Boundary Element Method in Generalized Thermoelasticity

Synonyms

BEM; Boundary element method; Boundary integral equation method; Boundary integral method

Overview

Many generalizations to the equations of the coupled theory of therrnoelasticity due to Biot [1] are presently available. The first to appear is the theory of generalized therrnoelasticity with one relaxation time introduced by Lord and Shulman [2]. In this theory, a modified law of heat conduction including both the heat flux and its time derivative replaces the conventional Fourier’s law. The heat equation associated with this theory is hyperbolic and, hence, automatically eliminates the paradox of infinite speeds of propagation inherent in the coupled theory of therrnoelasticity. This theory was extended [3] by Dhaliwal and Sherief to include the effects of anisotropy. The uniqueness of solution of these equations was proved by Ignaczak [4] and by Sherief [5]. The boundary element formulation was conducted by Anwar and Sherief in [6]. The fundamental solution was obtained by Sherief [7] for the spherical case and by Sherief and Anwar [8] for the cylindrical case.

The second generalization to the coupled theory, known as the theory of therrnoelasticity with two relaxation times, was obtained by Green and Lindsay [9]. In this theory, the classical Fourier law of heat conduction is not violated when the body under consideration has a center of symmetry. The uniqueness of solution for this theory was proved by Green [10]. The fundamental solution was obtained by Sherief [11]. The boundary element formulation was conducted by Anwar and Sherief in [12].

The boundary integral equation method (BEM) has been applied successfully in many branches of applied mathematics. The reason for this is its simplicity, efficiency, and ease of implementation compared to other numerical methods based on domain discretization such as the finite element method. Sladek and Sladek [13] set up the BEM formulation for coupled thermoelasticity.

In this work, a formulation of the boundary integral equation method for thermoelasticity with both one and two relaxation times is given. Fundamental solutions of the corresponding differential equations are obtained. A reciprocity theorem is derived. An outline of the implementation of the boundary element method is discussed for the solution of the above boundary equations.

The Mathematical Model

We shall consider a homogeneous isotropic thermoelastic solid occupying the region V and bounded by a smooth surface S. We shall also assume that the initial state of the medium is quiescent. Throughout this work, a comma denotes material derivatives and the summation convention is used. The governing equations for thermoelasticity with one–two relaxation times consist of [9]:

• 1. The equations of motion:

  $$\begin{array}{ll} \mu \,{u_{i,kk }} + (\lambda + \mu){u_{k,ki }} + \rho {F_i} - \gamma \,\left( {\,{T_{,i }} + \upsilon\frac{{\partial {T_{,i }}}}{{\partial \,t}}} \right) \\ = \rho\frac{{\partial^2 {u_i}}}{{\partial \,{t^2}}}\end{array}$$

  (1)

  where λ, μ are Lamé constants; ρ is the density; t is the time variable; T is the absolute temperature of the medium; u _i is the displacement component in the x_i-direction; F _i is the component of the body force per unit mass in the x_i-direction and γ = (3λ + 2 μ) α _t , α _t being the coefficient of linear thermal expansion; and v is a constant with the dimensions of time that acts as a relaxation time.

• 2. The equation of heat conduction:

  $$\begin{array}{ll} \chi T{,_{ii }} = \rho \,{c_E}\left({\frac{{\partial \,}}{{\partial \,t}} + {\tau_0}\frac{\partial^2}{{\partial \,{t^{^2 }}}}} \right)\,T + \left( {1 + \alpha\,{\tau_0}\frac{{\partial \,}}{{\partial \,t}}} \right)\cr \,\left({\gamma \,{T_0}\dot{e} - \rho Q} \right)\end{array}$$

  (2)

  where $\chi$ is the thermal conductivity of the medium, e is the cubical dilatation, c _E is the specific heat at constant strain, ${T_0}$ is a reference temperature assumed to be such that $\left| {\frac{{T - {T_0}}}{T_0 }} \right| < < 1$, τ ₀ is another relaxation time, and Q is the strength of the applied heat source per unit mass. For Lord-Shulman theory υ = 0, α = 1 and for Green-Lindsay theory, α = 0.

• 3. The constitutive relations:

  $${\sigma_\mathrm{{ij}}} = 2\mu {e_{ij }} + \lambda e{\delta_{ij }} - \gamma \left( {T - {T_0} + \nu \frac{\partial T }{\partial t }} \right){\delta_{ij }}$$

  (3)

  where e _ij are the strain components given by

  $${e_{ij }} = \frac{1}{2}({u_{i,j }} + {u_{j,i }}) \mathrm{\,\rm and\ } \ \mathrm{e} = {\mathrm{{e}}_\mathrm{{ii}}} = {\mathrm{{u}}_{\mathrm{{i},\mathrm{i}}}}$$

  (4)

Let us introduce the nondimensional variables

$$\begin{array}{ll} x_i^{*} = c\,\eta \,{x_i}, \,\,u_i^{*} =c\eta\,{u_i}, {t^{*}} = {c^2}\eta \,t,\cr \tau_0^{*} ={c^2}\,\eta\,{\tau_0}, \,\,{\upsilon^{*}} = {c^2}\,\eta \,\upsilon\cr\hfill\sigma_{ij}^{*} = \frac{{\sigma_{ij }}}{{\lambda + 2\mu}},\theta = \frac{{\gamma (T - {T_0})}}{{\lambda + 2\mu}},\cr\,\,\,\,{Q^{*}} = \frac{{\rho\,Q}}{{\chi\,{T_0}{\eta^2}{c^2}}},\,\,F_i^{*} = \frac{{\rho\,{F_i}}}{{\eta c(\lambda + 2\mu )}} \hfill \cr \end{array}$$

where ${c^2} = \frac{{\lambda + 2\mu }}{\rho }$ is the square of the velocity of longitudinal waves and $\eta = \frac{{\rho \,{c_E}}}{\chi }$ · In terms of these variables Equations (1–3), respectively, take the following forms (dropping the asterisks for convenience):

$$\begin{array}{ll} {u_{{i,\,kk}}} + ({\beta^2} - 1){u_{{k,\,ki}}} + {F_i} -{\beta^2}\left( {{\theta_{,i }} + \upsilon \frac{{\partial{\theta_{,i }}}}{\partial t }} \right) \cr ={\beta^2}\frac{{\partial^2 {u_i}}}{{\partial \,{t^2}}}\end{array}$$

(5)

$${\theta_{,ii }} = \frac{{\partial \theta }}{{\partial \,t}} + {\tau_0}\frac{{\partial^2 \theta }}{{\partial \,{t^2}}} + \varepsilon \,\left( {\frac{{\partial \,}}{{\partial \,t}} + \alpha \,{\tau_0}\frac{{\partial^2 \,}}{{\partial \,{t^{^2 }}}}} \right)\,e - \left( {1 + \alpha \,{\tau_0}\frac{{\partial \,}}{{\partial \,t}}} \right)Q$$

(6)

$$\begin{array}{ll}{\sigma_{ij }} = ({\beta^2} - 2)\,e\,{\delta_{ij}} + 2{e_{ij }} - {\beta^2}\left( {\theta + \upsilon \frac{{\partial\theta }}{\partial t }} \right){\delta_{ij }}\end{array}$$

(7)

where ${\beta^2} = \frac{{\lambda + 2\mu }}{\mu }$ and $\varepsilon = {T_0}{\gamma^2}/\rho \,{C_E}(\lambda + 2\mu )$.

The boundary conditions of the problem will be taken as follows:

• 1. Mechanical condition: The traction ${p_i} = {\sigma_{ij }}{n_j}$ is specified on a part ${\mathrm{{S}}_1}\subset S$ while the displacement u, is specified on $S - {S_1}$

• 2. Thermal conditions: The temperature increment θ is specified on a part ${\mathrm{{S}}_2}\subset S \, {\rm while\, the\, normal}$ derivative $\frac{{\partial \theta }}{\partial n }$ is specified on ${\mathrm{{S}}_2}\subset S$, where n(r) is the outward normal to the surface S. The above conditions can be stated as

$${\sigma_{ij }}{n_j} = {p_{i0 }}(r,t)\,\,\,on\,\,{S_1}$$

(8)

$${u_i} = {u_{i0 }}(r,t)\,\,\,on\,\,\overline {S - {S_1}}$$

(9)

$$\theta = {\theta_0}(r,t)\,\,\,on\,\,{S_2}$$

(10)

$$\theta {,_n} = {\theta_{0,n }}(r,t)\,\,\,on\,\,\overline {S - {S_2}}$$

(11)

Using (4) and (7), the traction p _i (r, t) can be written as

$$\begin{array}{ll}{ {p_i}(r,t) = \left[ {({\beta^2} -2){u_{j,j }} - {\beta^2}\,\left\{ {\theta + \nu \frac{{\partial\theta }}{\partial t }} \right\}} \right]\,{n_i}}\cr + ({u_{i,j }} +{u_{j,i }}){n_j}\end{array}$$

(12)

Formulation in the Laplace-Transform Domain

The Laplace transform of a function f (t) is defined by

$$\overline f (s) = \int\limits_0^{\infty } {{e^{- st }}f(t)\,dt}$$

Applying Laplace transform to both sides of (5)–(7) and using the homogeneous initial conditions, we obtain

$${\overline u_{i,kk }} + ({\beta^2} - 1){\overline u_{k,ki }} + {\overline F_i} - {\beta^2}\,(1 + \nu \,s){{\overline \theta}_{,i }} = {\beta^2}{s^2}{\overline u_i}$$

(13)

$$\begin{array}{ll}\overline \theta {,_{ii }} = s\,(1 +{\tau_0}\,s)\,\overline \theta + (1 + \alpha {\tau_0}s)\,(\epsilon\,s\,\,{\overline u_{k,k }} - \overline Q )\end{array}$$

(14)

$$\begin{array}{ll}{{\overline \sigma}_{ij }} = ({\beta^2} -2)\,{\overline u_{k,k }}\,{\delta_{ij }} + 2{e_{{i\,j}}} -{\beta^2}\,(1 + \nu \,s)\,\overline {\theta \,} {\delta_{ij}}\end{array}$$

(15)

Using the following Helmholtz decomposition for the displacement and body forces

$${u_i} = \varphi {,_i} + {\epsilon_{ijk }}\,{\psi_{j,k }}$$

(16)

$${F_i} = X{,_i} + {\epsilon_{{ij\,k}}}\,{Y_{{j,\,k}}}$$

(17)

Equations (13)–(15) yield

$${\beta^2}\left( {\nabla^2 - {s^2}} \right)\,\,\overline \varphi + \overline X = {\beta^2}(1 + \nu \,s)\overline {\,\theta }$$

(18)

$$({\nabla^2} - {\beta^2}\,{s^2})\,\overline {\psi_i } = - \overline {\,{Y_i}}$$

(19)

$$({\nabla^2} - s\,(1 + {\tau_0}\,s))\,\overline \theta - \epsilon \,s\,(1 + \alpha {\tau_0}s)\,{\nabla^2}\overline \varphi = - (1 + \alpha {\tau_0}s)\,\overline Q$$

(20)

Fundamental Solutions in the Laplace-Transform Domain

We shall consider two cases. The first is that of an instantaneous concentrated heat source acting at the point x = y in the absence of externally applied body forces. The second case is that when there is no heat source acting inside the medium with an instantaneous concentrated body force acting in the direction of one of the coordinate axes.

Case 1

We take

$$Q = \delta (x - y)\,\delta (t),\,\,\,\,\,\,\,{F_i} = 0$$

The fundamental solutions corresponding to this case will be denoted by primes. Substituting the above values in Equations (18–20), we get

$$\left( {\nabla^2 - {s^2}} \right)\,\overline {\varphi '} = (1 + \nu \,s)\overline {\,\theta '}$$

(21)

$$\left( {\nabla^2 - {\beta^2}{s^2}} \right)\,\overline {\psi_i^{'}} = 0$$

(22)

$$\begin{array}{ll} ({\nabla^2} - s\,(1 + {\tau_0}\,s))\,\overline{\theta '} - \epsilon \,s\,(1 + \alpha{\tau_0}s)\,{\nabla^2}\overline {\varphi '} \cr= - (1 + \alpha{\tau_0}s)\,\delta \,(x - y)\,\end{array}$$

(23)

Equation (22) immediately yields

$$\overline {\psi_i^{'}} = 0$$

(24)

while (21) and (23) give, upon elimination of $\overline {\theta '},$the following equation satisfied by $\overline {\varphi '}$

$\left\{ {\nabla^4 - \left[ {s^2 + s(1 + {\tau_0}\,s) + \varepsilon \,s(1 + \nu \,s)} \right]{\nabla^2} +{s^3}(1 + {\tau_0}\,s)} \right\}\,\overline {\varphi '} = - (1 + \nu \,s)\,\,(1 + \alpha {\tau_0}s)\,\delta (x - y)$ This equation can be factorized as

$$({\nabla^2} - k_1^2)({\nabla^2} - k_2^2)\,\overline {\varphi '} = - (1 + \nu \,s)\,(1 + \alpha {\tau_0}s)\,\delta (x - y)$$

(25)

where $k_1^2$ and $k_2^2$ are the roots of the characteristic equation

$$\begin{array}{ll}{k^4} - {k^2}\left[ {s^2} + \epsilon (s+{\tau_0}\,\alpha \,{s^2})(1 + \nu \,s) \right.\left. + (s+{\tau_0}\,{s^2}) \right] \\ + {s^3}(1 + {\tau_0}\,s) =0\end{array}$$

The sum and product of the roots of this equation satisfy the relations

$$k_1^2 + k_2^2 = ({s^2} + \epsilon (s + {\tau_0}\,\alpha \,{s^2})(1 + \nu \,s) + (s + {\tau_0}\,{s^2})$$

(26a)

$$k_1^2.\,k_2^2 = {s^3}(1 + {\tau_0}s)$$

(26b)

Using Helmholtz equations in space [14], namely,

$$\frac{1}{{\nabla^2 - {k^2}}}[\delta (r)] = \frac{ - 1 }{{4\pi \,r}}{e^{{- k\,r}}}$$

the solution of (25) takes the form

$$\overline {\varphi '} = \frac{{(1 + \nu \,s)(1 + \alpha {\tau_0}s)}}{{4\pi \,r(k_1^2 - k_2^2)}}\left[ {{e^{{- {k_1}\,r}}} - {e^{{- {k_2}\,r}}}} \right]$$

(27)

where r = $\sqrt {{({x_i} - {y_i})({x_i} - {y_i})}}$

From (16), (24), and (27), it follows that

$$\begin{array}{ll} \overline u_i^{'}(r,s) = \frac{{ - (1 + \nu\,s)\,(1 + \alpha {\tau_0}s)}}{{4\pi \,{r^2}(k_1^2 -k_2^2)}}r{,_i}\cr \left[ {({k_1}r + 1){e^{{- {k_1}r}}} - ({k_2}r +1){e^{{- {k_2}r}}}} \right]\end{array}$$

(28)

where r,_i = (x_i–y_i)/r

From Equations (21) and (27), we obtain

$$\begin{array}{ll}\overline {\theta '} (r,s) =\frac{1}{{4\pi \,r\,(k_1^2 - k_2^2)}}\left[\,(k_1^2 - {s^2}){e^{{-{k_1}r}}- (k_2^2 - {s^2}){e^{{- {k_2}r}}}} \right]\end{array}$$

(29)

The Laplace transform of the traction vector can be obtained from (15) as

$$\overline p_l^{'}(r,s) = \left[ {({\beta^2} - 2)\overline u_{k,k}^{'} - {\beta^2}(1 + \nu \,s){{{\overline \theta}}^{\prime }}} \right]{n_l} + (\overline u_{l,k}^{'} + \overline u_{k,l}^{'}){n_k}$$

It can easily be seen from (28) that

$$\overline u_{i,j}^{'} = \frac{{(1 + \nu \,s)\,(1 + \alpha {\tau_0}s)}}{{4\pi \,{r^3}(k_1^2 - k_2^2)}}\left\{ {r{,_i}r{,_j}\,{f_1} + (2r{,_i}r{,_j} - {\delta_{ij }}){f_2}} \right\}$$

(30)

where

$${f_1} = (k_1^2{r^2} + {k_1}r + 1){e^{{- {k_1}r}}} - (k_2^2{r^2} + {k_2}r + 1){e^{{- {k_2}r}}}$$

$${f_2} = ({k_1}r + 1){e^{{- {k_1}r}}} - ({k_2}r + 1){e^{{- {k_2}r}}}$$

Using the above expressions, the traction vector takes the form

$$\begin{array}{ll} \overline p_l^{'}(r,s) = \frac{{(1 + \nu\,s)\,(1 + \alpha {\tau_0}s)}}{{4\pi \,{r^3}(k_1^2 - k_2^2)}}\cr\left\{ {\left[ {({\beta^2} - 2)(\,{f_1} - {f_2}) -{\beta^2}{r^2}{f_3}} \right]{n_l}} \right. \hfill \cr \left. { +2\,\left[ {r{,_l}r{,_k}(\,{f_1} + 2{f_2}) - {\delta_{lk }}{f_2}}\right]{n_k}} \right\} \hfill \cr \end{array}$$

(31)

where

$${f_3} = (k_1^2 - {s^2}){e^{{- {k_1}r}}} - (k_2^2 - {s^2}){e^{{- {k_2}r}}}$$

The expression for $\frac{{\partial {{{\overline \theta}}^{\prime }}}}{\partial n }$ will be used later on in this entry. This is obtained from (29) as follows:

$$\frac{{\partial {{{\overline \theta}}^{\prime }}}}{\partial n } = - \overline \theta_{,i}^{'}\,\,{n_i} = \frac{{r{,_i}\,{n_i}}}{{4\pi (k_1^2 - k_2^2)\,{r^2}}}\,{f_4}$$

(32)

where

$$\begin{array}{ll} {f_4} = (k_1^2 - {s^2})({k_1}r + 1){e^{{- {k_1}r}}} - (k_2^2 - {s^2})({k_2}r + 1){e^{{- {k_2}r}}}\end{array}$$

Case 2

We take

$$Q = 0,\,\,\,\,\,\,\,\,\,\,{F_i} = F_i^{(\,j) } = {\delta_{ij }}\,\delta (t)\,\,\delta (x - y)$$

The fundamental solutions in this case will be denoted by a superscript (j).

The governing equations for this case take the form

$${\beta^2}\left( {\nabla^2 - {s^2}}\right)\,\,{{\overline \varphi}^{(\, j) }} + {\overline X^{(\, j) }}= {\beta^2}(1 + \nu \,s){{\overline {\,\theta}}^{(\, j) }} $$

(33)

$$({\nabla^2} - {\beta^2}\,{s^2})\,{{\overline {\psi_i}}^{(\, j) }} = - {{\overline {\,{Y_i}}}^{(\, j) }}$$

(34)

$$\begin{array}{ll} ({\nabla^2} - s\,(1 +{\tau_0}\,s))\,\overline \theta {\,^{(\, j) }} - \epsilon \,s\,(1 +\alpha {\tau_0}s)\,{\nabla^2}{{\overline \varphi}^{(\, j) }} =0\end{array}$$

(35)

The scalar and vector potentials in (17) have the forms

$$\begin{array}{ll}X = {X^{(\, j) }} = \frac{ - 1}{{4\pi }}\left( {\frac{1}{r}} \right)\,{,_j}\quad{Y_i} = Y_i^{(\,j) } = \frac{ - 1 }{{4\pi }}{\epsilon_{ilj }}\left( {\frac{1}{r}}\right)\,{,_l}\end{array}$$

(36)

Eliminating ${{\overline \theta}^{(\, j) }}$ from (33) and (35) and substituting in the resulting equationsfor ${X^{(\, j) }}$ given in (36), we get

$$({\nabla^2} - k_1^2)\,({\nabla^2} - k_2^2){{\overline {\,\varphi}}^{(\, j) }} = \frac{1}{{4\pi {\beta^2}}}\left[ {\nabla^2 - s(1 + {\tau_0}\,s)} \right]\,\,\left( {\frac{1}{r}} \right)\,{,_j}$$

(37)

To solve this equation, we make the change of variable ${{\overline {\,\varphi}}^{(\, j) }} = g{,_j}$ to arrive at

$$g = \frac{1}{{4\pi {\beta^2}(k_1^2 - k_2^2)}}\left[ {\,{\nabla^2} - s(1 + {\tau_0}\,s)} \right]({F_1} - {F_2})$$

where F _i is the solution of the equation

$$({\nabla^2} - k_i^2){F_i} = \frac{1}{r}, \,\,i = 1,\,2$$

The solution of these equations bounded both at the origin and at infinity is givenby

$${F_i} = \frac{1}{k_i^2r}\left[ {{e^{{- {k_i}r}}} - 1} \right], \,\,i = 1,\,2$$

Collecting the previous results and using (26a, b), we arrive at

$$\begin{array}{ll} {\overline \varphi}^{(\, j) } =\frac{{r{,_j}}}{{4\pi {\beta^2}{s^2}{r^2}}} \times \left[ {1 +\frac{1}{k_1^2 - k_2^2}\left\{ {(1 + {k_1}r)}\right.}\right.\cr\left.\left.(k_2^2 - {s^2}){e^{{- {k_1}r}}}- (1 + {k_2}r)(k_1^2 -{s^2}){e^{{- {k_2}r}}} \right\}\vphantom{\frac{1}{k_1^2 - k_2^2}}\right] \end{array}$$

(38)

Substituting from the second of (36) into (34) and using the substitution

$$\overline \psi_i^{(\, j) } = {\varepsilon_{ilj }}G{,_l}$$

we get as before

$${{\overline {\psi_l}}^{(\, j) }} = \frac{1}{{4\pi {\beta^2}{s^2}{r^2}}}{\epsilon_{lmj }}r{,_m}\left[ {(1 + \beta sr){e^{{- \beta sr}}} - 1} \right]$$

(39)

Substituting from (38) and (39) into (16) and using the relation

$$r{,_{ij }} = \frac{{{\delta_{ij }} - r{,_i}r{,_j}}}{r}$$

we get

$$\overline u_i^{(\, j) } = \frac{1}{r}\left[ {U_1 {\delta_{ij }} + {U_2}\,{r_{,i }}{r_{,j }}} \right]$$

(40)

where

$$\begin{array}{ll} {U_1} = \frac{1}{{4\pi{\beta^2}{s^2}{r^2}}}\left\{ {(1 + \beta sr){e^{{- \beta sr}}}}\right. \\ \left.+ {s^2}\sum\limits_{n = 1}^2 {{( - 1)^{n - 1}}{A_n}(1 + {k_n}r){e^{{- {k_n}r}}}} \right\} \hfill \cr {{U_2}} =\frac{ - 1 }{{4\pi {\beta^2}{s^2}{r^2}}} \hfill \times \left\{ {(3 +3\beta sr + {\beta^2}{s^2}{r^2}){e^{{- \beta sr}}}}\right.\cr\left.+ {s^2}\sum\limits_{n = 1}^2 {( - 1)^{n - 1 }}{A_n}(3 +3{k_n}r + k_n^2{r^2}){e^{{- {k_n}r}}} \right\} \hfill \cr\end{array}$$

where

$${A_n} = \frac{{k_n^2 - s(1 + {\tau_0}s)}}{k_n^2(k_2^2 - k_1^2) }, n = 1,\,2$$

Finally, to obtain the temperature function in this case, it can easily be checked that

$${{\overline \theta}^{(\, j) }} = \frac{{ - \epsilon \,s}}{{\beta^2 (1 + \nu \,s)}}\overline u_j^{'}$$

(41a)

This equation together with (28) give

$${{\overline \theta}^{(\, j) }} = \frac{{\,\epsilon \,s\,(1 + \alpha {\tau_0}s)}}{{4\pi \,{\beta^2}{r^2}(k_1^2 - k_2^2)}}r{,_j}\,{f_2}$$

(41b)

Equation (41b) leads to

$$\frac{{\partial {{{\overline \theta}}^{(\, j)}}}}{{\partial \,n}} = \frac{{\epsilon \,s\,\,(1 + \alpha{\tau_0}s)\,{n_i}}}{{4\pi \,{\beta^2}{r^3}\,(k_1^2 - k_2^2)}}\left[{{\delta_{ij }}{f_2} - r{,_i}r{,_j}\,{f_5}} \right]$$

(42)

where

$${f_5} = (k_1^2{r^2} + 3{k_1}r + 3){e^{{- {k_1}r}}} \\- (k_2^2{r^2} + 3{k_2}r + 3){e^{{- {k_2}r}}}$$

The Laplace transform of the traction vector for this case can also be obtained from (15) as

$$\overline p_l^{(\, j) }(r,s) = \left[ {({\beta^2} - 2)\,\overline u_{k,k}^{(\, j) } - {\beta^2}(1 + \nu \,s){{{\overline \theta}}^{(\, j) }}} \right]{n_l} + 2\left[ {\overline u_{{^{l,k}}}^{(\, j) } + \overline u_{k,l}^{(\, j) }} \right]\,{n_k}$$

From (40), we obtain

$$\begin{array}{ll}\overline u_{k,l}^{(\, j) } =\frac{1}{r^2}\left[ {U_3 {\delta_{kj }}r{,_l} + {U_2}({\delta_{kl}}r{,_j} + {\delta_{jl }}r{,_k}) + {U_4}r{,_j}r{,_k}r{,_l}}\right]\end{array}$$

(43)

and

$$\overline u_{k,k}^{(\, j) } = \frac{1}{r^2}\left[ {U_3 + 4{U_2} + {U_4}} \right]r{,_j}$$

(44)

where

${U_3} = r\frac{{\partial {U_1}}}{\partial r } - {U_1}\,\,\,\,\,$ and ${U_4} = r\frac{{\partial {U_2}}}{\partial r } - {U_2}\,\,\,\,\,$

Substituting from (43) and (44) into the expression for $\overline p_l^{(\, j) }(r,s)$, we get

$$\begin{array}{ll} \overline p_l^{(\, j) }(r,s) =\frac{1}{r^2}\left[ {({\beta^2} - 2)\left[ {U_3 + 4{U_2} + {U_4}}\right]+ 2{U_2 }}\right.\cr \left.+ (1 + \nu \,s){f_2}\right]\,r{,_j}{n_l}+ ({U_3} + {U_2})r{,_l}{n_j} \cr + \left[{({U_3} + {U_2}) + 2{U_4}r{,_j}r{,_l}}\right]r{,_k}{n_k}\end{array}$$

(45)

Reciprocity Theorem

This section is devoted to the derivation of a reciprocity theorem that shall be used later in this entry to obtain integral representations of the displacement and temperature distributions in terms of the boundary values of the problem.

Assume we have two systems of causes (heat sources and body forces) (Q, F _i ) and (Q’, F _i ). Due to the application of these causes, we get the effects (displacements and temperature) $({u_i},\theta )$ and $({u'_i},\theta ')$, respectively. Multiplying both sides of (15) by $\overline e_{ij}^{'}$ and integrating over the volume V, we obtain

$$\begin{array}{ll}\int\limits_V{{{{\overline\sigma}}_{ij}}\,\overline e_{ij}^{'}} dV=\int\limits_V {\left[{2\overline e_{ij}\overline e_{ij}^{'}+({\beta^2} - 2)\overline e\,\,{\overline e^{\prime}}}\right.} \cr\left.\qquad \qquad \qquad - {\beta^2}(1 +\nu\,s)\,\overline \theta \,\,{\overline e^{\prime}}\right]\,dV\end{array}$$

Subtracting this equation from its counterpart obtained by interchanging the two systems, we get upon using (4) and integration by parts

$$\begin{array}{ll} \int\limits_S {\left[ {{{{\overline\sigma}}_{ij}}\,\overline u_i^{'} -{{{\overline\sigma}}^{\prime}}_{ij}\,\overline u_i} \right]\,{n_j}dS-\int\limits_V {\left[ {{{{\overline \sigma}}_{ij,j }}\,\overline u_i^{'} - {{{\overline \sigma}}^{\prime}}_{ij,j}\,\overline u_i}\right]\,dV}} \cr = {\beta^2}(1 + \nu \,s)\int\limits_V{\left[{{{{\overline \theta}}^{\prime }}\,\overline e - \overline\theta\,{\overline e^{\prime }}} \right]\,dV} \end{array}$$

Using the transformed equations of motion in nondimensional form ${{\overline \sigma}_{ij,j }} = {\beta^2}{s^2}{\overline u_i} - {\overline F_i}$ and ${{\overline {\sigma '}}_{ij,j }} = {\beta^2}{s^2}{\overline u^{\prime}}_i - {{\overline {F'}}_i}$we obtain

$$\begin{array}{ll}\int\limits_S{\left[{{{{\overline\sigma}}_{ij}}\,\overline u_i^{'}-{{{\overline\sigma}}^{\prime}}_{ij}\,\overline u_i}\right]\,{n_j}dS+\int\limits_V {\left[ {\overline F_i\,\overline u_i^{'}-{\overline F^{\prime}}_i\,\overline u_i}\right]\,dV}}\cr={\beta^2}(1 + \nu\,s)\int\limits_V{\left[{{{{\overline\theta}}^{\prime }}\,\overline e - \overline\theta\,{\overline e^{\prime }}} \right]\,dV}\end{array}$$

(46)

Multiplying both sides of (14) by ${{\overline \theta}^{\prime }}$ and following the same procedure as above, we get

$$\begin{array}{ll} \int\limits_S {\left[ {{{{\overline\theta}}^{\prime }}\,\overline \theta \,{,_n} - \overline {\,\theta} \,\overline {\theta '} \,{,_n}} \right]\,dS = \big(1 + \alpha{\tau_0}s\big)}\cr \left\{ {\epsilon \,s\int\limits_V {\left[{{{{\overline \theta}}^{\prime }}\,\overline e - \overline \theta\,{\overline e^{\prime }}} \right]\,dV - \int\limits_V {\left[{{{{\overline \theta}}^{\prime }}\,\overline Q - \overline \theta\,{\overline Q^{\prime }}} \right]\,dV} } } \right\}\end{array}$$

(47)

Eliminating $\int\limits_V {\left[ {{{{\overline \theta}}^{\prime }}\,\overline e - \overline \theta \,{\overline e^{\prime }}} \right]\,dV}$ between (46) and (47), we get

$$\begin{array}{ll} \epsilon \,s\,(1 + \alpha{\tau_0}s)\int\limits_V {\left[ {\overline F_i \,\overline u_i^{'} -{{{\overline {F_i}}}^{\prime }}\,\overline u_i} \right]\,dV} \cr +{\beta^2}(1 + \nu \,s)\int\limits_V {\left[ {{\overline Q^{\prime}}\,\overline \theta - \overline Q \,{{{\overline \theta}}^{\prime}}} \right]\,dV} \hfill \cr = {\beta^2}(1 + \nu \,s)\int\limits_S{\left[ {{{{\overline \theta}}^{\prime }}\,\overline \theta \,{,_n}- \overline {\,\theta } \,\overline {\theta '} \,{,_n}} \right]\,dS}\cr - \epsilon \,s\,(1 + \alpha {\tau_0}s)\int\limits_S {\left[{{{{\overline \sigma}}_{ij }}\,\overline u_i^{'} - {{{\overline\sigma}}^{\prime}}_{ij}\,\overline u_i} \right]\,{n_j}dS} \hfill \cr\end{array}$$

(48)

Boundary Integral Equations

In order to obtain integral representations for the transformed displacement and temperature distributions, we shall take the causes and effects (Q, F _i ) and (u _i ,θ) of the previous section to denote those of a given distribution under consideration.

The second system of causes and effects will be first taken as those obtained above in Case 1, i.e., we take

$Q' = \delta (x - y)\delta (t),\,\,\,{F'_i} = 0$ where $y\in V\cup S$

The corresponding transformed displacement $\overline {u_i }$ and temperature $\overline {\theta '}$ are given by

Equations (28) and (29). It follows from (48) that

$$\begin{array}{ll} (1 + \nu \,s)\Delta (x)\,\overline \theta (x,s)= (1 + \nu \,s)\cr \int\limits_S {\left[ {{{{\overline\theta}}^{\prime }}\,\overline \theta \,{,_n} - \overline {\,\theta} \,\overline {\theta '} \,{,_n}} \right]\,dS} + (1 + \nu \,s)\cr\int\limits_V {{{{\overline \theta}}^{\prime }}} \overline Q \,dV\hfill - \frac{{\epsilon \,s\,\,(1 + \alpha {\tau_0}s)}}{\beta^2}\cr\left\{ {\int\limits_S {\left[ {{{{\overline \sigma}}_{ij}}\,\overline u_i^{'} - {{{\overline\sigma}}^{\prime}}_{ij}\,\overline u_i} \right]\,{n_j}dS} +\int\limits_V {\overline u_i^{'}{\overline F_i}dV} } \right\} \hfill\cr \end{array}$$

(49)

where $\Delta (x) = \left\{ \matrix{ 1\quad if\,\,x\in V\cr0\quad if\,x\,\notin V\cup S \cr \frac{1}{2}\,\,\,\,if\,x\in S \hfill \cr } \right.$

Next, we take second system of causes and effects to denote those of Case 2, i.e., we take

${Q^{(\, j) }} = 0,\,\,\,F_i^{(\, j) } = {\delta_{ij }}\delta (x - y)\delta (t)$ where $y\in V\cup S$***

The corresponding transformed displacement ${{\overline {u_i}}^{(\, j) }}$ and temperature ${{\overline {\theta '}}^{(\, j) }}$ are given by (41) and (40). It follows from (48) that

$\begin{array}{lll} & \epsilon \,s\,(1 + \alpha{\tau_0}s)\Delta(x)\,{\overline u_j}(x,s) = \epsilon \,s\,(1+\alpha{\tau_0}s)\\ & \left\{\int\limits_V {\overline{u_i}^{(\, j)}{\overline{F_i}dV + \int\limits_S {\left[{{{{\overline\sigma}}_{ij}}\,\overline u_i^{(\, j) } -\overline\sigma_{ij}^{(\,j)}\,{\overline u_i}}\right]\,{n_j}dS} } } \right\}\hfill \cr & - {\beta^2}(1 +\nu \,s)\left\{ {\int\limits_V{{{{\overline\theta}}^{(\,j) }}} \overline Q \,dV +\int\limits_S{{{{\overline\theta}}^{(\, j) }}} {{{\overline{\theta,}}}_n}-\overline \theta \,\overline {\theta,}_n^{(\,j)}dS}\right\}\end{array}$ Substituting from (41a) into the above equation, we get

$$\begin{array}{ll} \,(1 + \alpha{\tau_0}s)\Delta(x)\,{\overline u_j}(x,s) = (1 + \alpha{\tau_0}s)\cr \left\{{\int\limits_V{\overline u_i^{(\, j)}{\overline F_i}dV +\int\limits_S {\left[{{{{\overline \sigma}}_{ij}}\,\overline u_i^{(\, j) } - \overline\sigma_{ij}^{(\,j)}\,\overline u_i}\right]\,{n_j}dS} } } \right\}\hfill \cr +\int\limits_V {\overline u_j^{'}\,} \overline Q \,dV +\int\limits_S{(\overline u_j^{'}}{{\overline {\theta,}}_n} -\overline \theta\,\overline u_j^{'}{,_n})\,dS \end{array}$$

(50)

Using the convolution theorem of the Laplace transforms [15], namely,

$${L^{- 1 }}\left[ \,{\overline f_1 (s){\overline f_2}(s)} \right] = \int\limits_0^t {f_1 (z){f_2}(t - z)dz}$$

together with (49) yields

$$\begin{array}{ll} \Delta (x)\,\left( {1 + \nu \frac{\partial}{\partial t }} \right)\theta (x,t)= \int\limits_0^t {\int\limits_S{\left[ {\frac{{\partial \theta (\eta, z)}}{{\partial n(\eta)}}}\right.}}\cr {\left. - \theta (\eta, z)\frac{\partial}{{\partial n(\eta )}} \right] \left( {1 - \nu \frac{\partial}{\partial z }} \right)\theta '(x - \eta, t - z)d{S_n}dz \hfill}\cr{+ \int\limits_0^t {\int\limits_V {Q(y,z)} } \left( {1 - \nu\frac{\partial }{\partial z }} \right)\theta '(x - \eta, t -z)d{V_y}dz \hfill }\cr - \frac{{\epsilon\,}}{\beta^2}\int\limits_0^t {\int\limits_V {F_i (y,z)} }\left({\frac{\partial }{\partial z } - \alpha {\tau_0}\frac{\partial^2}{{\partial {z^2}}}} \right)\cr u_i^{'}(x - y,t - z){d{V_y}dz} \cr -\frac{{\epsilon \,}}{\beta^2}\int\limits_0^t {\int\limits_S} {\left[{{\sigma_{ij}}(\eta, z)\left( {\frac{\partial }{\partial z } -\alpha {\tau_0}\frac{\partial^2 }{{\partial {{z^2}}}} }\right)}\right.} \\ {\left.u_i^{'}(x - \eta, t - z)- u_i(\eta, z)\left({\frac{\partial }{\partial z } - \alpha {\tau_0}\frac{\partial^2}{{\partial {z^2}}}} \right)\right.} \\ \left.\sigma_{ij}^{'}(x -\eta, t - z) \vphantom{\frac{\partial }{\partial z}}\right]{n_j}d{S_n}dz \end{array}$$

(51)

Similarly, inverting the Laplace transforms in (50) gives***

$$\begin{array}{ll} \,(1 + \alpha{\tau_0}s)\Delta(x)\,{\overline u_j}(x,s) = (1 + \alpha{\tau_0}s)\cr\left\{{\int\limits_V{\overline u_i^{(\, j) }{\overline F_i}dV+\int\limits_S {\left[{{{{\overline \sigma}}_{ij }}\,\overline u_i^{(\, j) } - \overline\sigma_{ij}^{(\, j)}\,\overline u_i}\right]\,{n_j}dS} } } \right\}\hfill \cr +\int\limits_V{\overline u_j^{'}\,} \overline Q \,dV +\int\limits_S{(\overline u_j^{'}}{{\overline {\theta,}}_n} -\overline \theta\,\overline u_j^{'}{,_n})\,dS\end{array}$$

$$\begin{array}{ll}\Delta (x)\left( {1 +\alpha{\tau_0}\frac{\partial }{\partial t }} \right)\,{u_j}(x,t)=\int\limits_0^t {\int\limits_S {\left[ {\frac{{\partial\theta(\eta, z)}}{{\partial n(\eta )}} -\theta (\eta,z)\frac{\partial}{{\partial n(\eta )}}} \right]} }\cr \,\overline u_j^{'}(x - \eta,t - z)d{S_n}dz+ \int\limits_0^t {\int\limits_V{Q(y,z)} }\,\overline u_j^{'}(x - y,t - z)d{V_y}dz \cr +\int\limits_0^t{\int\limits_V {F_i (y,z)} \,\left( {1 -\alpha{\tau_0}\frac{\partial }{\partial z }} \right)\,} u_i^{(\, j)}(x -y,t - z)\,d{V_y}dz \hfill \cr + \int\limits_0^t{\int\limits_S{\left[ {{\sigma_{ij }}(\eta, z)\left( {1 -\alpha{\tau_0}\frac{\partial }{\partial z }} \right)\,u_i^{(\, j)}(x -\eta, t - z)\,{n_j}d{S_n}dz} \right.} } \hfill \cr-\int\limits_0^t {\int\limits_S {\left[ { - u_i(\eta, z)\left( {1-\alpha {\tau_0}\frac{\partial }{\partial z}}\right)\sigma_{ij}^{(i) }(x - \eta, t -z)\,{n_j}d{S_n}dz}\right]\,} } \end{array}$$

(52)

Equations (51) and (52) can be written more concisely as

$$\Delta (x)\,\left( {1 + \nu \frac{\partial }{\partial t }} \right)\,\theta (x,t) = {W_1}(x,t)$$

(53)

$$\Delta (x)\,\left( {1 + \alpha {\tau_0}\frac{\partial }{\partial t }} \right)\,{u_j}(x,t) = W_2^{(\, j) }(x,t)$$

(54)

where W₁(x, t) and $\mathrm{W}_2^{(\mathrm{j})}(\mathrm{x}, \ \mathrm{t})$ are the right-hand sides of (51) and (52), respectively.

The solutions of these equations have the forms:

$$\Delta (x)\theta (x,t) = \frac{1}{\nu }{e^{{- t/\nu }}}\int\limits_0^t {{e^{{z/\nu }}}{W_1}(x,z)dz}$$

(55)

$$\Delta (x){u_j}(x,t) = \frac{1}{{\alpha {\tau_0}}}{e^{{- t/\alpha {\tau_0}}}}\int\limits_0^t {{e^{{z/\alpha {\tau_0}}}}\,W_2^{(\, j) }(x,z)dz}$$

(56)

Letting $x\to \xi \in S$ in (55) and (56), we obtain

$$\theta (\xi, t) = \frac{2}{\nu }{e^{{- t/\nu }}}\int\limits_0^t {{e^{{z/\nu }}}{W_1}(\xi, z)dz}$$

(57)

$${u_j}(\xi, t) = \frac{2}{{\alpha {\tau_0}}}{e^{{- t/\alpha {\tau_0}}}}\int\limits_0^t {{e^{{z/\alpha {\tau_0}}}}\,W_2^{(\, j) }(\xi, z)\,dz}$$

(58)

Equations (57) and (58) together with the boundary conditions (8)–(11) and (31), (32), (42), and (45) can be used to set up the system of linear equations of the boundary integral equation method.