Memoirs on Differential Equations and Mathematical Physics
Volume 45, 2008, 75–83
Lamara Bitsadze
EXPLICIT SOLUTION OF THE FIRST BVP
OF THE ELASTIC MIXTURE FOR HALF-SPACE
Abstract. We consider the first BVP of elastic mixture theory for a
transversally-isotropic half-space. The solution of the first BVP for the
transversally-isotropic half-space is given in [1]. The present paper is an
attempt to use this result for the BVP of elastic mixture theory for a
transversally-isotropic elastic body. Using the potential method and the
theory of integral equations, the uniqueness theorem is proved for a halfspace and the first BVP previously is solved effectively (in quadratures),
which has not been solved.
2000 Mathematics Subject Classification. 74E30, 74G05.
Key words and phrases. Elastic mixture, uniqueness theorem, potential method, explicit solution.
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77
On the Explicit Solution of the First BVP
The first BVP and the uniqueness theorem for a half-space. Let
the plane ox1 x2 be the boundary of a half-space x3 > 0. Let the upper
half-space be denoted by D and the boundary of D by S. Let the axis ox3
be directed vertically upwards and the normal be n(0, 0, 1).
A basic homogeneous equation of statics of transversally-isotropic elastic
mixture theory can be written in the form [2]
(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)k3x3 are given in
the form
(j)
(j)
,
= Cqp
Cpq
j = 1, 2, 3;
p, q = 1, 2, 3,
2
2
∂
(j) ∂
(j) ∂
+ c66
+ c44
,
2
2
∂x1
∂x2
∂x23
∂2
(j)
(j)
(j)
,
C12 (∂x) = (c11 − c66 )
∂x1 ∂x2
∂2
(j)
(j)
(j)
Ck3 (∂x) = (c13 + c44 )
, k = 1, 2,
∂xk ∂x3
2
2
2
(j) ∂
(j) ∂
(j)
(j) ∂
+ c11
+ c44
,
C22 (∂x) = c66
2
2
∂x1
∂x2
∂x23
2
2
∂2
∂
(j) ∂
(j)
(j)
+ c33
+
,
C33 (∂x) = c44
2
2
∂x1
∂x2
∂x23
(j)
(j)
2
C11 (∂x) = c11
(k)
cpq are the constants characterizing physical properties of the mixture and
satisfying certain inequalities obtained due to positive definiteness of the
potential energy. U = U T (x) = (u0 , u00 ) is a six-dimensional displacement
vector-function, u0 (x) = (u01 , u02 , u03 ) and u00 (x) = (u001 , u002 , u003 ) are partial
displacement vectors. Throughout this paper “T ” denotes transposition.
Definition. A vector-function U (x) defined in the domain D is called
regular if it has integrable continuous second derivatives in D and U (x)
itself and its first derivatives are continuously extendable at every point of
the boundary of D , i.e. U (x) ∈ C 2 (D) ∩ C 1 (D) and satisfies the following
conditions at infinity
U (x) = O(|x|−1 ),
∂U
= O(|x|−2 ),
∂xk
|x|2 = x21 + x22 + x23 , k = 1, 2, 3.
For the equation (1) we pose the following BVP. Find a regular function
U (x) satisfying the equation (1) in D if on the boundary S the displacement
vector U is given in the form
U + = f (z),
z ∈ S.
(2)
78
L. Bitsadze
where (.)+ denotes the limiting value from D and f is a given vector.
q
|fk | < AR, R = z12 + z22 ≤ 1, |fk | < AR−α ,
α > 0,
R > 1,
k = 1, . . . , 6,
(3)
A = const > 0.
The Uniqueness Theorem. Let us prove that the first homogeneous BVP
has only a trivial solution. Note that if U is a regular solution of the equation
(1) and satisfies the following conditions at infinity
U (x) = O(|x|−α ),
P (∂x, n)U = O(|x|−1−α ),
α > 0,
then we have the formula
U (x) =
Z∞ Z∞
1
(P (∂y, n)Γ)∗ u+ −Γ(y−z)(P (∂y, n)u)+ dy1 dy2 , x ∈ D,
=
2π
(4)
−∞ −∞
where P (∂y, n)U is the generalized stress vector
0
(1) ∂uk
00
(3) ∂uk
∂u0
∂u00
+ δ (1) 3 + δ (3) 3 ,
∂x3
∂x3
∂x
∂xk
0
k 00
0
00
∂u
∂u
∂u
∂u
1
2
2
1
(1)
(3)
(P (∂y, n)U )3 = β
+
+
+β
+
∂x1 ∂x2
∂x1 ∂x2
00
0
(3) ∂u3
(1) ∂u3
+c33
,
+ c33
∂x3
∂x3
0
00
(3) ∂uk−3
(2) ∂uk−3
(P (∂y, n)U )k = c44
+ c44
+
∂x3
∂x3
∂u003
∂u03
+ δ (2)
, k = 4, 5,
+ δ (4)
∂xk−3
∂xk−3
0
∂u001 ∂u002
∂u1 ∂u02
(2)
(4)
+β
+
+
+
(P (∂y, n)U )6 = β
∂x1 ∂x2
∂x1 ∂x2
0
00
(3) ∂u3
(2) ∂u3
+ c33
+c33
,
∂x3
∂x3
(P (∂y, n)U )k = c44
(j)
+ c44
β (j) + δ (j) = α13 , j = 1, 2, 3,
(j)
(j)
k = 1, 2,
(5)
(3)
β (4) + δ (4) = α13 ,
(j)
c13 + c44 = α13 .
Γ(y−x) is the symmetric matrix of the fundamental solution of the equation
(1)
(1)
Γ
Γ(3)
Γ(x − y) =
,
(6)
Γ(3)T Γ(2)
where
Γ(j) (x − y) =
6
X
k=1
kΓj(k)
pq k3x3 , j = 1, 2, 3,
j(k)
Γj(k)
pq = Γqp ,
79
On the Explicit Solution of the First BVP
(k)
2
A11
(k) ∂ Φk
+ A12
, p = 1, 2; q = 1, 2;
rk
∂xp ∂xq
= 1, p = q, δpq = 0, p 6= q,
Γ1(k)
pq = δpq
δpq
1(k)
(k)
3(k)
(k)
Γp3 = A13
Γp3 = A16
∂ 2 Φk
,
∂xp ∂x3
(k)
=
A33
,
rk
∂ 2 Φk
, p = 1, 2,
∂xp ∂x3
Γ33
1(k)
Γ33
Γ3(k)
pq = δpq
(k)
3(k)
=
A36
,
rk
2(k)
=
A66
.
rk
(k)
2
A14
(k) ∂ Φk
+ A42
,
rk
∂xp ∂xq
3(k)
(k)
Γ3p = A34
∂ 2 Φk
,
∂xp ∂x3
(k)
Γ2(k)
pq = δpq
2(k)
2
A44
(k) ∂ Φk
+ A45
,
rk
∂xp ∂xq
(k)
Γp3 = A46
∂ 2 Φk
, p = 1, 2,
∂xp ∂x3
(k)
Γ33
(k)
The coefficients Apq are defined as follows
(k)
(2)
(2)
A11 = (−1)k (c44 − c66 ak )r00 ,
(k)
(k)
A12 =
(k)
A11
,
ak
(k)
(k)
A24 =
(1)
A14
,
ak
(k)
(3)
(3)
A14 = −(−1)k (c44 − c66 ak )r00 ,
(k)
(k)
A45 =
A44
,
ak
(1)
A44 = (−1)k (c44 − c66 ak )r00 , k = 1, 2, r00 = [r0 (a1 − a2 )]−1 ,
(k)
δk
(2)
(2)
[−q3 c44 + ak t12 − a2k t11 + c11 q4 a3k ],
ak
δk
(3)
(3)
=
[q3 c44 + ak t13 − a2k t22 − c11 q4 a3k ],
ak
δk
(1)
(1)
[−q3 c44 + ak t23 − a2k t33 + c11 q4 a3k ],
=
ak
A12 =
(k)
A42
(k)
A45
((k))
A33
(k)
(2)
(2)
(3)
(3)
(7)
= δk [q4 c33 − ak t42 + a2k t44 − c44 q1 a3k ],
A36 = δk [−q4 c33 − ak t62 + a2k t66 + c44 q1 a3k ],
(k)
(1)
(1)
A66 = δk [q4 c33 − ak t52 + a2k t55 − c44 q1 a3k ],
(k)
A13 = δk [v13 − v11 ak + v12 a2k ],
(k)
A34 = δk [v23 − v21 ak + v22 a2k ],
δk = dk (a1 − ak )(a2 − ak )b−1
0 ,
(k)
A16 = δk [w13 − w12 ak + w11 a2k ],
(k)
A46 = δk [w34 − w14 ak + w24 a2k ],
k = 3, . . . , 6,
where ak are the positive roots of the characteristic equations
(r0 a2 − c0 a + q4 )(b0 a4 − b1 a3 + b2 a2 − b3 a + b4 ) = 0,
(1) (2)
(3)2
r0 = c66 c66 − c66 ,
(1) (2)
(1) (2)
(3) (3)
c0 = c66 c44 + c44 c66 − 2c66 c44 .
The coefficients dk , bk , vij , wij , tij are given in [3]. The singular matrix
6
P
(k)
(Mpq )6x6 , which is obtained from P (∂x, n)Γ(x − y) by
[P (∂y, n)Γ]∗ =
k=1
80
L. Bitsadze
transposition of the columns and rows and the variables x and y, has the
form
6 X
M (1k)
[P (∂x)Γ(x − y)] =
M (4k)
M (3k)
M (2k)
∗
k=1
(8)
,
(jk)
where the elements of the matrix M (jk) = kMpq k3x3 , j = 1, 2, 3, 4, are
written as
∂ 3 Φk
∂ 1
(k)
+ R12
,
∂x3 rk
∂xp ∂xj ∂x3
= 1, p = j, δpj = 0, p 6= j, p, j = 1, 2,
(1k)
Mpj
δpj
(1k)
Mp3
(1k)
M33
(3k)
Mp3
(4k)
Mpj
(4k)
M3p
(2k)
Mpj
(2k)
M3p
(k)
= δpj R11
∂ 1
,
∂xp rk
(k) ∂ 1
= R33
,
∂x3 rk
(k) ∂ 1
= R61
,
∂xp rk
(1k)
(k)
= R31
M3p
(3k)
Mpj
(3k)
M3p
∂ 1
,
∂xp rk
∂ 3 Φk
(k) ∂ 1
(k)
= δpj R14
+ R24
,
∂x3 rk
∂xj ∂xp ∂x3
(k) ∂ 1
(3k)
(k) ∂ 1
= R43
, M33 = R63
,
∂xp rk
∂x3 rk
(k)
= R13
∂ 3 Φk
∂ 1
(k)
(4k)
(k) ∂ 1
+ R42
, Mp3 = R34
,
∂x3 rk
∂xj ∂xp ∂x3
∂xp rk
(4k)
(k) ∂ 1
(k) ∂ 1
, M33 = R36
,
= R16
∂xp rk
∂x3 rk
∂ 3 Φk
(k) ∂ 1
(k)
(2k)
(k) ∂ 1
= δpj µ44
+ R44
, Mp3 = R64
,
∂x3 rk
∂xp ∂xj ∂x3
∂xp rk
(2k)
(k) ∂ 1
(k) ∂ 1
, M33 = R66
, p = 1, 2.
= R46
∂xp rk
∂x3 rk
(k)
= δpj R41
(k)
The coefficients Rpq satisfy the following conditions
2
(k)
X
R
11
k=1
2
X
k=1
(k)
R41
ak
=
6
X
k=3
ak
(k)
R36
ak
=
6
(k)
X
R
33
k=3
2
X
ak
(k)
R14
k=1
6
X
=
k=1
ak
(k)
=
=
R24 =
6
(k)
X
R
66
k=3
6
X
k=1
6
X
k=3
ak
=
2
(k)
X
µ44
= 1,
ak
k=1
(k)
R12 = 0,
(k)
2
6
k=1
k=1
X (k) X (k)
R63
=
R44 =
R42 = 0
ak
81
On the Explicit Solution of the First BVP
(k)
(k)
and, after elementary calculations the coefficients R13 , . . . , R64 take the
form
(k)
(1)
(k)
(3)
(k)
(1)
(k)
(3)
(k)
(k)
(1)
(k)
(3)
(k)
(1)
(k)
(3)
(k)
R13 = δ0 A33 + δ0 A36 + c44 A13 + c44 A43 ,
R16 = δ0 A36 + δ0 A66 + c44 A16 + c44 A46 ,
(k)
(1)
(k)
(3)
(k)
(1)
(k)
(3)
(k)
(k)
(1)
(k)
(3)
(k)
(1)
(k)
(3)
(k)
R31 = −ak β0 A12 − ak β0 A42 + c33 A13 + c33 A16 ,
R34 = −ak β0 A42 − ak β0 A45 + c33 A43 + c33 A46 ,
(k)
(4)
(k)
(2)
(k)
(3)
(k)
(2)
(k)
(k)
(4)
(k)
(2)
(k)
(3)
(k)
(2)
(k)
(9)
R43 = δ0 A33 + δ0 A36 + c44 A13 + c44 A43 ,
R46 = δ0 A36 + δ0 A66 + c44 A16 + c44 A46 ,
(k)
(4)
(k)
(2)
(k)
(3)
(k)
(2)
(k)
(k)
(4)
(k)
(2)
(k)
(3)
(k)
(2)
(k)
R61 = −ak β0 A12 − ak β0 A42 + c33 A13 + c33 A16 ,
R64 = −ak β0 A42 − ak β0 A45 + c33 A43 + c33 A46 ,
k = 3, . . . , 6.
We can easily prove that every column of the matrix [P (∂x, n)Γ]∗ is a solution of the system (1) with respect to the point x if x 6= y and all elements
(k)
Mpq have a singularity of type |x|−2 .
(J)
(j)
We choose δ0 , β0 , j = 1, . . . , 4, so that
6
(k)
X
R13
√ = 0,
ak
6
(k)
X
R31
√ = 0,
ak
6
(k)
X
R16
√ = 0,
ak
6
(k)
X
R34
√ = 0,
ak
k=3
k=3
k=3
k=3
6
(k)
X
R43
√ = 0,
ak
6
(k)
X
R46
√ = 0,
ak
6
(k)
X
R61
√ = 0,
ak
6
(k)
X
R64
√ = 0,
ak
k=3
k=3
k=3
After some simplification, we find from (10) that
∆=
6
X
(k) √
A12 ak
(k) √
A45 ak
k=3
k=3
=
6
X

−
6
X
(k) √
A42 ak
k=3
6
(k)
(k) 6
X
√
A33 X A66
= a3 a4 a5 a6 
√
√ −
ak
ak
k=3
k=3
!2
k=3
=
6
(k)
X
A36
√
ak
k=3
!2 
=
√
B0 [(δ11 δ22 + b0 m1 m3 )q4 + q1 b4 + δ22 b0 m2 ]( a3 a4 a5 a6 )−1 +
b20
+ q1 (δ11 δ22 + b0 m1 m3 − k1 ) + b0 δ11 m2 ,
where
(1) (2)
(3)2
q1 = c11 c11 − c11 ,
m1 =
6
X
√
ak ,
k=3
m3 =
(10)
X √
p6=q6=j
(1) (2)
(3)2
q4 = c44 c44 − c44 ,
m2 =
X√
ap aq ,
p6=q
ap aq aj , p, q, j = 3, . . . , 6,
b0 = q 1 q4 ,
82
L. Bitsadze
(1) (2)
(1) (2)
(3) (3)
(1) (2)
(1) (2)
(3) (3)
δ11 = c11 c44 + c44 c11 − 2c11 c44 > 0,
δ22 = c33 c44 + c44 c33 − 2c33 c44 > 0,
(1) (2)
(3)2
(1)
(2)
(3)
k1 + k2 = 2(α13 α13 − α13 ) − α13 v11 − α13 w14 − α13 (w12 + v21 ),
i2
1 h (2)2 (3)
(2) (3) (3)
(3)2 (2)
(2)
k1 = (2)2 c44 c13 − 2c44 c44 c13 + c44 c13 + c44 +
c44
i2
q42 (2)2
2q4 h (2) (3)
(3) (2)
α13 ,
+ (2)2 c44 c13 − c44 c13 + (2)2
c44
c44
Y √
√
B0−1 =
( ap + aq ), p, q = 3, . . . , 6.
p6=q
Taking into account the inequalities obtained from the positive definiteness
(j)
(j)
the energy E(u, u), we conclude that ∆ 6= 0. When δ0 , β0 are solutions
of the system (10), we denote the vector P (∂y, n)U, by N (∂y, n)U . Then
from (4), when U + = 0, we have
1
U (x) = −
2π
Z∞ Z∞
−∞ −∞
Γ(y − x)N (∂y, n)U dy1 dy2 .
Hence for the vector N U as x(x1 , x2 , x3 ) −→ z(z1 , z2 , 0) we find
1
[N (∂z, n)U ] +
2π
+
Z∞ Z∞
−∞ −∞
N Γ(y − z)(N U )+ dy1 dy2 = 0.
Note that N Γ(z − y) = 0, z ∈ S. ATherefore (N U )+ = 0, and from (4)
we have U = 0, x ∈ D. Therefore the homogeneous equation has only the
trivial solution. Thus we formulate the following
Theorem. The first BVP has at most one regular solution.
The first BVP. A solution of the first BVP will be sought in the domain
D in terms of the double layer potential
1
U (x) =
2π
Z∞ Z∞
−∞ −∞
[N (∂y, n)Γ(y − x]∗ g(y)dy1 dy2 ,
(11)
where g is an unknown real vector. Taking into account the properties of
the double layer potential and the boundary condition for determining g,
we obtain the following Fredholm integral equation of second kind:
1
g(z) +
2π
Z∞ Z∞
−∞ −∞
[N (∂y, n)Γ(y − z)]∗ g(y)dy1 dy2 = f (z),
On the Explicit Solution of the First BVP
83
Taking into account the fact that [N Γ]∗ = 0, x3 = 0, from the latter equation we have g(z) = f (z) and (11) takes the form
Z∞ Z∞
1
[N (∂y, n)Γ(y − x)]∗ f (y)dy1 dy2 .
(12)
U (x) =
2π
−∞ −∞
Thus we have obtained the Poisson formula for the solution of the first BVP
for the half-space. Note that (12) is valid if and only if f ∈ C 1,α (S) and
satisfies the condition f = O( |x|A1+β ) at infinity, where A is a constant vector
and β > 0.
Acknowledgement
This project was fulfilled by the financial support from Georgian National
Science Foundation (Grant No. GNSF/ST06/3-033).
References
1. 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 the second Russian edition. Edited by V. D. Kupradze.
North-Holland Series in Applied Mathematics and Mechanics, 25. North-Holland
Publishing Co., Amsterdam–New York, 1979.
2. Ya. Ya. Rushchitski, Elements of Mixture Theory. (Russian) Naukova Dumka,
Kiev, 1991.
3. L. Bitsadze, The basic boundary value problems of statics of the theory of
elastic transversally isotropic mixtures. Semin. I. Vekua Inst. Appl. Math. Rep.
26/27(2000/01), 79–87.
(Received 22.11.2007)
Author’s address:
I. Javakhishvili Tbilisi State University
2, University St., Tbilisi 0186
Georgia
E-mail: lbits@viam.sci.tsu.ge