GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 1, 1999, 1-18

APPLICATION OF ANALOGUES OF GENERAL KOLOSOV–MUSKHELISHVILI REPRESENTATIONS IN

THE THEORY OF ELASTIC MIXTURES

M. BASHELEISHVILI

Abstract. The existence and uniqueness of a solution of the first, the second and the third plane boundary value problem are considered for the basic homogeneous equations of statics in the theory of elastic mixtures. Applying the general Kolosov–Muskhelishvili representations from [1], these problems can be splitted and reduced to the first and the second boundary value problem for an elliptic equation which structurally coincides with the equation of statics of an isotropic elastic body.

Introduction

Analogues of Kolosov–Muskhelishvili representation formulas were obtained in [1] for equations of statics in the theory of elastic mixtures. These formulas have various applications. In this paper they will be used to reduce the first, the second and the third boundary value problems of statics in the theory of elastic mixtures [2] to the first and the second boundary value problem for an elliptic equation which structurally coincides with an equation of statics of an isotropic elastic body. It will be shown that the theory and methods of solving the boundary value problems developed in [3] can be extended to the plane boundary value problems of statics in the theory of elastic mixtures.

§ 1. First Boundary Value Problem

As is known [2], the first boundary value problem is considered with a displacement vector given on the boundary. To split this problem we have

1 to use the general representations of displacement vector components [1] having the form

1991 Mathematics Subject Classification. 73C35.

Key words and phrases. Elasticity, elastic mixtures, general representations, Fredholm integral equations, boundary value problems, splitted boundary value problems.

1072-947X/99/0100-0001$15.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1999 Plenum Publishing Corporation

\begin{matrix}u_{1} + iu_{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)}, \\ u_{3} + iu_{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)}, \end{matrix} (1.1)

where u = (u_{1}, u_{2}, u_{3}, u_{4}) is a four-dimensional displacement vector; φ_{1}(z), φ_{2}(z), ψ_{1}(z), ψ_{2}(z) are arbitrary analytic functions, and

\begin{matrix}m_{1} = e_{1} + \frac{l_{4}}{2} , m_{2} = e_{2} + \frac{l_{5}}{2} , m_{3} = e_{3} + \frac{l_{6}}{2} , \\ l_{1} = \frac{a_{2}}{d_{2}} , l_{2} = −\frac{c}{d_{2}} , l_{3} = \frac{a_{1}}{d_{2}} , d_{2} = a_{1}a_{2} − c^{2} > 0, \\ 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}} , \\ d_{1} = (a_{1} + b_{1})(a_{2} + b_{2}) − (c + d)^{2} > 0 \end{matrix} (1.2)

the coefficients a_{1}, b_{1}, a_{2}, b_{2}, c, d are contained in the basic homogeneous equations of statics in the theory of elastic mixtures which are written as [1]:

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

(1.3)

where u′ = (u_{1}, u_{2}), u′′ = (u_{3}, u_{4}) are partial displacements of an elastic mixture and

θ′ = div u′, θ′′ = div u′′.  (1.4)

To reduce the first boundary value problem of statics in the theory of elastic mixtures to the first boundary value problem of statics of an isotropic elastic body we rewrite (1.1) as

u_{1} + Xu_{3} + i(u_{2} + Xu_{4}) = (m_{1} + Xm_{2})φ_{1}(z) + (m_{2} + Xm_{3})φ_{2}(z) +

+\frac{z}{2}[(l_{4} + Xl_{5})\overline{φ′_{1}(z)} + (l_{5} + Xl_{6})\overline{φ′_{2}(z)}] + \overline{ψ_{1}(z)} + X\overline{ψ_{2}(z)}, (1.5)

where X is an arbitrary real constant. We define the unknown X by the equation

\frac{m_{2} + Xm_{3}}{m_{1} + Xm_{2}} = \frac{l_{5} + Xl_{6}}{l_{4} + Xl_{5}} .  (1.6)

Using the formulas [1,2]

2Δ_{0}ε_{1} = l_{5}m_{2} − l_{4}m_{3}, 2Δ_{0}ε_{3} = l_{4}m_{2} − l_{5}m_{1},
2Δ_{0}ε_{2} = l_{6}m_{2} − l_{5}m_{3}, 2Δ_{0}ε_{4} = l_{5}m_{2} − l_{6}m_{1},  (1.7)
Δ_{0} = m_{1}m_{3} − m_{2}^{2} > 0

and their equivalent formulas

δ_{0}ε_{1} = 2(a_{2}b_{1} − cd) + b_{1}b_{2} − d^{2}, δ_{0}ε_{3} = 2(da_{2} − cb_{2}),
δ_{0}ε_{2} = 2(da_{1} − cb_{1}), δ_{0}ε_{4} = 2(a_{1}b_{2} − cd) + b_{1}b_{2} − d^{2},  (1.8)
δ_{0} = (2a_{1} + b_{1})(2a_{2} + b_{2}) − (2c + d)^{2} = 4Δ_{0}d_{1}d_{2} > 0,

by (1.6) we obtain the quadratic equation with respect to X

ε_{2}X^{2} − (ε_{4} − ε_{1})X − ε_{3} = 0.  (1.9)

Note that ε_{2} and ε_{3} do not vanish simultaneously. Indeed, if the equality ε_{2} = ε_{3} = 0 is satisfied, then by virtue of (1.8) we obtain

\frac{b_{1}}{a_{1}} = \frac{d}{c} = \frac{b_{2}}{a_{2}} = λ.  (1.10)

The constant λ ≠ 0, since for λ = 0 equality (1.10) implies b_{1} = d = b_{2} = 0 and (1.3) gives

a_{1}Δu′ + cΔu′′ = 0, cΔu′ + a_{2}Δu′′ = 0.

Hence, taking into account that d_{2} = a_{1}a_{2} − c^{2} > 0 [2], we have

Δu′ = 0, Δu′′ = 0.

Thus we have obtained a trivial case of an elastic mixture.

Now, substituting (1.10) into (1.3), we have

a_{1}(Δu′ + λ grad θ′) + c(Δu′′ + λ grad θ′′) = 0,
c(Δu′ + λ grad θ′) + a_{2}(Δu′′ + λ grad θ′′) = 0.

Hence, again taking into account that a_{1}a_{2} − c^{2} > 0, we find

Δu′ + λ grad θ′ = 0, Δu′′ + λ grad θ′′ = 0,  (1.11)

i.e., we have splitted the first boundary value problem. For u′ and u′′ the splitted problems are investigated by the same technique as the first boundary value problem of statics of an isotropic elastic body. Since 1+λ = \frac{a_{1}+b_{1}}{a_{1}} = \frac{a_{2}+b_{2}}{a_{2}} > 0, equations (1.11) are elliptic.

Thus we have shown that ε_{2} and ε_{3} do not vanish simultaneously.

In what follows it will be assumed that ε_{2} ≠ 0. This assumption can be made without loss of generality. Indeed, if ε_{3} ≠ 0 and ε_{2} = 0, then equation (1.9) has only one root, which is not sufficient for our further investigation. So we have to combine (1.1) and (1.5) as follows:

u_{3} + Y u_{1} + i(u_{4} + Y u_{2}).

By repeating the above arguments we obtain the quadratic equation

ε_{3}Y^{2} − (ε_{4} − ε_{1})Y = 0,

which yields

Y_{1} = 0, Y_{2} = \frac{ε_{4} − ε_{1}}{ε_{3}} .

Thus we have derived two roots, which enables us to accomplish our task. It is important that the roots of equation (1.9) be real values. The

discriminant of equation (1.9) can be written as

(ε_{4}−ε_{1})^{2}+4ε_{2}ε_{3} = \frac{4}{δ_{0}^{2}a_{1}a_{2}} \{[a_{2}(da_{1} −cb_{1})+a_{1}(da_{2} −cb_{2})]^{2}+d_{2}(a_{1}b_{2} −a_{2}b_{1})^{2}\}.

The latter expression vanishes only if conditions (1.10) are fulfilled. In what follows this trivial case will be omitted.

Thus equation (1.9) has two different real roots:

\begin{matrix}X_{1} = \frac{ε_{4} − ε_{1} + \sqrt{(ε_{4} − ε_{1})^{2} + 4ε_{2}ε_{3}}}{2ε_{2}} , \\ X_{2} = \frac{ε_{4} − ε_{1} − \sqrt{(ε_{4} − ε_{1})^{2} + 4ε_{2}ε_{3}}}{2ε_{2}} . \end{matrix}  (1.12)

Rewriting condition (1.5) for X_{1} and X_{2} separately, we have

\begin{aligned}u_{1} & + X_{1}u_{3} + i(u_{2} + X_{1}u_{4}) = (m_{1} + X_{1}m_{2})φ_{1}(z) + \\ & + (m_{2} + X_{1}m_{3})φ_{2}(z) − k_{1}z[(m_{1} + X_{1}m_{2})\overline{φ′_{1}(z)} + \\ & + (m_{2} + X_{1}m_{3})\overline{φ′_{2}(z)}] + \overline{ψ_{1}(z)} + X_{1}\overline{ψ_{2}(z)}, \\ u_{1} & + X_{2}u_{3} + i(u_{2} + X_{2}u_{4}) = (m_{1} + X_{2}m_{2})φ_{1}(z) + \\ & + (m_{2} + X_{2}m_{3})φ_{2}(z) − k_{2}z[(m_{1} + X_{2}m_{2})\overline{φ′_{1}(z)} + \\ & + (m_{2} + X_{2}m_{3})\overline{φ′_{2}(z)}] + \overline{ψ_{1}(z)} + X_{2}\overline{ψ_{2}(z)}, \end{aligned}  (1.13)

where we have introduced the notation

k_{j} = −\frac{l_{4} + X_{j}l_{5}}{2(m_{1} + X_{j}m_{2})}, j = 1, 2.  (1.14)

By virtue of (1.7) and (1.9) we readily obtain

k_{j} = ε_{1} + X_{j}ε_{2}, j = 1, 2.

Therefore k_{1} and k_{2} have the form

\begin{matrix}2k_{1} = ε_{1} + ε_{4} + \sqrt{(ε_{4} − ε_{1})^{2} + 4ε_{2}ε_{3}}, \\ 2k_{2} = ε_{1} + ε_{4} − \sqrt{(ε_{4} − ε_{1})^{2} + 4ε_{2}ε_{3}}. \end{matrix}  (1.15)

Introducing the notation

u_{1} + X_{j}u_{3} = v_{1}^{(j)}, u_{2} + X_{j}u_{4} = v_{2}^{(j)},  (1.16)
\begin{matrix}(m_{1} + X_{j}m_{2})φ_{1}(z) + (m_{2} + X_{j}m_{3})φ_{2}(z) = Φ_{j}(z), \\ ψ_{1}(z) + X_{j}ψ_{2}(z) = Ψ_{j}(z), j = 1, 2, \end{matrix}  (1.17)

we can rewrite (1.13) as

v_{1}^{(j)} + iv_{2}^{(j)} = Φ_{j}(z) − k_{j}z\overline{Φ′_{j}(z)} + \overline{Ψ_{j}(z)}, j = 1, 2,  (1.18)

where Φ_{1}(z), Φ_{2}(z), Ψ_{1}(z), Ψ_{2}(z) are new analytic functions.

Note that structurally (1.18) coincides with the general Kolosov–Muskhelishvili representation for displacement vector components.

It is obvious by (1.16) that if u_{1}, u_{2}, u_{3}, u_{4} are given, this will mean that v_{1}^{(j)} and v_{2}^{(j)}, j = 1, 2, are given, too. It is likewise obvious that if v_{1}^{(j)} and v_{2}^{(j)}, j = 1, 2, are found, then

\begin{matrix}u_{1} = \frac{−X_{2}v_{1}^{(1)} + X_{1}v_{1}^{(2)}}{X_{1} − X_{2}} , & u_{2} = \frac{−X_{2}v_{2}^{(1)} + X_{1}v_{2}^{(2)}}{X_{1} − X_{2}} \\ u_{3} = \frac{v_{1}^{(1)} − v_{1}^{(2)}}{X_{1} − X_{2}} , & u_{4} = \frac{v_{2}^{(1)} − v_{2}^{(2)}}{X_{1} − X_{2}} . \end{matrix}  (1.19)

(1.17) immediately implies that when Φ_{j}(z) and Ψ_{j}(z), j = 1, 2, are known, we can define φ_{1}(z), φ_{2}(z), ψ_{1}(z), ψ_{2}(z) uniquely and write them as

φ_{1}(z) = \frac{−(m_{2} + X_{2}m_{3})Φ_{1}(z) + (m_{2} + X_{1}m_{3})Φ_{2}(z)}{(X_{1} − X_{2})Δ_{0}},

φ_{2}(z) = \frac{(m_{1} + X_{2}m_{2})Φ_{1}(z) − (m_{1} + X_{1}m_{2})Φ_{2}(z)}{(X_{1} − X_{2})Δ_{0}},

ψ_{1}(z) = \frac{−X_{2}Ψ_{1} + X_{1}Ψ_{2}}{X_{1} − X_{2}} , ψ_{2}(z) = \frac{Ψ_{1} − Ψ_{2}}{X_{1} − X_{2}} .

(1.20)

Next, we shall show which equation is satisfied by the vector v^{(j)} = (v_{1}^{(j)}, v_{2}^{(j)}), j = 1, 2. Let this equation have the form

Δv^{(j)} + M_{j} grad div v^{(j)} = 0, j = 1, 2,  (1.21)

and define M_{j} depending on k_{j}.

By simple calculations we find from (1.18) that

div v^{j)} =2(1 − k_{j}) Re Φ′_{j}(z), Δv_{1}^{(j)} =−4k_{j} Re Φ′′_{j}(z), Δv_{2}^{(j)} =4k_{j} Im Φ′′_{j}(z).

After substituting the latter expression into (1.21) we obtain

M_{j} = \frac{2k_{j}}{1 − k_{j}} , j = 1, 2.  (1.22)

Therefore we have proved that the vector v^{(j)} = (v_{1}^{(j)}, v_{2}^{(j)}) defined by (1.18) satisfies equation (1.21) if M_{j} is given by (1.22). It will be shown below that |k_{j}| < 1, 1+M_{j} > 0, j = 1, 2. The latter inequality is a sufficient condition for system (1.21) to be elliptic.

Thus in the theory of elastic mixtures the first boundary value problem for an equation of statics is splitted, in the general case, into two problems for equation (1.21) which structurally coincides with an equation of statics of an isotropic elastic body.

Similarly to the first boundary value problem for an equation of statics of an isotropic elastic body [3] we can derive here an integral Fredholm equation of second order for equation (1.21) using the boundary conditions

v_{1}^{(j)} + iv_{2}^{(j)} = F_{j}(t), t ∈ S, j = 1, 2.  (1.23)

Indeed, if we define the functions Φ_{j}(z) and Ψ_{j}(z), j = 1, 2, from (1.18) by means of the potentials

\begin{matrix}Φ_{j}(z) = \frac{1}{2πi} ∫_{S} g_{j}(ζ)\frac{∂ ln σ}{∂s(y)}ds, \\ Ψ_{j}(z) = \frac{1}{2πi} ∫_{S} \overline{g}_{j}(ζ)\frac{∂ ln σ}{∂s(y)}ds − \frac{k_{j}}{2πi} ∫_{S} g_{j}(ζ)\frac{∂}{∂s(y)} \frac{\overline{ζ}}{σ} ds, \end{matrix} (1.24)

then

\overline{Φ′_{j}(z)} = −\frac{1}{2πi} ∫_{S} \overline{g_{j}(ζ)}\frac{∂}{∂s(y)} \frac{1}{\overline{σ}} ds

and

v_{1}^{(j)}+iv_{2}^{(j)} = \frac{1}{2πi} ∫_{S} g_{j}(ζ) \frac{∂}{∂s(y)} ln \frac{σ}{\overline{σ}} ds+ \frac{k_{j}}{2πi} ∫_{S} \overline{g_{j}(ζ)} \frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} ds. (1.25)

In (1.24) and (1.25) g_{j}(ζ) are the desired functions of the point ζ = y_{1} + iy_{2}, where y_{1} and y_{2} are the coordinates of the point y ∈ s, σ = z − ζ, \overline{σ} = \overline{z} − \overline{ζ}, z = x_{1} + ix_{2}, and

\frac{∂}{∂s(y)} = n_{1}(y)\frac{∂}{∂y_{2}} − n_{2}(y)\frac{∂}{∂y_{1}} ,  (1.26)

where n = (n_{1}(y), n_{2}(y)) is the external (with respect to the finite domain D^{+}) normal unit vector at the point y, while s is the Lyapunov curve.

Passing to the limit as z → t ∈ s, externally or internally, to define g_{j} we obtain the integral Fredholm equation of second kind

±g_{j}(t) + \frac{1}{2πi} ∫_{S} g_{j}(ζ)\frac{∂}{∂s(y)} ln \frac{t − ζ}{\overline{t} − \overline{ζ}} ds +
+ \frac{k_{j}}{2πi} ∫_{S} \overline{g_{j}(ζ)}\frac{∂}{∂s(y)} \frac{t − ζ}{\overline{t} − \overline{ζ}} ds = F_{j}(t),  (1.27)

where

F_{j}(t) = f_{1} + X_{j}f_{3} + i(f_{2} + X_{j}f_{4}), j = 1, 2,  (1.28)

and f_{k}(t) = (u_{k})^{±}, k = \overline{1, 4}.

Equations (1.27) have a simple form and are very helpful both for theoretical investigations and for an effective solution of the first boundary value problem. These equations actually coincide with the Sherman–Lauricella equation [3].

Let us investigate the parameters k_{1} and k_{2} in (1.27) which are defined by (1.15).

Formula (1.15) gives rise to

k_{1} + k_{2} = ε_{1} + ε_{4}, k_{1}k_{2} = ε_{1}ε_{4} − ε_{2}ε_{3},  (1.29)

which implies

\begin{matrix}1 − k_{1} + 1 − k_{2} = 2 − ε_{1} − ε_{4}, \\ (1 − k_{1})(1 − k_{2}) = 1 − ε_{1} − ε_{4} + ε_{1}ε_{4} − ε_{2}ε_{3}. \end{matrix}  (1.30)

By easy calculations it follows from (1.8) that

\begin{matrix}δ_{0}(ε_{1} + ε_{4}) = 2(a_{1}b_{2} + a_{2}b_{1} − 2cd + b_{1}b_{2} − d^{2}), \\ δ_{0}(ε_{1}ε_{4} − ε_{2}ε_{3}) = b_{1}b_{2} − d^{2}. \end{matrix}  (1.31)

Substituting the latter expression into (1.30) and performing some simple transformations, we obtain

\begin{matrix}1 − k_{1} + 1 − k_{2} = \frac{2}{δ_{0}} \{4Δ_{1} − 3λ_{5}(μ_{1} + μ_{2} + 2μ_{3}) + \\ +\frac{1}{a_{1}(b_{2} − λ_{5})}[a_{1}(b_{2} − λ_{5}) − c(d + λ_{3})]^{2} + \\ +\frac{d_{2}(b_{1} − λ_{5})(b_{2} − λ_{5}) + c^{2}[(b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{5})^{2}]}{a_{1}(b_{2} − λ_{5})}\}, \\ (1 − k_{1})(1 − k_{2})=1− \frac{2(a_{1}b_{2}+a_{2}b_{1}−2cd+b_{1}b_{2}−d^{2})}{δ_{0}} + \\ +\frac{b_{1}b_{2} − d^{2}}{δ_{0}} = \frac{4d_{2}}{δ_{0}} . \end{matrix} (1.32)

Since the potential energy of an elastic mixture is positive definite [2], we have

Δ_{1} = μ_{1}μ_{2} − μ_{3}^{2} > 0, λ_{5} < 0, μ_{1} + μ_{2} + 2μ_{3} > 0,
d_{2} = a_{1}a_{2} − c^{2} = Δ_{1} − λ_{5}(μ_{1} + μ_{2} + 2μ_{3}) > 0,  (1.33)
a_{1} > 0, b_{2} − λ_{5} > 0, (b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{5})^{2} > 0.

With (1.33) taken into account, (1.32) implies that the sum and the product of two values 1 − k_{1} and 1 − k_{2} are greater than zero. Hence we conclude that each value is positive.

Therefore

k_{1} < 1, k_{2} < 1.  (1.34)

In quite a similar manner, with (1.15) taken into account, we obtain

1 + k_{1} + 1 + k_{2} = 2 + ε_{1} + ε_{4},
(1 + k_{1})(1 + k_{2}) = 1 + ε_{1} + ε_{4} + ε_{1}ε_{4} − ε_{2}ε_{3}.

Now using (1.31) and performing cumbersome calculations, we derive

1 + k_{1} + 1 + k_{2} = \frac{2}{δ_{0}} \{4Δ_{1} − λ_{5}(μ_{1} + μ_{2} + 2μ_{3}) +

+2[(b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{5})^{2}] +

+\frac{1}{(3μ_{1} − λ_{5})(b_{2} − λ_{5})}[(3μ_{1} − λ_{5})(b_{2} − λ_{5}) − (3μ_{3} + λ_{5})(d + λ_{5})]^{2} +

+\frac{[(b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{5})^{2}](3μ_{1} − λ_{5})(3μ_{2} − λ_{5})}{(3μ_{1} − λ_{5})(b_{2} − λ_{5})} +

+3(d + λ_{5})^{2}[3Δ_{1} − λ_{5}(μ_{1} + μ_{2} + 2μ_{3})]},

(1 + k_{1})(1 + k_{2}) = \frac{4}{δ_{0}μ_{1}} \{(a_{1} + b_{1})Δ_{1} + \frac{μ_{1}(b_{2} − λ_{5}) + μ_{3}^{2}}{b_{2} − λ_{5}} ×

×[(b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{3})^{2}] + \frac{1}{b_{2} − λ_{5}} [μ_{1}(b_{2} − λ_{5}) − μ_{3}(d + λ_{5})]^{2}\}.

Hence, as above, we conclude that 1 + k_{1} and 1 + k_{2} are greater than zero, i. e., which together with (1.34) leads to

−1 < k_{j} < 1, j = 1, 2.  (1.35)

It is interesting to note that for the parameter k_{1} one can obtain a more narrow change interval. Indeed, by virtue of (1.32) formula (1.22) for j = 1 can be rewritten as

M_{1} = \frac{2k_{1}(1 − k_{2})δ_{0}}{4d_{2}} .

By virtue of (1.15) and (1.31) the latter formula takes the form

M_{1} = \frac{a_{1}b_{2} + a_{2}b_{1} − 2cd}{2d_{2}} + \frac{δ_{0}}{4d_{2}} \sqrt{(ε_{4} − ε_{1})^{2} + 4ε_{2}ε_{3}},

which implies

M_{1} + \frac{1}{2} = \frac{a_{1}(b_{2} − λ_{5}) + a_{2}(b_{1} − λ_{5}) − 2c(d + λ_{5})Δ_{1}}{2d_{2}} +

+ \frac{δ_{0}}{4d_{2}} \sqrt{(ε_{4} − ε_{1})^{2} + 4ε_{2}ε_{3}} > 0.

Thus we have obtained

M_{1} > −\frac{1}{2}.  (1.36)

Using now (1.22), it is easy to establish that k_{1} > −\frac{1}{3} and thus we obtain the interval

k_{1} ∈ ] − \frac{1}{3}, 1[.  (1.37)

Note that though inequality (1.36) does not hold for M_{2}, we can rewrite M_{2} similarly to M_{1} as follows:

M_{2} = \frac{a_{1}b_{2} + a_{2}b_{1} − 2cd}{2d_{2}} − \frac{δ_{0}}{4d_{2}} \sqrt{(ε_{4} − ε_{1})^{2} + 4ε_{2}ε_{3}}.

Applying (1.35), we find from (1.22) that

1 + M_{j} = \frac{1 + k_{j}}{1 − k_{j}} > 0, j = 1, 2.

Therefore equation (1.21) is elliptic.

Let us now rewrite (1.21) as

Δv + M grad div v = 0.  (1.38)

where v = v^{(j)}, j = 1, 2, when M = M_{j}.

For equation (1.38) we introduce the generalized stress vector [2]:

T^{κ}v = (1 + κ)\frac{∂v}{∂n} + (M − κ)n div v + κs(\frac{∂v_{2}}{∂x_{1}} − \frac{∂v_{1}}{∂x_{2}} ),

where n = (n_{1}, n_{2}) is an arbitrary unit vector, s = −(n_{2}, n_{1}), and κ an arbitrary constant.

For the generalized stress vector we choose a particular case with κ = \frac{M}{M+2}.

Then T^{κ} ≡ N and the Green formula for the finite domain D^{+} and the infinite domain D^{−} will respectively have the form

∫_{D^{+}} N(v, v)dy_{1}dy_{2} = ∫_{S} vNv ds,  (1.39)
∫_{D^{−}} N(v, v)dy_{1}dy_{2} = − ∫_{S} vNv ds,  (1.40)

where

N(v, v) = \frac{(1 + M)^{2}}{M + 2} (\frac{∂v_{1}}{∂y_{1}} + \frac{∂v_{2}}{∂y_{2}} )^{2} + \frac{1}{M + 2}(\frac{∂v_{2}}{∂y_{1}} − \frac{∂v_{1}}{∂y_{2}} )^{2} +

+\frac{M + 1}{M + 2}[(\frac{∂v_{1}}{∂y_{2}} + \frac{∂v_{2}}{∂y_{1}} )^{2} + (\frac{∂v_{1}}{∂y_{1}} − \frac{∂v_{2}}{∂y_{2}} )^{2}].  (1.41)

For equation (1.38) 1 + M > 0, which means that N(v, v) defined by (1.41) is positive definite.

Formulae (1.39) and (1.40) hold when v is a regular vector [2]. Moreover, for the infinite domain D^{−} the vector v satisfies the conditions

v = O(1), \frac{∂v}{∂y_{k}} = O(R^{−2}), k = 1, 2,

where R^{2} = y_{1}^{2} + y_{2}^{2}.

The operator N plays an important role in the investigation of the first boundary value problem.

Our further discussion and effective solution of the first boundary value problem proceed exactly in the same way as for an equation of statics of an isotropic elastic body.

§ 2. Second Boundary Value Problem

The second boundary value problem is investigated with the vector T u given on the boundary. The projections of this vector are defined as follows [1]:

\begin{aligned}(T u)_{2} & − i(T u)_{1} = \frac{∂}{∂s(x)}\{(A_{1} − 2)φ_{1}(z) + A_{2}φ_{2}(z) + \\ & + z[B_{1}\overline{φ′_{1}(z)} + B_{2}\overline{φ′_{2}(z)}] + 2μ_{1}\overline{ψ_{1}(z)} + 2μ_{3}\overline{ψ_{2}(z)}\}, \\ (T u)_{4} & − i(T u)_{3} = \frac{∂}{∂s(x)}\{A_{3}φ_{1}(z) + (A_{4} − 2)φ_{2}(z) + \\ & + z[B_{3}\overline{φ′_{1}(z)} + B_{4}\overline{φ′_{2}(z)}] + 2μ_{3}\overline{ψ_{1}(z)} + 2μ_{2}\overline{ψ_{2}(z)}\}, \end{aligned} (2.1)

where φ_{k}(z) and ψ_{k}(z) (k = 1, 2) are arbitrary analytic functions; the constants A_{k} and B_{k} (k = \overline{1, 4}) have the values:

\begin{matrix}A_{1} = 2(μ_{1}m_{1} + μ_{3}m_{2}) = 2 + B_{1} + 2λ_{5} \frac{a_{2} + c}{d_{2}}, B_{1} = μ_{1}l_{4} + μ_{3}l_{5}, \\ A_{2} = 2(μ_{1}m_{2} + μ_{3}m_{3}) = B_{2} − 2λ_{5} \frac{a_{1} + c}{d_{2}}, B_{2} = μ_{1}l_{5} + μ_{3}l_{6}, \\ A_{3} = 2(μ_{3}m_{1} + μ_{2}m_{2}) = B_{3} − 2λ_{5} \frac{a_{2} + c}{d_{2}}, B_{3} = μ_{3}l_{4} + μ_{2}l_{5}, \\ A_{4} = 2(μ_{3}m_{2} + μ_{2}m_{3}) = 2 + B_{4} + 2λ_{5} \frac{a_{1} + c}{d_{2}}, B_{4} = μ_{3}l_{5} + μ_{2}l_{6}, \end{matrix} (2.2)

the operator \frac{∂}{∂s(x)} is defined by (1.26).

Representations (2.1) can be rewritten equivalently as

\begin{aligned}F_{2} & − iF_{1} + c_{1} = (A_{1} − 2)φ_{1}(z) + A_{2}φ_{2}(z) + \\ & + z[B_{1}\overline{φ′_{1}(z)} + B_{2}\overline{φ′_{2}(z)}] + 2μ_{1}\overline{ψ_{1}(z)} + 2μ_{3}\overline{ψ_{2}(z)}, \\ F_{4} & − iF_{3} + c_{2} = A_{3}φ_{1}(z) + (A_{4} − 2)φ_{2}(z) + \\ & + z[B_{3}\overline{φ′_{1}(z)} + B_{4}\overline{φ′_{2}(z)}] + 2μ_{3}\overline{ψ_{1}(z)} + 2μ_{2}\overline{ψ_{2}(z)}, \end{aligned}  (2.3)

where c_{1} and c_{2} are arbitrary constants and

F_{k} = ∫_{0}^{S(x)}(T u)_{k}ds, k = \overline{1, 4}.  (2.4)

Now we combine (2.3) as follows:

F_{2} + XF_{4} − i(F_{1} + XF_{3}) + c_{1} + Xc_{2} = (A_{1} − 2 + XA_{3})φ_{1}(z) +

+[A_{2} + X(A_{4} − 2)]φ_{2}(z) + z[(B_{1} + XB_{3})\overline{φ′_{1}(z)} +
+(B_{2} + XB_{4})\overline{φ′_{2}(z)}] + 2(μ_{1} + Xμ_{3})\overline{ψ_{1}(z)} + 2(μ_{3} + Xμ_{2})\overline{ψ_{2}(z)}, (2.5)

where X is the unknown constant value.

Define X by the equation

\frac{B_{2} + XB_{4}}{B_{1} + XB_{3}} = \frac{A_{2} + X(A_{4} − 2)}{A_{1} − 2 + XA_{3}}.  (2.6)

Using the notation [2]

\begin{matrix}H_{1} = B_{1}(2 − A_{4}) + B_{2}A_{3}, H_{2} = B_{1}A_{2} + B_{2}(2 − A_{1}), \\ H_{3} = B_{3}(2 − A_{4}) + B_{4}A_{3}, H_{4} = B_{3}A_{2} + B_{4}(2 − A_{1}), \end{matrix}  (2.7)

we can rewrite (2.6) as

H_{3}X^{2} − (H_{4} − H_{1})X − H_{2} = 0.  (2.8)

By substituting the coefficients from (2.2) and (1.2) into (2.7) we obtain, for H_{k} (k = \overline{1, 4}), the new expressions

\begin{matrix}H_{1} = −Δ_{2} + \frac{2λ_{5}}{d_{2}} [(a_{1} + c)A_{3} + (a_{2} + c)(A_{4} − 2)], \\ H_{2} = \frac{2λ_{5}}{d_{2}} [(a_{1} + c)(2 − A_{1}) − (a_{2} + c)A_{2}], \\ H_{3} = \frac{2λ_{5}}{d_{2}} [(a_{2} + c)(2 − A_{4}) − (a_{1} + c)A_{3}], \\ H_{4} = −Δ_{2} − \frac{2λ_{5}}{d_{2}} [(a_{1} + c)(2 − A_{1}) − (a_{2} + c)A_{2}], \end{matrix}  (2.9)

where

\begin{matrix}Δ_{2} = (2 − A_{1})(2 − A_{4}) − A_{2}A_{3}, \\ H_{1} + H_{3} = −Δ_{2}, H_{2} + H_{4} = −Δ_{2}. \end{matrix}  (2.10)

We shall show that the condition Δ_{2} ≠ 0 is fulfilled. To this end, using (2.2), we rewrite Δ_{2} as

Δ_{2} = (B_{1} + 2λ_{5} \frac{a_{2} + c}{d_{2}})(B_{4} + 2λ_{5} \frac{a_{1} + c}{d_{2}}) −

−(B_{2} − 2λ_{5} \frac{a_{1} + c}{d_{2}})(B_{3} − 2λ_{5} \frac{a_{2} + c}{d_{2}}) =

=B_{1}B_{4}−B_{2}B_{3}+ \frac{2λ_{5}}{d_{2}} [(a_{1}+c)(B_{1}+B_{3})+(a_{2}+c)(B_{2}+B_{4})]. (2.11)

Note that, by virtue of (1.2), from (2.2) we readily obtain

B_{1}B_{4} − B_{2}B_{3} = Δ_{1} \frac{b_{1}b_{2} − d^{2}}{d_{1}d_{2}} ,  (2.12)

\begin{matrix}B_{1} + B_{3} = −\frac{1}{d_{1}} (b_{1}b_{2} − d^{2} + a_{2}b_{1} − cd + da_{2} − cb_{2}), \\ B_{2} + B_{4} = −\frac{1}{d_{1}} (b_{1}b_{2} − d^{2} + a_{1}b_{2} − cd + da_{1} − cb_{1}). \end{matrix}  (2.13)

Using the identity

(a_{1}+c)(a_{2}b_{1}−cd+da_{2}−cb_{2})+(a_{2}+c)(a_{1}b_{2}−cd+da_{1}−cb_{1}) = d_{2}(b_{1}+b_{2}+2d),

which is easy to prove, and taking into account (2.12) and (2.13), expression (2.11) can be rewritten as

Δ_{2}d_{1}d_{2} = [Δ_{1} − 2λ_{5}(a_{1} + a_{2} + 2c)](b_{1}b_{2} − d^{2}) − 2λ_{5}d_{2}(b_{1} + b_{2} + 2d) =

= [Δ_{1} − 2λ_{5}(a_{1} + a_{2} + 2c)][(b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{5})^{2}] −
−λ_{5}(b_{1} + b_{2} + 2d)Δ_{1}.  (2.14)

Since d_{1} > 0, d_{2} > 0, λ_{5} < 0, a_{1} + a_{2} + 2c ≡ μ_{1} + μ_{2} + 2μ_{3} > 0, (b_{1}−λ_{5})(b_{2}−λ_{5})−(d+λ_{5})^{2} > 0 and b_{1}+b_{2}+2d ≡ b_{1}−λ_{5}+b_{2}−λ_{5}+2(d+λ_{5}) > 0, we have Δ_{2} > 0.

Now we shall show that H_{2} and H_{3} do not vanish simultaneously. Indeed, if it is assumed that H_{2} and H_{3} vanish simultaneously, then for λ_{5} ≠ 0 and d_{2} ≠ 0 (2.9) will imply (a_{1} + c)(2 − A_{1}) − (a_{2} + c)A_{2} = 0, (a_{2} + c)(2 − A_{4}) − (a_{1} + c)A_{3} = 0, i. e.,

\frac{a_{2} + c}{a_{1} + c} = \frac{2 − A_{1}}{A_{2}} = \frac{A_{3}}{2 − A_{4}} .

Hence we conclude that Δ_{2} = 0, which is impossible.

Without loss of generality we can take H_{3} ≠ 0. This can be shown in the same manner as a similar statement in §1.

By solving equation (2.8) we obtain

X_{1} = 1, X_{2} = −\frac{H_{2}}{H_{3}} = H_{0}.  (2.15)

Rewriting now (2.5) for X_{1} and X_{2} separately, we have

F_{2}+F_{4}−i(F_{1}+F_{3})+c_{1}+c_{2} =Φ(z)+z\overline{Φ′(z)}+\overline{Ψ(z)},  (2.16)
F_{2}+H_{0}F_{4}−i(F_{1}+H_{0}F_{3})+c_{1}+H_{0}c_{2} =Φ_{0}(z)+k_{0}z\overline{Φ′_{0}(z)}+\overline{Ψ_{0}(z)}, (2.17)

where

k_{0} = \frac{B_{1} + H_{0}B_{3}}{A_{1} − 2 + H_{0}A_{3}} ,  (2.18)

and the functions Φ(z), Ψ(z), Φ_{0}(z), Ψ_{0}(z) are defined as

\begin{matrix}Φ(z) = (B_{1} + B_{3})φ_{1}(z) + (B_{2} + B_{4})φ_{2}(z), \\ Φ_{0}(z)=(A_{1}−2+H_{0}A_{3})φ_{1}(z)+(A_{2}+H_{0}A_{4}−2H_{0})φ_{2}(z), \end{matrix}  (2.19)

\begin{matrix}Ψ(z) = 2(μ_{1} + μ_{3})ψ_{1}(z) + 2(μ_{3} + μ_{2})ψ_{2}(z), \\ Ψ_{0}(z) = 2(μ_{1} + H_{0}μ_{3})ψ_{1}(z) + 2(μ_{3} + H_{0}μ_{2})ψ_{2}(z). \end{matrix}  (2.20)

We next substitute the value of H_{0} from (2.15) and the values of H_{2} and H_{3} from (2.7) into (2.18). After performing some simple transformations we obtain

k_{0} = \frac{Δ_{1}(b_{1}b_{2} − d^{2})}{Δ_{2}d_{1}d_{2}}.  (2.21)

Let us show that the value of k_{0} lies in the interval ]−1, 1[. Indeed, using (2.14), from (2.21) we have

1 − k_{0} = \frac{Δ_{2}d_{1}d_{2} − Δ_{1}(b_{1}b_{2} − d^{2})}{Δ_{2}d_{1}d_{2}} =

=−2λ_{5} \frac{(b_{1}+b_{2}+2d)Δ_{1}+(a_{1}+a_{2}+2c)[(b_{1}−λ_{5})(b_{2}−λ_{5})−(d+λ_{5})^{2}]}{Δ_{2}d_{1}d_{2}} >0.

Hence k_{0} < 1. (2.21) now readily implies

1 + k_{0} = \frac{2}{Δ_{2}d_{1}} [(b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{5})^{2}].

Thus we have found that

−1 < k_{0} < 1.  (2.22)

Note that H_{0} ≠ 1. Indeed, if H_{0} = 1, then (2.18) implies k_{0} = 1, which is impossible because (2.22) holds.

After Φ(z), Φ_{0}(z), Ψ(z) and Ψ_{0}(z) are found, using (2.19) and (2.20) we define φ_{k}(z) and ψ_{k}(z), k = 1, 2, uniquely, since the determinants of the respective transformations are equal to (H_{0} − 1)Δ_{2} and (H_{0} − 1)Δ_{1}.

Applying arguments similar to those used for representation (1.18), the desired functions Φ(z), Ψ(z), Φ_{0}(z) and Ψ_{0}(z) from (2.16) and (2.17) must be sought for in the form

Φ(z) = \frac{1}{2πi} ∫_{S} g(ζ)\frac{∂ ln σ}{∂s(y)}ds,

Ψ(z) = \frac{1}{2πi} ∫_{S} \overline{g(ζ)}\frac{∂ ln σ}{∂s(y)}ds + \frac{1}{2πi} ∫_{S} g(ζ)\frac{∂}{∂s(y)} \frac{\overline{ζ}}{σ} ds,

Φ_{0}(z) = \frac{1}{2πi} ∫_{S} g_{0}(ζ)\frac{∂ ln σ}{∂s(y)}ds,

Ψ_{0}(z) = \frac{1}{2πi} ∫_{S} \overline{g_{0}(ζ)}\frac{∂ ln σ}{∂s(y)}ds + \frac{k_{0}}{2πi} ∫_{S} g_{0}(ζ)\frac{∂}{∂s(y)} \frac{\overline{ζ}}{σ} ds.

In that case (2.16) and (2.17) can be rewritten as

\begin{aligned}F_{2} + F_{4} & − i(F_{1} + F_{3}) + c_{1} + c_{2} = \frac{1}{2πi} ∫_{S} g(ζ)\frac{∂}{∂s(y)} ln \frac{σ}{\overline{σ}} ds − \\ & − \frac{1}{2πi} ∫_{S} \overline{g(ζ)}\frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} ds, \\ F_{2} + H_{0} & F_{4} − i(F_{1} + H_{0}F_{3}) + c_{1} + H_{0}c_{2} = \\ & = \frac{1}{2πi} ∫_{S} g_{0}(ζ)\frac{∂}{∂s(y)}ln\frac{σ}{\overline{σ}} ds− \frac{k_{0}}{2πi} ∫_{S} \overline{g_{0}(ζ)}\frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} ds. \end{aligned} (2.23)

Now passing to the limit in (2.23) as z → t ∈ s, internally or externally, to define g and g_{0} we obtain the integral Fredholm equations of second order:

±g(t) + \frac{1}{2πi} ∫_{S} g(ζ)\frac{∂}{∂s(y)} ln \frac{t − ζ}{\overline{t} − \overline{ζ}} ds −

− \frac{1}{2πi} ∫_{S} \overline{g(ζ)}\frac{∂}{∂s(y)} \frac{t − ζ}{\overline{t} − \overline{ζ}} ds = f(t),  (2.24)
±g_{0}(t) + \frac{1}{2πi} ∫_{S} g_{0}(ζ)\frac{∂}{∂s(y)} ln \frac{t − ζ}{\overline{t} − \overline{ζ}} ds −
− \frac{k_{0}}{2πi} ∫_{S} \overline{g_{0}(ζ)}\frac{∂}{∂s(y)} \frac{t − ζ}{\overline{t} − \overline{ζ}} ds = F (t),  (2.25)

where

f(t) = [F_{2} + F_{4} − i(F_{1} + F_{3})]^{±} + c_{1} + c_{2},
F (t) = [F_{2} + H_{0}F_{4} − i(F_{1} + H_{0}F_{3})]^{±} + c_{1} + H_{0}c_{2}.

One can investigate equations (2.24) and (2.25) in the same manner as the equation of the basic biharmonic problem and the first boundary value problem of statics of an isotropic elastic body [3].

Thus the second boundary value problem of statics in the theory of elastic mixtures is reduced to the second and the first plane boundary value problem of statics of an isotropic elastic body.

§ 3. Third Boundary Value Problem

As is known [2], the third boundary value problem is considered with the values u_{3} − u_{1}, u_{4} − u_{2}, F_{1} + F_{3} + c_{1}, F_{1} + F_{4} + c_{2}, given on the boundary, where c_{1} and c_{2} are arbitrary constants; u_{1}, u_{2}, u_{3}, u_{4} are the projections of the four-dimensional vector u, and F_{k}(x) (k = \overline{1, 4}) is defined by (2.4).

By virtue of (1.1), (2.2) and (2.3) the conditions of the third boundary value problem can be written as follows:

u_{3} − u_{1} + i(u_{4} − u_{2}) = (m_{2} − m_{1})φ_{1}(z) + (m_{3} − m_{2})φ_{2}(z) +

+ \frac{z}{2}[(l_{5} − l_{4})\overline{φ′_{1}(z)} + (l_{6} − l_{5})\overline{φ′_{2}(z)}] + \overline{ψ_{2}(z)} − \overline{ψ_{1}(z)},

F_{2} + F_{4} − i(F_{1} + F_{3}) + c_{1} + c_{2} = (B_{1} + B_{3})φ_{1}(z) +  (3.1)

+ (B_{2} + B_{4})φ_{2}(z) + z[(B_{1} + B_{3})\overline{φ′_{1}(z)} +

+ (B_{2} + B_{4})\overline{φ′_{2}(z)}] + 2(μ_{1} + μ_{3})\overline{ψ_{1}(z)} + 2(μ_{2} + μ_{3})\overline{ψ_{2}(z)}.

Introducing the notattion

\begin{matrix}(m_{2} − m_{1})φ_{1}(z) + (m_{3} − m_{2})φ_{2}(z) = Φ_{3}(z), \\ ψ_{2}(z) − ψ_{1}(z) = Ψ_{3}(z), \\ (B_{1} + B_{3})φ_{1}(z) + (B_{2} + B_{4})φ_{2}(z) = Φ(z), \\ 2(μ_{1} + μ_{3})ψ_{1}(z) + 2(μ_{2} + μ_{3})ψ_{2}(z) = Ψ(z), \end{matrix}  (3.2)

16 M. BASHELEISHVILI

we obtain

φ_{1}(z) = \frac{(B_{2} + B_{4})Φ_{3} + (m_{2} − m_{3})Φ}{Δ_{3}},

φ_{2}(z) = \frac{−(B_{1} + B_{3})Φ_{3} + (m_{2} − m_{1})Φ}{Δ_{3}},

ψ_{1}(z) = −\frac{μ_{2} + μ_{3}}{β} Ψ_{3} + \frac{1}{2β} Ψ, ψ_{2}(z) = \frac{μ_{1} + μ_{3}}{β} Ψ_{3} + \frac{1}{2β} Ψ,

(3.3)

where

Δ_{3} = 2Δ_{0}(α − β), α = \frac{m_{1} + m_{3} − 2m_{2}}{Δ_{0}} , β = μ_{1} + μ_{2} + 2μ_{3}, (3.4)

and Δ_{0} > 0 is given by (1.7)

Since Δ_{0} > 0 and Δ_{1} = μ_{1}μ_{2} −μ_{3}^{2} > 0, the constants α and β are greater than zero. We shall show that Δ_{3} > 0. For this it is sufficient to prove that α − β > 0. By virtue of (3.4) and (1.2) we have

α − β = \frac{a_{1} + a_{2} + 2c}{2Δ_{0}d_{2}} + \frac{a_{1} + a_{2} + 2c + b_{1} + b_{2} + 2d}{2Δ_{0}d_{1}} − (a_{1} + a_{2} + 2c) =

= \frac{1}{δ_{0}} [2(a_{1} + a_{2} + 2c)(d_{1} + d_{2}) + 2(b_{1} + b_{2} + 2d)d_{2} − δ(a_{1} + a_{2} + 2c)],

where δ_{0} is defined by formula (1.8) which can be rewritten as

δ_{0} = d_{1} + d_{2} + a_{2}(a_{1} + b_{1}) + a_{1}(a_{2} + b_{2}) − 2c(c + d).

Substituting this value of δ_{0} into the preceding formula, we obtain

α − β = \frac{1}{δ_{0}} [2(b_{1} + b_{2} + 2d)d_{2} + (a_{1} + a_{2} + 2c)(b_{1}b_{2} − d^{2})].

Hence, after some simple transformations, we readily have

α − β = \frac{1}{δ_{0}} \{(a_{1} + a_{2} + 2c)[(b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{5})^{2}] +
+ (b_{1} + b_{2} + 2d)[2Δ_{1} − λ_{5}(a_{1} + a_{2} + 2c)]}.

Thus we have shown that Δ_{3} > 0.

Now, using (3.3), (1.7) and (2.2), we obtain

(l_{5} − l_{4})\overline{φ′_{1}(z)} + (l_{6} − l_{5})\overline{φ′_{2}(z)} = −2k_{3}\overline{Φ′_{3}(z)} −

−\frac{2Δ_{0}}{Δ_{3}} (ε_{1} + ε_{3} − ε_{2} − ε_{4})\overline{Φ′(z)},  (3.5)

where

k_{3} = \frac{β(b_{1}b_{2} − d^{2})}{2Δ_{3}d_{1}d_{2}}.  (3.6)

Let us show that the parameter k_{3} changes in the interval ]−1, 1[. Taking into account the fact that the inequality Δ_{3} > 0 holds, performing some obvious transformations and applying the formulae

1 − k_{3} = \frac{b_{1} + b_{2} + 2d}{Δ_{3}d_{1}} = \frac{(b_{1} + d)^{2} + (b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{5})^{2}}{(b_{1} − λ_{5})Δ_{3}d_{1}} > 0,

1 + k_{3} = \frac{β[(b_{1} − λ_{5})(b_{2} − λ_{5}) − (d + λ_{5})^{2}] + (b_{1} + b_{2} + 2d)Δ_{1}}{Δ_{3}d_{1}d_{2}} > 0,

we conclude that −1 < k_{3} < 1.

Taking into account (3.5) and substituting (3.2) into (3.1) we obtain

u_{3} − u_{1} + i(u_{4} − u_{2}) = Φ_{3}(z) − k_{3}z\overline{Φ′_{3}(z)} + \overline{Ψ_{3}(z)} −

− \frac{Δ_{0}}{Δ_{3}} (ε_{1} + ε_{3} − ε_{2} − ε_{4})z\overline{Φ′(z)},

F_{2} + F_{4} − i(F_{1} + F_{3}) + c_{1} + c_{2} = Φ(z) + z\overline{Φ′(z)} + \overline{Ψ(z)}.

(3.7)

Applying the arguments of the preceding paragraphs, we must seek for the desired functions having the form

Φ(z) = \frac{1}{2πi} ∫_{S} g(ζ)\frac{∂ ln σ}{∂s(y)}ds,
Ψ(z) = \frac{1}{2πi} ∫_{S} \overline{g(ζ)}\frac{∂ ln σ}{∂s(y)}ds + \frac{1}{2πi} ∫_{S} g(ζ)\frac{∂}{∂s(y)} \frac{\overline{ζ}}{σ} ds,
Φ_{3}(z) = \frac{1}{2πi} ∫_{S} g_{3}(ζ)\frac{∂ ln σ}{∂s(y)}ds,  (3.8)
Ψ_{3}(z) = \frac{1}{2πi} ∫_{S} \overline{g_{3}(ζ)}\frac{∂ ln σ}{∂s(y)}ds + \frac{k_{3}}{2πi} ∫_{S} g_{3}(ζ)\frac{∂}{∂s(y)} \frac{\overline{ζ}}{σ} ds,

+ \frac{Δ_{0}}{2πiΔ_{3}} (ε_{1} + ε_{3} − ε_{2} − ε_{4}) ∫_{S} g(ζ)\frac{∂}{∂s(y)} \frac{\overline{ζ}}{σ} ds.

After substituting (3.8) into (3.7) and performing some simple transformations we obtain

u_{3} − u_{1} + i(u_{4} − u_{2})= \frac{1}{2πi} ∫_{S} g_{3}(ζ)\frac{∂}{∂s(y)} ln \frac{σ}{\overline{σ}} ds+ \frac{k_{3}}{2πi} ∫_{S} \overline{g_{3}(ζ)}\frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} ds+

+ \frac{Δ_{0}}{2πiΔ_{3}} (ε_{1} + ε_{3} − ε_{2} − ε_{4}) ∫_{S} \overline{g(ζ)}\frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} ds,

F_{2} + F_{4} − i(F_{1} + F_{3}) + c_{1} + c_{2} = \frac{1}{2πi} ∫_{S} g(ζ)\frac{∂}{∂s(y)} ln \frac{σ}{\overline{σ}} ds −

− \frac{1}{2πi} ∫_{S} \overline{g(ζ)}\frac{∂}{∂s(y)} \frac{σ}{\overline{σ}} ds.

Passing to the limit in this formula as z → t ∈ s, internally or externally, to define g_{3} and g we obtain the integral Fredholm equations of second order

±g_{3}(t) + \frac{1}{2πi} ∫_{S} g_{3}(ζ)\frac{∂}{∂s(y)} ln \frac{t − ζ}{\overline{t} − \overline{ζ}} ds + \frac{k_{3}}{2πi} ∫_{S} \overline{g_{3}(ζ)}\frac{∂}{∂s(y)} \frac{t − ζ}{\overline{t} − \overline{ζ}} ds +
+ \frac{Δ_{0}}{2πiΔ_{3}} (ε_{1}+ε_{3}−ε_{2}−ε_{4})∫_{S}\overline{g(ζ)}\frac{∂}{∂s(y)} \frac{t − ζ}{\overline{t}−\overline{ζ}} ds=f(t),  (3.9)
±g(t) + \frac{1}{2πi} ∫_{S} g(ζ)\frac{∂}{∂s(y)} ln \frac{t − ζ}{\overline{t} − \overline{ζ}} ds −
− \frac{1}{2πi} ∫_{S} \overline{g(ζ)}\frac{∂}{∂s(y)} \frac{t − ζ}{\overline{t} − \overline{ζ}} ds = F (t),  (3.10)

where

f(t)=[u_{3}−u_{1}+i(u_{4} − u_{2})]^{±}, F (t)=[F_{2}+F_{4}−i(F_{1}+F_{3})]^{±}+c_{1}+c_{2}.

Equation (3.10) is the integral Fredholm equation of the basic biharmonic problem. By solving this equation and substituting the found value of g into (3.9) we obtain the integral Fredholm equation with respect to the desired function g_{3}. This equation is investigated as the equation of the first boundary value problem of statics of an isotropic elastic body.

Thus in the theory of elastic mixtures the third boundary value problem of statics is splitted into two boundary value problems, of which one is the basic biharmonic problem and the other is the first boundary value problem of statics of an isotropic elastic body.

References

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

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

3. N. I. Muskhelishvili, Some basic problems of the mathematical theory of elasticity. (Russian) Nauka, Moscow, 1966.

(Received 15.05.1997)

Author’s address:

I. Vekua Institute of Applied Mathematics

Tbilisi State University

2, University St., Tbilisi 380043

Georgia