Volume 13, 2009
1
MAIN ARTICLES
THE BASIC BVPs OF THE THEORY OF ELASTIC BINARY MIXTURES
FOR A HALF-PLANE WITH CURVILINEAR CUTS
Bitsadze L.
I. Vekua Institute of Applied Mathematics of
Iv. Javakhishvili Tbilisi State University
Phone: (+99532)18-92-39, E-mail: lamarabits@yahoo.com
Abstract. The first and second boundary value problems of the theory of
elastic binary mixtures for a transversally isotropic half-plane with curvilinear
cuts are investigated. The solvability of a system of singular integral equations
is proved by using the potential method and the theory of singular integral
equations.
Key words: Elastic mixture, uniqueness theorem, potential method, explicit solution.
MSC 2000 : 74E30, 74G05.
Introduction. In the present paper the first and second boundary value
problems (BVPs) of elastic binary mixture theory are investigated for a transversally-isotropic half plane with curvilinear cuts. The boundary value problems of elasticity for anisotropic media with cuts were considered in [1, 2]. In
this paper we extend this result to BVPs of elastic mixture for a transversallyisotropic elastic body. Here we shall be concerned with the plane problem
of elastic binary mixture theory (it is assumed that the second components
u02 and u002 of the three-dimensional partial displacement vectors u0 (u01 , u02 , u03 )
and u00 (u001 , u002 , u003 ) are equal to zero, while the components u01 , u03 , u001 , u003 depend
only on the variables x1 , x3 ). A solution of the first BVP is sought in the form
of a double-layer potential while a solution of the second BVP is sought in
the form of a single-layer potential. For the unknown density we obtain a system of singular integral equations. Using the potential method and the theory
of singular integral equations we rigorously prove the solvability of system of
singular integral equations, corresponding to the boundary value problems.
The basic homogeneous equations of statics of the transversally isotropic
elastic binary mixtures theory in the case of plane deformation can be written
in the form [3]
µ (1)
¶
C (∂x) C (3) (∂x)
C(∂x)U =
U = 0,
(1)
C (3) (∂x) C (2) (∂x)
(j)
where the components of the matrix C (j) (∂x) = kCpq (∂x)k2x2 are given in the
2
Bulletin of TICMI
form
(j)
(j)
(j)
(j)
(j)
(j)
Cpq (∂x) = Cqp (∂x), j = 1, 2, 3, p, q = 1, 2, C11 (∂x) = c11
(j)
C12 (∂x) = (c13 + c44 )
2
∂2
(j) ∂
+
c
,
44
∂x21
∂x23
2
2
∂2
(j) ∂
(j)
(j) ∂
, C22 (∂x) = c44 2 + c33 2 ,
∂x1 ∂x3
∂x1
∂x3
(k)
cpq -are constants, characterizing the physical properties of the mixture and satisfying certain inequalities caused by the positive definiteness of potential energy. U = U T (x) = (u0 , u00 ) is four-dimensional displacement vector-function,
u0 (x) = (u01 , u03 ) and u00 (x) = (u001 , u003 ) are partial displacement vectors depending on the variables x1 , x3 . Evrerywhere below by ”T” we denote transposition.
Let D be the half-plane x3 < 0 with the boundary x3 = 0 and suppose
that the boundary of the half-plane is fastened. Let’s assume that in D we
have p curvilinear cuts, lj = aj bj , j = 1, 2, ..., p, which are simple relatively
nonintersecting open Lyapunov arcs having no common points, and do not
intersect the boundary. The positive direction on lj is chosen from the point
aj to the point bj . The normal on lj is direct to the right with respect to the
p
S
positive motion of direction. l =
lj . We suppose that D is filled with binary
j=1
transversally-isotropic elastic mixture.
We introduce the notation z = x1 + ix3 , ζk = y1 + αk y3 , tk = t1 + αk t3 ,
σk = zk − ςk , zk = x1 + αk x3 , t = t1 + t3 .
The basic boundary value problems of static of the theory of elastic binary
mixtures are formulated as follows:
Problem 1. Find a regular solution of the equation (1) in D, when the
boundary values of the displacement vector are given on both sides of the
lj , j = 1, 2, ..., p and on the boundary x3 = 0. Let’s also assume, that the
principal vector of external force acting on l, stress vector and the rotation at
infinity are zero. It is required to define the deformed state of the plane.
If we denote by U + (U − ) the limits of U on l from the left (right), then the
boundary conditions of the problem will take the form
U + (t0 ) = f + (t0 ), U − (t0 ) = f − (t0 ), t0 ∈ l, U − = 0, x3 = 0,
(2)
where f + , and f − are the given functions on l satisfying Hölder’s conditions
on the cuts lj , having derivatives in the class H ∗ (for the definitions of the
classes H and H ∗ see[4]) and satisfying the following conditions on the ends
aj and bj of lj
f + (aj ) = f − (aj ), f + (bj ) = f − (bj ).
Problem 2. Find a regular solution of the equation (1) in D, when the stress
vector is given on both sides of the lj , j = 1, 2, ..., p and the boundary x3 = 0
is traction free. In addition it is assumed that the principal vector of external
force acting on l, stress vector and the rotation at infinity are zero. The
Volume 13, 2009
3
boundary conditions can be written as follows:
[T U ]+ (t0 ) = f + (t0 ),
[T U ]− (t0 ) = f − (t0 ), t0 ∈ l, [T U ]− = 0, x3 = 0, −∞ < x3 < +∞, (3)
where f + , and f − are the given known vector-functions on l of the Holder class
H, which have derivatives in the class H ∗ and satisfying at the ends aj and bj
of lj , the conditions
f + (aj ) = f − (aj ), f + (bj ) = f − (bj ).
Therefore, it is interesting to study the behavior of the solution of the problem
in the neighborhood of cuts.
The first BVP for the half-plane with curvilinear cuts. We seek
for a solution of the problem in the form [5],
Z
4
X
1
∂ln(tk − zk )
U (z) = Im
E(k)
[g(s) + ih(s)]ds+
π
∂s
k=1
l
Z
p
4
4 X
X
X
1
∂ln(tk − zk )
+ Im
E(k) Ej
Vj (z),
[g(s) − ih(s)]ds +
π
∂s
k=1 j=1
k=1
(4)
l
(k)
kApq A−1 k4x4
denotes a special matrix that reduces the first
where E(k) =
(k)
BVP to a Fredholm integral equation of second order, A(k) = kApq k4x4 . The
elements of the matrix E(k) and the matrix A−1 are defined as follows


A33 ∆2
0
−A13 ∆2
0
1 
0
A44 ∆1
0
−A24 ∆1 

,
A−1 =


−A13 ∆2
0
A11 ∆2
0
∆1 ∆2
0
−A24 ∆1
0
A22 ∆1
4
P
(k)
(k)
(k)
Aij =
Aij , Aij = Aji ,
(k)
k=1
(2)
(2)
A11 = dk [c11 q4 + αk2 t11 + αk4 t12 + c44 q3 αk6 ],
(k)
(2)
(2)
A22 = dk [q1 c44 + αk2 t44 + αk4 t42 + c33 q4 αk6 ],
(k)
(1)
(1)
A33 = dk [c11 q4 + αk2 t33 + αk4 t23 + c44 q3 αk6 ],
((k))
A44
(1)
(1)
= dk [q1 c44 + αk2 t55 + αk4 t52 + c33 q4 αk6 ],
(k)
A12 = dk αk [v12 + v11 αk2 + v13 αk4 ],
(k)
A14 = dk αk [w11 + w12 αk2 + w13 αk4 ], k = 3, 4,
(k)
A23 = dk αk [v22 αk + v21 αk2 + v23 αk4 ],
(k)
A34 = dk αk [w24 + w14 αk2 + w34 a4k ], k = 3, 4, 5, 6,
(k)
(3)
(3)
(k)
(3)
(3)
A24 = dk [−q1 c44 + αk2 t66 + αk4 t62 − c33 q4 αk6 ],
A13 = dk [−q4 c11 + αk2 t22 + αk4 t13 − c33 q3 αk6 ],
(5)
4
Bulletin of TICMI
∆1 =
√
a1 a2 a3 a4 ∆2 ,
∆2 = A22 A44 − A224 , ∆1 = A11 A33 − A213 ,
(2)
(2)
(3)2 (2)
(3)
(2)
(3)
(2)2 (1)
(2)
(2)
(3)2 (2)
(3)
(2)
(3)
(2)2 (1)
t11 = c11 δ22 + c44 q4 − α13 c44 + 2c44 α13 α13 − α13 c44 ,
t12 = c44 δ22 + c11 q3 − α13 c33 + 2c33 α13 α13 − α13 c33 ,
(3)
(3)
(1)
(3) (2)
(3)
(2)
(1)
(3)2
(2)
(3) (1)
(3)
(3)
(1)
(3) (2)
(3)
(2)
(1)
(3)2
(2)
(3) (1)
(1)
(3) (2)
t22 = −c11 δ22 − c44 q4 + α13 α13 c44 − c44 (α13 α13 + α13 ) + α13 α13 c44 ,
t13 = −c44 δ22 − c11 q3 + α13 α13 c33 − c33 (α13 α13 + α13 ) + α13 α13 c33 ,
(1)
(1)
(2)
(2)
(3)
(3)
t52 = t33 − c11 δ22 + c33 δ11 , t62 = t22 − c11 δ22 + c33 δ11 ,
t42 = t11 − c11 δ22 + c33 δ11 ,
(2)
(2)
(3)2 (1)
(3)
(2)
(3)
(3)2 (2)
t44 = c44 δ11 + c33 q1 − α13 c11 + 2c11 α13 α13 − α13 c11 ,
(3)
(3)
(2)
(3) (1)
(3)
(2)
(1)
(3)2
t66 = −c44 δ11 − c33 q1 + α13 α13 c11 − c11 (α13 α13 + α13 ) + α13 α13 c11 ,
(1)
(1)
(1)2 (2)
(3)
(1)
(3)
(3)2 (1)
(1)
(1)
(1)2 (2)
(3)
(1)
(3)
(3)2 (1)
(1)
(1)
(1)2 (2)
(3)
(1)
(3)
(3)2 (1)
t23 = c44 δ22 + c11 q3 − α13 c33 + 2c33 α13 α13 − α13 c33 ,
t33 = c11 δ22 + c44 q4 − α13 c44 + 2c44 α13 α13 − α13 c44 ,
t55 = c44 δ11 + c33 q1 − α13 c11 + 2c11 α13 α13 − α13 c11 ,
(2)
(2)
(1)
(3)2
(1)
(2)2
(2) (2)
(2)
(3)2
(3) (3)
v11 = α13 (α13 α13 − α13 ) − α13 (c44 + c11 c33 ) − α13 (c44 + c11 c33 )
(3) (2)
(2) (3)
(2) (3)
(3)
+α13 (2c44 c44 + c11 c33 + c11 c33 ),
(2)
(2)
(1)
(3)2
(3)
(1) (2)
(2) (1)
(3)2
(3) (3)
w12 = −α13 (α13 α13 − α13 ) − α13 (c44 c44 + c11 c33 + c44 + c11 c33 )
(2) (3)
(1)
(3) (3)
(1) (3)
(2)
(2) (3)
+α13 (c44 c44 + c11 c33 ) + α13 (c44 c44 + c11 c33 ),
(3)
(2)
(1)
(3)2
(3)
(2) (1)
(3) (3)
(3)2
(1) (2)
(3)2
(1) (3)
v21 = −α13 (α13 α13 − α13 ) − α13 (c44 c44 + c11 c33 + c44 + c11 c33 )
(2)
(1) (3)
(1) (3)
(1)
(2) (3)
(3) (2)
+α13 (c44 c44 + c11 c33 ) + α13 (c44 c44 + c11 c33 ),
(1)
(2)
(1)
(3)2
(3)
(1) (3)
(3) (1)
w14 = α13 (α13 α13 − α13 ) + α13 (2c44 c44 + c11 c33 + c44 + c11 c33 )
(1)
(3)2
(3) (3)
(2)
(1)2
(1) (1)
−α13 (c44 + c11 c33 ) − α13 (c44 + c11 c33 ),
(1) (2) (2)
(3)
(2) (3)
(3) (2)
(2) (3) (3)
(1) (2) (2)
(3)
(2) (3)
(3) (2)
(2) (3) (3)
v12 = −α13 c44 c11 + α13 (c11 c44 + c11 c44 ) − α13 c11 c44 ,
v13 = −α13 c44 c33 + α13 (c33 c44 + c33 c44 ) − α13 c33 c44 ,
(1) (3) (2)
(3)
(2) (1)
(3) (3)
(2) (3) (1)
(1) (2) (3)
(3)
(1) (2)
(3) (3)
(2) (3) (1)
w11 = α13 c44 c11 − α13 (c11 c44 + c11 c44 ) + α13 c11 c44 ,
w13 = α13 c44 c33 − α13 (c33 c44 + c33 c44 ) + α13 c44 c11 ,
(1) (2) (3)
(3)
(1) (2)
(3) (3)
(2) (1) (3)
(1) (3) (2)
(3)
(2) (1)
(3) (3)
(2) (3) (1)
v22 = α13 c44 c11 − α13 (c11 c44 + c11 c44 ) + α13 c11 c44 ,
v23 = α13 c44 c33 − α13 (c33 c44 + c33 c44 ) + α13 c33 c44 ,
(1) (3) (3)
(3)
(3) (1)
(1) (3)
(2) (1) (1)
w24 = −α13 c44 c11 + α13 (c11 c44 + c11 c44 ) − α13 c11 c44 ,
Volume 13, 2009
(1) (3) (3)
(3)
(1) (3)
(3) (1)
5
(2) (1) (1)
w34 = −α13 c44 c33 + α13 (c33 c44 + c33 c44 ) − α13 c33 c44 ,
αk , k = 1, 2, 3, 4, are the roots of the characteristic equation ([6]). g and h
are the unknown real vectors from the Holder class that have derivatives in
the class H ∗ . We write
(k)
(k)
(k)
(k)
4
X
(zk − bj )ln(zk − bj ) − (zk − aj )ln(zk − aj ) j
1
Vj (z) = Im
K ,
A(k)
(k)
(k)
π
bj − aj
k=1
(k)
(k)
aj = Re aj + αk Im aj , bj = Re bj + αk Im bj .
K (j) , j = 1, ..., p, are the unknown real constant vectors to be defined later on.
The vector Vj (z) satisfies the following conditions:
1. Vj (z) has the logarithmic singularity at infinity
4
X
1
Vj = Im
A(k) (− ln zk + 1)K j + O(zk−1 ).
π
k=1
2. Under Vj is supposed a branch, which is uniquely defined on the cut
plane along lj .
3. Vj is continuously continued function on lj from the left and right,
including the end points aj and bj , i. e., we have the equalities
Vj+ (aj ) = Vj− (aj ), Vj+ (bj ) = Vj− (bj ),
Vj+
−
Vj−
= 2 Re
4
P
(k)
A
(k)
k=1
τk − aj
(k)
bj
(k)
aj
−
K j , j = 1, ..., p.
It is easily to check U − = 0, x3 = 0, and
Z+∞
Z
−
(T U ) ds + [(T U )+ − (T U )− ]ds = 0.
−∞
l
To define the unknown density, we obtain by virtue of (4)-(2), the following
system of singular integral equation of the normal type
−
+
2g(τ ) = f − f − Re
p P
4
P
(k)
(k)
A
j=1 k=1
1
π
Z
l
h(t)dt 1
+
t − t0
π
τk − aj
(k)
bj
−
(k)
aj
Kj,
Z
K(t0 , t)dt = Ω(t0 ), t0 ∈ l,
l
where
K(t0 , t) = −iE
∂θ
+
∂s
(6)
6
Bulletin of TICMI
4 X
4
X
∂
t − t0
∂
), −Re
E(k) Eq ln(tq − −tk0 ),
+Re
E(k) ln(1 + λk
∂s
t − t0
∂s
k=1
k=1 q=1
p
1
1X +
(V + Vj− )
Ω(t0 ) = (f + + f − ) −
2
2 j=1 j
Z
4
X
1
∂θ
∂
t − t0
−
[E
+ Im
E(k)
ln(1 + λk
)
π
∂s(t)
∂s(t)
t
−
t
0
k=1
4
P
l
+Im
4
4 P
P
k=1 q=1
E(k) Eq
∂
ln(tq − tk0 )]g(t)ds,
∂s(t)
1 + iαk
, θ = arg(t − t0 ), tk0 = Re t0 + αk Im t0 , t = y1 + iy3 .
1 − iαk
Thus we defined vector g on l. It is not difficult to verify, that g ∈ H,
g 0 ∈ H ∗ , g(aj ) = g(bj ) = 0, Ω ∈ H, Ω0 ∈ H ∗ . Formula (6) is a system of
singular integral equations of the normal type with respect to the vector h.
The points aj and bj are nonsingular ones, while the summary index of the
class h2p is equal to −4p (for the definition of the class h2p see [4]).
If the solution of equation (6) on the class h2p exists , it will satisfy Holder’s
condition on l, vanishing on the points aj and bj and having to derivatives in
the class H ∗ .
Let’s prove that the homogeneous equation corresponding to the system of
equation (6) only has a trivial solution h0 in the class h2p . Let’s assume the
contrary. Let h0 be nontrivial solution of the homogeneous system in the class
h2p and construct the potential
Z
4
X
1
∂ ln (tk − zk )
U0 (z) = Re
E(k)
[g(t) + ih(t)]ds
π
∂s
k=1
l
Z
4
4
XX
∂ ln (tj − zk )
1
E(k) Ej
[g(t) − ih(t)]ds.
+ Im
π
∂s
k=1 j=1
λk =
l
U0+
U0−
It is clear
=
= 0, t ∈ l and on the basis of uniqueness theorem U0 = 0,
z ∈ D. Therefore T U0 = 0, z ∈ D, and
∂h0
(T U0 )+ − (T U0 )− = 2A−1
= 0.
∂s
Consequently h0 = 0, since h0 (aj ) = 0, that completes the proof. Thus the
corresponding adjoint homogeneous equation has 4p linearly independent solution σj , j = 1, ..., 4p in the adjoined class and the conditions of solvability
are
Z
Ωσj ds = 0.
l
From last equality we get 4p algebraic equation with respect to Kj . On the
basis of the uniqueness theorem it is not difficult to prove that the last system
is solvable.
Volume 13, 2009
7
The second BVP for the half-plane with curvilinear cuts. We
seek for a solution of the above formulated second BVP for the half-plane in
the form [5]
Z
4
X
1
T
U (z) = Im
R(k) iL( ln (tk − zk )[g(s) + ih(s)]ds
π
k=1
l
R
4
P
L(k) iL ln (zk − tj )[g(s) − ih(s)]ds),
j=1
l
where g and h are unknown real vector-functions.
R(k)
(k)
R1j
(k)
R2j
(k)
R3j
(k)
R4j
(k)
= kRpq k4x4 , p, q = 1, 2, 3, 4,
(1) (k)
(3) (k)
(1) (k)
(3) (k)
= αk (c44 A1j + c44 Aj3 ) + c44 Aj2 + c44 Aj4 ,
(k)
= −αk−1 R1j ,
(3) (k)
(2) (k)
(3) (k)
(2) (k)
= αk (c44 A1j + c44 Aj3 ) + c44 Aj2 + c44 Aj4 ,
(k)
= −αk−1 R3j , j = 1, 2, 3, 4,
(k)
(7)
Apq are given by (5) and L(k)T L denotes the complex conjugate matrix of
L(k)T L,


L33 ∆2
0
−L13 ∆2
0
1 
0
L44 ∆1
0
−L24 ∆1 

,
L=
(8)

0
L11 ∆2
0
∆1 ∆2  −L13 ∆2
0
−L24 ∆1
0
L22 ∆1
L11 = −∆q4 [a44 B1 + (b11 + 2a34 )A1 + a33 D1 ],
A1 = −B0 m3 ,
B 1 = Bo m 1 ,
L13 = ∆q4 [a24 B1 + (−b33 + a14 + a23 )A1 + a13 D1 ],
A ! + B1 m 2
,
α1 α2 α3 α4
= −∆q4 [a44 C1 + (b11 + 2a34 )B1 + a33 A1 ],
C1 = −
L22
D1 = −A1 m2 − B1 α1 α2 α3 α4 ,
L24 = ∆q4 [a24 C1 + (−b33 + a14 + a23 )B1 + a13 A1 ],
∆1 = L11 L33 − L213 ,
L33 = −∆q4 [a22 B1 + (b22 + 2a12 )A1 + a11 D1 ],
∆2 = L22 L44 − L224 ,
L44 = −∆q4 [a22 C1 + (b22 + 2a12 )B1 + a11 A1 ],
√
∆2 = [b4 (m1 m3 − 2 a1 a2 a3 a4 ) + q4 ∆m0 ]q4 ∆B0 > 0,
√
m0 = (a11 a44 + a33 a22 − 2a13 a24 ), ∆1 = a1 a2 a3 a4 ∆2 ,
8
Bulletin of TICMI
4
P
m1 =
αk , m2 = α1 α2 + α1 α3 + α1 α4 + α2 α3 + α2 α4 + α3 α4 ,
k=1
m3 = α1 α2 α3 + α1 α2 α4 + α1 α3 α4 + α2 α3 α4 ,
B0−1 = (α1 + α2 )(α1 + α3 )(α1 + α4 )(α2 + α3 )(α2 + α4 )(α3 + α4 ).
(j)
Between the coefficients apq , bpp and cpq there are the relations
(2)
(1) (2)2
(2) (3) (3)
(2) (2)2
a11 ∆ = c11 q3 − c33 c13 + 2c13 c13 c33 − c33 c13 > 0,
(2)
(1) (2)
(3)2
(1) (2) (2)
(2) (3) (3)
(3)
(2) (3)
(3) (2)
a12 ∆ = c13 (c13 c13 − c13 ) − c13 c11 c33 − c13 c11 c33 + c13 (c11 c33 + c11 c33 ),
(3)
(2) (1) (3)
(1) (2) (3)
(3)
(1) (2)
(3)2
a13 ∆ = −c11 q3 + c33 c13 c13 + c33 c13 c13 − c33 (c13 c13 + c13 ),
(3)
(1) (2)
(3)2
(1) (2) (3)
(2) (3) (1)
(3)
(2) (1)
(3) (3)
(3)
(1) (2)
(3)2
(1) (3) (2)
(2) (1) (3)
(3)
(1) (2)
(3) (3)
a14 ∆ = −c13 (c13 c13 − c13 ) + c13 c11 c33 + c13 c11 c33 − c13 (c11 c33 + c11 c33 ),
a23 ∆ = −c13 (c13 c13 − c13 ) + c13 c11 c33 + c13 c11 c33 − c13 (c11 c33 + c11 c33 ),
(2)
(1) (2)2
(2) (3) (3)
(2) (2)2
a22 ∆ = c33 q1 − c11 c13 + 2c13 c13 c11 − c11 c13 > 0,
(3)
(2) (1) (3)
(2) (3) (1)
(3)
(1) (2)
(3)2
a24 ∆ = −c33 q1 + c11 c13 c13 + c13 c13 c11 − c11 (c13 c13 + c13 ),
(1)
(2) (1)2
(1) (3) (3)
(1) (3)2
a33 ∆ = c11 q3 − c33 c13 + 2c13 c13 c33 − c33 c13 > 0,
(1)
(2) (1) (1)
(3)2
(1) (2)
(1) (3)
(3)
(1) (3) (3)
(3) (1)
a34 ∆ = c13 (c13 c13 − c13 ) − c13 c11 c33 − c13 c11 c33 + c13 (c11 c33 + c11 c33 ),
(1)
(2) (1)2
(1) (3) (3)
(1) (3)2
a44 ∆ = c33 q1 − c11 c13 + 2c13 c13 c11 − c11 c13 > 0,
(1)
(3)
(2)
(1) (2)
(3)2
∆ = (c11 a11 + c11 a33 + 2c11 a13 )∆ − q1 q3 + (c13 c13 − c13 ) > 0,
(j)
bjj = c44 q4−1 > 0, j = 1, 2, 3.

(k)
(k)
(k)
αk2 L22
− αk L22
αk2 L24

(k)
(2)
(k)
L22
− αk L24
 −αk L22
(k)
L =  2 (k)
(k)
(k)
 αk L24
− αk L24
αk2 L44
(k)
(k)
(k)
−αk L24
L24
− αk L44
(k)
− αk L24
(k)
L24
(k)
− αk L44
(k)
L44



,

(k)
L22 = −∆q4 dk [a44 + αk2 (b11 + 2a34 ) + a33 αk4 ],
(k)
L24 = ∆q4 dk [a24 + αk2 (−b33 + a14 + a23 ) + a13 αk4 ],
(k)
L44 = −∆q4 dk [a22 + αk2 (b22 + 2a12 ) + a11 αk4 ].
For the stress vector we get
Z
4
X
1
∂ ln(tk − zk )
T (∂x, n)U (z) = Im
L(k) iL[
[g(t) + ih(t)]ds+
π
∂s
k=1
l
Z
4
X
1
∂ ln (zk − tj )
Im
L(j) iL
[g(t) − ih(t)]ds].
π
∂s
j=1
l
Volume 13, 2009
9
It can be shown that the vector T U satisfies the condition (T U )− = 0, automatically;
Z
4
X
1
g(t) + ih(t))
[T (∂x, n)U (z)] = Im
L(k) iL[
ds
π
x1 − t j
k=1
l
Z
4
X
1
g(t) − ih(t)
+ Im
L(j) iL
ds] = 0.
π
x1 − t j
j=1
−
l
But on the arc ends lj , j = 1, 2, ..., p, we have
Z
4
X
1
∂ ln (tk0 − tk )
[T (∂x, n)U (z)] = ∓g(t0 ) + Im
[g(t) + ih(t)]ds
L(k) iL[
π
∂s
k=1
l
Z
4
X
∂ ln (tk0 − tj )
1
+ Im
L(j) iL
[g(t) − ih(t)]ds] = f ± (t0 ).
π
∂s
j=1
±
l
From last equation we get
2g(τ ) = f − − f + ,
4
h Z ∂ ln (t − t )
X
1
k0
k
P(k)
Re
h(t)dt
π
∂s
k=1
l
Z
4
i
X
∂ ln (tk0 − tj )
P(j)
−
h(t)dt = Ω(t0 ),
∂s
j=1
(9)
l
where
Z
4
X
1 −
1
∂ ln (tk0 − tk )
+
P(k) [
Ω(t0 ) = (f + f ) − Im
g(t)dt
2
π
∂s
k=1
l
Z
4
X
∂ ln(tk0 − tj )
P(j)
g(t)dt].
−
∂s
j=1
l
We must require in addition that the solution h of the system of singular
integral equations (9) satisfies the condition
Z
Z
h(t)ds = RL (f − − f + )ds,
(10)
l
where
4
P
k=1
l
R(k)T = −E + iR.
We seek for the solution of the system (9) in the class h0 ([4]). Therefore,
the total index in the class h0 is 4p.
10
Bulletin of TICMI
Let’s prove that the adjoint homogeneous system corresponding to the
system (9) has only the trivial solution in the adjoint class. The adjoint homogeneous system has the form
Z
4
X
1
∂ ln (tk − tk0 )
Re
ν(t)ds
iLL(k)
π
∂s
k=1
(11)
l
4
4 P
R ∂ ln(tj − tk0 )
P
−
LiL(j) iLLk
ν(t)ds = 0.
∂s
k=1 j=1
l
Note, that the solution of the system (11) will be vector, satisfying the Holder’s
conditions, vanishing at the end points aj and bj , and having derivative in the
class H ∗ ([4]).
Multiplying the system (11) by nonsingular matrix a = L−1 given by (8)
and taking into account the identity aLL(k) = P(k) a, we obtain
4
X
1
Re
P(k)
π
k=1
Z
∂ ln (tk − tk0 )
aν(t)ds
∂s
l
R ∂ ln(tj − tk0 )
−Re
P(j) Pk
aν(t)ds = 0.
∂s
k=1 j=1
l
4 P
4
P
(12)
Let ν0 be the nontrivial solution of the system (12) and construct the potential
Z
4
X
1
∂ ln (tk − zk )
T
R(k) iL
U0 (z) = Re
aν(t)ds
π
∂s
k=1
l
(13)
Z
4
4
XX
∂ ln (tj − tk )
P(j) Pk
aν(t)ds.
− Re
∂s
k=1 j=1
l
Then at every point of l, except may be aj and bj
[T (∂t, n)U0 ]± = 0
and on the basis of the uniqueness theorem we have U0 = 0.
Finally, from equality: U0+ − U0− = −2aν(t0 ) = 0, t0 ∈ l, it follows, that
ν0 = 0, which contradicts our assumption.
Thus, the system (9) is solvable in the class h0 for the arbitrary righthand side and the solution depended on the 4p arbitrary constants. These
constants are fixed by the conditions (10), given the system of algebraic 4p
linear equations.
Let’s prove that the determinant of this system is not zero. Indeed, let’s
take the homogeneous system, corresponding to the conditions f ± = 0. Sup(0)
(0)
posing the solution K1 , ..., K4p nontrivial, we construct the potential
Z
4
X
1
T
R(k) L ln (zk − tk )h(0) ds,
U0 (z) = Re
π
k=1
l
(14)
Volume 13, 2009
11
where h(0) is a linear combinations of solutions h(j)
h
(0)
=
4p
X
K (j) h(j) ,
k=1
and h(j) are linearly independent solutions of the homogeneous equation corresponding to (9). h(j) has satisfy the following condition
Z
h(0) ds = 0.
lj
Then the potential (14) is regular at infinity and by the uniqueness theorem
h(0) = 0. But we have the following equality
³ ∂u ´+
∂s
−
³ ∂u ´−
∂s
= Lh(0) = 0.
Whence we conclude that K (j) = 0, which contradict the assumption. Thus
the solvability of the problem is proved.
Acknowledgement. The designated project has been fulfilled by financial
support of Georgian National Science Foundation (Grant N GNSF/ST06/3033). Any idea in this publications is possessed by the author and may not
represent the opinion of Georgian National Science Foundation itself.
References
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4, No. 2 (1989), 95-98 (in Russian).
[2] Sh. Zazashvili. Contact Problem for two Anisotropic Half-plane with a Cut along
the Contact Line. Bull. Acad. Sci. Georgia, 145, No. 2 (1992), 283-285 (in Russian).
[3] Ya. Ya. Rushchitski. Elements of Mixture Theory. Naukova Dumka, Kiev, 1991 (in
Russian).
[4] N. I. Muskhelishvili. Singular Integral Equations. Boundary Problems of the Theory
of Functions and some their Applications in Mathematical Physics. 3rd ed. Nauka, Moscow,
1968 (in Russian); English Translation from 2nd Russian ed. (1946): P. Noordhoff, Groningen 1953, Corrected Reprint Dover Publications, Inc., N. Y., 1992.
[5] L. Bitsadze. The Basic BVPs of Statics of the Theory of Elastic Transversally
Isotropic Mixtures. Reports Seminar of I. Vekua Inst. of Appl. Math. Reports, 26-27, No.
1-3 (2000-2001), 79-87.
[6] L. Bitsadze. On Some Solutions of Statics of the Theory of Elastic Transversally
Isotropic Mixtures. Reports of Enlarged Session of the Seminar of I. Vekua Inst. of Appl.
Math., 16, No. 2 (2001), 41-45.
Received 12.05.2009; revised 6.10.2009; accepted 11.11.2009