GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 6, 1999, 517-524

THE THREE-DIMENSIONAL PROBLEM OF STATICS OF THE ELASTIC MIXTURE THEORY WITH DISPLACEMENTS GIVEN ON THE BOUNDARY

M. BASHELEISHVILI

Abstract. The first three-dimensional boundary value problem is considered for the basic equations of statics of the elastic mixture theory in the finite and infinite domains bounded by the closed surfaces. It is proved that this problem splits into two problems whose investigation is reduced to the first boundary value problem for an elliptic equation which structurally coincides with an equation of statics of an isotropic elastic body. Using the potential method and the theory of Fredholm integral equations of second kind, the existence and uniqueness of the solution of the first boundary value problem is proved for the split equation.

Basic homogeneous equations of statics of the elastic mixture theory have the form [1]

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)

where a_{1}, b_{1}, c, d, a_{2}, b_{2} are the coefficients kharacterizing the physical properties of an elastic mixture, u′ and u′′ are partial displacements.

The problem to be considered in this paper is formulated as follows: given a continuous displacement vector on the boundary S, in the domain D^{+} (or D^{−}) find a solution u(u′, u′′) ∈ C(\overline{D}^{±}) ∩ C^{2}(D^{±}) of equation (1). This problem is investigated in the space C^{1,α}(\overline{D}^{±}) ∩ C^{2}(D^{±}) in [1] by the method of potentials and the theory of singular integral equations.

Here we give a different technique of solving the above problem. Our investigation is carried out using Fredholm integral equations of second kind.

517 Instead of the vectors u′ and u′′ we introduce the vectors

1991 Mathematics Subject Classification. 73C02.

Key words and phrases. Theory of elastic mixtures, boundary value problem, Fredholm integral equations.

1072-947X/99/1100-0517$16.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1999 Plenum Publishing Corporation

v′ = u′ + X_{1}u′′, v′′ = u′ + X_{2}u′′,  (2)

where X_{1} and X_{2} are the roots of the quadratic equation

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

Here the coefficients ε_{1}, ε_{2}, ε_{3}, ε_{4}, are defined as follows [2]:

\begin{aligned}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}, Δ_{0} =m_{1}m_{3}−m_{2}^{2} >0, \\ d_{1} & = (a_{1} + b_{1})(a_{2} + b_{2}) − (c + d)^{2} > 0, d_{2} = a_{1}a_{2} − c^{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} , 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{aligned} (4)

If the equality ε_{2} = ε_{3} = 0 holds, then by (4) we obtain

\frac{b_{1}}{a_{1}} = \frac{d}{c} = \frac{b_{2}}{a_{2}} = λ.  (5)

The substitution of these values into (1) gives

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

Since a_{1}a_{2}−c^{2} >0, we now have Δu′+λ grad div u′ =0, Δu′′+λ grad div u′′ =0, i.e., the basic equations and the first boundary value problem split so that they can be investigated as the first three-dimensional boundary value problem of statics of an isotropic elastic body [3].

In what follows it will be assumed without loss of generality that ε_{2} ≠ 0. Then the roots of equation (3) can be expressed as

\begin{aligned}2ε_{2}X_{1} & = ε_{4} − ε_{1} + \sqrt{(ε_{1} − ε_{4})^{2} + 4ε_{2}ε_{3}}, \\ 2ε_{2}X_{2} & = ε_{4} − ε_{1} − \sqrt{(ε_{1} − ε_{4})^{2} + 4ε_{2}ε_{3}}. \end{aligned}  (6)

Since

(ε_{1} − ε_{4})^{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}} > 0,

the roots X_{1} and X_{2} are different real values. Note that conditions (5) are fulfilled if the discriminant of equation (3) is equal to zero. In addition to X_{1} and X_{2}, we also need the values

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

We prove that

−1 < k_{j} < 1, j = 1, 2,  (7)

and

(1 − k_{1})(1 − k_{2}) = 4d_{2}/δ_{0}.  (8)

Now from (2) we have

u′ = \frac{−X_{2}v′ + X_{1}v′′}{X_{1} − X_{2}} , u′′ = \frac{v′ − v′′}{X_{1} − X_{2}} .  (9)

After substituting these experssions into (1) and performing some simple transformations, we obtain

(c−a_{1}X_{2})(Δv′+M_{1} grad div v′)+(a_{1}X_{1}−c)(Δv′′+M_{2} grad div v′′)=0,

(a_{2}−cX_{2})(Δv′+M_{1} grad div v′)+(cX_{1}−a_{2})(Δv′′+M_{2} grad div v′′)=0,

(10)

where

M_{1} = \frac{d − b_{1}X_{2}}{c − a_{1}X_{2}} = \frac{b_{2} − dX_{2}}{a_{1} − cX_{2}} , M_{2} = \frac{b_{1}X_{1} − d}{a_{1}X_{1} − c} = \frac{dX_{1} − b_{2}}{cX_{1} − a_{2}} . (11)

Let us consider equations (10) as a system with respect to Δv′ + M_{1} grad div v′ and Δv′′ + M_{2} grad div v′′. Since

(c − a_{1}X_{2})(cX_{1} − a_{2}) − (a_{2} − cX_{2})(a_{1}X_{1} − c) = d_{2}(X_{2} − X_{1}) ≠ 0,

from (10) we have

Δv′ + M_{1} grad div v′ = 0,  (12)
Δv′′ + M_{2} grad div v′′ = 0.  (13)

Thus we have shown that the three-dimensional boundary value problem of statics of the theory of elastic mixtures with given displacements on the boundary splits in the general case.

Equations (12) and (13) can be combined as one equation

Δv + M grad div v = 0,  (14)

where v =v′ for M =M_{1} and v =v′′ for M =M_{2}. It is obvious that equation (14) is an elliptic system if 1 + M >0, i.e., 1+M_{1} >0 and 1+M_{2} >0.

Let us show that these conditions hold for M_{1} and M_{2}. To this end, we have to write the expressions of M_{1} and M_{2} in a different form. From (11) it obviously follows that M_{1} = \frac{(b_{2}−dX_{2})(cX_{1}−a_{2})}{(a_{2}−cX_{2})(cX_{1}−a_{2})}. Taking into account

X_{1} + X_{2} = (ε_{4} − ε_{1})/ε_{2}, X_{1}X_{2} = −ε_{3}/ε_{2},

and performing some obvious calculations, from equation (3) we obtain

M_{1} = (a_{2}b_{1} − cd + (da_{1} − cb_{1})X_{1})/d_{2}.  (15)

In a similar manner we have

M_{2} = (a_{2}b_{1} − cd + (da_{1} − cb_{1})X_{2})/d_{2}.  (16)

The substitution of the values X_{1} and X_{2} from (6) into (15) and (16) gives

\begin{matrix}M_{1} = \frac{a_{1}b_{2} + a_{2}b_{1} − 2cd}{2d_{2}} + \frac{δ_{0}}{4d_{2}} \sqrt{(ε_{1} − ε_{4})^{2} + 4ε_{2}ε_{3}}, \\ M_{2} = \frac{a_{1}b_{2} + a_{2}b_{1} − 2cd}{2d_{2}} − \frac{δ_{0}}{4d_{2}} \sqrt{(ε_{1} − ε_{4})^{2} + 4ε_{2}ε_{3}}. \end{matrix}  (17)

Now using (8) and carrying out some elementary transformations we have

M_{1} = 2k_{1}/(1 − k_{1}), M_{2} = 2k_{2}/(1 − k_{2}).  (18)

Hence by virtue of (7) we obtain

1 + M_{1} = \frac{1 + k_{1}}{1 − k_{1}} > 0, 1 + M_{2} = \frac{1 + k_{2}}{1 − k_{2}} > 0.  (19)

It has thus been shown that (14) is an elliptic system. Now setting

v = ΔΨ − M(M + 1)^{−1} grad div Ψ

in (14), we obtain the equation

ΔΔΨ = 0.  (20)

that defines the unknown vector Ψ.

For equation (14) we introduce the generalized stress vector

T^{κ}v = (1 + κ)\frac{∂v}{∂n} + (M − κ)n div v + κ[n rot v],

where κ is an arbitrary real constant. Now in (20) let

Ψ = Er/2,  (21)

where E is the three-dimensional unit matrix,

r = \sqrt{∑_{k=1}^{3}(x_{k} − y_{k})^{2}},  (22)

x_{1}, x_{2}, x_{3} and y_{1}, y_{2}, y_{3} are the coordinates of the points x and y, respectively. By virtue of (21) and (22) we can rewrite the basic fundamental matrix for equation (14) as Γ(x − y) = ∥Γ_{kj}∥_{3×3}, where

Γ_{kj} = \frac{δ_{kj}}{r} − \frac{M}{2(1 + M)} \frac{∂^{2}r}{∂x_{k}∂x_{j}} ≡ \frac{2 + M}{2(1 + M)} \frac{δ_{kj}}{r} + \frac{M}{2(1 + M)} \frac{1}{r} \frac{∂r}{∂x_{k}} \frac{∂r}{∂x_{j}} .

Let us now calculate, with respect to the coordinates of x, the generalized stress operator of the basic fundamental matrix. After some obvious

calculations we obtain T^{κ}_{x}Γ(x − y) = ∥(T^{κ}_{x}Γ^{(j)})_{k}∥_{3×3}, where

(T^{κ}_{x}Γ^{(j)})_{k} = [(1 + κ)\frac{2 + M}{2(1 + M)} − κ]δ_{kj} \frac{∂}{∂n(x)} \frac{1}{r} + (1 + κ)\frac{3M}{2(1 + M)} ×

×\frac{∂r}{∂x_{k}} \frac{∂r}{∂x_{j}} \frac{∂}{∂n(x)} \frac{1}{r} + \frac{1}{2(1 + M)}[κ(2 + M) − M](n_{j} \frac{∂}{∂x_{k}} − n_{k} \frac{∂}{∂x_{j}} )\frac{1}{r}.

For the matrix T^{κ}_{x}Γ(x − y) to contain only a weak singulary at x = y it is necessary and sufficient that κ(2 + M) − M = 0, i.e., that

κ = M/(M + 2).  (23)

When κ is defined by (23), the generalized stress operator will be denoted by N. We have

N_{x}Γ(x−y)= \frac{1}{2+M} ∥2δ_{kj} +3M \frac{∂r}{∂x_{k}} \frac{∂r}{∂x_{j}} ∥_{3×3}· \frac{∂}{∂n(x)} \frac{1}{r}, k, j =1, 2, 3.

It is obvious that

N_{y}Γ(y − x) = \frac{1}{2 + M} ∥2δ_{kj} + 3M \frac{∂r}{∂x_{k}} \frac{∂r}{∂x_{j}} ∥_{3×3} · \frac{∂}{∂n(y)} \frac{1}{r}.  (24)

By direct calculations it is proved that each column of matrix (24) is a solution of equation (14) with respect to x when x ≠ y. Taking into account formulas (18) and (19), we obtain

κ_{1} = k_{1} = M_{1}/(2 + M_{1}), κ_{2} = k_{2} = M_{2}/(2 + M_{2}).

Green’s formula for the operator N is obtained in a usual manner [2] and for the regular vector v has the form

∫_{D^{+}} N(v, v)dy_{1} dy_{2} = ∫_{S} vNv ds,  (25)

where v is a solution of equation (14) and

N(v, v)=(\frac{1−2k}{3} + \frac{2k}{1−k} )(div v)^{2}+ \frac{1+k}{3} [(\frac{∂v_{1}}{∂y_{1}} − \frac{∂v_{2}}{∂y_{2}} )^{2}+(\frac{∂v_{1}}{∂y_{1}} − \frac{∂v_{3}}{∂y_{3}} )^{2}+

+(\frac{∂v_{2}}{∂y_{2}} − \frac{∂v_{3}}{∂y_{3}} )^{2}]+ \frac{1+k}{2} [(\frac{∂v_{2}}{∂y_{1}} + \frac{∂v_{1}}{∂y_{2}} )^{2}+(\frac{∂v_{3}}{∂y_{1}} + \frac{∂v_{1}}{∂y_{3}} )^{2}+(\frac{∂v_{3}}{∂y_{2}} +

+\frac{∂v_{2}}{∂y_{3}} )^{2}]+ \frac{1−k}{2} [(\frac{∂v_{2}}{∂y_{1}} − \frac{∂v_{1}}{∂y_{2}} )^{2}+(\frac{∂v_{3}}{∂y_{2}} − \frac{∂v_{2}}{∂y_{3}} )^{2}+(\frac{∂v_{1}}{∂y_{3}} − \frac{∂v_{3}}{∂y_{1}} )^{2}]. (26)

For the infinite domain D^{−} we have

∫_{D^{−}} N(v, v)dy_{1} dy_{2} = − ∫_{S} vNv ds.  (27)

In this case the vector v satisfies at infinity the conditions

v = O(ρ^{−1}), \frac{∂v}{∂x_{k}} = O(ρ^{−2}), k = 1, 2, 3,

where ρ = \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}. To use formulas (25) and (27) in the proof of the uniqueness theorems it is necessary that expression (26) have a positively defined form both for k_{1} and k_{2}. Since (7) is fulfilled, all the terms in (26) except for the first one are positive. From (17) we obtain

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_{1} − ε_{4})^{2} + 4ε_{2}ε_{3}} > 0,

where Δ_{1} =μ_{1}μ_{2}−μ_{3}^{2} > 0 and a_{1}(b_{2}−λ_{5})+a_{2}(b_{1}−λ_{5})−2c(d+λ_{5})>0 [2]. By the first formula of (18) we have k_{1} >−\frac{1}{3}. Therefore −\frac{1}{3} <k_{1} <1. Then

\frac{1 − 2k_{1}}{3} + \frac{2k_{1}}{1 − k_{1}} = \frac{(2k_{1} + 1)(k_{1} + 1)}{3(1 − k_{1})} > 0.

For the expression \frac{1−2k_{2}}{3} + \frac{2k_{2}}{1−k_{2}} = \frac{(2k_{2}+1)(k_{2}+1)}{3(1−k_{2})} to be positive it is necessary and sufficient that (2k_{2} +1)(k_{2} +1) > 0. After obvious transformations the expanded form of this inequality is

\frac{1}{δ_{0}} [9(b_{1}b_{2} − d^{2}) + 6(a_{1}b_{2} + a_{2}b_{1} − 2cd) + 4d_{2}] ≡

≡ \frac{1}{δ_{0}} [(2a_{1} + 3b_{1})(2a_{2} + 3b_{2}) − (2c + 3d)^{2}] > 0,  (28)

where δ_{0} is given by (4). When (28) is fulfilled, the first term of formula (26) will be positive, too, for k =k_{2}. In what follows it will be assumed that (28) is valid.

Solutions of the first boundary value problem for equations (12) and (13) are sought for in the form

v′(x) = \frac{1}{2π} ∫_{S} N_{y}^{(1)}Γ^{(1)}(y − x)g′(y) ds,  (29)

v′′(x) = \frac{1}{2π} ∫_{S} N_{y}^{(2)}Γ^{(2)}(y − x)g′′(y) ds,  (30)

where g′(y) and g′′(y) are the unknown vector functions and

N_{y}^{(i)}Γ^{(i)}(y−x)=∥(1−k_{i})δ_{kj} +3k_{i} \frac{∂r}{∂x_{k}} \frac{∂r}{∂x_{j}} ∥_{3×3} · \frac{∂}{∂n(y)} \frac{1}{r}, i=1, 2.

In the case of the first internal problem, to define g′ and g′′, we obtain by virtue of the properties of potentials (29) and (30) [3] the Fredholm integral equation of second kind

\begin{matrix}− g′(z) + \frac{1}{2π} ∫_{S} N_{y}^{(1)}Γ^{(1)}(y − z)g′(y) ds = f^{(1)}(z), \\ − g′′(z) + \frac{1}{2π} ∫_{S} N_{y}^{(2)}Γ^{(2)}(y − z)g′′(y) ds = f^{(2)}(z), \end{matrix}  (31)

where the vectors f′(z) and f′′(z) given on the boundary S are the boundary values of the vectors v′ and v′′, respectively. Since it is assumed that condition (28) is fulfilled, the quadratic form (26) is positively defined both for k_{1} and k_{2}. Applying the method developed in [3] to (31), we readily conclude that these equations have unique solutions g′ and g′′ ∈ C^{1,α}(S) if f^{(i)} ∈ C^{1,α}(S) and S ∈ C^{2,β}, where i = 1, 2, 0 < β < α ≤ 1.

Similar arguments can be used in considering the first boundary value problem for the infinite domain D^{−} bounded by the closed surface S. In that case we are to seek for a solution the manner as follows:

\begin{matrix}v′(x)= \frac{1}{2π} ∫_{S}N_{y}^{(1)}Γ^{(1)}(y−x)g′(y)ds+ \frac{1}{4π} Γ^{(1)}(x) ∫_{S}N_{y}^{(1)}Γ^{(1)}(y)g′(y)ds, \\ v′′(x)= \frac{1}{2π} ∫_{S}N_{y}^{(2)}Γ^{(2)}(y−x)g′′(y)ds+ \frac{1}{4π} Γ^{(2)}(x) ∫_{S}N_{y}^{(2)}Γ^{(2)}(y)g′′(y)ds. \end{matrix} (32)

To define the unknown vectors g′(y) and g′′(y) we respectively obtain the Fredholm integral equations of second kind

\begin{aligned}g′(z) & + \frac{1}{2π} ∫_{S} N_{y}^{(1)}Γ^{(1)}(y − z)g′(y) ds + \\ & + \frac{1}{4π} Γ^{(1)}(z) ∫_{S} N_{y}^{(1)}Γ^{(1)}(y)g′(y) ds = f^{(1)}(z), \\ g′′(z) & + \frac{1}{2π} ∫_{S} N_{y}^{(2)}Γ^{(2)}(y − z)g′′(y) ds + \\ & + \frac{1}{4π} Γ^{(2)}(z) ∫_{S} N_{y}^{(2)}Γ^{(2)}(y)g′′(y) ds = f^{(2)}(z), \end{aligned}  (33)

where f^{(1)}, f^{(2)} and S satisfy the above conditions. Again applying the method developed in [3] we find that equations (33) have unique solutions and g(g′, g′′) ∈ C^{1,α}(s).

Thus we have proved that the vectors v′(x) and v′′(x) are uniquely defined both for the finite domain D^{+} and the infinite domain D^{−}. By virtue of (9) this means that u′ and u′′, i.e., the vector u(u′, u′′), are defined uniquely, too.

Using the methods and arguments from [4] and [5, Ch. II, §4], we can prove that the Fredholm equations (31) and (33) are also uniquely solvable in the space of continuous vectors (provided that f^{(i)} ∈ C(s), i = 1, 2), while the first three-dimensional problem for equations (1) and (14) (i.e.,

(12) and (13)) is uniquely solvable in the class C(\overline{D}^{±}) ∩ C^{2}(D^{±})). Since potentials (29), (30), and (32) with continuous densities are contin-

uous in the respective closed domains, the results obtained above prove the unique solvability of the considered problems in the class C(\overline{D}^{±}) ∩ C^{2}(D^{±}) when the boundary data are continuous.

The above reasoning also enables us to construct effective (explicit) solutions of the first three-dimensional boundary value problem of the elastic mixture theory for those specific domain for which we can effectively construct a solution of the elastostatic problem with given displacements on the boundary.

References

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

2. 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.

3. 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 Publishing Company, Amsterdam–New York–Oxford, 1979; Russian original: Nauka, Moscow, 1976.

4. G. Fichera. Il teorema del massimo modulo per l’equazione dell’elastostatica tridimensionalle. Arch. Rational Mech. Anal. 7(1961), 373–387.

5. D. G. Natroshvili, Estimates of Green’s tensors of the elasticity theory and some of their applications. Tbilisi University Press, Tbilisi, 1978.

(Received 10.10.1997; revised 10.11.1997)

Author’s address:

I. Vekua Institute of Applied Mathematics

Tbilisi State University

2, University St., Tbilisi 380043, Georgia