Synonyms

Boundary value problems

Overview

The method of boundary integral equations [1] is a powerful tool for solving boundary problems in the theory of elasticity. Compared with domain methods like the finite element method and the finite difference method in which the entire domain needs to be discretized, the boundary integral method (BIE) requires only the discretization of the boundary alone. Here and below are considered two classical problems of the two-dimensional theory of elasticity: the first, when tractions are prescribed at the boundary, and the second, when displacements are prescribed at the boundary. These problems are reduced to regular integral equations in regions with a smooth boundary.

Introduction

Recall that there are different ways for reducing boundary value problems to integral equations. One of them, usually used in practice, is based on the direct formulation, that is, the formulation directly dealing with the primitive variables (displacement or traction) prescribed at the boundary. For basic elastic problems, this method leads to singular integral equations of the first kind. Of course, if the posed problem has the zero index [2], it can be regularized and reduced to a system of Fredholm second-kind equations, but its discretization leads to an ill-conditioned system of linear algebraic equations. Therefore, it is desirable to use second-kind equations for solving basic elastic problems, since the resulting system of linear algebraic equations is well conditioned. We use here an indirect method, based on introduction of a fictitious and nonphysical double-layer density, fitting to solve a boundary value problem. It is natural to ask yourself whether it is possible to write the system of regular equations for the given problem without using the procedure of regularization. In the three-dimensional isotropic elasticity, it is known as the so-called antennae, H. Weyl’s potential [1], whose application immediately leads to regular integral equations for basic three-dimensional problems of the isotropic theory of elasticity. It was noted in [3] that it corresponds to a solution of an elastic problem, obtained by superposition of solutions, loaded at the surface by a concentrated force (the Boussinesq solution). In the two-dimensional elasticity, it is known as the similar solution [4] of the static isotropic traction problem. The complex system of second-kind regular integral equations, advanced by D. I. Sherman [5] to solve the traction anisotropic elastic problem, is well known in the two-dimensional elasticity. In reality that author studied its modification consisting in assigning at a boundary first-order derivatives of the stress function. This modification has been considered recently by this author by means of a new approach [6]. The author’s effort to perform the limiting transition to an isotropic material using his equations was useless. Therefore, this author has undertaken an effort to produce the correct derivation of required potentials. It was successful; it turned out that the difference of complex parameters of an elastic material was absent in Sherman’s equations in the denominator, forbidding the limiting transition to an isotropic case. His equations are simply a result of a guess. The system of equations of the two-dimensional theory of elasticity has two pairs of distinct complex characteristics; this author have shown [4, 6, 7], and [8] that simplicity of characteristic roots essentially simplifies the procedure of reducing an anisotropic elastic boundary value problem to the system of integral equations. It is adequate to the considered boundary value problem. Moreover, this approach lets us use minimum assumptions on smoothness of a boundary and boundary data. In distinction to the approach based on the knowledge of the fundamental solution, which reduces the considered problem to the system of singular integral equations, this approach immediately reduces the problem to the system of regular equations for regions with a smooth boundary. This author also wants to add that the explicit construction of the required potentials is a nontrivial computational problem, even in the simplest case of an orthotropic material, when reducing a considered boundary value problem to singular integral equations. The limiting transition to an isotropic material can be performed without any difficulty. A large number of works are devoted to the deformation of anisotropic cylinders with various symmetry properties of the material. In this entry we study the plain strain problem when the material is homogeneous and has a plane of elastic symmetry, normal to the axis of cylinder. Let Q be the cross section of the cylinder. Throughout this entry a rectangular Cartesian coordinate system \(O{x_k}(k=1,2,3)\) is used. The coordinate frame is chosen such that the x 3-axis is parallel to the generators of the cylinder. We shall employ the usual summation and differential conventions. Latin subscripts (unless otherwise stated) are understood to range over integers (1, 2). In this entry we consider the linear theory of elastostatics for anisotropic bodies [13]. It was S. G. Mikhlin [13] in 1936, who initiated the study of plane anisotropic elasticity in Russia. Up to now there are no correct set-ups of boundary value problems for anisotropic elastic solids. Some work for isotropic solids was done by authors of [12] and [14].

Preliminaries

The constitutive relations are

$$\begin{array}{ll} {\epsilon_{11 }} & ={a_{11 }}{\sigma_{11 }}+{a_{12}}{\sigma_{22 }}+{a_{16 }}{\sigma_{12 }} \cr {\epsilon_{22 }} &={a_{12 }}{\sigma_{11 }}+{a_{22 }}{\sigma_{22 }}+{a_{26}}{\sigma_{12 }} \cr 2{\epsilon_{12 }} & ={a_{16 }}{\sigma_{11}}+{a_{26 }}{\sigma_{22 }}+{a_{66 }}{\sigma_{12 }} \cr \end{array}$$

where

$${\epsilon_{ij }}=\frac{1}{2}\left(\frac{{\partial {u_i}}}{{\partial {x_j}}}+\frac{{\partial {u_j}}}{{\partial {x_i}}}\right)$$

are strains, \({\sigma_{i,j }}\) are stresses, and u 1,u 2 are displacements. The positive definiteness of the compliances’ matrix \(({a_{ij }}),i,j=1,2,6\) is assumed. Here and below, all considered functions are defined in Q. Excluding displacements u 1,u 2 by differentiation of strains, we derive the equation of the strain’s compatibility:

$$\frac{{{\partial^2}{\epsilon_{11 }}}}{{\partial x_2^2}}+\frac{{{\partial^2}{\epsilon_{22 }}}}{{\partial x_1^2}}-2\frac{{{\partial^2}{\epsilon_{12 }}}}{{\partial {x_1}\partial {x_2}}}=0$$

Now, substitute into this equation strains from the constitutive relations and represent stresses as second order derivatives of a function u(x 1,x 2):

$${\sigma_{11 }}=\frac{{{\partial^2}u}}{{\partial x_2^2}},\quad {\sigma_{22 }}=\frac{{{\partial^2}u}}{{\partial x_1^2}},\quad {\sigma_{12 }}=-\frac{{{\partial^2}u}}{{\partial {x_1}\partial {x_2}}}$$

In the absence of body forces, the equilibrium equations are identically satisfied, and the stress function u(x 1,x 2) satisfies the elliptic fourth-order equation:

$$\begin{array}{ll} L(u)=\ L\left(\frac{\partial }{{\partial {x_1}}},\frac{\partial }{{\partial {x_2}}}\right)u({x_1},{x_2})={a_{22 }}\frac{{{\partial^4}u}}{{\partial x_1^4}}\cr -2{a_{26 }}\frac{{{\partial^4}u}}{{\partial x_1^3\partial {x_2}}}+(2{a_{12 }}+{a_{66 }})\frac{{{\partial^4}u}}{{\partial x_1^2\partial x_2^2}}\cr -{2_{26 }}\frac{{{\partial^4}u}}{{\partial {x_1}\partial x_2^3}}+{a_{11 }}\frac{{{\partial^4}u}}{{\partial x_2^4}}=0 \end{array}$$
(1)

Following Lekhnitskii [9], associate with equation (1) the (characteristic) equation

$$\begin{array}{ll} {a_{11 }}{\mu^4}-2{a_{16 }}{\mu^3}+(2{a_{12 }}+{a_{66 }}){\mu^2}-2{a_{26 }}\mu +{a_{22 }}=0.\end{array}$$

It has two pairs of complex conjugate roots (in terms of Lekhnitskii, complex parameters of an elastic material) \({\mu_k}={\alpha_k}+i{\beta_k},{\beta_k}>0,\mathop{\bar{\mu}}\nolimits_k={\alpha_k}-i{\beta_k},k=1,2\), since (1) is an elliptic equation with real coefficients. Write u(x 1, x 2) as the sum \(u({x_1},{x_2})={w_1}({x_1},{x_2})+{w_2}({x_1},{x_2})\). Here w k (x 1,x 2) is a quasi-harmonic function, that is, a solution of a “quasi-harmonic” equation

$$(\beta_k^2+\alpha_k^2)\frac{{{\partial^2}{w_k}}}{{\partial x_1^2}}-2{\alpha_k}{\beta_k}\frac{{{\partial^2}{w_k}}}{{\partial {x_1}\partial {x_2}}}+\frac{{{\partial^2}{w_k}}}{{\partial x_2^2}}=0$$

as the change of independent variables \({y_1}={x_1}+{\alpha_k}{x_2},{y_2}={\beta_k}{x_2}\) reduces it to Laplace’s equation. The quotation marks at the term “quasi-harmonic” below are omitted. In a simply connected region, any quasi-harmonic function \({w_k}({x_1},{x_2}),k=1,2\) can be represented as the real part of a holomorphic function of the (complex) argument \({z_k}={x_1}+{\mu_k}{x_2}\), \({w_k}({x_1},{x_2})=Re{\varphi_k}({z_k})\). As result, the general solution of (1) is

$$\begin{array}{ll} u({x_1},{x_2})=Re\{{\varphi_1}({x_1}+{\mu_1}{x_2})+{\varphi_2}({x_1}+{\mu_2}{x_2})\}\end{array}$$

Here and below, prime denotes differentiation with respect to the argument in parentheses. Let Q be a simply connected, bounded, counterclockwise-oriented plane region with a Lyapunov boundary \(\partial\) Q, that is, it is assumed that \(\partial\) Q has a uniformly Hölder continuous inward normal field \(\nu (z)=({n_1},{n_2})\). It means that the functions \({x_k}(s)\) specifying the shape of a region are continuous and continuously differentiable, and, moreover, there is a positive α, 0 < α < 1, such that

$$|{x^{\prime}}(s)-{x^{\prime}}({s_0})|\ < \ c|s-{s_0}{|^{\alpha }}$$

Here s is the arc-length parameter, and \({C^k}(Q)\) and \({C^k}(\bar{Q})\) are, respectively, the spaces of real functions that are continuously differentiable up to the order k in Q and \(\bar{Q}\). \({C^{{0,\alpha }}}(\bar{Q})\) and \({C^{{0,\alpha }}}(\partial Q)\) are, respectively, the spaces of real continuous functions satisfying in \(\bar{Q}\) and ∂Q a uniform Hölder condition with some exponent \(\alpha, 0\ <\ \alpha\ <\ 1\). \({C^{{k,\alpha }}}(\bar{Q})\) is the subclass of \({C^k}(\bar{Q})\) consisting from functions u such that \({D^{\nu }}u\in {C^{{0,\alpha }}}(\bar{Q}),|\nu |=k\). Here ν is the multi-index \(\nu =({k_1},{k_2})\):

$${D^{\nu }}u=\frac{{{\partial^{\nu }}u}}{{\partial x_1^{{{k_1}}}\partial x_2^{{{k_2}}}}},\quad {k_1}+{k_2}=\nu$$

The spaces \({C^{{l,\alpha }}}(\partial Q)\) are defined in a similar way. It is also assumed that the origin of coordinates lies inside the region. Let \({u_1}({x_1},{x_2}),{u_2}({x_1},{x_2})\) be the displacements in the anisotropic elasticity; they can be written as

$$\begin{array}{ll} {u_1}=Re[{b_{11 }}{\Phi_1}({z_1})+{b_{12}}{\Phi_2}({z_2})]+\alpha {x_2}+{\delta_1} \\ {u_2}=Re[{b_{21}}{\Phi_1}({z_1})+{b_{22 }}{\Phi_2}({z_2})]-\alpha {x_1}+{\delta_2}\cr \end{array}$$
(2)

Here \({\Phi_k}({z_k})={{\varphi^{\prime}}_k}({z_k})\), terms \(\alpha {x_2}+{\delta_1},\) and \(-\alpha {x_1}+{\delta_2}\) answer for a rigid displacement of an elastic body:

$$\begin{array}{ll} {b_{1k }} ={a_{11 }}\mu_k^2+{a_{12 }}-{a_{16 }}{\mu_k} \cr{b_{2k }} ={a_{12 }}{\mu_k}+{a_{22 }}\mu_k^{-1 }-{a_{26 }}\end{array}$$
(3)

Then stresses \({\sigma_{ij }},i,j=1,2\) are then rewritten as derivatives of the analytical functions \({\Phi_k}({z_k}),{z_k}{x_1}+{\mu_k}{x_2},k=1,2\) as

$$\begin{array}{ll} {\sigma_{11}}=Re[\mu_1^2{{{\Phi^{\prime}}}_1}({z_1})+\mu_2^2{{{\Phi^{\prime}}}_2}({z_2})] \\{\sigma_{22}}=Re[{{{\Phi^{\prime}}}_1}({z_1})+{{{\Phi^{\prime}}}_2}({z_2})] \cr{\sigma_{12}}=-Re[{\mu_1}{{{\Phi^{\prime}}}_1}({z_1})+{\mu_2}{{{\Phi^{\prime}}}_2}({z_2})]\cr \end{array}$$
(4)

Consider the integral equation

$$f(s)+\lambda \int\limits_a^b\,K(s,{s_0})\,f({s_0})\,d\,{s_0}=g(s)$$

where λ, a, and b are real parameters and \(f(s),g(s),K(s,{s_0})\) are real functions. The function \(K(s,{s_0})\) is defined in the plane (x, s) in the square \(a\ <\ s,{s_0}\ < \ b\). According to the definition of S. G. Mikhlin [10], the equation above is a Fredholm equation of the second kind for the function f(s), if g(s) and \(K(s,{s_0})\) are square integrable in the square \(a\ <\ s,{s_0}\ <\ b\). Recall that Fredholm assumed continuity of the kernel \(K(s,{s_0})\) in the same square. If a boundary of a plane region is a Lyapunov curve, then the kernel of the integral operator

$$\frac{1}{{\pi i}}\int\limits_{{\partial Q}}\,\frac{{f(s)\,d\,t}}{t-z }$$

where f(s) is a real function,

$$\begin{array}{ll} t =t(s)={x_1}(s)+i{x_2}(s),\quad\,\cr d\,t={x^{\prime}}(s)+i{x^{\prime}}(s)\,d\,s,\quad z={x_1}+i{x_2}\end{array}$$

is a kernel with a weak singularity, that is, it can be written as a fraction

$$K(s-{s_0})=\frac{{a(s,{s_0})}}{{|s-{s_0}{|^{\alpha }}}},0\ <\ \alpha\ <\ 1$$

where \(a(s,{s_0})\) is a bounded function. It can be proved (Mikhlin [11]) that if a kernel of an integral operator has a weak singularity, all iterated kernels, beginning from some, are bounded. Hence, the equations with weakly singular kernels are Fredholm.

The Traction Boundary Value Problem

Let \(\mathbf{n}=({n_1},{n_2}),{n_1}=-{x^{\prime}}(s),{n_2}={{x^{\prime}}_1}(s)\) be the internal normal vector to a rectifiable Jordan boundary ∂Q of length L > 0 of a simply connected region Q in the plane R 2 with a boundary of the class \({C^{{1,\lambda }}}(0,L),0\,< \,\lambda\, <\, 1\). In another words, it is assumed that the functions \({x_1}(s),{x_2}(s)\), defining a boundary, are continuous and continuously differentiable, their derivatives satisfy the Hölder condition. Assume (without loss of generality) that the origin of coordinates belongs to a region Q. Tractions are written at a boundary as

$$\begin{array}{ll} {\sigma_{11 }}{n_1}+{\sigma_{12 }}{n_2}{|_{{\partial Q}}}={g_1}(s) \cr {\sigma_{12 }}{n_1}+{\sigma_{22}}{n_2}{|_{{\partial Q}}}={g_2}(s) \cr \end{array}$$
(5)

where s is the arc-length parameter. Put \({t_k}(s)={x_1}(s)+{\mu_k}{x_2}(s),k=1,2\); prime sign below denotes the derivative with respect to s. If stresses \({\sigma_{ij }},i,j=1,2\) are square integrable in Q, the solution of the traction problems (1) and (5) is unique up to a rigid displacement. Indeed, the volume strain energy

$$\begin{array}{ll}2a(u,u)= \int\limits_Q \left\{{{a_{11 }}\sigma_{11}^2+2{a_{12 }}{\sigma_{11 }}{a_{22 }}+2{a_{16}}{\sigma_{11 }}{\sigma_{12 }}}\right.\\ \qquad \qquad \,\left.+{a_{22}}\sigma_{22}^2+2{a_{26 }}{\sigma_{22 }}{\sigma_{12 }}+{a_{66}}\sigma_{12}^2 \right\}dx\end{array}$$

is positive defined and as a consequence there is a positive constant \(C>0\), such that

$$\begin{array}{ll} a(u,u)\geq \,C\int\limits_Q\,\left\{{\left(\frac{{{\partial^2}u}}{{\partial x_1^2}}\right)^2}+{{\left(\frac{{{\partial^2}u}}{{\partial x_2^2}}\right)^2}}\right. \\ \left.+{\left(\frac{{{\partial^2}u}}{{\partial {x_1}\partial{x_2}}}\right)^2}\right.\,d\,{x_1}\,d\,{x_2}\end{array}$$

If \(u({x_1},{x_2})\) is a solution of the homogeneous problem, stresses and strains vanish inside a region.

On the other hand, the proof of the uniqueness of a solution of problems (1) and (5), given above, uses only a piecewise smoothness of a boundary; it means that the Fredholm alternative is true for regions more general than bounded regions with a Lyapunov boundary. Boundary conditions (5) can be rewritten as

$$\begin{array}{ll}Re\{-{\mu_1}{{{t^{\prime}}}}({s_0}){{{\Phi^{\prime}}}{\hskip-1.5pt_1}}({t_1}({s_0}))-{\mu_2}{{{t^{\prime}}}{\hskip-1.7pt_2}({s_0}){{\Phi^{\prime}}}({t_2}({s_0}))\}}\cr={g_1}({s_0}) \\ Re\{{{{t^{\prime}}}}({s_0}){{{\Phi^{\prime}}}}({t_1}({s_0}))+{{{t^{\prime}}}}({s_0}){{{\Phi^{\prime}}}}({t_2}({s_0}))\}={g_2}({s_0})\cr \end{array}$$
(6)

Here \({t_k}({s_0})={x_1}({s_0})+{\mu_k}{x_2}({s_0}),k=1,2\). Write the functions \({{\Phi^{\prime}}{\hskip-1.5pt_1}}({z_1}),{{\Phi^{\prime}}{\hskip-1.5pt_2}}({z_2})\) as Cauchy-type integrals with unknown densities \({b_k}(s),k=1,2\):

$${{\Phi^{\prime}}_k}({z_k})=\frac{1}{{\pi i}}\int\limits_{{\partial Q}} \frac{{{b_k}(s){{{({t_k}(s))}}^{-1 }}ds}}{{{t_k}-{z_k}}},k=1,2$$

Densities \({b_k}(s),k=1,2\) are determined from the simple system of equations

$$-{\mu_1}{b_1}-{\mu_2}{b_2}={f_1}(s),\quad {b_1}+{b_2}={f_2}(s)$$
(7)

Here the functions \({f_k}(s),k=1,2\) are real. As result, the first-order derivatives are rewritten in terms of variables \({z_1},{z_2}\) in the required form. These computations can be done only if the inequality \({\mu_1}-{\mu_2}\ne 0\) is satisfied. The latter inequality is equivalent to the condition of Ya. B. Lopatinskii. Having solved it, we shall get that

$$\begin{array}{ll}{{{\Phi^{\prime}}}_1}({z_1})=-\frac{1}{{\pi i({\mu_1}-{\mu_2})}}\int\limits_{{\partial Q}}\frac{{({\,f_1}+{\mu_2}\,{f_2}){{{[{{{t^{\prime}}}{\hskip-2pt_1}}(s)]}}^{-1}}d{t_1}}}{{{t_1}-{z_1}}} \\{{{\Phi^{\prime}}}_2}({z_2})=\frac{1}{{\pi i({\mu_1}-{\mu_2})}}\int\limits_{{\partial Q}}\frac{{({f_1}+{\mu_1}{f_2}){{{[{{{t^{\prime}}}{\hskip-2pt_2}}(s)]}}^{-1}}d{t_2}}}{{{t_2}-{z_2}}} \cr \end{array}$$
(8)

Recall the formulae of Sokhotski-Plemelj [2]

$$\begin{array}{ll} \mathop{\lim}\limits_{{{z_j}\to{t_j}}}\frac{1}{{\pi i}}\int\limits_{{\partial Q}} \varphi(s)\frac{{d{t_j}}}{{{t_j}-{z_j}}}=\varphi ({s_0})+\frac{1}{{\pi i}}\int\limits_{{\partial Q}} \varphi(s)\frac{{d{t_j}}}{{{t_j}-{t_{j0 }}}}\end{array}$$
(9)

if \(z\in {Q_i}\), and

$$\begin{array}{ll} \mathop{\lim}\limits_{{{z_j}\to{t_j}}}\frac{1}{{\pi i}}\int\limits_{{\partial Q}}\,\varphi(s)\frac{{\,d\,{t_j}}}{{{t_j}-{z_j}}}= -\varphi (s)+\frac{1}{{\pi i}}\int\limits_{{\partial Q}}\,\varphi (s) \\\times\frac{{\,d\,{t_j}}}{{{t_j}-{t_{j0 }}}}\end{array}$$

if \(z\in {Q_e}\). Here \({Q_i}=Q\), and \({Q_e}\) is the region, external to Q, \({t_{j0 }}={t_j}({s_0})\). Of course, it is assumed that \(\varphi (s)\in {C^{{0,\lambda }}}(\partial Q)\). It follows that

$${\sigma_{kj }}{{(\mathbf{ u}(x,\mathbf{ f}){n_j})}_i}({s_0})-{\sigma_{kj }}{{(\mathbf{ u}(x,\mathbf{ f}){n_j})}_e}({s_0})=2{f_k}({s_0}),k=1,2$$

Here \({\sigma_{kj }}{{(\mathbf{ u}(x,\mathbf{ f}){n_j})}_i}({s_0})\) is the limiting value of the traction vector inside a region, and, correspondingly, \({\sigma_{kj }}{{(\mathbf{ u}(x,\mathbf{ f}){n_j})}_e}({s_0})\) is its limiting value outside a region, and \(\mathbf{ u}(x,\mathbf{ f})\) is the vectorial simple-layer potential defined below by formulas (12) and (13).

$$\begin{array}{ll}{f_1}({s_0})+Re\frac{{{\mu_1}{{{t^{\prime}}}{\hskip-2pt_1}}({s_0})}}{{\pi i({\mu_1}-{\mu_2})}}\int\limits_{{\partial Q}}\frac{{({\,f_1}+{\mu_2}\,{f_2}){{{[{{{t^{\prime}}}{\hskip-2pt_1}}(s)]}}^{-1}}d{t_1}}}{{{t_1}-{t_{10 }}}}+ \\-Re\frac{{{\mu_2}{{{t^{\prime}}}{\hskip-2pt_2}}({s_0})}}{{\pi i({\mu_1}-{\mu_2})}}\int\limits_{{\partial Q}}\frac{{({\,f_1}+{\mu_1}{f_2}){{{[{{{t^{\prime}}}{\hskip-2pt_2}}(s)]}}^{-1}}d{t_2}}}{{{t_2}-{t_{20 }}}}={g_1}({s_0})\cr\end{array}$$
(10)
$$\begin{array}{ll}{f_2}({s_0})-Re\frac{{{{{t^{\prime}}}{\hskip-2pt_1}}({s_0})}}{{\pi i({\mu_1}-{\mu_2})}}\int\limits_{{\partial Q}}\frac{{({\,f_1}+{\mu_2}\,{f_2}){{{[{{{t^{\prime}}}{\hskip-2pt_1}}(s)]}}^{-1}}d{t_1}}}{{{t_1}-{t_{10 }}}} \\ +Re\frac{{{{{t^{\prime}}}{\hskip-2pt_2}}({s_0})}}{{\pi i({\mu_1}-{\mu_2})}}\int\limits_{{\partial Q}}\frac{{({f_1}+{\mu_1}{f_2}){{{[{{{t^{\prime}}}{\hskip-2pt_2}}(s)]}}^{-1}}d{t_2}}}{{{t_2}-{t_{20 }}}}={g_2}({s_0}) \cr\end{array}$$
(11)

Here \({t_{k0 }}={x_1}({s_0})+{\mu_k}{x_2}({s_0}),k=1,2\)

Existence Results

Therefore, these mutually adjoint systems are Fredholm solvable. Now, for solubility of systems (10) and (11), it is necessary and sufficient that right-hand sides of equations (10) and (11) be orthogonal to all solutions of (12) and (13), and vice versa. Therefore, it is necessary to require vanishing of the principal vector and the principal vector of acting forces. In other words,

$$\begin{array}{ll} \int\limits_{{\partial Q}}\,{g_k}(s)\,d\,s=0,k=1,2\cr \int\limits_{{\partial Q}}\,({x_1}(s){g_2}(s)-{x_2}(s){g_1}(s))\,d\,s=0\end{array}$$

It is easy to verify its fulfillment. Indeed, multiply equations (10) and (11) on \(d{s_0}\) and integrate the result with respect to \({s_0}\), bearing in mind that

$$\int\limits_{{\partial Q}}\,\frac{{{{{t^{\prime}}}_{k0 }}\,d\,{s_0}}}{{{t_k}-{t_{k0 }}}}=-\pi i$$

Then left-hand sides of (10) and (11) become zero and we obtain two first equalities. In a similar way, multiply the first equation on \({x_1}({s_0})\), the second onto \(-{x_2}({s_0})\), change the order of integration, and integrate with respect to s 0. Then you shall get the third equality in the previous formula. Hence, these equalities are necessary for solubility of these equations. The solution of the traction problem is given by the simple-layer potential, which is written as

$$\begin{array}{ll} {u_1}({x_1},{x_2},f)=Re\frac{{{b_{11 }}}}{{\pi i({\mu_1}-{\mu_2})}}\cr \int\limits_{{\partial Q}}({f_1}+{\mu_2}\,{f_2})\ln ({z_1}-{t_1})ds-Re\frac{{{b_{12 }}}}{{\pi i({\mu_1}-{\mu_2})}}\cr \int\limits_{{\partial Q}}({f_1}+{\mu_1}{f_2})\ln ({z_2}-{t_2})ds\end{array}$$
(12)
$$\begin{array}{ll} {u_2}({x_1},{x_2},f)=Re\frac{{{b_{21 }}}}{{\pi i({\mu_1}-{\mu_2})}}\cr \int\limits_{{\partial Q}} ({f_1}+{\mu_2}\,{f_2})\ln ({z_1}-{t_1})ds-Re\frac{{{b_{22 }}}}{{\pi i({\mu_1}-{\mu_2})}}\cr \int\limits_{{\partial Q}} ({f_1}+{\mu_1}{f_2})\ln ({z_2}-{t_2})ds\end{array}$$
(13)

Fix, for example, the principal branch of logarithm. Consider in more detail properties of the simple-layer potential. For existence of integrals (14) and (15) is sufficient continuity of \({f_1},{f_2}\); then \({u_1},{u_2}\) are continuous in the whole plane but have the logarithmic growth at infinity. Now,

Lemma 1

The simple-layer potential with a continuous density \(\mathbf{ f}=({f_1},{f_2})\), satisfying the equation

$$\int\limits_{{\partial Q}}\,{f_i}\,d\,s=0,i=1,2$$

satisfies also the estimates

$$\begin{array}{ll} |{u_i}(x,f)|< \frac{c}{|x| },\quad \bigg|\frac{{\partial {u_i}}}{{\partial {x_j}}}\bigg|< \frac{{{c_1}}}{{|x{|^2}}},i,j=1,2,|x|\sqrt{{x_1^2+x_2^2}}.\end{array}$$

Indeed, consider the typical integral, entering into (14) and (15):

$$\int\limits_{{\partial Q}}\,f(s)\ln ({z_k}-{t_k})\,d\,s$$

Here f(s) is any density. Write it as a derivative:

$$f(s)=\frac{d}{ds }\left(\int\limits_0^s\,f(s)\,d\,s\right)=\frac{d}{ds}\tau (s)$$

After integration by parts we shall get that

$$\int\limits_{{\partial Q}}\,\frac{{d\tau }}{ds}\ln ({z_k}-{t_k})\,d\,s=\tau (s)\ln ({z_1}-{t_1}){|_{{\partial Q}}}-\int\limits_{{\partial Q}}\,\tau (s)\frac{{\,d\,{t_k}}}{{{t_k}-{z_k}}}$$

The first summand in the right-hand side vanishes if \(\tau (L)=0\), and the second summand is a single-valued Cauchy-type integral. The equality \(\tau (L)=0\) means that conditions of the lemma are satisfied. Other assertions are obvious. Therefore, functions \({v_1}({x_1},{x_2}),{v_2}({x_1},{x_2})\) are also single valued. Put \(\mathbf{ f}=({f_1},{f_2})\). Introduce now a short notation for a simple-layer potential: put

$$\begin{array}{ll}\mathbf{ u}(x,\mathbf{ f}) =({u_1}(x,\mathbf{ f}),{u_2}(x,\mathbf{ f})), \cr {u_i}(x,\mathbf{ f}) =\sum\limits_{j=1}^2\,\int\limits_{{\partial Q}}\,{G_{ij }}{f_j}(s)\,d\,s,i=1,2\end{array}$$

understanding by \({G_{ij }},i,j=1,2\) kernels of integral operators in (10) and (11). Or, quite shortly, put

$$\mathbf{ u}(x,\mathbf{ f})=\int\limits_{{\partial Q}}\,G(x-y)\mathbf{ f}\,d\,s$$

where G(x) is the matrix \(({G_{ij }}(x)),i,j=1,2\)

Lemma 2

The system of integral equations (10) and (11) has exactly three linearly independent solutions:

$${{\mathbf{ f}}_1}=(1,0),\quad {{\mathbf{ f}}_2}=(0,1),{{\mathbf{ f}}_3}=(-{x_2}(s),{x_1}(s))$$

Indeed, if it is not so, then the system of equations (12) and (13), adjoint to (10) and (11), would have more than three linearly independent solutions \({\varphi_k}(s),1\leq k,k>3\). To any of them answer a simple-layer potential \(\mathbf{ v}(x,{f_k})\), satisfying conditions \({\sigma_{ij }}(\mathbf{ v}(x,{f_k}){n_j}=0\) at the boundary. Let there be one more solution \({f_4}(s)\), linearly independent from others; then

$$f(s)={f_4}(s)-\sum\limits_{j=1}^3\,{c_j}\,{f_j}(s)$$

where \({c_j},j=1,2,3\) are arbitrary real constants, is also a solution of the homogeneous system of equations (10) and (11). Write up simple-layer potentials:

$$\mathbf{ u}(x,{f_4}),\mathbf{ u}(x,{f_j}),j=1,2,3$$

As solutions of homogeneous boundary value problems, they are rigid displacement vectors and so

$$\mathbf{ u}(x)=\mathbf{ u}(x,{f_4})-\sum\limits_{j=1}^3\,\mathbf{ u}(x,{f_j})$$

is a rigid displacement vector. Now choose constants \({c_1},{c_2},{c_3}\) to satisfy the conditions

$$\mathbf{ u}(0)=0,\;\frac{{\partial {u_2}}}{{\partial {x_1}}}-\frac{{\partial {u_1}}}{{\partial {x_2}}}=0$$

But then \({v^k}\) satisfies the system

$$\frac{{\partial {u_i}}}{{\partial {x_j}}}+\frac{{\partial {u_j}}}{{\partial {x_i}}}=0,i,j=1,2$$

Rewrite these conditions as

$$\sum\limits_{j=1}^3\,{c_j}\int\limits_{{\partial Q}}\,G({x_2}){\varphi_j}(s)\,d\,s=\int\limits_{{\partial Q}}\,G({x_2})\,d\,s$$
$$\sum\limits_{j=1}^3\,{c_j}\int\limits_{{\partial Q}}\,{G_0}({x_2}){\varphi_j}(s)\,d\,s=\int\limits_{{\partial Q}}\,{G_0}({x_2})\,d\,s$$

where

$${G_0}({x_2})=\frac{{\partial {G_2}}}{{\partial {x_2}}}-\frac{{\partial {G_1}}}{{\partial {x_1}}}$$

The determinant of this system is different from zero by linear independence of \({\varphi_j},j=1,2,3\). Solve this system with respect to \({c_j},j=1,2,3\). Then \(u({x_1},{x_2})=0,({x_1},{x_2})\in {Q_i}\). Continuity of a potential in the whole plane implies that \(u(x)=0\) at ∂Q. As \(u(x,)\) satisfies conditions of the lemma 1, then at infinity

$$|u({x_1},{x_2})|\; <\; \frac{c}{{\sqrt{{x_1^2+x_2^2}}}},|{\sigma_{ij }}(u(x))|\;<\; \frac{{{c_1}}}{{x_1^2+x_2^2}}$$

and so \(u({x_1},{x_2})=0,({x_1},{x_2})\in {Q_e}\). But then according to the Sokhotski-Plemelj formula, the jump of the traction vector is equal zero at the boundary and then \({f_4}=0\). This is a contradiction.

Denote a solution of the internal boundary value problem by \((I,{Q_i})\) and the solution of the external by \((I,{Q_e})\). Then

Lemma 3

The problem \((I,{Q_i})\) has unique solution (up to a linear combination) \({c_1}{\varphi_1}+{c_2}{\varphi_2}\;+{c_3}{\varphi_3}\), where

$${\varphi_1}=(0,1),{\varphi_2}=(1,0),{\varphi_3}=(-{x_2},{x_1})$$

Functions \({\varphi_k},k=1,2,3\) are a complete set of solutions of the system

$$\frac{{\partial {v_i}}}{{\partial {x_k}}}+\frac{{\partial {v_k}}}{{\partial {x_i}}}=0,i,k=1,2$$

Its proof follows from the integral identity

$$\int\limits_{{{Q_i}}}\,{\sigma_{ij }}(\mathbf{ v}){{\mathbf{ e}}_{\mathbf{ i}\mathbf{ j}}}(\mathbf{ v})\,\mathbf{ d}\,\mathbf{ x}=\int\limits_{{\partial \mathbf{ Q}}}\,{\sigma_{\mathbf{ i}\mathbf{ j}}}(\mathbf{ v}){{\mathbf{ v}}_{\mathbf{ i}}}{\nu_{\mathbf{ j}}}\,\mathbf{ d}\,\mathbf{ s}$$

Here ν is the external normal vector of Q i . For functions v, with properties

$$|\mathbf{ v}|\leq c,|\frac{{\partial {v_k}}}{{\partial {x_i}}}|\leq \frac{c}{{|x{|^2}}},i,k=1,2$$
(14)

for large |x| and satisfying the homogeneous system of elastic equations in the unbounded region, Q e holds the integral identity

$$\int\limits_{{{Q_e}}}\,{\sigma_{ij }}(\mathbf{ v}){e_{ij }}(\mathbf{ v})\,d\,x=-\int\limits_{{\partial Q}}\,{\sigma_{ij }}(\mathbf{ v}){v_i}{\nu_j}\,d\,s$$
(15)

The last equality is obtained from the previous integral identity for a bounded region. Indeed, consider the region lying between the boundary ∂Q and the circumference of sufficiently large radius R, with a center lying inside Q i . The previous equality holds for this region. The integral, taken along the external circumference, tends to zero, when \(R\to \infty\). Hence, equality (15) follows from the given above assertion and the absolute convergence of the integral with respect to Q e . This implies the following lemma.

Lemma 4

The solution of the problem \((I,{Q_e})\) in the class of functions, subject to (1.13), is unique up to a constant vector, and if \(\mathbf{ u}\to \mathbf{ 0}\), when \(|x|\to \infty\), then it is unique. Notice that the system of equations (10) and (11) can be modified to make it uniquely solvable. Indeed, introduce to equation (10) summands

$$\begin{array}{ll} -Re\frac{{{\mu_1}}}{{2\pi i({\mu_1}-{\mu_2})}}\frac{{{{{t^{\prime}}}{\hskip-2pt_1}}({s_0})}}{{{t_1}({s_0})}}\int\limits_{{\partial Q}}\,(\,{f_1}+{\mu_2}\,{f_2})\,d\,s\cr - Re\frac{{{\mu_2}}}{{2\pi i({\mu_1}-{\mu_2})}}\frac{{{{{t^{\prime}}}{\hskip-2pt_2}}({s_0})}}{{{t_2}({s_0})}}\int\limits_{{\partial Q}}\,(\,{f_1}+{\mu_1}{f_2})\,d\,s \cr -Re\frac{1}{{4\pi i}}\left\{-{\mu_1}\frac{\partial }{{\partial {s_0}}}\frac{1}{{{t_1}({s_0})}}-{\mu_2}\frac{\partial }{{\partial {s_0}}}\frac{1}{{{t_2}({s_0})}}\right\}M \cr \end{array}$$

and equation (11) by summands

$$\begin{array}{ll} -Re\frac{1}{{2\pi i({\mu_1}-{\mu_2})}}\frac{{{{{t^{\prime}}}{\hskip-2pt_1}}({s_0})}}{{{t_1}({s_0})}}\int\limits_{{\partial Q}}\,(\,{f_1}+{\mu_2}\,{f_2})\,d\,s \cr +Re\frac{1}{{2\pi i({\mu_1}-{\mu_2})}}\frac{{{{{t^{\prime}}}{\hskip-2pt_2}}({s_0})}}{{{t_2}({s_0})}}\int\limits_{{\partial Q}}\,(\,{f_1}+{\mu_1}{f_2})\,d\,s\cr - Re\frac{1}{{4\pi i}}\left\{\frac{\partial }{{\partial {s_0}}}\frac{1}{{{t_1}({s_0})}}+\frac{\partial }{{\partial {s_0}}}\frac{1}{{{t_2}({s_0})}}\right\}M \cr \end{array}$$

Here M is a real constant. As

$$\int\limits_{{\partial Q}}\,\frac{{d{t_k}({s_0})}}{{{t_k}({s_0})}}=2\pi i,k=1,2$$

(the origin of coordinates is inside a region), it follows that

$$\int\limits_{{\partial Q}}\,{f_k}(s)\,d\,s=\int\limits_{{\partial Q}}\,{g_k}(s)\,d\,s,k=1,2$$
$$M=\int\limits_{{\partial Q}}\,(-{x_2}(s){g_1}(s)+{x_1}(s){g_2}(s))\,d\,s$$

It is clear that if the principal vector and the principal moment of applied forces are equal to zero, the system of equations with these applied summands is equivalent to the system of equations (10) and (11). It is necessary to assume that \({g_k}(s)\in {C^{{0,\lambda }}}(\partial Q),k=1,2\). Then \({f_k}(s)\in {C^{{0,\lambda }}}(\partial Q),k=1,2\). Moreover, it is not quite obvious that it is the system of regular equations. Rewrite equations (10) and (11) as

$$\begin{array}{ll} {f_1}({s_0})+Re\frac{{{{{t^{\prime}}}{\hskip-2pt_1}}({s_0})}}{{\pi i}}\int\limits_{{\partial Q}} {f_1}(s)\frac{{{{{[{{{t^{\prime}}}}(s)]}}^{-1 }}d{t_1}}}{{{t_1}-{t_{10 }}}}\cr +Re\frac{{{\mu_2}}}{{\pi i({\mu_1}-{\mu_2})}}\int\limits_{{\partial Q}} ({f_1}+{\mu_1}{f_2})\Bigg\{\frac{{{{{t^{\prime}}}}({s_0})}}{{{t_2}-{t_{20 }}}}\cr -\frac{{{{t^{\prime}}}({s_0})}}{{t_1}-{t_{10 }}}\Bigg\}ds={g_1}({s_0})\end{array}$$
(16)
$$\begin{array}{ll} {f_2}({s_0})+Re\frac{{{{{t^{\prime}}}{\hskip-2pt_2}}({s_0})}}{{\pi i}}\int\limits_{{\partial Q}} {f_2}(s)\frac{{{{{({{{t^{\prime}}}{\hskip-2pt_2}}(s))}}^{-1 }}d{t_1}}}{{{t{_2}}-{t_{20 }}}}\cr +Re\frac{1}{{\pi i({\mu_1}-{\mu_2})}}\int\limits_{{\partial Q}} (\,{f_1}+{\mu_2}\,{f_2})\Bigg\{\frac{{{{{t^{\prime}}}{\hskip-2pt_1}}({s_0})}}{{{t_1}-{t_{10 }}}}\cr -\frac{{{{{t^{\prime}}}{\hskip-2pt_2}}({s_0})}}{{{t_2}-{t_{20 }}}}\Bigg\}ds={g_2}({s_0})\end{array}$$
(17)

Therefore, it is the system of regular equations. Formulate now the principal result of the previous investigation.

Theorem 1

Assume that

$$\begin{array}{ll}{g_k}(s)\in {C^{{0,\lambda }}}(\partial Q),k=1,2,\partial Q \\ \quad \quad \in {C^{{1,\lambda }}}(0,L),0\ <\ \lambda\ < \ 1.\end{array}$$

Under the given above assumptions, equality to zero of the principal vector and the principal moment of applied forces is the necessary and sufficient condition for existence of the solution of the system (10) and (11) belonging to \({C^{{0,\lambda }}}(0,L)\). Then \({u_k}(x)\in {C^{{1,\lambda }}}(\overline{Q}),k=1,2.\)

Indeed, let us seek the solution as a single-layer potential \(\mathbf{ u}=({u_1}(x,\mathbf{ f}),{u_2}(x,\mathbf{ f}))\). Then density \(\mathbf{ f}=({f_1},{f_2})\) satisfies systems (10) and (11). As two pairs of adjoint integral equations (10) and (11) and (12) and (13) are Fredholm solvable, the system of equations (10) and (11) is solvable if and only if the vector function \(\mathbf{ g}=({g_1},{g_2})\) is orthogonal to all solutions of the homogeneous system (12) and (13). By lemma 2 the general solution of the homogeneous system (12) and (13) is a linear combination of the three vector functions \(\varphi (x)={c_1}(1,0)+{c_2}(0,1)+{c_3}(-{x_2},{x_1})\). Conditions (1.10) are conditions of orthogonality of a solution of a homogeneous system (10) and (11) to the vector function \(\mathbf{ g}=({g_1},{g_2})\), and so they guaranty the existence of the system of (10) and (11). Necessity follows from Betti’s formula. Indeed, we have the conditions

$$0=\int\limits_{{\partial Q}}\,{\sigma_{ij }}{n_j}{\varphi_i}\,d\,s=\int\limits_{{\partial Q}}\,{g_i}{\varphi_i}\,d\,s$$

which coinciding with the equilibrium conditions. It is not hard to see that smoothness of solution grows up according to the growth of a boundary and boundary data. Now,

Theorem 2

Assume that

$$\begin{array}{ll}{g_k}(s)\in {C^{{l,\lambda }}}(\partial Q),k=1,2,\partial Q \\ \quad \quad \in {C^{{l+1,\lambda }}}(0,L),0\ <\ \lambda\ <\ 1,l\geq 1.\end{array}$$

Then \({C^{{l+1,\lambda }}}(\bar{Q})\). It follows immediately from the properties of the Cauchy-type integral (see theorem 1.10 from [11]).

The Case of an Isotropic Material

For an isotropic material,

$$\begin{array}{ll}{a_{11 }}={a_{22 }}=\frac{1}{E},{a_{12 }}=\frac{{-\nu }}{E},{a_{66 }}=\frac{1}{G},{a_{16 }}={a_{26 }}=0\end{array}$$

Here E is the Young’s modulus, G is the shear modulus, and ν is the Poisson’s ratio. The limiting transition, when \({\mu_1},{\mu_2}\to i\), is easily performed. As a result, we get the system of equations for an isotropic material:

$$\begin{array}{ll}{f_1}({s_0})+Re\frac{{t^{\prime}({s_0})}}{{\pi i}}\int\limits_{{\partial Q}} \frac{{{f_1}(s){{{[{t}^{\prime}(s)]}}^{-1 }}dt}}{{t-{t_0}}}\cr +Re\frac{1}{{2\pi i}}\int\limits_{{\partial Q}} ({\,f_1}+i{f_2})\frac{{(\bar{t}-\mathop{\bar{t}}\nolimits_0)dt-(t-{t_0})d\bar{t}}}{{{{{(t-{t_0})}}^2}}}\cr \quad={g_1}({s_0})\end{array}$$
(18)
$$\begin{array}{ll}{f_2}({s_0})+Re\frac{{t^{\prime}({s_0})}}{{\pi i}}\int\limits_{{\partial Q}} \frac{{{f_2}(s){{{[{t}^{\prime}(s)]}}^{-1 }}dt}}{{t-{t_0}}}\cr +Re\frac{i}{{2\pi i}}\int\limits_{{\partial Q}} ({\,f_1}+i{f_2})\frac{{(\bar{t}-\mathop{\bar{t}}\nolimits_0)dt-(t-{t_0})d\bar{t}}}{{{{{(t-{t_0})}}^2}}}\ ={g_2}({s_0})\end{array}$$
(19)

Here \(z={x_1}+i{x_2},{t_0}={x_1}({s_0})+i{x_2}({s_0})\). It is clear that this system can be reduced to one (complex) equation with the respect to the complex density \(\omega (s)={f_1}(s)+i{f_2}(s)\). In the complex notation it is equivalent to the system of equations from [5]. Denote by \(u_1^1(x),u_2^1(x)\) displacements for an isotropic material. Then

$$\begin{array}{ll} {u_1^1(x)=Re\Bigg\{\frac{2}{E}\frac{1}{\pi}\int\limits_{{\partial Q}}{f_1}(s)\ln (z-t)ds+ \frac{{1+\nu}}{E}\frac{1}{\pi}} \cr \hskip 30pt \int\limits_{{\partial Q}} {f_1}(s)\frac{{i(\eta-{x_2})}}{t-z }ds\bigg]+ Re\bigg[\frac{{1-\nu}}{E}\frac{1}{\pi}\cr \hskip 30pt\int\limits_{{\partial Q}} i{f_2}(s)\ln (z-t)ds+ \frac{{1+\nu }}{E}\frac{1}{\pi}\cr \hskip 30pt\int\limits_{{\partial Q}}{f_2}(s)\frac{{i(\eta -{x_2})}}{t-z }ds \cr \end{array}$$
(20)
$$\begin{array}{ll} u_2^1(x)= Re\Bigg\{\frac{{1-\nu }}{E}\frac{1}{{\pi_j}}\int\limits_{{\partial Q}} {f_1}(s)\ln (z-t)ds\cr - \frac{{1+\nu}}{E}\frac{1}{{\pi i}}\int\limits_{{\partial Q}}{f_1}(s)\frac{{i(\eta -{x_2})}}{t-z }ds\Bigg]+ Re\Bigg[\frac{2}{E}\frac{1}{{\pi i}}\cr \int\limits_{{\partial Q}}i{f_2}(s)\ln (z-t)ds- \frac{{1+\nu }}{E}\frac{1}{{\pi_j}}\int\limits_{{\partial Q}} i{f_2}(s)\frac{{i(\eta -{x_2})}}{t-z}ds\Bigg\} \cr \end{array}$$
(21)

Here \(\xi ={x_1}(s),\eta ={x_2}(s)\)

The Displacement Boundary Value Problem

Here we discuss (in short) the displacement boundary value problem. As in the previous section, consider a bounded simply connected region Q with the boundary ∂Q of the Lyapunov class and prescribe displacements at the boundary. Consider the boundary value problem of determination of the stress-strain state when displacements

$$\begin{array}{ll} Re\left\{ {{b_{11 }}{\Phi_1}({z_1})+{b_{12 }}{\Phi_2}({z_2})} \right\} ={g_1}({s_0})\cr Re\left\{ {{b_{21 }}{\Phi_1}({z_1})+{b_{22 }}{\Phi_2}({z_2})} \right\} ={g_2}({s_0}) \\ \qquad \quad {g_k}({s_0})\in {C^{{0,\alpha }}}(\partial Q), \ k =1,2\end{array}$$

are prescribed at the boundary. First-order derivatives of displacements should be square integrable in the closed region \(\bar{Q}\), as otherwise a solution of this problem is non-unique. Introduce the functions \({\Phi_k}({z_k})(k=1,2)\) as Cauchy-type integrals

$${\Phi_k}({z_k})=\frac{1}{{\pi i}}\int\limits_{{\partial Q}} \frac{{{\omega_k}(s)d{t_k}}}{{{t_k}-{z_k}}}k=1,2$$

where \({t_k}={x_1}(s)+{\mu_k}{x_2}(s)k=1,2\).

Let \({f_k}(s)(k=1,2)\) be two real functions. Solve the system of equations:

$$\begin{array}{ll} {b_{11 }}{\omega_1}+{b_{12 }}{\omega_2}={f_1}(s),\quad {b_{21 }}{\omega_1}+{b_{22 }}{\omega_2}={f_2}(s)\end{array}$$

Put

$$\begin{array}{ll} {u_1}{|_{{\partial Q}}}={g_1}(s),\quad {u_2}{|_{{\partial Q}}}={g_2}(s),{g_k}(s)\in {C^{{0,\alpha }}}(\partial Q)\end{array}$$
(22)
$${\Phi_k}({z_k})=\frac{1}{{\pi i}}\int\limits_{{\partial Q}}\,\frac{{{\omega_k}(s)\,d\,{t_k}}}{{{t_k}-{z_k}}}\quad k=1,2$$

Write \({\varphi_k}({z_k}),k=1,2\) as Cauchy-type integrals with complex densities \({m_k}(s)\):

$${\varphi_k}({z_k})=\frac{1}{{\pi i}}\int\limits_{{\partial Q}}\,\frac{{{\omega_k}(s)\,d\,{t_k}}}{{{t_k}-{z_k}}},k=1,2$$

Just as for the traction problem, by means of the Plemelj formula, we obtain the system of linear equations \({m_k}(s),k=1,2\),

$${b_{11 }}{\omega_1}+{b_{12 }}{\omega_2}={g_1},{b_{21 }}{\omega_1}+{b_{22 }}{\omega_2}={g_2}$$
(23)

and solve it by the Cramer’s rule. Its determinant is

$$\delta =\frac{{{\mu_1}-{\mu_2}}}{{{\mu_1}{\mu_2}}}[{\mu_1}{\mu_2}({a_{11 }}{a_{22 }}-a_{12}^2)-{a_{22 }}{a_{66 }}]$$

and it is distinct from zero if \({\mu_1}\ne {\mu_2}\). Put \(\delta =\xi ({\mu_1}-{\mu_2})\). Here ξ depends only on the symmetric functions of roots of the characteristic equation and, therefore, on elastic coefficients. It is assumed further that \(\xi \ne 0\). It is always different from zero in elastic problems by positivity of the stress energy. For example, for an orthotropic material \(({a_{16 }}=0,{a_{26 }}=0\), it is equal to

$$\begin{array}{ll} \xi ={{({\gamma_1}{\gamma_2})}^{-1 }}\left\{({a_{11 }}{a_{22 }}-a_{12}^2)\sqrt{{{a_{22 }}a_{11}^{-1 }}}+{a_{22 }}{a_{66 }}\right\}\end{array}$$

and it is always positive, as

$${a_{11 }}{a_{22 }}-a_{12}^2>0,{a_{11 }}>0,{a_{22 }}>0,{a_{66 }}>0$$

For an isotropic material \(\xi =(1+\nu )(3-\nu ){E^{-2 }}\), where E is the Young’s modulus and ν is the Poisson’s ratio.

Hence, \({u_k}({x_2},{x_2}),k=1,2\) can be written as

$$\begin{array}{ll} {u_1}({x_1},{x_2})=Re\frac{1}{{\pi i}}\int\limits_{{\partial Q}}\,\frac{{{f_1}(s)\,d\,{t_1}}}{{{t_1}-{z_1}}}\cr + Re\frac{{{b_{12 }}}}{{\pi i\delta }} \int\limits_{{\partial Q}}\,(-{b_{21 }}{f_1}(s)+{b_{11 }}{f_2}(s))\left( {\frac{{\,d\,{t_2}}}{{{t_2}-{z_2}}}-\frac{{\,d\,{t_1}}}{{{t_1}-{z_1}}}} \right) \end{array}$$
$$\begin{array}{ll} {u_2}({x_1},{x_2})=Re\frac{1}{{\pi i}}\int\limits_{{\partial Q}}\,\frac{{{f_2}(s)\,d\,{t_2}}}{{{t_2}-{z_2}}}\cr + Re\frac{{{b_{21 }}}}{{\pi i\delta }}\int\limits_{{\partial Q}}\,({b_{22 }}{f_1}(s)-{b_{12 }}{f_2}(s))\left( {\frac{{\,d\,{t_1}}}{{{t_1}-{z_1}}}-\frac{{\,d\,{t_2}}}{{{t_2}-{z_2}}}} \right) \cr \end{array}$$

As a result, we arrive to the system of regular integral equations:

$$\begin{array}{ll} {f_1}({s_0}) +Re\frac{1}{{\pi i}}\int\limits_{{\partial Q}} \frac{{{f_1}(s)d{t_1}}}{{{t_1}-{t_{10 }}}}+Re\frac{{{b_{12 }}}}{{\pi i\delta }}\int\limits_{{\partial Q}} (-{b_{21 }}{f_1}(s) \\ \quad \quad \quad +{b_{11 }}{f_2}(s))\left( {\frac{{d{t_2}}}{{{t_2}-{t_{20 }}}}-\frac{{d{t_1}}}{{{t_1}-{t_{10 }}}}} \right)={g_1}({s_0})\end{array}$$
(24)
$$\begin{array}{ll} {f_2}({s_0}) +Re\frac{1}{{\pi i}}\int\limits_{{\partial Q}} \frac{{{f_2}(s)d{t_2}}}{{{t_2}-{t_{20 }}}}+Re\frac{{{b_{21 }}}}{{\pi i\delta }}\int\limits_{{\partial Q}} ({b_{22 }}\,{f_1}(s) \\ \quad \quad \quad -{b_{12 }}\,{f_2}(s))\left( {\frac{{d{t_1}}}{{{t_1}-{t_{10 }}}}-\frac{{d{t_2}}}{{{t_2}-{t_{20 }}}}} \right)={g_2}({s_0})\end{array}$$
(25)

The Case of an Isotropic Solid

Now, in the limiting transition to an isotropic material due to the presence of the factor \({\mu_1}-{\mu_2}\) in the denominator δ, the difference

$$\frac{1}{{({\mu_1}-{\mu_2})}}\left( {\frac{{\,d\,{t_2}}}{{{t_2}-{z_2}}}-\frac{{\,d\,{t_1}}}{{{t_1}-{z_1}}}} \right)$$

tends in the limit to the difference

$$\frac{{{{{x^{\prime}}}_1}(s)({x_2}(s)-{x_2})-{{{x^{\prime}}}_2}(s)({x_1}(s)-{x_1})}}{{{(t-z)^2}}}$$

and the limiting values of displacements are expressed as

$$\begin{array}{ll} u_1^0= Re\frac{1}{{\pi i}}\int\limits_{{\partial Q}}\,\frac{{{f_1}(s)\,d\,t}}{t-z }- Re\frac{i}{{\pi i}}\frac{{1+\nu }}{{3-\nu }} \\ \qquad \int\limits_{{\partial Q}}\,({f_1}+i{f_2})\frac{{{{{x^{\prime}}}{\hskip-2pt_1}}(s)({x_2}(s)-{x_2})-{{{x^{\prime}}}{\hskip-2pt_2}}(s)({x_1}(s)-{x_1})}}{{{(t-z)^2}}}\,d\,s \cr \end{array}$$
$$\begin{array}{ll} u_2^0= Re\frac{1}{{\pi i}}\int\limits_{{\partial Q}}\,\frac{{{f_2}(s)\,d\,t}}{t-z }- Re\frac{1}{{\pi i}}\frac{{1+\nu }}{{3-\nu }} \\ \quad \quad \int\limits_{{\partial Q}}\,({f_1}+i{f_2})\frac{{{{{x^{\prime}}}{\hskip-2pt_1}}(s)({x_2}(s)-{x_2})-{{{x^{\prime}}}{\hskip-2pt_2}}(s)({x_1}(s)-{x_1})}}{{{(t-z)^2}}}\,d\,s \cr \end{array}$$

where the superscript 0 refers to the field of displacements in an isotropic material. We get the formulae and the integral equation of D. I. Sherman for an isotropic material. The author sincerely thanks the editor, Professor Dorin Iesan, for careful editing of this entry.