partial ϕ}{\partial y} = \frac{\partial T}{\partial x} / \frac{\partial T}{\partial y} \to 0

(70)

The expression above means that both the temperature and electric potential gradients at the crack tip are perpendicular to the crack surface, a quite remarkable result that is not anticipated due to the anisotropy and the nonlinearity in the constitutive equations and the arbitrariness of the loading direction. The direction of the singular flow field, on the other hand, is tilted to the crack direction, due to the anisotropy in the constitutive equations.

4.2. Circular cavity

For the special case of a circular cavity with $a = b$ , $R_{s} = (a - i μ_{s} a) / (a - i μ_{s} a) 2 2$ , $m_{s} = (a + i μ_{s} a) / (a + i μ_{s} a) (a - i μ_{s} a) (a - i μ_{s} a)$ , $R_{t} = (a - i μ_{t} a) / (a - i μ_{t} a) 2 2$ , $m_{t} = (a + i μ_{t} a) / (a + i μ_{t} a) (a - i μ_{t} a) (a - i μ_{t} a)$ , the non-zero coefficients derived in Section 3 can be simplified as

A_{- 1}^{(2)} = - \frac{σ_{y} {\bar{R}}_{t}}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty})

(71)

B_{- k}^{(2)} = \bar{b_{i} {R_{s}}^{i}} k = 1, 2

(72)

Thus, the distribution of temperature and electric potential can be written as

T = 2 Re (\sum_{k = 1}^{2} \bar{B_{- k}^{(2)}} ξ^{k} + \sum_{k = 1}^{2} B_{- k}^{(2)} ξ^{- k}) - \frac{κ_{y} λ_{t}^{2}}{σ_{y} λ_{s}^{2}} \int \int g' (z_{t}) \bar{g' (z_{t})} d z_{s} d {\bar{z}}_{s} + Cons

(73)

ϕ = 2 Re [- \frac{R_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + \bar{μ_{t}} J_{y}^{\infty}) w - \frac{σ_{y} {\bar{R}}_{t}}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty}) w^{- 1}]

(74)

Also, the electric current density, thermal flux and energy flux are as follows

J_{x} + {\bar{μ}}_{t} J_{y} = (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) - \frac{2 λ_{t}^{2}}{σ_{y}} [\frac{a (J_{x}^{\infty} + {iJ}_{y}^{\infty})}{2 λ_{t} R_{t} (w^{2} - m_{t})}]

(75)

J_{Qx} + {\bar{μ}}_{s} J_{Qy} = - \frac{2 λ_{s}^{2}}{κ_{y}} (\sum_{k = 1}^{2} k b_{k} {z_{s}}^{k} - \sum_{k = 1}^{2} \frac{{kB}_{- k}^{(2)}}{R_{s} (ξ^{k + 1} - m_{s} ξ^{k - 1})} + P_{1})

(76)

\begin{matrix} J_{Ex} = 2 Re [- \frac{R_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w - \frac{σ_{y} R_{t}}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty}) w^{- 1}] \\ \times 2 Re [i μ_{t} λ_{t} (- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + \frac{a (J_{x}^{\infty} + {iJ}_{y}^{\infty})}{2 λ_{t} R_{t} (w^{2} - m_{t})})] \\ + 2 Re [i μ_{s} λ_{s} (\sum_{k = 1}^{2} k b_{k} z_{s}^{k - 1} - \sum_{k = 1}^{2} \frac{{kB}_{- k}^{(2)}}{R_{s} (ξ^{k + 1} - m_{s} ξ^{k - 1})} + P_{1})] \end{matrix}

(77.a)

\begin{matrix} J_{Ey} = 2 Re [- \frac{R_{t} σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) w - \frac{σ_{y} R_{t}}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ_{t} J_{y}^{\infty}) w^{- 1}] \\ \times 2 Re [- i λ_{t} (- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + {\bar{μ}}_{t} J_{y}^{\infty}) + \frac{a (J_{x}^{\infty} + {iJ}_{y}^{\infty})}{2 λ_{t} R_{t} (w^{2} - m_{t})})] \\ - 2 Re [i λ_{s} (\sum_{k = 1}^{2} k b_{k} z_{s}^{k - 1} - \sum_{k = 1}^{2} \frac{{kB}_{- k}^{(2)}}{R_{s} (ξ^{k + 1} - m_{s} ξ^{k - 1})} + P_{1})] \end{matrix}

(77.b)

4.3. Equal degree of anisotropy

Now we assume that the thermal and electric fields have equal degrees of anisotropy, thus $σ_{xy} = κ_{xy} = 0$ , $σ_{x} / σ_{x} σ_{y} σ_{y} = κ_{x} / κ_{x} κ_{y} κ_{y}$ , and we have $R = R_{t} = R_{s}$ , $m = m_{t} = m_{s}$ , $η = ξ = w$ , $z = z_{s} = z_{t} = x + μ y$ and $μ = μ_{t} = μ_{s}$ . The non-zero coefficients under such circumstances can be simplified as

A_{- 1}^{(3)} = - \frac{σ_{y} \bar{R}}{2 κ_{t}^{2}} (J_{x}^{\infty} + μ J_{y}^{\infty})

(78)

B_{- k}^{(3)} = \bar{b_{k} R^{k}} k = 1, 2

(79)

The corresponding temperature and electrochemical potential are

T = 2 Re (\sum_{k = 1}^{2} \bar{B_{- k}^{(3)}} η^{k} + \sum_{k = 1}^{\infty} B_{- k}^{(3)} η^{- k}) - \frac{σ_{x}}{κ_{x}} g (z) \bar{g (z)} + Cons

(80)

ϕ = 2 Re [- \frac{R σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + \bar{μ} J_{y}^{\infty}) η - \frac{σ_{y} R}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ J_{y}^{\infty}) η^{- 1}] + Cons

(81)

and the electric current density, thermal flux and energy flux are as follows

J_{x} + \bar{μ} J_{y} = (J_{x}^{\infty} + \bar{μ} J_{y}^{\infty}) - \frac{2 κ_{t}^{2}}{σ_{y}} [\frac{a (J_{x}^{\infty} + {iJ}_{y}^{\infty})}{2 λ_{t} R (η^{2} - m)}]

(82)

J_{Qx} + \bar{μ} J_{Qy} = - \frac{2 λ_{s}^{2}}{κ_{y}} (\sum_{k = 1}^{2} k b_{k} z^{k - 1} - \sum_{k = 1}^{2} \frac{{kB}_{- k}^{(3)}}{R (η^{k + 1} - m η^{k - 1})} - \frac{σ_{x}}{κ_{x}} g (z) \bar{g (z)})

(83)

\begin{matrix} J_{Ex} = 2 Re [- \frac{R σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + \bar{μ} J_{y}^{\infty}) η - \frac{σ_{y} R}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ J_{y}^{\infty}) η^{- 1}] \\ \times 2 Re [i μ λ_{t} (- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + \bar{μ} J_{y}^{\infty}) + \frac{{bJ}_{x}^{\infty} + {iaJ}_{y}^{\infty}}{2 λ_{t} R (η^{2} - m)})] \\ + 2 Re [i μ λ_{s} (\sum_{k = 1}^{2} k b_{k} z^{k - 1} - \sum_{k = 1}^{2} \frac{{kB}_{- k}^{(3)}}{R (η^{k + 1} - m η^{k - 1})} - \frac{σ_{x}}{κ_{x}} g (z) \bar{g (z)})] \end{matrix}

(84.a)

\begin{matrix} J_{Ey} = 2 Re [- \frac{R σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + \bar{μ} J_{y}^{\infty}) η - \frac{σ_{y} R}{2 λ_{t}^{2}} (J_{x}^{\infty} + μ J_{y}^{\infty}) η^{- 1}] \\ \times 2 Re [- i λ_{t} (- \frac{σ_{y}}{2 λ_{t}^{2}} (J_{x}^{\infty} + \bar{μ} J_{y}^{\infty}) + \frac{{bJ}_{x}^{\infty} + {iaJ}_{y}^{\infty}}{2 λ_{t} R (η^{2} - m)})] \\ - 2 Re [i λ_{s} (\sum_{k = 1}^{2} k b_{k} z^{k - 1} - \sum_{k = 1}^{2} \frac{{kB}_{- k}^{(3)}}{R (η^{k + 1} - m η^{k - 1})} - \frac{σ_{x}}{κ_{x}} g (z) \bar{g (z)})] \end{matrix}

(84.b)

5. Numerical analysis and discussion

Numerical calculations were carried out to demonstrate the analysis, assuming that the matrix is $A l_{76} C o_{22} N i_{2}$ with the corresponding material constants shown in Table 1. The cavity with $a = 1 mm$ is considered, with b varying.

Table 1.

Material parameters of samples [25].

Sample	$σ_{x}$ [S/m]	$σ_{y}$ [S/m]	$σ_{xy}$ [S/m]	$κ_{x}$ [W/mK]	$κ_{y}$ [W/mK]	$κ_{xy}$ [W/mK]
$A l_{76} C o_{22} N i_{2}$	$1.77 \times 10^{6}$	$1.44 \times 10^{6}$	$0.65 \times 10^{6}$	10.54	7.28	3.59

We first consider the electric current and thermal flux at the right tip of the cavity, as shown in Figure 2, and examine how they depend on the aspect ratio of the cavity.

The remote non-zero thermal-electric loads are specified as $J_{y}^{\infty} = 10, 000 A / m^{2}$ , $J_{Qy}^{\infty} = 30, 000 W / m^{2}$ . For a linear crack with $b \to 0$ , it is observed in Figure 2(a) that the current density $J_{y}$ has a huge value, as expected. With the increase of the aspect ratio, $J_{y}$ gradually decreased, and eventually approach three times the value of $J_{y}^{\infty}$ specified at the boundary. Although the thermal flux $J_{Qy}$ is very complicated in Equation (51), its variation with the aspect ratio is similar to that of $J_{y}$ , as shown in Figure 2(b).

We then examine the contours of temperature/electric potential near an elliptic cavity in Figure 3. The parameters are the same as those used in Figure 2, except that $b = 0.6 mm$ is selected. From Figure 3 it is seen that both the temperature and electric potential vary nonlinearly with respect to the spatial coordinate, and neither temperature nor electric potential are symmetrical to the $y$ -axis. It is also observed that the gradients of temperature and electric potential near the cavity are higher than those far away from it, due to the electrical and thermal impermeable of the cavity. These results thus highlight the importance of fully anisotropic nonlinear analysis of conductive materials.

Figure 4 shows the flow fields of electric current as well as thermal flux near an elliptic cavity; it is observed that both the electric current and thermal flux flow around the cavity, as if they are repelled by it.

To further appreciate the thermal flux intensity factors at the crack tip, we also show in Figure 5 the thermal flux intensity factors $K_{JQx}$ , $K_{JQy}$ versus remote electric load $J_{y}^{\infty}$ (Figure 5(a)) and crack length $a$ (Figure 5(b)). As can be seen, both the thermal flux intensity factor components $K_{JQx}$ and $K_{JQy}$ exist under remote loading, which is significantly different from the isotropic cases. It is also observed that $K_{JQx}$ and $K_{JQy}$ are basically proportional to the square root of $a$ and to the square of $J_{y}^{\infty}$ . In addition, the ratio of $K_{JQx}$ to $K_{JQy}$ is constant, regardless of the change of loading condition and crack length.

Figure 6 shows the contours of electric potential and temperature at the crack tip. It is observed that both the temperature and electric potential gradients at the crack tip are indeed perpendicular to the crack surface, which confirms the results given by Equations (66)–(70). This also explains the phenomena observed in Figure 5: the thermal flux at the crack tip is inclined to the crack surface, due to the anisotropy in the constitutive equations.

6. Conclusions

The thermal-electric fields around an elliptic cavity in a two-dimension plate are analyzed in this paper. Using the complex variable method, the distributions of thermal and electric fields are obtained in closed-forms, including the temperature, electric potential, energy flux, heat flux and electric current. The field distributions around the cavity as well as the fields intensity factors at the crack tip are discussed in detail, and the results can be concluded as follows.

Theoretical results show that the electric current density components $J_{x}$ and $J_{y}$ as well as the thermal flux components $J_{Qx}$ and $J_{Qy}$ at the crack tip all have $r^{- 1 / 2}$ singularity. This is significantly different from the isotropic cases.

Both the temperature and electric potential gradients at the crack tip are perpendicular to the crack surface, a quite remarkable result that is not anticipated due to the anisotropy and the nonlinearity in the constitutive equations and the arbitrariness of loading direction.

Both the temperature and electric potential vary nonlinearly with respect to the spatial coordinate. Moreover, the energy and thermal fluxes flow around the cavity, as if they are repelled by it.

$K_{JQx}$ and $K_{JQy}$ are basically proportional to the square root of $a$ and to the square of $J_{y}^{\infty}$ . In addition, the ratio of $K_{JQx}$ to $K_{JQy}$ is constant, regardless of the change of loading condition and crack length.

Supplemental Material

Supplement_Information – Supplemental material for Electric and heat conduction across an elliptic cavity in an anisotropic medium

Supplemental material, Supplement_Information for Electric and heat conduction across an elliptic cavity in an anisotropic medium by Kunkun Xie, Haopeng Song and Cunfa Gao in Mathematics and Mechanics of Solids

Footnotes

Funding

The work was supported by the Fundamental Research Funds for the Central Universities (NS2016008 and NS2015014).

ORCID iDs

Haopeng Song

Cunfa Gao

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