Georgian Mathematical Journal Volume 8 (2001), Number 3, 427–446

A NEW METHOD OF SOLVING THE BASIC PLANE BOUNDARY VALUE PROBLEMS OF STATICS OF THE

ELASTIC MIXTURE THEORY

M. BASHELEISHVILI AND K. SVANADZE

Abstract. The basic plane boundary value problems of statics of the elastic mixture theory are considered when on the boundary are given: a displacement vector (the first problem), a stress vector (the second problem); differences of partial displacements and the sum of stress vector components (the third problem). A simple method of deriving Fredholm type integral equations of second order for these problems is given. The properties of the new operators are established. Using these operators and generalized Green formulas we investigate the above-mentioned integral equations and prove the existence and uniqueness of a solution of all the boundary value problems in a finite and an infinite domain.

2000 Mathematics Subject Classification: 74B05. Key words and phrases: Conjugate vectors, conjugate operators, Fredholm type integral equations, potentials.

1. Some Auxiliary Formulas and Operators

In the two-dimensional case, the basic homogeneous equations of statics of the elastic mixture theory have the form (see [1] and [2]):

a_{1}Δu′ + b_{1} grad div u′ + cΔu′′ + d grad div u′′ = 0,
cΔu′ + d grad div u′ + a_{2}Δu′′ + b_{2} grad div u′′ = 0,

(1.1)

where Δ is the two-dimensional Laplacian, grad and div are the principal operators of the field theory, u′ = (u′_{1}, u′_{2}) and u′′ = (u′′_{1}, u′′_{2}) are partial displacements, a_{k}, b_{k} (k = 1, 2), c, d are the known constants characterizing the physical properties of a mixture, and at that

\begin{matrix}a_{1} = μ_{1} − λ_{5}, a_{2} = μ_{2} − λ_{5}, c = μ_{3} + λ_{5}, \\ b_{1} = μ_{1} + λ_{1} + λ_{5} − ρ_{2}α_{2}/ρ, b_{2} = μ_{2} + λ_{2} + λ_{5} + ρ_{1}α_{2}/ρ, \\ d = μ_{3} + λ_{3} − λ_{5} − ρ_{1}α_{2}/ρ ≡ μ_{3} + λ_{4} − λ_{5} + ρ_{2}α_{2}/ρ, \\ ρ = ρ_{1} + ρ_{2}, α_{2} = λ_{3} − λ_{4}, \end{matrix}  (1.2)

where μ_{1}, μ_{2}, μ_{3}, λ_{1}, λ_{2}, λ_{3}, λ_{4}, λ_{5}, ρ_{1}, ρ_{2} are new constants also characterizing the physical properties of the mixture and satisfying the definite conditions (inequalities) [2].

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

In the theory of elastic mixtures, the displacement vector is usually denoted by u = (u′, u′′), while the four-dimensional vector by u = (u_{1}, u_{2}, u_{3}, u_{4}) or u_{1} = u′_{1}, u_{2} = u′_{2}, u_{3} = u′′_{1}, u_{4} = u′′_{2}.

The system of basic equations (1.1) can be rewritten (equivalently) as follows:

a_{1}Δu′ + cΔu′′ + b_{1} grad θ′ + d grad θ′′ = 0,
cΔu′ + a_{2}Δu′′ + d grad θ′ + b_{2} grad θ′′ = 0,

(1.3)

where

θ′ = \frac{∂u′_{1}}{∂x_{1}} + \frac{∂u′_{2}}{∂x_{2}} , θ′′ = \frac{∂u′′_{1}}{∂x_{1}} + \frac{∂u′′_{2}}{∂x_{2}} .  (1.4)

Let us consider the variables z = x_{1} + ix_{2}, z̄ = x_{1} − ix_{2}, by which x_{1} = \frac{z+z̄}{2}, x_{2} = \frac{z−z̄}{2i}. Then

\begin{matrix}\frac{∂}{∂x_{1}} = \frac{∂}{∂z} + \frac{∂}{∂z̄}, \frac{∂}{∂x_{2}} = i(\frac{∂}{∂z} − \frac{∂}{∂z̄}), \\ \frac{∂}{∂z} = \frac{1}{2}(\frac{∂}{∂x_{1}} − i\frac{∂}{∂x_{2}} ), \frac{∂}{∂z̄} = \frac{1}{2}(\frac{∂}{∂x_{1}} + i\frac{∂}{∂x_{2}} ). \end{matrix}  (1.5)

Elementary calculations give

Δ = 4 \frac{∂^{2}}{∂z∂z̄}, θ′ = \frac{∂U_{1}}{∂z} + \frac{∂\overline{U}_{1}}{∂z̄} , θ′′ = \frac{∂U_{2}}{∂z} + \frac{∂\overline{U}_{2}}{∂z̄} ,  (1.6)

where

U_{1} = u_{1} + iu_{2}, U_{2} = u_{3} + iu_{4}.  (1.7)

Using formulas (1.5), (1.6) and (1.7), system (1.3) can be written in terms of two complex equations

2a_{1} \frac{∂^{2}U_{1}}{∂z∂z̄} + 2c \frac{∂^{2}U_{2}}{∂z∂z̄} + b_{1} \frac{∂θ′}{∂z̄} + d \frac{∂θ′′}{∂z̄} = 0,

2c \frac{∂^{2}U_{1}}{∂z∂z̄} + 2a_{2} \frac{∂^{2}U_{2}}{∂z∂z̄} + d \frac{∂θ′}{∂z̄} + b_{2} \frac{∂θ′′}{∂z̄} = 0.

After substituting here the values of θ′ and θ′′ from (1.6) we obtain

(2a_{1} + b_{1}) \frac{∂^{2}U_{1}}{∂z∂z̄} + (2c + d) \frac{∂^{2}U_{2}}{∂z∂z̄} + b_{1} \frac{∂^{2}\overline{U}_{1}}{∂z̄^{2}} + d \frac{∂^{2}\overline{U}_{2}}{∂z̄^{2}} = 0,

(2c + d) \frac{∂^{2}U_{1}}{∂z∂z̄} + (2a_{2} + b_{2}) \frac{∂^{2}U_{2}}{∂z∂z̄} + d \frac{∂^{2}\overline{U}_{1}}{∂z̄^{2}} + b_{2} \frac{∂^{2}\overline{U}_{2}}{∂z̄^{2}} = 0.

Hence, by some elementary transformations, we have

\frac{∂^{2}U}{∂z∂z̄} + ε^{⊤} \frac{∂^{2}\overline{U}}{∂z̄^{2}} = 0,  (1.8)

where

\begin{matrix}ε^{⊤} = [\begin{matrix}ε_{1}, & ε_{3} \\ ε_{2}, & ε_{4} \end{matrix}] , \\ δ_{0}ε_{1} = 2(a_{2}b_{1} − cd) + b_{1}b_{2} − d^{2}, δ_{0}ε_{2} = 2(da_{1} − cb_{1}), \\ δ_{0}ε_{3} = 2(da_{2} − cb_{2}), δ_{0}ε_{4} = 2(a_{1}b_{2} − cd) + b_{1}b_{2} − d^{2}, \\ δ_{0} = (2a_{1} + b_{1})(2a_{2} + b_{2}) − (2c + d)^{2} ≡ 4Δ_{0}d_{1}d_{2} > 0, \\ Δ_{0} = m_{1}m_{3}−m_{2}^{2} > 0, m_{1} = l_{1}+ \frac{l_{4}}{2} , m_{2} = l_{2}+ \frac{l_{5}}{2} , m_{3} = l_{3}+ \frac{l_{6}}{2} , \\ d_{1} = (a_{1} + b_{1})(a_{2} + b_{2}) − (c + d)^{2} > 0, d_{2} = a_{1}a_{2} − c^{2} > 0, \\ l_{1} = \frac{a_{2}}{d_{2}} , l_{2} = −\frac{c}{d_{2}} , l_{3} = \frac{a_{1}}{d_{2}} , \\ l_{1} + l_{4} = \frac{a_{2} + b_{2}}{d_{1}} , l_{2} + l_{5} = −\frac{c + d}{d_{1}}, l_{3} + l_{6} = \frac{a_{1} + b_{1}}{d_{1}} . \end{matrix} (1.9)

Equation (1.8) represents basic homogeneous equations of statics of the elastic mixture theory in the complex-vector form. One can likewise esily verify the validity of the identity

ε^{⊤} = −\frac{1}{2} ℓm^{−1},  (1.10)

where

ℓ = [\begin{matrix}ℓ_{4}, & ℓ_{5} \\ ℓ_{5}, & ℓ_{6} \end{matrix}] , m^{−1} = \frac{1}{Δ_{0}} [\begin{matrix}m_{3}, & −m_{2} \\ −m_{2}, & m_{1} \end{matrix}] ,  (1.11)

ℓ_{k} (k = 4, 5, 6), Δ_{0} and m_{k} (k = 1, 2, 3) are defined from (1.9).

In addition to the vector

U = (\begin{matrix}u_{1} + iu_{2} \\ u_{3} + iu_{4} \end{matrix}) ,  (1.12)

using the formulas

iv_{1} − v_{2} = m_{1}φ_{1}(z) + m_{2}φ_{2}(z) − \frac{z}{2}[ ℓ_{4}\overline{φ′_{1}(z)} + ℓ_{5}\overline{φ′_{2}(z)} ] − \overline{ψ_{1}(z)},

iv_{3} − v_{4} = m_{2}φ_{1}(z) + m_{3}φ_{2}(z) − \frac{z}{2}[ ℓ_{5}\overline{φ′_{1}(z)} + ℓ_{6}\overline{φ′_{2}(z)} ] − \overline{ψ_{2}(z)},

(see [3], p.242), we write the vector V as

V = (\begin{matrix}v_{1} + iv_{2} \\ v_{3} + iv_{4} \end{matrix}) ,  (1.13)

where v_{1}, v_{2}, v_{3}, v_{4} are the components of the vector v. As is known from [3], U and V are the conjugate vectors, i.e., V, like U, satisfies equation (1.8).

Using analogues of the general Kolosov–Muskhelishvili representations from [3], we can write

U =mφ(z)+ \frac{ℓ}{2} z\overline{φ′(z)}+\overline{ψ(z)}, V =i[−mφ(z)+ \frac{ℓ}{2} z\overline{φ′(z)}+\overline{ψ(z)}], (1.14)

where φ(z) and ψ(z) are arbitrary analytic vectors

m = [\begin{matrix}m_{1}, & m_{2} \\ m_{2}, & m_{3} \end{matrix}] ,  (1.15)

m_{1}, m_{2}, m_{3} are defined from (1.9).

By (1.14) it is obvious that

U + iV = 2mφ(z).  (1.16)

Let us now introduce the vectors

T^{κ} U = (\begin{matrix}(T^{κ} u)_{2} − i(T^{κ} u)_{1} \\ (T^{κ} u)_{4} − i(T^{κ} u)_{3} \end{matrix}) , T^{κ} V = (\begin{matrix}(T^{κ} v)_{2} − i(T^{κ} v)_{1} \\ (T^{κ} v)_{4} − i(T^{κ} v)_{3} \end{matrix}) ,  (1.17)

where U and V are defined from (1.12), (1.13), (1.14), κ is an arbitrary constant matrix:

κ = [\begin{matrix}κ_{1}, & κ_{3} \\ κ_{3}, & κ_{2} \end{matrix}] .  (1.18)

Using the formula

(\begin{matrix}(T^{κ} u)_{2} − i(T^{κ} u)_{1} \\ (T^{κ} u)_{4} − i(T^{κ} u)_{3} \end{matrix}) = \frac{∂}{∂s(x)}[ − 2φ(z) + (2μ − κ)U]

(see [3], p. 236) and (1.16), we can rewrite (1.17) as follows:

\begin{aligned}T^{κ} U & = \frac{∂}{∂s(x)}[(A − 2E − κm)φ(z) \\ & + (2μ − κ)\frac{ℓ}{2} z\overline{φ′(z)} + (2μ − κ)\overline{ψ(z)}], \\ T^{κ} V & = i \frac{∂}{∂s(x)}[ − (A − 2E − κm)φ(z) \\ & + (2μ − κ)\frac{ℓ}{2} z\overline{φ′(z)} + (2μ − κ)\overline{ψ(z)}], \end{aligned}  (1.19)

where E is the unit matrix and

A = 2μm, μ = [\begin{matrix}μ_{1}, & μ_{3} \\ μ_{3}, & μ_{2} \end{matrix}] , \frac{∂}{∂s(x)} = n_{1} \frac{∂}{∂x_{2}} − n_{2} \frac{∂}{∂x_{1}} ,  (1.20)

n = (n_{1}, n_{2}) is an arbitrary unit vector.

If κ = 0, then T^{κ} ≡ T, where T is the stress operator. Now (1.19) can be rewritten as

\begin{matrix}T U = \frac{∂}{∂s(x)}[(A − 2E)φ(z) + Bz\overline{φ′(z)} + 2μ\overline{ψ(z)} ], \\ T V = i \frac{∂}{∂s(x)}[ − (A − 2E)φ(z) + Bz\overline{φ′(z)} + 2μ\overline{ψ(z)} ], \end{matrix}  (1.21)

where

B = μℓ.  (1.22)

In addition to the operator T, we will need from (1.19) the particular case of the matrix κ, where

κ = 2μ − m^{−1}.  (1.23)

In that case T^{κ} ≡ N, where N is the pseudostress operator which plays an important part in studying the first boundary value problem.

Taking into account (1.14), (1.19) and (1.23), for the operator N we obtain

NU = −im^{−1} \frac{∂V}{∂s(x)} , NV = im^{−1} \frac{∂U}{∂s(x)} .  (1.24)

These relations are important when investigating the basic plane boundary value problems of statics of an elastic mixture.

Put now in (1.19)

φ(z) = \frac{(A − 2E − κm)^{−1}}{2πi} ∫_{S} ln σg(y) dS,

\overline{φ′(z)} = −\frac{(A − 2E − κm)^{−1}}{2πi} ∫_{S} \frac{\overline{g}(y)}{\overline{σ}} dS,  (1.25)

\overline{ψ(z)} = −\frac{(2μ − κ)^{−1}}{2πi} ∫_{S} ln \overline{σ}g(y) dS + \frac{ℓ(A − 2E − κm)^{−1}}{4πi} ∫_{S} \frac{ζ}{\overline{σ}} \overline{g(y)} dS ,

where σ = z − ξ, \overline{σ} = \overline{z} − \overline{ξ}, ξ = y_{1} + iy_{2}, g(y) is the complex vector we seek for. It is assumed here that

det |A − 2E − κm| ≠ 0, det |2μ − κ| ≠ 0.  (1.26)

In the sequel we will see that these restrictions are actually fulfilled.

Substituting (1.25) into (1.19) and performing some elementary transformations, we obtain

\begin{aligned}T^{κ} U & = \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(x)} g(y) dS \\ & − \frac{(2μ − κ)ℓ(A − 2E − κm)^{−1}}{4πi} ∫_{S} \frac{∂}{∂s(x)} \frac{σ}{\overline{σ}} \overline{g(y)} dS, \\ T^{κ} V & = −\frac{1}{π} ∫_{S} \frac{∂ ln r}{∂s(x)} g(y) dS \\ & − \frac{(2μ − κ)ℓ(A − 2E − κm)^{−1}}{4π} ∫_{S} \frac{∂}{∂s(x)} \frac{σ}{\overline{σ}} \overline{g(y)} dS, \end{aligned}  (1.27)

where

r = \sqrt{(x_{1} − y_{1})^{2} + (x_{2} − y_{2})^{2}} , θ = arctg \frac{y_{2} − x_{2}}{y_{2} − x_{1}} .  (1.28)

By (1.25) and (1.14) we obtain

U = \frac{m(A − 2E − κm)^{−1}}{2πi} ∫_{S} ln σg(y) dS − \frac{(2μ − κ)^{−1}}{2πi} ∫_{S} ln \overline{σ}g(y) dS

− \frac{ℓ(A − 2E − κm)^{−1}}{4πi} ∫_{S} \frac{σ}{\overline{σ}} \overline{g(y)} dS,

V =−\frac{m(A−2E−κm)^{−1}}{2π} ∫_{S} ln σg(y) dS− \frac{(2μ−κ)^{−1}}{2π} ∫_{S} ln \overline{σ}g(y) dS

− \frac{ℓ(A − 2E − κm)^{−1}}{4π} ∫_{S} \frac{σ}{\overline{σ}} \overline{g(y)} dS.

(1.29)

Let us show that U and V satisfy equation (1.8) for any κ. Indeed, by (1.29)

\frac{∂^{2}U}{∂z∂z̄} = \frac{ℓ(A − 2E − κm)^{−1}}{4πi} ∫_{S} \frac{\overline{g(y)}}{\overline{σ}^{2}} dS,

\frac{∂^{2}U}{∂z̄^{2}} = \frac{m(A − 2E − κm)^{−1}}{2πi} ∫_{S} \frac{\overline{g(y)}}{\overline{σ}^{2}} dS.

Now by virtue of (1.10) we conclude that U satisfies (1.8). In a similar manner one can show that V, too, satisfies (1.8).

Next, using (1.29), we calculate the operator T^{κ_{0}}, where κ_{0} is an arbitrary real matrix (different from κ).

By (1.25) we have

T^{κ_{0}}U = \frac{∂}{∂s(x)}[\frac{(A − 2E − κ_{0}m)(A − 2E − κm)^{−1}}{2πi} ∫_{S} ln σg(y) dS

− \frac{(2μ − κ_{0})(2μ − κ)^{−1}}{2πi} ∫_{S} ln \overline{σ}g(y) dS

− \frac{(2μ − κ_{0})ℓ(A − 2E − κm)^{−1}}{4πi} ∫_{S} \frac{σ}{\overline{σ}} \overline{g(y)} dS],

T^{κ_{0}}V = \frac{∂}{∂s(x)}[\frac{−(A − 2E − κ_{0}m)(A − 2E − κm)^{−1}}{2π} ∫_{S} ln σg(y) dS

− \frac{(2μ − κ_{0})(2μ − κ)^{−1}}{2π} ∫_{S} ln \overline{σ}g(y) dS

− \frac{(2μ − κ_{0})ℓ(A − 2E − κm)^{−1}}{4π} ∫_{S} \frac{σ}{\overline{σ}} \overline{g(y)} dS].

Assume that κ and κ_{0} satisfy the equation

(A − 2E − κ_{0}m)(A − 2E − κm)^{−1} + (2μ − κ_{0})(2μ − κ_{0})^{−1} = 0. (1.30)

Then the preceding formulas can be rewritten as follows:

\begin{aligned}T^{κ_{0}}U & = −(2μ − κ_{0})(2μ − κ)^{−1}[\frac{1}{πi} ∫_{S} \frac{∂ ln r}{∂s(x)} g(y) dS \\ & + \frac{(2μ − κ)ℓ(A − 2E − κm)^{−1}}{4πi} ∫_{S} \frac{∂}{∂s(x)} \frac{σ}{\overline{σ}} \overline{g(y)} dS], \\ T^{κ_{0}}V & = (2μ − κ_{0})(2μ − κ)^{−1}[\frac{i}{π} ∫_{S} \frac{∂θ}{∂s(x)} g(y) dS \\ & − \frac{(2μ − κ)ℓ(A − 2E − κm)^{−1}}{4π} ∫_{S} \frac{∂}{∂s(x)} \frac{σ}{\overline{σ}} \overline{g(y)} dS]. \end{aligned}  (1.31)

Comparing (1.27) and (1.31), we obtain

T^{κ_{0}}U =−i(2μ−κ_{0})(2μ−κ)^{−1}T^{κ} V, T^{κ_{0}}V =i(2μ−κ_{0})(2μ−κ)^{−1}T^{κ} U. (1.32)

We introduce the following definition: if κ and κ_{0} satisfy equation (1.30), then the operators T^{κ_{0}} and T^{κ} are self-conjugate ones, i.e., identities (1.32) are valid. Let us consider some particular cases. Let κ = 0; then T^{κ} ≡ T and from (1.30) it follows that

κ_{0} = 2μ − 2(A − E)^{−1}μ.  (1.33)

Therefore T^{κ_{0}} ≡ L. Thus the operators T and L are self-conjugate ones. Formulas (1.32) take the form

T V = i(A − E)LU, T U = −i(A − E)LV.  (1.34)

If in (1.30) κ is defined from (1.23), then T^{κ} ≡ N and to obtain the conjugate operator we have the indefiniteness. But this does not mean that no conjugate operator exists for N. We use here formula (1.24) which implies that for N the conjugate operator is N. Let us rewrite formula (5.15) from [4] as

∫_{D^{+}} T^{κ} (u, u) dy_{1} dy_{2} = ∫_{S} uT^{κ} u dS ≡ Im ∫_{S} UT^{κ} \overline{U} dS,  (1.35)

where Im is the imaginary part, U and T^{κ} \overline{U} are defined from (1.12) and (1.17), D^{+} is the finite domain bounded by the closed contour S. From the latter formula we obtain two formulas to be used below. For κ = 0 and κ = 2μ−m^{−1}, (1.35) respectively yields

∫_{D^{+}} T (u, u) dy_{1} dy_{2} = ∫_{S} uT u dS ≡Im ∫_{S} UT \overline{U} dS,  (1.36)

∫_{D^{+}} N(u, u) dy_{1} dy_{2} = ∫_{S} uNu dS ≡Im ∫_{S} UN\overline{U} dS ≡Im ∫_{S} V N\overline{V} dS,  (1.37)

where T (u, u) and N(u, u) are defined in [4], pp. 75–76. Formulas (1.35), (1.36) and (1.37) hold for the infinite domain D^{−} = E_{2}\\overline{D^{+}} as well provided that conditions (5.22) from [4] are fulfilled. In that case we have

∫_{D^{−}} T^{κ} (u, u) dy_{1} dy_{2} =−∫_{S} uT^{κ} u dS ≡−Im ∫_{S} UT^{κ} \overline{U} dS,  (1.38)

∫_{D^{−}} T (u, u) dy_{1} dy_{2} =−∫_{S} uT u dS ≡−Im ∫_{S} UT \overline{U} dS,  (1.39)

∫_{D^{−}} N(u, u) dy_{1} dy_{2} =−∫_{S} uNu dS ≡−Im∫_{S} UN\overline{U}dS ≡− Im∫_{S} V N\overline{V} dS. (1.40)

These formulas play an essential role in investigating the basic plane boundary value problems of statics for an elastic mixture.

2. The First Boundary Value Problem

The first boundary value problem is formulated as follows: Find, in the domain D^{+}(D^{−}), a vector U which belongs to the class C^{2}(D^{+}) ∩ C^{1,α}(D^{+} ∪ S)[C^{2}(D^{−}) ∩ C^{1,α}(D^{−} ∪ S)], is a solution of equation (1.8) and satisfies the boundary condition

(u)^{±} = f(t),  (2.1)

where f is a given vector on the boundary, the signs + and −− denote the limits from inside and from outside. Note that for the infinite domain D^{−} the vector U additionally satisfies the following conditions at infinity:

U = O(1), \frac{∂U}{∂x_{k}} = O(ρ^{−2}), k = 1, 2,  (2.2)

where ρ^{2} = x_{1}^{2} + x_{2}^{2}.

The direction of the external normal is assumed to be the positive direction of the normal, i.e., the direction from D^{+} into D^{−}.

First we are to write a Fredholm integral equation for the first boundary value problem. Using formulas (1.14) and choosing φ(z) and \overline{ψ(z)} in the form

φ(z) = \frac{m^{−1}}{2πi} ∫_{S} \frac{∂ ln σ}{∂s(y)} g(y) dS,

\overline{ψ(z)} = −\frac{1}{2πi} ∫_{S} \frac{∂ ln \overline{σ}}{∂s(y)} g(y) dS + \frac{ℓm^{−1}}{4πi} ∫_{S} \frac{∂}{∂s(y)} \frac{ζ}{\overline{σ}} \overline{g(y)} dS,

after some simple transformations we have by (1.10) that

U = \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(y)} g(y) dS + \frac{ε^{⊤}}{2πi} ∫_{S} \frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} \overline{g(y)} dS,  (2.3)

V = −\frac{1}{π} ∫_{S} \frac{∂ ln r}{∂s(y)} g(y) dS + \frac{ε^{⊤}}{2π} ∫_{S} \frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} \overline{g(y)} dS,  (2.4)

where θ and r are defined from (1.28), g(y) is the complex vector we seek for, while the values of σ and \overline{σ} are given above (see §1).

Let us first investigate the first internal problem. Passing to the limit in (2.3) as x → t ∈ S and using the boundary condition (2.1), to define the vector g we obtain the following Fredholm integral equation of second order:

g(t) + \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(y)} g(y) dS + \frac{ε^{⊤}}{2πi} ∫_{S} \frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} \overline{g(y)} dS = f(t),  (2.5)

where f ∈ C^{1,β}(s) (β > 0) is a given vector on the boundary, and

θ = arctg \frac{y_{2} − x_{2}}{y_{1} − x_{1}} , σ = t − ζ, \overline{σ} = t̄− ζ̄, t = t_{1} + it_{2}.  (2.6)

From (2.5) we have

\frac{∂g}{∂s(t)} + \frac{1}{π} ∫_{S} \frac{∂^{2}θ}{∂s(t)∂s(y)} g(y) dS + \frac{ε^{⊤}}{2πi} ∫_{S} \frac{∂^{2}}{∂s(t)∂s(y)} \frac{σ}{\overline{σ}} \overline{g(y)} dS = \frac{∂f}{∂s(t)} .

Taking into account that s ∈ C^{2,α} (α > 0), the latter formula implies

∫_{S} \frac{∂^{2}θ}{∂s(t)∂s(y)} g(y) dS ∈ C^{0,α}(S) and ∫_{S} \frac{∂^{2}}{∂s(t)∂s(y)} \frac{σ}{\overline{σ}} \overline{g(y)} dS ∈ C^{0,α}(S)

and since f ∈ C^{1,β} (0 < β < α ≤ 1), from (2.5) we obtain g ∈ C^{1,β}(s).

Let us prove that the homogeneous equation correspending to (2.5) has only a trivial solution. Assume that it has a nontrivial solution denoted by g_{0}. By the same reasoning as above we can easily find that g_{0} ∈ C^{1,α}(s) and NU_{0} ∈ C^{0,α}(s). Applying (1.37), we have u_{0}(x) = C, x ∈ D^{+}, where C is an arbitrary constant vector. But since (U_{0})^{+} ≡ 0, we have U_{0}(x) = 0. Taking into account that for g_{0} ∈ C^{1,α}(s), (NU_{0})^{+} −(NU_{0})^{−} = 0, we obtain (NU_{0})^{−} = 0. Using now formula (1.40), we obtain U_{0}(x) = C x ∈ D^{−} for the domain D^{−}. Since the potential U_{0}(x) is equal to zero at infinity, we have U_{0}(x) = 0 for x ∈ D^{−} and, using the formula (U_{0})^{+} − (U_{0})^{−} = 2g_{0}, we obtain g_{0} = 0. Therefore the homogeneous equation has only a trivial solution. We have proved that, by the first Fredholm theorem, equation (2.5) is solvable for an arbitrary right-hand part f ∈ C^{1,β}(s) (β > 0).

One can prove that a solution of equation (2.5) exists iff s ∈ C^{1,α} and f ∈ C^{1,β}(s), 0 < β < α ≤ 1. The proof is the same as in [5].

Let us now consider the first external boundary value problem. Its solution is sought for in the form

W = U(x) + U(0),  (2.7)

where U is defined by formula (2.3), and

U(0) = \frac{1}{π} ∫_{S} \frac{∂}{∂s(y)} arctg \frac{y_{2}}{y_{1}} g(y) dS + \frac{ε^{⊤}}{2πi} ∫_{S} \frac{∂}{∂s(y)} \frac{ζ}{ζ̄} \overline{g(y)} dS. (2.8)

The origin is assumed to lie in the domain D^{+}.

Taking into account the boundary behavior of the potential U(x) and the boundary condition (2.1), to define the unknown vector g we obtain from (2.7) the Fredholm integral equation of second order

− g(t) + \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(y)} g(y) dS + \frac{ε^{⊤}}{2πi} ∫_{S} \frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} \overline{g(y)} dS + U(0) = f(t). (2.9)

where θ, σ, \overline{σ} are defined by (2.6). Quite in the same manner as above, one can prove that a solution of equation (2.9), if it exists, belongs to the class C^{1,β}(s) for s ∈ C^{2,α} and f^{1,β}(s), 0 < β < α ≤ 1.

Let us show now that equation (2.9) is always solvable. For this it is sufficient that the homogeneous equation corresponding to (2.9) have only a trivial solution. Denote the homogeneous equation (which we do not write) by (2.9)_{0} and assume that it has a solution different from zero which is denoted by g_{0}. Denote the corresponding potentials and values by W_{0}, U_{0} and U_{0}(0). Using (1.40), we obtain W_{0}(x) = α. But (W_{0})^{−} = 0 and therefore α = 0 and W_{0}(x) = 0, x ∈ D^{−}. Hence, for x → ∞, we obtain

U_{0}(0) = 0.  (2.10)

In that case, (2.7) implies W ≡ U_{0}(x) = 0 and NU_{0}(x) = 0, x ∈ D^{−}. Since under our restrictions (NU_{0})^{+} − (NU_{0})^{−} = 0, and (NU_{0})^{−} = 0, we obtain (NU_{0})^{+} = 0. Hence by (1.37) we have U_{0}(x) = β, x ∈ D^{+}. But by (2.10) U_{0}(0) = 0 and, obviously, β = 0. Thus U_{0}(x) = 0, x ∈ D^{+}. Taking into account that (U_{0})^{+} − (U_{0})^{−} = 2g_{0} and (U_{0})^{+} = 0, (U_{0})^{−} = 0 we obtain g_{0} = 0. Thus our assumption is not valid. Equation (2.9) has a solution for an arbitrary right-hand part.

As above, here, too, we can note that a solution of equation (2.9) exists iff s ∈ C^{1,α} and f ∈ C^{1,β}(s), 0 < β < α ≤ 1.

So far it has been assumed that the principal vector of external forces, stress components and rotation at infinity are equal to zero. The general case with these values given and different from zero is considered applying a reasoning similar to that used in [6]. When D^{+} is a multiply-connected finite domain, the proof of the existence of solutions for this domain is easy and carried out as in [6].

3. The Second Boundary Value Problem

The second boundary value problem is posed as folows: Find, in the domain D^{+}(D^{−}), a vector U which belongs to the class C^{2}(D^{+}) ∩ C^{1,α}(\overline{D}^{+})[C^{2}(D^{−}) ∩ C^{1,α}(D^{−}∪S)], is a solution of equation (1.8) and satisfies the boundary condition

(T U)^{±} = F (t),  (3.1)

where F is a given vector on the boundary. For an infinite domain we have conditions (2.2).

To derive Fredholm integral equations of second order for the second boundary value problem is not difficult. Indeed, after substituting κ = 0 into (1.25) and (1.27), we obtain

\begin{matrix}T U = \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(x)} g(y) dS − \frac{H}{2πi} ∫_{S} \frac{∂}{∂s(x)} \frac{σ}{\overline{σ}} \overline{g(y)} dS, \\ T V = −\frac{1}{π} ∫_{S} \frac{∂ ln r}{∂s(x)} g(y) dS − \frac{H}{2π} ∫_{S} \frac{∂}{∂s(x)} \frac{σ}{\overline{σ}} \overline{g(y)} dS, \end{matrix}  (3.2)

where U and V are defined as follows:

U = \frac{m(A − 2E)^{−1}}{2πi} ∫_{S} ln σg(y) dS − \frac{μ^{−1}}{4πi} ∫_{S} ln \overline{σ}g(y) dS

− \frac{ℓ(A − 2E)^{−1}}{4πi} ∫_{S} \frac{σ}{\overline{σ}} \overline{g(y)} dS,

V = −\frac{m(A − 2E)^{−1}}{2π} ∫_{S} ln σg(y) dS − \frac{μ^{−1}}{4π} ∫_{S} ln \overline{σ}g(y) dS

− \frac{ℓ(A − 2E)^{−1}}{4π} ∫_{S} \frac{σ}{\overline{σ}} \overline{g(y)} dS.

(3.3)

Here, as above, g(\overline{g}) is the complex vector we seek for,

H = [\begin{matrix}H_{1}, & H_{2} \\ H_{3}, & H_{4} \end{matrix}] ,

where

\begin{matrix}H_{1} = 1 − \frac{2λ_{5}}{Δ_{2}d_{2}} [(a_{1} + c)A_{3} + (a_{2} + c)(A_{4} − 2)], \\ H_{2} = \frac{2λ_{5}}{Δ_{2}d_{2}} [(a_{1} + c)(A_{2} − 2) + (a_{2} + c)A_{2}], \\ Δ_{2} = (A_{1}−2)(A_{4}−2)−A_{2}A_{3} > 0, H_{3} = 1−H_{1}, H_{4} =1−H_{2}, \\ (A − 2E)^{−1} = \frac{1}{Δ_{2}} [\begin{matrix}A_{4} − 2, & −A_{2} \\ −A_{3} & A_{1} − 2 \end{matrix}] , μ^{−1} = \frac{1}{Δ_{1}} [\begin{matrix}μ_{2}, & −μ_{3} \\ −μ_{3}, & μ_{1} \end{matrix}] ; \end{matrix} (3.4)

the other values contained in (3.2) and (3.3) are defined in the preceding paragraph.

Let us first consider the second boundary value problem for the domain D^{+}. By (3.1), to define the vector g we find from (3.2) that

− g(t) + \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(t)} g(y) dS − \frac{H}{2πi} ∫_{S} \frac{∂}{∂s(t)} \frac{σ}{\overline{σ}} \overline{g(y(} dS = F (t), (3.5)

where θ, σ and \overline{σ} are given from (2.6). We think that it is advisable to modify equation (3.5). To do so, we add to the left-hand side the expression

\frac{1}{2π} [\frac{∂θ}{∂s(t)} ∫_{S} g(y) dS − \frac{H}{2πi} \frac{∂}{∂s(t)} \frac{σ(t)}{\overline{σ(t)}} ∫_{S} \overline{g} dS] + \frac{1}{4πi} (\begin{matrix}1 \\ 1 \end{matrix}) \frac{∂}{∂s(t)} \frac{1}{t̄} · M,

where

M = [ − i \frac{∂}{∂z} (U_{1} + U_{2}) + i \frac{∂}{∂z̄}(\overline{U}_{1} + \overline{U}_{2})]_{x_{1}=x_{2}=0} ,  (3.6)

U_{1} and U_{2} and their conjugates are defined in (3.3). As above, it is assumed that the origin lies in the domain D^{+}.

Thus we obtain the equation

− g(t) + \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(t)} g(y) dS − \frac{H}{2πi} ∫_{S} \frac{∂}{∂s(t)} \frac{σ}{\overline{σ}} \overline{g(y)} dS + \frac{1}{2π} \frac{∂θ(t)}{∂s(t)} ∫_{S} g(y) dS

− \frac{H}{4πi} \frac{∂}{∂s(t)} \frac{σ(t)}{\overline{σ(t)}} ∫_{S} \overline{g(y)} dS + \frac{1}{4πi} (\begin{matrix}1 \\ 1 \end{matrix}) \frac{∂}{∂s(t)} \frac{1}{t̄} M = F (t).  (3.7)

Let us now show that if equation (3.7) has a solution, then it is necessary that

∫_{S} g dS = 0  (3.8)

and

M = 0,  (3.9)

provided that the principal vector and the principal moment of external forces are equal to zero.

Indeed, by some simple calculations, from (3.7) we obtain

∫_{S} g dS = ∫_{S} F dS, M = ∫_{S} Re t̄(F_{1} + F_{2})dS.  (3.10)

If the principal vector ∫_{S} F dS and the principal moment ∫_{S} Re \overline{t}(F_{1} + F_{2})dS are equal to zero, then ∫_{S} gdS = 0 and M = 0, which was required to be shown.

Thus if the requirement that the principal vector and the principal moment of external forces be equal to zero is fulfilled, then any solution g of equation (3.7) is simultaneously a solution of the initial equation (3.5).

Let us now prove that equation (3.7) is always solvable. To this end, consider the homogeneous equation obtained from (3.7) for F = 0 and prove that it has no solutions different from zero. Let g_{0} be any solution of this homogeneous equation. Since F = 0, it is obvious that conditions (3.8) and (3.9) are fulfilled for g_{0}. In that case the obtained homogeneous equation corresponds to the boundary condition

(T U_{0}(t))^{+} = 0,  (3.11)

where U_{0}(x) is obtained from (3.3) if g is replaced by g_{0}. Using (3.11) and (1.36) we obtain

U_{0}(x) = α + iβ_{1} (\begin{matrix}1 \\ 1 \end{matrix}) z, x ∈ D^{+},  (3.12)

where α is an arbitrary constant vector, and β_{1} is an arbitrary constant value. Since M_{0} = 0, (3.12) implies β_{1} = 0 and we obtain U_{0}(x) = α, x ∈ D^{+}. Hence, in view of (1.24), we have

NU_{0}(x) = −im^{−1} \frac{∂V_{0}(x)}{∂s(x)} = 0

and

V_{0}(x) = β, x ∈ D^{+},  (3.13)

where V_{0}(x) are defined from (3.3) if g = g_{0} and β is an arbitrary constant vector. By (3.2), from (3.13) we obtain T V_{0}(x) = 0, x ∈ D^{+}, and since (T V_{0}(t))^{+} − (T V_{0}(t))^{−} = 0, we have (T V_{0}(x))^{−} = 0. Applying now formula (1.39) we find

V_{0}(x) = α + iβ_{1} (\begin{matrix}1 \\ 1 \end{matrix}) z, x ∈ D^{−},

Hence, since V_{0}(x) is bounded at infinity, we have β_{1} = 0 and V_{0}(x) = α, x ∈ D^{−}. In that case U_{0}(x) = β and T U_{0}(x) = 0, x ∈ D^{−}. Taking into account that (T U_{0})^{−} − (T U_{0})^{+} = 2g_{0} and (T U_{0})^{+} = 0, (T U_{0})^{−} = 0 we obtain g_{0} = 0.

Thus we proved that the homogeneous equation corresponding to equation (3.7) has no solutions different from zero.

Therefore equation (3.7) has one and only one solution g. On substituting this value g into formula (3.3), we obtain a solution of the second boundary value problem provided that the requirement for the principal vector and the principal moment of external forces to be equal to zero is fulfilled. Displacements U are defined to within rigid disolacement, while stresses are defined precisely.

Let us now consider the second boundary value problem in the domain D^{−}. Its solution is to be sought for in the form

W (x) = U(x) − \frac{μ^{−1}}{8π} (\begin{matrix}1 \\ 1 \end{matrix}) \frac{1}{z̄} M,  (3.14)

where U are defined from (3.3), and

M = −i[\frac{∂(V_{1} + V_{2})}{∂z} − \frac{∂(\overline{V}_{1} + \overline{V}_{2})}{∂z̄} ]_{x_{1}=x_{2}=0} ,  (3.15)

V are given from (3.3).

From (3.14) we readily have T W = T U(x) − \frac{1}{4π} (\begin{matrix}1 \\ 1 \end{matrix}) \frac{∂}{∂s(x)} \frac{1}{z̄} · M.

Passing here to the limit as z → t ∈ S and taking into account (3.1), we obtain

g(t) + \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(t)} g(y) dS − \frac{H}{2πi} ∫_{S} \frac{∂}{∂s(t)} \frac{σ}{\overline{σ}} \overline{g(y)} dS

− \frac{1}{4π} (\begin{matrix}1 \\ 1 \end{matrix}) \frac{∂}{∂s(t)} \frac{1}{t̄} M = F (t).  (3.16)

Performing integration, from (3.16) we esily have 2 ∫_{S} g dS = ∫_{S} F dS.

So far it has been assumed that the principal vector of external forces is equal to zero. This means that

∫_{S} g dS = 0.  (3.17)

The latter condition implies that the vector U(x) from (3.3) is unique and bounded.

Now let us show that equation (3.16) is always solvable. To this end, consider the homogeneous equation which is obtained from (3.16) when F = 0. We have to prove that this homogeneous equation has no solutions different from zero. Assume the contrary and denote by g_{0} some solution of this homogeneous equation. Since F = 0, condition (3.17) is fulfilled for g_{0}.

Note that the homogeneous equation corresponds to the boundary condition (T W_{0}(t))^{−} = 0. Using formula (1.39), we have W_{0}(x) = 0, x ∈ D^{−}, or, by (3.14), we can write

U_{0}(x) − \frac{μ^{−1}}{8π} (\begin{matrix}1 \\ 1 \end{matrix}) \frac{1}{z̄} M_{0} = 0, x ∈ D^{−}.

Obviously, for the conjugate vector we have

V_{0}(x) − \frac{iμ^{−1}}{8π} (\begin{matrix}1 \\ 1 \end{matrix}) \frac{1}{z̄} M_{0} = c, x ∈ D^{−},  (3.18)

where c is an arbitrary constant vector. Calculating the stress vector and taking into account (3.2), from (3.18) we obtain

−\frac{1}{π} ∫_{S} \frac{∂ ln r}{∂s(t)} g_{0}(y) dS − \frac{H}{2π} ∫_{S} \frac{∂}{∂s(t)} \frac{σ}{\overline{σ}} \overline{g_{0}(y)} dS

− \frac{i}{4π} (\begin{matrix}1 \\ 1 \end{matrix}) \frac{∂}{∂s(t)} \frac{1}{t̄} · M_{0} = 0, t ∈ S.

Now we have to calculate the principal moment of this vector. After lengthy but obvious trasformations we have M_{0} = 0.

We have thus proved that V_{0}(x) = C for x ∈ D^{−}. But since (T V_{0})^{−} = (T V_{0})^{+} = 0, by (1.36) we obtain

V_{0}(x) = α + iβ_{1} (\begin{matrix}1 \\ 1 \end{matrix}) z, x ∈ D^{+},  (3.19)

where α is an arbitrary constant vector, and β_{1} is an arbitrary constant scalar. Since M_{0} = 0, (3.19) readily implies 0 = M_{0} = 4β_{1} = 0. In that case we have V_{0}(x) = α, x ∈ D^{+}, and U_{0}(x) = β, x ∈ D^{+}, where β is

a constant vector. Thus (T U_{0}(t))^{−} − (T U_{0}(t))^{+} = 2g_{0} = 0.

Therefore the homogeneous equation corresponding to equation (3.16) has no solutions different from zero. This means that equation (3.16) has one and only one solution when the principal vector of external forces is equal to zero.

Like in §2, here too we can note that a solution of the second boundary value exists iff the principal vector of external forces, and stress and rotation components are given values. It is understtod that in that case the displacement vector at infinity is unbounded.

The existence of solutions of the second boundary value problem can also be proved when D^{+} is a finite multiply-connected domain.

4. The Third Boundary Value Problem

In the case of the third boundary value problem the following values are given at the boundary:

\frac{∂}{∂s(x)}[u_{3} − u_{1} + i(u_{4} − u_{2})], (T u)_{2} + (T u)_{1} − i[(T u)_{4} + (T u)_{3}],

where u_{1}, u_{2}, u_{3}, u_{4} are the components of the four-dimensional vector U, and (T U)_{1}, (T U)_{2}, (T U)_{3}, (T U)_{4} are the components of the stress vector T U.

By virtue of (1.12), (1.14) and (1.21), these value can be rewritten as

\frac{∂}{∂s(x)} \{[\begin{matrix}m_{2} − m_{1}, & m_{3} − m_{2} \\ B_{1} + B_{3}, & B_{2} + B_{4} \end{matrix}] φ(z) + [\begin{matrix}\frac{ℓ_{5}−ℓ_{4}}{2} , & \frac{ℓ_{6}−ℓ_{5}}{2} \\ B_{1} + B_{3}, & B_{2} + B_{4} \end{matrix}] z\overline{φ′(z)}
+ [\begin{matrix}−1, & 1 \\ 2(μ_{1} + μ_{3}), & 2(μ_{2} + μ_{3}) \end{matrix}] \overline{ψ(z)}\} = F (x),  (4.1)

where

F (x) = (\begin{matrix}\frac{∂}{∂s(x)}[u_{3} − u_{1} + i(u_{4} − u_{2})] \\ (T u)_{2} + (T u)_{4} − i[(T u)_{1} + (T u)_{3}] \end{matrix}) ≡ (\begin{matrix}U_{2} − U_{1} \\ (T U)_{1} + (T U)_{2} \end{matrix}) , (4.2)

U_{1}, U_{2}, (T U)_{1} and (T U)_{2} are defined from (1.12) and (1.21).

Now we can formulate the thirs boundary value problem: Find, in the domain D^{+}(D^{−}), a vector U which belongs to the class C^{2}(D^{+}) ∩ C^{1,α}(\overline{D^{+}})[C^{2}(D^{−}) ∩ C^{1,α}(D^{−} ∪ S)], is a solution of equations (1.8), and satisfies the boundary condition

(F (t))^{±} = F (t),  (4.3)

where F (t) is a given vector on the boundary. Conditions (2.2) are fulfilled at infinity.

For this problem we need to derive Fredholm integral equations of second order.

Let

φ(z) = \frac{1}{2πiΔ_{3}} [\begin{matrix}B_{2} + B_{4}, & m_{2} − m_{3} \\ −(B_{1} + B_{3}), & m_{2} − m_{1} \end{matrix}] ∫_{S} ln σg(y) dS,  (4.4)

where g is the complex vector we seek for, and

Δ_{3} = 2(m_{1} + m_{3} − 2m_{2} − Δ_{0}a_{0}) > 0,  (4.5)

where a_{0} = μ_{1} + μ_{2} + 2μ_{3} ≡ a_{1} + a_{2} + 2c. Note that the proof for Δ_{3} > 0 is given in [7].

After lengthy but elementary calculations we obtain

\frac{1}{Δ_{3}} [\begin{matrix}\frac{ℓ_{5}−ℓ_{4}}{2} , & \frac{ℓ_{6}−ℓ_{5}}{2} \\ B_{1} + B_{3}, & B_{2} + B_{4} \end{matrix}] [\begin{matrix}B_{2} + B_{4}, & m_{2} − m_{3} \\ −(B_{1} + B_{3}), & m_{2} − m_{1} \end{matrix}] = [\begin{matrix}−K_{0}, & −α_{0} \\ 0, & 1 \end{matrix}] , (4.6)

where

K_{0} = \frac{a_{0}(b_{1}b_{2} − d^{2})}{2Δ_{3}d_{1}d_{2}}, α_{0} = \frac{Δ_{0}}{Δ_{3}} (ε_{1} + ε_{3} − ε_{2} − ε_{4}),  (4.7)

while ε_{1}, ε_{2}, ε_{3}, ε_{4} are defined from (1.9).

In view of (4.4) and (4.6), expression (4.1) takes the form

\frac{∂}{∂s(x)}\{\frac{1}{2πi} ∫_{S} ln σg(y) dS + \frac{K}{2πi} ∫_{S} \frac{z}{\overline{σ}} \overline{g(y)} dS

+ [\begin{matrix}−1 & 1 \\ 2(μ_{1} + μ_{3}), & 2(μ_{2} + μ_{3}) \end{matrix}] \overline{ψ(z)}\} = F (x),  (4.8)

where

K = [\begin{matrix}K_{0}, & α_{0} \\ 0, & −1 \end{matrix}]  (4.9)

Let us now choose \overline{ψ(z)} in (4.8) as follows:

[\begin{matrix}−1 & 1 \\ 2(μ_{1} + μ_{3}), & 2(μ_{2} + μ_{3}) \end{matrix}] \overline{ψ(z)} = −\frac{1}{2πi} ∫_{S} ln \overline{σ}g(y) dS − \frac{K}{2πi} ∫_{S} \frac{ζ}{\overline{σ}} g(y) dS,

Then (4.8) takes the final form

\frac{1}{π} ∫_{S} \frac{∂θ}{∂s(x)} g(y) dS + \frac{K}{2πi} ∫_{S} \frac{∂}{∂s(x)} \frac{σ}{\overline{σ}} \overline{g(y)} dS = F (x).  (4.10)

First we consider the third boundary value problem in the domain D^{+}. By the boundary condition, to define the unknown vector g we find from (4.10) the Fredholm integral equation of second order

− g(t) + \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(t)} g(y) dS + \frac{K}{2πi} ∫_{S} \frac{∂}{∂s(t)} \frac{σ}{\overline{σ}} \overline{g(y)} dS = F (t), (4.11)

where θ, σ and \overline{σ} are defined from (2.6).

To investigate (4.11), its advisable to consider, instead of (4.11), the equation

− g(t) + \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(x)} g(y) dS + \frac{K}{2πi} ∫_{S} \frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} \overline{g(y)} dS − \frac{i}{2π} (\begin{matrix}0 \\ 1 \end{matrix}) \frac{∂}{∂s(t)} \frac{1}{t̄} M

+ \frac{1}{2π} [\frac{∂θ(t)}{∂s(t)} ∫_{S} g dS − \frac{H}{2i} \frac{∂}{∂s(t)} \frac{t}{t̄} ∫_{S} \overline{g} dS] = F (t),  (4.12)

where

M = (−i)(\frac{∂U_{2}}{∂z} − \frac{∂\overline{U}_{2}}{∂z̄} )_{x_{1}=x_{2}=0} ,  (4.13)

and θ(t) are given from (2.6) at y_{1} = y_{2} = 0.

From (4.12) we readily obtain

∫_{S} g dS = ∫_{S} F dS, M = Re ∫_{S} t̄F_{2}(t) dS.  (4.14)

Let us prove that equation (4.12) is always solvable. Assume the contrary and denote any solution of the homogeneous equation corresponding to (4.12) by g_{0}. On account of (4.14) we have

∫_{S} g_{0} dS = 0, M_{0} = 0,  (4.15)

where M_{0} is obtained from (4.13) after replacing g by g_{0}.

In that case the homogeneous equations for (4.12) and (4.11) coincide, but to the homogeneous equation for (4.11) there corresponds the boundary condition (F (t))^{+} = 0.

Using (1.16) and taking into account that (U_{01})^{+} − (U_{02})^{+} = C and ∫_{S}(T \overline{U_{0}})^{+}dS = 0, we obtain

U_{0}(x) = α + iβ_{1} (\begin{matrix}1 \\ 1 \end{matrix}) z, x ∈ D^{+},

where α and β_{1} are arbitrary constants. By (4.15) we have U_{0}(x) = α, x ∈ D^{+}.

The conjugate vector V_{0}(x) defined from (1.14) has the form V_{0}(x) = β, x ∈ D^{+}, where β is another constant value. By repeating the arguments used above for the vector U_{0} we get

(V_{01} − V_{02})^{−} = (V_{01} − V_{02})^{+} = α_{1},
[(T V_{0})_{1} + (T V_{0})_{2}]^{−} = [(T V_{0})_{1} + (T V_{0})_{2}]^{+} = 0.

Using now (1.39) for the vector V_{0}, we have

V_{0}(x) = γ + iδ_{0} (\begin{matrix}1 \\ 1 \end{matrix}) z, z ∈ D^{−},

where γ is an arbitrary constant vector, and δ_{0} is an arbitrary constant scalar. Since the vector V_{0}(x) is equal to zero at infinity, we have δ_{0} = 0, γ = 0 and V_{0}(x) = 0, x ∈ D^{−}. In that case, the conjugate vector U_{0}(x) is U_{0}(x) = C, x ∈ D^{−}. Applying (4.15), we have

U_{0}(x) = 0, x ∈ D^{−}.  (4.18)

Taking into account (4.17) and (4.18) it is easy to obtain

\frac{∂}{∂s(t)}(U_{01} − U_{02})^{−} − \frac{∂}{∂s(t)}(U_{01} − U_{02})^{+} = 2g_{01} = 0,

[(T U_{0})_{1} + (T U_{0})_{2}]^{−} − [(T U_{0})_{1} + (T U_{0})_{2}]^{+} = 2g_{02} = 0.

This means that g_{0} = 0 and our assumption is not correct.

Thus we have proved that equation (4.12) is solvable for any right-hand part. By substitution this value of g into (4.11) we obtain a solution of the third boundary value problem when ∫_{S} F_{1}dS = 0 and the principal moment of external forces is equal to zero.

The investigation of the third internal value is finished.

Finally, let investigate the third boundary value problem in the domain D^{−}. Its solution will be sought for in the form

\frac{1}{π} ∫_{S} \frac{∂θ}{∂s(x)} g(y) dS + \frac{K}{2πi} ∫_{S} \frac{∂}{∂s(x)} \frac{σ}{\overline{σ}} \overline{g(y)} dS

− \frac{1}{2π} (\begin{matrix}0 \\ 1 \end{matrix}) \frac{∂}{∂s(x)} \frac{1}{z̄} M = F (x),  (4.19)

where

M = (−i)(\frac{∂V_{2}}{∂z} − \frac{∂\overline{V}_{2}}{∂z̄} )_{x_{1}=x_{2}=0} .  (4.20)

To define the vector g, from (4.19) we obtain the Fredholm integral equation of second order

g(t) + \frac{1}{π} ∫_{S} \frac{∂θ}{∂s(t)} g(y) dS + \frac{K}{2πi} ∫_{S} \frac{∂}{∂s(t)} \frac{σ}{\overline{σ}} \overline{g(y)} dS

− \frac{1}{2π} (\begin{matrix}0 \\ 1 \end{matrix}) \frac{∂}{∂s(t)} \frac{1}{t̄} M = F (t).  (4.21)

After obvious transformations, the latter equation gives

2 ∫_{S} g dS = ∫_{S} F dS.  (4.22)

Let us prove that equation (4.21) is always solvable. Assume that the homogeneous equation corresponding to (4.21) has a nonzero solution denoted by g_{0}. Since F = 0, (4.22) implies ∫_{S} g_{0} dS = 0.

This means that both U_{0} and the conjugate vector V_{0} are equal to zero at infinity. Applying (1.39), we obtain U_{0}(x) = 0, V_{0}(x) = 0.

On account of (4.21), for F = 0 we have

− \frac{1}{π} ∫_{S} \frac{∂ ln r}{∂s(t)} g_{0} dS− \frac{K}{2π} ∫_{S} \frac{∂}{∂s(t)} \frac{σ}{\overline{σ}} \overline{g}_{0} dS− \frac{i}{2π} (\begin{matrix}0 \\ 1 \end{matrix}) \frac{∂}{∂s(t)} \frac{1}{t̄} M_{0} =0. (4.23)

The latter equation implies M_{0} = 0. Now (4.23) takes the form (T V_{0}(t))^{+} = (T V_{0}(t))^{−} = 0. Applying (1.36) for the vector V_{0} we obtain

V_{0}(x) = α + iβ_{1} (\begin{matrix}1 \\ 1 \end{matrix}) z, x ∈ D^{+}.

Since M_{0} = 0, we have β_{1} = 0 and V_{0}(x) = α, from which it follows that U_{0}(x) = β, x ∈ D^{+}, where β is an arbitrary constant vector. Now by the boundary condition we have

\frac{∂}{∂s(t)}(U_{01} − U_{02})^{−} − \frac{∂}{∂s(t)}(U_{01} − U_{02})^{+} = 2g_{01} = 0,

[(T U_{0})_{1} + (T U_{0})_{2}]^{−} − [(T U_{0})_{1} + (T U_{0})_{2}]^{+} = 2g_{02} = 0.

Thus we have proved that the homogeneous equation corresponding to (4.21) has only a trivial solution. By choosing among solutions of (4.21) those satisfying the condition ∫_{S} F_{1}dS = 0 we obtain a solution of the third boundary value problem in the domain D^{−}.

References

1. T. R. Steel, Applications of a theory of interacting continua. Q. J. Mech. Appl. Math. 20(1967), No. 1, 57–72.

2. D. G. Natroshvili, A. Ya. Jagmaidze, and M. Z. Svanadze, Some problems of the linear theory of elastic mixtures. (Russian) Tbilisi Univ. Press, Tbilisi, 1986.

3. M. O. Basheleishvili, Analogues of the Kolosov–Muskhelishvili general representation formulas and Cauchy-Riemann conditions in the theory of elastic mixtures. Georgian Math. J. 4(1997), No. 3, 223–242.

4. M. O. Basheleishvili, Two-dimensional boundary-value problems of statics of the theory of elastic mixtures. Mem. Differential Equations Math. Phys. 6(1995), 59–105.

5. V. D. Kupradze, T. G. Gegelia, M. O. Basheleishvili, and T. V. Burchuladze, Three-dimensional problems of the mathematical theory of elasticity and thermoelasticity. (Translated from Russian) North-Holland Series in Applied Mathematics and Mechanics, v. 25, North-Holland Publ. Co., Amsterdam–New York–Oxford, 1979; Russian original: Nauka, Moscow, 1976.

6. M. O. Basheleishvili, Two-dimensional problems of elasticity of anisotropic bodies. Mem. Differential Equations Math. Phys. 16(1999), 9–140.

7. M. O. Basheleishvili, Application of analogues of general Kolosov–Muskhelishvili representations in the theory of elastic mixtures. Georgian Math. J. 6(1999). No. 1, 1–18.

(Received 25.12.2000)

Authors’ addresses:

M. Basheleishvili I. Vekua Institute of Applied Mathematics Tbilisi State University 2, University St., Tbilisi 380043 Georgia

K. Svanadze A. Tsereteli Kutaisi State University 59, Queen Tamar Ave., Kutaisi 384004 Georgia