Line Defects in One-Dimensional Hexagonal Quasicrystals

Abstract

Using the eight-dimensional framework of the integral formalism of one-dimensional quasicrystals, the analytical expressions for the displacement fields and stress functions of line defects, which are dislocations and line forces, in one-dimensional hexagonal quasicrystals of Laue class 10 are derived. The self-energy of a straight dislocation, the self-energy of a line force, the Peach–Koehler force between two straight dislocations, and the Cherepanov force between two straight line forces in one-dimensional hexagonal quasicrystals of Laue class 10 are calculated. In addition, the two-dimensional Green tensor of one-dimensional hexagonal quasicrystals of Laue class 10 is given within the framework of the integral formalism.

1. Introduction

Quasicrystals represent an interesting class of novel materials discovered by Shechtman et al. [1]. Quasicrystals are aperiodic crystals and possess long-range orientational order but no translational symmetry in quasiperiodic directions. Quasicrystals have unusual physical properties such as high hardness, low conductivity, resistivity that decreases with temperature, very low conductivity, small specific heat, low friction coefficients, and wear and oxidation resistance [2,3,4]. The basis of the continuum theory of solid quasicrystals consists of two elementary excitations, namely the phonons and the phasons [5,6]. In the atomistic picture, phonons are related to the translation of atoms, and phasons lead to local rearrangements of atoms in a cell. The generalized elasticity theory of quasicrystals including phonon and phason fields was developed by Ding et al. [7].

Dislocations are important line defects in quasicrystals that cause plasticity and influence the physical properties of quasicrystals (see, e.g., [6,8,9,10,11,12,13,14]). Dislocations have been observed in quasicrystals in many experimental studies (see, e.g., [15,16,17,18,19,20,21]). To understand the effects of dislocations on the physical properties of quasicrystals and to simulate the electron micrographs of dislocations in quasicrystals, the elastic fields of dislocations in quasicrystals are necessary. Therefore, the determination of the elastic fields of dislocations in quasicrystals is an important task in theory and experiment. For quasicrystals, the basic key equations of dislocations were given by Ding et al. [22,23] and Lazar and Agiasofitou [24]. The general expressions for the displacement fields induced by straight dislocations in quasicrystals in terms of the elastic Green tensor were given by Ding et al. [22,23]. Ding et al. [25] extended the Stroh formalism [26,27] and the integral formalism [28,29] for the displacement field of a dislocation in quasicrystals (see also [14,30]). Ding et al. [25] derived the integral formalism from the Stroh formalism for quasicrystals using the eigenvectors and eigenvalues of the Stroh formalism. Using the extended Stroh formalism, Li and Liu [31] studied a dislocation in an icosahedral quasicrystal and Radi and Mariano [32] investigated the steady-state propagation of dislocations in quasicrystals. Wang and Schiavone [33] have given the detailed structure of the extended Stroh formalism for quasicrystals. In the Stroh formalism, the fields of line defects are given as complex form solutions of an eigenvalue problem and in the integral formalism, the fields of line defects are given as real-form solutions of a matrix partial differential equation of the first order (see [34]).

Lazar [35] recently derived the extended integral formalism of line defects (straight dislocations and line forces) for the displacement fields and stress functions for quasicrystals in a straightforward manner without an unnecessary detour via the Stroh formalism. Lazar [35] extended the six-dimensional framework of the integral formalism for line defects in anisotropic elasticity towards a $2n$-dimensional integral formalism for line defects in quasicrystals ($n=4,5,6$ for one-, two-, three-dimensional quasicrystals). For generalized plane strain in quasicrystals, the appropriate two-dimensional matrix partial differential equation of the first order and its solution of the integral formalism have been given in [35]. For line defects in quasicrystals, the solution gives the $2n$-dimensional vector of the n-dimensional displacement vector $\mathit{U}$ and the n-dimensional stress function vector $\mathbf{\Phi }$ of a straight dislocation with Burgers vector $\mathit{b}$ and a line force with strength $\mathit{F}$. The integral formalism provides suitable expressions for both analytical and numerical modelling of line defects in quasicrystals.

Explicit analytical solutions require analytical expressions for the integrals in the integral formalism. In the general case, only numerical solutions are possible using numerical integration. For quasicrystals with high symmetry and symmetrical orientation of the defect line, the complexity is reduced and analytical solutions become possible because the integrals in the integral formalism can be computed analytically. In the case of lower symmetry of quasicrystals and of the orientation of the line defect, numerical treatment is required in the integral formalism. The successful derivation of analytic solutions is of particular value because it confirms and manifests the theoretical prediction. The form of the analytic solutions shows their functional dependence on all relevant parameters. Analytic solutions are always useful as benchmark solutions with which to verify numerical results, and they are useful for degenerate cases (special orientations or ratios of elastic constants) where the numerics may diverge [36].

The most suitable type of quasicrystals for testing the integral formalism are one-dimensional hexagonal quasicrystals of Laue class 10. The Laue class 10 is the Laue class with the highest symmetry for one-dimensional quasicrystals. A one-dimensional quasicrystal is defined as a three-dimensional body which is periodic in the $x_1{x}_2$-plane and quasiperiodic in the $x_3$-direction. One-dimensional quasicrystals are quasicrystals with crystallographic symmetries, namely 31 point groups and 10 Laue classes [13,30,37].

In this work, analytical solutions of line defects in one-dimensional quasicrystals are derived for the first time using the integral formalism. In particular, the closed-form expressions for the displacement fields and the stress functions of line defects in one-dimensional hexagonal quasicrystals of Laue class 10, where the defect line is parallel to the aperiodic direction, are the aim of this work. Within the framework of the integral formalism, the self-energies of a straight dislocation and a line force, the Peach–Koehler force between two straight dislocations, and the Cherepanov force between two line forces in one-dimensional hexagonal quasicrystals of Laue class 10 are calculated. In addition, the two-dimensional Green tensor of one-dimensional hexagonal quasicrystals of Laue class 10 is derived within the framework of the integral formalism.

2. Generalized Elasticity of One-Dimensional Quasicrystals

We consider one-dimensional quasicrystals which are periodic in the $x_1{x}_2$-plane and quasiperiodic in the $x_3$-direction. A one-dimensional quasicrystal can be obtained by projecting a four-dimensional periodic structure onto the three-dimensional physical space. The four-dimensional hyperspace $E^4$ can be decomposed into the direct sum of two orthogonal subspaces:

$$E^4=E_{\parallel }^3\oplus E_{\perp }^1,$$

(1)

where $E_{\parallel }^3$ is the three-dimensional physical or parallel space of the phonon fields and $E_{\perp }^1$ is the one-dimensional perpendicular space of the phason field with quasiperiodicity in the $x_3$-direction. Throughout the text, phonon fields will be denoted ${(·)}^{\parallel }$ and phason fields ${(·)}^{\perp }$. Note that all quantities (phonon and phason fields) depend on the so-called material space coordinates $\mathit{x}\in {\mathbb{R}}^3$. Indices in the parallel space are denoted by small Latin letters $i,j,k,l$ with $i,j,k,l=1,2,3$.

In the theory of generalized elasticity of one-dimensional quasicrystals, the (elastic) phonon and phason distortion tensors, ${\beta }_{kl}^{\parallel }$ and ${\beta }_{3l}^{\perp }$, are defined as the gradients of the phonon displacement vector $u_k^{\parallel }\in E_{\parallel }^3$ and the phason displacement field $u_3^{\perp }\in E_{\perp }^1$, respectively,

$${\beta }_{kl}^{\parallel }={\partial }_l{u}_k^{\parallel },$$

(2)

$${\beta }_{3l}^{\perp }={\partial }_l{u}_3^{\perp },$$

(3)

where ${\partial }_l=\partial /\partial x_l$ denotes the partial differentiation with respect to $x_l$. Therefore, the phason displacement vector has only one component $u_3^{\perp }$. The constitutive relations for a one-dimensional quasicrystal are given by

$${\sigma }_{ij}^{\parallel }=C_{ijkl}{\beta }_{kl}^{\parallel }+D_{ij3l}{\beta }_{3l}^{\perp },$$

(4)

$${\sigma }_{3j}^{\perp }=D_{kl3j}{\beta }_{kl}^{\parallel }+E_{3j3l}{\beta }_{3l}^{\perp },$$

(5)

where ${\sigma }_{ij}^{\parallel }$ and ${\sigma }_{3j}^{\perp }$ are the phonon and phason stress tensors, respectively. Note that the phonon stress tensor is symmetric, ${\sigma }_{ij}^{\parallel }={\sigma }_{ji}^{\parallel }$, whereas ${\sigma }_{3j}^{\perp }\ne {\sigma }_{j3}^{\perp }$ (see [24,38]). $C_{ijkl}$ is the tensor of the elastic moduli of phonons, $E_{3j3l}$ is the tensor of the elastic moduli of phasons, and $D_{ij3l}$ is the tensor of the elastic moduli of the phonon–phason coupling. The constitutive tensors possess the following symmetries [7,24]:

$$\begin{array}{c}\hfill C_{ijkl}=C_{klij}=C_{ijlk}=C_{jikl},D_{ij3l}=D_{ji3l},E_{3j3l}=E_{3l3j}.\end{array}$$

(6)

Using the hyperspace notation of quasicrystals given by Lazar and Agiasofitou [24], the phonon and phason fields can be unified in the corresponding extended fields in the hyperspace. The components of the extended fields will be denoted by capital letters, e.g., $I,K=1,2,3,4$. In the hyperspace notation, we have the extended displacement vector $U_K\in E_{\parallel }^3\oplus E_{\perp }^1$:

$$\begin{array}{c}\hfill U_K=\left\{\begin{array}{cc}{\displaystyle u_k^{\parallel },}\hfill & {\displaystyle K=1,2,3,}\hfill \\ {\displaystyle u_3^{\perp },}\hfill & {\displaystyle K=4,}\hfill \end{array}\right.\end{array}$$

(7)

the extended elastic distortion tensor $B_{Kl}\in \left(E_{\parallel }^3\oplus E_{\perp }^1\right)\otimes E_{\parallel }^3$:

$$\begin{array}{c}\hfill B_{Kl}=\left\{\begin{array}{cc}{\displaystyle {\beta }_{kl}^{\parallel },}\hfill & {\displaystyle K=1,2,3,}\hfill \\ {\displaystyle {\beta }_{3l}^{\perp },}\hfill & {\displaystyle K=4,}\hfill \end{array}\right.\end{array}$$

(8)

the extended stress tensor ${\mathrm{\Sigma }}_{Ij}\in \left(E_{\parallel }^3\oplus E_{\perp }^1\right)\otimes E_{\parallel }^3$:

$$\begin{array}{c}\hfill {\mathrm{\Sigma }}_{Ij}=\left\{\begin{array}{cc}{\displaystyle {\sigma }_{ij}^{\parallel },}\hfill & {\displaystyle I=1,2,3,}\hfill \\ {\displaystyle {\sigma }_{3j}^{\perp },}\hfill & {\displaystyle I=4,}\hfill \end{array}\right.\end{array}$$

(9)

and the tensor of the extended elastic moduli $C_{IjKl}\in \left(E_{\parallel }^3\oplus E_{\perp }^1\right)\otimes E_{\parallel }^3\otimes \left(E_{\parallel }^3\oplus E_{\perp }^1\right)\otimes E_{\parallel }^3$:

$$\begin{array}{c}\hfill C_{IjKl}=\left\{\begin{array}{ccc}{\displaystyle C_{ijkl},}\hfill & {\displaystyle I=1,2,3;}\hfill & {\displaystyle K=1,2,3,}\hfill \\ {\displaystyle D_{ij3l},}\hfill & {\displaystyle I=1,2,3;}\hfill & {\displaystyle K=4,}\hfill \\ {\displaystyle D_{kl3j},}\hfill & {\displaystyle I=4;}\hfill & {\displaystyle K=1,2,3,}\hfill \\ {\displaystyle E_{3j3l},}\hfill & {\displaystyle I=4;}\hfill & {\displaystyle K=4.}\hfill \end{array}\right.\end{array}$$

(10)

The tensor $C_{IjKl}$ retains the major symmetry:

$$\begin{array}{c}\hfill C_{IjKl}=C_{KlIj}.\end{array}$$

(11)

In matrix form, Equation (10) can be written as

$$\begin{array}{c}\hfill C_{IjKl}=\left(\begin{array}{cc}{C}_{ijkl}& D_{ij3l}\\ D_{kl3j}& E_{3j3l}\end{array}\right).\end{array}$$

(12)

Using the hyperspace notation, Equations (2) and (3) reduce to

$$\begin{array}{c}\hfill B_{Kl}={\partial }_l{U}_K\end{array}$$

(13)

and the constitutive relations (4) and (5) become

$$\begin{array}{c}\hfill {\mathrm{\Sigma }}_{Ij}=C_{IjKl}{B}_{Kl}=C_{IjKl}{\partial }_l{U}_K,\end{array}$$

(14)

which is the extended Hooke law for quasicrystals.

In the absence of external forces, the extended stress tensor (9) can be written in terms of an extended stress function tensor ${\mathsf{\Phi }}_{Im}\in \left(E_{\parallel }^3\oplus E_{\perp }^1\right)\otimes E_{\parallel }^3$:

$$\begin{array}{c}\hfill {\mathrm{\Sigma }}_{Ij}={ϵ}_{jkm}{\partial }_k{\mathsf{\Phi }}_{Im}\end{array}$$

(15)

with

$$\begin{array}{c}\hfill {\mathsf{\Phi }}_{Im}=\left\{\begin{array}{cc}{\displaystyle {\mathsf{\Phi }}_{im}^{\parallel },}\hfill & {\displaystyle I=1,2,3,}\hfill \\ {\displaystyle {\mathsf{\Phi }}_{3m}^{\perp },}\hfill & {\displaystyle I=4,}\hfill \end{array}\right.\end{array}$$

(16)

where ${\mathsf{\Phi }}_{im}^{\parallel }$ and ${\mathsf{\Phi }}_{3m}^{\perp }$ are the phonon and phason stress function tensors, respectively. Here, ${ϵ}_{jkm}$ denotes the three-dimensional Levi–Civita tensor. Substituting Equation (15) into the left hand side of the extended Hooke law (14), we obtain

$$\begin{array}{c}\hfill {ϵ}_{jkm}{\partial }_k{\mathsf{\Phi }}_{Im}=C_{IjKl}{\partial }_l{U}_K,\end{array}$$

(17)

which is nothing but the extended Hooke law (14) written in terms of $U_K$ and ${\mathsf{\Phi }}_{Im}$.

3. Integral Formalism for One-Dimensional Quasicrystals

The integral formalism for one-dimensional quasicrystals is based on the framework of generalized plane strain of one-dimensional quasicrystals. In generalized plane strain, all fields are independent of the variable $x_3$. The extended displacement fields depend only on $x_1$ and $x_2$, and $\mathit{x}\in {\mathbb{R}}^2$, but with index in the hyperspace $U_I(x_1,x_2)$, $I=1,2,3,4$.

We consider a Cartesian coordinate system in the parallel space. Using a unit vector $\mathit{m}$ in the $x_1$-direction and a unit vector $\mathit{n}$ in the $x_2$-direction ($\mathit{m}\equiv {\mathit{e}}_1$, $\mathit{n}\equiv {\mathit{e}}_2$) and the notation

$$\begin{array}{c}\hfill {\left(ab\right)}_{IK}=a_j{C}_{IjKl}{b}_l,\end{array}$$

(18)

which is a $4\times 4$ matrix in the hyperspace, it follows from the major symmetry (11) of the extended constitutive tensor $C_{IjKl}$ that

$$\begin{array}{c}\hfill {\left(ab\right)}_{IK}={\left(ba\right)}_{KI}.\end{array}$$

(19)

The extended constitutive relation (14) reads for generalized plane strain:

$$\begin{array}{cc}\hfill {\mathrm{\Sigma }}_{I1}& ={\left(mm\right)}_{IK}{B}_{K1}+{\left(mn\right)}_{IK}{B}_{K2}\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill {\mathrm{\Sigma }}_{I2}& ={\left(nm\right)}_{IK}{B}_{K1}+{\left(nn\right)}_{IK}{B}_{K2}\hfill \end{array}$$

(21)

with $I,K=1,2,3,4$. For a one-dimensional quasicrystal, the characteristic $4\times 4$ blocks of the constitutive tensor $C_{IjKl}$ entering the extended constitutive relations (20) and (21) are given by

$$\begin{array}{c}\hfill {\left(mm\right)}_{IK}=\left(\begin{array}{cccc}{C}_{1111}& C_{1121}& C_{1131}& D_{1131}\\ C_{2111}& C_{2121}& C_{2131}& D_{2131}\\ C_{3111}& C_{3121}& C_{3131}& D_{3131}\\ D_{1131}& D_{2131}& D_{3131}& E_{3131}\end{array}\right)=C_{I1K1}\end{array}$$

(22)

$$\begin{array}{c}\hfill {\left(mn\right)}_{IK}=\left(\begin{array}{cccc}{C}_{1112}& C_{1122}& C_{1132}& D_{1132}\\ C_{2112}& C_{2122}& C_{2132}& D_{2132}\\ C_{3112}& C_{3122}& C_{3132}& D_{3132}\\ D_{1231}& D_{2231}& D_{3231}& E_{3132}\end{array}\right)=C_{I1K2}\end{array}$$

(23)

$$\begin{array}{c}\hfill {\left(nn\right)}_{IK}=\left(\begin{array}{cccc}{C}_{1212}& C_{1222}& C_{1232}& D_{1232}\\ C_{2212}& C_{2222}& C_{2232}& D_{2232}\\ C_{3212}& C_{3222}& C_{3232}& D_{3232}\\ D_{1232}& D_{2232}& D_{3232}& E_{3232}\end{array}\right)=C_{I2K2}.\end{array}$$

(24)

Due to the major symmetry of $C_{IjKl}$, ${\left(mm\right)}_{IK}$ and ${\left(nn\right)}_{IK}$ are symmetric.

For generalized plane strain, the extended stress tensor can be written in terms of an extended stress function vector ${\mathsf{\Phi }}_{I3}\equiv {\mathsf{\Phi }}_I$, and Equation (15) reduces to

$${\mathrm{\Sigma }}_{I1}={\partial }_2{\mathsf{\Phi }}_I$$

(25)

$${\mathrm{\Sigma }}_{I2}=-{\partial }_1{\mathsf{\Phi }}_I.$$

(26)

Substituting Equations (13), (25) and (26) into the extended Hooke law (20) and (21), we obtain

$${\partial }_2{\mathsf{\Phi }}_I={\left(mm\right)}_{IK}{\partial }_1{U}_K+{\left(mn\right)}_{IK}{\partial }_2{U}_K$$

(27)

$$-{\partial }_1{\mathsf{\Phi }}_I={\left(nm\right)}_{IK}{\partial }_1{U}_K+{\left(nn\right)}_{IK}{\partial }_2{U}_K.$$

(28)

Equations (27) and (28), which have the meaning of the extended Hooke law (14) written in terms of $U_K$ and ${\mathsf{\Phi }}_I$ for generalized plane strain, are the two components of Equation (17) for generalized plane strain with $j=1,2$.

After rearrangement of Equations (27) and (28), we obtain the following matrix partial differential equation of the first order

$$\left[\mathit{I}{\partial }_2-\mathit{N}{\partial }_1\right]\left(\begin{array}{c}\mathit{U}(x_1,x_2)\\ \mathbf{\Phi }(x_1,x_2)\end{array}\right)=\left(\begin{array}{c}\mathbf{0}\\ \mathbf{0}\end{array}\right).$$

(29)

This is the matrix differential equation for generalized plane strain of one-dimensional quasicrystals. We have introduced the 8-vector of the extended displacement vector and the extended stress function vector for generalized plane strain:

$$\left(\begin{array}{c}\mathit{U}\\ \mathbf{\Phi }\end{array}\right)\equiv \left(\begin{array}{c}{U}_I\\ {\mathsf{\Phi }}_I\end{array}\right),I=1,2,3,4$$

(30)

and the $8\times 8$ real matrix is defined by its $4\times 4$ blocks

$$\begin{array}{c}\hfill \mathit{N}=-\left(\begin{array}{cc}{\left(nn\right)}^{-1}\left(nm\right)& {\left(nn\right)}^{-1}\\ \left(mn\right){\left(nn\right)}^{-1}\left(nm\right)-\left(mm\right)& \left(mn\right){\left(nn\right)}^{-1}\end{array}\right).\end{array}$$

(31)

Here, $\mathit{I}$ is the $8\times 8$ identity matrix. The $8\times 8$ matrix $\mathit{N}$ is the fundamental elasticity matrix for one-dimensional quasicrystals depending on the elastic constants of quasicrystals.

Now, we choose two orthogonal unit vectors $\mathit{m}$ and $\mathit{n}$, which are orthogonal to $\mathit{t}$ such that $(\mathit{m},\mathit{n},\mathit{t})$ forms a right-handed Cartesian basis in $E_{\parallel }^3$. This basis is rotated around $\mathit{t}$ by an angle $\varphi$ against another fixed $({\mathit{m}}_0,{\mathit{n}}_0,\mathit{t})$ basis in $E_{\parallel }^3$, such that

$$\left(\begin{array}{c}\mathit{m}\left(\varphi \right)\\ \mathit{n}\left(\varphi \right)\\ \mathit{t}\end{array}\right)=\left(\begin{array}{ccc}cos\varphi & sin\varphi & 0\\ -sin\varphi & cos\varphi & 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}{\mathit{m}}_0\\ {\mathit{n}}_0\\ \mathit{t}\end{array}\right)$$

(32)

as shown in Figure 1. It yields $\mathit{m}\equiv \mathit{m}\left(\varphi \right)$ and $\mathit{n}\equiv \mathit{n}\left(\varphi \right)$. Only the independent variable $x_j$ is transformed, but not the dependent variables $U_I$ and ${\mathsf{\Phi }}_I$. Using the rotation given in Equation (32) and polar coordinates $(r,\varphi )$, the matrix partial differential Equation (29) becomes

$$\left[\mathit{I}{\displaystyle \frac{1}{r}}{\partial }_{\varphi }-\mathit{N}\left(\varphi \right){\partial }_r\right]\left(\begin{array}{c}\mathit{U}(r,\varphi )\\ \mathbf{\Phi }(r,\varphi )\end{array}\right)=\left(\begin{array}{c}\mathbf{0}\\ \mathbf{0}\end{array}\right).$$

(33)

Here, the $8\times 8$ matrix $\mathit{N}\left(\varphi \right)$ is defined by contraction of the elastic constants with orthogonal unit vectors $\mathit{m}$ and $\mathit{n}$ according to Equation (18). As mentioned above, the vectors $\mathit{m}\left(\varphi \right)$ and $\mathit{n}\left(\varphi \right)$ are turned against the ${\mathit{m}}_0$, ${\mathit{n}}_0$ coordinate system by an angle $\varphi$ (see Figure 1), so that $\mathit{N}$ depends on the angle $\varphi$

$$\begin{array}{c}\hfill \mathit{N}\left(\varphi \right)=-\left(\begin{array}{cc}{\left(nn\right)}^{-1}\left(nm\right)& {\left(nn\right)}^{-1}\\ \left(mn\right){\left(nn\right)}^{-1}\left(nm\right)-\left(mm\right)& \left(mn\right){\left(nn\right)}^{-1}\end{array}\right).\end{array}$$

(34)

4. Line Defects in One-Dimensional Quasicrystals

Here, we consider a line defect in a one-dimensional quasicrystal, namely a straight dislocation with extended Burgers vector $\mathit{b}$ and a line force with extended strength $\mathit{F}$ in a one-dimensional quasicrystal located at the origin of the coordinate system. The defect line runs along the axis $\mathit{t}$, which is parallel to the quasiperiodic direction. The fields of the straight dislocation and line force are the extended displacement vector $\mathit{U}$ and the extended stress function vector $\mathbf{\Phi }$. In the hyperspace notation, the extended Burgers vector of a straight dislocation in a one-dimensional quasicrystal is given by

$$\begin{array}{c}\hfill b_I=\left\{\begin{array}{cc}{\displaystyle b_i^{\parallel },}\hfill & {\displaystyle I=1,2,3,}\hfill \\ {\displaystyle b_3^{\perp },}\hfill & {\displaystyle I=4,}\hfill \end{array}\right.\end{array}$$

(35)

where $b_i^{\parallel }$ is the phonon component and $b_3^{\perp }$ is the phason component of the extended Burgers vector, $b_I\in E_{\parallel }^3\oplus E_{\perp }^1$, and the extended body force vector of a line force in a one-dimensional quasicrystal reads

$$\begin{array}{c}\hfill F_I=\left\{\begin{array}{cc}{\displaystyle F_i^{\parallel },}\hfill & {\displaystyle I=1,2,3,}\hfill \\ {\displaystyle F_3^{\perp },}\hfill & {\displaystyle I=4,}\hfill \end{array}\right.\end{array}$$

(36)

where $F_i^{\parallel }$ is the phonon component and $F_3^{\perp }$ is the phason component of the extended strength $F_I\in E_{\parallel }^3\oplus E_{\perp }^1$.

For one-dimensional quasicrystals, the solution of Equation (33) for a straight dislocation with extended Burgers vector $\mathit{b}$ and a line force with extended strength $\mathit{F}$ reads (see [35])

$$\left(\begin{array}{c}\mathit{U}(r,\varphi )\\ \mathbf{\Phi }(r,\varphi )\end{array}\right)=-{\displaystyle \frac{1}{{\left(2\pi \right)}^2}}\left[\mathit{I}lnr+{\int }_0^{\varphi }\mathit{N}\left(\omega \right)\mathrm{d}\omega \right]{\int }_0^{2\pi }\mathit{N}\left(\omega \right)\mathrm{d}\omega \left(\begin{array}{c}\mathit{b}\\ -\mathit{F}\end{array}\right).$$

(37)

Equation (37) is the 8-vector of the solution of the extended displacement vector $\mathit{U}$ and the extended stress function vector $\mathbf{\Phi }$ for a straight dislocation with extended Burgers vector $\mathit{b}$ and a line force with extended strength $\mathit{F}$ in a one-dimensional quasicrystal. Using

$$\begin{array}{c}\hfill \tilde{\mathit{N}}\left(\varphi \right)={\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }\mathit{N}\left(\omega \right)\mathrm{d}\omega \end{array}$$

(38)

and $\tilde{\mathit{N}}\left(2\pi \right)=\tilde{\mathit{N}}(\varphi =2\pi )$, the solution (37) can be written in compact form as (see also [35,39])

$$\left(\begin{array}{c}\mathit{U}(r,\varphi )\\ \mathbf{\Phi }(r,\varphi )\end{array}\right)=-\left[{\displaystyle \frac{1}{2\pi }}\mathit{I}lnr+\tilde{\mathit{N}}\left(\varphi \right)\right]\tilde{\mathit{N}}\left(2\pi \right)\left(\begin{array}{c}\mathit{b}\\ -\mathit{F}\end{array}\right).$$

(39)

The $4\times 4$ matrices $S_{IK}$, $Q_{IK}$ and $B_{IK}$, which are tensors of rank two in the hyperspace, are the $4\times 4$ blocks of the $8\times 8$ matrix $\tilde{\mathit{N}}$, Equation (38), with the block structure (34)

$$\begin{array}{c}\hfill \tilde{\mathit{N}}\left(\varphi \right)=\left(\begin{array}{cc}\mathit{S}\left(\varphi \right)& \mathit{Q}\left(\varphi \right)\\ \mathit{B}\left(\varphi \right)& {\mathit{S}}^{\mathrm{T}}\left(\varphi \right)\end{array}\right)\end{array}$$

(40)

and

$$\mathit{Q}\left(\varphi \right)=-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }{\left(nn\right)}^{-1}\mathrm{d}\omega ,$$

(41)

$$\mathit{S}\left(\varphi \right)=-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }{\left(nn\right)}^{-1}\left(nm\right)\mathrm{d}\omega ,$$

(42)

$${\mathit{S}}^{\mathrm{T}}\left(\varphi \right)=-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }\left(mn\right){\left(nn\right)}^{-1}\mathrm{d}\omega ,$$

(43)

$$\mathit{B}\left(\varphi \right)=-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }\left[\left(mn\right){\left(nn\right)}^{-1}\left(nm\right)-\left(mm\right)\right]\mathrm{d}\omega ,$$

(44)

where $\mathit{S}$, $\mathit{Q}$, $\mathit{B}$ and ${\mathit{S}}^{\mathrm{T}}$ (the transpose of $\mathit{S}$) are $4\times 4$ matrices resulting from the four blocks in $\mathit{N}$ (see Equations (34) and (38)).

Equation (39) can be decomposed into its four pieces, which are four vectors in the four-dimensional hyperspace (see [35]):

• The extended displacement vector of a straight dislocation with extended Burgers vector $b_I$:

  $$U_K(r,\varphi )=-b_I\left[{\displaystyle \frac{1}{2\pi }}{S}_{KI}\left(2\pi \right)lnr+Q_{KM}\left(\varphi \right)B_{MI}\left(2\pi \right)+S_{KM}\left(\varphi \right)S_{MI}\left(2\pi \right)\right].$$

  (45)
• The extended displacement vector of a line force with extended strength $F_I$:

  $$U_K(r,\varphi )=F_I\left[{\displaystyle \frac{1}{2\pi }}{Q}_{KI}\left(2\pi \right)lnr+S_{KM}\left(\varphi \right)Q_{MI}\left(2\pi \right)+Q_{KM}\left(\varphi \right)S_{MI}^{\mathrm{T}}\left(2\pi \right)\right].$$

  (46)
• The extended stress function vector of a straight dislocation with extended Burgers vector $b_K$:

  $${\mathsf{\Phi }}_I(r,\varphi )=-b_K\left[{\displaystyle \frac{1}{2\pi }}{B}_{IK}\left(2\pi \right)lnr+B_{IM}\left(\varphi \right)S_{MK}\left(2\pi \right)+S_{IM}^{\mathrm{T}}\left(\varphi \right)B_{MK}\left(2\pi \right)\right].$$

  (47)
• The extended stress function vector of a line force with extended strength $F_K$:

  $${\mathsf{\Phi }}_I(r,\varphi )=F_K\left[{\displaystyle \frac{1}{2\pi }}{S}_{IK}^{\mathrm{T}}\left(2\pi \right)lnr+B_{IM}\left(\varphi \right)Q_{MK}\left(2\pi \right)+S_{IM}^{\mathrm{T}}\left(\varphi \right)S_{MK}^{\mathrm{T}}\left(2\pi \right)\right].$$

  (48)

Using the extended stress function vector of a straight dislocation (47) for $r=r_0/R$ and $\varphi =0$, the elastic self-energy of a straight dislocation per unit length reads [35]

$$\begin{array}{cc}\hfill \mathcal{E}& ={\displaystyle \frac{1}{2}}{b}_I{\mathsf{\Phi }}_I(r_0/R,0)\hfill \\ \hfill & ={\displaystyle \frac{1}{4\pi }}{b}_I{B}_{IK}\left(2\pi \right)b_{K}ln(R/r_0),\hfill \end{array}$$

(49)

where $r_0$ and R are the inner and outer cutoff radii. Using the extended displacement vector of a line force (48) for $r=r_0/R$ and $\varphi =0$, the elastic self-energy of a line force per unit length reads [35]

$$\begin{array}{cc}\hfill \mathcal{E}& ={\displaystyle \frac{1}{2}}{F}_K{U}_K(r_0/R,0)\hfill \\ \hfill & =-{\displaystyle \frac{1}{4\pi }}{F}_I{Q}_{IK}\left(2\pi \right)F_{K}ln(R/r_0).\hfill \end{array}$$

(50)

The Peach–Koehler force per unit length between a straight dislocation with Burgers vector $b_I^{\left(A\right)}$ at position $(r,\varphi )$ in the stress field ${\mathrm{\Sigma }}_{Ij}^{\left(B\right)}$ produced by another dislocation with Burgers vector $b_K^{\left(B\right)}$ located at the position $(0,0)$ is given by (see [35])

$$\begin{array}{cc}\hfill {\mathcal{F}}_l^{\mathrm{PK}\left(AB\right)}& ={ϵ}_{lj3}{\displaystyle \frac{b_I^{\left(A\right)}{b}_K^{\left(B\right)}}{2\pi r}}[n_j{B}_{IK}\left(2\pi \right)+m_j{\left[\left(mn\right){\left(nn\right)}^{-1}\left(nm\right)-\left(mm\right)\right]}_{IM}{S}_{MK}\left(2\pi \right)\hfill \\ \hfill & +m_j{\left[\left(mn\right){\left(nn\right)}^{-1}\right]}_{IM}{B}_{MK}\left(2\pi \right)].\hfill \end{array}$$

(51)

In polar coordinates, the two non-vanishing components of the Peach–Koehler force (51) read

$${\mathcal{F}}_r^{\mathrm{PK}\left(AB\right)}={\displaystyle \frac{b_I^{\left(A\right)}{b}_K^{\left(B\right)}}{2\pi r}}{B}_{IK}\left(2\pi \right),$$

(52)

$$\begin{array}{cc}\hfill {\mathcal{F}}_{\varphi }^{\mathrm{PK}\left(AB\right)}& =-{\displaystyle \frac{b_I^{\left(A\right)}{b}_K^{\left(B\right)}}{2\pi r}}[{\left[\left(mn\right){\left(nn\right)}^{-1}\left(nm\right)-\left(mm\right)\right]}_{IM}{S}_{MK}\left(2\pi \right)\hfill \\ \hfill & +{\left[\left(mn\right){\left(nn\right)}^{-1}\right]}_{IM}{B}_{MK}\left(2\pi \right)].\hfill \end{array}$$

(53)

The radial component (52) is connected with the matrix $\mathit{B}$, and the tangential component (53) is connected with the matrices $\mathit{S}$ and $\mathit{B}$. Note that the $\varphi$-dependence in Equation (53) is due to the $\varphi$-dependence of $\mathit{m}$ and $\mathit{n}$.

The Cherepanov force per unit length between a line force with strength $F_K^{\left(A\right)}$ at position $(r,\varphi )$ in the elastic distortion field $B_{Kl}^{\left(B\right)}$ produced by another line force with strength $F_I^{\left(B\right)}$ located the the position $(0,0)$ is given by (see [35])

$${\mathcal{F}}_l^{\mathrm{C}\left(AB\right)}={\displaystyle \frac{F_K^{\left(A\right)}{F}_I^{\left(B\right)}}{2\pi r}}\left[m_l{Q}_{KI}\left(2\pi \right)-n_l{\left(nn\right)}_{KN}^{-1}\left(S_{NI}^{\mathrm{T}}\left(2\pi \right)+{\left(nm\right)}_{NM}{Q}_{MI}\left(2\pi \right)\right)\right].$$

(54)

In polar coordinates, the two non-vanishing components of the Cherepanov force (54) read

$${\mathcal{F}}_r^{\mathrm{C}\left(AB\right)}={\displaystyle \frac{F_K^{\left(A\right)}{F}_I^{\left(B\right)}}{2\pi r}}{Q}_{KI}\left(2\pi \right),$$

(55)

$${\mathcal{F}}_{\varphi }^{\mathrm{C}\left(AB\right)}=-{\displaystyle \frac{F_K^{\left(A\right)}{F}_I^{\left(B\right)}}{2\pi r}}{\left(nn\right)}_{KN}^{-1}\left[S_{NI}^{\mathrm{T}}\left(2\pi \right)+{\left(nm\right)}_{NM}{Q}_{MI}\left(2\pi \right)\right].$$

(56)

The radial component (55) is connected with the matrix $\mathit{Q}$, and the tangential component (56) is connected with the matrices ${\mathit{S}}^{\mathrm{T}}$ and $\mathit{Q}$. The $\varphi$-dependence in Equation (56) is due to the $\varphi$-dependence of $\mathit{m}$ and $\mathit{n}$.

5. Line Defects in One-Dimensional Hexagonal Quasicrystals of Laue Class 10

We consider one-dimensional hexagonal quasicrystals of Laue class 10 characterized by the following elastic constants (see [37]), namely five elastic moduli of phonons:

$$\begin{array}{cc}\hfill & C_{1111}=C_{2222}=C_{11},C_{1122}=C_{2211}=C_{12},\hfill \\ \hfill & C_{1133}=C_{2233}=C_{3311}=C_{3322}=C_{13},C_{3333}=C_{33},\hfill \\ \hfill & C_{2323}=C_{2332}=C_{3223}=C_{3232}=C_{1313}=C_{1331}=C_{3113}=C_{3131}=C_{44},\hfill \\ \hfill & C_{1212}=C_{1221}=C_{2112}=C_{2121}=C_{66}=(C_{11}-C_{12})/2,\hfill \end{array}$$

(57)

three elastic moduli of phonon–phason coupling:

$$\begin{array}{c}\hfill D_{1133}=D_{2233}=R_1,D_{3333}=R_2,D_{3232}=D_{2332}=D_{3131}=D_{1331}=R_3\end{array}$$

(58)

and two elastic moduli of phasons:

$$\begin{array}{c}\hfill E_{3333}=K_1,E_{3131}=E_{3232}=K_2.\end{array}$$

(59)

The conditions of positive definiteness for the elastic constants of one-dimensional hexagonal quasicrystals of Laue class 10 to ensure a positive elastic energy density read [40,41]

$$\begin{array}{cc}\hfill & C_{11}>\left|C_{12}\right|,(C_{11}+C_{12})C_{33}-2C_{13}^2>0,C_{44}>0,\hfill \end{array}$$

(60)

$$\begin{array}{cc}\hfill & C_{44}{K}_2-R_3^2>0,\hfill \end{array}$$

(61)

$$\begin{array}{cc}\hfill & \left[(C_{11}+C_{12})C_{33}-2C_{13}^2\right]K_1+2(2C_{13}{R}_2-C_{33}{R}_1)R_1-(C_{11}+C_{12})R_2^2>0.\hfill \end{array}$$

(62)

A one-dimensional hexagonal quasicrystal of Laue class 10 is a transversely isotropic medium, concerning generalized elasticity of quasicrystals, namely, a one-dimensional hexagonal quasicrystal of Laue class 10 is isotropic in the basal $x_1{x}_2$-plane (see [38]). Laue class 10 consists of the following point groups [13]: ${62}_h{2}_h$, $6mm$, $\overline{6}m{2}_h$, $6/m_{h}mm$. Therefore, the Laue class 10 is the Laue class with the highest symmetry for one-dimensional quasicrystals. In order to keep equations as simple as possible, we may use $C_{66}=(C_{11}-C_{12})/2$ instead of $C_{12}$ in the formulas. For one-dimensional hexagonal quasicrystals of Laue class 10, Equations (22)–(24) reduce to

$${\left(mm\right)}_{IK}=\left(\begin{array}{cccc}{C}_{1111}& 0& 0& 0\\ 0& C_{2121}& 0& 0\\ 0& 0& C_{3131}& D_{3131}\\ 0& 0& D_{3131}& E_{3131}\end{array}\right)=\left(\begin{array}{cccc}{C}_{11}& 0& 0& 0\\ 0& C_{66}& 0& 0\\ 0& 0& C_{44}& R_3\\ 0& 0& R_3& K_2\end{array}\right)$$

(63)

$${\left(mn\right)}_{IK}=\left(\begin{array}{cccc}0& C_{1122}& 0& 0\\ C_{2112}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)=\left(\begin{array}{cccc}0& C_{12}& 0& 0\\ C_{66}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$$

(64)

$${\left(nn\right)}_{IK}=\left(\begin{array}{cccc}{C}_{1212}& 0& 0& 0\\ 0& C_{2222}& 0& 0\\ 0& 0& C_{3232}& D_{3232}\\ 0& 0& D_{3232}& E_{3232}\end{array}\right)=\left(\begin{array}{cccc}{C}_{66}& 0& 0& 0\\ 0& C_{11}& 0& 0\\ 0& 0& C_{44}& R_3\\ 0& 0& R_3& K_2\end{array}\right).$$

(65)

Therefore, for generalized plane strain, only three elastic moduli of phonons $C_{11}=C_{1111}=C_{2222}$, $C_{12}=C_{1122}$, $C_{44}=C_{3232}=C_{3131}$, $C_{66}=(C_{11}-C_{12})/2=(C_{1111}-C_{1122})/2=C_{2121}=C_{2112}=C_{1212}$; one elastic modulus of phonon–phason coupling $R_3=D_{3232}=D_{3131}$; and one elastic modulus of phasons $K_2=E_{3131}=E_{3232}$ are relevant.

The four $4\times 4$ blocks of the $8\times 8$ matrix $\mathit{N}$ read

$$\begin{array}{c}\hfill {\left(nn\right)}^{-1}=\left(\begin{array}{cccc}1/C_{66}& 0& 0& 0\\ 0& 1/C_{11}& 0& 0\\ 0& 0& {\displaystyle \frac{K_2}{C_{44}{K}_2-R_3^2}}& {\displaystyle -\frac{R_3}{C_{44}{K}_2-R_3^2}}\\ 0& 0& {\displaystyle -\frac{R_3}{C_{44}{K}_2-R_3^2}}& {\displaystyle -\frac{C_{44}}{C_{44}{K}_2-R_3^2}}\end{array}\right)\end{array}$$

(66)

$$\begin{array}{c}\hfill {\left(nn\right)}^{-1}\left(nm\right)=\left(\begin{array}{cccc}0& 1& 0& 0\\ C_{12}/C_{11}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\end{array}$$

(67)

$$\begin{array}{c}\hfill \left(mn\right){\left(nn\right)}^{-1}=\left(\begin{array}{cccc}0& C_{12}/C_{11}& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\end{array}$$

(68)

and

$$\begin{array}{c}\hfill [\left(mn\right){\left(nn\right)}^{-1}\left(nm\right)-\left(mm\right)]=-\left(\begin{array}{cccc}\alpha & 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& C_{44}& R_3\\ 0& 0& R_3& K_2\end{array}\right),\end{array}$$

(69)

where

$$\begin{array}{c}\hfill \alpha ={\displaystyle \frac{C_{11}^2-C_{12}^2}{C_{11}}}.\end{array}$$

(70)

The angular behaviour of the $4\times 4$ matrices $\mathit{Q}$, $\mathit{S}$, and $\mathit{B}$ is obtained by the transformations (see also [42,43])

$$\begin{array}{cc}\hfill \mathit{Q}\left(\varphi \right)& =-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }{\left(nn\right)}^{-1}\left(\omega \right)\mathrm{d}\omega \hfill \\ \hfill & =-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }{\mathit{R}}^{\mathrm{T}}\left(\omega \right){\left(nn\right)}^{-1}\left(0\right)\mathit{R}\left(\omega \right)\mathrm{d}\omega ,\hfill \end{array}$$

(71)

$$\begin{array}{cc}\hfill \mathit{S}\left(\varphi \right)& =-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }\left[{\left(nn\right)}^{-1}\left(nm\right)\right]\left(\omega \right)\mathrm{d}\omega \hfill \\ \hfill & =-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }{\mathit{R}}^{\mathrm{T}}\left(\omega \right)\left[{\left(nn\right)}^{-1}\left(nm\right)\right]\left(0\right)\mathit{R}\left(\omega \right)\mathrm{d}\omega ,\hfill \end{array}$$

(72)

$$\begin{array}{cc}\hfill \mathit{B}\left(\varphi \right)& =-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }[\left(mn\right){\left(nn\right)}^{-1}\left(nm\right)-\left(mm\right)]\left(\omega \right)\mathrm{d}\omega \hfill \\ \hfill & =-{\displaystyle \frac{1}{2\pi }}{\int }_0^{\varphi }{\mathit{R}}^{\mathrm{T}}\left(\omega \right)[\left(mn\right){\left(nn\right)}^{-1}\left(nm\right)-\left(mm\right)]\left(0\right)\mathit{R}\left(\omega \right)\mathrm{d}\omega ,\hfill \end{array}$$

(73)

where

$$\begin{array}{c}\hfill \mathit{R}=\left(\begin{array}{cccc}cos\omega & sin\omega & 0& 0\\ -sin\omega & cos\omega & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\end{array}$$

(74)

is a $4\times 4$ rotation matrix in the $x_1{x}_2$-plane.

Carrying out the integrations, we obtain

$$\begin{array}{c}\hfill \mathit{Q}\left(\varphi \right)=-{\displaystyle \frac{1}{4\pi }}\left(\begin{array}{cccc}{\displaystyle \beta \varphi +\gamma cos\varphi sin\varphi }& \gamma {sin}^2\varphi & 0& 0\\ \gamma {sin}^2\varphi & \beta \varphi -\gamma cos\varphi sin\varphi & 0& 0\\ 0& 0& {\displaystyle \frac{2K_2\varphi }{C_{44}{K}_2-R_3^2}}& {\displaystyle -\frac{2R_3\varphi }{C_{44}{K}_2-R_3^2}}\\ 0& 0& {\displaystyle -\frac{2R_3\varphi }{C_{44}{K}_2-R_3^2}}& {\displaystyle -\frac{2C_{44}\varphi }{C_{44}{K}_2-R_3^2}}\end{array}\right),\end{array}$$

(75)

where

$$\beta ={\displaystyle \frac{1}{C_{66}}}+{\displaystyle \frac{1}{C_{11}}}={\displaystyle \frac{C_{11}+C_{66}}{C_{11}{C}_{66}}}$$

(76)

$$\gamma ={\displaystyle \frac{1}{C_{66}}}-{\displaystyle \frac{1}{C_{11}}}={\displaystyle \frac{C_{11}-C_{66}}{C_{11}{C}_{66}}},$$

(77)

$$\begin{array}{c}\hfill \mathit{B}\left(\varphi \right)={\displaystyle \frac{1}{2\pi }}\left(\begin{array}{cccc}{\displaystyle \frac{\alpha }{2}(\varphi +cos\varphi sin\varphi )}& {\displaystyle \frac{\alpha }{2}{sin}^2\varphi }& 0& 0\\ {\displaystyle \frac{\alpha }{2}{sin}^2\varphi }& {\displaystyle \frac{\alpha }{2}(\varphi -cos\varphi sin\varphi )}& 0& 0\\ 0& 0& C_{44}\varphi & R_3\varphi \\ 0& 0& R_3\varphi & K_2\varphi \end{array}\right)\end{array}$$

(78)

and

$$\begin{array}{c}\hfill \mathit{S}\left(\varphi \right)={\displaystyle \frac{1}{2\pi }}\left(\begin{array}{cccc}{\displaystyle \frac{C_{11}-C_{66}}{C_{11}}{sin}^2\varphi }& {\displaystyle -\frac{C_{66}}{C_{11}}\varphi -\frac{C_{11}-C_{66}}{C_{11}}cos\varphi sin\varphi }& 0& 0\\ {\displaystyle \frac{C_{66}}{C_{11}}\varphi -\frac{C_{11}-C_{66}}{C_{11}}cos\varphi sin\varphi }& {\displaystyle -\frac{C_{11}-C_{66}}{C_{11}}{sin}^2\varphi }& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right).\end{array}$$

(79)

For $\varphi =2\pi$, the matrices (75), (78), and (79) reduce to

$$\begin{array}{c}\hfill \mathit{Q}\left(2\pi \right)=-{\displaystyle \frac{1}{2}}\left(\begin{array}{cccc}\beta & 0& 0& 0\\ 0& \beta & 0& 0\\ 0& 0& {\displaystyle \frac{2K_2}{C_{44}{K}_2-R_3^2}}& {\displaystyle -\frac{2R_3}{C_{44}{K}_2-R_3^2}}\\ 0& 0& {\displaystyle -\frac{2R_3}{C_{44}{K}_2-R_3^2}}& {\displaystyle \frac{2C_{44}}{C_{44}{K}_2-R_3^2}}\end{array}\right),\end{array}$$

(80)

$$\begin{array}{c}\hfill \mathit{B}\left(2\pi \right)=\left(\begin{array}{cccc}\alpha /2& 0& 0& 0\\ 0& \alpha /2& 0& 0\\ 0& 0& C_{44}& R_3\\ 0& 0& R_3& K_2\end{array}\right)\end{array}$$

(81)

and

$$\begin{array}{c}\hfill \mathit{S}\left(2\pi \right)=\left(\begin{array}{cccc}0& -C_{66}/C_{11}& 0& 0\\ C_{66}/C_{11}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right).\end{array}$$

(82)

These three matrices (80)–(82) obey the relations

$$\mathit{B}\left(2\pi \right)\mathit{S}\left(2\pi \right)+{\mathit{S}}^{\mathrm{T}}\left(2\pi \right)\mathit{B}\left(2\pi \right)=\mathbf{0}$$

(83)

$$\mathit{Q}\left(2\pi \right){\mathit{S}}^{\mathrm{T}}\left(2\pi \right)+\mathit{S}\left(2\pi \right)\mathit{Q}\left(2\pi \right)=\mathbf{0}$$

(84)

$$\mathit{B}\left(2\pi \right)\mathit{Q}\left(2\pi \right)+{\mathit{S}}^{\mathrm{T}}\left(2\pi \right){\mathit{S}}^{\mathrm{T}}\left(2\pi \right)=-\mathit{I}$$

(85)

as in anisotropic elasticity (see [44,45]).

Now, substituting Equations (75), (79)–(82) into Equation (45), the extended displacement vector of a straight dislocation with extended Burgers vector $b_I$ reads in matrix form:

$$2\pi \mathit{U}=\mathit{b}\varphi +\mathit{b}\left({\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}\right)\left(\begin{array}{cccc}sin\varphi cos\varphi & {sin}^2\varphi & 0& 0\\ {sin}^2\varphi & -sin\varphi cos\varphi & 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)-\mathit{b}{\displaystyle \frac{C_{66}}{C_{11}}}lnr\left(\begin{array}{cccc}0& -1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$$

(86)

and the four components of Equation (86) read explicitly:

$$\begin{array}{cc}\hfill u_1^{\parallel }& ={\displaystyle \frac{b_1^{\parallel }}{2\pi }}\left(arctan{\displaystyle \frac{x_2}{x_1}}+{\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}{\displaystyle \frac{x_1{x}_2}{r^2}}\right)+{\displaystyle \frac{b_2^{\parallel }}{2\pi }}\left({\displaystyle \frac{C_{66}}{C_{11}}}lnr+{\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}{\displaystyle \frac{x_2^2}{r^2}}\right),\hfill \end{array}$$

(87)

$$\begin{array}{cc}\hfill u_2^{\parallel }& =-{\displaystyle \frac{b_1^{\parallel }}{2\pi }}\left({\displaystyle \frac{C_{66}}{C_{11}}}lnr-{\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}{\displaystyle \frac{x_2^2}{r^2}}\right)+{\displaystyle \frac{b_2^{\parallel }}{2\pi }}\left(arctan{\displaystyle \frac{x_2}{x_1}}-{\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}{\displaystyle \frac{x_1{x}_2}{r^2}}\right),\hfill \end{array}$$

(88)

$$\begin{array}{cc}\hfill u_3^{\parallel }& ={\displaystyle \frac{b_3^{\parallel }}{2\pi }}arctan{\displaystyle \frac{x_2}{x_1}},\hfill \end{array}$$

(89)

$$\begin{array}{cc}\hfill u_3^{\perp }& ={\displaystyle \frac{b_3^{\perp }}{2\pi }}arctan{\displaystyle \frac{x_2}{x_1}}.\hfill \end{array}$$

(90)

Equations (87)–(90) are the displacement components of a straight dislocation with extended Burgers vector $(b_1^{\parallel },b_2^{\parallel },b_3^{\parallel },b_3^{\perp })$ in a one-dimensional hexagonal quasicrystal of Laue class 10. Equations (87) and (88) are the displacement components of an edge dislocation with Burgers vector $b_1^{\parallel }$ or $b_2^{\parallel }$. Equations (89) and (90) are the displacement components of a screw dislocation with Burgers vector components $b_3^{\parallel }$ and $b_3^{\perp }$. It can be seen that all four components of the Burgers vector are uncoupled. The displacement components of a straight dislocation in a one-dimensional hexagonal quasicrystal of Laue class 10 (87)–(90) obtained in the framework of the integral formalism are in agreement with the expressions given by Li and Fan [46], directly solving the field equations for the phonon and phason displacement fields (see also [47]). The only difference is an irrelevant constant displacement in Equation (88), leading to $x_1^2/r^2$ instead of $-x_2^2/r^2$ in [46]. The phonon displacement fields (87)–(89) take the same form as in hexagonal crystals.

Substituting Equations (75), (79), (80), and (82) into Equation (46), the extended displacement vector of a line force with extended strength $F_I$ reads in matrix form:

$$4\pi \mathit{U}=\mathit{F}\gamma \left(\begin{array}{cccc}-{sin}^2\varphi & sin\varphi cos\varphi & 0& 0\\ sin\varphi cos\varphi & {sin}^2\varphi & 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)-\mathit{F}lnr\left(\begin{array}{cccc}\beta & 0& 0& 0\\ 0& \beta & 0& 0\\ 0& 0& {\displaystyle \frac{2K_2}{C_{44}{K}_2-R_3^2}}& {\displaystyle -\frac{2R_3}{C_{44}{K}_2-R_3^2}}\\ 0& 0& {\displaystyle -\frac{2R_3}{C_{44}{K}_2-R_3^2}}& {\displaystyle \frac{2C_{44}}{C_{44}{K}_2-R_3^2}}\end{array}\right)$$

(91)

and the four components of Equation (91) read explicitly:

$$\begin{array}{cc}\hfill u_1^{\parallel }& =-{\displaystyle \frac{F_1^{\parallel }}{4\pi }}\left({\displaystyle \frac{C_{11}+C_{66}}{C_{11}{C}_{66}}}lnr+{\displaystyle \frac{C_{11}-C_{66}}{C_{11}{C}_{66}}}{\displaystyle \frac{x_2^2}{r^2}}\right)+{\displaystyle \frac{F_2^{\parallel }}{4\pi }}{\displaystyle \frac{C_{11}-C_{66}}{C_{11}{C}_{66}}}{\displaystyle \frac{x_1{x}_2}{r^2}},\hfill \end{array}$$

(92)

$$\begin{array}{cc}\hfill u_2^{\parallel }& ={\displaystyle \frac{F_1^{\parallel }}{4\pi }}{\displaystyle \frac{C_{11}-C_{66}}{C_{11}{C}_{66}}}{\displaystyle \frac{x_1{x}_2}{r^2}}-{\displaystyle \frac{F_2^{\parallel }}{4\pi }}\left({\displaystyle \frac{C_{11}+C_{66}}{C_{11}{C}_{66}}}lnr-{\displaystyle \frac{C_{11}-C_{66}}{C_{11}{C}_{66}}}{\displaystyle \frac{x_2^2}{r^2}}\right),\hfill \end{array}$$

(93)

$$\begin{array}{cc}\hfill u_3^{\parallel }& =-{\displaystyle \frac{F_3^{\parallel }{K}_2-F_3^{\perp }{R}_3}{2\pi (C_{44}{K}_2-R_3^2)}}lnr,\hfill \end{array}$$

(94)

$$\begin{array}{cc}\hfill u_3^{\perp }& ={\displaystyle \frac{F_3^{\parallel }{R}_3-F_3^{\perp }{C}_{44}}{2\pi (C_{44}{K}_2-R_3^2)}}lnr.\hfill \end{array}$$

(95)

Equations (92)–(95) are the displacement components of a line force with strength components $F_1^{\parallel }$, $F_2^{\parallel }$, $F_3^{\parallel }$ and $F_3^{\perp }$. In Equations (94) and (95), it can be seen that the phonon and phason components $F_3^{\parallel }$ and $F_3^{\perp }$ are coupled due to the elastic constant of the phonon–phason coupling $R_3$.

Substituting Equations (78), (79), (81), and (82) into Equation (47), the extended stress function vector of a straight dislocation with extended Burgers vector $b_K$ reads in matrix form:

$$2\pi \mathbf{\Phi }=\mathit{b}{\displaystyle \frac{\alpha }{2}}\left(\begin{array}{cccc}-{sin}^2\varphi & sin\varphi cos\varphi & 0& 0\\ sin\varphi cos\varphi & {sin}^2\varphi & 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)-\mathit{b}lnr\left(\begin{array}{cccc}\alpha /2& 0& 0& 0\\ 0& \alpha /2& 0& 0\\ 0& 0& C_{44}& R_3\\ 0& 0& R_3& K_2\end{array}\right)$$

(96)

and the four components of Equation (96) read explicitly:

$${\mathsf{\Phi }}_1^{\parallel }=-{\displaystyle \frac{C_{11}^2-C_{12}^2}{2C_{11}}}\left[{\displaystyle \frac{b_1^{\parallel }}{2\pi }}\left(lnr+{\displaystyle \frac{x_2^2}{r^2}}\right)-{\displaystyle \frac{b_2^{\parallel }}{2\pi }}{\displaystyle \frac{x_1{x}_2}{r^2}}\right],$$

(97)

$${\mathsf{\Phi }}_2^{\parallel }={\displaystyle \frac{C_{11}^2-C_{12}^2}{2C_{11}}}\left[{\displaystyle \frac{b_1^{\parallel }}{2\pi }}{\displaystyle \frac{x_1{x}_2}{r^2}}-{\displaystyle \frac{b_2^{\parallel }}{2\pi }}\left(lnr-{\displaystyle \frac{x_2^2}{r^2}}\right)\right],$$

(98)

$${\mathsf{\Phi }}_3^{\parallel }=-{\displaystyle \frac{b_3^{\parallel }{C}_{44}+b_3^{\perp }{R}_3}{2\pi }}lnr,$$

(99)

$${\mathsf{\Phi }}_3^{\perp }=-{\displaystyle \frac{b_3^{\parallel }{R}_3+b_3^{\perp }{K}_2}{2\pi }}lnr.$$

(100)

Equations (97)–(100) are the stress function components of a straight dislocation with extended Burgers vector $(b_1^{\parallel },b_2^{\parallel },b_3^{\parallel },b_3^{\perp })$ in a one-dimensional hexagonal quasicrystal of Laue class 10. Equations (97) and (98) are the stress function components of an edge dislocation with Burgers vector $b_1^{\parallel }$ or $b_2^{\parallel }$. Equations (99) and (100) are the stress function components of a screw dislocation with Burgers vector components $b_3^{\parallel }$ and $b_3^{\perp }$. In Equations (99) and (100), it can be seen that the phonon and phason components $b_3^{\parallel }$ and $b_3^{\perp }$ are coupled due to the elastic constant of the phonon–phason coupling $R_3$. It is interesting to note the similarity between the displacements of a line force (92)–(95) and the stress functions of a dislocation (97)–(100).

Substituting Equations (78)–(80) and (82) into Equation (48), the extended stress function vector of a line force with extended strength $F_K$ reads in matrix form:

$$2\pi \mathbf{\Phi }=-\mathit{F}\varphi -\mathit{F}\left({\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}\right)\left(\begin{array}{cccc}sin\varphi cos\varphi & {sin}^2\varphi & 0& 0\\ {sin}^2\varphi & -sin\varphi cos\varphi & 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)+\mathit{F}{\displaystyle \frac{C_{66}}{C_{11}}}lnr\left(\begin{array}{cccc}0& 1& 0& 0\\ -1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$$

(101)

and the four components of Equation (101) read explicitly:

$${\mathsf{\Phi }}_1^{\parallel }=-{\displaystyle \frac{F_1^{\parallel }}{2\pi }}\left(arctan{\displaystyle \frac{x_2}{x_1}}+{\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}{\displaystyle \frac{x_1{x}_2}{r^2}}\right)+{\displaystyle \frac{F_2^{\parallel }}{2\pi }}\left({\displaystyle \frac{C_{66}}{C_{11}}}lnr-{\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}{\displaystyle \frac{x_2^2}{r^2}}\right),$$

(102)

$${\mathsf{\Phi }}_2^{\parallel }=-{\displaystyle \frac{F_1^{\parallel }}{2\pi }}\left({\displaystyle \frac{C_{66}}{C_{11}}}lnr+{\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}{\displaystyle \frac{x_2^2}{r^2}}\right)-{\displaystyle \frac{F_2^{\parallel }}{2\pi }}\left(arctan{\displaystyle \frac{x_2}{x_1}}-{\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}{\displaystyle \frac{x_1{x}_2}{r^2}}\right),$$

(103)

$${\mathsf{\Phi }}_3^{\parallel }=-{\displaystyle \frac{F_3^{\parallel }}{2\pi }}arctan{\displaystyle \frac{x_2}{x_1}},$$

(104)

$${\mathsf{\Phi }}_3^{\perp }=-{\displaystyle \frac{F_3^{\perp }}{2\pi }}arctan{\displaystyle \frac{x_2}{x_1}}.$$

(105)

Equations (102)–(105) are the stress functions of a line force with strength components $F_1^{\parallel }$, $F_2^{\parallel }$, $F_3^{\parallel }$ and $F_3^{\perp }$. It can be seen that all four components of the strength components $F_1^{\parallel }$, $F_2^{\parallel }$, $F_3^{\parallel }$, and $F_3^{\perp }$ are uncoupled. The phonon stress functions (102)–(104) take the same form as in hexagonal crystals. It is interesting to note the similarity between the displacements of a dislocation (87)–(90) and the stress functions of a line force (102)–(105).

Substituting Equation (81) into Equation (49), the self-energy per unit length of a straight dislocation in a one-dimensional hexagonal quasicrystal of Laue class 10 is given by

$$\mathcal{E}={\displaystyle \frac{1}{4\pi }}\left[{\displaystyle \frac{\alpha }{2}}\left(b_1^{\parallel }{b}_1^{\parallel }+b_2^{\parallel }{b}_2^{\parallel }\right)+C_{44}{b}_3^{\parallel }{b}_3^{\parallel }+2R_3{b}_3^{\parallel }{b}_3^{\perp }+K_2{b}_3^{\perp }{b}_3^{\perp }\right]ln(R/r_0).$$

(106)

Substituting Equation (80) into Equation (50), the self-energy per unit length of a line force in a one-dimensional hexagonal quasicrystal of Laue class 10 is given by

$$\mathcal{E}={\displaystyle \frac{1}{8\pi }}\left[\beta \left(F_1^{\parallel }{F}_1^{\parallel }+F_2^{\parallel }{F}_2^{\parallel }\right)+{\displaystyle \frac{2}{C_{44}{K}_2-R_3^2}}\left(K_2{F}_3^{\parallel }{F}_3^{\parallel }-2R_3{F}_3^{\parallel }{F}_3^{\perp }+C_{44}{F}_3^{\perp }{F}_3^{\perp }\right)\right]ln(R/r_0).$$

(107)

Using Equations (81) and (82), the non-vanishing components of the Peach–Koehler force per unit length between two straight dislocations, Equations (52) and (53), read for a one-dimensional hexagonal quasicrystal of Laue class 10:

$${\mathcal{F}}_r^{\mathrm{PK}\left(AB\right)}={\displaystyle \frac{1}{2\pi r}}\left[{\displaystyle \frac{\alpha }{2}}\left(b_1^{\parallel \left(A\right)}{b}_1^{\parallel \left(B\right)}+b_2^{\parallel \left(A\right)}{b}_2^{\parallel \left(B\right)}\right)+C_{44}{b}_3^{\parallel \left(A\right)}{b}_3^{\parallel \left(B\right)}+R_3{b}_3^{\parallel \left(A\right)}{b}_3^{\perp \left(B\right)}+R_3{b}_3^{\perp \left(A\right)}{b}_3^{\parallel \left(B\right)}+K_2{b}_3^{\perp \left(A\right)}{b}_3^{\perp \left(B\right)}\right],$$

(108)

$${\mathcal{F}}_{\varphi }^{\mathrm{PK}\left(AB\right)}={\displaystyle \frac{\alpha }{4\pi r}}\left[\left(b_1^{\parallel \left(A\right)}{b}_1^{\parallel \left(B\right)}-b_2^{\parallel \left(A\right)}{b}_2^{\parallel \left(B\right)}\right)sin2\varphi -\left(b_1^{\parallel \left(A\right)}{b}_2^{\parallel \left(B\right)}+b_2^{\parallel \left(A\right)}{b}_1^{\parallel \left(B\right)}\right)cos2\varphi \right].$$

(109)

Equation (108) contains phonon, phason and phonon–phason-coupling contributions, whereas Equation (109) contains only phonon contributions. If we consider only two parallel screw dislocations, Equations (108) and (109) reduce to the Peach–Kohler force given in [38,46]. In the isotropic limit (see below), the Peach–Koehler force, given in Equations (108) and (109), reduces to the Peach–Koehler force between straight dislocations in isotropic elasticity given in [48,49,50,51].

Using Equations (80) and (82), the non-vanishing components of the Cherepanov force per unit length of between two line forces, Equations (55) and (56), read for a one-dimensional hexagonal quasicrystal of Laue class 10

$$\begin{array}{cc}\hfill {\mathcal{F}}_r^{\mathrm{C}\left(AB\right)}& =-{\displaystyle \frac{1}{4\pi r}}[\beta \left(F_1^{\parallel \left(A\right)}{F}_1^{\parallel \left(B\right)}+F_2^{\parallel \left(A\right)}{F}_2^{\parallel \left(B\right)}\right)\hfill \\ \hfill & +{\displaystyle \frac{2}{C_{44}{K}_2-R_3^2}}\left(K_2{F}_3^{\parallel \left(A\right)}{F}_3^{\parallel \left(B\right)}-R_3{F}_3^{\parallel \left(A\right)}{F}_3^{\perp \left(B\right)}-R_3{F}_3^{\perp \left(A\right)}{F}_3^{\parallel \left(B\right)}+C_{44}{F}_3^{\perp \left(A\right)}{F}_3^{\perp \left(B\right)}\right)],\hfill \end{array}$$

(110)

$${\mathcal{F}}_{\varphi }^{\mathrm{C}\left(AB\right)}=-{\displaystyle \frac{\beta }{4\pi r}}\left[\left(F_1^{\parallel \left(A\right)}{F}_1^{\parallel \left(B\right)}-F_2^{\parallel \left(A\right)}{F}_2^{\parallel \left(B\right)}\right)sin2\varphi -\left(F_1^{\parallel \left(A\right)}{F}_2^{\parallel \left(B\right)}+F_2^{\parallel \left(A\right)}{F}_1^{\parallel \left(B\right)}\right)cos2\varphi \right].$$

(111)

Equation (110) contains phonon, phason and phonon–phason-coupling contributions, whereas Equation (111) contains only phonon contributions. In the isotropic limit (see below), the Cherepanov force, given in Equations (110) and (111), reduces to the Cherepanov force between straight line forces in isotropic elasticity given in [52].

6. Green Tensor in the Integral Formalism

Within the framework of the integral formalism, the displacement field of a line force, Equations (92)–(95), gives the two-dimensional $4\times 4$ Green tensor for quasicrystals in the hyperspace, namely

$$\begin{array}{c}\hfill U_K=G_{KI}{F}_I\end{array}$$

(112)

with

$$\begin{array}{c}\hfill G_{KI}=\left(\begin{array}{cccc}{G}_{11}^{\parallel \parallel }& G_{12}^{\parallel \parallel }& G_{13}^{\parallel \parallel }& G_{14}^{\parallel \perp }\\ G_{12}^{\parallel \parallel }& G_{22}^{\parallel \parallel }& G_{23}^{\parallel \parallel }& G_{24}^{\parallel \perp }\\ G_{13}^{\parallel \parallel }& G_{23}^{\parallel \parallel }& G_{33}^{\parallel \parallel }& G_{34}^{\parallel \perp }\\ G_{14}^{\parallel \perp }& G_{24}^{\parallel \perp }& G_{34}^{\parallel \perp }& G_{44}^{\perp \perp }\end{array}\right).\end{array}$$

(113)

Therefore, the non-vanishing components of the two-dimensional Green tensor for one-dimensional hexagonal quasicrystals of Laue class 10 read

$$G_{11}^{\parallel \parallel }=-{\displaystyle \frac{1}{4\pi }}\left({\displaystyle \frac{C_{11}+C_{66}}{C_{11}{C}_{66}}}lnr+{\displaystyle \frac{C_{11}-C_{66}}{C_{11}{C}_{66}}}{\displaystyle \frac{x_2^2}{r^2}}\right),$$

(114)

$$G_{12}^{\parallel \parallel }={\displaystyle \frac{1}{4\pi }}{\displaystyle \frac{C_{11}-C_{66}}{C_{11}{C}_{66}}}{\displaystyle \frac{x_1{x}_2}{r^2}},$$

(115)

$$G_{22}^{\parallel \parallel }=-{\displaystyle \frac{1}{4\pi }}\left({\displaystyle \frac{C_{11}+C_{66}}{C_{11}{C}_{66}}}lnr-{\displaystyle \frac{C_{11}-C_{66}}{C_{11}{C}_{66}}}{\displaystyle \frac{x_2^2}{r^2}}\right),$$

(116)

$$G_{33}^{\parallel \parallel }=-{\displaystyle \frac{K_2}{2\pi (C_{44}{K}_2-R_3^2)}}lnr,$$

(117)

$$G_{34}^{\parallel \perp }={\displaystyle \frac{R_3}{2\pi (C_{44}{K}_2-R_3^2)}}lnr,$$

(118)

$$G_{44}^{\perp \perp }=-{\displaystyle \frac{C_{44}}{2\pi (C_{44}{K}_2-R_3^2)}}lnr.$$

(119)

Note that the three-dimensional Green tensor of one-dimensional hexagonal quasicrystals of Laue class 10 has been recently given by Lazar et al. [40].

7. Isotropic Limit

Now, we obtain the isotropic limit of the phonon displacement fields and stress functions of a straight dislocation and a straight line force:

$u_i^{\parallel }\to u_i$, ${\mathsf{\Phi }}_i^{\parallel }\to {\mathsf{\Phi }}_i$, $b_i^{\parallel }\to b_i$ and $F_i^{\parallel }\to F_i$.

To verify the isotropic limit of Equations (87)–(89), we need ${\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}\to {\displaystyle \frac{1}{2(1-\nu )}}$, ${\displaystyle \frac{C_{66}}{C_{11}}}\to {\displaystyle \frac{1-2\nu }{2(1-\nu )}}$, where $\nu$ denotes the Poisson ratio; we obtain for the three components of the displacement vector of a straight dislocation in an isotropic medium

$$u_1={\displaystyle \frac{b_1}{2\pi }}\left(arctan{\displaystyle \frac{x_2}{x_1}}+{\displaystyle \frac{1}{2(1-\nu )}}{\displaystyle \frac{x_1{x}_2}{r^2}}\right)+{\displaystyle \frac{b_2}{4\pi (1-\nu )}}\left((1-2\nu )lnr+{\displaystyle \frac{x_2^2}{r^2}}\right),$$

(120)

$$u_2=-{\displaystyle \frac{b_1}{4\pi (1-\nu )}}\left((1-2\nu )lnr-{\displaystyle \frac{x_2^2}{r^2}}\right)+{\displaystyle \frac{b_2}{2\pi }}\left(arctan{\displaystyle \frac{x_2}{x_1}}-{\displaystyle \frac{1}{2(1-\nu )}}{\displaystyle \frac{x_1{x}_2}{r^2}}\right),$$

(121)

$$u_3={\displaystyle \frac{b_3}{2\pi }}arctan{\displaystyle \frac{x_2}{x_1}},$$

(122)

which are in agreement with deWit [53] and Mura [54].

To verify the isotropic limit of Equations (92)–(94), we need $R_3\to 0$, ${\displaystyle \frac{C_{11}+C_{66}}{C_{11}}}\to {\displaystyle \frac{3-4\nu }{2(1-\nu )}}$, ${\displaystyle \frac{1}{C_{66}}}\to {\displaystyle \frac{1}{\mu }}$ and ${\displaystyle \frac{1}{C_{44}}}\to {\displaystyle \frac{1}{\mu }}$, where $\mu$ denotes the shear modulus, and we obtain for the three components of the displacement vector of a line force in an isotropic medium

$$u_1=-{\displaystyle \frac{F_1}{8\pi \mu (1-\mu )}}\left((3-4\nu )lnr+{\displaystyle \frac{x_2^2}{r^2}}\right)+{\displaystyle \frac{F_2}{8\pi \mu (1-\mu )}}{\displaystyle \frac{x_1{x}_2}{r^2}},$$

(123)

$$u_2={\displaystyle \frac{F_1}{8\pi \mu (1-\mu )}}{\displaystyle \frac{x_1{x}_2}{r^2}}-{\displaystyle \frac{F_2}{8\pi \mu (1-\mu )}}\left((3-4\nu )lnr-{\displaystyle \frac{x_2^2}{r^2}}\right),$$

(124)

$$u_3=-{\displaystyle \frac{F_3}{2\pi \mu }}lnr,$$

(125)

which are in agreement with Lazar and Agiasofitou [52].

To verify the isotropic limit of Equations (97)–(99), we need $R_3\to 0$, ${\displaystyle \frac{C_{11}^2-C_{12}^2}{2C_{11}}}\to {\displaystyle \frac{\mu }{1-\nu }}$, $C_{44}\to \mu$, and we obtain for the three stress functions of a straight dislocation in an isotropic medium

$${\mathsf{\Phi }}_1=-{\displaystyle \frac{\mu }{1-\nu }}\left[{\displaystyle \frac{b_1}{2\pi }}\left(lnr+{\displaystyle \frac{x_2^2}{r^2}}\right)-{\displaystyle \frac{b_2}{2\pi }}{\displaystyle \frac{x_1{x}_2}{r^2}}\right],$$

(126)

$${\mathsf{\Phi }}_2={\displaystyle \frac{\mu }{1-\nu }}\left[{\displaystyle \frac{b_1}{2\pi }}{\displaystyle \frac{x_1{x}_2}{r^2}}-{\displaystyle \frac{b_2}{2\pi }}\left(lnr-{\displaystyle \frac{x_2^2}{r^2}}\right)\right],$$

(127)

$${\mathsf{\Phi }}_3=-{\displaystyle \frac{\mu b_3}{2\pi }}lnr,$$

(128)

which are in agreement with Ni and Nemat-Nasser [55].

To verify the isotropic limit of Equations (102)–(104), we need ${\displaystyle \frac{C_{11}-C_{66}}{C_{11}}}\to {\displaystyle \frac{1}{2(1-\nu )}}$, ${\displaystyle \frac{C_{66}}{C_{11}}}\to {\displaystyle \frac{1-2\nu }{2(1-\nu )}}$, and we obtain for the three stress functions of a line force in an isotropic medium

$${\mathsf{\Phi }}_1=-{\displaystyle \frac{F_1}{2\pi }}\left(arctan{\displaystyle \frac{x_2}{x_1}}+{\displaystyle \frac{1}{2(1-\nu )}}{\displaystyle \frac{x_1{x}_2}{r^2}}\right)+{\displaystyle \frac{F_2}{4\pi (1-\nu )}}\left((1-2\nu )lnr-{\displaystyle \frac{x_2^2}{r^2}}\right),$$

(129)

$${\mathsf{\Phi }}_2=-{\displaystyle \frac{F_1}{4\pi (1-\nu )}}\left((1-2\nu )lnr+{\displaystyle \frac{x_2^2}{r^2}}\right)-{\displaystyle \frac{F_2}{2\pi }}\left(arctan{\displaystyle \frac{x_2}{x_1}}-{\displaystyle \frac{1}{2(1-\nu )}}{\displaystyle \frac{x_1{x}_2}{r^2}}\right),$$

(130)

$${\mathsf{\Phi }}_3=-{\displaystyle \frac{F_3}{2\pi }}arctan{\displaystyle \frac{x_2}{x_1}}.$$

(131)

Note that the components (129) and (130) are in agreement with Ni and Nemat-Nasser [55]. It is obvious that the component ${\mathsf{\Phi }}_3$ given in Ni and Nemat-Nasser [55] has a mistaken pre-factor.

8. Conclusions

Using the eight-dimensional integral formalism for one-dimensional quasicrystals, we have derived exact analytical solutions of the displacement fields and the stress functions of line defects, namely straight dislocations and line forces, parallel to the aperiodic direction in a one-dimensional hexagonal quasicrystal of Laue class 10. Moreover, we have calculated the self-energies of a straight dislocation and a line force, the Peach–Koehler force between two straight dislocations, and the Cherepanov force between two straight line forces. It has been shown that in the limit to isotropic elasticity, well-known results for the displacement fields and the stress functions of straight dislocations and line forces are recovered. The two-dimensional Green tensor of one-dimensional hexagonal quasicrystals of Laue class 10 has been derived within the framework of the integral formalism.