Memoirs on Differential Equations and Mathematical Physics
Volume 45, 2008, 85–115
L. Giorgashvili, G. Karseladze, G. Sadunishvili
SOLUTION OF A BOUNDARY VALUE PROBLEM
OF STATICS OF TWO-COMPONENT ELASTIC
MIXTURES FOR A SPACE WITH TWO
NONINTERSECTING SPHERICAL CAVITIES
Abstract. Using a general representation of solutions of a system of
homogeneous differential equations of statics of two-component elastic mixtures which is expressed by six harmonic functions, we study boundary value
problems of statics of two-component elastic mixtures for a space with two
nonintersecting spherical cavities when different boundary conditions are
given on the spherical surfaces. The uniqueness theorems are proved. The
solution of the considered problems is reduced to the investigation of an infinite system of linear algebraic equations. It is proved that such systems are
quasiregular. The question of regularity of the partial displacement vector
is studied.
2000 Mathematics Subject Classification. 35J55, 75H20, 74H25.
Key words and phrases. Elasticity theory, theory of mixtures, Green’s
formula, regular functions.
$
$
!
#
'
$
&
!
%
$
!
#
#
"
$
87
Solution of a Boundary Value Problem of Statics
Introduction
In recent years, considerable progress has been made in the investigation of physical and mechanical properties of composite materials, a stressstrained state of structural elements with regard for their structure. The
results of the investigation in the composite material technology area have
applications in various spheres of industry and engineering. The present
work is dedicated to the effective solution of static problems of the linear
theory of a mixture of two isotropic elastic materials. We use the fundamental equations derived in Green, Naghdi [6], Steel [14] and Green, Steel
[7]. Some of the problems touched upon in the present work were considered
in various aspects by other authors. For example, Atkin, Chadwick, Steel
[1], Knops, Steel [10], Natroshvili, Jagmaidze, Svanadze [13] are dedicated
to the uniqueness theorems of various linearized dynamic problems of the
mixture theory. Effective solutions of problems of statics and steady-state
oscillations of the elastic mixture theory are obtained for a ball in [4] and
[13]. In the present work, a new approach is proposed to the solution of
boundary value problems for concrete domains.
1. Some Notation, Auxiliary Formulas and Theorems
The three-dimensional Euclidean space is denoted by R3 , and the points
(vectors) of this space by x, y, z. The coordinates of these points in the
basis e1 = (1, 0, 0)> , e2 = (0, 1, 0)> , e3 = (0, 0, 1)> are denoted by x1 , x2 ,
x3 ; y1 , y2 , y3 ; z1 , z2 , z3 . > is the transposition symbol.
We denote by Ω+ a finite domain from R3 with the boundary ∂Ω, and
+
+
by Ω− the complement of the set Ω to R3 (Ω− = R3 \ Ω ).
If a function Φ(x) defined in Ω+ [Ω− ] is continuously extendable at a
point z ∈ ∂Ω, then we denote by [Φ(z)]+ ([Φ(z)]− ) the limit
[Φ(z)]+ = +lim Φ(x)
[Φ(z)]− = −lim Φ(x) .
Ω 3x→z
Ω 3x→z
We denote by r, ϑ, ϕ (0 ≤ r < +∞, 0 ≤ ϑ ≤ π, 0 ≤ ϕ < 2π) the spherical
coordinates of a point x ∈ R3 . Let us introduce the following vectors [12],
[16]:
(m)
Xmk (ϑ, ϕ) = er Yk
(ϑ, ϕ), k ≥ 0,
1
eϕ ∂ (m)
∂
Y
(ϑ, ϕ),
Ymk (ϑ, ϕ) = p
+
eϑ
∂ϑ sin ϑ ∂ϕ k
k(k + 1)
e
1
∂ (m)
∂
ϑ
Zmk (ϑ, ϕ) = p
Y
(ϑ, ϕ),
− eϕ
∂ϑ k
k(k + 1) sin ϑ ∂ϕ
where |m| ≤ k, er , eϑ , eϕ are the unit orthogonal vectors
>
er = cos ϕ sin ϑ, sin ϕ sin ϑ, cos ϑ ,
>
eϑ = cos ϕ cos ϑ, sin ϕ cos ϑ, − sin ϑ ,
k ≥ 1,
k ≥ 1,
(1.1)
88
L. Giorgashvili, G. Karseladze, G. Sadunishvili
eϕ = (− sin ϕ, cos ϕ, 0)> ,
s
2k + 1 (k − m)! (m)
(m)
Yk (ϑ, ϕ) =
·
P
(cos ϑ)eimϕ ,
4π
(k + m)! k
(1.2)
(m)
Pk (cos ϑ) is the first kind adjoint Legendre function of k-th degree and
m-th order [15]. In the sequel, under a vector we mean a one-column matrix.
On the sphere of unit radius the set
Xmk (ϑ, ϕ), Ymk (ϑ, ϕ), Zmk (ϑ, ϕ) |m|≤k, k=0,∞
forms a complete orthonormalized system of vector functions in the space L2 .
The following identities are valid [2]:
(m)
er · Xmk (ϑ, ϕ) = Yk
(ϑ, ϕ), er · Ymk (ϑ, ϕ) = 0, er · Zmk (ϑ, ϕ) = 0,
er × Xmk (ϑ, ϕ) = 0, er × Ymk (ϑ, ϕ) = −Zmk (ϑ, ϕ),
er × Zmk (ϑ, ϕ) = Ymk (ϑ, ϕ),
p
da(r)
k(k+1)
(m)
Xmk (ϑ, ϕ)+
a(r)Ymk (ϑ, ϕ),
grad a(r)Yk (ϑ, ϕ) =
dr
r
p
(m)
rot xa(r)Yk (ϑ, ϕ) = k(k + 1) a(r)Zmk (ϑ, ϕ),
(m)
rot rot xa(r)Yk (ϑ, ϕ) =
d 1
p
k(k+1)
a(r)Xmk (ϑ, ϕ)+ k(k+1)
+ a(r)Ymk (ϑ, ϕ),
=
r
dr r
d
2
(m)
div a(r)Xmk (ϑ, ϕ) =
a(r)Yk (ϑ, ϕ),
+
dr
r
p
a(r) (m)
div a(r)Ymk (ϑ, ϕ) = − k(k + 1)
Yk (ϑ, ϕ),
r
(1.3)
div a(r)Zmk (ϑ, ϕ) = 0,
p
a(r)
Zm k (ϑ, ϕ),
rot a(r)Xmk (ϑ, ϕ) = k(k + 1)
r
1
d
a(r)Zmk (ϑ, ϕ),
rot a(r)Ymk (ϑ, ϕ) = −
+
dr
r
rot a(r)Zmk (ϑ, ϕ) =
d
p
1
a(r)
= k(k + 1)
a(r)Ymk (ϑ, ϕ),
Xm k (ϑ, ϕ) +
+
r
dr
r
d d
2
grad div a(r)Xmk (ϑ, ϕ) =
a(r)Xmk (ϑ, ϕ)+
+
dr dr
r
p
k(k + 1) d
2
+
a(r)Ymk (ϑ, ϕ),
+
r
r
dr
grad div a(r)Ymk (ϑ, ϕ) =
p
k(k+1) d 1 k(k+1)
− a(r)Xmk (ϑ, ϕ)−
a(r)Ymk (ϑ, ϕ),
=−
r
dr r
r2
89
Solution of a Boundary Value Problem of Statics
where a(r) is a function of r, x = (x1 , x2 , x3 )> , a · b; a × b are the scalar
and the vector product of the vectors a and b, respectively.
Taking into account the recurrent relations of Legendre polynomials and
the orthogonality of the vectors (1.1), we obtain
∞ X
k
X
(m)
ej ak (r)Yk
k=0 m=−k
+
∞
X
h
k
X
k=1 m=−k
(j,1)
(ϑ, ϕ) = −a1 (r)η00 (A00 )X00 (ϑ, ϕ)+
i
(j,2)
(j,1)
ak−1 (r)ηmk (Amk ) − ak+1 (r)ηmk (Amk ) Xmk (ϑ, ϕ)+
h1
i
p
1
(j,2)
(j,1)
+ k(k+1) ak−1 (r)ηmk (Amk)+
ak+1 (r)ηmk (Amk) Ymk (ϑ, ϕ)+
k
k+1
a
(r)
k
(j,3)
j+1
p
+(−1)
ηmk (Amk )Zmk (ϑ, ϕ) , j = 1, 2,
k(k + 1)
∞ X
k
X
1
(ϑ, ϕ)Amk = √ a1 (r)A01 X00 (ϑ, ϕ)+
3
k=0 m=−k
r
h
∞ X
k
X
k 2 − m2
ak−1 (r)Amk−1 +
+
4k 2 − 1
k=1 m=−k
s
i
(k + 1)2 − m2
+
ak+1 (r)Amk+1 Xmk (ϑ, ϕ)+
(2k + 1)(2k + 3)
h 1 r k 2 − m2
p
+ k(k + 1)
ak−1 (r)Amk−1 −
k
4k 2 − 1
s
i
(k + 1)2 − m2
1
ak+1 (r)Amk+1 Ymk (ϑ, ϕ)+
−
k + 1 (2k + 1)(2k + 3)
im
ak (r)Amk Zmk (ϑ, ϕ) ,
+p
k(k + 1)
(m)
e3 ak (r)Yk
(1.4)
where Amk is a constant and ak (r) is a function of r,
s
δ1j + iδ2j
(k + m + 1)(k + m + 2)
=
Am+1k+1 +
2
(2k + 1)(2k + 3)
s
(k − m + 1)(k − m + 2)
j
Am−1k+1 ,
+ (−1)
(2k + 1)(2k + 3)
r
(k − m)(k − m − 1)
δ1j + iδ2j
(j,2)
Am+1k−1 +
ηmk (Amk ) =
2
4k 2 − 1
r
(k + m)(k + m − 1)
j
+ (−1)
(1.5)
Am−1k−1 ,
4k 2 − 1
(j,1)
ηmk (Amk )
90
L. Giorgashvili, G. Karseladze, G. Sadunishvili
δ2j + iδ1j hp
(k − m)(k + m + 1) Am+1k −
2
i
p
− (−1)j (k + m)(k − m + 1) Am−1k , j = 1, 2,
√
δ`j is the Kronecker symbol, i = −1.
Using the formulas (1.3) and taking into account the identity
(m)
(m)
ej × grad ak (r)Yk (ϑ, ϕ) = − rot ej ak (r)Yk (ϑ, ϕ) , j = 1, 2, 3,
(j,3)
ηmk (Amk ) =
from (1.4) we obtain
∞ X
k
X
(m)
∂j ak (r)Yk
(ϑ, ϕ)Amk =
k=0 m=−k
∞ X
k
X
ak (r) (j,3)
ηmk (Amk )Xmk (ϑ, ϕ)+
r
k=1 m=−k
d
1
1
(j,3)
ak (r)ηmk (Amk )Ymk (ϑ, ϕ)+
+p
+
r
k(k + 1) dr
p
k + 2
1 d
(j,1)
j
ak+1 (r)ηmk (Amk )+
+
+(−1) k(k + 1)
k + 1 dr
r
1d
k − 1
(j,2)
+
ak−1 (r)ηmk (Amk ) Zmk (ϑ, ϕ) , j = 1, 2,
(1.6)
−
k dr
r
∞ X
k
∞ X
k X
X
im
(m)
∂3 ak (r)Yk (ϑ, ϕ)Amk =
−
ak (r)Amk Xmk (ϑ, ϕ)−
r
k=0m=−k
k=1m=−k
d
im
1
−p
ak (r)Amk Ymk (ϑ, ϕ)+
+
r
k(k + 1) dr
r 2
p
k − 1
1 k − m2 d
+ k(k + 1)
ak−1 (r)Amk−1 −
−
2
k
4k − 1 dr
r
s
(k+1)2 −m2 d k+2 1
ak+1 (r)Amk+1 Zmk (ϑ, ϕ) ,
+
−
k+1 (2k+1)(2k+3) dr
r
=
(−1)j
(j,`)
where ∂j = ej × grad, ηmk , j = 1, 2, ` = 1, 2, 3, has form (1.5).
Using the formulas (1.3) and taking into account the identity
grad
∂ (m)
(m)
ak (r)Yk (ϑ, ϕ) = grad div ej ak (r)Yk (ϑ, ϕ) , j = 1, 2, 3,
∂xj
from (1.4) we obtain
∞ X
k
X
∂
(m)
ak (r)Yk (ϑ, ϕ)Amk =
∂xj
k=0 m=−k
d d
2
(j,1)
a1 (r)η00 (A00 )X00 (ϑ, ϕ)+
=−
+
dr dr
r
grad
Solution of a Boundary Value Problem of Statics
∞ X
k
X
91
d d
k + 2
(j,1)
ak+1 (r)ηmk (Amk )+
+
dr dr
r
k=1 m=−k
d d
k − 1
(j,2)
+
ak−1 (r)ηmk (Amk ) Xmk (ϑ, ϕ)+
−
dr dr
r
p
k(k + 1)
d
k + 2
(j,1)
+
−
ak+1 (r)ηmk (Amk )+
+
r
dr
r
d
k − 1
(j,2)
+
ak−1 (r)ηmk (Amk ) Ymk (ϑ, ϕ) , j = 1, 2,
(1.7)
−
dr
r
∞ X
k
X
1 d d 2
∂
(m)
ak (r)Yk (ϑ, ϕ)Amk = √
+ A01 X00 (ϑ, ϕ)+
grad
∂x3
3 dr dr r
k=0m=−k
hr 2
∞ X
k
X
k − m2 d d
k − 1
ak−1 (r)Amk−1 +
−
+
2
4k − 1 dr dr
r
k=1 m=−k
s
i
(k + 1)2 − m2 d d
k + 2
ak+1 (r)Amk+1 Xmk (ϑ, ϕ)+
+
+
(2k + 1)(2k + 3) dr dr
r
r
h k 2 − m2 1 d
p
k − 1
+ k(k + 1)
ak−1 (r)Amk−1 +
−
4k 2 − 1 r dr
r
s
i
k + 2
(k + 1)2 − m2 1 d
+
ak+1 (r)Amk+1 Ymk (ϑ, ϕ) .
+
(2k + 1)(2k + 3) r dr
r
+
−
Let the vector f (ϑ, ϕ) satisfy sufficient smoothness conditions under
which it can be represented as a Fourier series
f (ϑ, ϕ) =
∞ X
k
n
X
k=0 m=−k
+
where
αmk =
Z2π
0
βmk
dϕ
Zπ
0
αmk Xmk (ϑ, ϕ)+
h
io
p
k(k + 1) βmk Ymk (ϑ, ϕ) + γmk Zmk (ϑ, ϕ , (1.8)
f (ϑ, ϕ)X mk (ϑ, ϕ) sin ϑ dϑ, k ≥ 0,
1
=p
k(k + 1)
1
γmk = p
k(k + 1)
Z2π
dϕ
0
Zπ
f (ϑ, ϕ)Y mk (ϑ, ϕ) sin ϑ dϑ, k ≥ 1,
Z2π
dϕ
Zπ
f (ϑ, ϕ)Z mk (ϑ, ϕ) sin ϑ dϑ, k ≥ 1.
0
0
0
(1.9)
92
L. Giorgashvili, G. Karseladze, G. Sadunishvili
Note that in the sequel, in (1.8) and in analogous series the summation
index k in the summands containing Ymk (ϑ, ϕ) and Zmk (ϑ, ϕ) varies from
1 to +∞.
The following theorems are valid [2].
Theorem 1.1. If f (z) ∈ C (`) (∂Ω), then the coefficients αmk , βmk , γmk
defined by the formulas (1.9) admit the following estimates
αmk = O(k −` ), βmk = O(k −`−1 ), γmk = O(k −`−1 ), ` ≥ 1.
Theorem 1.2. The vectors Xmk (ϑ, ϕ), Ymk (ϑ, ϕ), Zmk (ϑ, ϕ) admit the
following estimates for any k ≥ 0
r
2k + 1
, k ≥ 0,
|Xmk (ϑ, ϕ)| ≤
4π
r
2k(k + 1)
(1.10)
, k ≥ 1,
|Ymk (ϑ, ϕ)| <
2k + 1
r
2k(k + 1)
|Zmk (ϑ, ϕ)| <
, k ≥ 1.
2k + 1
Moreover, as is known [4], [15],
(m)
|Yk (ϑ, ϕ)|
≤
r
2k + 1
, k ≥ 0.
4π
(1.11)
Definition 1.3. A vector u defined in the domain Ω will be called regular
if u ∈ C 2 (Ω) ∩ C 1 (Ω).
2. Statement of Boundary Value Problems. Uniqueness
Theorems
The system of homogeneous differential equations of statics of the threedimensional theory of mixture of two isotropic elastic materials is written
in the form [6], [7], [14]
a1 ∆u0 + b1 grad div u0 + c∆u00 + d grad div u00 = 0,
c∆u0 + d grad div u0 + a2 ∆u00 + b2 grad div u00 = 0,
(2.1)
where u0 = (u01 , u02 , u03 )> , u00 = (u001 , u002 , u003 )> are partial displacement vectors,
ρ2 0
a 1 = µ 1 − λ 5 , b1 = µ 1 + λ 5 + λ 1 −
α , a2 = µ2 − λ5 ,
ρ 2
ρ1 0
b2 = µ 2 + λ 2 + λ 5 +
α , c = µ3 + λ5 , α02 = λ3 − λ4 ,
ρ 2
ρ1 0
d = µ 3 + λ3 − λ5 −
α , ρ = ρ 1 + ρ2 ,
ρ 2
ρ1 , ρ2 are the partial densities of the mixture; λ1 , λ2 , . . . , λ5 , µ1 , µ2 , µ3 are
the elasticity moduli characterizing the mechanical properties of the mixture
93
Solution of a Boundary Value Problem of Statics
which satisfy the conditions [3]
ρ2 0
2
µ1 −
α > 0,
3
ρ 2
2
2
2
ρ2
ρ1
ρ 1 2
λ1 + µ1 − α02 λ2 + µ2 + α02 > λ3 + µ3 − α02 .
3
ρ
3
ρ
3
ρ
µ1 > 0, µ1 µ2 − µ23 > 0, λ5 < 0, λ1 +
(2.2)
From these inequalities it follows that
a1 a2 − c2 > 0, (a1 + b1 )(a2 + b2 ) − (c + d)2 > 0.
Let us introduce the matrix differential operator


..
(1)
(2)
T (∂x, n) . T (∂x, n)

T (∂x, n) = 
. . . . . . . . . . . . . . . . . . . . . . . . . . ,
.
T (3) (∂x, n) .. T (4) (∂x, n)
(2.3)
(2.4)
where
(`)
T (`) (∂x, n) = [Tkj (∂x, n)]3×3 , ` = 1, 2, 3, 4,
(1)
Tkj (∂x, n) = (µ1 −λ5 )δkj
(2)
Tkj (∂x, n) = (µ3 +λ5 )δkj
(3)
Tkj (∂x, n) = (µ3 +λ5 )δkj
(4)
Tkj (∂x, n) = (µ2 −λ5 )δkj
∂
+(µ1 +λ5 )nj
∂n
∂
+(µ3 −λ5 )nj
∂n
∂
+(µ3 −λ5 )nj
∂n
∂
+(µ2 +λ5 )nj
∂n
∂
ρ2
+ λ1 −
∂xk
ρ
∂
ρ1
+ λ3 −
∂xk
ρ
∂
ρ2
+ λ4 +
∂xk
ρ
∂
ρ1
+ λ2 +
∂xk
ρ
∂
,
α02 nk
∂xj
∂
,
α02 nk
∂xj
∂
,
α02 nk
∂xj
∂
.
α02 nk
∂xj
The operator T (∂x, n) defined by (2.4) is called the generalized stress
operator, δkj is the Kronecker symbol.
The vector form of the notation for the expressions T (`) (∂x, n), ` =
1, 2, 3, 4, where u is a three-component vector, looks like
T (`) (∂x, n)u = ξ`
∂u
+ η` n div u + ζ` [n × rot u],
∂n
(2.5)
where
ρ2
ρ
ρ1
ξ2 = 2µ3 , η2 = λ3 −
ρ
ρ2
ξ3 = 2µ3 , η3 = λ4 +
ρ
ρ1
ξ4 = 2µ2 , η4 = λ2 +
ρ
ξ1 = 2µ1 , η1 = λ1 −
α02 , ζ1 = µ1 + λ5 ,
α02 , ζ2 = µ3 − λ5 ,
α02 , ζ3 = µ3 − λ5 ,
α02 , ζ4 = µ2 + λ5 .
Let ∂Ωj , j = 1, 2, be a spherical surface with center at the origin Oj and
the radius Rj . Denote by Ωj , j = 1, 2, the ball bounded by the surface ∂Ωj ,
94
L. Giorgashvili, G. Karseladze, G. Sadunishvili
while Ω− = R3 \ (Ω1 ∪ ΩT
2 ). Assume that a homogeneous medium occupies
the domain Ω− and Ω1 Ω2 = ∅.
Problem. Find in the domain Ω− a regular solution U (x)=(u0 (x),u00 (x))>
of the system (2.1) which on the boundary ∂Ωj , j = 1, 2, satisfies one of the
following conditions:
[U (z)]− = f (j) (z), j = 1, 2, or
(2.6)
−
T (∂z, n)U (z) = f (j) (z), j = 1, 2, or
(2.7)
−
−
(j,1)
0
0
(j,1)
n(z) · u (z) = f4 (z), n(z) × rot u (z) = f
(z),
(2.8)
−
−
(j,2)
n(z) · u00 (z) = f4 (z), n(z) × rot u00 (z) = f (j,2) (z), j = 1, 2,
where f (j) (z) = (f (j,1) (z), f (j,2) (z))> is a six-component vector, f (j,`) (z) =
(j,`)
(j,`)
(j,`)
(f1 (z), f2 (z), f3 (z))> , `, j = 1, 2, is a three-component vector,
(j,`)
fk (z), `, j = 1, 2, k = 1, 2, 3, 4, are given functions on ∂Ωj , n(z) is
the outward normal vector with respect to Ωj at the point z ∈ ∂Ωj .
In the neighborhood of infinity the vector U (x) must satisfy the following
conditions:
u0j (x) = O(|x|−1 ),
∂u0j (x)
= o(|x|−1 ),
∂xk
u00j (x) = O(|x|−1 ),
∂u00j (x)
= o(|x|−1 ), k, j = 1, 2, 3.
∂xk
(2.9)
Denote by (I)− , (II)− and (III)− the problems containing the conditions
(2.6), (2.7) and (2.8), respectively.
Theorem 2.1. If ∂Ωj ∈ Λ1 (α), 0 < α ≤ 1, then Problems (I)− , (II)−
and (III)− admit at most one regular solution.
Proof. The theorem will be proved if we show that the homogeneous prob(j,`)
−
−
(j)
lems (I)−
(z) = 0, f4 (z) = 0, `, j = 1, 2) have
0 , (II)0 and (III)0 (f
only the trivial solution.
Let us introduce the matrix differential operator A(∂x):


..
(1)
(2)
A (∂x) . A (∂x)

A(∂x) = 
,
. . . . . . . . . . . . . . . . . . . . . .
.
A(3) (∂x) .. A(4) (∂x)
6×6
where
(`)
A(`) (∂x) = Akj (∂x) 3×3 , ` = 1, 2, 3, 4,
(1)
Akj (∂x) = a1 δkj ∆ + b1
(`)
Akj (∂x) = cδkj ∆ + d
∂2
,
∂xk ∂xj
∂2
, ` = 2, 3,
∂xk ∂xj
95
Solution of a Boundary Value Problem of Statics
(4)
Akj (∂x) = a2 δkj ∆ + b2
∂2
.
∂xk ∂xj
Using this notation, the system (2.1) can be written in the form
A(∂x)U (x) = 0.
Let us write Green’s formula in the domain Ω− for the system (2.1). We
obtain [13]
Z h
i
U (x) · A(∂x)U (x) + E(U, U ) dx =
Ω−
Z
=
∂Ω1 ∪∂Ω2
−
[U (z)]− · T (∂z, n)U (z) ds, (2.10)
where
ρ1 0 ρ2 0 α2 (div u0 )2 + λ2 +
α (div u00 )2 +
E(U, U ) = λ1 −
ρ
ρ 2
3
3 ∂u0j 2 µ2 X
∂u00j 2
µ1 X ∂u0k
∂u00k
+
+
+
+
+
2
∂xj
∂xk
2
∂xj
∂xk
k,j=1
+ µ3
k,j=1
3
X
k,j=1
∂u0
k
∂xj
+
∂u0j
∂xk
∂u00
k
∂xj
+
∂u00j +
∂xk
3
∂u0j
∂u00j 2
λ5 X ∂u0k
ρ1 0 ∂u00k
+ 2 λ3 −
.
−
−
+
α2 div u0 div u00 −
ρ
2
∂xj
∂xk
∂xj
∂xk
k,j=1
The quadratic form of E(U, U ) is provided by the inequality (2.2).
If in (2.10) we take into account the boundary conditions of the homo−
geneous problems (I)−
0 , (II)0 and A(∂x)U = 0, then we obtain
Z
E(U, U ) dx = 0.
Ω−
Hence it follows that E(U, U ) = 0. A solution of this equation has the
form [13]
u0 (x) = b0 + [a0 × x], u00 (x) = b00 + [a0 × x],
where a0 , b0 , b00 are arbitrary three-component constant vectors.
Taking into account the conditions at infinity (2.9), we conclude that
a0 = b0 = b00 = 0, i.e. u0 (x) = 0, u00 (x) = 0, x ∈ Ω− .
In the case of Problem (III)−
0 we need to consider the scalar derivative
U · A(∂x)U = (a1 u0 + cu00 )∆u0 + (cu0 + a2 u00 )∆u00 +
+ (b1 u0 + du00 ) grad div u0 + (du0 + b2 u00 ) grad div u00 . (2.11)
96
L. Giorgashvili, G. Karseladze, G. Sadunishvili
Let u = (u1 , u2 , u3 )> and v = (v1 , v2 , v3 )> be three-component vectors.
Then after some transformations we obtain
u · ∆v = div(u div v) − div u div v + div[u × rot v] − rot u · rot v,
u · grad div v = div(u div v) − div u div v.
Taking these equalities into account in (2.11), we obtain
h
U · A(∂x)U = div (a1 + b1 )u0 + (c + d)u00 div u0 +
+ (c + d)u0 + (a2 + b2 )u00 div u00 + a1 (u0 × rot u0 )+
i
e
+ c(u00 × rot u0 ) + c(u0 × rot u00 ) + a2 (u00 × rot u00 ) − E(U,
U ), (2.12)
where
i
2
1 h
(a1 + b1 ) div u0 + (c + d) div u00 + d1 (div u00 )2 +
a1 + b 1
i
2
1 h
+
(2.13)
a1 rot u0 + c rot u00 + d2 (rot u00 )2 ,
a1
d1 = (a1 + b1 )(a2 + b2 ) − (c + d)2 > 0, d2 = a1 a2 − c2 > 0.
e
E(U,
U) =
Denote by B(0, R) the ball bounded by the spherical surface S(0, R) with
−
the center at the origin and of the radius R. Let Ω−
R = Ω ∩ B(0, R), where
R > 0 is sufficiently large so that ∂Ωj ⊂ B(0, R), j = 1, 2.
Applying the Gauss–Ostrogradskiı̌ theorem, from (2.12) we obtain
Z
e
U · A(∂x)U + E(U,
U ) dx =
=−
Z
Ω−
R
−
[U (z)]− · P (∂z, n)U (z) ds+
∂Ω1 ∪∂Ω2
Z
U (x) · P (∂x, n)U (x) ds,
(2.14)
S(0,R)
where
U · P (∂z, n)U = (n · u0 ) (a1 + b1 ) div u0 + (c + d) div u00 +
+ (n · u00 ) (c + d) div u0 + (a2 + b2 ) div u00 −
− (a1 u0 + cu00 ) · [n × rot u0 ] − (cu0 + a2 u00 ) · [n × rot u00 ]. (2.15)
Here we have used the identity n · [u × rot v] = −u · [n × rot v].
If in (2.15) we use the estimates (2.9), then we have
U (x) · P (∂x · n)U (x) = o(R−2 ).
(2.16)
If in both parts of the equality (2.14) we pass to limit as R → ∞ and
take into account the estimate (2.16), then we obtain
Z
Z
−
e
[U (z)]− · P (∂z, n)U (z) ds. (2.17)
U · A(∂x)U +E(U, U ) dx = −
Ω−
∂Ω1 ∪∂Ω2
97
Solution of a Boundary Value Problem of Statics
Taking into account the boundary conditions of the problem (III)−
0 in
the formula (2.15), we have [U (z)]− ·[P (∂z, n)U (z)]− = 0, z ∈ ∂Ωj , j = 1, 2.
Since A(∂x)U = 0, from the formula (2.17), with the latter equality taken
into account, we obtain
Z
e
E(U,
U ) dx = 0.
(2.18)
Ω−
e
Using (2.3), from (2.13) we obtain E(U,
U ) ≥ 0. Taking this inequality
e
into account, from (2.18) we have E(U, U ) = 0, x ∈ Ω− .
Hence, with (2.13) taken into account, we obtain div u0 (x) = 0,
div u00 (x) = 0, rot u0 (x) = 0, rot u00 (x) = 0, x ∈ Ω− .
A solution of this system has the form
u0 (x) = grad Ψ1 (x),
u00 (x) = grad Ψ2 (x), x ∈ Ω− ,
(2.19)
where Ψj (x), j = 1, 2, is an arbitrary harmonic function.
Since [n(z)·u0 (z)]− = 0, [n(z)·u00 (z)]− = 0, the harmonic function Ψj (x),
j = 1, 2, on the boundary ∂Ωj satisfies the Neumann condition
h ∂Ψ (z) i−
j
= 0, z ∈ ∂Ω` , `, j = 1, 2.
∂n(z)
As is known, the Neumann homogeneous problem has the solution
Ψj (x) = cj = const, j = 1, 2, x ∈ Ω− . Inserting this value of the function Ψj (x) into (2.19), we have that u0 (x) = 0, u00 (x) = 0, x ∈ Ω− .
3. Solution of the Problem
(j) (j) (j)
Let us assume that the axes of the coordinate systems Oj x1 x2 x3 ,
j = 1, 2, are parallel and have the same orientation. We denote by x(j) =
(j)
(j)
(j)
(x1 , x2 , x3 ) and (rj , ϑj , ϕj ) the Cartesian and spherical coordinates of
(j) (j) (j)
the point x with respect to the coordinate systems Oj x1 x2 x3 , j =
1, 2. The spherical coordinates of the point O2 with respect to the sys(1) (1) (1)
tem O1 x1 x2 x3 are denoted by (h, ϑ0 , ϕ0 ).
Since the axes of the coordinate systems are parallel and have the same
orientation, the following equality is true:
x(1) = x(2) + he0 ,
(3.1)
where e0 = (cos ϕ0 sin ϑ0 , sin ϕ0 sin ϑ0 , cos ϑ0 )> .
Passing to a new origin of the coordinates, we can write the transformation formula of the spherical coordinates in the form [5], [8]
(m)
rq−k−1 Yk
=
∞
X
p
X
p=0 s=−p
(s,m)
(ϑq , ϕq ) =
Gpk,q (ϑ0 , ϕ0 )rjp Yp(s) (ϑj , ϕj ), q 6= j = 1, 2, rj < h,
(3.2)
98
L. Giorgashvili, G. Karseladze, G. Sadunishvili
where
1
(s,m)
Gpk,q (ϑ0 , ϕ0 ) = (−1)s (−1)p δ1q + (−1)k δ2q k+p+1 ×
h
i1/2
h
4π(2k+1)(k+p+m−s)!(k+p−m+s)!
m−s)
Yk+p (ϑ0 , ϕ0 ),
×
(2p+1)(2k+2p+1)(k+m)!(k−m)!(p+s)!(p−s)!
(m)
δkj is the Kronecker symbol, Yk
(ϑq , ϕq ) has the form (1.2).
If in the formula (3.2) we take into account that
rq2 = rj2 + h2 − (−1)q 2hrj cos γj , q 6= j = 1, 2,
where
cos γj = sin ϑj sin ϑ0 cos(ϕ − ϕ0 ) + cos ϑj cos ϑ0 ,
and use the recurrent relations of Legendre polynomials [5], [15]
(m−1)
(k − m + 1) sin ϑPk
(m−1)
(2k + 1) sin ϑPk
(m+1)
sin ϑPk
sin ϑ
(m)
(cos ϑ) = cos ϑPk
(m)
(m)
(cos ϑ) + Pk−1 (cos ϑ),
(m)
(cos ϑ) = Pk−1 (cos ϑ) − Pk+1 (cos ϑ),
(m)
(cos ϑ) = (k − m) cos ϑPk
(m)
(cos ϑ) − (k + m)Pk−1 (cos ϑ),
d (m)
(m+1)
(m)
P
(cos ϑ) = sin ϑPk
(cos ϑ) + m cos ϑPk (cos ϑ),
dϑ k
99
Solution of a Boundary Value Problem of Statics
then we obtain
(m)
rq−k+1 Yk
(ϑq , ϕq ) =
p h
∞ X
X
(s,m)
apk,q (ϑ0 , ϕ0 )rjp+2 +
p=0 s=−p
i
(s,m)
+ bpk,q (ϑ0 , ϕ0 )rjp Yp(s) (ϑj , ϕj ), q 6= j = 1, 2, (3.3)
where
(s,m)
(s,m)
(1,q)
apk,q (ϑ0 , ϕ0 ) = Gpk,q (ϑ0 , ϕ0 ) + (−1)q−1 hζmskp (ϑ0 , ϕ0 ),
(s,m)
(s,m)
(2,q)
bpk,q (ϑ0 , ϕ0 ) = h2 Gpk,q (ϑ0 , ϕ0 ) + (−1)q hζmskp (ϑ0 , ϕ0 ),
s
(p−s+1)(p−s+2) (s−1,m)
(1,q)
−iϕ0
Gp+1k,q (ϑ0 , ϕ0 )−
ζmskp (ϑ0 , ϕ0 ) = sin ϑ0 e
(2p+1)(2p+3)
s
(p + s + 1)(p + s + 2) (s+1,m)
iϕ0
−e
Gp+1k,q (ϑ0 , ϕ0 ) +
(2p + 1)(2p + 3)
s
(p + 1)2 − s2
(s,m)
G
(ϑ0 , ϕ0 ),
+ 2 cos ϑ0
(2p + 1)(2p + 3) p+1k,q
s
(p + s)(p + s − 1) (s−1,m)
(2,q)
ζmskp (ϑ0 , ϕ0 ) = sin ϑ0 e−iϕ0
Gp−1k,q (ϑ0 , ϕ0 )−
4p2 − 1
s
(p − s)(p − s − 1) (s+1,m)
iϕ0
Gp−1k,q (ϑ0 , ϕ0 ) −
−e
4p2 − 1
s
p2 − s2 (s,m)
− 2 cos ϑ0
G
(ϑ0 , ϕ0 ).
4p2 − 1 p−1k,q
A solution of the considered problem will be sought in the form [4]
2 X
∂
(j)
(j)
+1 α1 Φ2 (x(j) )+
u0 (x)=
grad Φ1 (x(j) )+grad rj2 rj
∂r
j
j=1
(j)
(j)
(j)
+β1 Φ3 (x(j) ) +rot rot x(j) rj2 Φ2 (x(j) ) +rot x(j) Φ5 (x(j) ) ,
2 X
∂
(j)
(j)
grad Φ4 (x(j) )+grad rj2 rj
u (x)=
+1 β2 Φ2 (x(j) )+
∂r
j
j=1
(3.4)
00
(j)
(j)
(j)
+α2 Φ3 (x(j) ) +rot rot x(j) rj2 Φ3 (x(j) ) +rot x(j) Φ6 (x(j) ) ,
(j)
where Φ` (x(j) ), j = 1, 2,, ` = 1, 2, . . . , 6 are scalar harmonic functions,
(j)
(j)
(j)
x(j) = (x1 , x2 , x3 )> , rj = |x(j) |, rj
∂
= x(j) · grad,
∂rj
100
L. Giorgashvili, G. Karseladze, G. Sadunishvili
1
d1
1
α2 =
d1
α1 =
1
c(c + d) − a1 (a2 + b2 ) , β1 = (a2 d − cb2 ),
d1
1
c(c + d) − a2 (a1 + b1 ) , β2 = (a1 d − cb1 ).
d1
(j)
The harmonic functions Φ` (x(j) ), j = 1, 2, ` = 1, 2, . . . , 6 will be sought in
the form
∞ X
k
R k+1
X
j
(m)
(j,`)
(j)
Yk (ϑj , ϕj )Amk ,
(3.5)
Φ` (x(j) ) =
rj
k=0 m=−k
j = 1, 2, ` = 1, 2, . . . , 6,
(j,`)
where Amk are unknown constants.
(j)
We require of the Φ` (x(j) ), j = 1, 2, ` = 2, 3, 5, 6, that
Z
(j)
Φ` (x(j) ) ds = 0, j = 1, 2, ` = 2, 3, 5, 6,
(3.6)
∂Ω0j
where ∂Ω0j is the sphere with center at the point Oj and of the radius Rj0
(Rq + h < Rj0 < +∞, q 6= j = 1, 2).
(j)
Inserting the value of the function Φ` (x(j) ), j = 1, 2, ` = 2, 3, 5, 6, from
(3.5) in (3.6) and taking into account the identity
( √
Z2πZπ
2 π, k = m = 0,
(m)
Yk (ϑ, ϕ) sin ϑ dϑ =
0,
for others k and m,
0
0
(j,`)
we have A00 = 0, j = 1, 2, ` = 2, 3, 5, 6.
Using the formulas (3.2), (3.3) and (3.5), we obtain:
(j)
Φ` (x(j) ) =
∞ X
k
X
(m)
rqk Yk
(q,`)
(ϑq , ϕq )Bmk , ` = 1, 4, 5, 6,
k=0 m=−k
∂
(j)
+ 1 Φ` (x(j) ) =
rj2 rj
∂rj
∞ X
k
h
i
X
(q,`)
(q,`)
(m)
=−
rqk+2 Bmk + rqk Cmk Yk (ϑq , ϕq ),
k=0 m=−k
∂
(j)
rj
+ 3 Φ` (x(j) ) =
∂rj
∞ X
k
h
i
X
(q,`)
(q,`)
(m)
=−
rqk+2 Dmk + rqk Emk Yk (ϑq , ϕq ), ` = 2, 3,
rj2
k=0 m=−k
∞ X
k
X
∂
(j)
(m)
(q,`)
+3 Φ` (x(j) ) = −
rqk Yk (ϑq , ϕq )Hmk , ` = 2, 3,
2rj
∂rj
k=0m=−k
(3.7)
Solution of a Boundary Value Problem of Statics
101
where q 6= j = 1, 2,
(q,`)
Bmk =
(q,`)
Bmk =
(q,`)
Bmk =
(q,`)
Cmk =
(q,`)
Dmk =
(q,`)
Emk =
(q,`)
Hmk =
p
∞ X
X
p=0 s=−p
p
∞ X
X
p=1 s=−p
p
∞ X
X
p=1 s=−p
p
∞ X
X
p=1 s=−p
p
∞ X
X
p=1 s=−p
p
∞ X
X
p=1 s=−p
p
∞ X
X
p=1 s=−p
(m,s)
Rjp+1 Gkp,j (ϑ0 , ϕ0 )A(j,`)
sp , ` = 1, 4,
(m,s)
Rjp+1 Gkp,j (ϑ0 , ϕ0 )A(j,`)
sp , ` = 5, 6,
(m,s)
pRjp+1 akp,j (ϑ0 , ϕ0 )A(j,`)
sp , ` = 2, 3,
(m,s)
pRjp+1 bkp,j (ϑ0 , ϕ0 )A(j,`)
sp , ` = 2, 3,
(3.8)
(m,s)
(p − 2)Rjp+1 akp,j (ϑ0 , ϕ0 )A(j,`)
sp , ` = 2, 3,
(m,s)
(p − 2)Rjp+1 bkp,j (ϑ0 , ϕ0 )A(j,`)
sp , ` = 2, 3,
(m,s)
(2p − 1)Rjp+1 Gkp,j (ϑ0 , ϕ0 )A(j,`)
sp , ` = 2, 3.
Taking the equalities (3.1) and (3.7) into account in (3.4) and using (1.5),
we obtain
∞ X
k
X
k + 1 Rj k+2 (j,1)
0 (j)
u (x ) =
−
Amk +
Rj
rj
k=0 m=−k
R k
Rj k (j,2)
j
(j,3)
Amk +β1 k(k−1)Rj
Amk +
+kRj (k−1)(α1 +1)+2
rj
rj
(j,1)
(j,2)
(j,3)
(j,2) + krjk−1 Bmk − α1 Cmk − β1 Cmk − Emk −
(j,2)
(j,3)
− rjk+1 α1 (k + 2)Bmk + β1 (k + 2)Bmk +
p
(j,2)
(j,2)
+ (k + 2)Dmk − 2Hmk
Xmk (ϑj , ϕj ) + k(k + 1)×
Rj k (j,2)
1 Rj k+2 (j,1)
Amk − Rj k(α1 + 1) − 2
Amk −
×
Rj rj
rj
R k
j
(j,3)
(j,1)
(j,2)
(j,3) − β1 kRj
Amk + rjk−1 Bmk − α1 Cmk − β1 Cmk −
rj
(j,2)
(j,3)
(j,2)
k+1
− rj
α1 Bmk + β1 Bmk + Dmk
Ymk (ϑj , ϕj )+
i
h R k+1
p
j
(j,5)
k (j,5)
Amk + rj Bmk Zmk (ϑj , ϕj +
+ k(k + 1)
rj
102
L. Giorgashvili, G. Karseladze, G. Sadunishvili
+ (−1)j 2he0
∞ X
k
X
(m)
rjk Yk
(j,2)
(ϑj , ϕj )Hmk +
k=0 m=−k
+ (−1)j he0 × grad
∞ X
k
X
∞ X
k
X
(m)
rjk Yk
(j,5)
(ϑj , ϕj )Bmk ,
(3.9)
k=0 m=−k
k + 1 Rj k+2 (j,4)
Amk +
Rj
rj
k=0 m=−k
R k
Rj k (j,3)
j
(j,2)
+β2 k(k−1)Rj
Amk +kRj (k−1)(α2 +1)+2
Amk +
rj
rj
(j,4)
(j,2)
(j,3)
(j,3) + krjk−1 Bmk − β2 Cmk − α2 Cmk − Emk −
(j,2)
(j,3)
− rjk+1 β2 (k + 2)Bmk + α2 (k + 2)Bmk +
(j,3)
(j,3)
+ (k + 2)Dmk − 2Hmk
Xmk (ϑj , ϕj )+
R k
p
1 Rj k+2 (j,4)
j
(j,2)
Amk − β2 kRj
Amk −
+ k(k + 1)
Rj rj
rj
R k
j
(j,3)
− Rj k(α2 + 1) − 2
Amk +
rj
(j,4)
(j,2)
(j,3)
(j,3) + rjk−1 Bmk − β2 Cmk − α2 Cmk − Emk −
(j,2)
(j,3)
(j,3) k+1
− rj
β2 Bmk + α2 Bmk + Dmk
Ymk (ϑj , ϕj )+
i
h R k+1
p
j
(j,6)
k (j,6)
Amk + rj Bmk Zmk (ϑj , ϕj +
+ k(k + 1)
rj
u00 (x(j) ) =
+ (−1)j 2he0
−
∞ X
k
X
(m)
rjk Yk
(j,3)
(ϑj , ϕj )Hmk +
k=0 m=−k
∞ X
k
X
(m)
(j,6)
+(−1)j he0 ×grad
rjk Yk (ϑj , ϕj )Bmk , j = 1, 2. (3.10)
k=0m=−k
Using the formulas (1) and (1.8), we have
(−1)j 2he0
∞ X
k
X
(m)
rjk Yk
(j,`)
(ϑj , ϕj )Hmk =
k=0 m=−k
=
∞
X
k
X
k=0 m=−k
h
p
+ k(k+1)
h
i
(j,1)
(j,`)
(j,2)
(j,`)
− rjk+1 σmk (Hmk ) + rjk−1 σmk (Hmk ) Xmk (ϑj , ϕj )+
i
1
1
(j,1)
(j,`)
(j,2)
(j,`)
rjk+1 σmk (Hmk )+ rjk−1 σmk (Hmk ) Ymk (ϑj , ϕj )+
k+1
k
103
Solution of a Boundary Value Problem of Statics
p
(j,3)
(j,`)
+ k(k + 1) rjk σmk (Hmk )Zmk (ϑj , ϕj ) , j = 1, 2, ` = 2, 3,
(−1)j he0 × grad
∞ X
k
X
(m)
rjk Yk
(3.11)
(j,`)
(ϑj , ϕj )Bmk =
k=0 m=−k
p
h
i
k(k+1)
=
Xmk (ϑj , ϕj )+
Ymk (ϑj , ϕj ) +
k
k=0m=−k
p
(j,5)
(j,`)
+ k(k + 1) rjk σmk (Bmk )Zmk (ϑj , ϕj ) , j = 1, 2, ` = 5, 6,
(3.12)
∞
X
k
X
(j,4)
(j,`)
rjk−1 σmk (Bmk )×
where
(j,1)
(j,`)
(1,1)
(j,`)
σmk (Hmk ) = (−1)j 2h cos ϕ0 sin ϑ0 ηmk (Hmk )+
s
(k+1)2 −m2
(j,`)
(2,1)
(j,`)
Hmk+1 ,
+sin ϕ0 sin ϑ0 ηmk (Hmk )−cos ϑ0
(2k+1)(2k+3)
(j,2)
(j,`)
(1,2)
(j,`)
σmk (Hmk ) = (−1)j 2h cos ϕ0 sin ϑ0 ηmk (Hmk )+
r
k 2 − m2 (j,`)
(2,2)
(j,`)
+ sin ϕ0 sin ϑ0 ηmk (Hmk ) + cos ϑ0
H
,
4k 2 − 1 mk−1
2h(−1)j
(1,3)
(j,`)
(j,3)
(j,`)
cos ϕ0 sin ϑ0 ηmk (Hmk ) − sin ϕ0 sin ϑ0 ×
σmk (Hmk ) =
k(k + 1)
(2,3)
(j,`)
(j,`)
× ηmk (Hmk )+im cos ϑ0 Hmk , j = 1, 2, ` = 2, 3,
(3.13)
(j,4)
(j,`)
(1,3)
(j,`)
σmk (Hmk ) = (−1)j h − cos ϕ0 sin ϑ0 ηmk (Hmk )+
(2,3)
(j,`)
(j,`)
+ sin ϑ0 sin ϕ0 ηmk (Bmk ) − im cos ϑ0 Bmk ,
(2k + 3)h (1,1) (j,`)
(2,1)
(j,`)
(j,5)
(j,`)
ηmk (Bmk ) + ηmk (Bmk )−
σmk (Hmk ) = (−1)j
k+1
s
(k + 1)2 − m2
(j,`)
Bmk+1 , j = 1, 2, ` = 5, 6.
−
(2k + 1)(2k + 3)
Taking the equalities (3.11)and (3.12) into account in (3.9) and (3.10),
we obtain
∞ X
k
X
(1)
0 (j)
u (x ) =
umk (rj )Xmk (ϑj , ϕj )+
k=0 m=−k
h
i
p
(1)
(1)
+ k(k + 1) vmk (rj )Ymk (ϑj , ϕj ) + wmk (rj )Zmk (ϑj , ϕj ) ,
(3.14)
104
L. Giorgashvili, G. Karseladze, G. Sadunishvili
u00 (x(j) ) =
∞ X
k
X
(2)
umk (rj )Xmk (ϑj , ϕj )+
k=0 m=−k
h
i
p
(2)
(2)
+ k(k+1) vmk (rj )Ymk (ϑj , ϕj )+wmk (rj )Zmk (ϑj , ϕj ) , j = 1, 2,
where
k + 1 Rj k+2 (j,1)
Amk +
Rj
rj
Rj k (j,2)
Amk +
+ kRj (k − 1)(α1 + 1) + 2
rj
R k
j
(j,3)
(j,1)
(j,2)
+ β1 k(k − 1)Rj
Amk + rjk+1 `mk + rjk−1 `mk ,
rj
Rj k (j,2)
1 Rj k+2 (j,1)
(1)
Amk − Rj k(α1 + 1) − 2
Amk −
vmk (rj ) =
Rj rj
rj
R k
1
j
(j,3)
(j,3)
(j,2)
Amk + rjk+1 `mk + rjk−1 `mk ,
− β1 kRj
rj
k
R k+1
j
(j,5)
(j,4)
(1)
Amk + rjk `mk ,
(3.15)
wmk (rj ) =
rj
R k
k + 1 Rj k+2 (j,4)
j
(2)
(j,2)
umk (rj ) = −
Amk + β2 k(k − 1)Rj
Amk +
Rj
rj
rj
R k
j
(j,3)
+ kRj (k − 1)(α2 + 1) + 2
Amk +
rj
(1)
umk (rj ) = −
(j,5)
(j,6)
+ rjk+1 `mk + rjk−1 `mk ,
R k
1 Rj k+2 (j,4)
j
(2)
(j,2)
vmk (rj ) =
Amk − β2 kRj
Amk −
Rj rj
rj
Rj k (j,3) k+1 (j,7) 1 k−1 (j,6)
Amk +rj `mk + rj `mk ,
−Rj k(α2 +1)−2
rj
k
R k+1
j
(j,6)
(j,8)
(2)
Amk + rjk `mk ;
wmk (rj ) =
rj
(j,1)
(j,2)
(j,3)
`mk = −α1 (k + 2)Bmk − β1 (k + 2)Bmk −
(j,2)
(j,2)
(j,2)
(j,1)
(j,2)
− (k + 2)Dmk + 2Hmk − σmk (Hmk ),
(j,1)
(j,2)
(j,3)
(j,2)
`mk = kBmk − α1 kCmk − β1 kCmk − kEmk +
(j,2)
(j,2)
(j,4)
(j,5)
+ σmk (Hmk ) + σmk (Bmk ),
(j,3)
(j,2)
(j,3)
(j,2)
(j,2)
(j,5)
`mk = −α1 Bmk − β1 Bmk − Dmk +
(j,4)
(j,5)
(j,3)
1
(j,1)
(j,2)
σ
(Hmk ),
k + 1 mk
(j,5)
`mk = Bmk + σmk (Hmk ) + σmk (Bmk ),
(j,5)
(j,3)
(j,2)
(3.16)
(j,3)
`mk = 2Hmk − β2 (k + 2)Bmk − α2 (k + 2)Bmk −
105
Solution of a Boundary Value Problem of Statics
(j,3)
(j,6)
(j,1)
(j,3)
− (k + 2)Dmk − σmk (Hmk ),
(j,4)
(j,2)
(j,3)
(j,3)
`mk = kBmk − kβ2 Cmk − kα2 Cmk − kEmk +
(j,2)
(j,3)
(j,4)
(j,6)
+ σmk (Hmk ) + σmk (Bmk ),
(j,7)
(j,2)
(j,3)
(j,3)
(j,3)
(j,5)
`mk = −β2 Bmk − α2 Bmk − Dmk +
(j,8)
(j,6)
(j,3)
1
(j,1)
(j,3)
σ
(Hmk ),
k + 1 mk
(j,6)
`mk = Bmk + σmk (Hmk ) + σmk (Bmk ).
Let us consider Problem (I)− . It will be assumed that in the system
(1.1) the vector function f (j,`) (z), `, j = 1, 2, is expanded into a series as
follows:
∞ X
k
X
(j,`)
f (j,`) (z) =
αmk Xmk (ϑj , ϕj )+
k=0 m=−k
h
p
(j,`)
(j,`)
+ k(k + 1) βmk Ymk (ϑj , ϕj ) + γmk Zmk (ϑj , ϕj )
i
, z ∈ ∂Ωj .
(3.17)
If in both parts of the equality (3) we pass to limit as x → z ∈ ∂Ωj
(rj → Rj ) and take into account the boundary conditions of the problem
(I)− and the formulas (3.15) and (3.17), then for the unknown constants
we obtain the following system of algebraic equations:
1 (j,1)
(j,1)
(j,2)
(j,1)
A
+ Rj `00 + Rj `00 = α00 ,
Rj 00
(3.18)
1 (j,4)
1 (j,6)
(j,5)
(j,2)
−
A00 + Rj `00 +
`00 = α00 , j = 1, 2;
Rj
Rj
(j,2)
k + 1 (j,1)
(j,3)
(j,1)
Amk +kRj (k−1)(α1 +1)+2 Amk +β1 k(k−1)Rj Amk = xmk ,
−
Rj
(j,2)
1 (j,1)
(j,3)
(j,2)
A
− Rj k(α1 + 1) − 2 Amk − β1 kRj Amk = xmk ,
Rj mk
k + 1 (j,4)
(j,2)
−
Amk + β2 k(k − 1)Rj Amk +
Rj
(j,3)
(j,3)
+kRj (k − 1)(α2 + 1) + 2 Amk = xmk ,
(3.19)
(j,3)
1 (j,4)
(j,2)
(j,4)
A
− β2 kRj Amk − Rj k(α2 + 1) − 2 Amk = xmk ,
Rj mk
−
(j,5)
(j,5)
(j,6)
(j,6)
Amk = xmk , Amk = xmk , k ≥ 1,
where
(j,1)
(j,1)
(j,1)
(j,2)
xmk = αmk − Rjk+1 `mk − Rjk−1 `mk ,
1
(j,2)
(j,1)
(j,3)
(j,2)
xmk = βmk − Rjk+1 `mk − Rjk−1 `mk ,
k
(j,6)
(j,5)
(j,3)
(j,2)
xmk = αmk − Rjk+1 `mk − Rjk−1 `mk ,
(3.20)
106
L. Giorgashvili, G. Karseladze, G. Sadunishvili
(j,4)
1 k−1 (j,6)
`mk ,
R
k j
(j,2)
(j,8)
= γmk − Rjk `mk , k ≥ 1, j = 1, 2.
(j,7)
(j,2)
xmk = βmk − Rjk+1 `mk −
(j,5)
(j,1)
(j,4)
xmk = γmk − Rjk `mk ,
(j,6)
xmk
If in the formulas (3) we use the notation of (3.8) and (3.13), then we
have
p
∞ X
h
i
X
(m,s)
(j,1)
(j,1)
Rqp+1 Csmpk A(q,2)
− β1 (k + 2)pakp,q (ϑ0 , ϕ0 )A(q,3)
`mk =
,
sp
sp
(j,2)
`mk =
p=1 s=−p
p
∞ X
X
p=1 s=−p
h
(m,s)
(j,2)
Rqp+1 kGkp,q (ϑ0 , ϕ0 )A(q,1)
+ Csmpk A(q,2)
sp
sp −
i
(m,s)
(j,4)
(m,s)
+
− β1 kpbkp,q (ϑ0 , ϕ0 )A(q,3)
+ σmk (Gkp,q )A(q,5)
sp
sp
(m,0)
(q,1)
+ kRq Gk0,q (ϑ0 , ϕ0 )A00 ,
p
∞ X
i
h
X
(j,3)
(m,s)
(j,3)
(q,3)
,
−
β
pa
(ϑ
,
ϕ
)A
Rqp+1 Csmpk A(q,2)
`mk =
1
0
0
sp
sp
kp,q
p=1 s=−p
p
∞ X
X
i
h
(j,3)
(m,s)
(j,4)
,
+ Csmpk A(q,5)
Rqp+1 (2p − 1)σmk (Gkp,q )A(q,2)
sp
sp
p=1s=−p
p
∞ X
X
h
(m,s)
(j,6)
+ Csmpk A(q,3)
Rqp+1 − β2 pkbkp,q (ϑ0 , ϕ0 )A(q,2)
sp
sp +
(j,4)
`mk =
p=1 s=−p
p
∞ X
h
i
X
(m,s)
(j,5)
(j,5)
(q,3)
Rqp+1 −β2 (k+2)pakp,q (ϑ0 , ϕ0 )A(q,2)
`mk =
, (3.21)
sp +Csmpk Asp
(j,6)
`mk =
p=1 s=−p
i
(m,s)
(j,4)
(m,s)
+
+ kGkp,q (ϑ0 , ϕ0 )A(q,4)
+ σmk (Gkp,q )A(q,6)
sp
sp
(m,0)
(q,4)
+ kRq Gk0,q (ϑ0, ϕ0 )A00 ,
p
∞ X
i
h
X
(m,s)
(j,7)
(j,7)
(q,3)
,
+
C
A
Rqp+1 − β2 pakp,q (ϑ0 , ϕ0 )A(q,2)
`mk =
sp
smpk sp
p=1 s=−p
(j,8)
`mk =
p
∞ X
X
p=1 s=−p
where
i
h
(j,3)
(m,s)
(j,8)
,
+ Csmpk A(q,6)
Rqp+1 (2p − 1)σmk (Gkp,q )A(q,3)
sp
sp
(m,s)
(j,1)
Csmpk = −(k + 2) p(α1 + 1) − 2 akp,q (ϑ0 , ϕ0 )+
(m,s)
(j,2)
Csmpk
(j,3)
Csmpk
(j,4)
(j,1)
(m,s)
+ 2(2p − 1)Gkp,q (ϑ0 , ϕ0 ) + (2p − 1)σmk (Gkp,q ),
(m,s)
(j,2)
(m,s)
= −k p(α1 + 1) − 2 bkp,q (ϑ0 , ϕ0 ) + (2p − 1)σmk (Gkp,q ),
(m,s)
2p − 1 (j,1) (m,s)
= − p(α1 + 1) − 2 akp,q (ϑ0 , ϕ0 ) +
σ
(Gkp,q ),
k + 1 mk
(m,s)
(j,5)
(m,s)
Csmpk = Gkp,q (ϑ0 , ϕ0 ) + σmk (Gkp,q ),
107
Solution of a Boundary Value Problem of Statics
(j,5)
(m,s)
Csmpk = 2(2p − 1)Gkp,q (ϑ0 , ϕ0 )−
(m,s)
(j,1)
(m,s)
− (k + 2) p(α2 + 1) − 2 akp,q (ϑ0 , ϕ0 ) − (2p − 1)σmk (Gkp,q ),
(j,6)
(m,s)
(j,2)
(m,s)
Csmpk = −k p(α2 + 1) − 2 bkp,q (ϑ0 , ϕ0 ) + (2p − 1)σmk (Gkp,q ),
(m,s)
2p − 1 (j,1) (m,s)
(j,7)
σ
(Gkp,q ),
Csmpk = − p(α2 + 1) − 2 akp,q (ϑ0 , ϕ0 ) +
k + 1 mk
(j,8)
(m,s)
(j,5)
(m,s)
Csmpk = Gkp,q (ϑ0 , ϕ0 ) + σmk (Gkp,q ), j 6= q = 1, 2.
With (3.22) taken into account, the system (3.18) implies
(j,1)
(j,1)
A00 = −Rj α00 +
p
∞
i
h
X X
(j,1)
(0,s)
(q,3)
,
−
2β
pa
(ϑ
,
ϕ
)A
Rj2 Rqp+1 Cs0p0 A(q,2)
+
1
0
0
sp
sp
0p,q
p=1 s=−p
(j,4)
A00
+
p
∞ X
X
p=1 s=−p
(j,2)
= −Rj α00 +
(3.22)
i
h
(0,s)
(j,5)
.
+ Cs0p0 A(q,3)
Rj2 Rqp+1 − 2β2 pa0p,q (ϑ0 , ϕ0 )A(q,2)
sp
sp
A solution of the system (3) has the form
Amk = H(k)xmk ,
(3.23)
where
(1,1)
(1,6)
(2,1)
(2,6) >
Amk = Amk , . . . , Amk , Amk , . . . , Amk
,
(3.24)
(1,1)
(1,6)
(2,1)
(2,6) >
xmk = xmk , . . . , xmk , xmk , . . . , xmk
,


..
(1)
0 
H (k) .
(j) .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . .
H(k) = 
, H (j) (k) = hi` (k) 6×6 , j = 1, 2,


..
0
. H (2) (k)
12×12
Rj − ∆(k) + 4(2k − 1)(α2 k − k − 1) ,
=
2∆(k)
h k(k − 1)d
i
2R
k
j
2
(j)
h12 (k) =
+ k(k + 1)α2 − (k 2 − 1)α1 − (k + 1)2 ,
∆(k)
d1
2β1 Rj k(2k 2 + k − 1)
2β
R
k(2k
− 1)
1 j
(j)
(j)
, h14 (k) = −
,
h13 (k) = −
∆(k)
∆(k)
(j)
h11 (k)
(j)
h1` (k) = 0, ` = 5, 6,
(j)
(j)
h23 (k) =
(j)
(j)
h21 (k) =
2h11 (k) + Rj
2h (k) + Rj (k − 1)
(j)
, h22 (k) = 12
,
2(2k − 1)Rj2
2(2k − 1)Rj2
(j)
(j)
h13 (k)
h14 (k)
(j)
(j)
, h24 (k) =
, h2` (k) = 0, ` = 5, 6,
(2k − 1)Rj2
(2k − 1)Rj2
108
L. Giorgashvili, G. Karseladze, G. Sadunishvili
h
(j)
i
1
(j)
,
h31 (k) = −
2
k(α
+1)−2k−1
h
(k)+R
k(α
+1)−2
1
j
1
11
2β1 Rj2 k(2k−1)
h
(j)
1
(j)
h32 (k) = −
2 k(α1 + 1) − 2k − 1 h12 (k)+
2
2β1 Rj k(2k − 1)
i
+Rj (k − 1)(α1 + 1) + 2 ,
(j)
h3` (k) = −
k(α1 + 1) − 2k − 1 (j)
(j)
h1` (k), ` = 3, 4, h3` (k) = 0, ` = 5, 6,
β1 Rj2 k(2k − 1)
2β2 Rj k(2k − 1)
2β2 Rj k(2k 2 − k + 1)
(j)
, h42 (k) = −
,
∆(k)
∆(k)
Rj (j)
− ∆(k) + 4(2k − 1)(kα1 − k − 1) ,
h43 (k) =
2∆(k)
i
2Rj k h k(k − 1)d2
(j)
+ k(k + 1)α1 − (k 2 − 1)α2 − (k + 1)2 ,
h44 (k) =
∆(k)
d1
(j)
h41 (k) = −
(j)
(j)
h4` (k) = 0, ` = 5, 6, h`` (k) = 1, ` = 5, 6,
(j)
(j)
h5` (k) = 0, ` = 1, 2, 3, 4, 6, h6` (k) = 0, ` = 1, 2, . . . , 5,
i
h
∆(k) = −4 (α1 α2 − β1 β2 )k 2 − (α1 + α2 )k(k + 1) + (k + 1)2 .
Since α1 α2 − β1 β2 =
d2
d1
> 0, α1 + α2 < 0, we have ∆(k) < 0.
(j,`)
A00 ,
If the values
` = 1, 4 from (3.22) are inserted into the expression
(j,`)
`mk , ` = 2, 6, contained in (3.21), then we obtain
(j,2)
(m,0)
(q,1)
`mk = − kRq2 Gk0,q (ϑ0 , ϕ0 )α00 +
p h
∞ X
X
(m,s)
p+1 (j,2)
kRqp+1 Gkp,q (ϑ0 , ϕ0 )A(q,1)
Csmpk A(q,2)
+
sp +Rq
sp −
p=1s=−p
(m,s)
(j,4)
(m,s)
p+1
−β1 kpRqp+1 bkp,q (ϑ0 , ϕ0 )A(q,3)
σmk (Gkp,q )A(q,5)
sp +Rq
sp +
(m,0)
(j,1)
+ kRq3 Rjp+1 Gk0,q (ϑ0 , ϕ0 )Cs0p0 A(j,2)
sp −
(j,6)
i
(m,0)
(0,s)
,
− 2β1 kpRq3 Rjp+1 Gk0,q (ϑ0 , ϕ0 )a0p,j (ϑ0 , ϕ0 )A(j,3)
sp
(m,0)
(q,2)
`mk = − kRq2 Gk0,q (ϑ0 , ϕ0 )α00 +
p h
∞ X
X
(m,s)
kRqp+1 Gkp,q (ϑ0 , ϕ0 )A(q,4)
+
sp −
p=1 s=−p
(m,s)
(j,6)
− β2 pkRqp+1 bkp,q (ϑ0 , ϕ0 )A(q,2)
+ Rqp+1 Csmpk A(q,3)
sp
sp +
(j,4)
(m,s)
+ Rqp+1 σmk (Gkp,q )A(q,6)
sp −
(m,0)
(0,s)
− 2β2 pkRq3 Rjp+1 Gk0,q (ϑ0 , ϕ0 )a0p,j (ϑ0 , ϕ0 )A(j,2)
sp +
i
(j,5) (j,3)
3 p+1 (m,0)
+ kRq Rj Gk0,q (ϑ0 , ϕ0 )Cs0p0 Asp .
(3.25)
109
Solution of a Boundary Value Problem of Statics
(j,`)
Substituting the values `mk , j = 1, 2, ` = 1, 2, . . . , 8, from (3.21) and
(3.24) into (3), we have
xmk = α
emk +
p
∞ X
X
N (s, m, p, k)Asp ,
(3.26)
p=1 s=−p
where Asp , xmk have the form (3.24), and


..
(1)
(2)
e
N (s, m, p, k) . N (s, m, p, k)

N (s, m, p, k) = 
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
e (1) (s, m, p, k) .. N (2) (s, m, p, k)
N
(`)
N (s, m, p, k) = Nij (s, m, p, k) 6×6 , ` = 1, 2,
(`)
e (`) (s, m, p, k) = N
e (s, m, p, k)
N
, ` = 1, 2,
ij
6×6
h
i>
(1,1)
(1,6)
(2,1)
(1,6)
α
emk = α
emk , . . . , α
emk , α
emk , . . . , α
emk
,
,
12×12
(`)
(j,2`−1)
α
emk
(j,2`)
α
emk
(j,5)
α
emk
(j)
=
(j,`)
(m,0)
(q,`)
= αmk + kRq2 Rjk−1 Gk0,q (ϑ0 , ϕ0 )α00 ,
(j,`)
(m,0)
(q,`)
= βmk + Rq2 Rjk−1 Gk0,q (ϑ0 , ϕ0 )α00 ,
(j,1)
γmk ,
(j,6)
α
emk
=
(j,2)
γmk ,
(3.27)
j 6= q = 1, 2, ` = 1, 2;
(m,0)
(j,1)
N12 (s, m, p, k) = −kRq3 Rjk+p Gk0,q (ϑ0 , ϕ0 )Cs0p0 ,
(j)
(m,0)
(0,s)
(j)
(m,0)
(0,s)
N13 (s, m, p, k) = 2β1 kpRq3 Rjk+p Gk0,q (ϑ0 , ϕ0 )a0p,j (ϑ0 , ϕ0 ),
N32 (s, m, p, k) = 2β2 kpRq3 Rjk+p Gk0,q (ϑ0 , ϕ0 )a0p,j (ϑ0 , ϕ0 ),
(j)
(m,0)
(j,5)
N33 (s, m, p, k) = −kRq3 Rjk+p Gk0,q (ϑ0 , ϕ0 )Cs0p0 ,
1 (j)
(j)
Ni` (s, m, p, k) = Ni−1` (s, m, p, k), i = 2, 4, ` = 2, 3,
k
(j)
Ni` (s, m, p, k) = 0, i = 1, 2, 3, 4, ` = 1, 4, 5, 6,
(j)
Ni` (s, m, p, k) = 0, i = 5, 6, ` = 1, 2, . . . , 6, j 6= q = 1, 2;
e (j) (s, m, p, k) = −kRk−1 Rqp+1 G(m,s) (ϑ0 , ϕ0 ),
N
11
j
kp,q
2 (j,1)
(j)
(j,2) k−1 p+1
e
N12 (s, m, p, k) = −Rj Rq
Rj Csmpk + Csmpk ,
e (j) (s, m, p, k) = β1 pRk−1 Rp+1 ×
N
q
13
j
2
(m,s)
(m,s)
× Rj (k + 2)akp,q (ϑ0 , ϕ0 ) + kbkp,q (ϑ0 , ϕ0 ) ,
e (j) (s, m, p, k) = −Rk−1 Rqp+1 σ (j,4) (G(m,s) ),
N
15
j
mk
kp,q
e (j) (s, m, p, k) = 0, ` = 4, 6,
N
1`
e (j) (s, m, p, k) = −Rk−1 Rp+1 G(m,s) (ϑ0 , ϕ0 ),
N
q
21
j
kp,q
(3.28)
110
L. Giorgashvili, G. Karseladze, G. Sadunishvili
h
i
e (j) (s, m, p, k) = −Rk−1 Rqp+1 Rj2 C (j,3) + 1 C (j,2) ,
N
22
j
smpk
k smpk
2 (m,s)
(m,s)
(j)
k−1 p+1
e
Rj akp,q (ϑ0 , ϕ0 ) + bkp,q (ϑ0 , ϕ0 ) ,
N23 (s, m, p, k) = β1 pRj Rq
e (j) (s, m, p, k) = − 1 Rk−1 Rqp+1 σ (j,4) (G(m,s) ),
N
25
mk
kp,q
k j
e (j) (s, m, p, k) = 0, ` = 4, 6,
N
2`
e (j) (s, m, p, k) = β2 pRk−1 Rqp+1 ×
N
32
j
(m,s)
(m,s)
× (k + 2)Rj2 akp,q (ϑ0 , ϕ0 ) + kbkp,q (ϑ0 , ϕ0 ) ,
e (j) (s, m, p, k) = −Rk−1 Rp+1 R2 C (j,5) + C (j,6) ,
N
j smpk
q
33
j
smpk
e (j) (s, m, p, k) = −Rk−1 Rqp+1 G(m,s) (ϑ0 , ϕ0 ),
N
34
j
kp,q
e (j) (s, m, p, k) = −Rk−1 Rqp+1 σ (j,4) (G(m,s) ),
N
36
j
mk
kp,q
e (j) (s, m, p, k) = 0, ` = 1, 5,
N
3`
e (j) (s, m, p, k) = β2 pRk−1 Rp+1 R2 a(m,s) (ϑ0 , ϕ0 ) + b(m,s) (ϑ0 , ϕ0 ) ,
N
j kp,q
q
42
j
kp,q
i
h
e (j) (s, m, p, k) = −Rk−1 Rp+1 R2 C (j,7) + 1 C (j,6) ,
N
j smpk
q
43
j
k smpk
(j)
k−1 p+1 (m,s)
e
N44 (s, m, p, k) = −Rj Rq Gkp,q (ϑ0 , ϕ0 ),
e (j) (s, m, p, k) = − 1 Rk−1 Rp+1 σ (j,4) (G(m,s) ),
N
q
46
mk
kp,q
k j
(j)
e (s, m, p, k) = 0, ` = 1, 5,
N
4`
e (j) (s, m, p, k) = −(2p − 1)Rk Rp+1 σ (j,3) (G(m,s) ),
N
j q
52
mk
kp,q
(3.29)
e (j) (s, m, p, k) = −Rk Rp+1 C (j,4) ,
N
j q
55
smpk
e (j) (s, m, p, k) = 0, ` = 1, 3, 4, 6,
N
5`
e (j) (s, m, p, k) = −(2p − 1)Rjk Rqp+1 σ (j,3) (G(m,s) ),
N
63
mk
kp,q
e (j) (s, m, p, k) = −Rjk Rqp+1 C (j,4) ,
N
65
smpk
e (j) (s, m, p, k) = 0, ` = 1, 2, 4, 6, j 6= q = 1, 2.
N
6`
If the values of the vector xmk from (3.26) are inserted into (3.23), then
for the unknown constants we obtain the following infinite system of linear
algebraic equations:
Amk = αmk +
p
∞ X
X
p=1 s=−p
Lsmpk Asp , k ≥ 1,
where Amk has the form (3.24),
αmk = H(k)e
αmk , Lsmpk = H(k)N (s, m, p, k),
(3.30)
111
Solution of a Boundary Value Problem of Statics
Lsmpk


..
(1)
(2)
L
.
L
smpk 
 smpk

=
. . . . . . . . . . . . . . . . . ,
.
(3)
.. L(4)
L
smpk
(1)
Lsmpk = H (1) (k)N (1) (s, m, p, k),
(3)
smpk
(2)
e (2) (s, m, p, k),
Lsmpk = H (1) (k)N
(4)
(2)
e (1) (s, m, p, k), L
Lsmpk = H (2) (k)N
(k)N (2) (s, m, p, k).
smpk = H
(3.31)
Let us investigate the system (3.30). For this we need estimates of the
elements of the matrix Lsmpk with respect to p and k.
The following estimate is valid [8]:
h (p + k + m − s)!(p + k − m + s)! i1/2
(k + m)!(k − m)!(p + s)!(p − s)!
≤
(p + k)!
, k ≥ 1, p ≥ 1. (3.32)
p!k!
Taking the inequalities (1.11) and (3.32) into account, we obtain
r
(m,s)
1
2p + 1 (p + k)!
G
,
kp,q (ϑ0 , ϕ0 ) ≤ k+p+1
d
2k + 1 p!k!
r
(m,s)
2p + 1 (p + k + 1)!
3
a
,
(3.33)
kp,q (ϑ0 , ϕ0 ) < k+p+1
d
2k + 1 p!(k + 1)!
r
(m,s)
2p + 1 (p + k)!
3
b
.
kp,q (ϑ0 , ϕ0 ) < k+p−1
d
2k + 1 p!k!
Using the inequalities (3.33), from (3.31), with the formulas (3.28) and
(3.29) taken into account, we obtain the following estimates:
k+p r 2p + 1
(1,`j)
2 R1
|Lsmpk | ≤ αk(p + 1)
,
d
2k + 1
R k R p+1
2
1
(2,`j)
×
|Lsmpk | ≤ β(p + 1)(k + 1)
d
d
r
(k + p + 1)! 2p + 1
×
,
p!(k + 1)!
2k + 1
R k R p+1
(3.34)
1
2
(3,`j)
×
|Lsmpk | ≤ γ(p + 1)(k + 1)
d
d
r
(k + p + 1)! 2p + 1
×
,
p!(k + 1)!
2k + 1
r
R k+p 2p + 1
2
(4,`j)
|Lsmpk | ≤ δk(p + 1)2
,
d
2k + 1
`, j = 1, 2, . . . , 6, k ≥ 1, p ≥ 1,
where α, β, γ, δ are positive constants not depending on p and k.
Let us show that the system (3.30) is quasiregular, which means that the
regularity condition is fulfilled only in the rows, starting from a certain one,
112
L. Giorgashvili, G. Karseladze, G. Sadunishvili
i.e., [9]
p X
12
∞ X
X
p=1 s=−p j=1
(`j)
(3.35)
(`j)
(3.36)
|Lsmpk | < 1, k = N +1, N +2, . . . , ` = 1, 2, . . . , 12,
and, moreover,
p X
12
∞ X
X
p=1 s=−p j=1
|Lsmpk | < +∞, k = 1, 2, . . . , N, ` = 1, 2, . . . , 12,
∞
X
|αmk,` | ≤ M 1 −
p X
12
X
p=N +1 s=−p j=1
(`j)
|Lsmpk | ,
(3.37)
k = N + 1, N + 2, . . . , ` = 1, 2, . . . , 12,
where M > 0 is a constant.
We first prove the existence of a natural number N such that for k > N
we have the following inequality:
p X
12
∞ X
X
p=1 s=−p j=1
(`j)
|Lsmpk | < 1 − ρ,
(3.38)
k = N + 1, N + 2, . . . , ` = 1, 2, . . . , 12,
where 0 < ε0 < ρ < ε < 1.
Taking into account the inequalities (3.34) and the identity [5]
∞
X
(k + p)!
p=0
k!p!
tp =
1 k+1
, 0 < t < 1,
1−t
we obtain
p X
12
∞ X
X
p=1 s=−p j=1
+σ3 (k + 1)4
(`j)
|Lsmpk | < σ1 k
R k
1
d
+ σ2 k
R k
2
d
+
d 5 R k
d 5 R1 k
2
+ σ4 (k + 1)4
<
d − R2
d − R2
d − R1
d − R1
R k
1
< σk 4
, k ≥ 1,
(3.39)
d − R2
where σ, σ` , ` = 1, 2, 3, 4, are positive constants not depending on k.
The following lemma is true [3].
Lemma 3.1. If 0 < α = const < 1, c = const > 0, m and k are natural
numbers, then the inequality k m ak < c is valid for any
n 2
o
1
k > N = max 2, 4a m c− m (1 − a1/m )−2 ,
where [◦] is the integer part of the number enclosed in the brackets.
113
Solution of a Boundary Value Problem of Statics
By virtue of this lemma we conclude that the inequality
R k
1
σk 4
<1−ρ
d − R2
holds for any k > N where
n h 1
p −2 io
1 p
1
N = max 2, 4σ 4 (1 − ρ)− 4 R12 4 d − R2 − 4 R1
.
Thus we have proved the inequality (3.38) and, along with it, the inequality
(3.35).
From the inequality (3.39) it follows that
p X
12
∞ X
X
p=1s=−p j=1
(`j)
|Lsmpk | < σ1 k 4 ≤ σN 4 < +∞, k = 1, 2, . . . , N, ` = 1, 2, . . . , 12.
The inequality (3.36) is thereby proved.
Let us choose a constant M such that
1
M = 0 max |αmk,` |, k ≥ 1, ` = 1, 2, . . . , 12.
ε
Hence, with (3.38) taken into account, it follows that
|αmk,` | ≤ max |αmk,` | = M ε0 < M ρ <
k≥1
< M 1−
∞
X
p X
12
X
p=1s=−p j=1
(`j)
|Lsmpk |
p X
12
∞
X
X
(`j)
|Lsmpk | ,
< M 1−
(3.40)
p=N +1s=−p j=1
k = N + 1, N + 2, . . . , ` = 1, 2, . . . , 12.
The inequality (3.37) is thereby proved.
Thus we have proved that the system (3.30) is quasiregular. As is known
[9], the problem of the existence of a solution of such a system reduces to
the problem of the existence of a solution of a finite system.
From the system (3) it follows that for k → ∞
(j,1)
(j,1)
(j,2)
(j,1)
(j,3)
(j,2)
(j,4)
(j,2)
|xmk | ∼ |αmk |, |xmk | ∼ |βmk |,
|xmk | ∼ |αmk |, |xmk | ∼ |βmk |,
(j,5)
(j,1)
(j,6)
(j,2)
|xmk | ∼ |γmk |, |xmk | ∼ |γmk |, j = 1, 2.
(j,`)
(j,`)
The series (3) containing the constants Bmk , ` = 1, 2, . . . , 6, Cmk ,
(j,`)
Hmk , ` = 2, 3, converge irrespective of the fact whether
x ∈ Ω or x ∈ ∂Ωj , j = 1, 2. The regularity of the vector U = (u0 , u00 ) depends on the convergence of the series (3) and their derivatives of first order
(j,`)
containing the constants Amk , j = 1, 2, ` = 1, 2, . . . , 6. If x ∈ Ω− , then
rj > Rj , j = 1, 2, and the above-mentioned series converge absolutely and
uniformly in Ω− . If x ∈ ∂Ωj , then, in view of the estimates (1.10), (1.11),
for the convergence of the series (3) and their first degree derivatives it is
(j,`)
(j,`)
Dmk , Emk ,
−
114
L. Giorgashvili, G. Karseladze, G. Sadunishvili
(j)
(j)
(j)
sufficient that the coefficients αmk , βmk , γmk , j = 1, 2, admit the following
estimates for k → ∞
(j)
αmk = O(k −4 ),
(j)
(j)
βmk = O(k −5 ), γmk = O(k −5 ), j = 1, 2.
(j)
(j)
(3.41)
(j)
From Theorem 1.1 it follows that the coefficients αmk , βmk , γmk , j = 1, 2,
admit the estimates (3.41) if the vector function f (j) (z) belongs to the class
f (j) (z) ∈ C 4 (∂Ωj ), j = 1, 2.
(j)
(j)
(j)
Note that since the coefficients αmk , βmk , γmk , j = 1, 2, tend to zero for
k → ∞, we can always choose a constant M > 0 such that the inequality
(3.40) be fulfilled.
With (3.5) taken into account, from (3.4) it follows that for |x| → ∞ we
have
u0 (x) = O(|x|−1 ),
∂u0j (x)
= O(|x|−2 ),
∂xk
u00 (x) = O(|x|−1 ),
∂u00j (x)
= O(|x|−2 ), k, j = 1, 2, 3.
∂xk
(3.42)
Problem (I)− is solved.
Problems (II)− and (III)− are solved analogously.
References
1. R. J. Atkin, P. Chadwick, and T. R. Steel, Uniqueness theorems for linearized
theories of interacting continua. Mathematika 14 (1967), 27–42.
2. L. Giorgashvili, Solution of the basic boundary value problems of stationary thermoelastic oscillations for domains bounded by spherical surfaces. Georgian Math. J.
4 (1997), No. 5, 421–438.
3. L. Giorgashvili and K. Skhvitaridze, Motion of a viscous incompressible fluid in
an infinite domain bounded by two spherical surfaces. (Russian) Sbornik Stat. Comt.
Math and Cyber. Fac. MGU im. M. V. Lomonososva, 2002, 14–29.
4. L. Giorgashvili and K. Skvitaridze, Solution of a nonclassical problem of oscillation of two-component mixtures. Georgian Math. J. 13 (2006), No. 1, 35–53.
5. E. W. Hobson, Theory of spherical and elipsoidal functions. Izdat. Inostr. Literaturi,
Moscow, 1952 (translated into Russian).
6. A. E. Green and P. M. Naghdi, On basic equations for mixtures. Q. J. Mech. Appl.
Math. 22 (1969), 427–438.
7. A. E. Green and T. R. Steel, Constitutive equations for interacting continua.
Internat. J. Engrg. Sci. 4 (1966), No. 4, 483–500.
8. A. N. Gooz and V. T. Golovchan, Diffraction of elastic waves in multiply connected
bodies. (Russian) Naukova Dumka, Kiev, 1972.
9. L. V. Kantorovich and V. I. Krilov, Approximate methods of higher analysis.
(Russian) Gos. Izdat. Fiz-Math. Literaturi, Moscow–Leningrad, 1962.
10. R. J. Knops and T. R. Steel, Uniqueness in the linear theory of a mixture of two
elastic solids. Internat. J. Engrg. Sci. 7 (1969), 571–577.
11. V. D. Kupradze, T. G. Gegelia, M. O. Basheleishvili, and T. V. Burchuladze,
Three-dimensional problems of the mathematical theory of elasticity and thermoelasticity. North-Holland Publishing Co., Amsterdam–New York, 1979.
12. F. M. Mors and G. Feshbah, Methods of theoretical physics, II. (Russian) Izdat.
Inostr. Literaturi, Moscow, 1960.
Solution of a Boundary Value Problem of Statics
115
13. D. G. Natroshvili, A. Ya. Jagmaidze, and M. Zh. Svanadze, Some problems in
the linear theory of elastic mixtures. (Russian) Tbilis. Gos. Univ., Tbilisi, 1986.
14. T. R. Steel, Applications of a theory of interacting continua. Q. J. Mech. Appl.
Math. 20 (1967), 57–72.
15. A. N. Tikhonov and A. A. Samarskiı̌, Equations of mathematical physics. (Russian) Fizmatgiz, Moscow, 1966.
16. A. F. Ulitko, The method of eigenvector functions in three-dimensional problems
of the theory of elasticity. (Russian) Naukova Dumka, Kiev, 1979.
(Received 23.07.2008)
Authors’ address:
Georgian Technical University
Department of Mathematics
77, M. Kostava St., Tbilisi 0175
Georgia
E-mail: lgiorgashvili@yahoo.com