An Extended Dielectric Crack Model for Fracture Analysis of a Thermopiezoelectric Strip

Abstract

An extended dielectric crack model is proposed to capture the effects of the physical properties of crack interior on crack-tip thermoelectroelastic fields. The typical crack-face boundary conditions can be retrieved by considering the limiting cases of electrical permeability and thermal conductivity inside a crack. Making use of the Fourier transform technique, the problem of a thermopiezoelectric strip weakened by a Griffith crack is investigated and transformed to solve the system of the second kind Fredholm integral equations with Cauchy kernel. The Lobatto–Chebyshev collocation method is used to form a nonlinear system of algebraic equations, which is solved by elaborating on an algorithm. The crack-tip thermoelectroelastic fields are determined by using the approximate solutions. Numerical simulations are carried out to show the variations of the fracture parameters of concern under applied thermoelectromechanical loads, the physical properties of the dielectric medium inside the crack and the geometry of the cracked thermopiezoelectric strip. Some comparisons with the experimental results are reported to reveal the effectiveness of the extended dielectric crack model.

Appendices

Appendix A

(I) The derivation process of Eq. (29) is given as follows: Using the Fourier transform technique, we express the solution of Eq. (15) as

$$\begin{aligned} \theta ^{j}(x,z)=\int _0^\infty {A^{j} (\xi )\frac{\cosh [-\lambda (\delta ^{j}z-h_{j} )\xi ]}{\sinh (\lambda \xi h_{j} )}\cos (\xi x)} \text{ d }\xi +A_{0} z \end{aligned}$$

(A.1)

where $A_{i}^{j} (\xi )(j=\mathrm{I},\mathrm{I}\mathrm{I})$ are the unknowns to be solved, $\delta ^{\mathrm{I}}=1$ and $\delta ^{\mathrm{I}\mathrm{I}}=-1$. Then, from Eqs. (14) and (A.1), the components of the heat flow $q_{x}^{j} (x,z)$ and $q_{z}^{j} (x,z)$ can be calculated as

$$\begin{aligned} q_{x}^{j} (x,z)= & {} \lambda _{x} \int _0^\infty {\xi A^{j} (\xi )\frac{\cosh [-\lambda (\delta ^{j}z-h_{j} )\xi ]}{\sinh (\lambda \xi h_{j} )}\sin (\xi x)} \text{ d }\xi \end{aligned}$$

(A.2)

$$\begin{aligned} q_{z}^{j} (x,z)= & {} \lambda _{z} \int _0^\infty {\delta ^{j}\xi A^{j} (\xi )\frac{\sinh [-\lambda (\delta ^{j}z-h_{j} )\xi ]}{\sinh (\lambda \xi h_{j} )}\cos (\xi x)} \text{ d }\xi -A_{0} \lambda _{z} \end{aligned}$$

(A.3)

Equation (18) and the second relation in Eq. (27) lead to

$$\begin{aligned} A^{\mathrm{I}}(\xi )=-A^{\mathrm{I}\mathrm{I}}(\xi ) \end{aligned}$$

(A.4)

According to Eq. (18), the first relation in Eqs. (27) and (A.4), we obtain the following dual integral equations

$$\begin{aligned} \int _0^\infty {A^{\mathrm{I}}(\xi )[\coth (\lambda h_{\mathrm{I}} \xi )+\coth (\lambda h_{\mathrm{I}\mathrm{I}} \xi )]} \cos (\xi x)\text{ d }\xi =0 \quad (\vert x\vert >a) \end{aligned}$$

(A.5)

$$\begin{aligned} \int _0^\infty {A^{\mathrm{I}}(\xi )} \xi \cos (\xi x)\text{ d }\xi =-\frac{q_{0} -q_{c} }{\lambda _{z} \lambda } \quad (\vert x\vert <a) \end{aligned}$$

(A.6)

The application of Eq. (28) yields

$$\begin{aligned} A^{\mathrm{I}}(\xi )=-\frac{{1}}{\pi }\frac{1}{\xi [\coth (\lambda h_{\mathrm{I}} \xi )+\coth (\lambda h_{\mathrm{I}\mathrm{I}} \xi )]}\int _{-a}^a {g(s)} \sin (\xi s)\text{ d }s \end{aligned}$$

(A.7)

Substituting Eq. (A.7) in Eq. (A.6), one arrives at Eq. (29).

(II) For convenience, the derivation procedure of Eq. (32) is given as follows:

$$\begin{aligned} 2\int _0^\infty {\sin (\xi s)} \cos (\xi x)\text{ d }\xi= & {} \frac{\partial }{\partial x}\int _0^\infty {\frac{2\sin (\xi s)\sin (\xi x)}{\xi }} \text{ d }\xi \\= & {} \frac{\partial }{\partial x}\int _0^\infty {\frac{\cos [(x-s)\xi ]-\cos [(x+s)\xi ]}{\xi }} \text{ d }\xi \\= & {} \frac{\partial }{\partial x}\left\{ {\lim \limits _{\delta \rightarrow 0} \left[ {\int _\delta ^\infty {\frac{\cos [\vert x-s\vert \xi ]}{\xi }} \text{ d }\xi -\int _\delta ^\infty {\frac{\cos [\vert x+s\vert \xi ]}{\xi }} \text{ d }\xi } \right] } \right\} \\= & {} \frac{\partial }{\partial x}\left[ {\lim \limits _{\delta \rightarrow 0} \left( {\int _{\vert x-s\vert \delta }^\infty {\frac{\cos t}{t}} \text{ d }t-\int _{\vert x+s\vert \delta }^\infty {\frac{\cos t}{t}} \text{ d }t} \right) } \right] \\= & {} \frac{\partial }{\partial x}\left( {\lim \limits _{\delta \rightarrow 0} \int _{\vert x-s\vert \delta }^{\vert x+s\vert \delta } {\frac{\cos t}{t}} \text{ d }t} \right) \text{= }\frac{\partial }{\partial x} \left( {\lim \limits _{\delta \rightarrow 0,\eta \rightarrow {0}} \cos \eta \cdot \int _{\vert x-s\vert \delta }^{\vert x+s\vert \delta } {\frac{\text{ d }t}{t}} } \right) \\= & {} \frac{\partial }{\partial x}\left( {\ln \left| {\frac{x+s}{x-s}} \right| } \right) =\frac{1}{s-x}+\frac{1}{s+x} \end{aligned}$$

(III) The detailed process of deriving Eqs. (64)–(66) is shown as follows: By considering the boundary condition of Eq. (25), it is suitable to give the component of electric field $E_{z}^{j} (x,z)$ as

$$\begin{aligned} E_{z}^{j} (x,z)= & {} \sum \limits _{i=1}^3 {\int _0^\infty {\eta _{4i} \alpha _{i} ^{2}\xi [A_{i}^{j} (\xi )} } \cosh (-\delta ^{j}\alpha _{i} z\xi )\nonumber \\&+\,B_{i}^{j} (\xi )\sinh (-\delta ^{j}\alpha _{i} z\xi )]\cos (\xi x)\text{ d }\xi -C_{2} \end{aligned}$$

(A.8)

Then, it follows that $C_{2} =-E_{0}$ and

$$\begin{aligned} \sum \limits _{i=1}^3 {\eta _{4i} \alpha _{i}^{2}[A_{i}^{j} (\xi )} \cosh (-\alpha _{i} h_{j} \xi )+B_{i}^{j} (\xi )\sinh (-\alpha _{i} h_{j} \xi )]=0 \end{aligned}$$

(A.9)

Moreover, it is seen that the application of Eqs. (24) and (26) cannot determine the solutions of $C_{0}$ and $C_{1}$. Here, we still follow [46] to obtain the component of electric displacement as

$$\begin{aligned} D_{0} =\frac{c_{11} e_{33} -c_{13} e_{31} }{c_{11} c_{33} -c_{13}^{2} }\sigma _{0} +\left( {\frac{c_{33} e_{31}^{2} +c_{11} e_{33}^{2} -2c_{13} e_{33} e_{31} }{c_{11} c_{33} -c_{13}^{2} }+\varepsilon _{33} } \right) E_{0} \end{aligned}$$

which is in accordance with the finding for an infinite piezoelectric material [47]. According to Eq. (24), we have

$$\begin{aligned} \sum \limits _{i=1}^3 {\gamma _{1i} } [A_{i}^{j} (\xi )\cosh (-\alpha _{i} h_{j} \xi )+B_{i}^{j} (\xi )\sinh (-\alpha _{i} h_{j} \xi )]=0 \end{aligned}$$

(A.10)

In terms of (26) and (61), it gives

$$\begin{aligned} \sum \limits _{i=1}^3 {\gamma _{0i} } [A_{i}^{j} (\xi )\cosh (-\alpha _{i} h_{j} \xi )+B_{i}^{j} (\xi )\sinh (-\alpha _{i} h_{j} \xi )]=0 \end{aligned}$$

(A.11)

Moreover, we consider the boundary conditions at the crack plane, i.e., Eqs. (16), (17), (19) and (49)–(51), and obtain the following relations:

$$\begin{aligned} \sum \limits _{i=1}^3 {\gamma _{1i} } [A_{i}^{\mathrm{I}} (\xi )-A_{i}^{\mathrm{I}\mathrm{I}} (\xi )]=0 \end{aligned}$$

(A.12)

$$\begin{aligned} \sum \limits _{i=1}^3 {\gamma _{2i} } [A_{i}^{\mathrm{I}} (\xi )-A_{i}^{\mathrm{I}\mathrm{I}} (\xi )]=0 \end{aligned}$$

(A.13)

$$\begin{aligned} \sum \limits _{i=1}^3 {\gamma _{3i} } [A_{i}^{\mathrm{I}} (\xi )-A_{i}^{\mathrm{I}\mathrm{I}} (\xi )]=0. \end{aligned}$$

(A.14)

It follows that

$$\begin{aligned}&\sum \limits _{i=1}^3 {[A_{i}^{\mathrm{I}} (\xi )-A_{i}^{\mathrm{I}\mathrm{I}} (\xi )]} =\frac{2}{\pi \xi }\int _0^a {\Phi _{1} (s)} \cos (\xi s)\text{ d }s \end{aligned}$$

(A.15)

$$\begin{aligned}&\sum \limits _{i=1}^3 {\eta _{3i} \alpha _{i} [B_{i}^{\mathrm{I}} (\xi )+B_{i}^{\mathrm{I}\mathrm{I}} (\xi )]} =-\frac{2}{\pi \xi }\int _0^a {\Phi _{2} (s)} \sin (\xi s)\text{ d }s \end{aligned}$$

(A.16)

$$\begin{aligned}&\sum \limits _{i=1}^3 {\eta _{4i} \alpha _{i} [B_{i}^{\mathrm{I}} (\xi )+B_{i}^{\mathrm{I}\mathrm{I}} (\xi )]} =-\frac{2}{\pi \xi }\int _0^a {\Phi _{3} (s)} \sin (\xi s)\text{ d }s \end{aligned}$$

(A.17)

where the Fourier inverse transform and the conditions of Eqs. (49)–(51) have been used. With the knowledge of Eqs. (A.10)–(A.14) and (A.15)–(A.17), the unknowns $A_{i}^{j} (\xi )$ and $B_{i}^{j} (\xi )$ can be computed as follows:

$$\begin{aligned} \Pi =[b_{ij} (\xi )]_{12\times 12} \left\{ {{\begin{array}{c} {\frac{2}{\pi \xi }\int _0^a {\Phi _{1} (s)} \cos (\xi s)\text{ d }s} \\ {-\frac{2}{\pi \xi }\int _0^a {\Phi _{2} (s)} \sin (\xi s)\text{ d }s} \\ {-\frac{2}{\pi \xi }\int _0^a {\Phi _{3} (s)} \sin (\xi s)\text{ d }s} \\ 0 \\ \vdots \\ 0 \\ \end{array} }} \right\} \end{aligned}$$

(A.18)

where

$$\begin{aligned} \Pi =\left\{ {A_{1}^{\mathrm{I}} (\xi )} \;\; {A_{2}^{\mathrm{I}} (\xi )} \;\; \cdots \;\; {A_{3}^{\mathrm{I}\mathrm{I}} (\xi )} \;\; {B_{1}^{\mathrm{I}} (\xi )} \;\; {B_{2}^{\mathrm{I}} (\xi )} \;\; \cdots \;\; {B_{3}^{\mathrm{I}\mathrm{I}} (\xi )} \right\} ^{\mathrm{T}}. \end{aligned}$$

Since it is easy to obtain the coefficient matrix $[b_{ij} (\xi )]_{12\times 12}$, the detailed expression has been omitted here.

In addition, the application of Eqs. (16), (17), (19) and (62) leads to

$$\begin{aligned}&\frac{2}{\pi }\sum \limits _{i=1}^3 {\gamma _{1i} \left[ {\int _0^a {\Phi _{1} (s)} \text{ d }s\int _0^\infty {b_{i1} (\xi )\cos (\xi s)\cos (\xi x)} \text{ d }\xi } \right. }\nonumber \\&\quad \left. -\,\sum \limits _{j=2}^3 {\int _0^a {\Phi _{j} (s)} \text{ d }s\int _0^\infty {b_{ij} (\xi )} \sin (\xi s)\cos (\xi x)\text{ d }\xi } \right] =-\sigma _{0} \end{aligned}$$

(A.19)

$$\begin{aligned}&\frac{2}{\pi }\sum \limits _{i=1}^3 {\gamma _{2i} \left[ {\int _0^a {\Phi _{1} (s)} \text{ d }s\int _0^\infty {b_{i1} (\xi )\cos (\xi s)\cos (\xi x)} \text{ d }\xi } \right. }\nonumber \\&\quad \left. -\,\sum \limits _{j=2}^3 {\int _0^a {\Phi _{j} (s)} \text{ d }s\int _0^\infty {b_{ij} (\xi )} \sin (\xi s)\cos (\xi x)\text{ d }\xi } \right] =-(D_{0} -D_{c} ) \end{aligned}$$

(A.20)

$$\begin{aligned}&\frac{2}{\pi }\sum \limits _{i=1}^3 {\gamma _{3i} \left[ {\int _0^a {\Phi _{1} (s)} \text{ d }s\int _0^\infty {b_{(i+6)1} (\xi )\cos (\xi s)\sin s(\xi x)} \text{ d }\xi } \right. }\nonumber \\&\quad \left. {-\,\sum \limits _{j=2}^3 {\int _0^a {\Phi _{j} (s)} \text{ d }s\int _0^\infty {b_{(i+6)j} (\xi )} \sin (\xi s)\cos (\xi x)\text{ d }\xi } } \right] =0 \end{aligned}$$

(A.21)

By considering Eq. (63) and the following limits

$$\begin{aligned}&\lim \limits _{\xi \rightarrow \infty } \sum \limits _{i=1}^3 {\gamma _{ki} b_{ij} (\xi )} =x_{kj} \quad (k=1,2;j=2,3) \end{aligned}$$

(A.22)

$$\begin{aligned}&\lim \limits _{\xi \rightarrow \infty } \sum \limits _{i=1}^3 {\gamma _{ki} b_{(i+6)j} (\xi )} =x_{kj} \quad (k=3;j=1), \end{aligned}$$

(A.23)

we can rewrite Eqs. (A.19)–(A.21) as Eqs. (64)–(66).

Appendix B

The constants $\alpha _{j} (j=1,2,3)(\hbox {Re}(\alpha _{j} )>0)$ are the roots of the following characteristic equation

$$\begin{aligned} \left| {{\begin{array}{ccc} {c_{11} -c_{44} \alpha ^{2}} &{} {-(c_{13} +c_{44} )\alpha } &{} {-(e_{31} +e_{15} )\alpha } \\ {-(c_{13} +c_{44} )\alpha } &{} {c_{33} \alpha ^{2}-c_{44} } &{} {e_{33} \alpha ^{2}-e_{15} } \\ {-(e_{31} +e_{15} )\alpha } &{} {e_{33} \alpha ^{2}-e_{15} } &{} {\varepsilon _{11} -\varepsilon _{33} \alpha ^{2}} \\ \end{array} }} \right| =0 \end{aligned}$$

The constants $\eta _{3j}$ and $\eta _{4j}$ can be obtained from the following relations:

$$\begin{aligned} \alpha _{j}^{2}= & {} \frac{c_{11} }{c_{44} +(c_{13} +c_{44} )\eta _{3j} +(e_{31} +e_{15} )\eta _{4j} }\\= & {} \frac{c_{13} +c_{44} +c_{44} \eta _{3j} +e_{15} \eta _{4j} }{c_{33} \eta _{3j} +e_{33} \eta _{4j} }\\= & {} \frac{e_{31} +e_{15} +e_{15} \eta _{3j} -\varepsilon _{11} \eta _{4j} }{e_{33} \eta _{3j} -\varepsilon _{33} \eta _{4j} } \end{aligned}$$

The constants $\gamma _{kj} (k=0,1,2,3,4)$ are

$$\begin{aligned} \gamma _{0j}= & {} c_{11} -(c_{33} \eta _{3j} +e_{31} \eta _{4j} )\alpha _{j}^{2}\\ \gamma _{1j}= & {} c_{13} -(c_{33} \eta _{3j} +e_{31} \eta _{4j} )\alpha _{j}^{2}\\ \gamma _{2j}= & {} e_{31} -(e_{33} \eta _{3j} -\varepsilon _{33} \eta _{4j} )\alpha _{j}^{2}\\ \gamma _{3j}= & {} -[c_{44} (1+\eta _{3j} )+e_{15} \eta _{4j} ]\alpha _{j}\\ \gamma _{4j}= & {} -[e_{15} (1-\eta _{3j} )+\varepsilon _{11} \eta _{4j} ]\alpha _{j} \end{aligned}$$

The constants $\kappa _{k} (k=0,1,2,3,4)$ are given as

$$\begin{aligned} \kappa _{0}= & {} c_{11} N_{1} -c_{12} N_{2} \lambda -e_{31} N_{3} \lambda -\beta _{11}\\ \kappa _{{1}}= & {} c_{1{3}} N_{1} -c_{{33}} N_{2} \lambda -e_{3{3}} N_{3} \lambda -\beta _{{33}}\\ \kappa _{{2}}= & {} e_{31} N_{1} -e_{{33}} N_{2} \lambda +\varepsilon _{3{3}} N_{3} \lambda +p_{z}\\ \kappa _{{3}}= & {} -(c_{44} N_{1} \lambda +c_{44} N_{2} +e_{15} N_{3} )\\ \kappa _{{4}}= & {} -(e_{15} N_{1} \lambda +e_{15} N_{2} -\varepsilon _{11} N_{3} ) \end{aligned}$$

The functions $k_{ik} ({\overline{s}},{\overline{x}} )(i=1,2,3,k=1,2,3)$ can be calculated as

$$\begin{aligned} k_{11} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\sum \limits _{i=1}^3 {\gamma _{1i} b_{i1} (\xi )\cos (a{\overline{s}} \xi )} } \cos (a{\overline{x}} \xi )\text{ d }\xi \\ k_{12} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{1i} b_{i2} (\xi )} -x_{12} } \right] } \sin (a{\overline{s}} \xi )\cos (a{\overline{x}} \xi )\text{ d }\xi \\ k_{13} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{1i} b_{i3} (\xi )} -x_{13} } \right] } \sin (a{\overline{s}} \xi )\cos (a{\overline{x}} \xi )\text{ d }\xi \\ k_{21} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\sum \limits _{i=1}^3 {\gamma _{2i} b_{i1} (\xi )\cos (a{\overline{s}} \xi )} } \cos (a{\overline{x}} \xi )\text{ d }\xi \\ k_{22} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{2i} b_{i2} (\xi )} -x_{22} } \right] } \sin (a{\overline{s}} \xi )\cos (a{\overline{x}} \xi )\text{ d }\xi \\ k_{23} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{2i} b_{i3} (\xi )} -x_{23} } \right] } \sin (a{\overline{s}} \xi )\cos (a{\overline{x}} \xi )\text{ d }\xi \\ k_{31} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{3i} b_{(i+6)1} (\xi )} -x_{31} } \right] } \cos (a{\overline{s}} \xi )\sin (a{\overline{x}} \xi )\text{ d }\xi \\ k_{32} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\sum \limits _{i=1}^3 {\gamma _{3i} b_{(i+6)2} (\xi )\sin (a{\overline{s}} \xi )} } \sin (a\overline{x} \xi )\text{ d }\xi \\ k_{33} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\sum \limits _{i=1}^3 {\gamma _{3i} b_{(i+6)3} (\xi )\sin (a{\overline{s}} \xi )} } \sin (a\overline{x} \xi )\text{ d }\xi \end{aligned}$$

The functions ${\overline{k}}_{ik} ({\overline{s}},{\overline{x}} )(i=1,2,3,k=1,2,3,4)$ are given as follows :

$$\begin{aligned} {\overline{k}}_{11} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\sum \limits _{i=1}^3 {\gamma _{1i} {\overline{b}}_{i1} (\xi )\cos (a\overline{s} \xi )} } \cos (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{12} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{1i} {\overline{b}}_{i2} (\xi )} -x_{12} } \right] } \sin (a{\overline{s}} \xi )\cos (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{13} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{1i} {\overline{b}}_{i3} (\xi )} -x_{13} } \right] } \sin (a{\overline{s}} \xi )\cos (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{21} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\sum \limits _{i=1}^3 {\gamma _{2i} {\overline{b}}_{i1} (\xi )\cos (a\overline{s} \xi )} } \cos (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{22} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{2i} {\overline{b}}_{i2} (\xi )} -x_{22} } \right] } \sin (a{\overline{s}} \xi )\cos (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{23} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{2i} {\overline{b}}_{i3} (\xi )} -x_{23} } \right] } \sin (a{\overline{s}} \xi )\cos (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{31} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\left[ {\sum \limits _{i=1}^3 {\gamma _{3i} {\overline{b}}_{(i+6)1} (\xi )} -x_{31} } \right] } \cos (a{\overline{s}} \xi )\sin (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{32} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\sum \limits _{i=1}^3 {\gamma _{3i} {\overline{b}}_{(i+6)2} (\xi )\sin (a{\overline{s}} \xi )} } \sin (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{33} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {\sum \limits _{i=1}^3 {\gamma _{3i} {\overline{b}}_{(i+6)3} (\xi )\sin (a{\overline{s}} \xi )} } \sin (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{{14}} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {{\overline{b}}_{{1}} (\xi )\sin (a{\overline{s}} \xi )} \cos (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{{24}} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {{\overline{b}}_{{2}} (\xi )\sin (a{\overline{s}} \xi )} \cos (a{\overline{x}} \xi )\text{ d }\xi \\ {\overline{k}}_{{34}} ({\overline{s}},{\overline{x}} )= & {} a\int _0^\infty {{\overline{b}}_{{3}} (\xi )\sin (a{\overline{s}} \xi )} \sin (a{\overline{x}} \xi )\text{ d }\xi \end{aligned}$$